Maths In Focus Preliminary 3 Unit

Mathematics Extension 1 Preliminary Course

maths

Mathematics Extension 1 Preliminary Course

maths Margaret Grove

Text © 2010 Grove and Associates Pty Ltd Illustrations and design © 2010 McGraw-Hill Australia Pty Ltd Additional owners of copyright are acknowledged in on-page credits Every effort has been made to trace and acknowledge copyrighted material. The authors and publishers tender their apologies should any infringement have occurred. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the institution (or the body that administers it) has sent a Statutory Educational notice to Copyright Agency Limited (CAL) and been granted a licence. For details of statutory educational and other copyright licences contact: Copyright Agency Limited, Level 15, 233 Castlereagh Street, Sydney NSW 2000. Telephone: (02) 9394 7600. Website: www.copyright.com.au Reproduction and communication for other purposes Apart from any fair dealing for the purposes of study, research, criticism or review, as permitted under the Act, no part of this publication may be reproduced, distributed or transmitted in any form or by any means, or stored in a database or retrieval system, without the written permission of McGraw-Hill Australia including, but not limited to, any network or other electronic storage. Enquiries should be made to the publisher via www.mcgraw-hill.com.au National Library of Australia Cataloguing-in-Publication Data Author: Grove, Margaret. Title: Maths in focus: mathematics extension preliminary course/Margaret Grove. Edition: 2nd ed. ISBN: 9780070278585 (pbk.) Target Audience: For secondary school age. Subjects: Mathematics–Problems, exercises, etc. Mathematics–Textbooks. Dewey Number: 510.76 Published in Australia by McGraw-Hill Australia Pty Ltd Level 2, 82 Waterloo Road, North Ryde NSW 2113 Publisher: Eiko Bron Managing Editor: Kathryn Fairfax Production Editor: Natalie Crouch Editorial Assistant: Ivy Chung Art Director: Astred Hicks Cover and Internal Design: Simon Rattray, Squirt Creative Cover Image: Corbis Proofreaders: Terence Townsend and Ron Buck CD-ROM Preparation: Nicole McKenzie Typeset in ITC Stone serif, 10/14 by diacriTech Printed in China on 80 gsm matt art by iBook

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Contents PREFACE

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ACKNOWLEDGEMENTS

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CREDITS

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FEATURES OF THIS BOOK

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SYLLABUS MATRIX

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STUDY SKILLS

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Chapter 1: Basic Arithmetic

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INTRODUCTION REAL NUMBERS DIRECTED NUMBERS FRACTIONS, DECIMALS AND PERCENTAGES POWERS AND ROOTS ABSOLUTE VALUE TEST YOURSELF 1 CHALLENGE EXERCISE 1 Chapter 2: Algebra and Surds INTRODUCTION SIMPLIFYING EXPRESSIONS BINOMIAL PRODUCTS FACTORISATION COMPLETING THE SQUARE ALGEBRAIC FRACTIONS SUBSTITUTION SURDS TEST YOURSELF 2 CHALLENGE EXERCISE 2 Chapter 3: Equations INTRODUCTION SIMPLE EQUATIONS SUBSTITUTION INEQUATIONS EQUATIONS AND INEQUATIONS INVOLVING ABSOLUTE VALUES EXPONENTIAL EQUATIONS QUADRATIC EQUATIONS FURTHER INEQUATIONS QUADRATIC INEQUATIONS SIMULTANEOUS EQUATIONS TEST YOURSELF 3 CHALLENGE EXERCISE 3

3 3 9 12 19 37 41 43 44 45 45 51 55 69 71 73 76 90 93 94 95 95 100 103 107 114 118 125 129 132 138 139

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Chapter 4: Geometry 1

140

INTRODUCTION NOTATION TYPES OF ANGLES PARALLEL LINES TYPES OF TRIANGLES CONGRUENT TRIANGLES SIMILAR TRIANGLES PYTHAGORAS’ THEOREM TYPES OF QUADRILATERALS POLYGONS AREAS TEST YOURSELF 4 CHALLENGE EXERCISE 4

141 141 142 149 153 159 163 171 177 184 188 195 197

Practice Assessment Task Set 1 Chapter 5: Functions and Graphs INTRODUCTION FUNCTIONS GRAPHING TECHNIQUES LINEAR FUNCTION QUADRATIC FUNCTION ABSOLUTE VALUE FUNCTION THE HYPERBOLA CIRCLES AND SEMI-CIRCLES OTHER GRAPHS LIMITS AND CONTINUITY FURTHER GRAPHS REGIONS TEST YOURSELF 5 CHALLENGE EXERCISE 5 Chapter 6: Trigonometry INTRODUCTION TRIGONOMETRIC RATIOS RIGHT-ANGLED TRIANGLE PROBLEMS APPLICATIONS EXACT RATIOS ANGLES OF ANY MAGNITUDE TRIGONOMETRIC EQUATIONS TRIGONOMETRIC IDENTITIES NON-RIGHT-ANGLED TRIANGLE RESULTS APPLICATIONS AREA TRIGONOMETRY IN THREE DIMENSIONS SUMS AND DIFFERENCES OF ANGLES FURTHER TRIGONOMETRIC EQUATIONS TEST YOURSELF 6 CHALLENGE EXERCISE 6

199 204 205 205 216 224 228 234 242 246 254 260 264 277 287 288 290 291 291 299 308 318 322 336 342 347 358 362 365 367 374 385 387

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Chapter 7: Linear Functions INTRODUCTION DISTANCE MIDPOINT GRADIENT EQUATION OF A STRAIGHT LINE PARALLEL AND PERPENDICULAR LINES INTERSECTION OF LINES PERPENDICULAR DISTANCE ANGLE BETWEEN TWO LINES RATIOS TEST YOURSELF 7 CHALLENGE EXERCISE 7 Chapter 8: Introduction to Calculus INTRODUCTION GRADIENT DIFFERENTIATION FROM FIRST PRINCIPLES SHORT METHODS OF DIFFERENTIATION TANGENTS AND NORMALS FURTHER DIFFERENTIATION AND INDICES COMPOSITE FUNCTION RULE PRODUCT RULE QUOTIENT RULE ANGLE BETWEEN 2 CURVES TEST YOURSELF 8 CHALLENGE EXERCISE 8 Practice Assessment Task Set 2 Chapter 9: Properties of the Circle INTRODUCTION PARTS OF A CIRCLE ARCS, ANGLES AND CHORDS CHORD PROPERTIES CONCYCLIC POINTS TANGENT PROPERTIES TEST YOURSELF 9 CHALLENGE EXERCISE 9 Chapter 10: The Quadratic Function INTRODUCTION GRAPH OF A QUADRATIC FUNCTION QUADRATIC INEQUALITIES THE DISCRIMINANT QUADRATIC IDENTITIES SUM AND PRODUCT OF ROOTS EQUATIONS REDUCIBLE TO QUADRATICS TEST YOURSELF 10 CHALLENGE EXERCISE 10

390 391 391 396 398 408 412 417 422 426 430 434 435 438 439 440 449 465 471 476 478 482 485 487 490 491 494 498 499 499 500 512 519 525 537 539 542 543 543 549 555 562 566 571 575 576

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Chapter 11: Locus and the Parabola INTRODUCTION LOCUS CIRCLE AS A LOCUS PARABOLA AS A LOCUS GENERAL PARABOLA TANGENTS AND NORMALS PARAMETRIC EQUATIONS OF THE PARABOLA CHORDS, TANGENTS AND NORMALS PROPERTIES OF THE PARABOLA LOCUS PROBLEMS TEST YOURSELF 11 CHALLENGE EXERCISE 11 Practice Assessment Task Set 3 Chapter 12: Polynomials 1 INTRODUCTION DEFINITION OF A POLYNOMIAL DIVISION OF POLYNOMIALS REMAINDER AND FACTOR THEOREMS GRAPH OF A POLYNOMIAL ROOTS AND COEFFICIENTS OF POLYNOMIAL EQUATIONS TEST YOURSELF 12 CHALLENGE EXERCISE 12 Chapter 13: Permutations and Combinations INTRODUCTION FUNDAMENTAL COUNTING PRINCIPLE PERMUTATIONS COMBINATIONS TEST YOURSELF 13 CHALLENGE EXERCISE 13 Practice Assessment Task Set 4 Answers

578 579 579 587 591 610 625 627 634 643 648 652 653 655 662 663 663 667 672 681 706 713 714 716 717 717 730 740 746 747 749 756

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PREFACE This book covers the Preliminary syllabus for Mathematics and Extension 1. The extension material is easy to see as it has green headings and there is green shading next to all extension question and answers. The syllabus is available through the NSW Board of Studies website on www.boardofstudies. nsw.edu.au. You can also access resources, study techniques, examination technique, sample and past examination papers through other websites such as www.math.nsw.edu.au and www.csu.edu. au. Searching the Internet generally will pick up many websites supporting the work in this course. Each chapter has comprehensive fully worked examples and explanations as well as ample sets of graded exercises. The theory follows a logical order, although some topics may be learned in any order. Each chapter contains Test Yourself and Challenge exercises, and there are several practice assessment tasks throughout the book. If you have trouble doing the Test Yourself exercises at the end of a chapter, you will need to go back into the chapter and revise it before trying them again. Don’t attempt to do the Challenge exercises until you are confident that you can do the Test Yourself exercises, as these are more difficult and are designed to test the more able students who understand the topic really well.

ACKNOWLEDGEMENTS Thanks go to my family, especially my husband Geoff, for supporting me in writing this book.

CREDITS Fairfax Photos: p 327 Istockphoto: p 101, p 171 Margaret Grove: p 37, p 163, p 206, p 246, p 260, p 291, p308 (bottom), p 310, p 311, p 313, p 316, p 391, p 499, p 543, p 591, p 717, p 719, p 726, p 729, p 730, p 739 Photolibrary: p 205 Shutterstock: p 74, p 164, p 229, p 308 (top), p 580

FEATURES OF THIS BOOK This second edition retains all the features of previous Maths in Focus books while adding in new improvements. The main feature of Maths in Focus is in its readability, its plentiful worked examples and straightforward language so that students can understand it and use it in self-paced learning. The logical progression of topics, the comprehensive fully worked examples and graded exercises are still major features. A wide variety of questions is maintained, with more comprehensive and more difficult questions included in each topic. At the end of each chapter is a consolidation set of exercises (Test yourself) in no particular order that will test whether the student has grasped the concepts contained in the chapter. There is also a challenge set for the more able students. The four practice assessment tasks provide a comprehensive variety of mixed questions from various chapters. These have been extended to contain questions in the form of sample examination questions, including short answer, free response and multiple-choice questions that students may encounter in assessments. The second edition also features a short summary of general study skills that students will find useful, both in the classroom and when doing assessment tasks and examinations. These study skills are also repeated in the HSC book.

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A syllabus matrix is included to show where each syllabus topic fits into the book. Topics are generally arranged in a logical order. For example, arithmetic and algebra are needed in most, if not all other topics, so these are treated at the beginning of the book. Some teachers like to introduce particular topics before others, e.g. linear functions before more general functions. However, part of the work on gradient requires some knowledge of trigonometry and the topic of angles of any magnitude in trigonometry needs some knowledge of functions. So the order of most chapters in the book have been carefully thought out. Some chapters, however, could be covered in a different order, such as geometry which is covered in Chapter 4, and quadratic functions and locus, which are near the end of the book.

SYLLABUS MATRIX This matrix shows how the syllabus is organised in the chapters of this book.

Mathematics (2 Unit) Basic arithmetic and algebra (1.1 – 1.4)

Chapter 1: Basic arithmetic Chapter 2: Algebra and surds Chapter 3: Equations

Real functions (4.1 – 4.4)

Chapter 5: Functions and graphs

Trigonometric ratios (5.1 – 5.5)

Chapter 6: Trigonometry

Linear functions (6.1 – 6.5, 6.7)

Chapter 7: Linear functions

The quadratic polynomial and the parabola (9.1 – 9.5)

Chapter 10: The quadratic function Chapter 11: Locus and the parabola

Plane geometry (2.1 – 2.4)


Tangent to a curve and derivative of a function (8.1 – 8.9)

Chapter 8: Introduction to calculus

Extension 1 Other inequalities (1.4E)

Chapter 3: Equations

Circle geometry (2.6 – 2.10E)

Chapter 9: Properties of the circle

Further trigonometry (5.6 – 5.9E)


Angles between two lines (6.6E)


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Internal and external division of lines into given ratios (6.7E)


Parametric representation (9.6E)

Chapter 11: Locus and the parabola

Permutations and combinations (18.1E)

Chapter 13: Permutations and combinations

Polynomials (16.1 – 16.3E)

Chapter 12: Polynomials 1

STUDY SKILLS You may have coasted through previous stages without needing to rely on regular study, but in this course many of the topics are new and you will need to systematically revise in order to build up your skills and to remember them. The Preliminary course introduces the basics of topics such as calculus that are then applied in the HSC course. You will struggle in the HSC if you don’t set yourself up to revise the preliminary topics as you learn new HSC topics. Your teachers will be able to help you build up and manage good study habits. Here are a few hints to get you started. There is no right or wrong way to learn. Different styles of learning suit different people. There is also no magical number of hours a week that you should study, as this will be different for every student. But just listening in class and taking notes is not enough, especially when learning material that is totally new. You wouldn’t go for your driver’s licence after just one trip in the car, or enter a dance competition after learning a dance routine once. These skills take a lot of practice. Studying mathematics is just the same. If a skill is not practised within the first 24 hours, up to 50% can be forgotten. If it is not practised within 72 hours, up to 85–90% can be forgotten! So it is really important that whatever your study timetable, new work must be looked at soon after it is presented to you. With a continual succession of new work to learn and retain, this is a challenge. But the good news is that you don’t have to study for hours on end!

In the classroom In order to remember, first you need to focus on what is being said and done. According to an ancient proverb:

‘I hear and I forget I see and I remember I do and I understand’

If you chat to friends and just take notes without really paying attention, you aren’t giving yourself a chance to remember anything and will have to study harder at home.

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If you have just had a fight with a friend, have been chatting about weekend activities or myriad other conversations outside the classroom, it helps if you can check these at the door and don’t keep chatting about them once the lesson starts. If you are unsure of something that the teacher has said, the chances are that others are also not sure. Asking questions and clarifying things will ultimately help you gain better results, especially in a subject like mathematics where much of the knowledge and skills depends on being able to understand the basics. Learning is all about knowing what you know and what you don’t know. Many students feel like they don’t know anything, but it’s surprising just how much they know already. Picking up the main concepts in class and not worrying too much about other less important parts can really help. The teacher can guide you on this. Here are some pointers to get the best out of classroom learning: ■ Take control and be responsible for your own learning ■ Clear your head of other issues in the classroom ■ Active, not passive, learning is more memorable ■ Ask questions if you don’t understand something ■ Listen for cues from the teacher ■ Look out for what are the main concepts Note taking varies from class to class, but there are some general guidelines that will help when you come to read over your notes later on at home: ■ Write legibly ■ Use different colours to highlight important points or formulae ■ Make notes in textbooks (using pencil if you don’t own the textbook) ■ Use highlighter pens to point out important points ■ Summarise the main points ■ If notes are scribbled, rewrite them at home

At home You are responsible for your own learning and nobody else can tell you how best to study. Some people need more revision time than others, some study better in the mornings while others do better at night, and some can work at home while others prefer a library. There are some general guidelines for studying at home: ■ Revise both new and older topics regularly ■ Have a realistic timetable and be flexible ■ Summarise the main points ■ Revise when you are fresh and energetic ■ Divide study time into smaller rather than longer chunks

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■ Study in a quiet environment ■ Have a balanced life and don’t forget to have fun! If you are given exercises out of a textbook to do for homework, consider asking the teacher if you can leave some of them till later and use these for revision. It is not necessary to do every exercise at one sitting, and you learn better if you can spread these over time. People use different learning styles to help them study. The more variety the better, and you will find some that help you more than others. Some people (around 35%) learn best visually, some (25%) learn best by hearing and others (40%) learn by doing. Here are some ideas to give you a variety of ways to study: ■ Summarise on cue cards or in a small notebook ■ Use colourful posters ■ Use mindmaps and diagrams ■ Discuss work with a group of friends ■ Read notes out aloud ■ Make up songs and rhymes ■ Do exercises regularly ■ Role play teaching someone else

Assessment tasks and exams Many of the assessment tasks for maths are closed book examinations. You will cope better in exams if you have practised doing sample exams under exam conditions. Regular revision will give you confidence and if you feel well prepared, this will help get rid of nerves in the exam. You will also cope better if you have had a reasonable night’s sleep before the exam. One of the biggest problems students have with exams is in timing. Make sure you don’t spend too much time on questions you’re unsure about, but work through and find questions you can do first. Divide the time up into smaller chunks for each question and allow some extra time to go back to questions you couldn’t do or finish. For example, in a 2 hour exam with 6 questions, allow around 15 minutes for each question. This will give an extra half hour at the end to tidy up and finish off questions. Here are some general guidelines for doing exams: ■ Read through and ensure you know how many questions there are ■ Divide your time between questions with extra time at the end ■ Don’t spend too much time on one question ■ Read each question carefully, underlining key words ■ Show all working out, including diagrams and formulae ■ Cross out mistakes with a single line so it can still be read ■ Write legibly

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And finally… Study involves knowing what you don’t know, and putting in a lot of time into concentrating on these areas. This is a positive way to learn. Rather than just saying, ‘I can’t do this’, say instead, ‘I can’t do this yet’, and use your teachers, friends, textbooks and other ways of finding out. With the parts of the course that you do know, make sure you can remember these easily under exam pressure by putting in lots of practice. Remember to look at new work ■ today ■ tomorrow ■ in a week ■ in a month Some people hardly ever find time to study while others give up their outside lives to devote their time to study. The ideal situation is to balance study with other aspects of your life, including going out with friends, working and keeping up with sport and other activities that you enjoy.

Good luck with your studies!

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Basic Arithmetic TERMINOLOGY Absolute value: The distance of a number from zero on the number line. Hence it is the magnitude or value of a number without the sign Directed numbers: The set of integers or whole numbers f -3, -2, -1, 0, 1, 2, 3, f Exponent: Power or index of a number. For example 23 has a base number of 2 and an exponent of 3 Index: The power of a base number showing how many times this number is multiplied by itself e.g. 2 3 = 2 # 2 # 2. The index is 3

Indices: More than one index (plural) Recurring decimal: A repeating decimal that does not terminate e.g. 0.777777 … is a recurring decimal that can be written as a fraction. More than one digit can recur e.g. 0.14141414 ... Scientific notation: Sometimes called standard notation. A standard form to write very large or very small numbers as a product of a number between 1 and 10 and a power of 10 e.g. 765 000 000 is 7.65 # 10 8 in scientific notation

Chapter 1 Basic Arithmetic

INTRODUCTION THIS CHAPTER GIVES A review of basic arithmetic skills, including knowing the

correct order of operations, rounding off, and working with fractions, decimals and percentages. Work on significant figures, scientific notation and indices is also included, as are the concepts of absolute values. Basic calculator skills are also covered in this chapter.

Real Numbers Types of numbers Unreal or imaginary numbers Real numbers

Rational numbers

Irrational numbers

Integers

Integers are whole numbers that may be positive, negative or zero. e.g. - 4, 7, 0, -11 a Rational numbers can be written in the form of a fraction b • 3 where a and b are integers, b ! 0. e.g. 1 , 3.7, 0. 5, - 5 4 a Irrational numbers cannot be written in the form of a fraction (that b is, they are not rational) e.g. 2 , r

EXAMPLE Which of these numbers are rational and which are irrational? • 3 r 3 , 1. 3, , 9 , , - 2.65 4 5

Solution r are irrational as they cannot be written as fractions (r is irrational). 4 • 3 13 1 1. 3 = 1 , 9 = and - 2.65 = - 2 so they are all rational. 3 1 20 3 and

3

4

Maths In Focus Mathematics Extension 1 Preliminary Course

Order of operations 1. Brackets: do calculations inside grouping symbols first. (For example, a fraction line, square root sign or absolute value sign can act as a grouping symbol.) 2. Multiply or divide from left to right. 3. Add or subtract from left to right.

EXAMPLE Evaluate 40 - 3 ] 5 + 4 g .

Solution 40 - 3 (5 + 4) = 40 - 3# 9 = 40 - 27 = 13

BRACKETS KEYS Use ( and ) to open and close brackets. Always use them in pairs. For example, to evaluate 40 - 3 ] 5 + 4 g press 40 - 3 #

( 5 + 4 ) = = 13 5.67 - 3.49 correct to 1 decimal place 1.69 + 2.77

To evaluate press :

(

( 5.67 - 3.49 )

'

( 1.69 + 2.77 )

)

=

= 0.7 correct to 1 decimal place

PROBLEM What is wrong with this calculation? 19 - 4 1+2 Press 19 - 4 ' 1 + 2 = 19 - 4 '1 + 2

Evaluate

What is the correct answer?

17


MEMORY KEYS Use STO to store a number in memory. There are several memories that you can use at the same time—any letter from A to F, or X, Y and M on the keypad. To store the number 50 in, say, A press 50 STO A To recall this number, press ALPHA A = To clear all memories press SHIFT CLR

X -1 KEY Use this key to find the reciprocal of x. For example, to evaluate 1 - 7.6 # 2.1 -1 = press ( (-) 7.6 # 2.1 ) x = - 0.063 (correct to 3 decimal places)

Rounding off Rounding off is often done in everyday life. A quick look at a newspaper will give plenty of examples. For example in the sports section, a newspaper may report that 50 000 fans attended a football match. An accurate number is not always necessary. There may have been exactly 49 976 people at the football game, but 50 000 gives an idea of the size of the crowd.

EXAMPLES 1. Round off 24 629 to the nearest thousand.

Solution This number is between 24 000 and 25 000, but it is closer to 25 000.

` 24 629 = 25 000 to the nearest thousand

CONTINUED

Different calculators use different keys so check the instructions for your calculator.

5

6


2. Write 850 to the nearest hundred.

Solution This number is exactly halfway between 800 and 900. When a number is halfway, we round it off to the larger number. ` 850 = 900 to the nearest hundred

In this course you will need to round off decimals, especially when using trigonometry or logarithms. To round a number off to a certain number of decimal places, look at the next digit to the right. If this digit is 5 or more, add 1 to the digit before it and drop all the other digits after it. If the digit to the right is less than 5, leave the digit before it and drop all the digits to the right.

EXAMPLES 1. Round off 0.6825371 correct to 1 decimal place. Add 1 to the 6 as the 8 is greater than 5.

Solution 0.6825371 # ` 0.6825371 = 0.7 correct to 1 decimal place 2. Round off 0.6825371 correct to 2 decimal places.

Drop off the 2 and all digits to the right as 2 is smaller than 5.

Solution 0.6825371 # ` 0.6825371 = 0.68 correct to 2 decimal places 3. Evaluate 3.56 ' 2.1 correct to 2 decimal places.

Check this on your calculator. Add 1 to the 69 as 5 is too large to just drop off.

Solution 3.56 ' 2.1 = 1.69 # 5238095

= 1.70 correct to 2 decimal places


FIX KEY Use MODE or SET UP to fix the number of decimal places (see the instructions for your calculator). This will cause all answers to have a fixed number of decimal places until the calculator is turned off or switched back to normal.

While using a fixed number of decimal places on the display, the calculator still keeps track internally of the full number of decimal places.

EXAMPLE Calculate 3.25 ' 1.72 # 5.97 + 7.32 correct to 2 decimal places.

Solution 3.25 ' 1.72 # 5.97 + 7.32 = 1.889534884 # 5.97 + 7.32 = 11.28052326 + 7.32 = 18.60052326 = 18.60 correct to 2 decimal places If the FIX key is set to 2 decimal places, then the display will show 2 decimal places at each step. 3.25 ' 1.72 # 5.97 + 7.32 = 1.89 # 5.97 + 7.32 = 11.28 + 7.32 = 18.60 If you then set the calculator back to normal, the display will show the full answer of 18.60052326.

The calculator does not round off at each step. If it did, the answer might not be as accurate. This is an important point, since some students round off each step in calculations and then wonder why they do not get the same answer as other students and the textbook.

1.1 Exercises 1.

State which numbers are rational and which are irrational. (a) 169 (b) 0.546 (c) -17 r (d) 3

•

(e) 0.34 (f)

218

(g) 2 2 1 (h) 27 (i) 17.4% 1 (j) 5

Don’t round off at each step of a series of calculations.

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8


2.

3.

Evaluate (a) 20 - 8 ' 4 (b) 3 # 7 - 2 # 5 (c) 4 # ] 27 ' 3 g ' 6 (d) 17 + 3 # - 2 (e) 1.9 - 2 # 3.1 14 ' 7 (f) -1 + 3 3 1 2 (g) 2 - # 5 5 3 3 1 1 4 8 (h) 5 6 5 5 ' 8 6 (i) 1 1 + 4 8 1 7 3 5 10 (j) 1 1 1 4 2

7.

A crowd of 10 739 spectators attended a tennis match. Write this figure to the nearest thousand.

8.

A school has 623 students. What is this to the nearest hundred?

9.

A bank made loans to the value of $7 635 718 last year. Round this off to the nearest million.

Evaluate correct to 2 decimal places. (a) 2.36 + 4.2 ' 0.3 (b) ] 2.36 + 4.2 g ' 0.3 (c) 12.7 # 3.95 ' 5.7 (d) 8.2 ' 0.4 + 4.1# 0.54 (e) ] 3.2 - 6.5 g # ] 1.3 + 2.7 g 1 (f) 4.7 + 1.3 1 (g) 4.51 + 3.28

13. Round off 32.569148 to the nearest unit.

0.9 + 1.4 (h) 5.2 - 3.6 5.33 + 2.87 (i) 1.23 - 3.15 (j) 4.

1.7 2 + 8.9 2 - 3.94 2

Round off 1289 to the nearest hundred.

5.

Write 947 to the nearest ten.

6.

Round off 3200 to the nearest thousand.

10. A company made a profit of $34 562 991.39 last year. Write this to the nearest hundred thousand. 11. The distance between two cities is 843.72 km. What is this to the nearest kilometre? 12. Write 0.72548 correct to 2 decimal places.

14. Round off 3.24819 to 3 decimal places. 15. Evaluate 2.45 # 1.72 correct to 2 decimal places. 16. Evaluate 8.7 ' 5 correct to 1 decimal place. 17. If pies are on special at 3 for $2.38, find the cost of each pie. 18. Evaluate 7.48 correct to 2 decimal places. 6.4 + 2.3 correct to 8 1 decimal place.

19. Evaluate

20. Find the length of each piece of material, to 1 decimal place, if 25 m of material is cut into 7 equal pieces.


21. How much will 7.5 m 2 of tiles cost, at $37.59 per m2?

3.5 + 9.8 5.6 + 4.35 15.9 + 6.3 - 7.8 (d) 7.63 - 5.12 1 (e) 6.87 - 3.21

(c)

22. Divide 12.9 grams of salt into 7 equal portions, to 1 decimal place. 23. The cost of 9 peaches is $5.72. How much would 5 peaches cost?

9.91 - ] 9.68 - 5.47 g 5.39 2 correct to 1 decimal place.

25. Evaluate

24. Evaluate correct to 2 decimal places. (a) 17.3 - 4.33 # 2.16 (b) 8.72 # 5.68 - 4.9 # 3.98

DID YOU KNOW? In building, engineering and other industries where accurate measurements are used, the number of decimal places used indicates how accurate the measurements are. For example, if a 2.431 m length of timber is cut into 8 equal parts, according to the calculator each part should be 0.303875 m. However, a machine could not cut this accurately. A length of 2.431 m shows that the measurement of the timber is only accurate to the nearest mm (2.431 m is 2431 mm). The cut pieces can also only be accurate to the nearest mm (0.304 m or 304 mm). The error in measurement is related to rounding off, as the error is half the smallest measurement. In the above example, the measurement error is half a millimetre. The length of timber could be anywhere between 2430.5 mm and 2431.5 mm.

Directed Numbers Many students use the calculator with work on directed numbers (numbers that can be positive or negative). Directed numbers occur in algebra and other topics, where you will need to remember how to use them. A good understanding of directed numbers will make your algebra skills much better.

^ - h KEY Use this key to enter negative numbers. For example, press (-) 3

=

9

10


Adding and subtracting

To add: move to the right along the number line To subtract: move to the left along the number line -4

-3

-2

-1

0

1

Subtract

2

3

4

Add

EXAMPLES You can also do these on a calculator, or you may have a different way of working these out.

Evaluate 1. - 4 + 3

Solution Start at - 4 and move 3 places to the right.

-4

-3

-2

-1

0

1

2

3

4

2

3

4

- 4 + 3 = -1 2. -1 - 2

Solution Start at -1 and move 2 places to the left.

-4

-3

-2

-1

0

1

-1 - 2 = -3

Multiplying and dividing To multiply or divide, follow these rules. This rule also works if there are two signs together without a number in between e.g. 2 - -3

Same signs = + + + =+ - - =+ Different signs = + - =- + =-


11

EXAMPLES Evaluate 1. - 2 #7

Solution Different signs (- 2 and + 7) give a negative answer. - 2 # 7 = -14 2. -12 ' - 4

Solution Same signs (-12 and - 4) give a positive answer. -12 ' - 4 = 3 3. -1 - - 3

Solution The signs together are the same (both negative) so give a positive answer. -

-1 - 3 = -1 + 3 =2

1.2 Exercises Evaluate 1.

-2 + 3

11. 5 - 3 # 4

2.

-7 - 4

12. - 2 + 7 # - 3

3.

8 # -7

13. 4 - 3 # - 2

4.

7 - ]-3 g

14. -1 - -2

5.

28 ' -7

15. 7 + - 2

6.

- 4 . 9 + 3 .7

16. 2 - ] -1 g

7.

- 2.14 - 5.37

17. - 2 + 15 ' 5

8.

4.8 # -7.4

18. - 2 # 6 # - 5

9.

1.7 - ] - 4.87 g

19. - 28 ' -7 # - 5

10. -

3 2 -1 5 3

20. ] - 3 g2

Start at -1 and move 3 places to the right.

12


Fractions, Decimals and Percentages Conversions You can do all these conversions on your calculator using the b a or S + D key. c

EXAMPLES 1. Write 0.45 as a fraction in its simplest form.

Solution 45 5 ' 5 100 9 = 20

0.45 =

3 means 3 ' 8. 8

2. Convert

3 to a decimal. 8

Solution 0.375 8 g 3.000 3 So = 0.375 8 3. Change 35.5% to a fraction.

Solution 35.5 2 # 100 2 71 = 200

35.5% =

4. Write 0.436 as a percentage.

Solution Multiply by 100% to change a fraction or decimal to a percentage.

0.436 = 0.436 #100% = 43.6% 5. Write 20 g as a fraction of 1 kg in its simplest form.

Solution 1 kg = 1000 g 20 g 20 g = 1000 g 1 kg 1 = 50


13

6. Find the percentage of people who prefer to drink Lemon Fuzzy, if 24 out of every 30 people prefer it.

Solution 24 100% # = 80% 30 1

Sometimes decimals repeat, or recur. Example • 1 = 0.33333333 f = 0. 3 3 There are different methods that can be used to change a recurring decimal into a fraction. Here is one way of doing it. Later you will discover another method when studying series. (See HSC Course book, Chapter 8.)

EXAMPLES

A rational number is any number that can be written as a fraction.

•

1. Write 0. 4 as a rational number.

Solution Let n = 0.44444 f Then 10n = 4.44444 f (2) - (1): 9n = 4 4 n= 9

( 1) ( 2)

Check this on your calculator by dividing 4 by 9.

• •

2. Change 1.329 to a fraction.

Solution n = 1.3292929 f Let Then 100n = 132.9292929 f (2) - (1): 99n = 131.6 131.6 10 n= # 99 10 1316 = 990 163 =1 495

( 1) ( 2)

CONTINUED

Try multiplying n by 10. Why doesn’t this work?

14


Another method Let n = 1.3292929 f Then 10n = 13.2929292 f and 1000n = 1329.292929 f (2) - (1): 990n = 1316 1316 n= 990 163 =1 495

This method avoids decimals in the fraction at the end.

(1 ) (2 )

1.3 Exercises 1.

2.

3.

Write each decimal as a fraction in its lowest terms. (a) 0.64 (b) 0.051 (c) 5.05 (d) 11.8 Change each fraction into a decimal. 2 (a) 5 7 (b) 1 8 5 (c) 12 7 (d) 11 Convert each percentage to a fraction in its simplest form. (a) 2% (b) 37.5% (c) 0.1% (d) 109.7%

4.

Write each percentage as a decimal. (a) 27% (b) 109% (c) 0.3% (d) 6.23%

5.

Write each fraction as a percentage. 7 20 1 (b) 3 (a)

4 15 1 (d) 1000

(c) 2

6.

Write each decimal as a percentage. (a) 1.24 (b) 0.7 (c) 0.405 (d) 1.2794

7.

Write each percentage as a decimal and as a fraction. (a) 52% (b) 7% (c) 16.8% (d) 109% (e) 43.4% 1 (f) 12 % 4

8.

Write these fractions as recurring decimals. 5 (a) 6 7 (b) 99 13 (c) 99 1 (d) 6 2 (e) 3


5 33 1 (g) 7 2 (h) 1 11

31 99 13 + 6 (e) 7+4 (d) 1 -

(f)

9.

Express as fractions in lowest terms. •

(a) 0. 8 (b) (c) (d) (e) (f) (g)

•

0. 2 • 1. 5 • 3. 7 • • 0. 67 • • 0. 54 • 0.15 •

(h) 0.216 • • (i) 0.2 19 • • (j) 1.074 10. Evaluate and express as a decimal. 5 (a) 3+6 (b) 8 - 3 ' 5 4+7 (c) 12 + 3

11. Evaluate and write as a fraction. (a) 7.5 ' ] 4.1 + 7.9 g 15.7 - 8.9 (b) 4.5 - 1.3 6.3 + 1.7 (c) 12.3 - 8.9 + 7.6 4 .3 (d) 11.5 - 9.7 64 (e) 8100 12. Angel scored 17 out of 23 in a class test. What was her score as a percentage, to the nearest unit? 13. A survey showed that 31 out of 40 people watched the news on Monday night. What percentage of people watched the news? 14. What percentage of 2 kg is 350 g? 15. Write 25 minutes as a percentage of an hour.

Investigation Explore patterns in recurring decimals by dividing numbers by 3, 6, 9, 11, and so on. Can you predict what the recurring decimal will be if a fraction has 3 in the denominator? What about 9 in the denominator? What about 11? Can you predict what fraction certain recurring decimals will be? What denominator would 1 digit recurring give? What denominator would you have for 2 digits recurring?

Operations with fractions, decimals and percentages You will need to know how to work with fractions without using a calculator, as they occur in other areas such as algebra, trigonometry and surds.

15

16


The examples on fractions show how to add, subtract, multiply or divide fractions both with and without the calculator. The decimal examples will help with some simple multiplying and the percentage examples will be useful in Chapter 8 of the HSC Course book when doing compound interest. Most students use their calculators for decimal calculations. However, it is important for you to know how to operate with decimals. Sometimes the calculator can give a wrong answer if the wrong key is pressed. If you can estimate the size of the answer, you can work out if it makes sense or not. You can also save time by doing simple calculations in your head.

DID YOU KNOW? Some countries use a comma for the decimal point—for example, 0,45 for 0.45. This is the reason that our large numbers now have spaces instead of commas between digits—for example, 15 000 rather than 15,000.

EXAMPLES 1. Evaluate 1

3 2 - . 5 4

Solution 1

3 3 2 7 - = 5 4 5 4 28 15 = 20 20 13 = 20

2. Evaluate 2

1 ' 3. 2

Solution 2

3 5 1 '3 = ' 2 2 1 5 1 = # 2 3 5 = 6

3. Evaluate 0.056 # 100. Move the decimal point 2 places to the right.

Solution 0.056 #100 = 5.6


17

4. Evaluate 0.02 # 0.3. Multiply the numbers and count the number of decimal places in the question.

Solution 0.02 # 0.3 = 0.006 5. Evaluate

8.753 . 10

Solution

Move the decimal point 1 place to the left.

8.753 ' 10 = 0.8753 1 6. The price of a $75 tennis racquet increased by 5 %. Find the new 2 price.

Solution 1 5 % = 0.055 2 1 ` 5 % of $75 = 0.055#$75 2 = $4.13

1 or 105 % of $75 = 1.055#$75 2 = $79.13

So the price increases by $4.13 to $79.13. 7. The price of a book increased by 12%. If it now costs $18.00, what did it cost before the price rise?

Solution The new price is 112% (old price 100%, plus 12%) $18.00 ` 1% = 112 $18.00 100 100% = # 112 1 = $16.07 So the old price was $16.07.

1.4 Exercises 1.

Write 18 minutes as a fraction of 2 hours in its lowest terms.

2.

Write 350 mL as a fraction of 1 litre in its simplest form.

3.

Evaluate 3 1 (a) + 5 4

2 7 -2 5 10 3 2 (c) #1 5 4 3 (d) ' 4 7 3 2 (e) 1 ' 2 5 3 (b) 3

18


3 of $912.60. 5

4.

Find

5.

5 Find of 1 kg, in grams correct 7 to 1 decimal place.

6.

Trinh spends sleeping,

1 of her day 3

7 1 at work and 24 12

eating. What fraction of the day is left? 7.

I get $150.00 a week for a casual 1 job. If I spend on bus fares, 10 2 1 on lunches and on outings, 15 3 how much money is left over for savings?

8.

John grew by

9.

17 of his height 200 this year. If he was 165 cm tall last year, what is his height now, to the nearest cm? Evaluate (a) 8.9 + 3 (b) 9 - 3.7 (c) 1.9 #10 (d) 0.032 #100 (e) 0.7 # 5 (f) 0.8 # 0.3 (g) 0.02 # 0.009 (h) 5.72 #1000 8.74 (i) 100 (j) 3.76 # 0.1

10. Find 7% of $750. 11. Find 6.5% of 845 mL. 12. What is 12.5% of 9217 g? 13. Find 3.7% of $289.45. 14. If Kaye makes a profit of $5 by selling a bike for $85, find the profit as a percentage of the selling price.

15. Increase 350 g by 15%. 1 16. Decrease 45 m by 8 %. 2 17. The cost of a calculator is now $32. If it has increased by 3.5%, how much was the old cost? 18. A tree now measures 3.5 m, which is 8.3% more than its previous year’s height. How high was the tree then, to 1 decimal place?

19. This month there has been a 4.9% increase in stolen cars. If 546 cars were stolen last month, how many were stolen this month? 20. George’s computer cost $3500. If it has depreciated by 17.2%, what is the computer worth now?


19

PROBLEM If both the hour hand and minute hand start at the same position at 12 o’clock, when is the first time, correct to a fraction of a minute, that the two hands will be together again?

Powers and Roots A power (or index) of a number shows how many times a number is multiplied by itself.

EXAMPLES 1. 4 3 = 4 # 4 # 4 = 64 2. 2 5 = 2 # 2 # 2 # 2 # 2 = 32 A root of a number is the inverse of the power.

EXAMPLES 1.

36 = 6 since 6 2 = 36

2.

3

8 = 2 since 2 3 = 8

3.

6

64 = 2 since 2 6 = 64

DID YOU KNOW? Many formulae use indices (powers and roots). For example the compound interest formula that you will study in Chapter 8 of the HSC n Course book is A = P ^ 1 + r h 4 Geometry uses formulae involving indices, such as V = rr 3. Do you know what this 3 formula is for? In Chapter 7, the formula for the distance between 2 points on a number plane is d=

2

(x 2 - x 1) + (y 2 - y 1)

2

See if you can find other formulae involving indices.

In 4 3 the 4 is called the base number and the 3 is called the index or power.

20


POWER AND ROOT KEYS Use the x 2 and x 3 keys for squares and cubes. y Use the x or ^ key to find powers of numbers.

key for square roots.

Use the

These laws work for any m and n, including fractions and negative numbers.

Use the

3

key for cube roots.

Use the

x

for other roots.

Index laws There are some general laws that simplify calculations with indices.

am # an = am + n

Proof a m # a n = (a # a #f# a) # (a # a #f# a) 14444244443 14444244443 m times n times =a # # f # a a 14444244443 m + n times = am + n

am ' an = am - n

Proof am an a # a #f# a (m times) = a # a #f# a (n times) a # a #f# a (m - n times) = 1 = am - n

am ' an =

(a m)n = a mn

Proof (a m) n = a m # a m # a m #f# a m = am + m + m + f + m = a mn

(n times) (n times)


(ab) n = a n b n

Proof (ab) n = ab # ab # ab #f# ab (n times) = (a # a #f# a) # (b # b #f# b) 14444244443 14444244443 n times n times = an bn

a n an c m = n b b

Proof a n a a a a c m = # # #f# b b b b b a # a # a #f # a = b # b # b #f # b an = n b

(n times) (n times) (n times)

EXAMPLES Simplify 1. m 9 # m 7 ' m 2

Solution m9 #m7 ' m2 = m9 + 7 - 2 = m 14 2. (2y 4)3

Solution (2y 4) 3 = 2 3 (y 4) 3 = 23 y4 # 3 = 8y 12

CONTINUED

21

22


3.

(y 6) 3 # y - 4 y5

Solution (y 6) 3 # y - 4 y5

= = =

y 18 # y - 4 y5

y 18 + (- 4) y5 y

14

y5 = y9

1.5 Exercises 1.

Evaluate without using a calculator. (a) 5 3 # 2 2 (b) 3 4 + 8 2 1 3 (c) c m 4 (d) (e)

2.

3.

3 4

(h) (i) (j) (k)

5

x2 p y9 w6 # w7 (m) w3 2 p #(p 3) 4 (n) p9 6 x ' x7 (o) x2 2 a # ( b 2) 6 (p) a4 # b9 (x 2) - 3 #(y 3) 2 (q) x -1 # y 4 (l) f

27 16

Evaluate correct to 1 decimal place. (a) 3.7 2 (b) 1.06 1.5 (c) 2.3 - 0.2 (d) 3 19 (e) 3 34.8 - 1.2 # 43.1 1 (f) 3 0.99 + 5.61 Simplify (a) a 6 # a 9 # a 2 (b) y 3 # y - 8 # y 5 (c) a -1 # a -3 1

1

(d) w 2 # w 2 (e) x 6 ' x (f) p 3 ' p - 7 y 11 (g) 5 y

(x 7) 3 (2x 5) 2 (3y - 2) 4 a3 #a5 ' a7

4.

Simplify (a) x 5 # x 9 (b) a -1 # a - 6 m7 (c) m3 (d) k 13 # k 6 ' k 9 (e) a - 5 # a 4 # a - 7 2

3

(f) x 5 # x 5 m5 # n4 (g) 4 m # n2


1

1

p2 # p2

(h)

10. (a) Simplify

p (i) (3x 11) 2 (x 4) 6 (j) x3

5.

2

1

2 6 11. Evaluate (a ) when a = c m . 3 12. Evaluate b=

(2m 7) 3

m4 xy 3 #(xy 2) 4 (f) xy 8 4 (2k ) (g) (6k 3) 3 y 12 7 (h) _ 2y 5 i # 8

y=

Evaluate a3b2 when a = 2 and 3 b= . 4

7.

If x = of

8.

9.

2 1 and y = , find the value 3 9

x3 y2 xy 5

.

1 1 1 , b = and c = , 4 2 3 a2 b3 evaluate 4 as a fraction. c

If a =

a b . a8 b7 11

(a) Simplify

8

(b) Hence evaluate a=

a 11 b 8 when a8 b7

5 2 and b = as a fraction. 5 8

x5 y5

when x =

1 and 3

14. Evaluate

k-5 1 when k = . 3 k-9

15. Evaluate

a4 b6 3 when a = and 3 2 2 4 a (b )

b=

6.

x4 y7

2 . 9

-3

p

a3 b6 1 when a = and 2 b4

2 . 3

13. Evaluate

a6 # a4 o a 11 3 5xy 9 x8 # y3

p5 q8 r4

4 3

(d) (7a5b)2

(j) f

.

as a p4 q6 r2 7 2 fraction when p = , q = and 8 3 3 r= . 4

a 8 (b) c m b 4a 3 (c) d 4 n b

(i) e

p4 q6 r2

(b) Hence evaluate

Simplify (a) (pq 3) 5

(e)

p5 q8 r4

1 . 9

a6 # b3 as a fraction a5 # b2 3 1 when a = and b = . 4 9

16. Evaluate

a2 b7 as a fraction in a3 b 2 4 index form when a = c m and 5 5 3 b=c m. 8

17. Evaluate

18. Evaluate

(a 3) 2 b 4 c

as a fraction a (b 2) 4 c 3 6 1 7 when a = , b = and c = . 7 3 9

23

24


Negative and zero indices

Class Investigation Explore zero and negative indices by looking at these questions. For example simplify x 3 ' x 5 using (i) index laws and (ii) cancelling. (i) x 3 ' x 5 = x - 2 by index laws 3 x# x# x (ii) x = 5 x x# x# x# x # x 1 = 2 x 1 So x - 2 = 2 x Now simplify these questions by (i) index laws and (ii) cancelling. (a) x 2 ' x 3 (b) x 2 ' x 4 (c) x 2 ' x 5 (d) x 3 ' x 6 (e) x 3 ' x 3 (f) x 2 ' x 2 (g) x ' x 2 (h) x 5 ' x 6 (i) x 4 ' x 7 (j) x ' x 3 Use your results to complete: x0 = x-n =

x0 = 1

Proof xn ' xn = xn - n = x0 xn xn ' xn = n x =1 `

x0 = 1


x-n =

1 xn

Proof x0 ' xn = x0 - n = x-n x0 x0 ' xn = n x 1 = n x 1 ` x-n = n x

EXAMPLES 0

1. Simplify e

Solution

ab 5 c o . abc 4

0

e

ab 5 c o =1 abc 4

2. Evaluate 2 - 3 .

Solution 1 23 1 = 8

2-3 =

3. Write in index form. 1 x2 3 (b) 5 x 1 (c) 5x 1 (d) x +1 (a)

CONTINUED

25

26


Solution 1 = x-2 x2 3 (b) 5 = 3# 15 x x -5 = 3x 1 1 1 = #x (c) 5x 5 1 -1 = x 5 1 1 = (d) x +1 (x + 1) 1 = ] x + 1 g-1 (a)

4. Write a−3 without the negative index.

Solution a-3 =

1 a3

1.6 Exercises 1.

Evaluate as a fraction or whole number. (a) 3 - 3 (b) 4 - 1 (c) 7 - 3 (d) 10 - 4 (e) 2 - 8 (f) 60 (g) 2 - 5 (h) 3 - 4 (i) 7 - 1 (j) 9 - 2 (k) 2 - 6 (l) 3 - 2 (m) 40 (n) 6 - 2 (o) 5 - 3 (p) 10 - 5 (q) 2 - 7 (r) 2 0 (s) 8 - 2 (t) 4 - 3

2.

Evaluate (a) 2 0 1 -4 (b) c m 2 2 -1 (c) c m 3 5 -2 (d) c m 6 x + 2y 0 p (e) f 3x - y 1 -3 (f) c m 5 3 -1 (g) c m 4 1 -2 (h) c m 7 2 -3 (i) c m 3 1 -5 (j) c m 2 3 -1 (k) c m 7


8 0 (l) c m 9 6 -2 (m)c m 7 9 -2 (n) c m 10 6 0 (o) c m 11 1 -2 (p) c - m 4 2 -3 (q) c - m 5 2 -1 (r) c - 3 m 7 3 0 (s) c - m 8 1 -2 (t) c - 1 m 4 3.

Change into index form. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

1 m3 1 x 1 p7 1 d9 1 k5 1 x2 2 x4 3 y2 1 2z 6 3 5t 8 2 7x

5 2m 6 2 (m) 7 3y (l)

1 (3x + 4) 2 1 (o) ( a + b) 8 1 (p) x-2 (n)

1 (5p + 1) 3 2 (r) (4t - 9) 5 1 (s) 4 (x + 1) 11 5 (t) 9 ( a + 3 b) 7 (q)

4.

Write without negative indices. (a) t - 5 (b) x - 6 (c) y - 3 (d) n - 8 (e) w - 10 (f) 2x -1 (g) 3m - 4 (h) 5x - 7 (i) ]2xg- 3 (j) ] 4n g-1 (k) ] x + 1 g- 6 (l) ^ 8y + z h-1 (m) ]k - 3g- 2 (n) ^ 3x + 2y h- 9 1 -5 (o) b x l 1 -10 (p) c y m 2 -1 (q) d n p 1 -2 m a+b x + y -1 (s) e x - y o (r) c

(t) e

2w - z - 7 o 3x + y

27

28


Fractional indices

Class Investigation Explore fractional indices by looking at these questions. 1 2

For example simplify (i) ` x 2 j and (ii) ^ x h . 1 2

(i) ` x 2 j = x 1 =x 2 (ii) ^ x h = x

2

^ by index laws h

1 2

So ` x 2 j = ^ x h = x 2

1

`

x2 =

x

Now simplify these questions. 1

(a) ^ x 2 h 2 x2

(b)

1 3

(c) ` x 3 j

1

(d) ^ x 3 h 3 3 (e) ^ 3 x h

(f)

3

x3 1 4

4 (g) ` x j

1

(h) ^ x 4 h 4 4 (i) ^ 4 x h

(j)

4

x4

Use your results to complete: 1

xn =

1 n

a =n a

Proof 1 n

àn j = a ^ n a hn = a 1 n

` a =n a

^ by index laws h


EXAMPLES 1. Evaluate (a) 49

1 2 1

(b) 27 3

Solution 1 2

(a) 49 = 49 =7 1 3

(b) 27 = 3 27 =3 2. Write

3x - 2 in index form.

Solution 1

3x - 2 = (3x - 2) 2 1

3. Write (a + b) 7 without fractional indices.

Solution 1

( a + b) 7 = 7 a + b

Putting the fractional and negative indices together gives this rule.

a

1 -n

=

n

1 a

Here are some further rules.

m n

a = n am = (n a ) m

Proof m

1 m

m n

1 n

n n a = à j m = ^n a h

a = ^ am h = n am

29

30


a -n b n c m = bal b

Proof a -n 1 c m = b a n c m b 1 = n a bn

an bn bn =1# n a bn = n a b n = bal =1'

EXAMPLES 1. Evaluate 4

(a) 8 3 (b) 125

-

1 3

2 -3 (c) c m 3

Solution 4

(a) 8 3 = (3 8 ) 4 (or 3 8 4 ) = 24 = 16 (b) 125

-

1 3

=

1 1

125 3 1 =3 125 1 = 5


-3

(c) c 2 m 3

3 3 =c m 2 27 = 8 3 =3 8

2. Write in index form. x5

(a) (b)

1 (4x - 1) 2 2

3

Solution 5

x5 = x 2 1

(a) (b)

(4x - 1) 2

3

2

=

1 2

(4x 2 - 1) 3 -

= (4x 2 - 1) 3. Write r

-

3 5

2 3

without the negative and fractional indices.

Solution r

-

3 5

= =

1 3

r5 1 5

r3

DID YOU KNOW? Nicole Oresme (1323–82) was the first mathematician to use fractional indices. John Wallis (1616–1703) was the first person to explain the significance of zero, negative and fractional indices. He also introduced the symbol 3 for infinity. Do an Internet search on these mathematicians and find out more about their work and backgrounds. You could use keywords such as indices and infinity as well as their names to find this information.

31

32


1.7 Exercises 1.

3.

Evaluate (a) 81

1 2

Write without fractional indices. 1

(a) y 3

1

2

(b) y 3

(b) 27 3 1

(c) x

(c) 16 2 1

-

1 2 1

(d) (2x + 5) 2

(d) 8 3 1

(e) (3x - 1)

(e) 49 2 1

-

1 2

1

(f) (6q + r) 3

(f) 1000 3 1

(g) (x + 7)

(g) 16 4

-

2 5

1

(h) 64 2 (i) 64 (j) 1

4.

1 3

(l) 32

t

(a)

1 7

(k) 81

Write in index form. (b)

5

x3

(c)

1 4

(d) (e)

1 5

3

1

(m) 0 8

(f)

(n) 125

1 3

(g)

1 1 1

(r) 9 (s) 8

(i)

(t) 64 2.

(x - 2) 2 1 (j) 2 y+7 5 (k) 3 x+4 2 (l) 3 y2 - 1 3 (m) 5 4 (x 2 + 2) 3

3 2 -

1 3 -

2 3

Evaluate correct to 2 decimal places. 1

(a) 23 4 (b) 4 45.8 (c) (d) (e)

7

5

8

5 .9 # 3 .7 8.79 - 1.4

4

(f)

1.24 + 4.3 2 1 12.9 3 .6 - 1 .4 1 .5 + 3 .7

(3x + 1) 5 1

(h)

(q) 256 4

9-x 4s + 1 1 2t + 3 1 (5x - y) 3

(o) 343 3 (p) 128 7

y

5.

3

Write in index form and simplify. (a) x x x (b) x x (c) 3 x x2 (d) 3 x (e) x 4 x


6.

Expand and simplify, and write in index form.

7.

(a) ( x + x) 2 (b) (3 a + 3 b ) (3 a - 3 b ) 1 2 (c) f p + p p 1 2 ) x x ( x 2 - 3x + 1 )

(d) ( x + (e)

x3

33

Write without fractional or negative indices. (a) (a - 2b) (b) (y - 3)

-

-

1 3

2 3

-

4 7

-

2 9

(c) 4 (6a + 1) ( x + y) (d) 3

-

5 4

6 (3 x + 8 ) (e) 7

Scientific notation (standard form) Very large or very small numbers are usually written in scientific notation to make them easier to read. What could be done to make the figures in the box below easier to read?

DID YOU KNOW? The Bay of Fundy, Canada, has the largest tidal changes in the world. About 100 000 000 000 tons of water are moved with each tide change. The dinosaurs dwelt on Earth for 185 000 000 years until they died out 65 000 000 years ago. The width of one plant cell is about 0.000 06 m. In 2005, the total storage capacity of dams in Australia was 83 853 000 000 000 litres and households in Australia used 2 108 000 000 000 litres of water.

A number in scientific notation is written as a number between 1 and 10 multiplied by a power of 10.

EXAMPLES 1. Write 320 000 000 in scientific notation.

Solution 320 000 000 = 3.2 #10 8

Write the number between 1 and 10 and count the decimal places moved.

2. Write 7.1#10 -5 as a decimal number.

Solution 7.1#10

-5

= 7.1 ' 10 = 0.000 071 5

Count 5 places to the left.

34


SCIENTIFIC NOTATION KEY Use the EXP or #10 x key to put numbers in scientific notation. For example, to evaluate 3.1#10 4 ' 2.5 #10 - 2, press 3.1 EXP 4 ' 2.5 EXP (-) 2 = = 1 240 000

DID YOU KNOW? Engineering notation is similar to scientific notation, except the powers of 10 are always multiples of 3. For example, 3.5 # 10

3

15.4 # 10

-6

SIGNIFICANT FIGURES The concept of significant figures is related to rounding off. When we look at very large (or very small) numbers, some of the smaller digits are not significant. For example, in a football crowd of 49 976, the 6 people are not really significant in terms of a crowd of about 50 000! Even the 76 people are not significant. When a company makes a profit of $5 012 342.87, the amount of 87 cents is not exactly a significant sum! Nor is the sum of $342.87. To round off to a certain number of significant figures, we count from the first non-zero digit. In any number, non-zero digits are always significant. Zeros are not significant, except between two non-zero digits or at the end of a decimal number. Even though zeros may not be significant, they are still necessary. For example 31, 310, 3100, 31 000 and 310 000 all have 2 significant figures but are very different numbers! Scientific notation uses the significant figures in a number.

EXAMPLES 12 000 = 1.2 #10 4 0.000 043 5 = 4.35#10 - 5 0.020 7 = 2.07 #10 - 2

(2 significant figures) (3 significant figures) (3 significant figures)

When rounding off to significant figures, use the usual rules for rounding off.


35

EXAMPLES 1. Round off 4 592 170 to 3 significant figures.

Solution 4 592 170 = 4 590 000 to 3 significant figures 2. Round off 0.248 391 to 2 significant figures.

Solution 0.248 391 = 0.25 to 2 significant figures 3. Round off 1.396 794 to 3 significant figures.

Solution 1.396 794 = 1.40 to 3 significant figures

1.8 Exercises 1.

Write in scientific notation. (a) 3 800 (b) 1 230 000 (c) 61 900 (d) 12 000 000 (e) 8 670 000 000 (f) 416 000 (g) 900 (h) 13 760 (i) 20 000 000 (j) 80 000

3.

Write as a decimal number. (a) 3.6 #10 4 (b) 2.78 #10 7 (c) 9.25#10 3 (d) 6.33#10 6 (e) 4 #10 5 (f) 7.23#10 - 2 (g) 9.7 #10 - 5 (h) 3.8 # 10 - 8 (i) 7 #10 - 6 (j) 5#10 - 4

2.

Write in scientific notation. (a) 0.057 (b) 0.000 055 (c) 0.004 (d) 0.000 62 (e) 0.000 002 (f) 0.000 000 08 (g) 0.000 007 6 (h) 0.23 (i) 0.008 5 (j) 0.000 000 000 07

4.

Round these numbers to 2 significant figures. (a) 235 980 (b) 9 234 605 (c) 10 742 (d) 0.364 258 (e) 1.293 542 (f) 8.973 498 011 (g) 15.694 (h) 322.78 (i) 2904.686 (j) 9.0741

Remember to put the 0’s in!

36


5.

Evaluate correct to 3 significant figures. (a) 14.6 # 0.453 (b) 4.8 ' 7 (c) 4.47 + 2.59 #1.46 1 (d) 3.47 - 2.7

6.

Evaluate 4.5#10 4 # 2.9 #10 5, giving your answer in scientific notation.

7.

Calculate

8.72 #10 - 3 and write 1.34 #10 7 your answer in standard form correct to 3 significant figures.

Investigation A logarithm is an index. It is a way of finding the power (or index) to which a base number is raised. For example, when solving 3 x = 9, the solution is x = 2. The 3 is called the base number and the x is the index or power. You will learn about logarithms in the HSC course. The a is called the base number and the x is the index or power.

If a x = y then log a y = x

1. The expression log7 49 means the power of 7 that gives 49. The solution is 2 since 7 2 = 49. 2. The expression log2 16 means the power of 2 that gives 16. The solution is 4 since 2 4 = 16. Can you evaluate these logarithms? 1. log3 27 2. log5 25 3. log10 10 000 4. log2 64 5. log4 4 6. log7 7 7. log3 1 8. log4 2 1 9. log 3 3 1 10. log 2 4


37

Absolute Value Negative numbers are used in maths and science, to show opposite directions. For example, temperatures can be positive or negative.

But sometimes it is not appropriate to use negative numbers. For example, solving c 2 = 9 gives two solutions, c = !3. However when solving c 2 = 9, using Pythagoras’ theorem, we only use the positive answer, c = 3, as this gives the length of the side of a triangle. The negative answer doesn’t make sense. We don’t use negative numbers in other situations, such as speed. In science we would talk about a vehicle travelling at –60k/h going in a negative direction, but we would not commonly use this when talking about the speed of our cars!

Absolute value definitions We write the absolute value of x as x

x =)

We can also define x as the distance of x from 0 on the number line. We will use this in Chapter 3.

x when x $ 0 - x when x 1 0

EXAMPLES 1. Evaluate 4 .

Solution 4 = 4 since 4 $ 0

CONTINUED

38


2. Evaluate - 3 .

Solution -3 = - ] - 3 g since - 3 1 0 =3

The absolute value has some properties shown below.

Properties of absolute value

| ab | = | a |#| b |

e.g. | 2 # - 3 | = | 2 |#| - 3 | = 6

|a | = a

e.g. | - 3 | 2 = ] - 3 g2 = 9

2

2

a2 = | a | |- a | = | a | |a - b | = | b - a | | a + b |#| a | + | b |

e.g. 5 2 = | 5 | = 5 e.g. | -7 | = | 7 | = 7 e.g. | 2 - 3 | = | 3 - 2 | = 1 e.g. | 2 + 3 | = | 2 | + | 3 | but | - 3 + 4 | 1 | - 3 | + | 4 |

EXAMPLES 1. Evaluate 2 - -1 + - 3 2.

Solution 2 - -1 + - 3 2 = 2 - 1 + 3 2 =2 -1 + 9 = 10 2. Show that a + b # a + b when a = - 2 and b = 3.

Solution LHS means Left Hand Side.

LHS = a + b = -2 + 3 = 1 =1


RHS means Right Hand Side.

RHS = a + b = -2 + 3 = 2+3 =5 Since 11 5 a+b # a + b 3. Write expressions for 2x - 4 without the absolute value signs.

Solution 2x - 4 = 2x - 4 when 2x - 4 $ 0 i.e. 2x $ 4 x$2 2x - 4 = - ] 2x - 4 g when 2x - 4 1 0 = - 2x + 4 i.e.

2x 1 4 x12

Class Discussion Are these statements true? If so, are there some values for which the expression is undefined (values of x or y that the expression cannot have)?

2.

x =1 x 2x = 2x

3.

2x = 2 x

4.

x + y = x+y

5.

2 x = x2

6. 7.

3 x = x3 x +1 = x +1

1.

3x - 2 =1 3x - 2 x 9. =1 x2 10. x $ 0 8.

Discuss absolute value and its definition in relation to these statements.

39

40


1.9 Exercises 1.

2.

3.

Evaluate (a) 7 (b) - 5 (c) - 6 (d) 0 (e) 2 (f) -11 (g) - 2 3 (h) 3 - 8 2 (i) - 5 (j) - 5 3 Evaluate (a) 3 + - 2 (b) - 3 - 4 (c) - 5 + 3 (d) 2 #-7 (e) - 3 + -1 2 (f) 5 - - 2 # 6 (g) - 2 + 5# -1 (h) 3 - 4 (i) 2 - 3 - 3 - 4 (j) 5 - 7 + 4 - 2

(i) (j)

Show that a + b # a + b when (a) a = 2 and b = 4 (b) a = -1 and b = - 2 (c) a = - 2 and b = 3 (d) a = - 4 and b = 5 (e) a = -7 and b = - 3.

6.

Show that x 2 = x when (a) x = 5 (b) x = - 2 (c) x = - 3 (d) x = 4 (e) x = - 9.

7.

Use the definition of absolute value to write each expression without the absolute value signs (a) x + 5 (b) b - 3 (c) a + 4 (d) 2y - 6 (e) 3x + 9 (f) 4 - x (g) 2k + 1 (h) 5x - 2 (i) a + b (j) p - q

8.

Find values of x for which x = 3.

9.

n Simplify n where n ! 0.

a = 5 and b = 2 a = -1 and b = 2 a = - 2 and b = - 3 a = 4 and b = 7 a = -1 and b = - 2.

Write an expression for (a) a when a 2 0 (b) (c) (d) (e) (f) (g)

a when a 1 0 a when a = 0 3a when a 2 0 3a when a 1 0 3a when a = 0 a + 1 when a 2 -1

x - 2 when x 2 2 x - 2 when x 1 2.

5.

Evaluate a - b if (a) (b) (c) (d) (e)

4.

(h) a + 1 when a 1 -1

x-2 and state which x-2 value x cannot be.

10. Simplify


Test Yourself 1 1.

2.

Convert (a) 0.45 to a fraction (b) 14% to a decimal 5 (c) to a decimal 8 (d) 78.5% to a fraction (e) 0.012 to a percentage 11 (f) to a percentage 15

(c) 9

-

(b) (c) (d) (e) 7.

1 2

4.5 2 + 7.6 2

(e) 6 4.

5.

2

(e) 8 3 (f) - 2 - 1

1.3#10 9 3.8 #10 6 -

(g) 49

2 3

-

1 2

as a fraction

1 4

Evaluate (a) |-3 | -| 2 | (b) | 4 - 5 | (c) 7 + 4 # 8 (d) [(3 + 2)#(5 - 1) - 4] ' 8 (e) - 4 + 3 - 9 (f) - 2 - -1 (g) - 24 ' - 6

(h) 16 (i) ] -3 g0 (j) 4 - 7 2 - -2 - 3 8.

(a) x 5 # x 7 ' x 3 (b) (5y 3) 2 (a 5) 4 b 7 (c) a9 b 3 2x 6 n (d) d 3 0

ab 4 o a5 b6

Simplify (a) a 14 ' a 9 6 (b) _ x 5 y 3 i (c) p 6 # p 5 ' p 2 4 (d) ^ 2b 9h (2x 7) 3 y 2 (e) x 10 y

Simplify

(e) e

Evaluate (a) - 4 (b) 36 2 (c) - 5 2 - 2 3 (d) 4 - 3 as fraction

(b) 4.3 0.3 2 (c) 3 5.7 (d)

3 7 5 8 6 2 #3 7 3 3 9' 4 2 1 +2 5 10 5 15# 6

1

Evaluate correct to 3 significant figures. (a)

Evaluate (a) 1

Evaluate as a fraction. (a) 7 - 2 (b) 5 -1

3.

6.

9.

Write in index form. n 1 (b) 5 x 1 (c) x+y (a)

(d)

4

x +1

41

42


(e)

7

(c) If he spends 3 hours watching TV, what fraction of the day is this? (d) What percentage of the day does he spend sleeping?

a+b

2 (f) x 1 (g) 2x 3 (h)

3

x4

(i)

7

(5x + 3) 9 1

4

m3

(j)

10. Write without fractional or negative indices. (a) a - 5 1

(b) n 4

1

(c) (x + 1) 2 (d) (x - y) -1 (e) (4t - 7) - 4 1

(f) (a + b) 5 (g) x

3

(h) b 4 (j) x

-

17. Rachel scored 56 out of 80 for a maths test. What percentage did she score? 18. Evaluate 2118, and write your answer in scientific notation correct to 1 decimal place. 19. Write in index form. (a) x 1 (b) y x+3 1 (d) (2x - 3) 11

1 3

(i) (2x + 3)

16. The price of a car increased by 12%. If the car cost $34 500 previously, what is its new price?

4 3

3 2

11. Show that a + b # a + b when a = 5 and b = - 3. 9 2 12. Evaluate a b when a = and b = 1 . 25 3 2 4

3 1 4 13. If a = c m and b = , evaluate ab 3 as a 4 3 fraction. 14. Increase 650 mL by 6%. 1 of his 24-hour day 3 1 sleeping and at work. 4 (a) How many hours does Johan spend at work? (b) What fraction of his day is spent at work or sleeping?

15. Johan spends

(c)

6

(e)

3

y7

20. Write in scientific notation. (a) 0.000 013 (b) 123 000 000 000 21. Convert to a fraction. • (a) 0. 7 • • (b) 0.124 22. Write without the negative index. (a) x - 3 (b) (2a + 5)- 1 a -5 (c) c m b 23. The number of people attending a football match increased by 4% from last week. If there were 15 080 people at the match this week, how many attended last week? 24. Show that | a + b | # a + b when a = - 2 and b = - 5.


Challenge Exercise 1 3 2 2 7 + 3 m ' c4 - 1 m. 4 5 3 8

1.

Simplify c 8

2.

3 5 149 7 Simplify + + . 5 12 180 30

3.

Arrange in increasing order of size: 51%, • 51 0.502, 0. 5, . 99

4.

1 1 of his day sleeping, 3 12 1 of the day eating and of the day 20 watching TV. What percentage of the day is left?

5.

Write 64

6.

Express 3.2 ' 0.014 in scientific notation correct to 3 significant figures.

7.

Mark spends

-

2 3

as a rational number.

11. Show that 2 (2 k - 1) + 2 k + 1 = 2 (2 k + 1 - 1) . 12. Find the value of

3 2 2 4 1 3 a = c m , b = c - m and c = c m . 5 5 3 13. Which of the following are rational • 3 numbers: 3 , - 0.34, 2, 3r, 1. 5, 0, ? 7 14. The percentage of salt in 1 L of water is 10%. If 500 mL of water is added to this mixture, what percentage of salt is there now? 15. Simplify

25

1 out of 20 for a maths 2 1 test, 19 out of 23 for English and 55 2 out of 70 for physics. Find his average score as a percentage, to the nearest whole percentage.

Vinh scored 17

• • •

8.

Write 1.3274 as a rational number.

9.

The distance from the Earth to the moon is 3.84 #10 5 km. How long would it take a rocket travelling at 2.13#10 4 km h to reach the moon, to the nearest hour? 8.3# 4.1 correct to 0.2 + 5.4 ' 1.3 3 significant figures.

10. Evaluate 3

a in index form if b3 c2

|x + 1 | x2 - 1

for x ! !1.

4.3 1.3 - 2.9 correct to 2.4 3 + 3.31 2 2 decimal places.

16. Evaluate 6

17. Write 15 g as a percentage of 2.5 kg. 18. Evaluate 2.3 1.8 + 5.7 #10 - 2 correct to 3 significant figures. - 3.4 #10 - 3 + 1.7 #10 - 2 and (6.9 #10 5) 3 express your answer in scientific notation correct to 3 significant figures.

19. Evaluate

20. Prove | a + b | # | a | + | b | for all real a, b.

43

2 Algebra and Surds TERMINOLOGY Binomial: A mathematical expression consisting of two terms such as x + 3 or 3x - 1 Binomial product: The product of two binomial expressions such as (x + 3) (2x - 4) Expression: A mathematical statement involving numbers, pronumerals and symbols e.g. 2x - 3 Factorise: The process of writing an expression as a product of its factors. It is the reverse operation of expanding brackets i.e. take out the highest common factor in an expression and place the rest in brackets e.g. 2y - 8 = 2 (y - 4) Pronumeral: A letter or symbol that stands for a number

Rationalising the denominator: A process for replacing a surd in the denominator by a rational number without altering its value Surd: From ‘absurd’. The root of a number that has an irrational value e.g. 3 . It cannot be expressed as a rational number Term: An element of an expression containing pronumerals and/or numbers separated by an operation such as + , - , # or ' e.g. 2x, - 3 Trinomial: An expression with three terms such as 3x 2 - 2x + 1

Chapter 2 Algebra and Surds

45

INTRODUCTION THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions, removing grouping symbols, factorising, completing the square and simplifying algebraic fractions. Operations with surds, including rationalising the denominator, are also studied in this chapter.

DID YOU KNOW? One of the earliest mathematicians to use algebra was Diophantus of Alexandria. It is not known when he lived, but it is thought this may have been around 250 AD. In Baghdad around 700–800 AD a mathematician named Mohammed Un-Musa Al-Khowarezmi wrote books on algebra and Hindu numerals. One of his books was named Al-Jabr wa’l Migabaloh, and the word algebra comes from the first word in this title.

Simplifying Expressions Addition and subtraction

EXAMPLES Simplify 1. 7x - x

Solution

Here x is called a pronumeral.

7x - x = 7x - 1 x = 6x 2. 4x 2 - 3x 2 + 6x 2

Solution 4x 2 - 3x 2 + 6x 2 = x 2 + 6 x 2 = 7x 2

CONTINUED

46


3. x 3 - 3x - 5x + 4 Only add or subtract ‘like’ terms. These have the same pronumeral (for example, 3x and 5x).

Solution x 3 - 3 x - 5x + 4 = x 3 - 8 x + 4 4. 3a - 4b - 5a - b

Solution 3a - 4b - 5a - b = 3a - 5a - 4b - b = - 2a - 5b

2.1 Exercises Simplify 1.

2x + 5x

16. 7b + b - 3b

2.

9a - 6a

17. 3b - 5b + 4b + 9b

3.

5z - 4z

18. - 5x + 3x - x - 7x

4.

5a + a

19. 6x - 5y - y

5.

4b - b

20. 8a + b - 4b - 7a

6.

2r - 5r

21. xy + 2y + 3xy

7.

- 4y + 3y

22. 2ab 2 - 5ab 2 - 3ab 2

8.

- 2x - 3x

23. m 2 - 5m - m + 12

9.

2a - 2a

24. p 2 - 7p + 5p - 6

10. - 4k + 7k

25. 3x + 7y + 5x - 4y

11. 3t + 4t + 2t

26. ab + 2b - 3ab + 8b

12. 8w - w + 3w

27. ab + bc - ab - ac + bc

13. 4m - 3m - 2m

28. a 5 - 7x 3 + a 5 - 2x 3 + 1

14. x + 3x - 5x

29. x 3 - 3xy 2 + 4x 2 y - x 2 y + xy 2 + 2y 3

15. 8h - h - 7h

30. 3x 3 - 4x 2 - 3x + 5x 2 - 4x - 6


47

Multiplication EXAMPLES Simplify 1. - 5x # 3y # 2x

Solution - 5x # 3y # 2x = - 30xyx = - 30x 2 y 2. - 3x 3 y 2 # - 4xy 5

Solution

Use index laws to simplify this question.

- 3x 3 y 2 # - 4xy 5 = 12x 4 y 7


5 # 2b

5 11. ^ 2x 2h

2.

2x # 4y

12. 2ab 3 # 3a

3.

5p # 2p

13. 5a 2 b # - 2ab

4.

- 3z # 2w

14. 7pq 2 # 3p 2 q 2

5.

- 5a # - 3b

15. 5ab # a 2 b 2

6.

x # 2y # 7z

16. 4h 3 # - 2h 7

7.

8ab # 6c

17. k 3 p # p 2

8.

4d # 3d

4 18. ^ - 3t 3 h

9.

3a # 4a # a

19. 7m 6 # - 2m 5

10. ^ - 3y h3

20. - 2x 2 # 3x 3 y # - 4xy 2

48


Division Use cancelling or index laws to simplify divisions.

EXAMPLES Simplify 1. 6v 2 y ' 2vy

Solution By cancelling, 6v 2 y ' 2vy = =

6v 2 y 2vy 63 # v # v1 # y1 21 # v # y1

= 3v Using index laws, 6v 2 y ' 2vy = 3v 2 - 1 y 1 - 1 = 3v 1 y 0 = 3v 2.

5a 3 b 15ab 2

Solution 5a 3 b = 1 a3 -1 b1- 2 3 15ab 2 = 1 a 2 b -1 3 a2 = 3b


30x ' 5

2.

2y ' y

3. 4. 5.

8a 2

6.

xy 2x

7.

12p 3 ' 4p 2

8.

3a 2 b 2 6ab

9.

20x 15xy

10.

- 9x 7 3x 4

2

8a 2 a 8a 2 2a


11. -15ab ' - 5b 12.

2ab 6a 2 b 3

13.

- 8p 4pqs

16.

7pq 3

17. 5a 9 b 4 c - 2 ' 20a 5 b -3 c -1 2 ^ a -5 h b 4 2

18.

14. 14cd 2 ' 21c 3 d 3 15.

42p 5 q 4

-1

4a - 9 ^ b 2 h

19. - 5x 4 y 7 z ' 15xy 8 z - 2

2xy 2 z 3

20. - 9 ^ a 4 b -1 h ' -18a -1 b 3 3

4x 3 y 2 z

Removing grouping symbols The distributive law of numbers is given by

a ] b + c g = ab + ac

EXAMPLE 7 # (9 + 11) = 7 # 20 = 140 Using the distributive law, 7 # (9 + 11) = 7 # 9 + 7 # 11 = 63 + 77 = 140

This rule is used in algebra to help remove grouping symbols.

EXAMPLES Expand and simplify. 1. 2 ] a + 3 g

Solution 2 (a + 3) = 2 # a + 2 # 3 = 2a + 6

CONTINUED

49

50


2. - ] 2x - 5 g

Solution -(2x - 5) = -1 (2x - 5) = -1 # 2x - 1 # - 5 = - 2x + 5 3. 5a 2]4 + 3ab - c g

Solution 5a 2 (4 + 3ab - c) = 5a 2 # 4 + 5a 2 # 3ab - 5a 2 # c = 20a 2 + 15a 3 b - 5a 2 c 4. 5 - 2 ^ y + 3 h

Solution 5 - 2 (y + 3 ) = 5 - 2 # y - 2 # 3 = 5 - 2y - 6 = - 2y - 1 5. 2 ] b - 5 g - ] b + 1 g

Solution 2 (b - 5) - (b + 1) = 2 # b + 2 # - 5 - 1 # b -1 # 1 = 2b - 10 - b - 1 = b - 11

2.4 Exercises Expand and simplify 1.

2]x - 4 g

7.

ab ] 2a + b g

2.

3 ] 2h + 3 g

8.

5n ] n - 4 g

3.

-5 ] a - 2 g

9.

3x 2 y _ xy + 2y 2 i

4.

x ^ 2y + 3 h

10. 3 + 4 ] k + 1 g

5.

x]x - 2 g

11. 2 ] t - 7 g - 3

6.

2a ] 3a - 8 b g

12. y ^ 4y + 3 h + 8y


13. 9 - 5 ] b + 3 g

20. 2ab ] 3 - a g - b ] 4a - 1 g

14. 3 - ] 2x - 5 g

21. 5x - ] x - 2 g - 3

15. 5] 3 - 2m g + 7 ] m - 2 g

22. 8 - 4 ^ 2y + 1 h + y

16. 2 ] h + 4 g + 3 ] 2h - 9 g

23. ] a + b g - ] a - b g

17. 3 ] 2d - 3 g - ] 5d - 3 g

24. 2 ] 3t - 4 g - ] t + 1 g + 3

18. a ] 2a + 1 g - ^ a 2 + 3a - 4 h

25. 4 + 3 ] a + 5 g - ] a - 7 g

51

19. x ] 3x - 4 g - 5 ] x + 1 g

Binomial Products A binomial expression consists of two numbers, for example x + 3. A set of two binomial expressions multiplied together is called a binomial product. Example: ] x + 3 g ] x - 2 g. Each term in the first bracket is multiplied by each term in the second bracket.

] a + b g ^ x + y h = ax + ay + bx + by

Proof ]a + bg]c + d g = a ]c + d g + b ]c + d g = ac + ad + bc + bd

EXAMPLES Expand and simplify 1. ^ p + 3h^ q - 4h

Solution ^ p + 3 h ^ q - 4 h = pq - 4p + 3q - 12 2. ]a + 5g2

Solution ] a + 5 g2 = (a + 5)(a + 5) = a 2 + 5a + 5a + 25 = a 2 + 10a + 25

Can you see a quick way of doing this?

52


The rule below is not a binomial product (one expression is a trinomial), but it works the same way.

] a + b g ^ x + y + z h = ax + ay + az + bx + by + bz

EXAMPLE Expand and simplify ] x + 4 g ^ 2x - 3y - 1 h .

Solution (x + 4) (2x - 3y - 1) = 2x 2 - 3xy - x + 8x - 12y - 4 = 2x 2 - 3xy + 7x - 12y - 4


]a + 5g]a + 2g

17. ]a + 2bg]a - 2bg

2.

]x + 3g]x - 1g

18. ^ 3x - 4y h^ 3x + 4y h

3.

^ 2y - 3h^ y + 5h

19. ]x + 3g]x - 3g

4.

]m - 4g]m - 2g

20. ^ y - 6h^ y + 6h

5.

]x + 4g]x + 3g

21. ] 3a + 1 g ] 3a - 1 g

6.

^ y + 2h^ y - 5h

22. ]2z - 7g]2z + 7g

7.

]2x - 3g]x + 2g

23. ]x + 9g^ x - 2y + 2h

8.

]h - 7g]h - 3g

24. ] b - 3 g ] 2a + 2b - 1 g

9.

]x + 5g]x - 5g

25. ]x + 2g^ x 2 - 2x + 4h

10. ] 5a - 4 g ] 3a - 1 g

26. ]a - 3g^ a 2 + 3a + 9h

11. ^ 2y + 3h^ 4y - 3h

27. ]a + 9g2

12. ]x - 4g^ y + 7h

28. ]k - 4g2

13. ^ x 2 + 3h]x - 2g

29. ]x + 2g2

14. ]n + 2g]n - 2g

30. ^ y - 7h2

15. ]2x + 3g]2x - 3g

31. ]2x + 3g2

16. ^ 4 - 7y h^ 4 + 7y h

32. ]2t - 1g2


33. ]3a + 4bg2

37. ] a + b g2

34. ^ x - 5y h2

38. ] a - b g2

35. ]2a + bg2

39. ] a + b g ^ a 2 - ab + b 2 h

36. ] a - b g ] a + b g

40. ] a - b g ^ a 2 + ab + b 2 h

Some binomial products have special results and can be simplified quickly using their special properties. Binomial products involving perfect squares and the difference of two squares occur in many topics in mathematics. Their expansions are given below.

Difference of 2 squares ] a + b g ] a - b g = a2 - b2

Proof (a + b) (a - b) = a 2 - ab + ab - b 2 = a2 - b2

Perfect squares

] a + b g2 = a 2 + 2ab + b 2

Proof ] a + b g2 = (a + b) (a + b) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2

]a - bg2 = a 2 - 2ab + b 2

Proof ] a - b g2 = (a - b) (a - b) = a 2 - ab - ab + b 2 = a 2 - 2ab + b 2

53

54


EXAMPLES Expand and simplify 1. ]2x - 3g2

Solution ] 2x - 3 g2 = ] 2x g2 - 2 (2x) 3 + 3 2 = 4x 2 - 12x + 9 2. ^ 3y - 4h^ 3y + 4h

Solution (3y - 4) (3y + 4) = ^ 3y h2 - 4 2 = 9y 2 - 16


]t + 4g2

16. ^ p + 1 h ^ p - 1 h

2.

]z - 6g2

17. ]r + 6g]r - 6g

3.

] x - 1 g2

18. ] x - 10 g ] x + 10 g

4.

^ y + 8h2

19. ]2a + 3g]2a - 3g

5.

^ q + 3h2

20. ^ x - 5y h^ x + 5y h

6.

]k - 7g2

21. ] 4a + 1 g ] 4a - 1 g

7.

] n + 1 g2

22. ]7 - 3xg]7 + 3xg

8.

]2b + 5g2

23. ^ x 2 + 2h^ x 2 - 2h

9.

]3 - xg2

2 24. ^ x 2 + 5h

10. ^ 3y - 1 h2

25. ]3ab - 4cg]3ab + 4c g

11. ^ x + y h2

2 2 26. b x + x l

12. ] 3a - b g2 13. ]4d + 5eg2

1 1 27. b a - a lb a + a l

14. ]t + 4g]t - 4g

28. _ x + 6 y - 2 @ i _ x - 6 y - 2 @ i

15. ] x - 3 g ] x + 3 g

29. 6]a + bg + c @2


30. 7 ] x + 1 g - y A

36. ] x - 4 g3

2

55

Expand (x - 4) (x - 4) 2 .

31. ] a + 3 g2 - ] a - 3 g2

1 2 1 2 37. b x - x l - b x l + 2

32. 16 - ]z - 4g]z + 4g

38. _ x 2 + y 2 i - 4x 2 y 2

33. 2x + ]3x + 1g2 - 4

39. ]2a + 5g3

34. ^ x + y h2 - x ^ 2 - y h

40. ] 2x - 1 g ] 2x + 1 g ] x + 2 g2

2

35. ] 4n - 3 g ] 4n + 3 g - 2n 2 + 5

PROBLEM Find values of all pronumerals that make this true. a b d f e i i i h i i c c

c e b g b

#

Try c = 9.

Factorisation Simple factors Factors are numbers that exactly divide or go into an equal or larger number, without leaving a remainder.

EXAMPLES The numbers 1, 2, 3, 4, 6, 8, 12 and 24 are all the factors of 24. Factors of 5x are 1, 5, x and 5x.

To factorise an expression, we use the distributive law.

ax + bx = x ] a + b g

56


EXAMPLES Factorise 1. 3x + 12

Solution Divide each term by 3 to find the terms inside the brackets.

The highest common factor is 3. 3x + 12 = 3 ] x + 4 g 2. y 2 - 2y

Solution Check answers by expanding brackets.

The highest common factor is y. y 2 - 2y = y ^ y - 2 h 3. x 3 - 2x 2

Solution x and x2 are both common factors. We take out the highest common factor which is x2. x 3 - 2x 2 = x 2 ] x - 2 g 4. 5] x + 3 g + 2y ] x + 3 g

Solution The highest common factor is x + 3. 5 ] x + 3 g + 2y ] x + 3 g = ] x + 3 g ^ 5 + 2 y h 5. 8a 3 b 2 - 2ab 3

Solution There are several common factors here. The highest common factor is 2ab2. 8a 3 b 2 - 2ab 3 = 2ab 2 ^ 4a 2 - bh


2.7 Exercises Factorise 1.

2y + 6

19. x ] m + 5 g + 7 ] m + 5 g

2.

5x - 10

20. 2 ^ y - 1 h - y ^ y - 1 h

3.

3m - 9

21. 4^ 7 + y h - 3x ^ 7 + y h

4.

8x + 2

22. 6x ]a - 2g + 5]a - 2g

5.

24 - 18y

23. x ] 2t + 1 g - y ] 2t + 1 g

6.

x 2 + 2x

7.

m 2 - 3m

24. a ] 3x - 2 g + 2b ] 3x - 2 g - 3c ] 3x - 2 g

8.

2y 2 + 4y

9.

15a - 3a 2

25. 6x 3 + 9x 2 26. 3pq 5 - 6q 3 27. 15a 4 b 3 + 3ab

10. ab 2 + ab

28. 4x 3 - 24x 2

11. 4x 2 y - 2xy

29. 35m 3 n 4 - 25m 2 n

12. 3mn 3 + 9mn

30. 24a 2 b 5 + 16ab 2

13. 8x 2 z - 2xz 2 14. 6ab + 3a - 2a

31. 2rr 2 + 2rrh

2

32. ]x - 3g2 + 5]x - 3g

15. 5x 2 - 2x + xy

33. y 2 ]x + 4g + 2]x + 4g

16. 3q 5 - 2q 2

34. a ] a + 1 g - ] a + 1 g2

17. 5b 3 + 15b 2

35. 4ab ^ a 2 + 1 h - 3 ^ a 2 + 1 h

18. 6a 2 b 3 - 3a 3 b 2

Grouping in pairs If an expression has 4 terms, it may be factorised in pairs.

ax + bx + ay + by = x(a + b) + y (a + b) = ( a + b) ( x + y)

57

58


EXAMPLES Factorise 1. x 2 - 2x + 3x - 6

Solution x 2 - 2x + 3x - 6 = x (x - 2) + 3 (x - 2) = (x - 2) (x + 3) 2. 2x - 4 + 6y - 3xy

Solution 2x - 4 + 6y - 3xy = 2 (x - 2) + 3y (2 - x) = 2 ( x - 2) - 3y ( x - 2 ) = (x - 2) (2 - 3y) or 2x - 4 + 6y - 3xy = 2 (x - 2) - 3y (- 2 + x) = 2 ( x - 2) - 3y ( x - 2 ) = (x - 2) (2 - 3y)


2x + 8 + bx + 4b

12. m - 2 + 4y - 2my

2.

ay - 3a + by - 3b

13. 2x 2 + 10xy - 3xy - 15y 2

3.

x 2 + 5x + 2x + 10

14. a 2 b + ab 3 - 4a - 4b 2

4.

m 2 - 2m + 3m - 6

15. 5x - x 2 - 3x + 15

5.

ad - ac + bd - bc

16. x 4 + 7x 3 - 4x - 28

6.

x 3 + x 2 + 3x + 3

17. 7x - 21 - xy + 3y

7.

5ab - 3b + 10a - 6

18. 4d + 12 - de - 3e

8.

2xy - x 2 + 2y 2 - xy

19. 3x - 12 + xy - 4y

9.

ay + a + y + 1

20. 2a + 6 - ab - 3b

10. x 2 + 5x - x - 5

21. x 3 - 3x 2 + 6x - 18

11. y + 3 + ay + 3a

22. pq - 3p + q 2 - 3q


23. 3x 3 - 6x 2 - 5x + 10

27. 4x 3 - 6x 2 + 8x - 12

24. 4a - 12b + ac - 3bc

28. 3a 2 + 9a + 6ab + 18b

25. xy + 7x - 4y - 28

29. 5y - 15 + 10xy - 30x

26. x 4 - 4x 3 - 5x + 20

30. rr 2 + 2rr - 3r - 6

59

Trinomials A trinomial is an expression with three terms, for example x 2 - 4x + 3. Factorising a trinomial usually gives a binomial product. x 2 + ] a + b g x + ab = ] x + a g ] x + b g

Proof x 2 + (a + b) x + ab = x 2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a) (x + b)

EXAMPLES Factorise 1. m 2 - 5m + 6

Solution a + b = - 5 and ab = + 6 -2 +6 ' -3 -5 Numbers with sum - 5 and product + 6 are - 2 and - 3. ` m 2 - 5m + 6 = [m + ] - 2 g] [m + ] - 3 g] = ]m - 2g]m - 3g

Guess and check by trying - 2 and - 3 or -1 and - 6.

2. y 2 + y - 2

Solution a + b = + 1 and ab = - 2 +2 -2 ' -1 +1 Two numbers with sum + 1 and product - 2 are + 2 and -1. ` y2 + y - 2 = ^ y + 2 h ^ y - 1 h

Guess and check by trying 2 and -1 or - 2 and 1.

60



x 2 + 4x + 3

14. a 2 - 4a + 4

2.

y 2 + 7y + 12

15. x 2 + 14x - 32

3.

m 2 + 2m + 1

16. y 2 - 5y - 36

4.

t 2 + 8t + 16

17. n 2 - 10n + 24

5.

z2 + z - 6

18. x 2 - 10x + 25

6.

x 2 - 5x - 6

19. p 2 + 8p - 9

7.

v 2 - 8v + 15

20. k 2 - 7k + 10

8.

t 2 - 6t + 9

21. x 2 + x - 12

9.

x 2 + 9x - 10

22. m 2 - 6m - 7

10. y 2 - 10y + 21

23. q 2 + 12q + 20

11. m 2 - 9m + 18

24. d 2 - 4d - 5

12. y 2 + 9y - 36

25. l 2 - 11l + 18

13. x 2 - 5x - 24

The result x 2 + ] a + b g x + ab = ] x + a g ] x + b g only works when the coefficient of x 2 (the number in front of x 2) is 1. When the coefficient of x 2 is not 1, for example in the expression 5x 2 - 2x + 4, we need to use a different method to factorise the trinomial. There are different ways of factorising these trinomials. One method is the cross method. Another is called the PSF method. Or you can simply guess and check.

EXAMPLES Factorise 1. 5y 2 - 13y + 6

Solution—guess and check For 5y2, one bracket will have 5y and the other y: ^ 5y h ^ y h . Now look at the constant (term without y in it): + 6.


The two numbers inside the brackets must multiply to give + 6. To get a positive answer, they must both have the same signs. But there is a negative sign in front of 13y so the numbers cannot be both positive. They must both be negative. ^ 5y - h ^ y - h To get a product of 6, the numbers must be 2 and 3 or 1 and 6. Guess 2 and 3 and check: ^ 5y - 2 h ^ y - 3 h = 5y 2 - 15y - 2y + 6 = 5y 2 - 17y + 6 This is not correct. Notice that we are mainly interested in checking the middle two terms, -15y and - 2y. Try 2 and 3 the other way around: ^ 5y - 3 h ^ y - 2 h . Checking the middle terms: -10y - 3y = -13y This is correct, so the answer is ^ 5y - 3 h ^ y - 2 h . Note: If this did not check out, do the same with 1 and 6.

Solution—cross method Factors of 5y 2 are 5y and y. Factors of 6 are -1 and - 6 or - 2 and - 3. Possible combinations that give a middle term of -13y are 5y

-2

5y

-3

5y

-1

5y

-6

y

-3

y

-2

y

-6

y

-1

By guessing and checking, we choose the correct combination. -3 5y # - 2 = -10y 5y y

-2

y # - 3 = - 3y -13y

` 5y 2 - 13y + 6 = ^ 5y - 3 h ^ y - 2 h

Solution—PSF method P: Product of first and last terms S: Sum or middle term F: Factors of P that give S - 3y 30y 2 ) -10y -13y

30y 2 -13y - 3y, -10y

` 5y 2 - 13y + 6 = 5y 2 - 3y - 10y + 6 = y ^ 5y - 3 h - 2 ^ 5 y - 3 h = ^ 5y - 3 h ^ y - 2 h

CONTINUED

61

62


2. 4y 2 + 4y - 3

Solution—guess and check For 4y2, both brackets will have 2y or one bracket will have 4y and the other y. Try 2y in each bracket: ^ 2y h ^ 2y h . Now look at the constant: - 3. The two numbers inside the brackets must multiply to give - 3. To get a negative answer, they must have different signs. ^ 2y - h ^ 2y + h To get a product of 3, the numbers must be 1 and 3. Guess and check: ^ 2y - 3 h ^ 2 y + 1 h Checking the middle terms: 2y - 6y = - 4y This is almost correct, as the sign is wrong but the coefficient is right (the number in front of y). Swap the signs around: ^ 2y - 1 h ^ 2 y + 3 h = 4y 2 + 6 y - 2 y - 3 = 4y 2 + 4y - 3 This is correct, so the answer is ^ 2y - 1 h ^ 2y + 3 h .

Solution—cross method Factors of 4y 2 are 4y and y or 2y and 2y. Factors of 3 are -1 and 3 or - 3 and 1. Trying combinations of these factors gives 3 2y 2y # - 1 = - 2 y 2y

-1

2y # 3 =

6y 4y

` 4y 2 + 4y - 3 = ^ 2 y + 3 h ^ 2 y - 1 h

Solution—PSF method P: Product of first and last terms -12y 2 S: Sum or middle term 4y F: Factors of P that give S + 6y, - 2y 2 + 6y -12y ) -2y + 4y ` 4y 2 + 4y - 3 = 4 y 2 + 6 y - 2 y - 3 = 2y ^ 2y + 3 h - 1 ^ 2 y + 3 h = ^ 2y + 3 h ^ 2y - 1 h


2.10

Exercises

Factorise 1.

2a 2 + 11a + 5

16. 4n 2 - 11n + 6

2.

5y 2 + 7y + 2

17. 8t 2 + 18t - 5

3.

3x 2 + 10x + 7

18. 12q 2 + 23q + 10

4.

3x 2 + 8x + 4

19. 8r 2 + 22r - 6

5.

2b 2 - 5b + 3

20. 4x 2 - 4x - 15

6.

7x 2 - 9x + 2

21. 6y 2 - 13y + 2

7.

3y 2 + 5y - 2

22. 6p 2 - 5p - 6

8.

2x 2 + 11x + 12

23. 8x 2 + 31x + 21

9.

5p 2 + 13p - 6

24. 12b 2 - 43b + 36

10. 6x 2 + 13x + 5

25. 6x 2 - 53x - 9

11. 2y 2 - 11y - 6

26. 9x 2 + 30x + 25

12. 10x 2 + 3x - 1

27. 16y 2 + 24y + 9

13. 8t 2 - 14t + 3

28. 25k 2 - 20k + 4

14. 6x 2 - x - 12

29. 36a 2 - 12a + 1

15. 6y 2 + 47y - 8

30. 49m 2 + 84m + 36

Perfect squares You have looked at some special binomial products, including ]a + bg2 = a 2 + 2ab + b 2 and ]a - bg2 = a 2 - 2ab + b 2 . When factorising, use these results the other way around.

a 2 + 2ab + b 2 = ] a + b g2 a 2 - 2ab + b 2 = ] a - b g2

63

64


EXAMPLES In a perfect square, the constant term is always a square number.

Factorise 1. x 2 - 8x + 16

Solution x 2 - 8x + 16 = x 2 - 2 (4) x + 4 2 = ] x - 4 g2 2. 4a 2 + 20a + 25

Solution 4a 2 + 20a + 25 = ] 2a g2 + 2 (2a) (5) + 5 2 = ] 2a + 5 g2

2.11

Exercises

Factorise 1.

y 2 - 2y + 1

12. 16k 2 - 24k + 9

2.

x 2 + 6x + 9

13. 25x 2 + 10x + 1

3.

m 2 + 10m + 25

14. 81a 2 - 36a + 4

4.

t 2 - 4t + 4

15. 49m 2 + 84m + 36

5.

x 2 - 12x + 36

16. t 2 + t +

6.

4x 2 + 12x + 9

7.

16b 2 - 8b + 1

8.

9a 2 + 12a + 4

4x 4 + 3 9 6y 1 18. 9y 2 + + 5 25

9.

25x 2 - 40x + 16

19. x 2 + 2 +

10. 49y 2 + 14y + 1 11. 9y 2 - 30y + 25

1 4

17. x 2 -

1 x2

20. 25k 2 - 20 +

4 k2


Difference of 2 squares A special case of binomial products is ] a + b g ] a - b g = a 2 - b 2. a2 - b2 = ] a + b g ] a - b g

EXAMPLES Factorise 1. d 2 - 36

Solution d 2 - 36 = d 2 - 6 2 = ]d + 6 g]d - 6 g 2. 9b 2 - 1

Solution 9b 2 - 1 = ] 3b g2 - 1 2 = ( 3 b + 1) ( 3 b - 1 ) 3. (a + 3) 2 - (b - 1) 2

Solution ] a + 3 g2 - ] b - 1 g2 = [(a + 3) + (b - 1)] [(a + 3) - (b - 1)] = (a + 3 + b - 1) ( a + 3 - b + 1)

= ( a + b + 2 ) (a - b + 4 )

2.12

Exercises

Factorise 1.

a2 - 4

7.

1 - 4z 2

2.

x2 - 9

8.

25t 2 - 1

3.

y2 - 1

9.

9t 2 - 4

4.

x 2 - 25

10. 9 - 16x 2

5.

4x 2 - 49

11. x 2 - 4y 2

6.

16y 2 - 9

12. 36x 2 - y 2

65

66


13. 4a 2 - 9b 2

20.

14. x 2 - 100y 2 15. 4a - 81b 2

21. ] x + 2 g2 - ^ 2y + 1 h2

2

22. x 4 - 1

16. ]x + 2g2 - y 2 17. ] a - 1 g - ] b - 2 g 2

2

18. z - ] 1 + w g 2

19. x 2 -

y2 -1 9

2

1 4

23. 9x 6 - 4y 2 24. x 4 - 16y 4 25. a 8 - 1

Sums and differences of 2 cubes

a 3 + b 3 = ] a + b g ^ a 2 - ab + b 2 h

Proof (a + b) (a 2 - ab + b 2) = a 3 - a 2 b + ab 2 + a 2 b - ab 2 + b 3 = a3 + b3 a 3 - b 3 = ] a - b g ^ a 2 + ab + b 2 h

Proof (a - b) (a 2 + ab + b 2) = a 3 + a 2 b + ab 2 - a 2 b - ab 2 - b 3 = a3 - b3

EXAMPLES Factorise 1. 8x 3 + 1

Solution 8x 3 + 1 = ] 2x g3 + 1 3 = (2x + 1) [] 2x g2 - (2x) (1) + 1 2] = (2x + 1 ) (4 x 2 - 2 x + 1 )


2. 27a 3 - 64b 3

Solution 27a 3 - 64b 3 = ] 3a g3 - ] 4b g3 = (3a - 4b) [] 3a g2 + (3a) (4b) + ] 4b g2] = (3a - 4b) (9a 2 + 12ab + 16b 2)

2.13

Exercises

Factorise 1.

b3 - 8

2.

x 3 + 27

3.

12.

x3 - 27 8

t3 + 1

13.

1000 1 + 3 3 a b

4.

a 3 - 64

14. ] x + 1 g3 - y 3

5.

1 - x3

15. 125x 3 y 3 + 216z 3

6.

8 + 27y 3

16. ]a - 2g3 - ]a + 1g3

7.

y 3 + 8z 3

8.

x 3 - 125y 3

9.

8x 3 + 27y 3

10. a 3 b 3 - 1 11. 1000 + 8t 3

17. 1 -

x3 27

18. y 3 + ]3 + xg3 19. ] x + 1 g3 + ^ y - 2 h3 20. 8]a + 3g3 - b 3

Mixed factors Sometimes more than one method of factorising is needed to completely factorise an expression.

EXAMPLE Factorise 5x 2 - 45.

Solution 5x 2 - 45 = 5 (x 2 - 9) = 5 (x + 3) (x - 3)

(using simple factors) (the difference of two squares)

67

68


2.14

Exercises

Factorise 1.

2x 2 - 18

16. x 3 - 3x 2 - 10x

2.

3p 2 - 3p - 36

17. x 3 - 3x 2 - 9x + 27

3.

5y 3 - 5

18. 4x 2 y 3 - y

4.

4a 3 b + 8a 2 b 2 - 4ab 2 - 2a 2 b

19. 24 - 3b 3

5.

5a 2 - 10a + 5

20. 18x 2 + 33x - 30

6.

- 2x 2 + 11x - 12

21. 3x 2 - 6x + 3

7.

3z 3 + 27z 2 + 60z

22. x 3 + 2x 2 - 25x - 50

8.

9ab - 4a 3 b 3

23. z 3 + 6z 2 + 9z

9.

x3 - x

24. 4x 4 - 13x 2 + 9

10. 6x 2 + 8x - 8

25. 2x 5 + 2x 2 y 3 - 8x 3 - 8y 3

11. 3m - 15 - 5n + mn

26. 4a 3 - 36a

12. ] x - 3 g2 - ] x + 4 g2

27. 40x - 5x 4

13. y 2 ^ y + 5 h - 16 ^ y + 5 h

28. a 4 - 13a 2 + 36

14. x 4 - x 3 + 8x - 8

29. 4k 3 + 40k 2 + 100k

15. x 6 - 1

30. 3x 3 + 9x 2 - 3x - 9

DID YOU KNOW? Long division can be used to find factors of an expression. For example, x - 1 is a factor of x 3 + 4x - 5. We can find the other factor by dividing x 3 + 4x - 5 by x - 1. x2 + x + 5 x - 1 x3 + 4x - 5

g

x3

-

x2 x 2 + 4x x2

You will study this in Chapter 12.

-

x 5x - 5 5x - 5

0 So the other factor of x 3 + 4x - 5 is x 2 + x + 5 ` x 3 + 4x - 5 = (x - 1) (x 2 + x + 5)


69

Completing the Square Factorising a perfect square uses the results a 2 ! 2ab + b 2 = ] a ! b g2

EXAMPLES 1. Complete the square on x 2 + 6x.

Solution Using a 2 + 2ab + b 2: a=x 2ab = 6x Substituting a = x: 2xb = 6x b=3

Notice that 3 is half of 6.

To complete the square: a 2 + 2ab + b 2 = ] a + b g2 2 x + 2x ] 3 g + 3 2 = ] x + 3 g2 x 2 + 6x + 9 = ] x + 3 g2 2. Complete the square on n 2 - 10n.

Solution Using a 2 - 2ab + b 2: a=n 2ab = 10x Substituting a = n: 2nb = 10n b=5

Notice that 5 is half of 10.

To complete the square: a 2 - 2ab + b 2 = ] a - b g2 n 2 - 2n ] 5 g + 5 2 = ] n - 5 g2 n 2 - 10n + 25 = ] n - 5 g2

To complete the square on a 2 + pa, divide p by 2 and square it. p 2 p 2 a 2 + pa + d n = d a + n 2 2

70


EXAMPLES 1. Complete the square on x 2 + 12x.

Solution Divide 12 by 2 and square it: x 2 + 12x + c

12 2 m = x 2 + 12x + 6 2 2 = x 2 + 12x + 36 = ]x + 6g2

2. Complete the square on y 2 - 2y.

Solution Divide 2 by 2 and square it: 2 2 y 2 - 2y + c m = y 2 - 2 y + 1 2 2 = y 2 - 2y + 1 = ^ y - 1 h2

2.15

Exercises

Complete the square on 1.

x 2 + 4x

12. y 2 + 3y

2.

b 2 - 6b

13. x 2 - 7x

3.

x 2 - 10x

14. a 2 + a

4.

y 2 + 8y

15. x 2 + 9x

5.

m 2 - 14m

16. y 2 -

6.

q 2 + 18q

5y 2

7.

x 2 + 2x

17. k 2 -

11k 2

8.

t 2 - 16t

18. x 2 + 6xy

9.

x 2 - 20x

19. a 2 - 4ab

10. w 2 + 44w 11. x 2 - 32x

20. p 2 - 8pq


71

Algebraic Fractions Simplifying fractions EXAMPLES Simplify 4x + 2 2

1.

Solution 2 ] 2x + 1 g 4x + 2 = 2 2 = 2x + 1

Factorise first, then cancel.

2x 2 - 3x - 2 x3 - 8

2.

Solution ] 2x + 1 g ] x - 2 g 2x 2 - 3x - 2 = 3 ] x - 2 g ^ x 2 + 2x + 4 h x -8 2x + 1 = 2 x + 2x + 4

2.16

Exercises

Simplify 1.

5a + 10 5

2. 3. 4.

9.

b3 - 1 b2 - 1

6t - 3 3

10.

8y + 2 6

2p 2 + 7p - 15 6p - 9

11.

a2 - 1 a + 2a - 3

8 4d - 2 2

5.

6.

x 5x 2 - 2x y-4

12.

13.

y - 8y + 16

2

3 ]x - 2g + y ]x - 2g x3 - 8 x 3 + 3x 2 - 9x - 27 x 2 + 6x + 9

2

7.

2ab - 4a 2 a 2 - 3a

8.

s2 + s - 2 s 2 + 5s + 6

14.

15.

2p 2 - 3p - 2 8p 3 + 1 ay - ax + by - bx 2ay - by - 2ax + bx

72


Operations with algebraic fractions

EXAMPLES Simplify 1.

x+3 x-1 5 4

Solution Do algebraic fractions the same way as ordinary fractions.

4 ]x - 1 g - 5 ]x + 3 g x -1 x +3 = 5 4 20 4x - 4 - 5x - 15 = 20 - x - 19 = 20

2.

2a 2 b + 10ab a 2 - 25 ' 3 4b + 12 b + 27

Solution 2a 2 b + 10ab a 2 - 25 2a 2 b + 10ab 4b + 12 ' = # 2 4b + 12 b 3 + 27 b 3 + 27 a - 25 2ab ] a + 5 g 4 ]b + 3 g = # 2 ] a + 5 g]a - 5 g ] b + 3 g ^ b - 3b + 9 h 8ab = ] a - 5 g ^ b 2 - 3b + 9 h

3.

2 1 + x-5 x+2

Solution 2 ]x + 2g + 1 ]x - 5g 2 1 + = x-5 x+2 ]x - 5g]x + 2g 2x + 4 + x - 5 = ]x - 5g]x + 2g 3x - 1 = ]x - 5g]x + 2g


2.17 1.

2.

Exercises

Simplify x 3x (a) + 4 2 y + 1 2y (b) + 5 3 a+2 a (c) 4 3 p-3 p+2 (d) + 6 2 x-5 x-1 (e) 2 3 4.

Simplify 3 b 2 + 2b # (a) b + 2 6a - 3

1 1 + x+1 x-3

(g)

3 2 x 2 + x -4

(h)

1 1 + a 2 + 2a + 1 a + 1

(i)

5 2 1 + y+2 y+3 y-1

(j)

2 7 x 2 - 16 x 2 - x - 12

2

Simplify (a)

y2 - 9 3x 2 x 2 - 2x - 8 # # 4y - 12 6x - 24 y 3 + 27

q3 + 1 (b) 2 # q + 2q + 1 p + 2

(b)

2 a 2 - 5a 3a - 15 y - y - 2 ' # 5ay y 2 - 4y + 4 y2 - 4

3ab 2 12ab - 6a (c) ' 2 5xy x y + 2xy 2

(c)

3 x 2 + 3x 2x + 8 + 2 # x-3 4x - 16 x -9

(d)

5b b2 b ' 2 2b + 6 b 1 + b +b-6

(e)

x 2 - 8x + 15 x 2 - 9 x 2 + 5x + 6 ' # 2 2x - 10 5x + 10x 10x 2

p2 - 4

(d)

ax - ay + bx - by x2 - y2

#

x3 + y3 ab 2 + a 2 b

x 2 - 6x + 9 x 2 - 5x + 6 (e) ' x 2 - 25 x 2 + 4x - 5 3.

(f)

5.

Simplify 2 3 (a) x + x

Simplify (a)

1 2 4 + x 2 - 7x + 10 x 2 - 2x - 15 x 2 + x - 6

1 2 x-1 x

(b)

3 5 2 + 2 2 x x x -4

(c) 1 +

3 a+b

(c)

3 2 + p 2 + pq pq - q 2

(d) x -

x2 x+2

(d)

a b 1 + a + b a - b a2 - b2

(b)

(e) p - q +

1 p+q

2

x+y y x (e) x - y + y - x - 2 y - x2

Substitution Algebra is used in writing general formulae or rules. For example, the formula A = lb is used to find the area of a rectangle with length l and breadth b. We can substitute any values for l and b to find the area of different rectangles.

73

74


EXAMPLES 1. P = 2l + 2b is the formula for finding the perimeter of a rectangle with length l and breadth b. Find P when l = 1.3 and b = 3.2.

Solution P = 2 l + 2b = 2 ] 1 . 3 g + 2 ] 3 .2 g = 2 .6 + 6 . 4 =9 2. V = rr 2 h is the formula for finding the volume of a cylinder with radius r and height h. Find V (correct to 1 decimal place) when r = 2.1 and h = 8.7.

Solution V = rr 2 h = r ] 2.1 g2 (8.7) = 120.5 correct to 1 decimal place

9C + 32 is the formula for changing degrees Celsius ] °C g into 5 degrees Fahrenheit ] °F g find F when C = 25. 3. If F =

Solution 9C + 32 5 9 ] 25 g = + 32 5 225 = + 32 5 225 + 160 = 5 385 = 5 = 77 This means that 25°C is the same as 77°F. F=


2.18 1.

Exercises

Given a = 3.1 and b = - 2.3 find, correct to 1 decimal place. (a) ab (b) 3b (c) 5a 2 (d) ab 3 (e) ]a + bg2 (f)

a-b

(g) - b 2 2.

T = a + ] n - 1 g d is the formula for finding the term of an arithmetic series. Find T when a = - 4, n = 18 and d = 3.

3.

Given y = mx + b, the equation of a straight line, find y if m = 3, x = - 2 and b = - 1.

4.

If h = 100t - 5t 2 is the height of a particle at time t, find h when t = 5.

5.

Given vertical velocity v = - gt, find v when g = 9.8 and t = 20.

6.

If y = 2 x + 3 is the equation of a function, find y when x = 1.3, correct to 1 decimal place.

7.

S = 2r r ] r + h g is the formula for the surface area of a cylinder. Find S when r = 5 and h = 7, correct to the nearest whole number.

8.

A = rr 2 is the area of a circle with radius r. Find A when r = 9.5, correct to 3 significant figures.

9.

n-1

Given u n = ar is the nth term of a geometric series, find u n if a = 5, r = - 2 and n = 4.

10. Given V = 1 lbh is the volume 3 formula for a rectangular pyramid, find V if l = 4.7, b = 5.1 and h = 6.5. 11. The gradient of a straight line is y2 - y1 given by m = x - x . Find m 2 1 if x 1 = 3, x 2 = -1, y 1 = - 2 and y 2 = 5. 12. If A = 1 h ] a + b g gives the area 2 of a trapezium, find A when h = 7, a = 2.5 and b = 3.9. 13. Find V if V = 4 rr 3 is the volume 3 formula for a sphere with radius r and r = 7.6, to 1 decimal place.

14. The velocity of an object at a certain time t is given by the formula v = u + at. Find v when u = 1 , a = 3 and t = 5 . 4 5 6 a 15. Given S = , find S if a = 5 1-r and r = 2 . S is the sum to infinity 3 of a geometric series. 16. c = a 2 + b 2 , according to Pythagoras’ theorem. Find the value of c if a = 6 and b = 8. 17. Given y = 16 - x 2 is the equation of a semicircle, find the exact value of y when x = 2.

75

76


18. Find the value of E in the energy equation E = mc 2 if m = 8.3 and c = 1.7. 19. A = P c 1 +

20. If S =

a geometric series, find S if a = 3, r = 2 and n = 5.

r n m is the formula 100

for finding compound interest. Find A when P = 200, r = 12 and n = 5, correct to 2 decimal places.

a ^rn - 1h is the sum of r -1

21. Find the value of

a3 b2 if c2

2 3 1 4 a = c 3 m , b = c 2 m and c = c m . 4 3 2

Surds An irrational number is a number that cannot be written as a ratio or fraction (rational). Surds are special types of irrational numbers, such as 2, 3 and 5 . Some surds give rational values: for example, 9 = 3. Others, like 2 , do not have an exact decimal value. If a question involving surds asks for an exact answer, then leave it as a surd rather than giving a decimal approximation.

Simplifying surds

Class Investigations 1. Is there an exact decimal equivalent for 2 ? 2. Can you draw a line of length exactly 2 ? 3. Do these calculations give the same results? (a) 9 # 4 and 9 # 4 (b)

4

and

4 9

(c)

9 9 + 4 and

9 +

4

(d)

9 - 4 and

9 -

4

Here are some basic properties of surds.

a# b =

ab

a' b =

a

^ x h2 =

b

=

x2 = x

a b


77

EXAMPLES 1. Express in simplest surd form

45 .

45 also equals 3 # 15 but this will not simplify. We look for a number that is a perfect square.

Solution 45 = 9 # 5 = 9 # 5 =3# 5 =3 5 2. Simplify 3 40 .

Solution Find a factor of 40 that is a perfect square.

3 40 = 3 4 # 10 = 3 # 4 # 10 = 3 # 2 # 10 = 6 10 3. Write 5 2 as a single surd.

Solution 5 2 = =

2.19 1.

25 # 2 50

Exercises

Express these surds in simplest surd form.

(k)

112

(l)

300

(a)

12

(b)

63

(c)

24

(d)

50

(e)

72

(f)

200

(g)

48

(h)

75

(i)

32

(a) 2 27

(j)

54

(b) 5 80

(m) 128

2.

(n)

243

(o)

245

(p)

108

(q)

99

(r)

125

Simplify

78


(c) 4 98

(g) 3 13

(d) 2 28

(h) 7 2

(e) 8 20

(i) 11 3

(f) 4 56

(j) 12 7

(g) 8 405

4.

(h) 15 8

(a)

(i) 7 40

x =3 5

(b) 2 3 =

x

(c) 3 7 =

x

Write as a single surd.

(d) 5 2 =

x

(a) 3 2

(e) 2 11 =

(b) 2 5

(f)

(c) 4 11

(g) 4 19 =

(d) 8 2

(h)

(e) 5 3

(i) 5 31 =

(f) 4 10

(j)

(j) 8 45 3.

Evaluate x if

x

x =7 3 x

x = 6 23 x

x = 8 15

Addition and subtraction Calculations with surds are similar to calculations in algebra. We can only add or subtract ‘like terms’ with algebraic expressions. This is the same with surds.

EXAMPLES 1. Simplify 3 2 + 4 2 .

Solution 3 2+4 2 =7 2 2. Simplify

3 - 12 .

Solution First, change into ‘like’ surds. 3 - 12 = 3 - 4 # 3 = 3 -2 3 =- 3 3. Simplify 2 2 - 2 + 3 .

Solution 2 2- 2+ 3=

2+ 3


2.20

79

Exercises

Simplify 1.

5 +2 5

14.

50 -

32

2.

3 2 -2 2

15.

28 +

63

3.

3 +5 3

16. 2 8 -

18

4.

7 3 -4 3

5.

5 -4 5 4 6 -

6.

17. 3 54 + 2 24 18.

90 - 5 40 - 2 10

19. 4 48 + 3 147 + 5 12

6

7.

2 -8 2

20. 3 2 + 8 - 12

8.

5 +4 5 +3 5

21.

63 - 28 - 50

9.

2 -2 2 -3 2

22.

12 - 45 - 48 - 5

10.

5 +

45

23.

150 + 45 + 24

11.

8 -

2

24.

32 - 243 - 50 + 147

12.

3 +

48

25.

80 - 3 245 + 2 50

13.

12 -

27

Multiplication and division To get a b # c d = ac bd , multiply surds with surds and rationals with rationals.

a # b = ab a b # c d = ac bd a# a =

a b

=

a2 = a

a b

EXAMPLES Simplify 1. 2 2 #- 5 7

Solution 2 2 #- 5 7 = -10 14

CONTINUED

80


2. 4 2 # 5 18

Solution 4 2 # 5 18 = 20 36 = 20 # 6 = 120

3.

2 14 4 2

Solution 2 14 4 2

=

2 2 # 7 2

=

4.

7

4 2

3 10 15 2

Solution 3 10 15 2

=

3# 5 # 2 15 2

5 = 5

5. d

2

10 n 3

Solution 2 ^ 10 h 10 n = 3 ^ 3 h2 10 = 3 =31 3

2

d


2.21

Exercises

Simplify 1.

7 #

2.

3# 5

3.

2 #3 3

4.

5 7 #2 2

3

5.

-3 3 #2 2

6.

5 3 #2 3

7.

- 4 5 # 3 11

8.

2 7# 7

9.

2 3 # 5 12

10.

6# 2

11.

8 #2 6

23.

24.

25.

26.

27.

28.

5 8 10 2 16 2 2 12 10 30 5 10 2 2 6 20 4 2 8 10 3 3 15 2

29.

8

12. 3 2 # 5 14 13.

10 # 2 2

14. 2 6 #-7 6 15. ^ 2 h

2

2 16. ^ 2 7 h

17.

31.

32.

3 15 6 10 5 12 5 8 15 18 10 10

3# 5# 2

18. 2 3 # 7 #- 5 19.

30.

2 # 6 #3 3

33.

15 2 6 2n 3

35. d

5n 7

20. 2 5 # - 3 2 # - 5 5 21.

22.

4 12 2 2

2

34. d

2

12 18 3 6

Expanding brackets The same rules for expanding brackets and binomial products that you use in algebra also apply to surds.

81

82


Simplifying surds by removing grouping symbols uses these general rules.

a^ b + ch=

ab + ac

Proof a^ b + ch = =

a# b + ab + ac

a# c

Binomial product:

^ a + b h^ c + d h =

ac +

ad +

bc +

bd

Proof ^ a + b h^ c + d h = a # c + a # d + b # c + b # d = ac + ad + bc + bd Perfect squares:

^ a + b h2 = a + 2 ab + b

Proof ^ a + b h2 = ^ a + b h ^ a + b h = a 2 + ab + ab + b 2 = a + 2 ab + b ^ a - b h2 = a - 2 ab + b

Proof ^ a - b h2 = ^ a - b h ^ a - b h = a 2 - ab - ab + b 2 = a - 2 ab + b Difference of two squares:

^ a + b h^ a - b h = a - b

Proof ^ a + b h ^ a - b h = a 2 - ab + ab - b 2 =a-b


83

EXAMPLES Expand and simplify 1. 2 ^ 5 + 2 h

Solution 2( 5 +

2) = = =

2# 5 + 10 + 4 10 + 2

2# 2

2. 3 7 ^ 2 3 - 3 2 h

Solution 3 7 (2 3 - 3 2 ) = 3 7 # 2 3 - 3 7 # 3 2 = 6 21 - 9 14 3. ^ 2 + 3 5 h ^ 3 -

2h

Solution ( 2 + 3 5)( 3 -

2) = =

2# 3 - 2# 2 +3 5# 3 -3 5# 2 6 - 2 + 3 15 - 3 10

4. ^ 5 + 2 3 h ^ 5 - 2 3 h

Solution ( 5 + 2 3 ) ( 5 - 2 3 ) = 5 # 5 - 5 #2 3 + 2 3 # 5 - 2 3 #2 3 = 5 - 2 15 + 2 15 - 4#3 = 5 - 12 = -7 Another way to do this question is by using the difference of two squares. 2 2 ( 5 + 2 3)( 5 - 2 3) = ^ 5 h - ^2 3 h = 5 - 4#3 = -7

Notice that using the difference of two squares gives a rational answer.

84


2.22 1.

Exercises (m)^ 2 11 + 5 2 h^ 2 11 - 5 2 h

Expand and simplify (a)

2^ 5 + 3h

(b)

3 ^2 2 - 5 h

(n) ^ 5 + 2 h

2

2 (o) ^ 2 2 - 3 h

(c) 4 3 ^ 3 + 2 5 h (d)

2 (p) ^ 3 2 + 7 h

7 ^5 2 - 2 3 h

2 (q) ^ 2 3 + 3 5 h

(e) - 3 ^ 2 - 4 6 h (f)

2 (r) ^ 7 - 2 5 h

3 ^ 5 11 + 3 7 h

2 (s) ^ 2 8 - 3 5 h

(g) - 3 2 ^ 2 + 4 3 h (h)

5^ 5 - 5 3h

(i)

3 ^ 12 + 10 h

2 (t) ^ 3 5 + 2 2 h

3.

If a = 3 2 , simplify (a) a2 (b) 2a3 (c) (2a)3 (d) ]a + 1g2 (e) ] a + 3 g ] a – 3 g

4.

Evaluate a and b if 2 (a) ^ 2 5 + 1h = a + b

(j) 2 3 ^ 18 + 3 h (k) - 4 2 ^ 2 - 3 6 h (l) - 7 5 ^ - 3 20 + 2 3 h (m) 10 3 ^ 2 - 2 12 h (n) - 2 ^ 5 + 2 h (o) 2 3 ^ 2 - 12 h 2.

(b) ^ 2 2 - 5 h ^ 2 - 3 5 h = a + b 10

Expand and simplify (a) ^ 2 + 3h^ 5 + 3 3 h

5.

Expand and simplify (a) ^ a + 3 - 2 h ^ a + 3 + 2 h 2 (b) _ p - 1 - p i

6.

Evaluate k if ^ 2 7 - 3 h ^ 2 7 + 3 h = k.

(g) ^ 7 + 3 h^ 7 - 3 h

7.

Simplify _ 2 x + y i _ x - 3 y i .

(h) ^ 2 - 3 h^ 2 + 3 h

8.

If ^ 2 3 - 5 h = a - b , evaluate a and b.

9.

Evaluate a and b if ^ 7 2 - 3 h2 = a + b 2 .

(b) ^ 5 - 2 h^ 2 - 7 h (c) ^ 2 + 5 3 h^ 2 5 - 3 2 h (d) ^ 3 10 - 2 5 h^ 4 2 + 6 6 h (e) ^ 2 5 - 7 2 h^ 5 - 3 2 h (f) ^ 5 + 6 2 h^ 3 5 - 3 h

(i) ^ 6 + 3 2 h^ 6 - 3 2 h (j) ^ 3 5 + 2 h^ 3 5 - 2 h (k) ^ 8 - 5 h^ 8 + 5 h (l) ^ 2 + 9 3 h^ 2 - 9 3 h

2

10. A rectangle has sides 5 + 1 and 2 5 - 1. Find its exact area.

Rationalising the denominator Rationalising the denominator of a fractional surd means writing it with a rational number (not a surd) in the denominator. For example, after 3 5 3 rationalising the denominator, becomes . 5 5


85

DID YOU KNOW? A major reason for rationalising the denominator used to be to make it easier to evaluate the fraction (before calculators were available). It is easier to divide by a rational number than an irrational one; for example, 3 = 3 ' 2.236 5 3

5 5

This is hard to do without a calculator.

This is easier to calculate.

= 3 # 2.236 ' 5

Squaring a surd in the denominator will rationalise it since ^ x h = x. 2

Multiplying by

b a b a # = b b b

b

b is the same as multiplying by 1.

Proof b a b a # = b b b2 a b = b

EXAMPLES 1. Rationalise the denominator of

3 . 5

Solution 5 3 5 3 # = 5 5 5 2. Rationalise the denominator of

Solution

2 5 3

. Don’t multiply by 5

2 5 3

#

3 3

=

2 3

5 9 2 3 = 5# 3 2 3 = 15

3

as it takes 5 3 longer to simplify.

86


When there is a binomial denominator, we use the difference of two squares to rationalise it, as the result is always a rational number.

To rationalise the denominator of

a+ b c+ d

, multiply by

Proof a+ b c+ d

^ a + b h^ c - d h c- d ^ c + d h^ c - d h ^ a + b h^ c - d h = ^ c h2 - ^ d h2 ^ a + b h^ c - d h = c-d c- d

#

=

EXAMPLES 1. Write with a rational denominator 5 2 -3 Multiply by the conjugate surd 2 + 3.

.

Solution 5 2 -3

2 +3

#

2 +3

=

5 ^ 2 + 3h

^ 2 h2 - 3 2 10 + 3 5 = 2-9 10 + 3 5 = -7 10 + 3 5 =7

2. Write with a rational denominator 2 3+ 5 3+4 2

.

Solution 2 3 +

5

3 +4 2

#

3 -4 2 3 -4 2

=

^2 3 + 5 h^ 3 - 4 2 h

^ 3 h2 - ^ 4 2 h2 2 # 3 - 8 6 + 15 - 4 10 = 3 - 16 # 2

c- d c- d


6 - 8 6 + 15 - 4 10 - 29 - 6 + 8 6 - 15 + 4 10 = 29 =

3. Evaluate a and b if

3 3 3- 2

= a + b.

Solution 3 3 3- 2

#

3+ 2 3+ 2

=

3 3^ 3 + 2h

^ 3 - 2 h^ 3 + 2 h 3 9+3 6 = ^ 3 h2 - ^ 2 h2 3#3+3 6 3-2 9+3 6 = 1 =9+3 6 =

=9+ 9# 6 = 9 + 54 So a = 9 and b = 54. 4. Evaluate as a fraction with rational denominator 2 + 3+2

5 3-2

.

Solution 2 + 3+2

5 3 -2

=

2^ 3 - 2h + 5 ^ 3 + 2h

^ 3 + 2h ^ 3 - 2h 2 3 - 4 + 15 + 2 5 = ^ 3 h2 - 2 2 2 3 - 4 + 15 + 2 5 3-4 2 3 - 4 + 15 + 2 5 = -1 = - 2 3 + 4 - 15 - 2 5 =

87

88


2.23 1.

Express with rational denominator (a) (b) (c) (d) (e)

2.

Exercises 3.

1 7

(a)

3

(b)

2 2 2 3

(c)

5 6 7

(d)

5 2 1+

2 3

6 -5

(g)

5 +2 2

8+3 2

(j)

4 3 -2 2

(f)

1 5 +

2

2 -

7

2 +

3

2 +3

4 5 (j)

7 5 (k)

4 3 +

(l) 2

3

4.

2 -7 5 +2 6 3 -4 3 +4 3 3 3 +

(b) (c)

2 +5 2 2

2 5 +3 2

3 2 +

+

3 3 2 -

#

3

6 -

3

2 3 2 +3 5 6 +2 2 +7 4+

2 3 +

3 -2

3

6 +

1 3

+

2

-

2

2

3 -

2

(d) (e)

2 5 3 4 2

2

2 -1

+

+

5 -

3

5

2

3

3 5 3 2 4-

3

2+

3

3 +1

Find a and b if (a)

2 3

-

1 where z = 1 + z2

(h) (i)

1 2 -1

1 where t = t

3 2 +4

2 7

Express with rational denominator

(e)

3

(g)

5

(i)

(d)

2 -

2

3 2 -4

(c)

2

(f) z 2 -

(h)

(b)

1 + 2 +1

(e) t +

(f)

(a)

Express as a single fraction with rational denominator

=

a b

=

a 6 b

2 =a+b 5 5 +1 2 7 7 -4 2 +3 2 -1

=a+b 7 =a+

b

2 -

2 6 -1


5.

2 -1

Show that

2 +1

+

4 is 2

7.

If x =

(b) x 2 +

2

+

1 5 -

2

-

as a single fraction with 3 rational denominator.

3 + 2, simplify

1 (a) x + x

2 5 +

5 +1

rational. 6.

Write

8.

1 x2

Show that

8 2 + is 3+2 2 2

rational. 2

1 (c) b x + x l

9.

1 If 2 + x = 3 , where x ! 0, find x as a surd with rational denominator.

10. Rationalise the denominator of b +2 ]b ! 4 g b -2

89

90


Test Yourself 2 1.

2.

3.

4.

Simplify (a) 5y - 7y 3a + 12 (b) 3 (c) - 2k 3 # 3k 2 y x (d) + 5 3 (e) 4a - 3b - a - 5b (f) 8 + 32 (g) 3 5 - 20 + 45 Factorise (a) x 2 - 36 (b) a 2 + 2a - 3 (c) 4ab 2 - 8ab (d) 5y - 15 + xy - 3x (e) 4n - 2p + 6 (f) 8 - x 3 Expand and simplify (a) b + 3 ] b - 2 g (b) ] 2x - 1 g ] x + 3 g (c) 5 ] m + 3 g - ] m - 2 g (d) ]4x - 3g2 (e) ^ p - 5h^ p + 5h (f) 7 - 2 ] a + 4 g - 5a (g) 3 ^ 2 2 - 5 h (h) ^ 3 + 7 h^ 3 - 2h Simplify 4a - 12 10b (a) # 3 5b 3 a - 27 (b)

5.

5m + 10 m2 - 4 ' 2 m - m - 2 3m + 3

The volume of a cube is V = s 3. Evaluate V when s = 5.4.

6.

(a) Expand and simplify ^ 2 5 + 3 h ^ 2 5 - 3 h. (b) Rationalise the denominator of 3 3 . 2 5+ 3

7.

Simplify

8.

If a = 4, b = - 3 and c = - 2, find the value of (a) ab 2 (b) a - bc (c) a (d) ]bcg3 (e) c ] 2a + 3b g

9.

Simplify 3 12 (a) 6 15 (b)

3 1 2 + - 2 . x-2 x+3 x +x-6

4 32 2 2

10. The formula for the distance an object falls is given by d = 5t 2 . Find d when t = 1.5. 11. Rationalise the denominator of 2 (a) 5 3 (b)

1+ 3 2

12. Expand and simplify (a) ^ 3 2 - 4h^ 3 - 2 h 2 (b) ^ 7 + 2h 13. Factorise fully (a) 3x 2 - 27 (b) 6x 2 - 12x - 18 (c) 5y 3 + 40


14. Simplify 3x 4 y (a) 9xy 5 (b)

5 15x - 5

15. Simplify 2 (a) ^ 3 11 h 3 (b) ^ 2 3 h 16. Expand and simplify (a) ] a + b g ] a - b g (b) ] a + b g 2 (c) ] a - b g 2 17. Factorise (a) a 2 - 2ab + b 2 (b) a 3 - b 3 1 18. If x = 3 + 1, simplify x + x and give your answer with a rational denominator. 19. Simplify 4 3 (a) a + b (b)

x-3 x-2 5 2

20. Simplify

2 3 , writing 5+2 2 2-1

your answer with a rational denominator. 21. Simplify (a) 3 8 (b) - 2 2 # 4 3 (c) 108 - 48 (d)

23. Rationalise the denominator of 3 (a) 7 (b)

2

5 3 2 (c) 5 -1 (d) (e)

2 2 3 2+ 3 5+ 2 4 5-3 3

24. Simplify 3x x-2 (a) 5 2 a+2 2a - 3 (b) + 7 3 1 2 (c) 2 1 x + x -1 4 1 (d) 2 + k + 2k - 3 k + 3 (e)

3 2+ 5

-

5 3- 2

25. Evaluate n if (a) 108 - 12 = (b)

112 + 7 =

n n

8 6

(c) 2 8 + 200 =

2 18

(d) 4 147 + 3 75 = n 180 (e) 2 245 + = n 2

(e) 5a # - 3b # - 2a (f)

22. Expand and simplify (a) 2 2 ^ 3 + 2 h (b) ^ 5 7 - 3 5 h^ 2 2 - 3 h (c) ^ 3 + 2 h^ 3 - 2 h (d) ^ 4 3 - 5 h^ 4 3 + 5 h 2 (e) ^ 3 7 - 2 h

2m 3 n 6m 2 n 5

(g) 3x - 2y - x - y

n

91

92


26. Evaluate x 2 +

1+2 3 1 if x = 2 x 1-2 3

27. Rationalise the denominator of

3

2 7 (there may be more than one answer). 21 (a) 28 2 21 (b) 28 21 (c) 14 21 (d) 7 x-3 x +1 . 5 4 -]x + 7 g 20 x+7 20 x + 17 20 - ] x + 17 g 20

28. Simplify (a) (b) (c) (d)

(a) (b) (c) (d)

32. Simplify 5ab - 2a 2 - 7ab - 3a 2 . (a) 2ab + a 2 (b) - 2ab - 5a 2 (c) - 13a 3 b (d) - 2ab + 5a 2 33. Simplify (a) (b) (c)

29. Factorise x 3 - 4x 2 - x + 4 (there may be more than one answer). (a) ^ x 2 - 1 h ] x - 4 g (b) ^ x 2 + 1 h ] x - 4 g (c) x 2 ] x - 4 g (d) ] x - 4 g ] x + 1 g ] x - 1 g 30. Simplify 3 2 + 2 98 . (a) 5 2 (b) 5 10 (c) 17 2 (d) 10 2

3 2 1 + . x-2 x+2 x2 - 4 x+5 ]x + 2g]x - 2g x+1 ]x + 2g]x - 2g x+9 ]x + 2g]x - 2g x-3 ]x + 2g]x - 2g

31. Simplify

(d)

80 . 27

4 5 3 3 4 5 9 3 8 5 9 3 8 5 3 3

34. Expand and simplify ^ 3x - 2y h2 . (a) 3x 2 - 12xy - 2y 2 (b) 9x 2 - 12xy - 4y 2 (c) 3x 2 - 6xy + 2y 2 (d) 9x 2 - 12xy + 4y 2 35. Complete the square on a 2 - 16a. (a) a 2 - 16a + 16 = ^ a - 4 h2 (b) a 2 - 16a + 64 = ^ a - 8 h2 (c) a 2 - 16a + 8 = ^ a - 4 h2 (d) a 2 - 16a + 4 = ^ a - 2 h2


Challenge Exercise 2 1.

2.

Expand and simplify (a) 4ab ] a - 2b g - 2a 2 ] b - 3a g (b) _ y 2 - 2 i_ y 2 + 2 i (c) ] 2x - 5 g3 Find the value of x + y with rational denominator if x = 3 + 1 and 1 y= . 2 5-3 2 3

2x + y x-y 3x + 2y . + - 2 x-3 x+3 x +x-6

12. (a) Expand ^ 2x - 1 h3. 6x 2 + 5x - 4 (b) Simplify . 8x 3 - 12x 2 + 6x - 1 13. Expand and simplify ] x - 1 g ^ x - 3 h2. 14. Simplify and express with rational 2 +

5

-

5 3

3.

Simplify

4.

b Complete the square on x 2 + a x.

15. Complete the square on x 2 + 2 x. 3

Factorise (a) (x + 4)2 + 5 (x + 4) (b) x 4 - x 2 y - 6y 2 (c) 125x 3 + 343 (d) a 2 b - 2a 2 - 4b + 8

16. If x =

5.

6. 7.

8.

9.

7 6 - 54

.

11. Simplify

denominator

Simplify

d=

4x 2 - 16x + 12

| ax 1 + by 1 + c |

.

Simplify

10. Factorise

â + 1h a3 + 1

.

a2 4 - 2. 2 x b

.

lx 1 + kx 2

17. Find the exact value with rational 1 denominator of 2x 2 - 3x + x if x = 2 5 . 18. Find the exact value of 1+2 3 1 (a) x 2 + 2 if x = x 1-2 3 (b) a and b if

is the formula for

a2 + b2 the perpendicular distance from a point to a line. Find the exact value of d with a rational denominator if a = 2, b = -1, c = 3, x 1 = - 4 and y 1 = 5. 3

2 -1

, find the value of x when k+l k = 3, l = - 2, x 1 = 5 and x 2 = 4.

Complete the square on 4x 2 + 12x. 2xy + 2x - 6 - 6y

3 +4

3 -4 2+3 3

=a+b 3

19. A = 1 r 2 i is the area of a sector of a 2 circle. Find the value of i when A = 12 and r = 4. 20. If V = rr 2 h is the volume of a cylinder, find the exact value of r when V = 9 and h = 16. 21. If s = u + 1 at 2, find the exact value of s 2 when u = 2, a = 3 and t = 2 3 .

93

3

Equations TERMINOLOGY Absolute value: the distance of a number from zero on a number line.

pronumeral that is solved to find values that make the statement true e.g. 2x - 3 2 4

Equation: A mathematical statement that has a pronumeral or unknown number and an equal sign. An equation can be solved to find the value of the unknown number e.g. 2x - 3 = 5

Quadratic equation: An equation involving x 2 as the highest power of x that may have two, one or no solutions

Exponential equation: Equation where the unknown pronumeral is the power or index e.g. 2 x = 8 Inequation: A mathematical statement involving an inequality sign, 1, 2, # or $ that has an unknown

Simultaneous equations: Two or more independent equations that can be solved together to produce a solution that makes each equation true at the same time. The number of equations required is the same as the number of unknowns

Chapter 3 Equations

95

INTRODUCTION EQUATIONS ARE FOUND IN most branches of mathematics. They are also

important in many other fields, such as science, economics, statistics and engineering. In this chapter you will revise basic equations and inequations. Equations involving absolute values, exponential equations, quadratic equations and simultaneous equations are also covered here.

DID YOU KNOW? Algebra was known in ancient civilisations. Many equations were known in Babylonia, although general solutions were difficult because symbols were not used in those times. Diophantus, around 250 AD, first used algebraic notation and symbols (e.g. the minus sign). He wrote a treatise on algebra in his Arithmetica, comprising 13 books. Only six of these books survived. About 400 AD, Hypatia of Alexandria wrote a commentary on them. Hypatia was the daughter of Theon, a mathematician who ensured that she had the best education. She was the first female mathematician on record, and was a philosopher and teacher. She was murdered for her philosophical views by a fanatical Christian sect. In 1799 Carl Friedrich Gauss proved the Fundamental Theorem of Algebra: that every algebraic equation has a solution.

PROBLEM The age of Diophantus at his death can be calculated from this epitaph: Diophantus passed one-sixth of his life in childhood, one-twelfth in youth, and one-seventh more as a bachelor; five years after his marriage a son was born who died four years before his father at half his father’s final age. How old was Diophantus?

Simple Equations Here are the four rules for changing numbers or pronumerals from one side of an equation to the other.

• • • •

ch3.indd 95

If a number is added, subtract it from both sides If a number is subtracted, add it to both sides If a number is multiplied, divide both sides by the number If a number is divided, multiply both sides by the number

Do the opposite operation to take a number to the other side of an equation.

8/11/09 10:59:40 AM

96


EXAMPLES Solve 1. 3x + 5 = 17

Solution 3x + 5 = 17 3x + 5 - 5 = 17 - 5 3x = 12 3x 12 = 3 3 x=4 You can check the solution by substituting the value into the equation. LHS = 3x + 5 = 3 ( 4) + 5 = 12 + 5 = 17 = RHS Since LHS = RHS, x = 4 is the correct solution. 2. 4y - 3 = 8y + 21

Solution 4y - 3 4 y - 4y - 3 -3 - 3 - 21 - 24

`

= 8y + 21 = 8y - 4y + 21 = 4y + 21 = 4y + 21 - 21 = 4y 4y - 24 = 4 4 -6 = y y = -6

3. 2 ] 3x + 7 g = 6 - ] x - 1 g Check these solutions by substituting them into the equation.

Solution 2 (3 x + 7 ) = 6 - ( x - 1 ) 6x + 14 = 6 - x + 1 =7-x 6x + x + 14 = 7 - x + x 7x + 14 = 7

Chapter 3 Equations

7x + 14 - 14 7x 7x 7 x

= 7 - 14 = -7 -7 7 = -1 =

3.1 Exercises Solve 1.

t + 4 = -1

2.

z + 1.7 = -3.9

3.

y - 3 = -2

4.

w - 2 .6 = 4 .1

18. 3x + 5 = 17

5.

5 = x -7

19. 4a + 7 = - 21

6.

1.5x = 6

20. 7y - 1 = 20

7.

5y = 1 3

8.

b =5 7

9.

-2 =

10.

r 2 = 6 3

16.

x -3 =7 2

17.

m + 7 = 11 5

21. 8b - 4 = - 36 22. 3 (x + 2) = 15 23. -2 (3a + 1) = 8

n 8

11. 2y + 1 = 19 12. 33 = 4k + 9 13. 7d - 2 = 12 14. -2 = 5x - 27 y 15. +4=9 3

24. 7t + 4 = 3t - 12 25. x - 3 = 6x - 9 26. 2 (a - 2) = 4 - 3a 27. 5b + 2 = - 3(b - 1) 28. 3 (t + 7) = 2 (2t - 9) 29. 2 + 5( p - 1) = 5p - ( p - 2) 30. 3.7x + 1.2 = 5.4x - 6.3

A S TA R T L I N G FA C T ! Half full = half empty ` full = empty

97

98


Equations involving fractions There are different ways to solve this type of equation. One way is to multiply both sides of the equation by the common denominator of the fractions.

EXAMPLES Solve m 1 1. -4= 3 2

Solution Multiply by the common denominator, 6.

m 1 -4 = 3 2 m m - 6 (4) = 6 c 1 m 2 3 2m - 24 = 3 2m - 24 + 24 = 3 + 24 2m = 27 6c

2m 27 = 2 2 27 m= 2 = 13 1 2 2.

x+1 x + =5 4 3

Solution The common denominator of 3 and 4 is 12.

x +1 x + =5 4 3 x +1 x m + 12 c m = 12 (5) 12 c 4 3 4 (x + 1) + 3x = 60 4x + 4 + 3x = 60 7x + 4 = 60 7x + 4 - 4 = 60 - 4 7x = 56 7x 56 = 7 7 x=8

Chapter 3 Equations

3.

99

y +1 y-2 5 = 5 3 6

Solution y +1 y-2 5 = 5 3 6 y +1 y -2 o - 30 e o = 30 c 5 m 30 e 5 3 6 6 (y + 1) - 10 (y - 2) = 25 6y + 6 - 10y + 20 = 25 - 4y + 26 = 25 - 4y + 26 - 26 = 25 - 26 - 4y = -1 - 4y -1 = -4 -4 y=1 4 When there is a fraction on either side of the equation, multiplying by the common denominator is the same as cross multiplying.

EXAMPLES 5 8 1. Solve x = (x ! 0 ) 3

Solution 5 8 x =3 8x = 15 8x 15 = 8 8 7 x=1 8 2. Solve

3 8 ^n ! 0h = 5 2n

Solution 3 8 = 5 2n 16n = 15 16n 15 = 16 16 15 n= 16

The common denominator of 5, 3 and 6 is 30.

100



b 2 = 5 3

14.

3 x x - = 5 2 10

2.

7 1 x = 5 (x ! 0 )

15.

x+4 x + =1 3 2

3.

9 4 y = 10 (y ! 0)

16.

p-3 2p + =2 2 3

4.

5x 11 = 4 7

17.

t +3 t -1 + =4 7 3

5.

9 4 = ( k ! 0) 5 2k

18.

x+5 x+2 =1 5 9

6.

x -4=8 3

19.

q-1 q-2 =2 4 3

7.

3 5t = 4 4

20.

x+3 x +7 +2= 5 2

8.

5+x 2 = 7 7

21.

3b 1 b - = 4 5 2

9.

y 3 =5 2

22.

a 3 5 + = 4 3 8

10.

x 2 - =7 9 3

23.

3 5 =x x+2

^ x ! 0, -2 h

11.

w-3 =5 2

24.

1 1 = y +1 3y - 1

c y ! -1,

12.

2t t - =2 5 3

25.

2 1 + = 0 ^ t ! 3, - 4 h t-3 t+4

13.

x 1 + =4 4 2

1 m 3

Substitution Sometimes substituting values into a formula involves solving an equation.

Investigation Body mass index (BMI) is a formula that is used to measure body fatness and is used by health professionals to screen for weight categories that may lead to health problems.

Chapter 3 Equations

This is not the only measure that is used when looking for health problems, however. For example, there are other factors in cardiac (heart) disease. Research these to find out what other things doctors look for. The BMI is used in a different way with children and teens, and is taken in relation to the child’s age. w The formula for BMI is BMI = 2 where w is weight in kg and h is height h in metres. For adults over 20, a BMI under 18.5 means that the person is underweight and over 25 is overweight. Over 30 is obese. The BMI may not always be reliable in measuring body fat. Can you think of some reasons? Is it important where the body fat is stored? Does it make a difference if it is on the hips or the stomach? Research these questions and find out more about BMI generally.

EXAMPLES 1. The formula for the surface area of a rectangular prism is given by S = 2 (lb + bh + lh) . Find the value of b when S = 180, l = 9 and h = 6.

Solution S = 2 (lb + bh + lh) 180 = 2 (9b + 6b + 9 # 6) = 2 (15b + 54) = 30b + 108 180 - 108 = 30b + 108 - 108 72 = 30b 30b 72 = 30 30 2. 4 = b

Another way of doing this would be to change the subject of the formula first.

CONTINUED

101

102


2. The volume of a cylinder is given by V = rr 2 h. Evaluate the radius r, correct to 2 decimal places, when V = 350 and h = 6.5.

Solution V = rr 2 h 350 = rr 2 (6.5) r r 2 ( 6 .5 ) 350 = 6 .5 r 6.5r 350 = r2 6 .5 r 350 = r2 6 .5 r 350 =r 6 .5 r 4.14 = r

3.3 Exercises 1.

Given that v = u + at is the formula for the velocity of a particle at time t, find the value of t when u = 17.3, v = 100.6 and a = 9.8.

7.

The area of a rhombus is given by the formula A = 1 xy where x and 2 y are its diagonals. Find the value of x correct to 2 decimal places when y = 7.8 and A = 25.1.

2.

The sum of an arithmetic series is n given by S = (a + l ) . Find l if 2 a = 3, n = 26 and S = 1625.

8.

The simple interest formula is Pr n . Find n if r = 14.5, I= 100 P = 150 and I = 326.25.

3.

The formula for finding the area of a triangle is A = 1 bh. Find b 2 when A = 36 and h = 9.

9.

The gradient of a straight y2 - y1 line is given by m = x - x . 2 1

4.

The area of a trapezium is given by A = 1 h (a + b) . Find 2 the value of a when A = 120, h = 5 and b = 7.

5.

Find the value of y when x = 3, given the straight line equation 5x - 2y - 7 = 0.

6.

The area of a circle is given by A = rr 2 . Find r correct to 3 significant figures if A = 140.

Find y 1 when m = - 5 , 6 y 2 = 7, x 2 = - 3 and x 1 = 1. 10. The surface area of a cylinder is given by the formula S = 2rr ] r + h g . Evaluate h correct to 1 decimal place if S = 232 and r = 4.5.

Chapter 3 Equations

11. The formula for body mass index w is BMI = 2 . Evaluate h (a) the BMI when w = 65 and h = 1.6 (b) w when BMI = 21.5 and h = 1.8 (c) h when BMI = 19.7 and w = 73.8.

16. If the surface area of a sphere is S = 4rr 2, evaluate r to 3 significant figures when S = 56.3.

12. A formula for depreciation is D = P ] 1 - r g n . Find r if D = 12 000, P = 15 000 and n = 3.

18. If y =

13. The x-value of the midpoint is x1 + x2 given by x = . Find x1 2 when x = - 2 and x 2 = 5.

19. Given y = 2x + 5 , evaluate x when y = 4.

14. Given the height of a particle at time t is h = 5t 2, evaluate t when h = 23.

15. If y = x 2 + 1, evaluate x when y = 5.

17. The area of a sector of a circle 1 is A = r 2 i. Evaluate r when 2 A = 24.6 and i = 0.45. 2 , find the value of x x3 - 1 when y = 3.

20. The volume of a sphere is 4 V = rr 3. Evaluate r to 1 decimal 3 place when V = 150.

Inequations

• • • •

2 means greater than 1 means less than $ means greater than or equal to # means less than or equal to

In order to solve inequations, we need to see what effect one operation applied to both sides has on the inequality sign.

If a 2 b then a + c 2 b + c for all c

For example, 3 2 2 and 3 + 1 2 2 + 1 are both true.

If a 2 b then a - c 2 b - c for all c

For example, 3 2 2 and 3 - 1 2 2 - 1 are both true.

103

There are two solutions to this question.

104


If a 2 b then ac 2 bc for all c 2 0

For example, 3 2 2 and 3 # 2 2 2 # 2 are both true.

If a 2 b then ac 1 bc for all c 1 0

For example, 3 2 2 but 3 # -2 1 2 # -2.

If a 2 b then a ' c 2 b ' c for all c 2 0

For example, 6 2 4 and 6 ' 2 2 4 ' 2 are both true.

If a 2 b then a ' c 1 b ' c for all c 1 0

For example, 6 2 4 but 6 ' -2 1 4 ' -2.

1 1 If a 2 b then a 1 for all positive numbers a and b b

For example, 3 2 2 but

1 1 1 . 3 2

The inequality sign reverses when: • multiplying by a negative • dividing by a negative • taking the reciprocal of both sides

On the number plane, we graph inequalities using arrows and circles (open for greater than and less than and closed in for greater than or equal to and less than or equal to) 1 2 # $

Chapter 3 Equations

105

EXAMPLES Solve and show the solutions on a number line 1. 5x + 7 $ 17

Solution 5x + 7 $ 17 5x + 7 - 7 $ 17 - 7 5x $ 10 5x 10 $ 5 5 x$2 -4

-3

-2

-1

0

1

2

3

4

2. 3t - 2 2 5t + 4

Solution 3t - 2 2 5t + 3t - 3t - 2 2 5t -2 2 2t + - 2 - 4 2 2t + -6 2 2t 2t -6 2 2 2 -3 2 t

4 3t + 4 4 4-4

or 3t - 2 3t - 5t - 2 -2t - 2 - 2t - 2 + 2 -2t -2t -2 t -4

2 5t + 4 2 5t - 5t + 4 24 24+2 26 6 2 -2 1 -3 -3

-2

Remember to change the inequality sign when dividing by -2.

-1

0

1

2

3

4

CONTINUED

106


3. Solve 1 1 2z + 7 # 11.

Solution Method 1: Separate into two separate questions. 1 1 2z + 7 (i) 1 - 7 1 2z + 7 - 7 - 6 1 2z -6 2z 1 2 2 -3 1 z (ii)

2z + 7 # 11 2z + 7 - 7 # 11 - 7 2z # 4 2z 4 # 2 2 z #2

Putting these together gives the solution -3 1 z # 2. Method 2: Do as a single question. 1 1 2z + 7 # 11 1 - 7 1 2z + 7 - 7 # 11 - 7 -6 1 2z # 4 -6 2z 4 # 1 2 2 2 -3 1 z # 2

Solving this inequation as a single question is quicker than splitting it into two parts. Notice that the circle is not filled in for 1 and filled in for #.

-4

-3

-2

-1

0

1

2

3

4

3.4 Exercises 1.

Solve and plot the solution on a number line (a) x + 4 2 7 (b) y - 3 # 1

2.

Solve (a) 5t 2 35 (b) 3x - 7 $ 2 (c) 2 (p + 5) 2 8 (d) 4 - (x - 1) # 7 (e) 3y + 5 2 2y - 4 (f) 2a - 6 # 5a - 3 (g) 3 + 4y $ - 2 (1 - y)

(h) 2x + 9 1 1 - 4 (x + 1) a (i) # - 3 2 2y (j) 8 2 3 b (k) + 5 1 - 4 2 x (l) - 4 2 6 3 x 1 (m) + # 1 4 5 (n)

m 2 -3 2 4 3

Chapter 3 Equations

2b 1 - $6 5 2 r-3 (p) # -6 2 z+1 (q) +223 9 w 2w + 5 (r) + 14 6 3 (o)

(s)

x+1 x-2 $7 2 3

(t)

t+3 t+2 #2 7 2

(u)

q-2 3q 12+ 4 3

3.

(v)

2x x -1 2 2 3 2 9

(w)

2b - 5 b+6 +3# 8 12

Solve and plot the solutions on a number line (a) 3 1 x + 2 1 9 (b) -4 # 2p 1 10 (c) 2 1 3x - 1 1 11 (d) -6 # 5y + 9 # 34 (e) -2 1 3 (2y - 1) 1 7

PROBLEM Find a solution for this sum. Is it a unique solution? CR OS S +RO A DS DANGE R

Equations and Inequations Involving Absolute Values On a number line, x means the distance of x from zero in either direction.

EXAMPLES Plot on a number line and evaluate x 1. x = 2

Solution x = 2 means the distance of x from zero is 2 (in either direction). 2

-4

-3

-2

-1

2

0

1

2

3

4

x = !2

CONTINUED

107

108


2. x # 2

Solution x # 2 means the distance of x from zero is less than or equal to 2 (in either direction). 2

-4 The solution of | x | 1 2 would be - 2 1 x 1 2.

-3

2

-1

-2

1

0

2

3

4

Notice that there is one region on the number line. We can write this as the single statement - 2 # x # 2. 3. x 2 2

Solution x 2 2 means the distance of x from zero is greater than 2 (in either direction). 2

-4 The solution of | x | $ 2 would be x # - 2, x $ 2.

-3

-1

-2

2

0

1

2

3

4

There are two regions on the number line, so we write two separate inequalities x 1 - 2, x 2 2.

x = a means x = ! a x 1 a means -a 1 x 1 a x 2 a means x 2 a, x 1 -a

Class Discussion What does a - b mean as a distance along the number line? Select different values of a and b to help with this discussion.

We use absolute value as a distance on a number line to solve equations and inequations involving absolute values.

Chapter 3 Equations

109

EXAMPLES Solve 1. x + 4 = 7

Solution This means that the distance from x + 4 to zero is 7 in either direction. So x + 4 = ! 7. x+4 =7 x+4=7 or x + 4 = -7 x+4-4=7-4 x + 4 - 4 = -7 - 4 x=3 x = -11 2. 2y - 1 1 5

Solution This means that the distance from 2y - 1 to zero is less than 5 in either direction. So it means - 5 1 2y - 1 1 5. - 5 1 2y - 1 1 5 - 5 + 1 1 2y - 1 + 1 1 5 + 1 2y 6 -4 1 1 2 2 2 -2 1 y 1 3

You could solve these as two separate inequations.

3. 5b - 7 $ 3

Solution 5b - 7 $ 3 means that the distance from 5b - 7 to zero is greater than or equal to 3 in either direction. 5b - 7 # - 3

5b - 7 $ 3

5b - 7 + 7 # -3 + 7 5b # 4 5b 4 # 5 5 4 b # 5 4 So b # , b $ 2. 5

5b - 7 + 7 $ 3 + 7 5b $ 10 5b 10 $ 5 5 b$2

These must be solved and written as two separate inequations.

110


While it is always a good habit to check solutions to equations and inequations by substituting in values, in these next examples it is essential to check, as some of the solutions are impossible!

EXAMPLES Solve 1. 2x + 1 = 3x - 2

Solution 2x + 1 = 3x - 2 means that 2x + 1 is at a distance of 3x - 2 from zero. 2x + 1 = ! ] 3x - 2 g This question is impossible if 3x - 2 is negative. Can you see why? If 2x + 1 is equal to a negative number, this is impossible as the absolute value is always positive. Case (i) 2x + 1 = 3x - 2 2x - 2x + 1 = 3x - 2x - 2 1=x-2 1+2=x-2+2 3=x Check solution is possible: Substitute x = 3 into 2x + 1 = 3x - 2. LHS = 2 # 3 + 1 = 7 =7 RHS = 3 # 3 - 2 =9-2 =7 Since LHS = RHS, x = 3 is a solution. Case (ii) 2 x + 1 = - ( 3x - 2 ) = - 3x + 2 2 x + 3x + 1 = - 3 x + 3x + 2 5x + 1 = 2 5x + 1 - 1 = 2 - 1 5x = 1 5x 1 = 5 5 1 x= 5

Chapter 3 Equations

Check: 1 Substitute x = into 2x + 1 = 3x - 2. 5 1 LHS = 2 # + 1 5 2 = 1 5 2 =1 5 1 RHS = 3 # - 2 5 3 = -2 5 2 = -1 5 1 Since LHS ! RHS, x = is not a solution. 5 So the only solution is x = 3.

It is often easier to solve these harder equations graphically. You will do this in Chapter 5.

2. 2x - 3 + x + 1 = 9

Solution In this question it is difficult to use distances on the number line, so we use the definition of absolute value. 2x - 3 2x - 3 = ' - (2 x - 3) +1 x + 1 = ' -(xx + 1)

when 2x - 3 $ 0 when 2x - 3 1 0 when x + 1 $ 0 when x + 1 1 0

This gives 4 cases: (i) (2x - 3) + (x + 1) = 9 (ii) (2x - 3) - (x + 1) = 9 (iii) -(2x - 3) + (x + 1) = 9 (iv) -(2x - 3) - (x + 1) = 9 Case (i) ( 2x - 3 ) + ( x + 1 ) = 9 2x - 3 + x + 1 = 9 3x - 2 = 9 3x - 2 + 2 = 9 + 2 3x = 11 3x 11 = 3 3 2 x=3 3 Check by substituting x = 3

111

2 into 2x - 3 + x + 1 = 9. 3 CONTINUED

112


2 2 -3 + 3 +1 3 3 1 2 = 4 + 4 3 3 1 2 =4 +4 3 3 =9 = RHS 2 So x = 3 is a solution. 3 Case (ii) ( 2 x - 3 ) - (x + 1 ) = 9 2x - 3 - x - 1 = 9 x-4=9 x-4+4=9+4 x = 13 Check by substituting x = 13 into 2x - 3 + x + 1 = 9. LHS = 2 # 13 - 3 + 13 + 1 = 23 + 14 = 23 + 14 = 37 ! RHS So x = 13 is not a solution. Case (iii) -(2x - 3) + (x + 1) = 9 - 2x + 3 + x + 1 = 9 -x + 4 = 9 -x + 4 - 4 = 9 - 4 -x = 5 -x 5 = -1 -1 x = -5 LHS = 2 # 3

Check by substituting x = - 5 into 2x - 3 + x + 1 = 9. LHS = 2 # - 5 - 3 + - 5 + 1 = - 13 + - 4 = 13 + 4 = 17 ! RHS So x = - 5 is not a solution. Case (iv) - (2x - 3) - (x + 1) = 9 - 2x + 3 - x - 1 = 9 - 3x + 2 = 9 - 3x + 2 - 2 = 9 - 2 - 3x = 7

Chapter 3 Equations

113

- 3x 7 = -3 -3 1 3 1 Check by substituting x = - 2 into 2x - 3 + x + 1 = 9. 3 1 1 LHS = 2 # - 2 - 3 + - 2 + 1 3 3 2 1 = -7 + -1 3 3 2 1 = 7 +1 3 3 =9 = RHS 1 So x = - 2 is a solution. 3 2 1 So solutions are x = 3 , - 2 . 3 3 x = -2

While you should always check solutions, you can see that there are some cases where this is really important.

You will learn how to solve equations involving absolute values graphically in Chapter 5. With graphical solutions it is easy to see how many solutions there are.

3.5 Exercises 1.

Solve

3.

Solve (a) x + 2 = 5x - 3 (b) 2a - 1 = a + 2 (c) b - 3 = 2b - 4 (d) 3k - 2 = k - 4 (e) 6y + 23 = y - 7 (f) 4x + 3 = 5x - 4 (g) 2m - 5 = m (h) 3d + 1 = d + 6 (i) 5 - y = 4y + 1 (j) 2t - 7 = 3 - t

4.

Solve

(a) x = 5 (b) y = 8 (c) a 1 4 (d) k $ 1 (e) x 2 6 (f) p # 10 (g) x = 0 (h) a 2 14 (i) y 1 12 (j) b $ 20 2.

Solve

(a) x + 3 = 3x - 1

(a) x + 2 = 7

(b) 2y - 5 = y - 2 (c) 3a + 1 = 2a - 9

(b) n - 1 = 3

(d) 2x + 5 + x = 17

(c) 2a 2 4

(e) 3d - 2 + d + 4 = 18

(d) x - 5 # 1 (e) 9 = 2x + 3 (f) 7x - 1 = 34 (g) 4y + 3 1 11 (h) 2x - 3 $ 15 x (i) =4 3 a (j) -3 #2 2

5.

(a) Solve 4t - 3 + t - 1 = 11. (b) By plotting the solutions on a number line and looking at values in between the solutions, solve 4t - 3 + t - 1 1 11.

Remember to check solutions in questions 3, 4 and 5.

114


Exponential Equations An exponential equation involves an unknown index or power e.g. 2 x = 8. We can also solve other equations involving indices. In order to solve these, you need to understand their relationship. For example, squares and square roots are the reverse of each other (we call them inverse operations). Similarly cubes and cube roots are inverses, and this extends to all indices. To solve equations, use inverse operations: For squares, take the square root For cubes, take the cube root For square roots, take the square For cube roots, take the cube You have previously used these rules when substituting into formulae involving squares and cubes.

EXAMPLES Solve 1. x 2 = 9 There are two possible solutions for x – one positive and one negative since 3 2 = 9 and (- 3) 2 = 9.

Solution x2 = 9 x2 = ! 9 ` x= !3 2. 5n 3 = 40

Solution

There is only one answer for this question since 2 3 = 8 but (- 2) 3 = -8.

5n 3 = 40 5n 3 40 = 5 5 3 n =8 3

n3 = 3 8 n=2

Chapter 3 Equations

2

3. a 3 = 4

Solution 2 3

3 2

3 2

2 3

We use the fact that ` a j = ` a j = a. 2

a3 = 4 2 3

3 2

3

à j = 4 2 3

`

a= 42 3 a = ^ 4h = 23 =8

Investigation Investigate equations of the type x n = k where k is a constant, for example, x n = 9. Look at these questions: 1. 2. 3. 4. 5. 6.

What is the solution when n = 0? What is the solution when n = 1? How many solutions are there when n = 2? How many solutions are there when n = 3? How many solutions are there when n is even? How many solutions are there when n is odd?

In other types of equations, the pronumeral (or unknown variable) is in the index. We call these exponential equations, and we use the fact that if the base numbers are equal, then the powers (or indices or exponents) must be equal.

EXAMPLES Solve 1. 3 x = 81

Solution 3 x = 81 Equating indices: 3x = 34 `x=4 CONTINUED

115

116


2. 5 2k - 1 = 25

Solution 5 2k - 1 = 25 5 2k - 1 = 5 2 ` 2k - 1 = 2 2k - 1 + 1 = 2 + 1 2k = 3 3 2k = 2 2 1 k=1 2

We can check this solution 1 by substituting k = 1 into 2 2k -1 the equation 5 = 25.

3. 8 n = 4

Solution It is hard to write 8 as a power of 4 or 4 as a power of 8, but both can be written as powers of 2. 8n = 4 (2 ) = 2 2 2 3n = 2 2 ` 3n = 2 3n 2 = 3 3 2 n= 3 3 n

3.6 Exercises 1.

Solve (a) x 3 = 27 (b) y 2 = 64 (c) n 4 = 16 (d) x 2 = 20 (give the exact answer) (e) p 3 = 1000 (f) 2x 2 = 50 (g) 6y 4 = 486 (h) w 3 + 7 = 15 (i) 6n 2 - 4 = 92 (j) 3q 3 + 20 = - 4

2.

Solve and give the answer correct to 2 decimal places. (a) p 2 = 45 (b) x 3 = 100 (c) n 5 = 240 (d) 2x 2 = 70 (e) 4y 3 + 7 = 34 d4 (f) = 14 3 k2 (g) -3=7 2 x3 - 1 (h) =2 5 (i) 2y 2 - 9 = 20 (j) 7y 3 + 9 = 200

Chapter 3 Equations

3.

Solve

6.

Solve (a) 2 n = 16 (b) 3 y = 243 (c) 2 m = 512 (d) 10 x = 100 000 (e) 6 m = 1 (f) 4 x = 64 (g) 4 x + 3 = 19 (h) 5 (3 x ) = 45 (i) 4 x = 4 6k (j) = 18 2

7.

Solve (a) 3 2x = 81 (b) 2 5x - 1 = 16 (c) 4 x + 3 = 4 (d) 3 n - 2 = 1 (e) 7 2x + 1 = 7 (f) 3 x - 3 = 27 (g) 5 3y + 2 = 125 (h) 7 3x - 4 = 49 (i) 2 4x = 256 (j) 9 3a + 1 = 9

8.

Solve (a) 4 m = 2 (b) 27 x = 3 (c) 125 x = 5

2 3

(a) n = 9 3

(b) t 4 = 8 2

(c) x 5 = 4 4

(d) t 3 = 16 3

(e) p 5 = 27 3

(f) 2m 4 = 250 2

(g) b 3 + 3 = 39 4

(h) 5y 3 = 405 2

(i) 3a 7 - 2 = 10 3 4

(j) 4.

5.

t =9 3

Solve (all pronumerals ! 0) (a) x - 1 = 5 (b) a - 3 = 8 (c) y - 5 = 32 (d) x - 2 + 1 = 50 (e) 2n - 1 = 3 1 (f) a - 3 = 8 1 -2 (g) x = 4 1 (h) b - 1 = 9 1 (i) x - 2 = 2 4 16 (j) b - 4 = 81

1 k m =7 49 1 k m = 100 (e) c 1000 (f) 16 n = 8 (g) 25 x = 125 (h) 64 n = 16 (d) c

Solve (all pronumerals ! 0) (a) x

-

1 3

-

3 2

-

1 4

-

3 4

(b) x (c) a

(d) k

(e) 3x

-

3 2

=8 =

8 125

=3 = 125 2 3

= 12

1 8 2 1 3 (g) y = 4 2 4 (h) n 5 = 9 (f) x

(i) b

-

(j) m

5 3

2 3

1 3k (i) c m = 2 4 (j) 8 x - 1 = 4

=

= =

1 32 36 49

9.

Solve (a) 2 4x + 1 = 8 x (b) 3 5x = 9 x - 2 (c) 7 2k + 3 = 7 k - 1 (d) 4 3n = 8 n + 3 (e) 6 x - 5 = 216 x (f) 16 2x - 1 = 4 x - 4 (g) 27 x + 3 = 3 x 1 x 1 2x + 3 m (h) c m = c 2 64

117

118


3 x 27 2x - 3 m (i) c m = c 4 64 1 x-9 m (j) ] 5 g- x = c 25 10. Solve (a) 4 m =

2

9 k+3 m (b) c = 25 1 (c) = 4 2x - 5 2

3 5

(d) 3 k = 3 3 (e) c

3 1 3n + 1 m = 27 81

5 -n 2 3n + 1 (f) c m =c m 5 2 1 (g) 32 - x = 16 (h) 9 2b + 5 = 3 b 3 (i) 81 x + 1 =

3x

1 3m - 5 (j) 25 - m = c m 5

PUZZLE Test your logical thinking and that of your friends. 1. How many months have 28 days? 2. If I have 128 sheep and take away all but 10, how many do I have left? 3. A bottle and its cork cost $1.10 to make. If the bottle costs $1 more than the cork, how much does each cost? 4. What do you get if you add 1 to 15 four times? 5. On what day of the week does Good Friday fall in 2016?

Quadratic Equations A quadratic equation is an equation involving a square. For example, x 2 - 4 = 0.

Solving by factorisation When solving quadratic equations by factorising, we use a property of zero.

For any real numbers a and b, if ab = 0 then a = 0 or b = 0

EXAMPLES Solve 1. x 2 + x - 6 = 0

Solution x2 + x - 6 = 0 (x + 3) (x - 2) = 0

Chapter 3 Equations

`

x+3=0 or x-2=0 x+3-3=0-3 x-2+2 =0 +2 x = -3 or x= 2

So the solution is x = - 3 or 2. 2. y 2 - 7y = 0

Solution y 2 - 7y = 0 y ( y - 7) = 0 ` y=0

or

y-7=0

y-7+7=0+7 y=7 So the solution is y = 0 or 7. 3. 3a 2 - 14a = - 8

Solution 3a 2 - 14a = - 8 3a 2 - 14a + 8 = - 8 + 8 3a 2 - 14a + 8 = 0 (3a - 2) (a - 4) = 0 ` 3a - 2 = 0 or 3a - 2 + 2 = 0 or 3a = 2 3a 2 = 3 3 2 a= 3 2 So the solution is a = or 4. 3

a-4 =0 a-4+4 =0+4 a=4


y2 + y = 0

4.

t 2 - 5t = 0

2.

b2 - b - 2 = 0

5.

x 2 + 9x + 14 = 0

3.

p 2 + 2p - 15 = 0

6.

q2 - 9 = 0

119

120


7.

x2 - 1 = 0

17. 5x - x 2 = 0

8.

a 2 + 3a = 0

18. y 2 = y + 2

9.

2x 2 + 8x = 0

19. 8n = n 2 + 15

10. 4x 2 - 1 = 0

20. 12 = 7x - x 2

11. 3x 2 + 7x + 4 = 0

21. m 2 = 6 - 5m

12. 2y 2 + y - 3 = 0

22. x (x + 1) (x + 2) = 0

13. 8b 2 - 10b + 3 = 0

23. (y - 1) (y + 5) (y + 2) = 0

14. x 2 - 3x = 10

24. (x + 3) (x - 1) = 32

15. 3x 2 = 2x

25. (m - 3) (m - 4) = 20

16. 2x 2 = 7x - 5

Application 1 2 at where u is the 2 initial velocity and a is the acceleration. Find the time when the displacement will be zero, given u = - 12 and a = 10. A formula for displacement s at time t is given by s = ut +

2 s = ut + 1 at 2 2 0 = -12t + 1 (10) t 2

= -12t + 5t

2

= t (-12 + 5t ) ` t = 0 or

-12 + 5t = 0

-12 + 12 + 5t = 0 + 12 5t = 12 5t 12 = 5 5 t = 2.4 So displacement will be zero when t = 0 or 2.4.

Solving by completing the square Not all trinomials will factorise, so other methods need to be used to solve quadratic equations.

Chapter 3 Equations

121

EXAMPLES Solve 1. x 2 = 7

Solution x2 = 7 x=! 7 = ! 2.6 2. ] x + 3 g2 = 11

Solution ] x + 3 g2 = 11

Take the square root of both sides.

x + 3 = ! 11 x + 3 - 3 = ! 11 - 3 x = ! 11 - 3 = 0.3, - 6.3

3. ^ y - 2 h2 = 7

Solution ^ y - 2 h2 = 7 y-2=! 7 y-2+2=! 7+2 y=! 7+2 = 4.6, - 0.6

To solve a quadratic equation like x 2 - 6x + 3 = 0, which will not factorise, we can use the method of completing the square.

You learnt how to complete the square in Chapter 2.

EXAMPLES Solve by completing the square 1. x 2 - 6x + 3 = 0 (give exact answer)

Solution x 2 - 6x + 3 = 0 x 2 - 6x = - 3

Halve 6, square it and add to both sides of the equation.

2

c 6 m = 32 = 9 2 CONTINUED

122


x 2 - 6x + 9 = - 3 + 9 ] x - 3 g2 = 6 `

x-3=! 6 x-3+3=! 6+3 x=! 6+3

2. y 2 + 2y - 7 = 0 (correct to 3 significant figures)

Solution y 2 + 2y - 7 = 0 y 2 + 2y = 7

2

c 2 m = 12 = 1 2

y 2 + 2y + 1 = 7 + 1 ^ y + 1 h2 = 8 `

y+1=! 8 y + 1 - 1 = ! 8 -1 y = ! 8 -1 = !2 2 - 1 y = 1.83 or - 3.83

3.8 Exercises 1.

Solve by completing the square, giving exact answers in simplest surd form (a) x 2 + 4x - 1 = 0 (b) a 2 - 6a + 2 = 0 (c) y 2 - 8y - 7 = 0 (d) x 2 + 2x - 12 = 0 (e) p 2 + 14p + 5 = 0 (f) x 2 - 10x - 3 = 0 (g) y 2 + 20y + 12 = 0 (h) x 2 - 2x - 1 = 0 (i) n 2 + 24n + 7 = 0 (j) y 2 - 3y + 1 = 0

2.

Solve by completing the square and write your answers correct to 3 significant figures (a) x 2 - 2x - 5 = 0 (b) x 2 + 12x + 34 = 0 (c) q 2 + 18q - 1 = 0 (d) x 2 - 4x - 2 = 0 (e) b 2 + 16b + 50 = 0 (f) x 2 - 24x + 112 = 0 (g) r 2 - 22r - 7 = 0 (h) x 2 + 8x + 5 = 0 (i) a 2 + 6a - 1 = 0 (j) y 2 - 40y - 3 = 0

Solving by formula Completing the square is difficult with harder quadratic equations, for example 2x 2 - x - 5 = 0. Completing the square on a general quadratic equation gives the following formula.

Chapter 3 Equations

For the equation ax 2 + bx + c = 0 x=

-b !

b 2 - 4ac 2a

Proof Solve ax 2 + b + c = 0 by completing the square. ax 2 + bx + c = 0 ax 2 bx c 0 a + a +a=a bx c x2 + a + a = 0 c c bx c x2 + a + a - a = 0 - a bx c x2 + a = - a

2 2 2 b b ' 2l = c b m = b 2 a 2a 4a

bx c b2 b2 x2 + a + 2 = - a + 2 4a 4a c b2 b 2 cx + m = -a + 2 2a 4a - 4ac + b 2 = 4a 2 - 4ac + b 2 b x+ =! 2a 4a 2 2 b - 4ac =! 2a b 2 - 4ac b b b x+ =! 2a 2a 2a 2a b 2 - 4ac -b x= ! 2a 2a 2 - b ! b - 4ac = 2a

EXAMPLES 1. Solve x 2 - x - 2 = 0 by using the quadratic formula.

Solution a = 1, b = -1, c = - 2 b 2 - 4ac 2a - (-1) ! (-1) 2 - 4 (1) (-2) = 2 (1 ) 1! 1+8 = 2

x=

-b !

CONTINUED

123

124


1! 9 2 1!3 = 2 = 2 or - 1

1! 3 gives two 2 1+ 3 separate solutions, 2 1- 3 and . 2

=

x =

2. Solve 2y 2 - 9y + 3 = 0 by formula and give your answer correct to 2 decimal places.

Solution a = 2, b = -9, c = 3 -b ! b 2 - 4ac 2a - ] -9 g ! ] -9 g2 - 4 ] 2 g ] 3 g y= 2] 2 g 9 ! 81 - 24 = 4 9 ! 57 = 4 Z 4.14 or 0.36

x=

These solutions are irrational.

3.9 1.

Exercises

Solve by formula, correct to 3 significant figures where necessary (a) y 2 + 6y + 2 = 0 (b) 2x 2 - 5x + 3 = 0 (c) b 2 - b - 9 = 0 (d) 2x 2 - x - 1 = 0 (e) - 8x 2 + x + 3 = 0 (f) n 2 + 8n - 2 = 0 (g) m 2 + 7m + 10 = 0 (h) x 2 - 7x = 0 (i) x 2 + 5x = 6 (j) y 2 = 3y - 1

2.

Solve by formula, leaving the answer in simplest surd form (a) x 2 + x - 4 = 0 (b) 3x 2 - 5x + 1 = 0 (c) q 2 - 4q - 3 = 0 (d) 4h 2 + 12h + 1 = 0 (e) 3s 2 - 8s + 2 = 0 (f) x 2 + 11x - 3 = 0 (g) 6d 2 + 5d - 2 = 0 (h) x 2 - 2x = 7 (i) t 2 = t + 1 (j) 2x 2 + 1 = 7x

Class Investigation Here is a proof that 1 = 2. Can you see the fault in the proof? x2 - x2 = x2 - x2 x(x - x) = (x + x) (x - x) x=x+x x = 2x 1=2 `

Chapter 3 Equations

125

Further Inequations Inequations involving pronumerals in the denominator can be solved in several ways. Here is one method. You will use a different method in Chapter 10.

EXAMPLES 1 1. Solve x 1 3.

Solution 1 is undefined. 0

x!0 1 Solve x = 3. 1 x #x=3#x 1 = 3x 3x 1 = 3 3 1 =x 3 1 1 is not a solution of the inequation x 1 3. 3 1 Place x = 0 and x = on a number plane and test x values on either side 3 of these values in the inequation. x=

-3

-2

-1

0 1 3

1

Test for x 1 0, say x = -1 Substitute into the inequation: 1 x 13 1 13 -1 -1 1 3 So x 1 0 is part of the solution. 1 1 Test for 0 1 x 1 , say x = 3 10 1 13 1 10 10 1 3 1 So 0 1 x 1 is not part of the solution. 3 1 Test for x 2 , say x = 1 3 Substitute into the inequation:

2

3

4

5

(true)

(false)

CONTINUED

Circle these values as they are not included in the solution.

126


1 13 1 113

(true)

1 is part of the solution. 3 1 Solution is x 1 0, x 2 . 3

So x 2

-2

-3

2. Solve

-1

0 1 3

1

2

3

4

5

6 $ 1. x+3

Solution 6 is undefined. 0

x ! -3 Solve

6 = 1. x+3

6 # (x + 3) = 1 # (x + 3) x+3 6 =x+3 6-3 =x+ 3-3 3=x

Circle x = - 3 and fill in x = 3 since it is a part of the solution.

6 $ 1. x+3 Place x = - 3 and x = 3 on a number plane and test values on either side in the inequation. x = 3 is a solution of the inequation

-3

-2

-1

0

1

2

3

4

5

Test for x 1 - 3, say x = - 4 Substitute into the inequation: 6 $1 x+3 6 $1 -4 + 3 -6 $ 3

(false)

So x 1 - 3 is not part of the solution. Test for - 3 1 x # 3, say x = 0 6 $1 0+3 2$1

(true)

So - 3 1 x # 3 is part of the solution. Test for x $ 3, say x = 4 Substitute into the inequation: 6 $1 4+3 6 $1 7 So x $ 3 is not part of the solution.

(false)

Chapter 3 Equations

Solution is - 3 1 x # 3 -2

-3

3. Solve

-1

0

1

2

3

4

5

y2 - 6 # 1. y

Solution y!0 y2 - 6 = 1. y 2 y -6 y #y=1#y y2 - 6 = y y2 - y - 6 = y - y y2 - y - 6 = 0 ^y - 3h^y + 2h = 0 y - 3 = 0, y+2 =0 y - 3 + 3 = 0 + 3, y + 2 - 2 = 0 - 2 y = 3, y = -2 Solve

Sketch these on a number line and test values on either side. -3

-2

-1

0

1

Test for y # - 2, say y = - 3 Substitute into the inequation: y2 - 6 #1 y 2 ]-3 g - 6 #1 -3 -1 # 1

2

3

4

5

(true)

So y # - 2 is part of the solution. Test for - 2 # y 1 0, say y = -1 ] -1 g2 - 6 #1 -1 5#1

(false)

So - 2 # y 1 0 is not part of the solution. Test 0 1 y # 3, say y = 1 12 - 6 #1 1 -5 # 1 So 0 1 y # 3 is part of the solution.

(true)

Test y $ 3, say y = 4 CONTINUED

127

128


42 - 6 #1 4 1 2 #1 2 So y $ 3 is not part of the solution. The solution is y # - 2, 0 1 y # 3 -3

3.10

-2

-1

0

(false)

2

1

3

Exercises

Solve 1.

1 y 11

16.

3x + 1 1 $ x-4 3

2.

1 x 22

17.

8p + 7 25 2p - 9

3.

3 x 12

18.

3 x-2 # 5x + 1 4

4.

2 m $7

19.

7t + 4 $ -1 3t - 8

5.

3 x 2 -5

20.

5m + 4 1 1 4 2m

6.

2 # -1 b

21.

x2 - 5 1 -4 x

7.

1 24 x -1

22.

n2 + 8 $6 n

8.

1 1 -5 z+3

23.

x 2 - 15 22 x

9.

3 $4 x-2

24.

m2 - 8 #4 m +1

10.

-1 16 2-x

25.

4 $x x-3

11.

5 # -9 x+4

26.

2x 2 # -1 3x - 2

12.

2 25 3x - 4

27.

3 #x x-2

13.

-3 12 2a + 5

28.

n+5 2n n-3

14.

x 25 2x - 1

29.

3x 2 1 -2 7x + 4

15.

y 12 y +1

30.

2 x ( x - 4) #7 x -1

4

5

Chapter 3 Equations

129

Quadratic Inequations Solving quadratic inequations is similar to solving quadratic equations, but you need to do this in two stages. The first is to solve the equation and then the second step is to look at either the number line or the number plane for the inequality.

To solve a quadratic inequation: 1. Factorise and solve the quadratic equation 2. Test values in the inequality

In Chapter 10 you will look at how to use the number plane to solve these quadratic inequations. Here are some examples of solving quadratic inequations using the number line.

EXAMPLES Solve 1. x 2 + x - 6 2 0

Solution Be careful: x 2 + x - 6 2 0 does not mean x - 2 2 0 and x + 3 2 0.

First solve x + x - 6 = 0 (x - 2 ) (x + 3 ) = 0 ` x = 2 or -3 2

Now look at the number line. -4

-3

-2

-1

0

1

2

3

4

Choose a number between - 3 and 2, say x = 0. Substitute x = 0 into the inequation. x2 + x - 6 2 0 02 + 0 - 6 2 0 -6 2 0

(false)

So the solution is not between -3 and 2. ` the solution lies either side of -3 and 2. Check by choosing a number on either side of the two numbers. Choose a number on the RHS of 2, say x = 3. CONTINUED

130


Substitute x = 3 into the inequation. 32 + 3 - 6 2 0 620 So the solution is on the RHS of 2. Choose a number on the LHS of -3, say x = -4 Substitute x = -4 into the inequation

(true)

(- 4) 2 + ( - 4) - 6 2 0 620 So the solution is on the LHS of -3. -4

-3

-2

-1

(true)

0

1

2

3

4

1

2

3

4

This gives the solution x 1 -3, x 2 2. 2. 9 - x 2 $ 0

Solution First solve 9 - x 2 = 0 (3 - x) (3 + x) = 0 ` x = !3 -4

-3

-2

-1

0

Choose a number between -3 and 3, say x = 0. Substitute x = 0 into the inequation. Check numbers on the RHS and LHS to verify this.

9 - x2 $ 0 9 - 02 $ 0 9$0

(true)

So the solution is between -3 and 3, that is -3 # x # 3. On the number line: -4

-3

-2

-1

0

1

2

3

4

Earlier in the chapter you learned how to solve inequations with the unknown in the denominator. Some people like to solve these using quadratic inequations. Here are some examples of how to do this.

Chapter 3 Equations

131

EXAMPLES Solve 1 1. x 1 3

Solution x 2 is positive, so the inequality sign does not change.

x!0 First, multiply both sides by x 2 . 1 x 13 x 1 3x 2 0 1 3x 2 - x Now, solve

3x 2 - x = 0 x(3x - 1) = 0 x = 0 or -2

1 3 -1

0 1 3

1

2

By checking on each side of 0 and 1 , for 0 1 3x 2 - x, the solution is 3 x 1 0, x 2 1 . 3 2.

3 $2 x+5

Solution 2

(x + 5) is positive, so the inequality sign does not change.

x ! -5 First, multiply both sides by (x + 5)2 . 3 $2 x+5 3 ( x + 5 ) $ 2 ( x + 5) 2 0 $ 2 ( x + 5 ) 2 - 3 ( x + 5) 0 $ ( x + 5 ) [ 2 ( x + 5) - 3 ] 0 $ ( x + 5 ) ( 2 x + 7) Now, solve (x + 5) (2x + 7) = 0 ` x + 5 = 0 or 2x + 7 = 0 x = -5 -6

-5

Check this factorisation carefully.

x cannot be -5 as this would give 0 in the denominator.

x = -3 1 2 -4 -3 1 -3 2

-2

1 Check by choosing a number on each side of -5 and -3 for 2 1 0 $ (x + 5) (2x + 7) that the solution is -5 1 x # -3 . 2

132


3.11

Exercises

Solve 1.

x 2 + 3x 1 0

21. x 2 1 2x

2.

y 2 - 4y 1 0

22. 2a 2 - 5a + 3 # 0

3.

n2 - n $ 0

23. 5y 2 + 6y $ 8

4.

x2 - 4 $ 0

24. 6m 2 2 15 - m

5.

1 - n2 1 0

25. 3x 2 # 7x - 4

6.

n 2 + 2n - 15 # 0

1 26. x 2 2

7.

c2 - c - 2 2 0

8.

x + 6x + 8 # 0

9.

x 2 - 9x + 20 1 0

2

3 27. x # 6 28.

1 15 y+1

29.

1 $2 n-3

30.

3 $ -1 x+5

31.

1 17 5x - 2

32.

4 $ -5 x-5

33.

x #5 x+1

34.

2x + 1 21 x-2

35.

2x - 3 $6 5x + 3

10. 4b 2 + 10b + 4 $ 0 11. 1 - 2a - 3a 1 0 2

12. 2y 2 - y - 6 2 0 13. 3x 2 - 5x + 2 $ 0 14. 6 - 13b - 5b 1 0 2

15. 6x 2 + 11x + 3 # 0 16. y 2 + y # 12 17. x 2 2 16 18. a 2 # 1 19. x 2 1 x + 6 20. x $ 2x + 3 2

Simultaneous Equations Two equations, each with two unknown pronumerals, can be solved together to find one solution that satisfies both equations. There are different ways of solving simultaneous equations. The elimination method adds or subtracts the equations. The substitution method substitutes one equation into the other.

Chapter 3 Equations

Linear equations These equations can be solved by either method. Many students prefer the elimination method.

EXAMPLES Solve simultaneously 1. 3a + 2b = 5 and 2a - b = -6

Solution

] 2 g # 2: ] 1 g + (3):

3a + 2b = 5 2a - b = -6

(1) (2)

4a - 2b = -12 3a + 2b = 5 7a = - 7 a = -1

(3) (1)

Substitute a = -1 in (1) 3 (-1) + 2b = 5 -3 + 2b = 5 2b = 8 b=4 ` solution is a = -1, b = 4 2. 5x - 3y = 19 and 2x - 4y = 16

Solution

(1) # 4: ( 2 ) # 3: (3) - (4):

5x - 3y = 19 2x - 4y = 16 20x - 12y = 76 6x - 12y = 48 14x = 28 x=2

Substitute x = 2 in (2) 2 ( 2) - 4 y 4 - 4y - 4y y

= 16 = 16 = 12 = -3

( 1) ( 2) (3) (4 )

133

134


3.12

Exercises

Solve simultaneously 1.

a - b = -2 and a + b = 4

12. 3a - 4b = -16 and 2a + 3b = 12

2.

5x + 2y = 12 and 3x - 2y = 4

3.

4p - 3q = 11 and 5p + 3q = 7

13. 5p + 2q + 18 = 0 and 2p - 3q + 11 = 0

4.

y = 3x - 1 and y = 2x + 5

5.

2x + 3y = -14 and x + 3y = -4

6.

7t + v = 22 and 4t + v = 13

16. 5s - 3t - 13 = 0 and 3s - 7t - 13 = 0

7.

4x + 5y + 2 = 0 and 4x + y + 10 = 0

17. 3a - 2b = - 6 and a - 3b = - 2

8.

2x - 4y = 28 and 2x - 3y = -11

18. 3k - 2h = -14 and 2k - 5h = -13

9.

5x - y = 19 and 2x + 5y = -14

10. 5m + 4n = 22 and m - 5n = -13 11. 4w 1 + 3w 2 = 11 and 3w 1 + w 2 = 2

14. 7x 1 + 3x 2 = 4 and 3x 1 + 5x 2 = - 2 15. 9x - 2y = -1 and 7x - 4y = 9

19. 2v 1 + 5v 2 - 16 = 0 and 7v 1 + 2v 2 + 6 = 0 20. 1.5x + 3.4y = 7.8 and 2 . 1 x - 1 . 7y = 1 . 8

PROBLEM A group of 39 people went to see a play. There were both adults and children in the group. The total cost of the tickets was $939, with children paying $17 each and adults paying $29 each. How many in the group were adults and how many were children? (Hint: let x be the number of adults and y the number of children.)

Non-linear equations In questions involving non-linear equations there may be more than one set of solutions. In some of these, the elimination method cannot be used. Here are some examples using the substitution method.

Chapter 3 Equations

EXAMPLES Solve simultaneously 1. xy = 6 and x + y = 5

Solution xy = 6 x+y=5 From (2): y=5-x Substitute (3) in (1) x (5 - x) = 6

( 1) (2 ) (3 )

5x - x 2 = 6 0 = x 2 - 5x + 6 0 = (x - 2 ) (x - 3 ) ` x - 2 = 0 or x - 3 = 0 x = 2 or x = 3 Substitute x = 2 in (3) y=5-2=3 Substitute x = 3 in (3) y=5-3=2 ` solutions are x = 2, y = 3 and x = 3, y = 2 2. x 2 + y 2 = 16 and 3x - 4y - 20 = 0

Solution x 2 + y 2 = 16 3x - 4y - 20 = 0 From ] 2 g: 3x - 20 = 4y 3x - 20 =y 4 Substitute (3) into (1) 3x - 20 2 m = 16 x2 + c 4 9x 2 - 120x + 400 n = 16 x2 + d 16 16x 2 + 9x 2 - 120x + 400 = 256 25x 2 - 120x + 144 = 0 (5x - 12)2 = 0 ` 5x - 12 = 0 5x = 12 x = 2.4 Substitute x = 2.4 into ] 3 g 3 (2.4) - 20 4 = -3.2 So the solution is x = 2.4, y = -3.2. y=

(1) ( 2)

(3)

135

136


3.13

Exercises

Solve the simultaneous equations. 1.

y = x 2 and y = x

11. y = x - 1 and y = x 2 - 3

2.

y = x 2 and 2x + y = 0

12. y = x 2 + 1 and y = 1 - x 2

3.

x 2 + y 2 = 9 and x + y = 3

13. y = x 2 - 3x + 7 and y = 2x + 3

4.

x - y = 7 and xy = -12

14. xy = 1 and 4x - y + 3 = 0

5.

y = x 2 + 4x and 2x - y - 1 = 0

15. h = t 2 and h = ] t + 1 g2

6.

y = x 2 and 6x - y - 9 = 0

16. x + y = 2 and 2x 2 + xy - y 2 = 8

7.

x = t 2 and x + t - 2 = 0

17. y = x 3 and y = x 2 + 6x

8.

m 2 + n 2 = 16 and m + n + 4 = 0

18. y = | x | and y = x 2

9.

xy = 2 and y = 2x

19. y = x 2 - 7x + 6 and 24x + 4y - 23 = 0

10. y = x 3 and y = x 2

20. x 2 + y 2 = 1 and 5x + 12y + 13 = 0

Equations with 3 unknown variables Four unknowns need 4 equations, and so on.

Three equations can be solved simultaneously to find 3 unknown pronumerals.

EXAMPLE Solve simultaneously a - b + c = 7, a + 2b - c = -4 and 3a - b - c = 3.

Solution a-b +c=7 a + 2b - c = - 4 3a - b - c = 3 (1) + (2): a-b+c=7 a + 2b - c = - 4 2a + b =3 (1) + (3): a- b+c=7 3a - b - c = 3 4a - 2b = 10 or 2a - b =5 (4) + (5): 2a + b =3 4a =8 a=2

(1 ) (2) (3)

( 4)

(5)

Chapter 3 Equations

Substitute a = 2 in (4) 2 ( 2) + b = 3 4+b=3 b = -1 Substitute a = 2 and b = -1 in (1) 2 - (-1) + c = 7 2 +1 + c = 7 3+c=7 c=4 ` solution is a = 2, b = -1, c = 4

3.14

You will solve 3 simultaneous equations in later topics (for example, in Chapter 10).

Exercises

Solve the simultaneous equations. 1.

x = - 2, 2x - y = 4 and x - y + 6z = 0

2.

a = - 2, 2a - 3b = -1 and a - b + 5c = 9

3.

2a + b + c = 1, a + b = - 2 and c = 7

4.

a + b + c = 0, a - b + c = - 4 and 2a - 3b - c = -1

5.

x + y - z = 7, x + y + 2z = 1 and 3x + y - 2z = 19

6.

x - y - z = 1, 2x + y - z = -9 and 2x - 3y - 2z = 7

137

7.

2p + 5q - r = 25, 2p - 2q - r = -24 and 3p - q + 5r = 4

8.

2x - y + 3z = 9, 3x + y - 2z = -2 and 3x - y + 5z = 14

9.

3h + j - k = -3, h + 2j + k = -3 and 5h - 3j - 2k = -13

10. 2a - 7b + 3c = 7, a + 3b + 2c = -4 and 4a + 5b - c = 9

138


Test Yourself 3 1.

Solve (a) 8 = 3b - 22 a a+2 (b) =9 4 3 (c) 4 (3x + 1) = 11x - 3 -4 (d) #3 x+3 (e) 3p + 1 # p + 9

2.

3.

The compound interest formula is r n m . Find correct to 2 A = P c1 + 100 decimal places. (a) A when P = 1000, r = 6 and n = 4 (b) P when A = 12 450, r = 5.5 and n = 7 Complete the square on (a) x 2 - 8x (b) k 2 + 4k

9.

Solve -2 1 3y + 1 # 10, and plot your solution on a number line.

10. Solve correct to 3 significant figures (a) x 2 + 7x + 2 = 0 (b) y 2 - 2y - 9 = 0 (c) 3n 2 + 2n - 4 = 0 11. The surface area of a sphere is given by A = 4rr 2 . Evaluate to 1 decimal place (a) A when r = 7.8 (b) r when A = 102.9 12. Solve

x-3 3 - 2 9. 7 4

13. Solve x 2 - 11x + 18 2 0. 14. Solve the simultaneous equations x 2 + y 2 = 16 and 3x + 4y - 20 = 0. 4 3 rr . 3 Evaluate to 2 significant figures (a) V when r = 8 (b) r when V = 250

15. The volume of a sphere is V =

4.

Solve these simultaneous equations. (a) x - y + 7 = 0 and 3x - 4y + 26 = 0 (b) xy = 4 and 2x - y - 7 = 0

5.

Solve (a) 3 x + 2 = 81 (b) 16 y = 2

6.

Solve (a) 3b - 1 = 5

(a) x 2 - 6x + 9 = 0

(b) 5g - 3 = 3g + 1

(c) x - 2 = 7 - x

(c) 2x - 7 $ 1

(d) x 2 - x + 4 = 0

7.

8.

The area of a trapezium is given by A = 1 h (a + b). Find 2 (a) A when h = 6, a = 5 and b = 7 (b) b when A = 40, h = 5 and a = 4. Solve 2x 2 - 3x + 1 = 0 by (a) factorisation (b) quadratic formula.

16. Which of the following equations has (i) 2 solutions (ii) 1 solution (iii) no solutions? (b) 2x - 3 = 7

(e) 2x + 1 = x - 2 17. Solve simultaneously a + b = 5, 2 a + b + c = 4, a - b - c = 5. 18. Solve 3n + 5 2 5, and plot the solution on a number line. 19. Solve

3 4 =x x+1

^ x ! 0, -1 h .

Chapter 3 Equations

20. Solve 9 2x + 1 = 27 x .

(k) 27 2x - 1 = 9 (l) 4b - 3 # 5 (m) 3x + 2 = 2x - 3 (n) 4t - 5 = t + 2 (o) x 2 1 2x + 3 (p) m 2 + m $ 6 2t - 3 (q) 15 t y+1 (r) 22 y-1 n (s) $3 2n - 4 3x - 2 (t) # -1 2x + 1

21. Solve (a) 2 ^ 3y - 5 h 2 y + 5 (b) n 2 + 3n # 0 (c) 3 2x - 1 = 27 (d) 5x 3 - 1 = 39 (e) 5x - 4 = 11 (f) 2t + 1 $ 3 (g) x 2 + 2x - 8 # 0 (h) 8 x + 1 = 4 x (i) y 2 - 4 2 0 (j) 1 - x 2 # 0

Challenge Exercise 3 1 . a2

1.

Find the value of y if a 3y - 5 =

2.

Solve x 2 a .

3.

The solutions of x 2 - 6x - 3 = 0 are in the form a + b 3 . Find the values of a and b.

4.

2 1 = 1 correct to 3 x -1 x +1 significant figures. (x ! ! 1) y2 - 6 Solve # 1. y

5. 6. 7. 8.

9.

2

2

Solve

11. Solve ] x - 4 g ] x - 1 g # 28. 3

12. Solve x 2 =

1 . 8

13. The volume of a sphere is given by 4 V = rr 3 . Find the value of r when 3 V = 51.8 (correct to three significant figures). 14. Solve x - 3 + x + 4 = x - 2 . 15. Find the solutions of x 2 - 2ax - b = 0 by completing the square.

Factorise x 5 - 9x 3 - 8x 2 + 72. Hence solve x 5 - 9x 3 - 8x 2 + 72 = 0.

16. Solve

Solve simultaneous equations y = x 3 + x 2 and y = x + 1.

17. Given A = P c 1 +

Find the value of b if x 2 - 8x + b 2 is a perfect square. Hence solve x 2 - 8x - 1 = 0 by completing the square.

18. Solve 3x 2 = 8 (2x - 1) and write the solution in the simplest surd form.

Considering the definition of absolute x-3 value, solve = x, where x ! 3. 3-x

10. Solve t + 2 + 3t - 1 1 5.

6y 2 # - 3. 3y - 2

r n m , find P 100 correct to 2 decimal places when A = 3281.69, r = 1.27 and n = 30.

19. Solve

5x + 3 2 2 x. x+4

20. Solve 3y - 1 + 2y + 3 2 5.

139

4

Geometry 1 TERMINOLOGY Altitude: Height. Any line segment from a vertex to the opposite side of a polygon that is perpendicular to that side

Polygon: General term for a many sided plane figure. A closed plane (two dimensional) figure with straight sides

Congruent triangles: Identical triangles that are the same shape and size. Corresponding sides and angles are equal. The symbol is /

Quadrilateral: A four-sided closed figure such as a square, rectangle, trapezium etc.

Interval: Part of a line including the endpoints

Similar triangles: Triangles that are the same shape but different sizes. The symbol is zy

Median: A line segment that joins a vertex to the opposite side of a triangle that bisects that side

Vertex: The point where three planes meet. The corner of a figure

Perpendicular: A line that is at right angles to another line. The symbol is =

Vertically opposite angles: Angles that are formed opposite each other when two lines intersect

Chapter 4 Geometry 1

INTRODUCTION GEOMETRY IS USED IN many areas, including surveying, building and graphics.

These fields all require a knowledge of angles, parallel lines and so on, and how to measure them. In this chapter, you will study angles, parallel lines, triangles, types of quadrilaterals and general polygons. Many exercises in this chapter on geometry need you to prove something or give reasons for your answers. The solutions to geometry proofs only give one method, but other methods are also acceptable.

DID YOU KNOW? Geometry means measurement of the earth and comes from Greek. Geometry was used in ancient civilisations such as Babylonia. However, it was the Greeks who formalised the study of geometry, in the period between 500 BC and AD 300.

Notation In order to show reasons for exercises, you must know how to name figures correctly. •B The point is called B.

The interval (part of a line) is called AB or BA.

If AB and CD are parallel lines, we write AB < CD.

This angle is named +BAC or +CAB. It can sometimes be named +A. ^

Angles can also be written as BAC or BAC.

This triangle is named 3ABC.

141

142


To name a quadrilateral, go around it: for example, BCDA is correct, but ACBD is not.

Producing a line is the same as extending it.

This quadrilateral is called ABCD.

Line AB is produced to C.

+ABD and +DBC are equal.

DB bisects +ABC.

AM is a median of D ABC.

AP is an altitude of D ABC.

Types of Angles Acute angle

0c1 xc1 90c


Right angle

A right angle is 90c. Complementary angles are angles whose sum is 90c.

Obtuse angle

90c1 xc1180c

Straight angle

A straight angle is 180c. Supplementary angles are angles whose sum is 180c.

Reflex angle

180c1 xc1 360c

Angle of revolution

An angle of revolution is 360c.

Vertically opposite angles

+AEC and +DEB are called vertically opposite angles. +AED and +CEB are also vertically opposite angles.

143

144


Vertically opposite angles are equal. That is, +AEC = +DEB and +AED = +CEB.

Proof Let +AEC = xc Then +AED = 180c - xc (+CED straight angle, 180c) Now +DEB = 180c - (180c - xc) (+AEB straight angle, 180c) = xc Also +CEB = 180c - xc (+CED straight angle, 180c) ` +AEC = +DEB and +AED =+CEB

EXAMPLES Find the values of all pronumerals, giving reasons. 1.

Solution x + 154 = 180 (+ABC is a straight angle, 180c) x + 154 - 154 = 180 - 154 ` x = 26 2.

Solution 2x + 142 + 90 = 360 (angle of revolution, 360c ) 2x + 232 = 360 2x + 232 - 232 = 360 - 232 2x = 128 2x 128 = 2 2 x = 64


3.

Solution y + 2y + 30 = 90 (right angle, 90c) 3y + 30 = 90 3y + 30 - 30 = 90 - 30 3y = 60 3y 60 = 3 3 y = 20 4.

Solution x + 50 = 165 x + 50 - 50 = 165 - 50 x = 115 y = 180 - 165 = 15 w = 15

(+WZX and +YZV vertically opposite)

(+XZY straight angle, 180c) (+WZY and +XZV vertically opposite)

5.

CONTINUED

145

146


Solution a = 90 b + 53 + 90 = 180 b + 143 = 180 b + 143 - 143 = 180 - 143 b = 37 d = 37 c = 53

(vertically opposite angles) (straight angle, 180c)

(vertically opposite angles) (similarly)

6. Find the supplement of 57c 12l.

Solution Supplementary angles add up to 180c. So the supplement of 57c 12l is 180c - 57c 12l = 122c 48l. 7. Prove that AB and CD are straight lines. A

D

(x + 30)c C

(6x + 10)c

(2x 2 + 10)c E (5x + 30)c B

Solution 6x + 10 + x + 30 + 5x + 30 + 2x + 10 = 360 ^ angle of revolution h 14x + 80 - 80 = 360 - 80 14x = 280 14x 280 = 14 14 x = 20 +AEC = (20 + 30)c = 50c +DEB = (2 # 20 + 10)c = 50c These are equal vertically opposite angles. ` AB and CD are straight lines


4.1 Exercises 1.

Find values of all pronumerals, giving reasons. (a)

yc

(i)

133c

(b)

(j)

(c) 2.

Find the supplement of (a) 59c (b) 107c 31l (c) 45c 12l

3.

Find the complement of (a) 48c (b) 34c 23l (c) 16c 57l

4.

Find the (i) complement and (ii) supplement of (a) 43c (b) 81c (c) 27c (d) 55c (e) 38c (f) 74c 53l (g) 42c 24l (h) 17c 39l (i) 63c 49l (j) 51c 9l

5.

(a) Evaluate x. (b) Find the complement of x. (c) Find the supplement of x.

(d)

(e)

(f)

(g)

(h) (2x + 30)c 142c

147

148


6.

Find the values of all pronumerals, giving reasons for each step of your working.

8.

(a) Prove that CD bisects +AFE. 9.

Prove that AC is a straight line. D C

(b) (3x + 70)c (110 - 3x)c B

(c) A

10. Show that +AED is a right angle. A

(d)

B

(50 - 8y)c

(e)

C

(5y - 20)c

E

(f)

7.

Prove that AC and DE are straight lines.

(3y + 60)c

D


149

Parallel Lines When a transversal cuts two lines, it forms pairs of angles. When the two lines are parallel, these pairs of angles have special properties.

Alternate angles

Alternate angles form a Z shape. Can you find another set of alternate angles?

If the lines are parallel, then alternate angles are equal.

Corresponding angles

Corresponding angles form an F shape. There are 4 pairs of corresponding angles. Can you find them?

If the lines are parallel, then corresponding angles are equal.

150


Cointerior angles Cointerior angles form a U shape. Can you find another pair?

If the lines are parallel, cointerior angles are supplementary (i.e. their sum is 180c).

Tests for parallel lines

If alternate angles are equal, then the lines are parallel.

If +AEF = +EFD, then AB < CD.

If corresponding angles are equal, then the lines are parallel.

If +BEF = +DFG, then AB < CD.

If cointerior angles are supplementary, then the lines are parallel.

If +BEF + +DFE = 180c, then AB < CD.


If 2 lines are both parallel to a third line, then the 3 lines are parallel to each other. That is, if AB < CD and EF < CD, then AB < EF.

EXAMPLES 1. Find the value of y, giving reasons for each step of your working.

Solution +AGF = 180c - 125c = 55c

(+FGH is a straight angle)

`

(+AGF, +CFE corresponding angles, AB < CD)

y = 55c

2. Prove EF < GH.

Solution +CBF = 180c - 120c (+ABC is a straight angle) = 60c ` +CBF = +HCD = 60c But +CBF and +HCD are corresponding angles ` EF < GH

Can you prove this in a different way?

151

152


Think about the reasons for each step of your calculations.

4.2 Exercises 1.

Find values of all pronumerals. (a)

(h)

(i)

(b) (j)

(c)

2.

Prove AB < CD. (a)

(d)

(b) (e)

(c)

A

(f)

(g)

B

104c

C 76c

D

E


A

(d)

(e)

B 138c

B

52c

E C

C

E 128c

D

23c

F 115c

G

H

F

Types of Triangles Names of triangles A scalene triangle has no two sides or angles equal.

A right (or right-angled) triangle contains a right angle.

The side opposite the right angle (the longest side) is called the hypotenuse. An isosceles triangle has two equal sides. The angles (called the base angles) opposite the equal sides in an isosceles triangle are equal.

An equilateral triangle has three equal sides and angles.

A D

153

154


All the angles are acute in an acute-angled triangle.

An obtuse-angled triangle contains an obtuse angle.

Angle sum of a triangle

The sum of the interior angles in any triangle is 180c, that is, a + b + c = 180

Proof

Let +YXZ = ac, +XYZ = bc and +YZX = cc Draw line AB < YZ Then +BXZ = cc (+BXZ, +XZY alternate angles, AB < YZ) +AXY = bc (similarly) +YXZ + +AXY + +BXZ = 180c (+AXB is a straight angle) ` a + b + c = 180


Class Investigation 1. 2. 3. 4. 5.

Could you prove the base angles in an isosceles triangle are equal? Can there be more than one obtuse angle in a triangle? Could you prove that each angle in an equilateral triangle is 60c? Can a right-angled triangle be an obtuse-angled triangle? Can you find an isosceles triangle with a right angle in it?

Exterior angle of a triangle

The exterior angle in any triangle is equal to the sum of the two opposite interior angles. That is, x+y=z

Proof

Let +ABC = xc , +BAC = yc and +ACD = zc Draw line CE < AB zc = +ACE + +ECD +ECD = xc +ACE = yc ` z=x+y

(+ECD,+ABC corresponding angles, AB < CE) (+ACE,+BAC alternate angles, AB < CE)

EXAMPLES Find the values of all pronumerals, giving reasons for each step. 1.

CONTINUED

155

156


Solution x + 53 + 82 = 180 (angle sum of D 180c) x + 135 = 180 x + 135 - 135 = 180 - 135 x = 45 2.

Solution +A = +C = x x + x + 48 = 180 2x + 48 = 180 2x + 48 - 48 = 180 - 48 2x = 132 132 2x = 2 2 x = 66

(base angles of isosceles D) (angle sum in a D 180c)

3.

Solution y + 35 = 141 (exterior angle of D) y + 35 - 35 = 141 - 35 ` y = 106 This example can be done using the interior sum of angles. +BCA = 180c - 141c = 39c y + 39 + 35 = 180 y + 74 = 180 y + 74 - 74 = 180 - 74 ` y = 106

(+BCD is a straight angle 180c) (angle sum of D 180c)


Think of the reasons for each step of your calculations.

4.3 Exercises 1.

Find the values of all pronumerals. (a)

(h)

(b) (i)

(j) (c)

(d)

(k)

(e)

(f)

(g)

157

2.

Show that each angle in an equilateral triangle is 60c.

3.

Find +ACB in terms of x.

158


4.

5.

6.

Prove AB < ED.

(d)

8.

Prove D IJL is equilateral and D JKL is isosceles.

9.

In triangle BCD below, BC = BD. Prove AB ED.

Show D ABC is isosceles.

Line CE bisects +BCD. Find the value of y, giving reasons.

A B C

46c E 88c

D

7.

Evaluate all pronumerals, giving reasons for your working. (a)

10. Prove that MN QP . 32c

M

(b) 75c

O

73c

Q

(c)

P

N


Congruent Triangles Two triangles are congruent if they are the same shape and size. All pairs of corresponding sides and angles are equal. For example:

We write D ABC / D XYZ.

Tests To prove that two triangles are congruent, we only need to prove that certain combinations of sides or angles are equal.

Two triangles are congruent if • SSS: all three pairs of corresponding sides are equal • SAS: two pairs of corresponding sides and their included angles are equal • AAS: two pairs of angles and one pair of corresponding sides are equal • RHS: both have a right angle, their hypotenuses are equal and one other pair of corresponding sides are equal

EXAMPLES 1. Prove that DOTS / DOQP where O is the centre of the circle.

CONTINUED

The included angle is the angle between the 2 sides.

159

160


Solution S: A: S:

OS = OQ +TOS = +QOP OT = OP

`

by SAS, DOTS / DOQP

(equal radii) (vertically opposite angles) (equal radii)

2. Which two triangles are congruent?

Solution To find corresponding sides, look at each side in relation to the angles. For example, one set of corresponding sides is AB, DF, GH and JL. D ABC / D JKL (by SAS) 3. Show that triangles ABC and DEC are congruent. Hence prove that AB = ED.

Solution A: +BAC = +CDE A: +ABC = +CED S: AC = CD

(alternate angles, AB < ED) (similarly) (given)

` by AAS, D ABC / D DEC ` AB = ED

(corresponding sides in congruent D s)


4.4 Exercises 1.

Are these triangles congruent? If they are, prove that they are congruent. (a)

2.

Prove that these triangles are congruent. (a)

B

(b)

Y 4.7

m

110c

2.3

4.7

m

m

Z

110 c C

A 2

.3 m

(b)

X

(c)

(c)

(d)

(d) (e)

(e)

161

162


3.

A

Prove that (a) Δ ABD is congruent to Δ ACD (b) AB bisects BC, given D ABC is isosceles with AB = AC.

D

B

4.

Prove that triangles ABD and CDB are congruent. Hence prove that AD = BC.

C

(a) Prove that TABC and TADC are congruent. (b) Show that +ABC = +ADC. 7.

The centre of a circle is O and AC is perpendicular to OB. A

5.

In the circle below, O is the centre of the circle. A

O

D

B O

C

B

C

(a) Prove that TOAB and TOCD are congruent. (b) Show that AB = CD. 6.

(a) Show that TOAB and TOBC are congruent. (b) Prove that +ABC = 90c. 8.

ABCF is a trapezium with AF = BC and FE = CD. AE and BD are perpendicular to FC.

In the kite ABCD, AB = AD and BC = DC.

F

A

B

E

D

C

(a) Show that TAFE and TBCD are congruent. (b) Prove that +AFE = +BCD.


9.

The circle below has centre O and OB bisects chord AC.

10. ABCD is a rectangle as shown below. A

B

D

C

C O B

A

(a) Prove that TOAB is congruent to TOBC. (b) Prove that OB is perpendicular to AC.

(a) Prove that TADC is congruent to TBCD. (b) Show that diagonals AC and BD are equal.

Investigation The triangle is used in many structures, for example trestle tables, stepladders and roofs. Find out how many different ways the triangle is used in the building industry. Visit a building site, or interview a carpenter. Write a report on what you find.

Similar Triangles Triangles, for example ABC and XYZ, are similar if they are the same shape but different sizes. As in the example, all three pairs of corresponding angles are equal. All three pairs of corresponding sides are in proportion (in the same ratio).

163

164


We write: D ABC <; D XYZ D XYZ is three times larger than D ABC. 6 XY = =3 AB 2 XZ 12 = =3 4 AC 15 YZ = =3 5 BC XY XZ YZ ` = = AB AC BC

This shows that all 3 pairs of sides are in proportion.

Application Similar figures are used in many areas, including maps, scale drawings, models and enlargements.

EXAMPLE 1. Find the values of x and y in similar triangles CBA and XYZ.

Solution First check which sides correspond to one another (by looking at their relationships to the angles). YZ and BA, XZ and CA, and XY and CB are corresponding sides. `

XZ XY = CA CB y 5.4 = 4.9 3.6 3.6y = 4.9 # 5.4


165

4 . 9 # 5 .4 3 .6 = 7.35 XY = CB 5 .4 = 3.6 = 2 . 3 # 5 .4 2 . 3 # 5 .4 = 3 .6 = 3.45

y= YZ BA x 2 .3 3 .6x x

Tests There are three tests for similar triangles. Two triangles are similar if: • three pairs of corresponding angles are equal • three pairs of corresponding sides are in proportion • two pairs of sides are in proportion and their included angles are equal

EXAMPLES 1. (a) Prove that triangles ABC and ADE are similar. (b) Hence find the value of y, to 1 decimal place.

Solution (a) +A is common +ABC = +ADE +ACB = +AED ` D ABC <; D ADE (b)

(corresponding angles, BC < DE) (similarly) (3 pairs of angles equal)

CONTINUED

If 2 pairs of angles are equal then the third pair must also be equal.

166


AE = 2.4 + 1.9 = 4 .3 DE AE = BC AC y 4 .3 = 3 .7 2.4 2 .4 y = 3 . 7 # 4. 3 3 .7 # 4 .3 y= 2 .4 = 6 .6 2. Prove D XYZ <; DWVZ.

Solution XZ ZV YZ ZW XZ ` ZV +XZY

3 15 = 7 35 3 6 = = 7 14 YZ = ZW = +WZV =

(vertically opposite angles)

` since two pairs of sides are in proportion and their included angles are equal the triangles are similar

Ratio of intercepts The following result comes from similar triangles.

When two (or more) transversals cut a series of parallel lines, the ratios of their intercepts are equal. That is, AB : BC = DE : EF AB DE or = EF BC


Proof Draw DG and EH parallel to AC.

Then Also `

`

DG = AB EH = BC DG AB = EH BC +GDE = +HEF +DEG = +EFH +DGE = +EHF

(opposite sides of a parallelogram) (similarly) (1) (corresponding +s, DG < EH) (corresponding +s, BE < CF) (angle sum of Ds)

So D DGE <; D EHF DG DE = ` EH EF From (1) and (2):

(2)

AB DE = EF BC

EXAMPLES 1. Find the value of x, to 3 significant figures.

Solution x 1.5 = ^ ratios of intercepts on parallel lines h 8.9 9.3 9.3x = 8.9 # 1.5 8.9 # 1.5 x= 9.3 = 1.44

CONTINUED

167

168


2. Evaluate x and y, to 1 decimal place.

Solution Use either similar triangles or ratios of intercepts to find x. You must use similar triangles to find y.

Why?

x 2.7 = 5.8 3. 4 2.7 # 5.8 x= 3.4 = 4.6 y 2.7 + 3.4 = 7.1 3.4 6.1 # 7.1 y= 3.4 = 12.7

These ratios come from intercepts on parallel lines.

These ratios come from similar triangles.

4.5 Exercises 1.

Find the value of all pronumerals, to 1 decimal place where appropriate. (a)

(c)

(d)

(e) (b)


(f) 46 c

xc

11

8.9 25.7

9.1

1.3

5c

c

1.82 E

14.3

19

c

4.

52c

A

4.2

B

4.9 5.88

yc

6.86

F

The diagram shows two concentric circles with centre O. (a) Prove that DOAB <; D OCD. (b) If radius OC = 5.9 cm and radius OB = 8.3 cm, and the length of CD = 3.7 cm, find the length of AB, correct to 2 decimal places.

7.

(a) Prove that D ABC <; D ADE. (b) Find the values of x and y, correct to 2 decimal places.

8.

ABCD is a parallelogram, with CD produced to E. Prove that D ABF <; DCEB.

Evaluate a and b to 2 decimal places.

Show that D ABC and DCDE are similar.

EF bisects +GFD. Show that D DEF and D FGE are similar.

C

6.

(g)

3.

D 87c

y

46

a

2.

Show that D ABC and D DEF are similar. Hence find the value of y.

5.

169

170


9.

Show that D AED <; D ABC. Find the value of m.

10. Prove that D ABC and D ACD are similar. Hence evaluate x and y.

(e)

12. Show that AB AF (a) = BC FG AB AF (b) = AC AG BD DF (c) = CE EG

11. Find the values of all pronumerals, to 1 decimal place. (a) 13. Evaluate a and b correct to 1 decimal place.

(b)

14. Find the value of y to 2 significant figures. (c)

(d) 15. Evaluate x and y correct to 2 decimal places.


Pythagoras’ Theorem DID YOU KNOW? The triangle with sides in the proportion 3:4:5 was known to be right angled as far back as ancient Egyptian times. Egyptian surveyors used to measure right angles by stretching out a rope with knots tied in it at regular intervals. They used the rope for forming right angles while building and dividing fields into rectangular plots. It was Pythagoras (572–495 BC) who actually discovered the relationship between the sides of the right-angled triangle. He was able to generalise the rule to all right-angled triangles. Pythagoras was a Greek mathematician, philosopher and mystic. He founded the Pythagorean School, where mathematics, science and philosophy were studied. The school developed a brotherhood and performed secret rituals. He and his followers believed that the whole universe was based on numbers. Pythagoras was murdered when he was 77, and the brotherhood was disbanded.

The square on the hypotenuse in any right-angled triangle is equal to the sum of the squares on the other two sides. That is, c2 = a2 + b2 or

c=

a2 + b2

171

172


Proof

Draw CD perpendicular to AB Let AD = x, DB = y Then x + y = c In D ADC and D ABC, +A is common +ADC = +ACB = 90c `

D ADC <; D ABC (equal corresponding +s) AC AD = AB AC x b c =b b 2 = xc D BDC <; D ABC Similarly, BC DB = AB BC y a a= c a 2 = yc Now a 2 + b 2 = yc + xc = c ^y + xh = c ]c g = c2 If c 2 = a 2 + b 2, then D ABC must be right angled

EXAMPLES 1. Find the value of x, correct to 2 decimal places.

Solution c2 = a2 + b2 x2 = 72 + 42 = 49 + 16 = 65


173

x = 65 = 8.06 to 2 decimal places 2. Find the exact value of y.

Solution c2 = a2 + b2 82 = y2 + 42 64 = y 2 + 16 48 = y 2 `

Leave the answer in surd form for the exact answer.

y = 48 = 16 # 3 =4 3

3. Find the length of the diagonal in a square with sides 6 cm. Answer to 1 decimal place.

Solution

6 cm

6 cm

c =a +b = 62 + 62 = 72 2

2

2

c = 72 = 8 .5 So the length of the diagonal is 8.5 cm.

CONTINUED

174


4. A triangle has sides 5.1 cm, 6.8 cm and 8.5 cm. Prove that the triangle is right angled.

Solution

5.1 cm

8.5 cm

6.8 cm

Let c = 8.5 (largest side) and a and b the other two smaller sides. a 2 + b 2 = 5 . 1 2 + 6. 8 2 = 72.25 c 2 = 8. 5 2 = 72.25 ` c2 = a2 + b2 So the triangle is right angled.

4.6 Exercises 1.

Find the value of all pronumerals, correct to 1 decimal place. (a)

2.

Find the exact value of all pronumerals. (a)

(b) (b)

(c) (c)

(d) (d)


3.

Find the slant height s of a cone with diameter 6.8 m and perpendicular height 5.2 m, to 1 decimal place.

4.

Find the length of CE, correct to 1 decimal place, in this rectangular pyramid. AB = 8.6 cm and CF = 15.9 cm.

5.

Prove that D ABC is a right-angled triangle.

6.

7.

Show that AC = 2 BC.

8.

(a) Find the length of diagonal AC in the figure. (b) Hence, or otherwise, prove that AC is perpendicular to DC.

9.

Find the length of side AB in terms of b.

Show that D XYZ is a right-angled isosceles triangle. X

XY in YZ terms of x and y in D XYZ.

10. Find the exact ratio of 2

1

Y

1

Z

175

176


11. Show that the distance squared between A and B is given by d 2 = 13t 2 - 180t + 625.

16. A ramp is 4.5 m long and 1.3 m high. How far along the ground does the ramp go? Answer correct to one decimal place. 12. An 850 mm by 1200 mm gate is to have a diagonal timber brace to give it strength. To what length should the timber be cut, to the nearest mm?

4.5 m

1.3 m

17. The diagonal of a television screen is 72 cm. If the screen is 58 cm high, how wide is it? 18. A property has one side 1.3 km and another 1.1 km as shown with a straight road diagonally through the middle of the property. If the road is 1.5 km long, show that the property is not rectangular.

13. A rectangular park has a length of 620 m and a width of 287 m. If I walk diagonally across the park, how far do I walk? 14. The triangular garden bed below is to have a border around it. How many metres of border are needed, to 1 decimal place?

1.5 km

1.3 km

1.1 km

15. What is the longest length of stick that will fit into the box below, to 1 decimal place?

19. Jodie buys a ladder 2 m long and wants to take it home in the boot of her car. If the boot is 1.2 m by 0.7 m, will the ladder fit?


20. A chord AB in a circle with centre O and radius 6 cm has a perpendicular line OC as shown 4 cm long.

(a) By finding the lengths of AC and BC, show that OC bisects the chord. (b) By proving congruent triangles, show that OC bisects the chord.

O A

4 cm

6 cm

C B

Types of Quadrilaterals A quadrilateral is any four-sided figure

In any quadrilateral the sum of the interior angles is 360c

Proof Draw in diagonal AC +ADC + +DCA + +CAD = 180c (angle sum of D) +ABC + +BCA + +CAB = 180c (similarly) ` +ADC + +DCA + +CAD + +ABC + +BCA + +CAB = 360c That is, +ADC + +DCB + +CBA + +BAD = 360c

177

178


EXAMPLE Find the value of i.

Solution i + 120 + 56 + 90 = 360 ^ angle sum of quadrilateral h i + 266 = 360 i = 94

Parallelogram

A parallelogram is a quadrilateral with opposite sides parallel PROPERTIES

• • • •

These properties can all be proven.

opposite sides of a parallelogram are equal opposite angles of a parallelogram are equal diagonals in a parallelogram bisect each other each diagonal bisects the parallelogram into two congruent triangles

TESTS A quadrilateral is a parallelogram if: • both pairs of opposite sides are equal • both pairs of opposite angles are equal • one pair of sides is both equal and parallel • the diagonals bisect each other


179

Rectangle

A rectangle is a parallelogram with one angle a right angle

If one angle is a right angle, then you can prove all angles are right angles.

PROPERTIES • the same as for a parallelogram, and also • diagonals are equal

TEST A quadrilateral is a rectangle if its diagonals are equal

Application Builders use the property of equal diagonals to check if a rectangle is accurate. For example, a timber frame may look rectangular, but may be slightly slanting. Checking the diagonals makes sure that a building does not end up like the Leaning Tower of Pisa!

Rhombus

A rhombus is a parallelogram with a pair of adjacent sides equal PROPERTIES • the same as for parallelogram, and also • diagonals bisect at right angles • diagonals bisect the angles of the rhombus

It can be proved that all sides are equal.

180


TESTS

A quadrilateral is a rhombus if: • all sides are equal • diagonals bisect each other at right angles

Square

A square is a rectangle with a pair of adjacent sides equal PROPERTIES

• the same as for rectangle, and also • diagonals are perpendicular • diagonals make angles of 45c with the sides

Trapezium

A trapezium is a quadrilateral with one pair of sides parallel

Kite

A kite is a quadrilateral with two pairs of adjacent sides equal


EXAMPLES 1. Find the values of i, x and y, giving reasons.

Solution i = 83c x = 6.7 cm y = 2.3 cm

(opposite +s in < gram) (opposite sides in < gram) (opposite sides in < gram)

2. Find the length of AB in square ABCD as a surd in its simplest form if BD = 6 cm.

Solution Let AB = x Since ABCD is a square, AB = AD = x (adjacent sides equal) Also, +A = 90c (by definition) By Pythagoras’ theorem: c2 = a2 + b2 62 = x2 + x2 36 = 2x 2 18 = x 2 ` x = 18 = 3 2 cm

CONTINUED

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182


3. (a) Two equal circles have centres O and P respectively. Prove that OAPB is a rhombus. (b) Hence, or otherwise, show that AB is the perpendicular bisector of OP.

Solution OA = OB (equal radii) PA = PB (similarly) Since the circles are equal, OA = OB = PA = PB ` since all sides are equal, OAPB is a rhombus (b) The diagonals in any rhombus are perpendicular bisectors. Since OAPB is a rhombus, with diagonals AB and OP, AB is the perpendicular bisector of OP.

(a)

4.7 Exercises 1.

Find the value of all pronumerals, giving reasons. (a)

(e)

(f) (b)

(g) (c)

(d)


2.

Given AB = AE, prove CD is perpendicular to AD.

(c)

(d)

3.

(a) Show that +C = xc and +B = +D = (180 - x)c. (b) Hence show that the sum of angles of ABCD is 360c.

(e)

(f) 4.

5.

Find the value of a and b.

7

3x

x+

6

y

6.

In the figure, BD bisects +ADC. Prove BD also bisects +ABC.

7.

(a)

Prove that each figure is a parallelogram. (a)

(b)

(b)

Find the values of all pronumerals, giving reasons.

183

184


(c)

(d)

(e) (d)

8.

Evaluate all pronumerals. (a)

9.

The diagonals of a rhombus are 8 cm and 10 cm long. Find the length of the sides of the rhombus.

10. ABCD is a rectangle with +EBC = 59c . Find +ECB, +EDC and +ADE. (b)

(c)

11. The diagonals of a square are 8 cm long. Find the exact length of the side of the square. 12. In the rhombus, +ECB = 33c. Find the value of x and y.

ABCD is a kite

Polygons A polygon is a closed plane figure with straight sides

A regular polygon has all sides and all interior angles equal


EXAMPLES

3-sided (equilateral triangle)

4-sided (square)

5-sided (pentagon)

6-sided (hexagon)

8-sided (octagon)

10-sided (decagon)

DID YOU KNOW? Carl Gauss (1777–1855) was a famous German mathematician, physicist and astronomer. When he was 19 years old, he showed that a 17-sided polygon could be constructed using a ruler and compasses. This was a major achievement in geometry. Gauss made a huge contribution to the study of mathematics and science, including correctly calculating where the magnetic south pole is and designing a lens to correct astigmatism. He was the director of the Göttingen Observatory for 40 years. It is said that he did not become a professor of mathematics because he did not like teaching.

The sum of the interior angles of an n-sided polygon is given by S = 180n - 360 or S = (n - 2) # 180c

Proof Draw any n-sided polygon and divide it into n triangles as shown. Then the total sum of angles is n # 180c or 180n. But this sum includes all the angles at O. So the sum of interior angles is 180n - 360c . That is, S = 180n - 360 = ] n - 2 g #180c

The sum of the exterior angles of any polygon is 360c

Proof Draw any n-sided polygon. Then the sum of both the exterior and interior angles is n #180c. Sum of exterior angles = n #180c - sum of interior angles = 180n - ] 180n - 360c g = 180n - 180n + 360c = 360c

185

186


EXAMPLES 1. Find the sum of the interior angles of a regular polygon with 15 sides. How large is each angle?

Solution n = 15 S = (n - 2)#180c = (15 - 2)#180c = 13#180c = 2340c Each angle has size 2340c' 15 = 156c. 2. Find the number of sides in a regular polygon whose interior angles are 140c.

Solution Let n be the number of sides Then the sum of interior angles is 140n But S = (n - 2)#180c So 140n = (n - 2)#180c = 180n - 360 360 = 40n 9=n So the polygon has 9 sides.

There are n sides and so n angles, each 140c.

4.8 Exercises 1.

2.

Find the sum of the interior angles of (a) a pentagon (b) a hexagon (c) an octagon (d) a decagon (e) a 12-sided polygon (f) an 18-sided polygon Find the size of each interior angle of a regular (a) pentagon (b) octagon (c) 12-sided polygon (d) 20-sided polygon (e) 15-sided polygon

3.

Find the size of each exterior angle of a regular (a) hexagon (b) decagon (c) octagon (d) 15-sided polygon

4.

Calculate the size of each interior angle in a regular 7-sided polygon, to the nearest minute.

5.

The sum of the interior angles of a regular polygon is 1980c. (a) How many sides has the polygon? (b) Find the size of each interior angle, to the nearest minute.


6.

Find the number of sides of a regular polygon whose interior angles are 157c 30l.

7.

Find the sum of the interior angles of a regular polygon whose exterior angles are 18c.

8.

9.

A regular polygon has interior angles of 156c. Find the sum of its interior angles.

13. A regular octagon has a quadrilateral ACEG inscribed as shown. B

D

H

Find the size of each interior angle in a regular polygon if the sum of the interior angles is 5220c.

10. Show that there is no regular polygon with interior angles of 145c. 11. Find the number of sides of a regular polygon with exterior angles (a) 40c (b) 30c (c) 45c (d) 36c (e) 12c

C

A

G

E

F

Show that ACEG is a square. 14. In the regular pentagon below, show that EAC is an isosceles triangle. A

E

B

12. ABCDEF is a regular hexagon. A

B D

F

C

E

D

(a) Show that triangles AFE and BCD are congruent. (b) Show that AE and BD are parallel.

C

15. (a) Find the size of each exterior angle in a regular polygon with side p. (b) Hence show that each interior 180 (p - 2) . angle is p

187

188


Areas Most areas of plane figures come from the area of a rectangle.

Rectangle

A = lb

Square

A square is a special rectangle.

A = x2

Triangle

The area of a triangle is half the area of a rectangle.

A=

1 bh 2

Proof

h

b

Draw rectangle ABCD, where b = length and h = breadth.


189

area = bh

`

1 1 area AEFD and area DCEF = area EBCF 2 2 1 area DCDE = area ABCD 2 1 That is, A = bh 2

Area D DEF = `

Parallelogram

A = bh

Proof In parallelogram ABCD, produce DC to E and draw BE perpendicular to CE. Then ABEF is a rectangle.

Area ABEF = bh In D ADF and D BCE, +AFD = +BEC = 90c AF = BE = h (opposite sides of a rectangle) AD = BC (opposite sides of a parallelogram) ` by RHS, D ADF / D BCE ` area D ADF = area D BCE So area ABCD = area ABEF = bh

Rhombus

1 xy 2 (x and y are lengths of diagonals) A=

The area of a parallelogram is the same as the area of two triangles.

190


Proof

Let AC = x and BD = y By properties of a rhombus, 1 1 AE = EC = x and DE = EB = y 2 2 Also +AEB = 90c Area D ABC = = Area D ADC = = ` total area of rhombus = =

1 1 x: y 2 2 1 xy 4 1 1 x: y 2 2 1 xy 4 1 1 xy + xy 4 4 1 xy 2

Trapezium

A=

Proof

Let DE = x Then DF = x + a ` FC = b - ] x + a g =b-x-a

1 h ( a + b) 2


Area trapezium = area D ADE + area rectangle ABFE + area D BFC 1 1 = xh + ah + (b - x - a) h 2 2 1 = h (x + 2a + b - x - a) 2 1 = h (a + b) 2

Circle

You will study the circle in more detail in Chapter 9.

A = rr 2

EXAMPLES 1. Find the area of this trapezium.

Solution 1 h ( a + b) 2 1 = ( 4) ( 7 + 5) 2 = 2 # 12

A=

= 24 m 2

4.2 cm

8.9 cm

3.7 cm

2. Find the area of the shaded region in this figure.

12.1 cm CONTINUED

191


Solution Area large rectangle = lb = 8.9 #12.1 = 107.69 cm 2 Area small rectangle = lb = 3 . 7 # 4 .2 `

= 15.54 cm 2 shaded area = 107.69 - 15.54 = 92.15 cm 2

3. A park with straight sides of length 126 m and width 54 m has semicircular ends as shown. Find its area, correct to 2 decimal places. 126 m

54 m

192

Solution Area of 2 semi-circles = area of 1 circle 54 r = 2 = 27 A = rr 2 = r (27) 2 = 2290.22 m2 Area rectangle = 126 # 54 = 6804 Total area = 2290.22 + 6804 = 9094.22 m2

4.9 Exercises 1.

Find the area of each figure. (a)

(b)


(c)

(b)

(c)

(d)

(e) (d)

(e)

(f)

6

cm

2 cm

(g) 4.

2.

Find the area of a rhombus with diagonals 2.3 m and 4.2 m.

3.

Find each shaded area. (a)

Find the area of each figure. (a)

(b)

193

194


(c)

(d)

8.

The dimensions of a battleaxe block of land are shown below. (a) Find its area. (b) A house in the district where this land is can only take up 55% of the land. How large (to the nearest m 2) can the area of the house be? (c) If the house is to be a rectangular shape with width 8.5 m, what will its length be?

9.

A rhombus has one diagonal 25 cm long and its area is 600 cm 2 . Find the length of (a) its other diagonal and (b) its side, to the nearest cm.

(e)

5.

Find the exact area of the figure.

6.

Find the area of this figure, correct to 4 significant figures. The arch is a semicircle.

7.

Jenny buys tiles for the floor of her bathroom (shown top next column) at $45.50 per m 2 . How much do they cost altogether?

10. The width w of a rectangle is a quarter the size of its length. If the width is increased by 3 units while the length remains constant, find the amount of increase in its area in terms of w.


195

Test Yourself 4 1.

Find the values of all pronumerals (a)

2.

Prove that AB and CD are parallel lines.

3.

Find the area of the figure, to 2 decimal places.

4.

(a) Prove that triangles ABC and ADE are similar. (b) Evaluate x and y to 1 decimal place.

5.

Find the size of each interior angle in a regular 20-sided polygon.

6.

Find the volume of a cylinder with radius 5.7 cm and height 10 cm, correct to 1 decimal place.

7.

Find the perimeter of the triangle below.

(b)

(c)

(d)

x

(e) (O is the centre of the circle.)

(f)

(g)

The perimeter is the distance around the outside of the figure.

196


8.

(a) Prove triangles ABC and ADC are congruent in the kite below.

(b) Prove triangle AOB and COD are congruent. (O is the centre of the circle.)

12. Triangle ABC is isosceles, and AD bisects BC.

(a) Prove triangles ABD and ACD are congruent. (b) Prove AD and BC are perpendicular. 13. Triangle ABC is isosceles, with AB = AC. Show that triangle ACD is isosceles.

9.

Find the area of the figure below.

14. Prove that opposite sides in any parallelogram are equal.

10. Prove triangle ABC is right angled.

15. A rhombus has diagonals 6 cm and 8 cm. (a) Find the area of the rhombus. (b) Find the length of its side. 16. The interior angles in a regular polygon are 140c . How many sides has the polygon? 17. Prove AB and CD are parallel.

11. Prove

AF AB . = AG AC


18. Find the area of the figure below. 6 cm 5 cm 8 cm

20. (a) Prove triangles ABC and DEF are similar. (b) Evaluate x to 1 decimal place.

2 cm

10 cm

19. Prove that z = x + y in the triangle below.


Find the value of x.

4.

Given +BAD =+DBC, show that D ABD and D BCD are similar and hence find d.

2.

Evaluate x, y and z.

5.

Prove that ABCD is a parallelogram. AB = DC.

3.

Find the sum of the interior angles of a regular 11-sided polygon. How large is each exterior angle?

6.

Find the shaded area.

197

198


7.

Prove that the diagonals in a square make angles of 45c with the sides.

8.

Prove that the diagonals in a kite are perpendicular.

9.

Prove that MN is parallel to XY.

12. Find the values of x and y correct to 1 decimal place.

13. Find the values of x and y, correct to 2 decimal places.

10. Evaluate x.

11. The letter Z is painted on a billboard.

(a) Find the area of the letter. (b) Find the exact perimeter of the letter.

14. ABCD is a square and BD is produced to 1 E such that DE = BD. 2 (a) Show that ABCE is a kite. 2x (b) Prove that DE = units when 2 sides of the square are x units long.

5 Functions and Graphs TERMINOLOGY Arc of a curve: Part or a section of a curve between two points

Even function: An even function has line symmetry (reflection) about the y-axis, and f ] - x g = - f ] x g

Asymptote: A line towards which a curve approaches but never touches

Function: For each value of the independent variable x, there is exactly one value of y, the dependent variable. A vertical line test can be used to determine if a relationship is a function

Cartesian coordinates: Named after Descartes. A system of locating points (x, y) on a number plane. Point (x, y) has Cartesian coordinates x and y Curve: Another word for arc. When a function consists of all values of x on an interval, the graph of y = f ] x g is called a curve y = f ] x g Dependent variable: A variable is a symbol that can represent any value in a set of values. A dependent variable is a variable whose value depends on the value chosen for the independent variable Direct relationship: Occurs when one variable varies directly with another i.e. as one variable increases, so does the other or as one variable decreases so does the other Discrete: Separate values of a variable rather than a continuum. The values are distinct and unrelated Domain: The set of possible values of x in a given domain for which a function is defined

Independent variable: A variable is independent if it may be chosen freely within the domain of the function Odd function: An odd function has rotational symmetry about the origin (0, 0) and where f ] - x g = - f ] x g Ordered pair: A pair of variables, one independent and one dependent, that together make up a single point in the number plane, usually written in the form (x, y) Ordinates: The vertical or y coordinates of a point are called ordinates Range: The set of real numbers that the dependent variable y can take over the domain (sometimes called the image of the function) Vertical line test: A vertical line will only cut the graph of a function in at most one point. If the vertical line cuts the graph in more than one point, it is not a function

Chapter 5 Functions and Graphs

INTRODUCTION FUNCTIONS AND THEIR GRAPHS are used in many areas, such as mathematics, science and economics. In this chapter you will study functions, function notation and how to sketch graphs. Some of these graphs will be studied in more detail in later chapters.

DID YOU KNOW? The number plane is called the Cartesian plane after Rene Descartes (1596–1650). He was known as one of the first modern mathematicians along with Pierre de Fermat (1601–1665). Descartes used the number plane to develop analytical geometry. He discovered that any equation with two unknown variables can be represented by a line. The points in the number plane can be called Cartesian coordinates. Descartes used letters at the beginning of the alphabet to stand for numbers that are known, and letters near the end of the alphabet for unknown numbers. This is why we still use x and y so often! Do a search on Descartes to find out more details of his life and work. Descartes

Functions Definition of a function Many examples of functions exist both in mathematics and in real life. These occur when we compare two different quantities. These quantities are called variables since they vary or take on different values according to some pattern. We put these two variables into a grouping called an ordered pair.

205

206


EXAMPLES 1. Eye colour

Name

Anne

Colour Blue

Jacquie Donna Hien

Marco

Russell

Trang

Brown

Green

Brown

Brown

Grey

Brown

Ordered pairs are (Anne, Blue), (Jacquie, Brown), (Donna, Grey), (Hien, Brown), (Marco, Green), (Russell, Brown) and (Trang, Brown). 2. y = x + 1 x

1

2

3

4

y

2

3

4

5

The ordered pairs are (1, 2), (2, 3), (3, 4) and (4, 5). 3. A

1

B

2

C 3 D 4

E

The ordered pairs are (A, 1), (B, 1), (C, 4), (D, 3) and (E, 2).

Notice that in all the examples, there was only one ordered pair for each variable. For example, it would not make sense for Anne to have both blue and brown eyes! (Although in rare cases some people have one eye that’s a different colour from the other.) A relation is a set of ordered points (x, y) where the variables x and y are related according to some rule. A function is a special type of relation. It is like a machine where for every INPUT there is only one OUTPUT. INPUT

PROCESS

OUTPUT

The first variable (INPUT) is called the independent variable and the second (OUTPUT) the dependent variable. The process is a rule or pattern.


For example, in y = x + 1, we can use any number for x (the independent variable), say x = 3. When x = 3 y=3+1 =4 As this value of y depends on the number we choose for x, y is called the dependent variable. A function is a relationship between two variables where for every independent variable, there is only one dependent variable. This means that for every x value, there is only one y value.

Investigation When we graph functions in mathematics, the independent variable (usually the x-value) is on the horizontal axis while the dependent variable (usually the y-value) is on the vertical axis. In other areas, the dependent variable goes on the horizontal axis. Find out in which subjects this happens at school by surveying teachers or students in different subjects. Research different types of graphs on the Internet to find some examples.

Here is an example of a relationship that is NOT a function. Can you see the difference between this example and the previous ones? A B

1 2

C 3 D E

4

In this example the ordered pairs are (A, 1), (A, 2), (B, 1), (C, 4), (D, 3) and (E, 2). Notice that A has two dependent variables, 1 and 2. This means that it is NOT a function.

While we often call the independent variable x and the dependent variable y, there are other pronumerals we could use. You will meet some of these in this course.

207

208


Here are two examples of graphs on a number plane. 1.

y

x

2.

y

x

There is a very simple test to see if these graphs are functions. Notice that in the first example, there are two values of y when x = 0. The y-axis passes through both these points. y

x


There are also other x values that give two y values around the curve. If we drew a vertical line anywhere along the curve, it would cross the curve in two places everywhere except one point. Can you see where this is? In the second graph, a vertical line would only ever cross the curve in one place. So when a vertical line cuts a graph in more than one place, it shows that it is not a function.

If a vertical line cuts a graph only once anywhere along the graph, the graph is a function. y

x

If a vertical line cuts a graph in more than one place anywhere along the graph, the graph is not a function. y

x

209

210


EXAMPLES 1. Is this graph a function?

Solution

You will learn how to sketch these graphs later in this chapter.

A vertical line only cuts the graph once. So the graph is a function. 2. Is this circle a function?

Solution

A vertical line can cut the curve in more than one place. So the circle is not a function.


3. Does this set of ordered pairs represent a function? ^ - 2 , 3 h , ^ - 1, 4 h , ^ 0 , 5 h , ^ 1 , 3 h , ^ 2 , 4 h

Solution For each x value there is only one y value, so this set of ordered pairs is a function. 4. Is this a function? y

3

x

Solution y

3

x

Although it looks like this is not a function, the open circle at x = 3 on the top line means that x = 3 is not included, while the closed circle on the bottom line means that x = 3 is included on this line. So a vertical line only touches the graph once at x = 3. The graph is a function.

211

212


5.1 Exercises Which of these curves are functions? 1.

6.

2.

7.

3.

8.

4.

9.

^ 1, 3 h, ^ 2, -1 h, ^ 3, 3 h, ^ 4, 0 h

10. ^ 1, 3 h, ^ 2, -1 h, ^ 2, 7 h, ^ 4, 0 h 11.

5. 12.

13.

1

1

2

2

3

3

4

4

5

5

1

1

2

2

3

3

4

4

5

5

1

1

2

2

3

3

4

4

5

5


14. Name Ben Paul Pierre Hamish Jacob Lee Pierre Lien Sport Tennis Football Tennis Football Football Badminton Football Badminton 15. A

3

B

4

C

7

D

3

E

5

F

7

G

4

Function notation If y depends on what value we give x in a function, then we can say that y is a function of x. We can write this as y = f ] x g.

EXAMPLES 1. Find the value of y when x = 3 in the equation y = x + 1.

Solution When x = 3: y = x +1 = 3+1 =4 2. If f ] x g = x + 1, evaluate f (3).

Solution f ]x g = x + 1 f ]3 g = 3 + 1 =4

Notice that these two examples are asking for the same value and f (3) is the value of the function when x = 3.

If y = f ] x g then f (a) is the value of y at the point on the function where x = a

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214


EXAMPLES 1. If f ] x g = x 2 + 3x + 1, find f ] - 2 g .

Solution This is the same as finding y when x = - 2.

f ( - 2) = ] - 2 g 2 + 3 (- 2) + 1 =4-6+1 = -1 2. If f ] x g = x 3 - x 2, find the value of f ] - 1 g .

Solution f (x) = x 3 - x 2 f (- 1) = ] - 1 g 3 - ] - 1 g 2 = -1 - 1 = -2 3. Find the values of x for which f ] x g = 0, given that f ] x g = x 2 + 3x - 10.

Solution f (x) = 0 Putting f (x) = 0 is different from finding f (0) . Follow this example carefully.

i.e.

2

x + 3x - 10 = 0 ( x + 5 ) ( x - 2) = 0 x + 5 = 0, x-2=0 x = -5 x=2

4. Find f ] 3 g, f ] 2 g, f ] 0 g and f ] - 4 g if f ] x g is defined as 3x + 4 when x $ 2 f ]x g = ) - 2x when x 1 2. Use f (x) = 3x + 4 when x is 2 or more, and use f (x) = - 2x when x is less than 2.

Solution f (3 ) = 3 ( 3) + 4 = 13 f (2 ) = 3 ( 2) + 4 = 10 f (0) = - 2 (0) =0 f (- 4) = - 2 ( - 4) =8 5. Find the value of x2 g ] x g = * 2x - 1 5

since

3$2

since

2$2

since

012

since -4 1 2

g ] 1 g + g ] - 2 g - g ] 3 g if when x 2 2 when - 1 # x # 2 when x 1 - 1


215

Solution g (1 ) = 2 ( 1) - 1 =1 g (- 2) = 5

since -1 # 1 # 2 since - 2 1 - 1

2

g (3) = 3 since 3 2 2 =9 So g (1) + g (- 2) - g (3) = 1 + 5 - 9 = -3

DID YOU KNOW? Leonhard Euler (1707–83), from Switzerland, studied functions and invented the term f (x) for function notation. He studied theology, astronomy, medicine, physics and oriental languages as well as mathematics, and wrote more than 500 books and articles on mathematics. He found time between books to marry and have 13 children, and even when he went blind he kept on having books published.

5.2 Exercises 1.

Given f ] x g = x + 3, find f ] 1 g and f ]-3 g.

10. If f ] x g = 2x - 9, find f ^ p h and f ]x + h g.

2.

If h ] x g = x 2 - 2, find h ] 0 g, h ] 2 g and h ] - 4 g .

11. Find g ] x - 1 g when g ] x g = x 2 + 2x + 3.

3.

If f ] x g = - x 2, find f ] 5 g, f ] - 1 g, f ] 3 g and f ] - 2 g .

12. If f ] x g = x 3 - 1, find f ] k g as a product of factors.

4.

Find the value of f ] 0 g + f ] - 2 g if f ] x g = x 4 - x 2 + 1.

5.

Find f ] - 3 g if f ] x g = 2x 3 - 5x + 4.

13. Given f ] t g = t 2 + 2t + 1, find t when f ] t g = 0. Also find any values of t for which f ] t g = 9.

6.

If f ] x g = 2x - 5, find x when f ] x g = 13.

7.

Given f ] x g = x + 3, find any values of x for which f ] x g = 28.

15. f ] x g = )

8.

If f ] x g = 3 x, find x when 1 f ]x g = . 27

9.

Find values of z for which f ] z g = 5 given f ] z g = 2z + 3 .

Z 2x - 4 if x $ 1 ] 16. f ] x g = [x + 3 if -1 1 x 1 1 ] 2 x if x # -1 \ Find the values of

2

14. Given f ] t g = t 4 + t 2 - 5, find the value of f ] b g - f ] - b g . x3 for x 2 1 x for x # 1 Find f ] 5 g, f ] 1 g and ] - 1 g .

f ] 2 g - f ] - 2 g + f ] -1 g .

We can use pronumerals other than f for functions.

216


17. Find g ] 3 g + g ] 0 g + g ] - 2 g if when x $ 0 x+1 g ]x g = ) - 2x + 1 when x 1 0 18. Find the value of f ] 3 g - f ] 2 g + 2f ] - 3 g when x for x 2 2 f ] x g = * x2 4

for -2 # x # 2 for x 1 -2

19. Find the value of f ] - 1 g - f ] 3 g if f (x) = *

x3 - 1 2x 2 + 3x - 1

for x $ 2 for x 1 2

x 2 - 2x - 3 x-3 (a) evaluate f (2) (b) explain why the function does not exist for x = 3 (c) by taking several x values close to 3, find the value of y that the function is moving towards as x moves towards 3.

20. If f ] x g =

21. If f ] x g = x 2 – 5x + 4, find f ] x + h g - f ] x g in its simplest form. f ]x + h g - f ]x g where 22. Simplify h 2 ] g f x = 2x + x 23. If f ] x g = 5x - 4, find f ] x g - f ] c g in its simplest form. 24. Find the value of f ^ k 2 h if 3x + 5 for x $ 0 f ]x g = * 2 x for x 1 0 Z 3 25. If when x $ 3 ]x f ] x g = [5 when 0 1 x 1 3 ] 2 x - x + 2 when x # 0 \ evaluate (a) f (0) (b) f ] 2 g - f ] 1 g (c) f ^ - n 2 h

Graphing Techniques You may have previously learned how to draw graphs by completing a table of values and then plotting points. In this course, you will learn some other techniques that will allow you to sketch graphs by showing their important features.

Intercepts One of the most useful techniques is to find the x- and y-intercepts.

Everywhere on the x-axis, y = 0 and everywhere on the y-axis x = 0 .

For x-intercept, y = 0 For y-intercept, x = 0


217

EXAMPLE Find the x- and y-intercepts of the function f ] x g = x 2 + 7x - 8.

Solution

This is the same as y = x 2 + 7x - 8.

For x-intercept: y = 0 0 = x 2 + 7x - 8 = ]x + 8g]x - 1g x + 8 = 0, x- 1=0 x = - 8, x=1 For y-intercept: x = 0

You will use the intercepts to draw graphs in the next section in this chapter.

y = ] 0 g2 + 7 ] 0 g - 8 = -8

Domain and range You have already seen that the x-coordinate is called the independent variable and the y-coordinate is the dependent variable. The set of all real numbers x for which a function is defined is called the domain. The set of real values for y or f (x) as x varies is called the range (or image) of f.

EXAMPLE Find the domain and range of f ] x g = x 2 .

Solution You can see the domain and range from the graph, which is the parabola y = x 2 . y

x

CONTINUED

218


Notice that the parabola curves outwards gradually, and will take on any real value for x. However, it is always on or above the x-axis. Domain: {all real x} Range: {y: y $ 0} You can also find the domain and range from the equation y = x 2. Notice that you can substitute any value for x and you will find a value of y. However, all the y-values are positive or zero since squaring any number will give a positive answer (except zero).

Odd and even functions When you draw a graph, it can help to know some of its properties, for example, whether it is increasing or decreasing on an interval or arc of the curve (part of the curve lying between two points). If a curve is increasing, as x increases, so does y, and the curve is moving upwards, looking from left to right.

If a curve is decreasing, then as x increases, y decreases and the curve moves downwards from left to right.


219

EXAMPLES 1. State the domain over which each curve is increasing and decreasing. y

x2

x1

x3

x

Solution The left-hand side of the parabola is decreasing and the right side is increasing. So the curve is increasing for x 2 x2 and the curve is decreasing when x 1 x2.

The curve isn’t increasing or decreasing at x2. We say that it is stationary at that point. You will study stationary points and further curve sketching in the HSC Course.

2. y

x1

x2

x3

x

Solution The left-hand side of the curve is increasing until it reaches the y-axis (where x = 0). It then turns around and decreases until x3 and then increases again. So the curve is increasing for x 1 0, x 2 x 3 and the curve is decreasing for 0 1 x 1 x 3 .

Notice that the curve is stationary at x = 0 and x = x 3 .

220


As well as looking at where the curve is increasing and decreasing, we can see if the curve is symmetrical in some way. You have already seen that the parabola is symmetrical in earlier stages of mathematics and you have learned how to find the axis of symmetry. Other types of graphs can also be symmetrical. Functions are even if they are symmetrical about the y-axis. They have line symmetry (reflection) about the y-axis. This is an even function: y

x

For even functions, f ] x g = f ] - x g for all values of x. Functions are odd if they have point symmetry about the origin. A graph rotated 180° about the origin gives the original graph. This is an odd function: y

x

For odd functions, f ] - x g = - f ] x g for all values of x in the domain.


221

EXAMPLES 1. Show that f ] x g = x 2 + 3 is an even function.

Solution f ] - x g = ] - x g2 + 3 = x2 + 3 = f ]x g ` f ] x g = x 2 + 3 is an even function 2. Show that f ] x g = x 3 - x is an odd function.

Solution f ] - x g = ] - x g3 - ] - x g = -x3 + x = - ^ x3 - x h = -f ]x g ` f ] x g = x 3 - x is an odd function

Investigation Explore the family of graphs of f ] x g = x n. For what values of n is the function even? For what values of n is the function odd? Which families of functions are still even or odd given k? Let k take on different values, both positive and negative. 1. f ] x g = kx n 2. f ] x g = x n + k 3. f ] x g = ] x + k gn

5.3 Exercises 1.

Find the x- and y-intercept of each function. (a) y = 3x - 2 (b) 2x - 5y + 20 = 0 (c) x + 3y - 12 = 0

(d) (e) (f) (g) (h)

f ] x g = x 2 + 3x f ] x g = x2 - 4 p ] x g = x 2 + 5x + 6 y = x 2 - 8x + 15 p ] x g = x3 + 5

k is called a parameter. Some graphics calculators and computer programs use parameters to show how changing values of k change the shape of graphs.

222


x+3 ]x ! 0 g x 2 (j) g ] x g = 9 - x (i) y =

2.

Show that f ] x g = f ] - x g where f ] x g = x 2 - 2. What type of function is it?

3.

If f ] x g = x 3 + 1, find (a) f ^ x 2 h (b) 6 f (x) @ 2 (c) f ] - x g (d) Is it an even or odd function?

4.

Show that g ] x g = x 8 + 3x 4 - 2x 2 is an even function.

5.

Show that f (x) is odd, where f ] x g = x.

6.

Show that f ] x g = x 2 - 1 is an even function.

7.

Show that f ] x g = 4x - x 3 is an odd function.

8.

Prove that f ] x g = x 4 + x 2 is an even function and hence find f ]x g - f ]-x g.

9.

Are these functions even, odd or neither? x3 (a) y = 4 x - x2 1 (b) y = 3 x -1 3 (c) f ] x g = 2 x -4 x-3 (d) y = x+3 x3 (e) f ] x g = 5 x - x2

10. If n is a positive integer, for what values of n is the function f ] x g = xn (a) even? (b) odd? 11. Can the function f ] x g = x n + x ever be (a) even? (b) odd?

12. For the functions below, state (i) the domain over which the graph is increasing (ii) the domain over which the graph is decreasing (iii) whether the graph is odd, even or neither. y (a)

x

(b)

y

4

x

y

(c)

-2

2

x


(d)

(e)

y

y

4 2

-2

-1

1

2

x

x

-2 -4

Investigation Use a graphics calculator or a computer with graphing software to sketch graphs and explore what effect different constants have on each type of graph. If your calculator or computer does not have the ability to use parameters (this may be called dynamic graphing), simply draw different graphs by choosing several values for k. Make sure you include positive and negative numbers and fractions for k. Alternatively, you may sketch these by hand. 1. Sketch the families of graphs for these graphs with parameter k. (a) y = kx (b) y = kx 2 (c) y = kx 3 (d) y = kx 4 k (e) y = x What effect does the parameter k have on these graphs? Could you give a general comment about y = k f ] x g? 2. Sketch the families of graphs for these graphs with parameter k. (a) y = ] x + k g 2 (b) y = x 2 + k (c) y = x 3 + k (d) y = x 4 + k 1 (e) y = x + k What effect does the parameter k have on these graphs? Could you give a general comment about y = f ] x g + k? CONTINUED

223

224


3. Sketch the families of graphs for these graphs with parameter k. (a) y = x + k (b) y = ] x + k g2 (c) y = ] x + k g3 (d) y = ] x + k g4 1 (e) y = x+k What effect does the parameter k have on these graphs? Could you give a general comment about y = f ] x + k g?

When k 2 0 , the graph moves to the left and when k 1 0 , the graph moves to the right.

For the family of functions y = k f ] x g, as k varies, the function changes its slope or steepness. For the family of functions y = f ] x g + k, as k varies, the graph moves up or down (vertical translation). For the family of functions y = f ] x + k g, as k varies, the graph moves left or right (horizontal translation). Notice that the shape of most graphs is generally the same regardless of the parameter k. For example, the parabola still has the same shape even though it may be narrower or wider or upside down. This means that if you know the shape of a graph by looking at its equation, you can sketch it easily by using some of the graphing techniques in this chapter rather than a time-consuming table of values. It also helps you to understand graphs more and makes it easier to find the domain and range. You have already sketched some of these graphs in previous years.

Linear Function A linear function is a function whose graph is a straight line. Gradient form: y = mx + b has gradient m and y-intercept b General form: ax + by + c = 0

Investigation Are straight line graphs always functions? Can you find an example of a straight line that is not a function? Are there any odd or even straight lines? What are their equations?


225

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a linear function, or choose different values of k (both positive and negative). Sketch the families of graphs for these graphs with parameter k 1. y = kx 2. y = x + k 3. y = mx + b where m and b are both parameters What effect do the parameters m and b have on these graphs?

EXAMPLE Sketch the function f ] x g = 3x - 5 and state its domain and range.

Solution This is a linear function. It could be written as y = 3x - 5. Find the intercepts y For x-intercept: y = 0 6 0 = 3x - 5 5 5 = 3x 4

2 =x 3 For y-intercept: x = 0 1

y = 3 ]0 g - 5 = -5

3 2 1 -4 -3 -2 -1

-1

1 23 1

2

3

4

x

-2 -3 -4 -5

Notice that the line extends over the whole of the number plane, so that it covers all real numbers for both the domain and range. Domain: {all real x} Range: {all real y}

The linear function ax + by + c = 0 has domain {all real x} and range {all real y} where a and b are non-zero

Special lines Horizontal and vertical lines have special equations.

Notice too, that you can substitute any real number into the equation of the function for x, and any real number is possible for y.

226


EXAMPLES 1. Sketch y = 2 on a number plane. What is its domain and range?

Solution x can be any value and y is always 2. Some of the points on the line will be (0, 2), (1, 2) and (2, 2). This gives a horizontal line with y-intercept 2. y

5 4 3 2 1 -4

-3

-2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

Domain: " all real x , Range: " y: y = 2 , 2. Sketch x = - 1 on a number plane and state its domain and range.

Solution y can be any value and x is always - 1. Some of the points on the line will be ^ - 1, 0 h, ^ - 1, 1 h and ^ - 1, 2 h . This gives a vertical line with x-intercept - 1. y 5 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 -5

Domain: " x: x = - 1 , Range: " all real y ,

1

3

4

x


x = a is a vertical line with x-intercept a Domain: ! x: x = a + Range: {all real y} y = b is a horizontal line with y-intercept b Domain: {all real x} Range: " y: y = b ,

5.4 Exercises 1.

Find the x- and y-intercepts of each function. (a) y = x - 2 (b) f ] x g = 2x + 3 (c) 2x + y - 1 = 0 (d) x - y + 3 = 0 (e) 3x - 6y - 2 = 0

2.

Draw the graph of each straight line. (a) x = 4 (b) x - 3 = 0 (c) y = 5 (d) y + 1 = 0 (e) f ] x g = 2x - 1 (f) y = x + 4 (g) f ] x g = 3x + 2 (h) x + y = 3 (i) x - y - 1 = 0 (j) 2x + y - 3 = 0

3.

Find the domain and range of (a) 3x - 2y + 7 = 0 (b) y = 2 (c) x = - 4 (d) x - 2 = 0 (e) 3 - y = 0

4.

Which of these linear functions are even or odd? (a) y = 2x (b) y = 3 (c) x = 4 (d) y = - x (e) y = x

5.

By sketching x - y - 4 = 0 and 2x + 3y - 3 = 0 on the same set of axes, find the point where they meet.

227

228


Quadratic Function The quadratic function gives the graph of a parabola.

f ] x g = ax 2 + bx + c is the general equation of a parabola. If a 2 0 the parabola is concave upwards

If a 1 0 the parabola is concave downwards

The pronumeral a is called the coefficient of x 2.

Applications The parabola shape is used in many different applications as it has special properties that are very useful. For example if a light is placed inside the parabola at a special place (called the focus), then all light rays coming from this light and bouncing off the parabola shape will radiate out parallel to each other, giving a strong light. This is how car headlights work. Satellite dishes also use this property of the parabola, as sound coming in to the dish will bounce back to the focus.


The lens in a camera and glasses are also parabola shaped. Some bridges look like they are shaped like a parabola, but they are often based on the catenary. Research the parabola and catenary on the Internet for further information.

Investigation Is the parabola always a function? Can you find an example of a parabola that is not a function? Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a quadratic function, or choose different values of k (both positive and negative). Sketch the families of graphs for these graphs with parameter k. 1. y = kx 2 2. y = x 2 + k 3. y = ] x + k g2 4. y = x 2 + kx What effect does the parameter k have on these graphs? Which of these families are even functions? Are there any odd quadratic functions?

229

230


EXAMPLES 1. (a) Sketch the graph of y = x 2 - 1, showing intercepts. (b) State the domain and range.

Solution (a) This is the graph of a parabola. Since a 2 0, it is concave upward For x-intercept: y = 0 0=x -1 1 = x2 !1 = x For y-intercept: x = 0 2

y = 02 - 1 = -1

y 5 4 3 2 1 -4 -3 -2 -1 -1

1

3

4

5

-2 -3 -4 -5 -6

(b) From the graph, the curve is moving outwards and will extend to all real x values. The minimum y value is - 1. Domain: " all real x , Range: " y: y $ -1 , 2. Sketch f ] x g = ] x + 1 g 2.

Solution This is a quadratic function. We find the intercepts to see where the parabola will lie. Alternatively, you may know from your work on parameters that f ] x g = ] x + a g 2 will move the function f ] x g = x 2 horizontally a units to the left. So f ] x g = ] x + 1 g 2 moves the parabola f ] x g = x 2 1 unit to the left. For x-intercept: y = 0 0 = ]x + 1 g2 x+1=0 x = -1 For y-intercept: x = 0 y = ]0 + 1 g2 =1

x


231

y 5 4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

3. For the quadratic function f ] x g = x 2 + x - 6 (a) Find the x- and y-intercepts (b) Find the minimum value of the function (c) State the domain and range (d) For what values of x is the curve decreasing?

Solution (a) For x-intercept: y = 0 This means f ] x g = 0 0 = x2 + x - 6 = ]x + 3 g]x - 2 g x + 3 = 0, x - 2 = 0 x = - 3, x = 2 For y-intercept: x = 0 f ] 0 g = ] 0 g2 + ] 0 g - 6 = -6 (b) Since a 2 0, the quadratic function has a minimum value. Since the parabola is symmetrical, this will lie halfway between the x-intercepts. Halfway between x = - 3 and x = 2: -3 + 2 1 =2 2 1 Minimum value is f c - m 2 1 1 2 1 f c- m = c- m + c- m - 6 2 2 2 1 1 = - -6 4 2 1 = -6 4

1 So the minimum value is - 6 . 4 CONTINUED

You will learn more about this in Chapter 10.

232


(c) Sketching the quadratic function gives a concave upward parabola. y 5 4 3 2 1 -4 -3 -2 -1 -1

1

3

4

5

x

-2 -3 -4 -5 -6 -1 , -6 14 2

From the graph, notice that the parabola is gradually going outwards and will include all real x values. Since the minimum value is - 6 1 , all y values are greater than this. 4 Domain: " all real x , 1 Range: ' y: y $ -6 1 4 (d) The curve decreases down to the minimum point and then 1 increases. So the curve is decreasing for all x 1 - . 2 4. (a) Find the x- and y-intercepts and the maximum value of the quadratic function f ] x g = - x 2 + 4x + 5. (b) Sketch the function and state the domain and range. (c) For what values of x is the curve increasing?

Solution (a) For x-intercept: y = 0 So f ]x g = 0 0 = - x 2 + 4x + 5 x 2 - 4x - 5 = 0 ]x - 5 g]x + 1 g = 0 x - 5 = 0, x + 1 = 0 x = 5, x = -1 For y-intercept: x = 0 f ] 0 g = - ] 0 g2 + 4 ] 0 g + 5 =5


Since a 1 0, the quadratic function is concave downwards and has a maximum value halfway between the x-intercepts x = - 1 and x = 5. -1 + 5 =2 2 f ]2 g = -]2 g 2+ 4 ]2 g + 5 =9 So the maximum value is 9. (b) Sketching the quadratic function gives a concave downward parabola. y

9 8 7 6 5 4 3 2 1 -4

-3

-2

-1 -1 -2 -3 -4 -5

1

2

3

4

5

6

x

From the graph, the function can take on all real numbers for x, but the maximum value for y is 9. Domain: " all real x , Range: " y: y # 9 , (c) From the graph, the function is increasing on the left of the maximum point and decreasing on the right. So the function is increasing when x 1 2.

5.5 Exercises 1.

Find the x- and y-intercepts of each function. (a) y = x 2 + 2x (b) y = - x 2 + 3x (c) f ] x g = x 2 - 1 (d) y = x 2 - x - 2 (e) y = x 2 - 9x + 8

2.

Sketch (a) y = x 2 + 2 (b) y = - x 2 + 1 (c) f ] x g = x 2 - 4 2 (d) y = x + 2x (e) y = - x 2 - x (f) f ] x g = ] x - 3 g 2

233

234


(g) (h) (i) (j) 3.

f ] x g = ] x + 1 g2 y = x 2 + 3x - 4 y = 2x 2 - 5 x + 3 f ] x g = - x 2 + 3x - 2

For each parabola, find (i) the x- and y-intercepts (ii) the domain and range (a) y = x 2 – 7x + 12 (b) f ] x g = x 2 + 4x (c) y = x 2 - 2x - 8 (d) y = x 2 - 6x + 9 (e) f ] t g = 4 - t 2

4.

Find the domain and range of (a) y = x 2 - 5 (b) f ] x g = x 2 - 6x (c) f ] x g = x 2 - x - 2 (d) y = - x 2 (e) f ] x g = ] x - 7 g 2

5.

Find the range of each function over the given domain. (a) y = x 2 for 0 # x # 3 (b) y = - x 2 + 4 for -1 # x # 2 (c) f ] x g = x 2 - 1 for -2 # x # 5 (d) y = x 2 + 2x - 3 for -2 # x # 4 (e) y = - x 2 - x + 2 for 0 # x # 4

6.

Find the domain over which each function is (i) increasing (ii) decreasing (a) y = x 2 (b) y = - x 2 (c) f ] x g = x 2 - 9 (d) y = - x 2 + 4x (e) f ] x g = ] x + 5 g2

7.

Show that f ] x g = - x 2 is an even function.

8.

State whether these functions are even or odd or neither. (a) y = x 2 + 1 (b) f ] x g = x 2 - 3 (c) y = -2x 2 (d) f ] x g = x 2 - 3x (e) f ] x g = x 2 + x (f) y = x 2 - 4 (g) y = x 2 - 2x - 3 (h) y = x 2 - 5x + 4 (i) p ] x g = ] x + 1 g 2 (j) y = ] x - 2 g 2

Absolute Value Function You may not have seen the graphs of absolute functions before. If you are not sure about what they look like, you can use a table of values or look at the definition of absolute value.

EXAMPLES 1. Sketch f ] x g = x - 1 and state its domain and range.

Solution Method 1: Table of values When sketching any new graph for the first time, you can use a table of values. A good selection of values is -3 # x # 3 but if these don’t give enough information, you can find other values.


e.g. When x = -3: y = | -3 | -1 =3-1 =2 x

-3

-2

-1

y

2

1

0

0

1

2

3

-1

0

1

2

This gives a v-shaped graph. y 5 4 3 2 1 -4

-3

-2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

Method 2: Use the definition of absolute value when x $ 0 x-1 y = | x | - 1 = &x - 1 when x 1 0 This gives 2 straight line graphs: y = x - 1 ]x $ 0 g y

5 y=x-1

4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

CONTINUED

235

236


y = -x - 1 ] x 1 0 g y y=-x-1

5 4 3 2 1

-4 -3 -2 -1 -1

1

2

3

x

4

-2 -3 -4 -5

Draw these on the same number plane and then disregard the dotted lines to get the graph shown in method 1. y y = -x - 1

5

y=x-1

4 3 2 1 -4 -3

-2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

Method 3: If you know the shape of the absolute value functions, find the intercepts. For x-intercept: y = 0 So f ] x g = 0 0 = | x |- 1 1 =| x | ` x = !1 For y-intercept: x = 0 f (0) = | 0 | - 1 = -1


The graph is V-shaped, passing through these intercepts. y 5 4 3 2 1 -4 -3

-2 -1 -1

1

2

3

4

5

x If you already know how to sketch the graph of y = | x | , translate the graph of y = | x | - 1 down 1 unit, giving it a y-intercept of -1.

-2 -3 -4 -5

From the graph, notice that x values can be any real number while the minimum value of y is - 1. Domain: {all real x} Range: {y: y $ -1} 2. Sketch y = | x + 2 | .

Solution Method 1: Use the definition of absolute value. +2 when x + 2 $ 0 y = | x + 2 | = 'x - (x + 2) when x + 2 1 0 This gives 2 straight lines: y = x + 2 when x + 2 $ 0 x $ -2 y 5

y=x+2

4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

CONTINUED

237

238


y = - ] x + 2 g when x + 2 1 0 i.e. y = - x - 2 when x 1 -2 y 5 4

y = -x - 2

3 2 1

-4

-3

-2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

Draw these on the same number plane and then disregard the dotted lines. y 5 y = -x - 2

y=x+2

4 3 2 1

-4 -3

-2 -1 -1 -2 -3 -4 -5

Method 2: Find intercepts For x-intercept: y = 0 So f ] x g = 0 There is only one solution for the equation | x + 2 | = 0. Can you see why?

0 =| x + 2 | 0=x+2 -2 = x For y-intercept: x = 0 f (0) = | 0 + 2 | =2

1

2

3

4

x


The graph is V-shaped, passing through these intercepts. y 5 4 3 2 1 -4

-3 -2 -1 -1

1

2

3

4

x

-2 -3 -4 -5

Investigation Are graphs that involve absolute value always functions? Can you find an example of one that is not a function? Can you find any odd or even functions involving absolute values? What are their equations? Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on an absolute value function, or choose different values of k (both positive and negative). Sketch the families of graphs for these graphs with parameter k 1. f ] x g = k | x | 2. f ] x g = | x | + k 3. f ] x g = | x + k | What effect does the parameter k have on these graphs?

The equations and inequations involving absolute values that you studied in Chapter 3 can be solved graphically.

If you know how to sketch the graph of y = | x | , translate it 2 places to the left for the graph of y = | x + 2 | .

239

240


EXAMPLES Solve 1. | 2x - 1 | = 3

Solution Sketch y = | 2x - 1 | and y = 3 on the same number plane.

The solution of | 2x - 1 | = 3 occurs at the intersection of the graphs, that is, x = -1, 2. 2. | 2x + 1 | = 3x - 2

Solution Sketch y = | 2x + 1 | and y = 3x - 2 on the same number plane.

The graph shows that there is only one solution. Algebraically, you need to find the 2 possible solutions and then check them.

The solution is x = 3. 3. | x + 1 | 1 2

Solution Sketch y = | x + 1 | and y = 2 on the same number plane.


The solution of | x + 1 | 1 2 is where the graph y = | x + 1 | is below the graph y = 2, that is, - 3 1 x 1 1.

5.6 Exercises 1.

2.

3.

Find the x- and y-intercepts of each function. (a) y = | x | (b) f ] x g = | x | + 7 (c) f ] x g = | x | - 2 (d) y = 5 | x | (e) f ] x g = - | x | + 3 (f) y = | x + 6 | (g) f ] x g = | 3x - 2 | (h) y = | 5x + 4 | (i) y = | 7x - 1 | (j) f ] x g = | 2x | + 9 Sketch each graph on a number plane. (a) y = | x | (b) f ] x g = | x | + 1 (c) f ] x g = | x | - 3 (d) y = 2 | x | (e) f ] x g = -| x | (f) y = | x + 1 | (g) f ] x g = -| x - 1 | (h) y = | 2x - 3 | (i) y = | 4x + 2 | (j) f ] x g = | 3x | + 1 Find the domain and range of each function. (a) y = | x - 1 | (b) f ] x g = | x | - 8

(c) (d) (e)

f ] x g = | 2x + 5 | y = 2 | x |- 3 f ] x g = -| x - 3 |

4.

Find the domain over which each function is (i) increasing (ii) decreasing (a) y = | x - 2 | (b) f ] x g = | x | + 2 (c) f ] x g = | 2x - 3 | (d) y = 4 | x | - 1 (e) f ] x g = - | x |

5.

For each domain, find the range of each function. (a) y = | x | for - 2 # x # 2 (b) f ] x g = - | x | - 4 for -4 # x # 3 (c) f ] x g = | x + 4 | for -7 # x # 2 (d) y = | 2x - 5 | for -3 # x # 3 (e) f ] x g = -| x | for - 1 # x # 1

6.

For what values of x is each function increasing? (a) y = | x + 3 | (b) f ] x g = - | x | + 4 (c) f ] x g = | x - 9 | (d) y = | x - 2 | - 1 (e) f ] x g = - | x + 2 |

241

242


7.

Solve graphically (a) | x | = 3 (b) | x | 2 1 (c) | x | # 2 (d) | x + 2 | = 1 (e) | x - 3 | = 0 (f) | 2x - 3 | = 1 (g) | x - 1 | 1 4 (h) | x + 1 | # 3 (i) | x - 2 | 2 2 (j) | x - 3 | $ 1

(k) | 2x + 3 | # 5 (l) | 2x - 1 | $ 1 (m) | 3x - 1 | = x + 3 (n) | 3x - 2 | = x - 4 (o) | x - 1 | = x + 1 (p) | x + 3 | = 2x + 2 (q) | 2x + 1 | = 1 - x (r) | 2x - 5 | = x - 3 (s) | x - 1 | = 2x (t) | 2x - 3 | = x + 3

The Hyperbola a A hyperbola is a function with its equation in the form xy = a or y = x .

EXAMPLE 1 Sketch y = x .

Solution 1 y = x is a discontinuous curve since the function is undefined at x = 0. Drawing up a table of values gives: x

-3

y

-

1 3

1 2

-2

-1

-

1 2

-1

-2

-

1 4

0

1 4

1 2

1

2

3

-4

—

4

2

1

1 2

1 3

-

Class Discussion What happens to the graph as x becomes closer to 0? What happens as x becomes very large in both positive and negative directions? The value of y is never 0. Why?


To sketch the graph of a more general hyperbola, we can use the domain and range to help find the asymptotes (lines towards which the curve approaches but never touches). The hyperbola is an example of a discontinuous graph, since it has a gap in it and is in two separate parts.

Investigation Is the hyperbola always a function? Can you find an example of a hyperbola that is not a function? Are there any families of odd or even hyperbolas? What are their equations? Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a hyperbola, or choose different values of k (both positive and negative). Sketch the families of graphs for these graphs with parameter k k 1. y = x 1 2. y = x + k 3. y =

1 x+k

What effect does the parameter k have on these graphs?

EXAMPLES 3 . x-3 (b) Hence sketch the graph of the function.

1. (a) Find the domain and range of f ] x g =

Solution This is the equation of a hyperbola. To find the domain, we notice that x - 3 ! 0. So x ! 3 Also y cannot be zero (see example on page 242). Domain: {all real x: x ! 3} Range: {all real y: y ! 0} The lines x = 3 and y = 0 (the x-axis) are called asymptotes. CONTINUED

The denominator cannot be zero.

243

244


To make the graph more accurate we can find another point or two. The easiest one to find is the y-intercept. For y-intercept, x = 0 3 y= 0-3 = -1 y 5 4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

-2 -3

Notice that this graph is 3 a translation of y = x three units to the right.

5

y=0

x

Asymptotes x=3

-4 -5

2. Sketch y = -

1 . 2x + 4

Solution This is the equation of a hyperbola. The negative sign turns the hyperbola around so that it will be in the opposite quadrants. If you are not sure where it will be, you can find two or three points on the curve. To find the domain, we notice that 2x + 4 ! 0. 2x ! - 4 x ! -2 For the range, y can never be zero. Domain: {all real x: x ! -2} Range: {all real y: y ! 0} So there are asymptotes at x = -2 and y = 0 (the x-axis). To make the graph more accurate we can find the y-intercept. For y-intercept, x = 0 1 2 ( 0) + 4 1 =4

y=-


y

x

-2 - 14

a is a hyperbola with bx + c c domain & all real x: x ! - 0 and b

The function f ] x g =

range {all real y: y ! 0}

5.7 Exercises 1.

For each graph (i) State the domain and range. (ii) Find the y-intercept if it exists. (iii) Sketch the graph. 2 (a) y = x 1 (b) y = - x 1 (c) f ] x g = x+1 3 (d) f ] x g = x-2 1 (e) y = 3x + 6 2 (f) f ] x g = x-3 4 (g) f ] x g = x-1

2 x+1 2 (i) f ] x g = 6x - 3 6 (j) y = x+2

(h) y = -

2. 3.

2 Show that f ] x g = x is an odd function. Find the range of each function over the given domain. 1 (a) f ] x g = for -2 # x # 2 2x + 5 1 (b) y = for -2 # x # 0 x+3 5 (c) f ] x g = for - 3 # x # 1 2x - 4

245

246


3 for - 3 # x # 3 x-4 2 for 0 # x # 5 (e) y = 3x + 1

1 1 for -1 # y # 7 x-1 3 for (d) f ] x g = 2x + 1 1 -1 # y # 3 6 1 (e) y = for 1 # y # 6 3x - 2 2 (c) f ] x g =

(d) f ] x g = -

4.

Find the domain of each function over the given range. 3 (a) y = x for 1 # y # 3 2 1 (b) y = - x for - 2 # y # 2

Circles and Semi-circles The circle is used in many applications, including building and design.

Circle gate

A graph whose equation is in the form x 2 + ax + y 2 + by + c = 0 has the shape of a circle. There is a special case of this formula:

The graph of x 2 + y 2 = r 2 is a circle, centre ^ 0, 0 h and radius r

Proof y

(x, y) r x

y x


247

Given the circle with centre (0, 0) and radius r: Let (x, y) be a general point on the circle, with distances from the origin x on the x-axis and y on the y-axis as shown. By Pythagoras’ theorem: c2 = a2 + b2 ` r2 = x2 + y2

EXAMPLE (a) Sketch the graph of x 2 + y 2 = 4. Is it a function? (b) State its domain and range.

Solution The radius is

(a) This is a circle with radius 2 and centre (0, 0). y

2

-2

2

x

-2

The circle is not a function since a vertical line will cut it in more than one place. y

2

-2

2

x

-2

CONTINUED

4.

248


(b) Notice that the x-values for this graph lie between - 2 and 2 and the y-values also lie between - 2 and 2. Domain: {x: -2 # x # 2} Range: {y: -2 # y # 2}

The circle x 2 + y 2 = r 2 has domain: ! x: -r # x # r + and range: " y: -r # y # r ,

We can use Pythagoras’ theorem to find the equation of a more general circle.

The equation of a circle, centre (a, b) and radius r is ] x – a g2 + ^ y – b h2 = r 2

Proof Take a general point on the circle, (x, y) and draw a right-angled triangle as shown. y

(x, y)

y r b

y-b

x-a

(a, b) a

x

x

Notice that the small sides of the triangle are x – a and y – b and the hypotenuse is r, the radius. By Pythagoras’ theorem: c2 = a2 + b2 r 2 = ] x – a g2 + ^ y – b h2


EXAMPLES 1. (a) Sketch the graph of x 2 + y 2 = 81. (b) State its domain and range.

Solution (a) The equation is in the form x 2 + y 2 = r 2. This is a circle, centre (0, 0) and radius 9. y

9

-9

9

x

-9

(b) From the graph, we can see all the values that are possible for x and y for the circle. Domain: {x: -9 # x # 9} Range: {y: -9 # y # 9} 2. (a) Sketch the circle ] x – 1 g2 + ^ y + 2 h2 = 4. (b) State its domain and range.

Solution (a) The equation is in the form ] x – a g2 + ^ y – b h2 = r 2. ] x – 1 g 2 + ^ y + 2 h2 = 4 ] x – 1 g 2 + _ y – ] - 2 g i2 = 2 2

So a = 1, b = - 2 and r = 2 CONTINUED

249

250


This is a circle with centre ^ 1, - 2 h and radius 2. To draw the circle, plot the centre point ^ 1, - 2 h and count 2 units up, down, left and right to find points on the circle. y 5 4 3 2 1 x

-4 -3 -2

-1 -1 -2 -3

1

2

3

4

(1, -2)

-4 -5

(b) From the graph, we can see all the values that are possible for x and y for the circle. Domain: {x: -1 # x # 3} Range: {y: -4 # y # 0} 3. Find the equation of a circle with radius 3 and centre ^ -2, 1 h in expanded form.

Solution This is a general circle with equation ] x – a g2 + ^ y – b h2 = r 2 where a = - 2, b = 1 and r = 3. Substituting: ] x – a g2 + ^ y – b h2 = r 2 You may need to revise this in Chapter 2.

] x - ] - 2 g g2 + ^ y – 1 h2 = 3 2 ] x + 2 g2 + ^ y – 1 h2 = 9 Remove the grouping symbols. ] a + b g2 = a 2 + 2ab + b 2 So ] x + 2 g2 = x 2 + 2 ] x g ] 2 g + 2 2 = x 2 + 4x + 4 2 ] a – b g = a 2 - 2ab + b 2 So ^ y – 1 h2 = y 2 - 2 ^ y h ] 1 g + 1 2 = y 2 - 2y + 1 The equation of the circle is: x 2 + 4x + 4 + y - 2y + 1 = 9 x 2 + 4 x + y - 2y + 5 = 9 x 2 + 4 x + y – 2y + 5 - 9 = 9 - 9 x 2 + 4x + y - 2y - 4 = 0


Investigation The circle is not a function. Could you break the circle up into two functions? Change the subject of this equation to y. What do you notice when you change the subject to y? Do you get two functions? What are their domains and ranges? If you have a graphics calculator, how could you draw the graph of a circle?

By rearranging the equation of a circle, we can also find the equations of semi-circles. The equation of the semi-circle above the x-axis with centre (0, 0) and radius r is y = r 2 - x 2 The equation of the semi-circle below the x-axis with centre (0, 0) and radius r is y = - r 2 - x 2

Proof x2 + y2 = r2 y2 = r2 – x2 y = ! r2 - x2 This gives two functions:

y = r 2 - x 2 is the semi-circle above the x-axis since its range is y $ 0 for all values. y

r

-r

r

x

The domain is {x: -r # x # r } and the range is {y: 0 # y # r }

251

252


y = - r 2 - x 2 is the semi-circle above the x-axis since its range is y # 0 for all values. y

-r

r

x

-r

The domain is {x: - r # x # r } and the range is {y: -r # y # 0}

EXAMPLES Sketch each function and state the domain and range. 1. f ] x g =

9 - x2

Solution This is in the form f ] x g = r 2 - x 2 where r = 3. It is a semi-circle above the x-axis with centre (0, 0) and radius 3. y

3

-3

Domain: {x: -3 # x # 3} Range: {y: 0 # y # 3}

3

x


2. y = - 4 - x 2

Solution This is in the form y = - r 2 - x 2 where r = 2. It is a semi-circle below the x-axis with centre (0, 0) and radius 2. y

-2

2

x

-2

Domain: {x: -2 # x # 2} Range: {y: -2 # y # 0}

5.8 Exercises 1.

2.

For each of the following (i) sketch each graph (ii) state the domain and range. (a) x 2 + y 2 = 9 (b) x 2 + y 2 - 16 = 0 (c) ] x – 2 g2 + ^ y – 1 h2 = 4 (d) ] x + 1 g2 + y 2 = 9 (e) ] x + 2 g2 + ^ y – 1 h2 = 1 For each semi-circle (i) state whether it is above or below the x-axis (ii) sketch the function (iii) state the domain and range.

(a) (b) (c) (d) (e) 3.

y = - 25 - x 2 y = 1 - x2 y = 36 - x 2 y = - 64 - x 2 y = - 7 - x2

Find the length of the radius and the coordinates of the centre of each circle. (a) x 2 + y 2 = 100 (b) x 2 + y 2 = 5 (c) ] x – 4 g2 + ^ y – 5 h2 = 16 (d) ] x – 5 g2 + ^ y + 6 h2 = 49 (e) x 2 + ^ y – 3 h2 = 81

253

254


4.

Find the equation of each circle in expanded form (without grouping symbols). (a) Centre (0, 0) and radius 4 (b) Centre (3, 2) and radius 5 (c) Centre ^ -1, 5 h and radius 3 (d) Centre (2, 3) and radius 6

(e) (f) (g) (h) (i) (j)

Centre ^ -4, 2 h and radius 5 Centre ^ 0, -2 h and radius 1 Centre (4, 2) and radius 7 Centre ^ -3, -4 h and radius 9 Centre ^ -2, 0 h and radius 5 Centre ^ -4, -7 h and radius 3

Other Graphs There are many other different types of graphs. We will look at some of these graphs and explore their domain and range. You will meet these graphs again in the HSC Course.

Exponential and logarithmic functions EXAMPLES 1. Sketch the graph of f ] x g = 3 x and state its domain and range.

Solution If you do not know what this graph looks like, draw up a table of values. You may need to revise the indices that you studied in Chapter 1. e.g. When x = 0: y = 3c =1 When x = -1: y = 3-1 1 = 1 3 1 = 3 x y

-3 1 27

-2 1 9

-1 1 3

0

1

2

3

1

3

9

27

If you already know what the shape of the graph is, you can draw it just using 2 or 3 points to make it more accurate.


You learned about exponential graphs in earlier stages of maths.

This is an exponential function with y-intercept 1. We can find one other point. When x = 1 y = 31 =3

y

3 2 1 x

1

From the graph, x can be any real value (the equation shows this as well since any x value substituted into the equation will give a value for y). From the graph, y is always positive, which can be confirmed by substituting different values of x into the equation. Domain: " all real x , Range: " y: y 2 0 , 2. Sketch f ] x g = log x and state the domain and range.

Solution Use the LOG key on your calculator to complete the table of values. Notice that you can’t find the log of 0 or a negative number. x

−2

−1

0

0.5

1

2

3

4

y

#

#

#

−0.3

0

0.3

0.5

0.6

y

2 1

-1

1

2

3

4

x

From the graph and by trying different values on the calculator, y can be any real number while x is always positive. Domain: ! x: x 2 0 + Range: " all real y ,

255

256


The exponential function y = a x has domain {all real x} and range {y: y 2 0} The logarithmic function y = log a x has domain ! x: x 2 0 + and range {all real y}

Cubic function A cubic function has an equation where the highest power of x is x 3 .

EXAMPLE 1. Sketch the function f ] x g = x 3 + 2 and state its domain and range.

Solution Draw up a table of values. x

−3

−2

−1

0

1

2

3

y

−25

−6

1

2

3

10

29

y 5 4 3 2 1

-4

-3

-2

-1

-1

1

-2 If you already know the shape of y = x 3, f (x) = x 3 + 2 has the same shape as f (x) = x 3 but it is translated 2 units up (this gives a y-intercept of 2).

-3 -4 -5

The function can have any real x or y value: Domain: " all real x , Range: " all real y ,

2

3

4

x


Domain and range Sometimes there is a restricted domain that affects the range of a function.

EXAMPLE 1. Find the range of f ] x g = x 3 + 2 over the given domain of -1 # x # 4.

Solution The graph of f ] x g = x 3 + 2 is the cubic function in the previous example. From the graph, the range is {all real y}. However, with a restricted domain of -1 # x # 4 we need to see where the endpoints of this function are. f ] -1 g = ] -1 g3 + 2 = -1 + 2 =1 f ] 4 g = ] 4 g3 + 2 = 64 + 2 = 66 Sketching the graph, we can see that the values of y all lie between these points. y

(4, 66)

(-1, 1) x

Range: " y: 1 # y # 66 ,

257

258


You may not know what a function looks like on a graph, but you can still find its domain and range by looking at its equation. When finding the domain, we look for values of x that are impossible. For example, with the hyperbola you have already seen that the denominator of a fraction cannot be zero. For the range, we look for the results when different values of x are substituted into the equation. For example, x2 will always give zero or a positive number.

EXAMPLE Find the domain and range of f ] x g =

x - 4.

Solution We can only find the square root of a positive number or zero. So x – 4 $ 0 x$4 When you take the square root of a number, the answer is always positive (or zero). So y $ 0 Domain: ! x: x $ 4 + Range: " y: y $ 0 ,

5.9 Exercises 1.

Find the domain and range of (a) y = 4x + 3 (b) f ] x g = -4

(c) f ] x g = | 2x - 3 | (d) y = | x | - 2 (e) f ] x g = - 2x + 5

(c) x = 3 (d) f ] x g = 4x 2 – 1 (e) p ] x g = x 3 – 2

You may like to simplify the function by dividing by x.

(f) y = 5 - | x | (g) y = 2 x (h) y = -5 x x+1 (i) f ] x g = x 4x - 3 (j) y = 2x

(f) f ] x g = 12 - x - x 2 (g) x 2 + y 2 = 64 3 t-4 2 (i) g (z) = + 5 z (j) f ] x g = | x |

(h) f ] t g =

2.

Find the domain and range of (a) y = x (b) y =

x-2

3.

Find the x-intercepts of (a) y = x ] x - 5 g2 (b) f ] x g = ] x – 1 g ] x – 2 g ] x + 3 g (c) y = x 3 - 6x 2 + 8x (d) g ] x g = x 4 - 16x 2 (e) x 2 + y 2 = 49


4.

(a) Solve 1 - x 2 $ 0. (b) Find the domain of f ] x g = 1 - x2 .

5.

Find the domain of (a) y = x 2 - x - 2 (b) g ] t g = t 2 + 6t

6.

Each of the graphs has a restricted domain. Find the range in each case. (a) y = 2x - 3 in the domain -3 # x # 3 (b) y = x 2 in the domain -2 # x # 3 (c) f ] x g = x 3 in the domain

9.

x Given the function f ] x g = x (a) find the domain of the function (b) find its range. Draw each graph on a number plane (a) f ] x g = x 4 (b) y = - x 3 (c) y = x 4 - 3 (d) p ] x g = 2x 3 (e) g ] x g = x 3 + 1 (f) x 2 + y 2 = 100 (g) y = 2 x + 1

-2 # x # 1 1 (d) y = x in the domain 1# x #5

10. (a) Find the domain and range of y = x - 1. (b) Sketch the graph of y = x - 1 .

(e) y = | x | in the domain 0#x#4

11. Sketch the graph of y = 5 x .

(f) y = x 2 - 2x in the domain -3 # x # 3

12. For each function, state (i) its domain and range (ii) the domain over which the function is increasing (iii) the domain over which the function is decreasing. (a) y = 2x - 9 (b) f ] x g = x 2 - 2 1 (c) y = x (d) f ] x g = x 3 (e) f ] x g = 3 x

(g) y = - x 2 in the domain -1 # x # 1 (h) y = x 2 - 1 in the domain -2 # x # 3 (i) y = x 2 - 2x - 3 in the domain -4 # x # 4 (j) y = - x 2 + 7x - 6 in the domain 0 # x # 7 7.

8.

(a) Find the domain for the 3 function y = . x+1 (b) Explain why there is no x- intercept for the function. (c) State the range of the function.

13. (a) Solve 4 - x 2 $ 0. (b) Find the domain and range of (i) y = 4 - x 2 (ii) y = - 4 - x 2 .

259

260


DID YOU KNOW? A lampshade can produce a hyperbola where the light meets the flat wall. • Can you find any other shapes made by a light?

Lamp casting its light

Limits and Continuity Limits

A line that a graph approaches but never touches is called an asymptote.

The exponential function and the hyperbola are examples of functions that approach a limit. The curve y = a x approaches the x-axis when x approaches very large negative numbers, but never touches it. That is, when x " - 3, a x " 0. Putting a - 3 into index form gives 1 a-3 = 3 a 1 =3 Z0 We say that the limit of a x as x approaches -3 is 0. In symbols, we write lim a x = 0. x " -3

EXAMPLES 1. Find lim x "0

x 2 + 5x . x

Solution 0 , which is undefined. 0 Factorising and cancelling help us find the limit. x 1 ]x + 5 g x 2 + 5x lim lim = x x "0 x "0 x1 = lim (x + 5) Substituting x = 0 into the function gives

x "0

=5


2. Find lim x "2

x-2 . x2 - 4

Solution Substituting x = 2 into the function gives

0 , which is undefined. 0

x-2 x-2 = lim 2 1 x " 2 x -4 ^x + 2h _x - 2i 1 = lim x "2 x + 2 1 = 4 1

lim x "2

3. Find lim h "0

2h 2 x + hx 2 - 7h . h

Solution lim h "0

h ^ 2hx + x 2 - 7 h 2h 2 x + hx 2 - 7h = lim h "0 h h = lim 2hx + x 2 - 7 h "0

= x2 - 7

Continuity Many functions are continuous. That is, they have a smooth, unbroken curve (or line). However, there are some discontinuous functions that have gaps in their graphs. The hyperbola is an example. If a curve is discontinuous at a certain point, we can use limits to find the value that the curve approaches at that point.


y=

x2 - 1 and hence describe the domain and range of the curve x-1

x -1 . Sketch the curve. x-1 2

Solution Substituting x = 1 into

x2 - 1 0 gives x-1 0 CONTINUED

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262


lim x "1

]x + 1 g]x - 1 g x2 - 1 = lim x-1 x-1 x-1 = lim (x + 1) x "1

=2 y=

x2 - 1 is discontinuous at x = 1 since y is undefined at that point. x-1

This leaves a gap in the curve. The limit tells us that y " 2 as x " 1, so the gap is at ^ 1, 2 h . Domain: " x: all real x, x ! 1 , Range: " y: all real y, y ! 2 , y= =

x2 - 1 x-1 ^x + 1h ^x - 1h

x-1 =x+1 the graph is y = x + 1 where x ! 1

Remember that x ! 1.

`

2. Find lim

x " -2

x2 + x - 2 x2 + x - 2 and hence sketch the curve y = . x+2 x+2

Solution Substituting x = -2 into lim

x " -2

x2 + x - 2 0 gives x+2 0

^x - 1h ^x + 2h x2 + x - 2 = lim x " 2 x+2 ^x + 2h = lim ^ x - 1 h x " -2 = -3

y= y=

x2 + x - 2 is discontinuous at x = - 2 x+2 ^x + 2h ^x - 1h

x+2 =x-1 So the function is y = x - 1 where x ! -2. It is discontinuous at ^ -2, -3 h .


5.10 Exercises 1.

Find (a) lim x 2 + 5

2.

x "4

(b) lim t - 7 t " -3

(c) lim x 3 + 2x - 4 x "2

(d) lim

x 2 + 3x x

(e) lim

h2 - h - 2 h-2

(f) lim

y 3 - 125 y-5

(g) lim

x 2 + 2x + 1 x+1

(h) lim

x 2 + 2x - 8 x+4

x "0

h "2

y "5

x "-1

x " -4

Determine which of these functions are discontinuous and find x values for which they are discontinuous. (a) y = x 2 - 3 1 (b) y = x+1 (c) f ] x g =

x-1 1 (d) y = 2 x +4 1 (e) y = 2 x -4 3.

Sketch these functions, showing any points of discontinuity. (a) y =

x 2 + 3x x

(i) lim

c-2 c2 - 4

(b) y =

(j) lim

x-1 x2 - x

x 2 + 3x x+3

(c) y =

x 2 + 5x + 4 x+1

(k) lim

h 3 + 2h 2 - 7h h

(l) lim

hx 2 - 3hx + h 2 h

(m) lim

2hx 3 - h 2 x 2 + 3hx - 5h h

c "2

x "1

h "0

h "0

h "0

x3 - c3 (n) lim x "c x - c

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Further Graphs Graphs of functions with asymptotes can be difficult to sketch. It is important to find the limits as the function approaches the asymptotes.

1 A special limit is lim x = 0 x "3

EXAMPLES 1. Find lim

x "3

3x 2 . x 2 - 2x + 3

Solution 3x 2 3x x2 (dividing by the highest power of x) = lim 2 lim 2 x " 3 x - 2x + 3 x "3 x 3 2x + x2 x2 x2 3 = lim x "3 3 2 1-x+ 2 x 3 = 1-0+0 =3 2

2. Find (a) lim

x "3

(b)

x x 2 + 4x + 4

lim

x " -3

x x + 4x + 4 2

Solution x x x2 (a) lim 2 = lim 2 x " 3 x + 4x + 4 x "3 x 4x 4 + + x2 x2 x2 1 x = lim x "3 4 4 1+x+ 2 x 0 1+0+0 =0 =


265

1 Since x " 0 from the positive side when x " + 3, we can write x lim = 0+ x " 3 x 2 + 4x + 4 1 x x (b) lim 2 = lim x " -3 x + 4x + 4 x " -3 4 4 1+x+ 2 x =0 1 Since x " 0 from the negative side when x " - 3, we can write x lim = 0x " -3 x 2 + 4x + 4 3. Find lim

x "3

3x 2 . x-1

Solution 3 Dividing by x 2 will give . 0 Divide by x. 3x 2 3x x lim = lim x "3 x - 1 x "3 x 1 x-x 3x = lim x "3 1 1-x 3x = 1-0 = 3x 2

General graphs It is not always appropriate to sketch graphs, for example, a hyperbola or circle, from a table of values. By restricting the table of values, important features of a graph may be overlooked.

Other ways of exploring the shape of a graph include: • intercepts The x-intercept occurs when y = 0. The y-intercept occurs when x = 0. • even and odd functions Even functions 6 f ^ - x h = f (x) @ are symmetrical about the y-axis. Odd functions 6 f ^ - x h = - f (x) @ are symmetrical about the origin.

1+

4 4 + is positive x x2

whether x is + or -. Can you see why?

266


• asymptotes Vertical asymptotes occur when f ] x g ! 0 and h ] x g = 0, given g ]x g f ]x g = . h ]x g Horizontal and other asymptotes are found (if they exist) when finding lim f ] x g . x "!3 • domain and range The domain is the set of all possible x values for a function. The range is the set of all possible y values for a function.

EXAMPLES 1. Sketch y =

Solution

x2 . x -9 2

Intercepts: For x-intercept, y = 0 x2 x2 - 9 0 = x2 0=x 0=

So the x-intercept is 0. For y-intercept, x = 0 02 0-2 =0

y=

So the y-intercept is 0 Type of function: f (- x) =

] - x g2

] - x g2 - 9 x2 = 2 x -9 = f (x)

The function is even so it is symmetrical about the y-axis. Vertical asymptotes: x2 - 9 ! 0 ]x + 3 g]x - 3 g ! 0 x + 3 ! 0, x - 3 ! 0 x ! -3, x!3 So there are asymptotes at x = !3.


267

As x " 3 from LHS: ^ 3 - h2

-

f (3 ) =

^ 3 - h2 - 9

+ ==So y " - 3 As x " 3 from RHS: f (3 +) =

^ 3 + h2 ^ 3 + h2 - 9

+ + =+ So y " 3 =

You could substitute values close to 3 on either side into the equation, say 2.9 on LHS and 3.1 on RHS.

As x " - 3 from LHS: f (- 3 -) =

^ - 3 - h2 ^ - 3 - h2 - 9

+ + =+ =

So y " 3 As x " - 3 from RHS: ^ - 3 + h2 f (- 3 +) = ^ - 3 + h2 - 9 + ==So y " - 3

You could substitute values close to - 3 on either side into the equation, say - 3.1 on LHS and - 2.9 on RHS.

Horizontal asymptotes: 2

x2 x2

x = lim 9 x2 - 9 x " 3 x2 - 2 2 x x 1 = lim x "3 9 1- 2 x 1 = 1-0 =1 As x " 3 32 f ( 3) = 2 3 -9 21 So as x " 3, y " 1 from above lim

x "3

CONTINUED

268


As x " - 3 You could substitute values such as 1000 and -1000 to see what y does as x approaches !3.

f (-3) =

] -3 g 2 ] -3 g 2 - 9

21

So as x " -3, y " 1 from above Domain: {x: all real x ! !3} Range: When x 2 3, y 2 1 When - 3 1 x 1 3, y # 0 When x 1 - 3, y 2 1 So the range is {y: y 2 1, y # 0}. All this information put together gives the graph below.

2. Sketch f (x) =

x2 . x-2

Solution Intercepts: For x-intercept, y = 0 x2 0= x-2 0 = x2 0=x So the x-intercept is 0 For y-intercept, x = 0 02 0-2 =0

y=

So the y-intercept is 0. Type of function: ] - x g2 f (- x) = ]-x g - 2 x2 = -x - 2 x2 =x+2 ! - f (x) The function is neither even nor odd.


269

Vertical asymptotes: x-2!0 x!2 So there is an asymptote at x = 2. As x " 2 from LHS: ^ 2 - h2

f (2 -) =

2- - 2 + ==So y " - 3 As x " 2 from RHS: ^ 2 + h2

f (2 +) =

2+ - 2 + = + =+ So y " 3 You could substitute values close to 2 on either side into the equation, say 1.9 on LHS and 2.1 on RHS. e.g. When x = 2.1 ] 2.1 g2 f (2.1) = 2 .1 - 2 = 44.1 Horizontal asymptotes: x2 x

2

x = lim x - 2 x "3 x 2 x-x x = lim x "3 2 1-x x = 1-0 =x This means that as x approaches !3, the function approaches y = x. As x " 3 lim

x "3

32 3- 2 2x

f ( 3) =

So as x " 3, y " x from above. As x " -3 f (-3) =

Note: If we divide everything 1 . Divide by x. 0

by x 2, we get

] -3 g 2

-3 - 2 1x So as x " -3, y " x from above. CONTINUED

270


This is not easy to see, so substitute values such as 1000 and -1000 to see what y does as x approaches ±3. e.g. When x = - 1000 ] -1000 g2 -1000 - 2 = - 998

f (-1000) =

The point ^ -1000, -998 h is just above the line y = x. Domain: {x: all real x ! 2} Range: When x 2 2 we find that an approximate range is y 2 35 (substituting different values of x) When x 1 2, y # 0 So the range is {y: y 2 35, y # 0} Putting all this information together gives the graph below. y

x 2

There is a method that combines all these features to make sketching easier.

EXAMPLES 1. Sketch y =

Solution

x2 . x2 - 9

First find the critical points (x-intercepts and vertical asymptotes). x#x y= ]x + 3g]x - 3g


271

x-intercepts: y = 0 x2 x -9 0 = x2 0=x asymptotes: (x + 3) (x - 3) = 0 x = !3 0=

2

These critical points, x = 0, !3, divide the number plane into four regions.

Then sketch y = x, y = x + 3 and y = x - 3 on your graph.

These are straight lines at the critical points.

A graph is positive if it is above the x-axis.

Look at the sign of the curve in each region. Re gion 1: y = x + y=x+3 + y=x-3 + x#x y= ` ]x + 3 g]x - 3 g +#+ = +#+ =+ Re gion 2: y = x + y=x+3 + y=x-3 x#x y= ` ]x + 3 g]x - 3 g +#+ = +#+ ==-

The curve is above the x-axis in this region.

The curve is below the x-axis in this region.

CONTINUED

272


The curve is below the x-axis in this region.

The curve is above the x-axis in this region.

Re gion 3: y = x y=x+3 + y=x-3 x#x y= ` ]x + 3 g]x - 3 g -#= +#+ ==Re gion 4: y = x y=x+3 y=x-3 x#x y= ` ]x + 3 g]x - 3 g -#= -#+ = + =+ Find any horizontal asymptotes. x2 1 lim 2 = lim x "3 x - 9 x "3 9 1- 2 x = 1 from above

Check these!

lim

x " -3

x2 = 1 from above x2 - 9

All this information put together gives the following graph.

2. Sketch y =

Solution

x+1 . ]x + 2 g]x - 1 g

Find the critical points. x = -1 (x - intercept) x = - 2 0 vertical asymptotes ^ h x=1


273

Use these to divide the number plane into 4 regions and sketch y = x + 1, y = x + 2 and y = x - 1.

x+1 ]x + 2 g]x - 1 g + = +#+ =+

Re gion 1: y =

x+1 ]x + 2 g]x - 1 g + = +#=-

Re gion 2: y =

x+1 ]x + 2 g]x - 1 g = +#=+

Re gion 3: y =

x+1 ]x + 2 g]x - 1 g = -#=-

Re gion 4: y =

For horizontal asymptotes x+1 x+1 = xlim "3 ]x + 2 g]x - 1 g x2 + x - 2 1 1 x + x2 = xlim "3 1 2 1+x- 2 x = 0+ x+1 = 0lim x " -3 ] x + 2 g]x - 1 g lim

x "3

All this information put together gives the following graph.

The y-intercept is -

1 . 2

274


Class Investigation You can explore graphs of this type on a graphical calculator or by using computer software designed to draw graphs.

5.11 Exercises 1.

Find

4.

x (a) lim 2 x "3 x (b) lim

x "3

2 x+4 5.

x2 (e) lim 2 x " 3 x + 7x + 1

(g) lim

x "3

(h) lim

x "3

(i) lim

x "3

5

6x x 5 - 2x - 7 3

2x - 3x - 6 3x 3 + 1 x2 4x 3 + 27x - 9 2

5x x+2 3 x

(j) lim

x-1

x "3

2.

(a) Show that 2

x +x+3 3 1 =1+x+ 2 x2 x x2 + x + 3 (b) Find lim x "3 x2 2 x +x+3 (c) Find lim x " -3 x2 3.

Find (a) lim

2x x+5

(b) lim

2x x+5

x "3

x " -3

x4 3x 3 + 7x

(b) lim

5x 3 4x + 3

x "3

2x 3 (d) lim 3 x "3 x - x

x "3

(a) lim

x "3

5x (c) lim 2 x "3 x + 1

(f) lim

Find

Sketch (a) y =

1 x +1

(b) y =

1 x2 - 1

(c) y =

x x+1

(d) y =

x2 x +1

(e) y =

x2 x2 - 4

2

2

(f) y = 1 +

x x2 + 1

(g) y =

x+2 x2 - 4

(h) y =

4 - x2 4 + x2

1 (i) y = x + x (j) y =

3 x -4 2


Solving inequations graphically There are different methods of solving inequations involving pronumerals in the denominator. You learned how to solve these using the number line in Chapter 3.

EXAMPLES 1. Solve

1 $ 2 graphically. x+1

Solution Sketch y =

1 and y = 2 on the same number plane. x+1

The hyperbola has domain {all real x: x ! -1} and range {all real y: y ! 0}. For y-intercept: x = 0 1 y= 0+1 =1 y

y=2 2 1

1 -1 - 2

-2

1

2

x

-1

y=

1 x+1

-2

1 1 is on or $ 2 occurs when the hyperbola y = x+1 x+1 above the line y = 2.

The solution of

CONTINUED

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276


The two graphs intersect where

1 = 2. x+1

1 # (x + 1) = 2 # (x + 1) x+1 1 = 2 ]x + 1 g = 2x + 2 -1 = 2x 2x -1 = 2 2 1 - =x 2 1 The solution is -1 1 x # (from the graph). 2

2. Solve

1 1 1 graphically. x-2

Solution Sketch y =

1 and y = 1 on the same number plane. x-2

The hyperbola has domain {all real x: x ! 2} and range {all real y: y ! 0}. For y-intercept: x = 0 1 y= 0-2 1 =2 y

2 y=1

-2

1

-1

- 12

1

2

3

4

x

-1 -2

The solution of the line y = 1.

y= 1 x-2

1 1 is below 1 1 occurs when the hyperbola y = x-2 x-2


The two graphs intersect where

1 =1 x-2

1 # ( x - 2) = 1 # ( x - 2 ) x-2 1=x-2 3=x The solution is x 1 2, x 2 3

5.12 Exercises Solve graphically and algebraically. 1.

1 x 1 -2

2.

1 x 23

3.

1 x $1

4.

1 -x $2

5.

1 23 x-1

1 x+2 2 7. x-2 3 8. x+3 -1 9. x-1 x 10. x+2 6.

#1 $5 2 -1 #3 $4

Regions Class Investigation How many solutions are there for y $ x + 2? How would you record them all?

Inequalities can be shown as regions in the Cartesian plane. You can shade regions on a number plane that involve either linear or non-linear graphs. This means that we can have regions bounded by a circle or a parabola, or any of the other graphs you have drawn in this chapter. Regions can be bounded or unbounded. A bounded region means that the line or curve is included in the region.

277

278


EXAMPLE Sketch the region x # 3.

Solution

Remember that x = 3 is a vertical line with x-intercept 3.

x # 3 includes both x = 3 and x 1 3 in the region. Sketch x = 3 as an unbroken or filled in line, as it will be included in the region. Shade in all points where x 1 3 as shown. y 5 4 3 2 1 -4 -3 -2

-1 -1

1

2

3

4

x

-2 -3 -4 -5

x=3

An unbounded region means that the line or curve is not included in the region.

EXAMPLE Sketch the region y 2 -1.

Solution y 2 -1 doesn’t include y = -1. When this happens, it is an unbounded region and we draw the line y = -1 as a broken line to show it is not included.


Sketch y = -1 as a broken line and shade in all points where y 2 -1 as shown. y 5 4 3 2 1 -4 -3 -2 -1 y = -1 -1

1

2

3

4

x

-2 -3 -4 -5

For lines that are not horizontal or vertical, or for curves, we need to check a point to see if it lies in the region.

EXAMPLES Find the region defined by 1. y $ x + 2

Solution First sketch y = x + 2 as an unbroken line. On one side of the line, y 2 x + 2 and on the other side, y 1 x + 2. To find which side gives y 2 x + 2, test a point on one side of the line (not on the line). For example, choose ^ 0, 0 h and substitute into y$x+2 0$0+2 0$2 (false) CONTINUED

279

Remember that y = -1 is a horizontal line with y-intercept -1.

280


This means that ^ 0, 0 h does not lie in the region y $ x + 2. The region is on the other side of the line.

Any point in the region will make the inequality true. Test one to see this.

2. 2x - 3y 1 6

Solution First sketch 2x - 3y = 6 as a broken line, as it is not included in the region. To find which side of the line gives 2x - 3y 1 6, test a point on one side of the line. For example, choose ^ 0, 1 h and substitute into 2x - 3y 1 6 2 ] 0 g - 3 (1 ) 1 6 -3 1 6 (true) This means that ^ 0, 1 h lies in the region 2x - 3y 1 6.

2x - 3y = 6

3. x 2 + y 2 2 1

Solution The equation x 2 + y 2 = 1 is a circle, radius 1 and centre ^ 0, 0 h . Draw x 2 + y 2 = 1 as a broken line, since the region does not include the curve.


Choose a point inside the circle, say ^ 0, 0 h x2 + y2 2 1 02 + 02 2 1 0 2 1 (false) So the region lies outside the circle.

4. y $ x 2

Solution The equation y = x 2 is a parabola. Sketch this as an unbroken line, as it is included in the region. Choose a point inside the parabola, say ^ 1, 3 h. y $ x2 3 2 12 3 2 1 (true) So ^ 1, 3 h lies in the region. y = x2

Sometimes a region includes two or more inequalities. When this happens, sketch each region on the number plane, and the final region is where they overlap (intersect).

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282


EXAMPLE Sketch the region x # 4, y 2 -2 and y # x 2 .

Solution Draw the three regions, either separately or on the same set of axes, and see where they overlap. .

Put the three regions together.

If you are given a region, you should also be able to describe it algebraically.


EXAMPLES Describe each region. 1.

y

6 5 4 3 2 1 -4 -3 -2

1

-1 -1

2

3

x

4

-2 -3 -4

Solution The shaded area is below and including y = 6 so can be described as y # 6. It is also to the left of, but not including the line x = 4, which can be described as x 1 4. The region is the intersection of these two regions: y # 6 and x 1 4 y

2. 2

-2

2

x

-2

CONTINUED

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284


Solution The shaded area is the interior of the circle, centre (0, 0) and radius 2 but it does not include the circle. The equation of the circle is x 2 + y 2 = 2 2 or x 2 + y 2 = 4. You may know (or guess) the inequality for the inside of the circle. If you are unsure, choose a point inside the circle and substitute into the equation e.g. (0, 0). LHS = x 2 + y 2 = 02 + 02 =0 1 4 ] RHS g So the region is x 2 + y 2 1 4.

5.13 Exercises 1.

2.

Shade the region defined by (a) x # 2 (b) x 2 1 (c) y $ 0 (d) y 1 5 (e) y # x + 1 (f) y $ 2x - 3 (g) x + y 2 1 (h) 3x - y - 6 1 0 (i) x + 2y - 2 $ 0 (j) 2x - 1 1 0

(b)

y 6 5 4 3 2 1

-4 -3 -2 -1 -1

2

3

x

4

-2

Write an inequation to describe each region. (a)

-3 -4

(c)

y

6

y

6

5

5

4

y=x+1

4

3 2

3

1

2

-4 -3 -2 -1 -1

1

1

2

3

4

x

1

-2

-4 -3 -2 -1 -1

-3

-2

-4

-3 -4

1

2

3

4

x


(d)

y 5

5.

Shade the region (a) ] x - 2 g2 + y 2 # 4 (b) ] x - 1 g2 + ^ y - 2 h2 # 1 (c) ] x + 2 g2 + ^ y - 1 h2 2 9

6.

Shade the intersection of these regions. (a) x # 3, y $ -1 (b) x $ -3, y 2 x -3 (c) y # 1, y $ 3x - 5 (d) y 2 x + 1, y # 3 - x (e) y # 1, x 2 + y 2 # 9 (f) x 2 -1, x 2 + y 2 1 4 (g) y # 4, y $ x 2 (h) x 1 -2, y # 3, y 2 x 3 (i) y # 0, x 2 + y 2 $ 1 (j) -1 1 x - y # 2

7.

Shade the region bounded by (a) the curve y = x 2, the x-axis and the lines x = 1 and x = 3 (b) the curve y = x 3, the y-axis and the lines y = 0 and y = 1 (c) the curve x 2 + y 2 = 4, the x-axis and the lines x = 0 and x = 1 in the first quadrant 2 (d) the curve y = x , the x-axis and the lines x = 1 and x = 4

y = x2 - 4

4 3 2 1 -4 -3 -2 -1 -1

1

2

3

4

5

-2 -3 -4 -5

(e)

y y = 2x 3 2 1 1

3.

4.

x

Shade each region described. (a) y 2 x 2 – 1 (b) x 2 + y 2 # 9 (c) x 2 + y 2 $ 1 (d) y # x 2 (e) y 1 x 3 Describe as an inequality (a) the set of points that lie below the line y = 3x - 2 (b) the set of points that lie inside the parabola y = x 2 + 2 (c) the interior of a circle with radius 7 and centre (0, 0) (d) the exterior of a circle with radius 9 and centre (0, 0) (e) the set of points that lie to the left of the line x = 5 and above the line y = 2

1 , the x+2 x-axis and the lines x = 0 and

(e) the curve y = x=2 8.

Shade the regions bounded by the intersection of (a) x 1 2, y 1 5 and y # x 2 (b) x 1 3, y $ -1, y # x - 2 (c) y # 1 - x, y # 2x + 1, 2x - 3y # 6 (d) x $ -3, y # 2, x 2 + y 2 $ 9 (e) x 1 2, y # 3, y $ | x |

The first quadrant is where x and y values are both positive.

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286


Application Regions are used in business applications to find optimum profit. Two (or more) equations are graphed together, and the region where a profit is made is shaded. The optimum profit occurs at the endpoints (or vertices) of the region.

EXAMPLE A company makes both roller skates (X ) and ice skates (Y ). Roller skates make a $25 profit, while ice skates make a profit of $21. Each pair of roller skates spends 2 hours on machine A (available 12 hours per day) and 2 hours on machine B (available 8 hours per day). Each pair of ice skates spends 3 hours on machine A and 1 hour on machine B. How many skates of each type should be made each day to give the greatest profit while making the most efficient use of the machines?

SOLUTION Profit P = $25 X + $21Y Machine A: 2X + 3Y # 12 Machine B: 2X + Y # 8 Sketch the regions and find the point of intersection of the lines.

The shaded area shows all possible ways of making a profit. Optimum profit occurs at one of the endpoints of the regions. (0, 4): P = $25 ] 0 g + $21 ] 4 g = $84 (4, 0): P = $25 ] 4 g + $21 ] 0 g = $100 (3, 2): P = $25 ] 3 g + $21 ] 2 g = $117

^ 3, 2 h gives the greatest profit, so 3 pairs of roller skates and 2 pairs of ice skates each day gives optimum profit.


Test Yourself 5 1.

If f ] x g = x 2 - 3x - 4, find (a) f ] -2 g (b) f ] a g (c) x when f ] x g = 0

2.

Sketch each graph (a) y = x 2 - 3x - 4 (b) f ] x g = x 3 (c) x 2 + y 2 = 1 (d) y = 1 - x 2 (e) y = - 1 - x 2 2 (f) y = x (g) 2x - 5y + 10 = 0 (h) y = | x + 2 |

(b)

3.

Find the domain and range of each graph in question 2.

4.

If f ] x g = *

2x

11. Describe each region (a)

if x $ 1

2

x -3

if x 1 1

find f ] 5 g - f ] 0 g + f ] 1 g 3

if x 2 3

5.

Given f ] x g = * x if 1 # x # 3 2 - x if x 1 1 find (a) f ] 2 g (b) f ] -3 g (c) f ] 3 g (d) f ] 5 g (e) f ] 0 g

6.

Shade the region y $ 2x + 1.

7.

Shade the region where x 1 3 and y $ -1.

8.

Shade the region given by x 2 + y 2 $ 1.

9.

Shade the region given by 2x + 3y - 6 # 0 and x $ -2.

2

10. Shade the region y 2 x + 1 and x + y # 2.

(c)

12. (a) Write down the domain and range of 2 the curve y = . x-3 2 (b) Sketch the graph of y = . x-3

287

288


13. (a) Sketch the graph y = | x + 1 |. (b) From the graph, solve (i) | x + 1 | = 3 (ii) | x + 1 | 1 3 (iii) | x + 1 | 2 3 14. If f ] x g = 3x - 4, find (a) f ] 2 g (b) x when f ] x g = 7 (c) x when f ] x g = 0 15. Find the x- and y-intercepts of (a) 2x - 5y + 20 = 0 (b) y = x 2 - 5x - 14 16. State which functions are (i) even (ii) odd (iii) neither even nor odd. (a) y = x 2 - 1 (b) y = x + 1 (c) y = x 3 (d) y = x 4 (e) y = 2 x 17. Find (a) lim

x 2 - 2x - 3 x-3

(b) lim

2x x 2 + 5x

x "3

x "0

(c) lim

x " -1

(d) lim h "0

x3 + 1 x2 - 1

2xh 2 + 3h h

18. Sketch y = 10 x, y = log x and y = x on the same number plane. 19. (a) State the domain and range of y = 2x - 4 . (b) Sketch the graph of y = 2x - 4 . 20. Show that (a) f ] x g = x 4 + 3x 2 - 1 is even (b) f ] x g = x 3 - x is odd. 21. Sketch the following graphs showing asymptotes and other features. 5 (a) y = 2 x +5 x x2 - 1 2x 2 (c) f ] x g = 2 x - 16 x2 (d) y = 2 x +3 x+2 (e) f ] x g = 2 x - 2x - 8 (b) y =

Challenge Exercise 5 Find the values of b if f ] x g = 3x 2 - 7x + 1 and f ] b g = 7.

5.

2.

Sketch y = ] x + 2 g2 - 1 in the domain -3 # x # 0.

6.

3.

Sketch the curve y =

4.

Sketch the region y 2 quadrant.

1.

2x 3 . x2 - 4 4 - x 2 in the first

7.

Draw the graph of y = | x | + 3x - 4. Z 2x + 3 when x 2 2 ] f ] x g = [1 when -2 # x # 2 ] 2 x when x 1 -2 \ Find f ] 3 g, f ] -4 g, f ] 0 g and sketch the curve. Find the domain and range of 1 y= 2 . x -1


8. 9.

Sketch the region x 1 y, x + 2y 1 6, x + 2y - 4 $ 0. Find the domain and range of x 2 = y in the first quadrant.

10. If f ] x g = 2x 3 - 2x 2 - 12x, find x when f ] x g = 0. 1 11. Sketch the region defined by y 2 x+2 in the first quadrant. 1 - t2 if t 2 1 2 t -1 if t # 1 ] find the value of h 2 g + h ] -1 g - h ] 0 g and sketch the curve.

12. If h ] t g = )

13. Sketch y =

1 - x 2 in the first quadrant.

(b) Find the domain and range of y=

2x + 7 . x+3

(c) Hence sketch the graph of y=

2x + 7 . x+3

18. Sketch y = 2 x - 1 . 19. Sketch y =

|x |

. x2 20. Find the domain and range of f ] x g = 2x - 6 . 21. What is the domain of y = 22. Sketch f ] x g = 1 -

14. Sketch the region y $ x - 5, y 1 x 2 + x. 15. If f ] x g = 2x - 1, show that f ^ a 2 h = f _ (-a)2 i for all real a. 16. Find the values of x for which f ] x g = 0 when f ] x g = 2x 2 - x - 5 (give exact answers). 17. (a) Show that

2x + 7 1 =2+ . x+3 x+3

23. (a) Find lim

x "3

2 x . x-1

(b) Sketch y = 24. Sketch y =

1 . x2

2 x . x-1

x+1 . x2 - 1

1 4 - x2

?

289

6

Trigonometry TERMINOLOGY Angle of depression: The angle between the horizontal and the line of sight when looking down to an object below Angle of elevation: The angle between the horizontal and the line of sight when looking up to an object above Angles of any magnitude: Angles can be measured around a circle at the centre to find the trigonometric ratios of angles of any size from 0c to 360c and beyond Bearing: The direction relative to north. Bearings may be written as true bearings (clockwise from North) or as compass bearings (using N, S, E and W)

Complementary angles: Two or more angles that add up to 90c Cosecant: The reciprocal ratio of sine (sin). It is the hypotenuse over the opposite side in a right triangle Cotangent: The reciprocal ratio of tangent (tan). It is the adjacent over the opposite side in a right triangle Secant: The reciprocal ratio of cosine (cos). It is the hypotenuse over the adjacent side in a right triangle Trigonometric identities: A statement that is true for all trigonometric values in the domain. Relationships between trigonometric ratios

Chapter 6 Trigonometry

291

INTRODUCTION TRIGONOMETRY IS USED IN many fields, such as building, surveying and navigating. Wave theory also uses trigonometry. This chapter revises basic right-angled triangle problems and applies them to real-life situations. Some properties of trigonometric ratios, angles greater than 90c and trigonometric equations are introduced. You will also study trigonometry in non-right-angled triangles.

DID YOU KNOW? Ptolemy (Claudius Ptolemaeus), in the second century, wrote He¯ mathe¯matike¯ syntaxis (or Almagest as it is now known) on astronomy. This is considered to be the first treatise on trigonometry, but was based on circles and spheres rather than on triangles. The notation ‘chord of an angle’ was used rather than sin, cos or tan. Ptolemy constructed a table of sines from 0c to 90c in steps of a quarter of a degree. He also calculated a value of r to 5 decimal places, and established the relationship for sin (X ! Y ) and cos (X ! Y ) .

Trigonometric Ratios In similar triangles, pairs of corresponding angles are equal and sides are in proportion. For example:

You studied similar triangles in Geometry in Chapter 4.

In any triangle containing an angle of 30c, the ratio of AB:AC = 1:2. Similarly, the ratios of other corresponding sides will be equal. These ratios of sides form the basis of the trigonometric ratios. In order to refer to these ratios, we name the sides in relation to the angle being studied:

• the hypotenuse is the longest side, and is always opposite the right angle • the opposite side is opposite the angle marked in the triangle • the adjacent side is next to the angle marked

ch6.indd 291

8/11/09 10:58:08 AM

292


The opposite and adjacent sides vary according to where the angle is marked. For example:

The trigonometric ratios are

You can learn these by their initials SOH, CAH, TOA.

What about Some Old Hags Can’t Always Hide Their Old Age?

Sine

sin i =

Cosine

cos i =

Tangent tan i =

opposite hypotenuse adjacent hypotenuse opposite adjacent

As well as these ratios, there are three inverse ratios,

Cosecant cosec i =

1 sin i

sec i =

1 cos i

Cotangent cot i =

1 tan i

Secant

hypotenuse p opposite hypotenuse f= p adjacent adjacent f= p opposite f=

DID YOU KNOW? Trigonometry, or triangle measurement, progressed from the study of geometry in ancient Greece. Trigonometry was seen as applied mathematics. It gave a tool for the measurement of planets and their motion. It was also used extensively in navigation, surveying and mapping, and it is still used in these fields today. Trigonometry was crucial in the setting up of an accurate calendar, since this involved measuring the distances between the Earth, sun and moon.


293

EXAMPLES 1. Find sin a, tan a and sec a.

Solution AB = hypotenuse = 5 BC = opposite side = 3 AC = adjacent side = 4 opposite sin a = hypotenuse 3 = 5 opposite tan a = adjacent 3 = 4 1 sec a = cos a hypotenuse = adjacent 5 = 4 2. If sin i =

2 , find the exact ratios of cos i, tan i and cot i. 7

Solution To find the other ratios you need to find the adjacent side.

By Pythagoras’ theorem: c2 = a2 + b2 72 = a2 + 22 49 = a 2 + 4 45 = a 2 à=

45 CONTINUED

294


cos i = = tan i =

adjacent hypotenuse 45 7 opposite

adjacent 2 = 45 1 cot i = tan i 45 = 2

Complementary angles

In D ABC, if+B = i, then +A = 90c - i

(angle sum of a Δ) a (90c - i) = c b cos (90c - i) = c a tan (90c - i) = b c sec (90c - i) = b c cosec (90c - i) = a b cot (90c - i) = a

b i= c a cos i = c b tan i = a c sec i = a c cosec i = b a cot i = b

sin

sin

From these ratios come the results.

sin i = cos (90° - i) cos i = sin (90° - i) sec i = cosec (90° - i) cosec i = sec (90° - i) tan i = cot (90° - i) cot i = tan (90° - i)


295

EXAMPLES 1. Simplify tan 50c - cot 40c.

Solution tan 50c = cot ] 90c - 50c g = cot 40c ` tan 50c - cot 40c = tan 50c - tan 50c =0

Check this answer on your calculator.

2. Find the value of m if sec 55c = cosec ] 2m - 15 g c.

Solution sec 55c = cosec ] 90c - 55c g = cosec 35c ` 2m - 15 = 35 2m = 50 m = 25

Check this by substituting m into the equation.

6.1 Exercises 1.

Write down the ratios of cos i, sin i and tan i.

2.

Find sin b, cot b and sec b.

3.

Find the exact ratios of sin b, tan b and cos b.

4.

Find exact values for cos x, tan x and cosec x.

296


5. 6.

7.

Hint: Change 0.7 to a fraction.

8.

9.

4 , find cos i and sin i. 3 2 If cos i = , find exact values for 3 tan i, sec i and sin i. If tan i =

1 If sin i = , find the exact ratios 6 of cos i and tan i. If cos i = 0.7, find exact values for tan i and sin i. D ABC is a right-angled isosceles triangle with +ABC = 90c and AB = BC = 1. (a) Find the exact length of AC. (b) Find +BAC. (c) From the triangle, write down the exact ratios of sin 45c, cos 45c and tan 45c.

10.

(c) Write down the exact ratios of sin 60c, cos 60c and tan 60c. 11. Show sin 67c = cos 23c. 12. Show sec 82c = cosec 8c. 13. Show tan 48c = cot 42c. 14. Simplify (a) cos 61c + sin 29c (b) sec i - cosec ] 90c - i g (c) tan 70c + cot 20c - 2 tan 70c (d)

sin 55c cos 35c

(e)

cot 25c + tan 65c cot 25c

15. Find the value of x if sin 80c = cos ] 90 - x g c. 16. Find the value of y if tan 22c = cot ^ 90 - y h c. 17. Find the value of p if cos 49c = sin ^ p + 10 h c. 18. Find the value of b if sin 35c = cos ] b + 30 g c. 19. Find the value of t if cot ] 2t + 5 g c = tan ] 3t - 15 g c.

(a) Using Pythagoras’ theorem, find the exact length of AC. (b) Write down the exact ratios of sin 30c, cos 30c and tan 30c.

20. Find the value of k if tan ] 15 - k g c = cot ] 2k + 60 g c.

Trigonometric ratios and the calculator Angles are usually given in degrees and minutes. In this section you will practise rounding off angles and finding trigonometric ratios on the calculator. Angles are usually given in degrees and minutes in this course. The calculator uses degrees, minutes and seconds, so you need to round off. 60 minutes = 1 deg ree (60l = 1c) 60 sec onds = 1 min ute (60m = 1l) In normal rounding off, you round up to the next number if the number to the right is 5 or more. Angles are rounded off to the nearest degree by rounding up if there are 30 minutes or more. Similarly, angles are rounded off to the nearest minute by rounding up if there are 30 seconds or more.


297

EXAMPLES Round off to the nearest minute. 1. 23c 12l 22m

Solution 23c 12l 22m = 23c 12l 2. 59c 34l 41m

Solution 59c 34l 41m = 59c 35l 3. 16c 54l 30m Because 30 seconds is half a minute, we round up to the next minute.

Solution 16c 54l 30m = 16c 55l

% , ,,

KEY Some calculators have

This key changes decimal angles into degrees, minutes and seconds

deg or dms keys.

and vice versa.

EXAMPLES 1. Change 58c 19l into a decimal.

Solution Press 58 % , ,, 19 % , ,, = % , ,, So 58c 19l = 58.31666667 2. Change 45.236c into degrees and minutes.

Solution Press 45.236 = SHIFT % , ,, So 45.236c = 45c14l

If your calculator does not give these answers, check the instructions for its use.

298


In order to use trigonometry in right-angled triangle problems, you need to find the ratios of angles on your calculator.

EXAMPLES 1. Find cos 58c 19l, correct to 3 decimal places.

Solution

If your calculator doesn't give this answer, check that it is in degree mode.

Press COS 58 % , ,, 19 % , ,, = So cos 58c19l = 0.525 2. Find sin 38c14l, correct to 3 decimal places.

Solution Press SIN 38 % , ,, 14 % , ,, = So sin 38c 14l = 0.619 3. If tani = 0.348, find i in degrees and minutes.

Solution This is the reverse of finding trigonometric ratios. To find the angle, given the ratio, use the inverse key ^ tan - 1 h . Press SHIFT TAN - 1 0.348 = SHIFT % , ,, tan i = 0.348 i = tan - 1 (0.348) = 19c11l 4. Find i in degrees and minutes if cos i = 0.675.

Solution Press SHIFT COS - 1 0.675 = SHIFT % , ,, cos i = 0.675 i = cos - 1 (0.675) = 47c 33l

6.2 Exercises 1.

Round off to the nearest degree. (a) 47° 13l 12m (b) 81° 45l 43m (c) 19° 25l 34m (d) 76° 37l 19m (e) 52° 29l 54m

2.

Round off to the nearest minute. (a) 47° 13l 12m (b) 81° 45l 43m (c) 19° 25l 34m (d) 76° 37l 19m (e) 52° 29l 54m


3.

Change to a decimal. (a) 77c45l (b) 65c30l (c) 24c51l (d) 68c21l (e) 82c31l

5.

Find correct to 3 decimal places. (a) sin 39c25l (b) cos 45c 51l (c) tan18c43l (d) sin 68c06l (e) tan 54c20l

4.

Change into degrees and minutes. (a) 59.53c (b) 72.231c (c) 85.887c (d) 46.9c (e) 73.213c

6.

Find i in degrees and minutes if (a) sin i = 0.298 (b) tan i = 0.683 (c) cos i = 0.827 (d) tan i = 1.056 (e) cos i = 0.188

Right-angled Triangle Problems Trigonometry is used to find an unknown side or angle of a triangle.

Finding a side We can use trigonometry to find a side of a right-angled triangle.

EXAMPLES 1. Find the value of x, correct to 1 decimal place.

Solution cos i =

adjacent

hypotenuse x cos 23° 49l = 11.8 x 11.8 # cos 23° 49l = 11.8 # 11.8 11.8 cos 23° 49l = x `

10.8 cm = x ^ to 1 decimal point h CONTINUED

299

300


2. Find the value of y, correct to 3 significant figures.

Solution sin i =

opposite

hypotenuse 9.7 sin 41c 15l = y 9.7 y # sin 41c 15l = y # y y sin 41c 15l = 9.7 y sin 41c 15l 9.7 = sin 41c 15l sin 41c 15l 9.7 y= sin 41c 15l = 14.7 m ^ to 3 significant figures h

6.3 Exercises 1.

Find the values of all pronumerals, correct to 1 decimal place. (a)

(c)

(b) (d)


(e) (l)

4.7 cm

x

(f)

37c22l 72c18l

(m) x

6.3 cm

(g) (n) 63c14l

23 mm

x

(o)

39c47l

(h) 3.7 m

(i) (p) k

(j)

46c5l

14.3 cm

(q)

5.4 cm

(k)

31c12l

y

h

x

4.8 m

74c29l

301

302


68c41l

(r)

d

0.45 m

6.2 cm 73c

4.

x

(s)

5.75 cm

19c17l

17.3 m

(t)

Hamish is standing at an angle of 67c from a goalpost and 12.8 m away as shown. How far does he need to kick a football for it to reach the goal?

x

12.8 m

67c b

5.

6c3l

Square ABCD with side 6 cm has line CD produced to E as shown so that +EAD = 64c 12l. Evaluate the length, correct to 1 decimal place, of (a) CE (b) AE E

2.

A roof is pitched at 60c. A room built inside the roof space is to have a 2.7 m high ceiling. How far in from the side of the roof will the wall for the room go? 64c12l

D

A

2.7 m 60c

x

3.

A diagonal in a rectangle with breadth 6.2 cm makes an angle of 73c with the vertex as shown. Find the length of the rectangle correct to 1 decimal place.

C

6.

6 cm

B

A right-angled triangle with hypotenuse 14.5 cm long has one interior angle of 43c 36l. Find the lengths of the other two sides of the triangle.


7.

8.

9.

A right-angled triangle ABC with the right angle at A has +B = 56c44l and AB = 26 mm. Find the length of the hypotenuse. A triangular fence is made for a garden inside a park. Three holes A, B and C for fence posts are made at the corners so that A and B are 10.2 m apart, AB and CB are perpendicular, and angle CAB is 59c 54l. How far apart are A and C? Triangle ABC has +BAC = 46c and +ABC = 54c. An altitude is drawn from C to meet AB at point D. If the altitude is 5.3 cm long, find, correct to 1 decimal place, the length of sides (a) AC (b) BC (c) AB

(a) Find the length of the side of the rhombus. (b) Find the length of the other diagonal. 11. Kite ABCD has diagonal BD = 15.8 cm as shown. If +ABD =57c29l and +DBC = 72c51l, find the length of the other diagonal AC. A

B

57c29l

72c51l

D

15.8 cm

C

10. A rhombus has one diagonal 12 cm long and the diagonal makes an angle of 28c 23l with the side of the rhombus.

Finding an angle Trigonometry can also be used to find one of the angles in a right-angled triangle.

EXAMPLES 1. Find the value of i, in degrees and minutes.

CONTINUED

303

304


Solution cos i =

adjacent

hypotenuse 5.8 = 7.3 5.8 ` i = cos - 1 c m 7.3 = 37c 23l

2. Find the value of a, in degrees and minutes.

Solution tan a =

opposite

adjacent 4 = .9 2 .1 4 .9 ` a = tan - 1 c m 2 .1 = 66c 48l

6.4 Exercises 1.

Find the value of each pronumeral, in degrees and minutes. (a)

(b)


(c)

(i)

(j) (d)

(e)

3.8 cm

(k)

2.4 cm

a

(l)

i

(f) 8.3 cm

5.7 cm

(m)

i 6.9 mm

(g)

11.3 mm

(n)

(h)

i

3m

7m

305

306


(o)

b 20 m

5.1 cm

i

11.6 cm

3.

(p) 15 m

A field is 13.7 m wide and Andre is on one side. There is a gate on the opposite side and 5.6 m along from where Andre is. At what angle will he walk to get to the gate?

a

Andre

13 m

(q)

4.4 cm

i

12.3 m

i

7.6 cm 13.7 m

(r)

a

5.6 m

14.3 cm

8.4 cm

(s)

4.

i

Gate

A 60 m long bridge has an opening in the middle and both sides open up to let boats pass underneath. The two parts of the bridge floor rise up to a height of 18 m. Through what angle do they move?

3m

18 m

5m

(t)

c

i 18.9 cm

10.3 cm

2.

A kite is flying at an angle of i above the ground as shown. If the kite is 12.3 m above the ground and has 20 m of string, find angle i.

5.

60 m

An equilateral triangle ABC with side 7 cm has an altitude AD that is 4.5 cm long. Evaluate the angle the altitude makes with vertex A ]+DAB g.


6.

7.

Rectangle ABCD has dimensions 18 cm # 7 cm. A line AE is drawn so that E bisects DC. (a) How long is line AE? (Answer to 1 decimal place). (b) Evaluate +DEA. A 52 m tall tower has wire stays on either side to minimise wind movement. One stay is 61.3 m long and the other is 74.5 m long as shown. Find the angles that the tower makes with each stay. a b 61.3 m

A

B

5 cm

D 1 cm

C

E

(a) Find +BEC. (b) Find the length of the rectangle. 10. A diagonal of a rhombus with side 9 cm makes an angle of 16c with the side as shown. Find the lengths of the diagonals.

74.5 m

16c

52 m

9 cm

8.

(a) The angle from the ground up to the top of a pole is 41c when standing 15 m on one side of it. Find the height h of the pole, to the nearest metre. (b) If Seb stands 6 m away on the other side, find angle i.

h

11. (a) Kate is standing at the side of a road at point A, 15.9 m away from an intersection. She is at an angle of 39c from point B on the other side of the road. What is the width w of the road? (b) Kate walks 7.4 m to point C. At what angle is she from point B? B

i 6m

9.

15 m

41c

Rectangle ABCD has a line BE drawn so that +AEB = 90c and DE = 1 cm. The width of the rectangle is 5 cm.

w A

39c 7.4 m

C

i 15.9 m

307

308


Applications DID YOU KNOW? The Leaning Tower of Pisa was built as a belfry for the cathedral nearby. Work started on the tower in 1174, but when it was only half completed the soil underneath one side of it subsided. This made the tower lean to one side. Work stopped, and it wasn’t until 100 years later that architects found a way of completing the tower. The third and fifth storeys were built close to the vertical to compensate for the lean. Later a vertical top storey was added.

The tower is about 55 m tall and 16 m in diameter. It is tilted about 5 m from the vertical, and tilts by an extra 0.6 cm each year.

Class Investigation Discuss some of the problems with the Leaning Tower of Pisa. • Find the angle at which it is tilted from the vertical. • Work out how far it will be tilted in 10 years. • Use research to find out if the tower will fall over, and if so, when.

Angle of elevation The angle of elevation is used to measure the height of tall objects that cannot be measured directly, for example a tree, cliff, tower or building.


Class Exercise Stand outside the school building and look up to the top of the building. Think about which angle your eyes pass through to look up to the top of the building.

The angle of elevation, i, is the angle measured when looking from the ground up to the top of the object. We assume that the ground is horizontal.

EXAMPLE The angle of elevation of a tree from a point 50 m out from its base is 38c 14l. Find the height of the tree, to the nearest metre.

Solution We assume that the tree is vertical!

A clinometer is used to measure the angle of elevation or depression.

tan 38c 14l =

h 50

50 # tan 38c 14l = 50 #

h 50

50 tan 38c 14l = h 39 Z h So the tree is 39 m tall, to the nearest metre.

309

310


Angle of depression The angle of depression is the angle formed when looking down from a high place to an object below.

Class Exercise If your classroom is high enough, stand at the window and look down to something below the window. If the classroom is not high enough, find a hill or other high place. Through which angle do your eyes pass as you look down?

The angle of depression, i, is the angle measured when looking down from the horizontal to an object below.

EXAMPLES 1. The angle of depression from the top of a 20 m building to a boy below is 61c 39l. How far is the boy from the building, to 1 decimal place?

Solution


+DAC = +ACB = 61c 39l 20 tan 61c 39l = x 20 x # tan 61c 39l = x # x x tan 61c 39l = 20 x tan 61c 39l 20 = tan 61c 39l tan 61c 39l 20 x= tan 61c 39l Z 10.8

(alternate angles, AD < BC)

So the boy is 10.8 m from the building. 2. A bird sitting on top of an 8 m tall tree looks down at a possum 3.5 m out from the base of the tree. Find the angle of elevation to the nearest minute.

Solution B

A

i

8m

C

3.5 m

The angle of depression is i Since AB < DC +BDC = i tan i =

8 3.5

8 m 3 .5 = 66c 22l

` i = tan - 1 c

] horizontal lines g ^ alternate angles h

D

311

312


Bearings Bearings can be described in different ways: For example, N70c W:

Start at north and measure 70o around towards the west.

True bearings measure angles clockwise from north

EXAMPLES We could write 315o T for true bearings.

1. Sketch the diagram when M is on a bearing of 315c from P.

Solution

Measure clockwise, starting at north.

2. X is on a bearing of 030c from Y. Sketch this diagram.

Solution

All bearings have 3 digits so 30° becomes 030° for a bearing.

3. A house is on a bearing of 305c from a school. What is the bearing of the school from the house?


Solution The diagram below shows the bearing of the house from the school. North House

School 305c

To find the bearing of the school from the house, draw in North from the house and use geometry to find the bearing as follows: N2

N1 H

S 305c

The bearing of the school from the house is +N 2 HS. +N 1 SH = 360c - 305c = 55c

^ angle of revolution h

+N 2 HS = 180c - 55c = 125c

(cointerior angles, N 2 H < N 1 S)

So the bearing of the school from the house is 125c.

4. A plane leaves Sydney and flies 100 km due east, then 125 km due north. Find the bearing of the plane from Sydney, to the nearest degree.

CONTINUED

313

314


Solution

125 100 = 1.25

tan x =

x = tan - 1 (1.25) (to the nearest degree) = 51c i = 90c - xc = 90c - 51c = 39c So the bearing of the plane from Sydney is 039°. 5. A ship sails on a bearing of 140° from Sydney for 250 km. How far east of Sydney is the ship now, to the nearest km?

Solution

A navigator on a ship uses a sextant to measure angles.

Could you use a different triangle for this question?

i = 140c - 90c = 50c x cos 50c = 250 x 250 # cos 50c = 250 # 250 250 cos 50c = x 161 Z x So the ship is 161 km east of Sydney, to the nearest kilometre.


6.5 Exercises 1.

2.

Draw a diagram to show the bearing in each question. (a) A boat is on a bearing of 100c from a beach house. (b) Jamie is on a bearing of 320c from a campsite. (c) A seagull is on a bearing of 200c from a jetty. (d) Alistair is on a bearing of 050c from the bus stop. (e) A plane is on a bearing of 285c from Broken Hill. (f) A farmhouse is on a bearing of 012c from a dam. (g) Mohammed is on a bearing of 160c from his house. (h) A mine shaft is on a bearing of 080c from a town. (i) Yvonne is on a bearing of 349c from her school. (j) A boat ramp is on a bearing of 280c from an island.

North

(b)

Y

West

East 35c X

South

(c)

North

X 10c

Y

West

East

South

(d) X

Find the bearing of X from Y in each question in 3 figure (true) bearings. North (a)

North

23c West

Y

East

South

(e) 112c

North

Y

West X

Y

X South

East

315

316


3.

Jack is on a bearing of 260c from Jill. What is Jill’s bearing from Jack?

4.

A tower is on a bearing of 030c from a house. What is the bearing of the house from the tower?

5.

Tamworth is on a bearing of 340c from Newcastle. What is the bearing of Newcastle from Tamworth?

6.

7.

8.

The angle of elevation from a point 11.5 m away from the base of a tree up to the top of the tree is 42c 12l. Find the height of the tree to one decimal point. Geoff stands 25.8 m away from the base of a tower and measures the angle of elevation as 39c 20l. Find the height of the tower to the nearest metre. A wire is suspended from the top of a 100 m tall bridge tower down to the bridge at an angle of elevation of 52c. How long is the wire, to 1 decimal place?

10. A plane leaves Melbourne and flies on a bearing of 065c for 2500 km. (a) How far north of Melbourne is the plane? (b) How far east of Melbourne is it? (c) What is the bearing of Melbourne from the plane? 11. The angle of elevation of a tower is 39c 44l when measured at a point 100 m from its base. Find the height of the tower, to 1 decimal place. 12. Kim leaves his house and walks for 2 km on a bearing of 155c . How far south is Kim from his house now, to 1 decimal place? 13. The angle of depression from the top of an 8 m tree down to a rabbit is 43c 52l. If an eagle is perched in the top of the tree, how far does it need to fly to reach the rabbit, to the nearest metre? 14. A girl rides a motorbike through her property, starting at her house. If she rides south for 1.3 km, then rides west for 2.4 km, what is her bearing from the house, to the nearest degree? 15. A plane flies north from Sydney for 560 km, then turns and flies east for 390 km. What is its bearing from Sydney, to the nearest degree?

9.

A cat crouches at the top of a 4.2 m high cliff and looks down at a mouse 1.3 m out from the foot (base) of the cliff. What is the angle of depression, to the nearest minute?

16. Find the height of a pole, correct to 1 decimal place, if a 10 m rope tied to it at the top and stretched out straight to reach the ground makes an angle of elevation of 67c13l.


17. The angle of depression from the top of a cliff down to a boat 100 m out from the foot of the cliff is 59c42l. How high is the cliff, to the nearest metre? 18. A group of students are bushwalking. They walk north from their camp for 7.5 km, then walk west until their bearing from camp is 320c. How far are they from camp, to 1 decimal place? 19. A 20 m tall tower casts a shadow 15.8 m long at a certain time of day. What is the angle of elevation from the edge of the shadow up to the top of the tower at this time?

20 m

15.8 m

20. A flat verandah roof 1.8 m deep is 2.6 m up from the ground. At a certain time of day, the sun makes an angle of elevation of 72c 25l. How much shade is provided on the ground by the verandah roof at that time, to 1 decimal place?

21. Find the angle of elevation of a 15.9 m cliff from a point 100 m out from its base. 22. A plane leaves Sydney and flies for 2000 km on a bearing of 195c. How far due south of Sydney is it? 23. The angle of depression from the top of a 15 m tree down to a pond is 25c41l. If a bird is perched in the top of the tree, how far does it need to fly to reach the pond, to the nearest metre? 24. A girl starting at her house, walks south for 2.7 km then walks east for 1.6 km. What is her bearing from the house, to the nearest degree? 25. The angle of depression from the top of a tower down to a car 250 m out from the foot of the tower is 38c19l. How high is the tower, to the nearest metre? 26. A hot air balloon flies south for 3.6 km then turns and flies east until it is on a bearing of 127c from where it started. How far east does it fly? 27. A 24 m wire is attached to the top of a pole and runs down to the ground where the angle of elevation is 22c 32l. Find the height of the pole.

317

318


28. A train depot has train tracks running north for 7.8 km where they meet another set of tracks going east for 5.8 km into a station. What is the bearing of the depot from the station, to the nearest degree? 29. Jessica leaves home and walks for 4.7 km on a bearing of 075c. She then turns and walks for 2.9 km on a bearing of 115c and she is then due east of her home. (a) How far north does Jessica walk? (b) How far is she from home?

30. Builder Jo stands 4.5 m out from the foot of a building and looks up at to the top of the building where the angle of elevation is 71c. Builder Ben stands at the top of the building looking down at his wheelbarrow that is 10.8 m out from the foot of the building on the opposite side from where Jo is standing. (a) Find the height of the building. (b) Find the angle of depression from Ben down to his wheelbarrow.

Exact Ratios A right-angled triangle with one angle of 45° is isosceles. The exact length of its hypotenuse can be found.

Pythagoras’ theorem is used to find the length of the hypotenuse.

c2 = a2 + b2 AC 2 = 1 2 + 1 2 =2 AC =

2

This means that the trigonometric ratios of 45c can be written as exact ratios.

1 2 1 cos 45c = 2 tan 45c = 1 sin 45c =


319

This angle is commonly used; for example, 45° is often used for the pitch of a roof. The triangle with angles of 60° and 30° can also be written with exact sides.

Halve the equilateral triangle to get TABD.

AD 2 = 2 2 - 1 2 =3 AD =

3

3 2 1 cos 60° = 2 tan 60° = 3

sin 30c =

sin 60° =

1 2

3 2 1 tan 30c = 3

It may be easier to remember the triangle rather than all these ratios.

cos 30c =

DID YOU KNOW? The ratios of all multiples of these angles follow a pattern: A

0c

30c

45c

60c

90c

120c

135c

150c

sin A

0 2

1 2

2 2

3 2

4 2

3 2

2 2

1 2

cos A

4 2

3 2

2 2

1 2

0 2

- 1 2

- 2 2

- 3 2

The rules of the pattern are: • for sin A, when you reach 4, reverse the numbers • for cos A, when you reach 0, change signs and reverse

320


EXAMPLES 1. Find the exact value of sec 45°.

Solution 1 cos 45° 1 = 1 2 = 2

sec 45° =

2. A boat ramp is to be made with an angle of 30c and base length 5 m. What is the exact length of the surface of the ramp?

Solution 5 cos 30c = x x cos 30c = 5 5 cos 30c 5 = 3 2 2 =5# 3 10 = 3 10 3 = 3

x=

So the exact length of the ramp is

10 3 m. 3


321

6.6 Exercises Find the exact value in all questions, with rational denominator where relevant. 1.

(b)

Evaluate (a) sin 60c + cos 60c (b) cos 2 45c + sin 2 45c (c) cosec 45c (d) 2 sec 60c (e) cot 30c + cot 60c

cos 2 45c = (cos 45c) 2

(c)

(f) tan 60c - tan 30c (g) sin 2 60c + sin 2 45c (h) sin 45c cos 30c + cos 45c sin 30c (i) 3 tan 30c tan 45c + tan 60c (j) 1 - tan 45c tan 60c

3.

(k) cos 30c cos 60c - sin 30c sin 60c 4. (l) cos 2 30c + sin 2 30c (m) 2 sec 45c - cosec 30c 2 sin 60c sin 45c (o) 1 + tan 2 30c (n)

(p)

1 - cos 45c 1 + cos 45c

(q)

cot 30c sec 60c

(r) sin 2 45c - 1

2.

If the tent in the previous question was pitched at an angle of 60c, how high would the pole need to be?

6.

The angle of elevation from a point 10 m out from the base of a tower to the top of the tower is 30c. Find the exact height of the tower, with rational denominator.

2 - tan 60c sec 2 45c

Find the exact value of all pronumerals (a)

A 2-person tent is pitched at an angle of 45c. Each side of the tent is 2 m long. A pole of what height is needed for the centre of the tent?

5.

(s) 5 cosec 2 60c (t)

A 2.4 m ladder reaches 1.2 m up a wall. At what angle is it resting against the wall?

322


7.

the floor. How far out from the wall is it?

The pitch of a roof is 45c and spans a length of 12 m.

Find the exact length of AC.

9.

(a) What is the length l of the roof? (b) If a wall is placed inside the roof one third of the way along from the corner, what height will the wall be? 8.

A 1.8 m ladder is placed so that it makes a 60c angle where it meets

10. The angle of depression from the top of a 100 m cliff down to a boat at the foot of the cliff is 30c. How far out from the cliff is the boat?

Angles of Any Magnitude The angles in a right-angled triangle are always acute. However, angles greater than 90c are used in many situations, such as in bearings. Negative angles are also used in areas such as engineering and science. We can use a circle to find trigonometric ratios of angles of any magnitude (size) up to and beyond 360c.

Investigation 1. (a) Copy and complete the table for these acute angles (between 0c and 90c). x

0c

10c

20c

30c

40c

50c

60c

70c

80c

90c

sin x cos x tan x (b) Copy and complete the table for these obtuse angles (between 90c and 180c). x sin x cos x tan x

100c

110c

120c

130c

140c

150c

160c

170c

180c


(c) Copy and complete the table for these reflex angles (between 180c and 270c). x

190c

200c

210c

220c

230c

240c

250c

260c

270c

350c

360c

sin x cos x tan x (d) Copy and complete the table for these reflex angles (between 270c and 360c). x

280c

290c

300c

310c

320c

330c

340c

sin x cos x tan x 2. What do you notice about their signs? Can you see any patterns? Could you write down any rules for the sign of sin, cos and tan for different angle sizes? 3. Draw the graphs of y = sin x, y = cos x and y = tan x for 0c # x # 360c. For y = tan x, you may need to find the ratios of angle close to and either side of 90c and 270c.

Drawing the graphs of the trigonometric ratios can help us to see the change in signs as angles increase. We divide the domain 0c to 360c into 4 quadrants:

1st quadrant: 0c to 90c 2nd quadrant: 90c to 180c 3rd quadrant: 180c to 270c 4th quadrant: 270c to 360c

EXAMPLES 1. Describe the sign of sin x in each section (quadrant) of the graph y = sin x.

Solution We can sketch the graph using the table below or using the values from the tables in the investigation above for more accuracy. x

0c

90c

180c

y

0

1

0

270c -1

360c 0 CONTINUED

323

324


y

1 y = sin x

90c

180c

270c

360c

x

-1

The graph is above the x-axis for the first 2 quadrants, then below for the 3rd and 4th quadrants. This means that sin x is positive in the 1st and 2nd quadrants and negative in the 3rd and 4th quadrants. 2. Describe the sign of cos x in each section (quadrant) of the graph of y = cos x.


0c

90c

y

1

0

180c -1

270c

360c

0

1

y

y = cos x

1

90c

180c

270c

360c

x

-1

The graph is above the x-axis in the 1st quadrant, then below for the 2nd and 3rd quadrants and above again for the 4th quadrant.


325

This means that cos x is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd quadrants. 3. Describe the sign of tan x in each section (quadrant) of the graph y = tan x.


0c

90c

180c

270c

360c

y

0

No result

0

No result

0

Neither tan 90c nor tan 270c exists (we say that they are undefined). Find the tan of angles close to these angles, for example tan 89c 59l and tan 90c 01l, tan 279c 59l and tan 270c 01l. There are asymptotes at 90c and 270c. On the left of 90c and 270c, tan x is positive and on the right, the ratio is negative. y

90c

180c

270c

360c

x

y = tan x

The graph is above the x-axis in the 1st quadrant, below for the 2nd, above for the 3rd and below for the 4th quadrant. This means that tan x is positive in the 1st and 3rd quadrants and negative in the 2nd and 4th quadrants.

To show why these ratios have different signs in different quadrants, we look at angles around a unit circle (a circle with radius 1 unit). We use congruent triangles when finding angles of any magnitude. Page 326 shows an example of congruent triangles all with angles of 20c inside a circle with radius 1 unit.

You will see why these ratios are undefined later on in this chapter.

326


y

1 unit

1 unit 20c 20c

20c 20c 1 unit

x

1 unit

If we divide the circle into 4 quadrants, we notice that the x- and y-values have different signs in different quadrants. This is crucial to notice when looking at angles of any magnitude and explains the different signs you get when finding sin, cos and tan for angles greater than 90c.

Quadrant 1 Looking at the first quadrant (see diagram below), notice that x and y are both positive and that angle i is turning anticlockwise from the x-axis. y First quadrant The angle at the x-axis is 0 and the angle at the y-axis is 90c, with all other angles in this quadrant between these two angles.

(x, y) 1 unit

y

i x

x

Point (x, y) forms a triangle with sides 1, x and y, so we can find the trigonometric ratios for angle i.


327

y 1 =y

sin i =

x 1 =x

cos i =

y tan i = x

Investigation Since cos i = x and sin i = y, we can write the point (x, y) as (cos i, sin i). The polar coordinates (cos i, sin i) give a circle. The polar coordinates 6 A sin ] ai + c g, B sin ] bi g @ form a shape called a Lissajous figure. These are sometimes called a Bowditch curve and they are often used as logos, for example the ABC logo. Use the Internet to research these and other similar shapes. Use a graphics calculator or a computer program such as Autograph to draw other graphs with polar coordinates using variations of sin i and cos i.

Quadrant 2 In the second quadrant, angles are between 90c and 180c. If we take the 1st quadrant coordinates (x, y), where x 2 0 and y 2 0 and put them in the 2nd quadrant, we notice that all x values are negative in the second quadrant and y values are positive. So the point in the 2nd quadrant will be (-x, y) y 90c Second quadrant (-x, y) y 180c

1 unit

x

i

180c- i

0c

x

These are called polar coordinates.

328


Since cos i = x, cos i will negative in the 2nd quadrant. Since sin i = y , sin i will be positive in the 2nd quadrant. y tan i = x so it will be negative (a positive number divided by a negative number). To have an angle of i in the triangle, the angle around the circle is 180c - i.

Quadrant 3 In the third quadrant, angles are between 180c and 270c. y

90c

180c

180c + i

y

0c

i

x

x

1 unit

(-x, -y) Third quadrant

270c

Notice that x and y are both negative in the third quadrant, so cos i and sin i will be both negative. y tan i = x so will be positive (a negative divided by a negative number). To have an angle of i in the triangle, the angle around the circle is 180c + i.

Quadrant 4 In the fourth quadrant, angles are between 270c and 360c. y 90c

180c

i 360c - i

0c

x

1 unit

360c y (x, -y)

270c

Fourth quadrant

x


329

While y remains negative in the fourth quadrant, x is positive again, so sin i is negative and cos i is positive. y tan i = x so will be negative (a negative divided by a positive number) For an angle i in the triangle, the angle around the circle is 360c - i.

ASTC rule Putting all of these results together gives a rule for all four quadrants that we usually call the ASTC rule. y 90c

2nd quadrant

1st quadrant

180c - i

You could remember this rule as All Stations To Central or A Silly Trigonometry Concept, or you could make up your own!

i

S

A 0c

180c

360c

T

180c + i

C

3rd quadrant

x

360c - i 4th quadrant

270c

A: ALL ratios are positive in the 1st quadrant S: Sin is positive in the 2nd quadrant (cos and tan are negative) T: Tan is positive in the 3rd quadrant (sin and cos are negative) C: Cos is positive in the 4th quadrant (sin and tan are negative) This rule also works for the reciprocal trigonometric ratios. For example, where cos is positive, sec is also positive, where sin is positive, so is cosec and where tan is positive, so is cot. We can summarise the ASTC rules for all 4 quadrants:

First quadrant: Angle i: sin i is positive cos i is positive tan i is positive

330


Second quadrant: Angle 180c - i: sin ] 180c - i g = sin i cos ] 180c - i g = - cos i tan ] 180c - i g = - tan i Third quadrant: Angle 180c + i: sin ] 180c + i g = - sin i cos ] 180c + i g = - cos i tan ] 180c + i g = tan i Fourth quadrant: Angle 360c - i: sin ] 360c - i g = - sin i cos ] 360c - i g = cos i tan ] 360c - i g = - tan i

EXAMPLES 1. Find all quadrants where (a) sin i 2 0 (b) cos i 1 0 (c) tan i 1 0 and cos i 2 0

Solution (a) sin i 2 0 means sin i is positive. Using the ASTC rule, sin i is positive in the 1st and 2nd quadrants. (b) cos i is positive in the 1st and 4th quadrants, so cos i is negative in the 2nd and 3rd quadrants. (c) tan i is positive in the 1st and 3rd quadrants so tan i is negative in the 2nd and 4th quadrants. Also cos i is positive in the 1st and 4th quadrants. So tan i 1 0 and cos i 2 0 in the 4th quadrant.


331

2. Find the exact ratio of tan 330c.

Solution First we find the quadrant that 330c is in. It is in the 4th quadrant. y

330c

30c

x

The angle inside the triangle in the 4th quadrant is 30c and tan is negative in the 4th quadrant.

Notice that 360c - 30c = 330c.

tan 330c = - tan 30c 1 =3

30c

:3

2

60c

1

3. Find the exact value of sin 225c.

Solution The angle in the triangle in the 3rd quadrant is 45c and sin is negative in the 3rd quadrant. CONTINUED

Notice that 180c + 45c = 225c.

332


y

225c

x

45c

sin 225c = - sin 45c 1 =2

45c

:2

45c

1

1

4. Find the exact value of cos 510c.

Solution To find cos 510c, we move around the circle more than once. y

30c

150c 510c

510c - 360c = 150c So 510c = 360c + 150c

x


The angle is in the 2nd quadrant where cos is negative. The triangle has 30c in it.

30c

cos 510c = - cos 30c =-

333

3 2

:3

2

60c

Notice that 180c - 30c = 150c.

1

5. Simplify cos (180c + x).

Solution 180c + x is an angle in the 3rd quadrant where cos is negative. So cos ] 180c + x g = - cos x 6. If sin x = -

3 and cos x 2 0, find the value of tan x and sec x. 5

Solution sin x 1 0 in the 3rd and 4th quadrants and cos x 2 0 in the 1st and 4th quadrants. So sin x 1 0 and cos x 2 0 in the 4th quadrant. This means that tan x 1 0 and sec x 2 0. sin x =

sec x is the reciprocal of cos x so is positive in the 4th quadrant.

opposite hypotenuse

So the opposite side is 3 and the hypotenuse is 5. y

x

x 5

3

This is a 3-4-5 triangle.

By Pythagoras’ theorem, the adjacent side is 4. CONTINUED

334


3 4 1 sec x = cos x 5 = 4

So tan x = -

The ASTC rule also works for negative angles. These are measured in the opposite way (clockwise) from positive angles as shown. y -270c

2nd quadrant -(180c+ i )

S

1st quadrant

A

-360c 0

-180c

-(180c- i )

-(360c- i )

T

C

3rd quadrant

x

-i 4th quadrant

-90c

The only difference with this rule is that the angles are labelled differently.

EXAMPLE Find the exact value of tan (-120c).

Solution Notice that - (180c - 60c) = -120c.

Moving around the circle the opposite way, the angle is in the 3rd quadrant, with 60c in the triangle. y

60c

120c

x


Tan is positive in the 3rd quadrant. tan ] -120c g = tan 60c =

3 30c

:3

2

60c

1

6.7 Exercises 1.

Find all quadrants where (a) cos i 2 0 (b) tan i 2 0 (c) sin i 2 0 (d) tan i 1 0 (e) sin i 1 0 (f) cos i 1 0 (g) sin i 1 0 and tan i 2 0 (h) cos i 1 0 and tan i 2 0 (i) sin i 2 0 and tan i 1 0 (j) sin i 1 0 and tan i 1 0

2.

(a) Which quadrant is the angle 240c in? (b) Find the exact value of cos 240c.

3.

(a) Which quadrant is the angle 315c in? (b) Find the exact value of sin 315c.

4.

(a) Which quadrant is the angle 120c in? (b) Find the exact value of tan 120c .

5.

(a) Which quadrant is the angle -225c in? (b) Find the exact value of sin (-225c).

6.

(a) Which quadrant is the angle -330c in? (b) Find the exact value of cos (-330c).

7.

Find the exact value of each ratio. (a) tan 225c (b) cos 315c (c) tan 300c (d) sin 150c (e) cos 120c (f) sin 210c (g) cos 330c (h) tan 150c (i) sin 300c (j) cos 135c

8.

Find the exact value of each ratio. (a) cos (-225c) (b) cos (-210c) (c) tan (-300c) (d) cos (-150c) (e) sin (-60c) (f) tan (-240c) (g) cos (-300c) (h) tan (-30c) (i) cos (-45c) (j) sin (-135c)

335

336


9.

Find the exact value of (a) cos 570c (b) tan 420c (c) sin 480c (d) cos 660c (e) sin 690c (f) tan 600c (g) sin 495c (h) cos 405c (i) tan 675c (j) sin 390c 3 and cos i 1 0, find 4 sin i and cos i as fractions.

10. If tan i =

Use Pythagoras’ theorem to find the third side.

4 11. Given sin i = and tan i 1 0, 7 find the exact value of cos i and tan i. 5 12. If sin x 1 0 and tan x = - , find 8 the exact value of cos x and cosec x. 2 and tan x 1 0, 5 find the exact value of cosec x, cot x and tan x.

13. Given cos x =

14. If cos x 1 0 and sin x 1 0, find cos x and sin x in surd form with 5 rational denominator if tan x = . 7

4 and 9 270c 1 i 1 360c, find the exact

15. If sin i = -

value of tan i and sec i. 3 and 8 180° 1 i 1 270°, find the exact value of tan x, sec x and cosec x.

16. If cos i = -

17. Given sin x = 0.3 and tan x 1 0, (a) express sin x as a fraction (b) find the exact value of cos x and tan x. 18. If tan a = - 1.2 and 270° 1 i 1 360°, find the exact values of cot a, sec a and cosec a. 19. Given that cos i = - 0.7 and 90c 1 i 1 180c , find the exact value of sin i and cot i. 20. Simplify (a) sin ] 180c - i g (b) cos ] 360c - x g (c) (d) (e) (f) (g) (h)

tan ^ 180c + b h sin ] 180c + a g tan ] 360c - i g sin ] - i g cos ] - a g tan ] - x g

Trigonometric Equations This is called the principle solution.

Whenever you find an unknown angle in a triangle, you solve a trigonometric equation e.g. cos x = 0.34. You can find this on your calculator. Now that we know how to find the trigonometric ratios of angles of any magnitude, there can be more than one solution to a trigonometric equation if we look at a larger domain.


337

EXAMPLES 1. Solve cos x =

3 in the domain 0° # x # 360°. 2

Solution 3 is a positive ratio and cos is positive in the 1st and 4th quadrants. 2 So there are two possible answers. In the 1st quadrant, angles are in the form of i and in the 4th quadrant angles are in the form of 360c - i. cos 30c =

3 2

30c

But there is also a solution in the 4th quadrant where the angle is 360c - i. 3 2 x = 30c , 360c - 30c = 30c , 330c

:3

2

For cos x =

60c

1

2. Solve 2 sin 2 x - 1 = 0 for 0c # x # 360c.

Solution 2 sin 2 x - 1 = 0 2 sin 2 x = 1 1 sin 2 x = 2 sin x = !

This is called the principle solution.

1

2 1 =! 2 Since the ratio could be positive or negative, there are solutions in all 4 quadrants. 1st quadrant: angle i 2nd quadrant: angle 180c - i 3rd quadrant: angle 180c + i 4th quadrant: angle 360c - i CONTINUED

338


1 2 x = 45c , 180c - 45c , 180c + 45c , 360c - 45c = 45c , 135c , 225c , 315c

sin 45c =

45c

:2

45c

3. Solve tan x =

1

1

3 for - 180c # x # 180c.

Solution 3 is a positive ratio and tan is positive in the 1st and 3rd quadrants. So there are two possible answers. In the domain - 180c # x # 180c, we use positive angles for 0c # x # 180c and negative angles for - 180c # x # 0c. y 90c

2nd quadrant

1st quadrant

180c - i

i

S

A

180c

0c

-180c -(180c - i)

x

0c

T

C

3rd quadrant

-i 4th quadrant

-90c

In the 1st quadrant, angles are in the form of i and in the 3rd quadrant angles are in the form of - ^ 180c - i h . tan 60c = 3 But there is also a solution in the 3rd quadrant where the angle is - ^ 180c - i h . For tan x = 3 x = 60c , - ] 180c - 60c g = 30c , -120c

30c

:3

2

60c

1


339

4. Solve 2 sin 2x - 1 = 0 for 0c # x # 360c.

Solution Notice that the angle is 2x but the domain is for x. If 0c # x # 360c then we multiply each part by 2 to get the domain for 2x. 0c # 2x # 720c This means that we can find the solutions by going around the circle twice!

30c

2 sin 2x - 1 = 0 2 sin 2x = 1

:3

2

1 2 1 sin 30c = 2 sin 2x =

60c

1

Sin is positive in the 1st and 2nd quadrants. First time around the circle, 1st quadrant is i and the 2nd quadrant is 180c - i. Second time around the circle, we add 360c to the angles. So 1st quadrant answer is 360c + i and the 2nd quadrant answer is 360c + ] 180c - i g or 540c - i. So 2x = 30c , 180c - 30c, 360c + 30c , 540c - 30c = 30c , 150c , 390c , 510c ` x = 15c , 75c , 195c , 255c

The trigonometric graphs can also help solve some trigonometric equations.

EXAMPLE Solve cos x = 0 for 0c # x # 360c. cos 90c = 0 However, looking at the graph of y = cos x shows that there is another solution in the domain 0c # x # 360c. y

1

90c -1

For cos x = 0 x = 90c, 270c

180c

270c 360c

x

Notice that these solutions lie inside the original domain of 0c # x # 360c.

340


Investigation Here are the 3 trigonometric graphs that you explored earlier in the chapter. y = sin x

y = cos x

y = tan x

Use the values in the sin, cos and tan graphs to find values for the inverse trigonometric functions in the tables below and then sketch the inverse trigonometric functions. For example sin 270° = -1 1 So cosec 270c = -1 = -1 Some values will be undefined, so you will need to find values near them in order to see where the graph goes. y = cosec x x sin x cosec x

0c

90c

180c

270c

360c


y = sec x x

0c

90c

180c

270c

360c

0c

90c

180c

270c

360c

cos x sec x y = cot x x tan x cot x

Here are the graphs of the inverse trigonometric functions. y = cosec x

y = sec x

y = cot x

y y = cotx 1

0 -1

90c

180c 270c

360c

x 360c

341

342


6.8 Exercises 1.

Solve for 0c # i # 360c. (a) sin i = 0.35 1 (b) cos i = 2 (c) tan i = - 1 3 (d) sin i = 2 1 (e) tan i = 3 (f) 2 cos i = 3 (g) tan 2i =

3

(h) 2 sin 3i = - 1 (i) 2 cos 2i - 1 = 0 (j) tan 2 3i = 1 2.

Solve for -180c # i # 180c. (a) cos i = 0.187 1 (b) sin i = 2 (c) tan i = 1 3 (d) sin i = 2 1 (e) tan i = 3 (f) 3 tan 2 i = 1 (g) tan 2i = 1 (h) 2 sin 2 3i = 1 (i) tan i + 1 = 0 (j) tan 2 2i = 3

3.

Sketch y = cos x for 0c # x # 360c .

4.

Evaluate sin 270c .

5.

Sketch y = tan x for 0c # x # 360c .

6.

Solve tan x = 0 for 0c # x # 360c .

7.

Evaluate cos 180c .

8.

Find the value of sin 90c .

9.

Solve cos x = 1 for 0c # x # 360c .

10. Sketch y = sin x for -180c # x # 180c . 11. Evaluate cos 270c. 12. Solve sin x + 1 = 0 for 0c # x # 360c . 13. Solve cos 2 x = 1 for 0c # x # 360c . 14. Solve sin x = 0 for 0c # x # 360c . 15. Solve sin x = 1 for - 360c # x # 360c . 16. Sketch y = sec x for 0c # x # 360c . 17. Sketch y = cot x for 0c # x # 360c .

Trigonometric Identities Trigonometric identities are statements about the relationships of trigonometric ratios. You have already met some of these—the reciprocal ratios, complementary angles and the rules for the angle of any magnitude.


Reciprocal ratios

1 sin i 1 sec i = cos i 1 cot i = tan i

cosec i =

Complementary angles

sin i = cos ] 90c - i g cosec i = sec ] 90c - i g tan i = cot ] 90c - i g

Angles of any magnitude

sin ] 180c - i g = sin i cos ] 180c - i g = - cos i tan ] 180c - i g = - tan i sin (180c + i) = - sin i cos (180c + i) = - cos i tan (180c + i) = tan i sin (360c - i) = - sin i cos (360c - i) = cos i tan (360c - i) = - tan i sin (- i) = - sin i cos (- i) = cos i tan (- i) = - tan i

In this section you will learn some other identities, based on the unit circle. In the work on angles of any magnitude, we defined sin i as the y-coordinate of P and cos i as the x-coordinate of P.

343

344


y tan i = x sin i = cos i

tan i =

sin i cos i

cot i =

cos i sin i

1 tan i cos i = sin i

cot i =

Pythagorean identities The circle has equation x 2 + y 2 = 1. Substituting x = cos i and y = sin i into x 2 + y 2 = 1 gives Remeber that cos 2 i means (cos i) 2.

cos 2 i + sin 2 i = 1

This is an equation so can be rearranged to give sin 2 i = 1 - cos 2 i cos 2 i = 1 - sin 2 i There are two other identities that can be derived from this identity.

1 + tan 2 i = sec 2 i


345

Proof cos 2 i + sin 2 i = 1 cos 2 i sin 2 i 1 + = cos 2 i cos 2 i cos 2 i 1 + tan 2 i = sec 2 i This identity can be rearranged to give tan 2 i = sec 2 i - 1 1 = sec 2 i - tan 2 i cot 2 i + 1 = cosec 2 i

Proof cos 2 i + sin 2 i = 1 cos 2 i sin 2 i 1 + = 2 2 sin i sin i sin 2 i 2 cot i + 1 = cosec 2 i This identity can be rearranged to give

These are called Pythagorean identities since the equation of the circle comes from Pythagoras’ rule (see Chapter 5).

cot 2 i = cosec 2 i - 1 1 = cosec 2 i - cot 2 i

EXAMPLES 1. Simplify sin i cot i.

Solution sin i cot i = sin i # = cos i

cos i sin i

2. Simplify sin ^ 90c - b h sec b where b is an acute angle.

Solution sin ^ 90c - b h sec b = cos b # =1

1 cos b

CONTINUED

346


3. Simplify

sin 4 i + sin 2 i cos 2 i .

Solution sin 4 i + sin 2 i cos 2 i = sin 2 i ^ sin 2 i + cos 2 i h = sin 2 i ] 1 g = sin 2 i = sin i 4. Prove cot x + tan x = cosec x sec x.

Solution LHS = cot x + tan x cos x sin x = + sin x cos x cos 2 x + sin 2 x sin x cos x 1 = sin x cos x 1 1 = # cos x sin x = cosec x sec x = RHS =

` cot x + tan x = cosec x sec x 5. Prove that

1 - cos x 1 = . 1 + cos x sin 2 x

Solution 1 - cos x sin 2 x 1 - cos x = 1 - cos 2 x 1 - cos x = ] 1 + cos x g ] 1 - cos x g 1 = 1 + cos x = RHS

LHS =

`

1 - cos x 1 = 2 1 cos x + sin x


6.9 Exercises 1.

2.

Simplify (a) sin ] 90c - i g (b) tan ] 360c - i g (c) cos ] - i g (d) cot ] 90c - i g (e) sec ] 180c + a g

= cosec 2 x - cot 2 x (e) ] sin x - cos x g3 = sin x - cos x - 2 sin 2 x cos x +2 sin x cos 2 x (f) cot i + 2 sec i 1 - sin 2 i + 2 sin i sin i cos i (g) cos 2 ] 90c - i g cot i

Simplify (a) tan i cos i (b) tan i cosec i (c) sec x cot x (d) 1 - sin 2 x 1 - cos a cot 2 x + 1 1 + tan 2 x sec 2 i - 1 5 cot 2 i + 5 1 (j) cosec 2 x (k) sin 2 a cosec 2 a (l) cot i - cot i cos 2 i (e) (f) (g) (h) (i)

3.

(d) sec 2 x - tan 2 x

=

= sin i cos i (h) (cosec x + cot x) (cosec x - cot x) = 1

2

Prove that (a) cos 2 x - 1 = - sin 2 x 1 + sin i (b) sec i + tan i = cos i 3 2 (c) 3 + 3 tan a = 1 - sin 2 a

1 - sin 2 i cos 2 i cos 2 i 2 = tan i + cos 2 i 1 + cot b (j) - cos b cosec b sec b = tan b + cot b (i)

4.

If x = 2 cos i and y = 2 sin i, show that x 2 + y 2 = 4.

5.

Show that x 2 + y 2 = 81 if x = 9 cos i and y = 9 sin i.

Non-right-angled Triangle Results A non-right-angled triangle is named so that its angles and opposite sides have the same pronumeral. There are two rules in trigonometry that refer to nonright-angled triangles. These are the sine rule and the cosine rule.

347

348


Sine rule

sin A sin B sin C a = b = c

Use this rule for finding an angle.

Use this rule for finding a side.

a c b = = sin A sin B sin C

or

Proof

In TABC, draw perpendicular AD and call it h. From TABD, h sin B = c ` h = c sin B

(1)

From TACD, h b h = b sin C

sin C = `

From (1) and (2), c sin B = b sin C sin B sin C = c b Similarly, drawing a perpendicular from C it can be proven that sin A sin B a = b .

(2)


349

EXAMPLES The sine rule uses 2 sides and 2 angles, with 1 unknown.

1. Find the value of x, correct to 1 decimal place.

Solution Name the sides a and b, and angles A and B. a b = sin A sin B 10.7 x = sin 43c 21l sin 79c 12l 10.7 x sin 43c 21l # = sin 43c 21l # sin 43c 21l sin 79c 12l 10.7 sin 43c 21l x= sin 79c 12l Z 7.5 cm 2. Find the value of y, to the nearest whole number. You can rename the triangle ABC or just make sure you put sides with their opposite angles together.

Solution +Y = 180c - (53c + 24c ) = 103c

You need to find +Y first, as it is opposite y.

a b = sin A sin B y 8 = sin 103c sin 53c y 8 = sin 103c # sin 103c # sin 103c sin 53c 8 sin 103c y= sin 53c Z 10

CONTINUED

350


3. Find the value of i, in degrees and minutes.

Solution sin A sin B a = b sin i sin 86c 11l = 6.7 8.3 sin i sin 86c 11l = 6.7 # 6.7 # 6 .7 8.3 6.7 sin 86c 11l sin i = 8 .3 - 1 6.7 sin 86c 11l i = sin c m 8 .3 Z 53c39l

Since sin x is positive in the first 2 quadrants, both acute angles (between 0c and 90c) and obtuse angles (between 90c and 180c) give positive sin ratios. e.g. sin 50c = 0.766 and sin 130c = 0.766 This affects the sine rule, since there is no way of distinguishing between an acute angle and an obtuse angle. When doing a question involving an obtuse angle, we need to use the 2nd quadrant angle of 180c - i rather than relying on the calculator to give the correct answer.

EXAMPLE Angle i is obtuse. Find the value of i, in degrees and minutes.


Solution sin A sin B a = b sin i sin 15c 49l = 5.4 11.9 sin i sin 15c 49l = 11.9 # 11.9 # 5.4 11.9 11.9 sin 15c 49l sin i = 5.4 - 1 11.9 sin 15c 49l m i = sin c 5.4 = 36c 55l ^ acute angle h But i is obtuse ` i = 180c - 36c 55l = 143c 05l

6.10 1.

Exercises

Evaluate all pronumerals, correct to 1 decimal place.

(c)

(a)

(d)

(b)

(e)

351

352


2.

BC = 4.6 cm and +ACB = 33c 47l.

Find the value of all pronumerals, in degrees and minutes. (a)

4.

Triangle EFG has +FEG = 48c , +EGF = 32c and FG = 18.9 mm. Find the length of (a) the shortest side (b) the longest side..

5.

Triangle XYZ has +XYZ = 51c , +YXZ = 86c and XZ = 2.1 m. Find the length of (a) the shortest side (b) the longest side.

6.

Triangle XYZ has XY = 5.4 cm, +ZXY = 48c and +XZY = 63c. Find the length of XZ.

7.

Triangle ABC has BC = 12.7 m, +ABC = 47c and +ACB = 53c as shown. Find the lengths of (a) AB (b) AC.

The shortest side is opposite the smallest angle and the longest side is opposite the largest angle.

(b)

(c)

(d)

A

B

(e) (i is obtuse)

C

9.

Triangle ABC is isosceles with AB = AC. BC is produced to D as shown. If AB = 8.3 cm, +BAC = 52c and +ADC = 32c find the length of

i

Triangle ABC has an obtuse angle at A. Evaluate this angle to the nearest minute if AB = 3.2 cm,

53c

Triangle PQR has sides PQ = 15 mm, QR = 14.7 mm and +PRQ = 62c 29l. Find to the nearest minute (a) +QPR (b) +PQR.

3.7

3.

12.7 m

8.

4.9 21c31l

47c


(a) AD (b) BD. A 52c

8.3 cm

B

32c

C

10. Triangle ABC is equilateral with side 63 mm. A line is drawn from A to BC where it meets BC at D and +DAB = 26c 15l. Find the length of (a) AD (b) DC. D

Cosine rule c 2 = a 2 + b 2 - 2ab cos C

Similarly a 2 = b 2 + c 2 - 2bc cos A b 2 = a 2 + c 2 - 2ac cos B

Proof A

c

b

C

x

p

D

a-x

B

In triangle ABC, draw perpendicular CD with length p and let CD = x. Since BC = a, BD = a - x From triangle ACD b2 = x2 + p2 x cos C = b ` b cos C = x

(1)

(2)

From triangle DAB c2 = p2 + ] a - x g 2 = p 2 + a 2 - 2ax + x 2 = p 2 + x 2 + a 2 - 2ax

(3)

353

354


Substitute (1) into (3): c 2 = b 2 + a 2 - 2ax

(4)

Substituting (2) into (4): c 2 = b 2 + a 2 - 2a ] b cos C g = b 2 + a 2 - 2ab cos C

DID YOU KNOW? Pythagoras’ theorem is a special case of the cosine rule when the triangle is right angled. c 2 = a 2 + b 2 - 2ab cos C When C = 90c c 2 = a 2 + b 2 - 2ab cos 90c = a 2 + b 2 - 2ab ] 0 g = a2 + b2

EXAMPLE Find the value of x, correct to the nearest whole number. The cosine rule uses 3 sides and 1 angle, with 1 unknown.

Solution c 2 = a 2 + b 2 - 2ab cos C x 2 = 5.6 2 + 6.4 2 - 2 (5.6) (6.4) cos 112c 32l Z 99.79 x = 99.79 Z 10 Press 5.6 x 2 + 6.4 x 2 - 2 # 5.6 # 6.4 # cos 112 % , ,, 32 % , ,, =

=


355

When finding an unknown angle, it is easier to change the subject of this formula to cos C. c 2 = a 2 + b 2 - 2ab cos C c 2 + 2ab cos C = a 2 + b 2 - 2ab cos C + 2ab cos C c 2 + 2ab cos C = a 2 + b 2 c 2 - c 2 + 2ab cos C = a 2 + b 2 - c 2 2ab cos C = a 2 + b 2 - c 2 2ab cos C a2 + b2 - c2 = 2ab 2ab cos C =

a2 + b2 - c2 2ab

Similarly cos A =

b2 + c2 - a2 2bc

cos B =

a +c -b 2ac 2

2

Subtract the square of the side opposite the unknown angle.

2

EXAMPLES 1. Find i, in degrees and minutes.

Solution a2 + b2 - c2 2ab 52 + 62 - 32 cos i = 2 ]5 g]6 g 52 = 60 52 m i = cos - 1 c 60 Z 29c 56l

cos C =

2. Evaluate +BAC in degrees and minutes. A 6.1 cm

4.5 cm B

8.4 cm

C

CONTINUED

356


Solution a2 + b2 - c2 2ab 4.5 2 + 6.1 2 - 8.4 2 cos +BAC = 2 ] 4. 5 g ] 6 .1 g = - 0.2386 cos C =

Notice that the negative sign tells us that the angle will be obtuse.

+BAC = cos- 1 ] - 0.2386 g = 103c 48l

6.11 1.

Exercises

Find the value of all pronumerals, correct to 1 decimal place.

(e)

(a) 2.

Evaluate all pronumerals correct to the nearest minute (a)

(b)

(b) (c)

(c) (d)


YZ = 5.9 cm. Find the value of all angles, to the nearest minute.

(d)

7.

Isosceles trapezium MNOP has MP = NO = 12 mm, MN = 8.9 mm, OP = 15.6 mm and +NMP = 119c 15l. (a) Find the length of diagonal NP. (b) Find +NOP.

8.

Given the figure below, find the length of (a) AC (b) AD.

(e)

3.

Kite ABCD has AB = 12.9 mm, CD = 23.8 mm and +ABC = 125c as shown. Find the length of diagonal AC.

B

42 c8 l 8.4 cm

B 12.9 mm

125 c

A

101 c38 l

C

3.7 cm

C

A

23.8 mm

9.9 cm

D

4.

5.

6.

Parallelogram ABCD has sides 11 cm and 5 cm, and one interior angle 79c 25l. Find the length of the diagonals. Quadrilateral ABCD has sides AB = 12 cm, BC = 10.4 cm, CD = 8.4 cm and AD = 9.7 cm with +ABC = 63c 57l. (a) Find the length of diagonal AC (b) Find +DAC (c) Find +ADC. Triangle XYZ is isosceles with XY = XZ = 7.3 cm and

D

9.

In a regular pentagon ABCDE with sides 8 cm, find the length of diagonal AD.

10. A regular hexagon ABCDEF has sides 5.5 cm. (a) Find the length of AD. (b) Find +ADF.

357

358


Applications The sine and cosine rules can be used in solving problems.

Use the sine rule to find: 1. a side, given one side and two angles 2. an angle, given two sides and one angle Use the cosine rule to find: 1. a side, given two sides and one angle 2. an angle, given three sides

EXAMPLES 1. The angle of elevation of a tower from point A is 72c. From point B, 50 m further away from the tower than A, the angle of elevation is 47c. (a) Find the exact length of AT. (b) Hence, or otherwise, find the height h of the tower to 1 decimal place.

Solution

Use TBTA to find AT.

(a) +BAT = 180c - 72c = 108c +BTA = 180c - ] 47c + 108c g = 25c a b = sin A sin B 50 AT = sin 47c sin 25c 50 sin 47c AT = ` sin 25c

^ straight angle h (angle sum of T)


h AT h = AT sin 72c 50 sin 47c = # sin 72c sin 25c Z 82.3 m

359

(b) sin 72c = `

Use right-angled TATO to find h. Do not use the sine rule.

2. A ship sails from Sydney for 200 km on a bearing of 040c , then sails on a bearing of 157c for 345 km. (a) How far from Sydney is the ship, to the nearest km? (b) What is the bearing of the ship from Sydney, to the nearest degree?

Solution

(a) +SAN = 180c - 40c = 140c

^ cointerior angles h

` +SAB = 360c - (140c + 157c) = 63c

^ angle of revolution h

c 2 = a 2 + b 2 - 2ab cos C x 2 = 200 2 + 345 2 - 2 (200) (345) cos 63c Z 96374.3 x = 96374.3 Z 310 So the ship is 310 km from Sydney. sin A sin B a = b sin i sin 63c = 345 310 345 sin 63c ` sin i = 310 Z 0.99 i Z 82c ( b)

The bearing from Sydney = 40c + 82c = 122c

To find the bearing, measure +TSB.

360


6.12

Exercises

1.

Find the lengths of the diagonals of a parallelogram with adjacent sides 5 cm and 8 cm and one of its angles 32c 42l.

2.

A car is broken down to the north of 2 towns. The car is 39 km from town A and 52 km from town B. If A is due west of B and the 2 towns are 68 km apart, what is the bearing of the car from (a) town A (b) town B, to the nearest degree?

3.

7.

A boat is sinking 1.3 km out to sea from a marina. Its bearing is 041c from the marina and 324c from a rescue boat. The rescue boat is due east of the marina. (a) How far, correct to 2 decimal places, is the rescue boat from the sinking boat? (b) How long will it take the rescue boat, to the nearest minute, to reach the other boat if it travels at 80 km/h?

8.

The angle of elevation of the top of a flagpole from a point a certain distance away from its base is 20c. After walking 80 m towards the flagpole, the angle of elevation is 75c. Find the height of the flagpole, to the nearest metre.

9.

A triangular field ABC has sides AB = 85 m and AC = 50 m. If B is on a bearing of 065c from A and C is on a bearing of 166c from A, find the length of BC, correct to the nearest metre.

The angle of elevation to the top of a tower is 54c 37l from a point 12.8 m out from its base. The tower is leaning at an angle of 85c 58l as shown. Find the height of the tower.

54 c37 l

4.

from one post and 11 m from the other, find the angle within which the ball must be kicked to score a goal, to the nearest degree.

12.8 m

85c58 l

A triangular park has sides 145.6 m, 210.3 m and 122.5 m. Find the size of the largest interior angle of the park.

5.

A 1.5 m high fence leans outwards from a house at an angle of 102c. A boy sits on top of the fence and the angle of depression from him down to the house is 32c 44l . How far from the fence is the house?

6.

Football posts are 3.5 m apart. If a footballer is standing 8 m

10. (a) Find the exact value of AC in the diagram. (b) Hence, or otherwise, find the angle i, correct to the nearest minute.


11. Find the value of h, correct to 1 decimal place.

16. Rhombus ABCD with side 8 cm has diagonal BD 11.3 cm long. Find +DAB. 17. Zeke leaves school and runs for 8.7 km on a bearing of 338c, then turns and runs on a bearing of 061c until he is due north of school. How far north of school is he?

12. A motorbike and a car leave a service station at the same time. The motorbike travels on a bearing of 080c and the car travels for 15.7 km on a bearing of 108c until the bearing of the motorbike from the car is 310c. How far, correct to 1 decimal place, has the motorbike travelled? 13. A submarine is being followed by two ships, A and B, 3.8 km apart, with A due east of B. If A is on a bearing of 165c from the submarine and B is on a bearing of 205c from the submarine, find the distance from the submarine to both ships. 14. A plane flies from Dubbo on a bearing of 139c for 852 km, then turns and flies on a bearing of 285cuntil it is due west of Dubbo. How far from Dubbo is the plane, to the nearest km? 15. A triangular roof is 16.8 m up to its peak, then 23.4 m on the other side with a 125c angle at the peak as shown. Find the length of the roof.

125 c 16.8 m

23.4 m

18. A car drives due east for 83.7 km then turns and travels for 105.6 km on a bearing of 029c. How far is the car from its starting point? 19. The figure below shows the diagram that a surveyor makes to measure a triangular piece of land. Find its perimeter.

13.9 m 58 c1l 11.4 m

14.3 m 132 c31l

20. A light plane leaves Sydney and flies for 1280 km on a bearing of 050c. It then turns and flies for 3215 km on a bearing of 149c. How far is the plane from Sydney, to the nearest km? 21. Trapezium ABCD has AD ; BC, with AB = 4.6 cm, BC = 11.3 cm, CD = 6.4 cm, +DAC = 23c 30l and +ABC = 78c . (a) Find the length of AC. (b) Find +ADC to the nearest minute. 22. A plane leaves Adelaide and flies for 875 km on a bearing of 056c. It then turns and flies on a bearing of i for 630 km until it is due east of Adelaide. Evaluate i to the nearest degree.

361

362


23. Quadrilateral ABCD has AB = AD = 7.2 cm, BC = 8.9 cm and CD = 10.4 cm, with +DAB = 107c (a) Find the length of diagonal BD. (b) Find +BCD. 24. Stig leaves home and travels on a bearing of 248c for 109.8 km. He then turns and travels for 271.8 km on a bearing of 143c. Stig then turns and travels home on a bearing of a. (a) How far does he travel on the final part of his journey? (b) Evaluate a.

25. A wall leans inwards and makes an angle of 88c with the floor. (a) A 4 m long ladder leans against the wall with its base 2.3 m out from the wall. Find the angle that the top of the ladder makes with the wall. (b) A longer ladder is placed the same distance out from the wall and its top makes an angle of 31c with the wall. (i) How long is this ladder? (ii) How much further does it reach up the wall than the first ladder?

Area To find the area of a triangle, you need to know its perpendicular height. Trigonometry allows us to find this height in terms of one of the angles in the triangle.

A=

Similarly, 1 ac sin B 2 1 A = bc sin A 2 A=

Proof From D BCD, h sin C = a ` h = a sin C 1 bh 2 1 = ba sin C 2

A=

1 ab sin C 2


363

EXAMPLE Find the area of D ABC correct to 2 decimal places.

To find the area, use 2 sides and their included angle.

Solution 1 ab sin C 2 1 = (4.3) (5.8) sin 112c 34l 2 Z 11.52 units 2

A=

6.13 1.

Exercises

Find the area of each triangle correct to 1 decimal place. (a)

(c)

(d)

(b)

(e)

364


2.

3.

Calculate the exact area of D ABC.

Find the area of DOAB correct to 1 decimal place (O is the centre of the circle).

7.

Find the area of a regular hexagon with sides 4 cm, to the nearest cm 2 .

8.

Calculate the area of a regular pentagon with sides 12 mm.

9.

The figure below is made from a rectangle and isosceles triangle with AE = AB as shown. A 84c

E

4.

5.

Find the area of a parallelogram with sides 3.5 cm and 4.8 cm, and one of its internal angles 67c 13l, correct to 1 decimal place.

B

14.3 cm

D

Find the area of kite ABCD, correct to 3 significant figures.

10.5 cm

C

(a) Find the length of AE. (b) Find the area of the figure. 10. Given the following figure, A 58c

B

6.

Find the area of the sail, correct to 1 decimal place.

44c

9.4 cm

C

36c 6.7 cm

D

(a) Find the length of AC (b) Find the area of triangle ACD (c) Find the area of triangle ABC.


365

Trigonometry in Three Dimensions EXAMPLES 1. From point X, 25 m due south of the base of a tower, the angle of elevation is 47c. Point Y is 15 m due east of the tower. Find: (a) the height, h, of the tower, correct to 1 decimal place (b) the angle of elevation, i, of the tower from point Y.

Solution (a) From D XTO h 25 25 tan 47c = h 26.8 = h tan 47c =

So the tower is 26.8 m high. (b) From DYTO 26.8 tan i = 15 `

26.8 m 15 = 60c 46l

i = tan - 1 c

So the angle of elevation from Y is 60c 46l. 2. A cone has a base diameter of 18 cm and a slant height of 15 cm. Find the vertical angle at the top of the cone.

Solution The radius of the base is 9 cm. 9 sin i = 15 9 ` i = sin - 1 c m 15 = 36c 52l Vertical angle = 2i = 73c 44l

Use the full value of 26.80921775 for a more accurate answer to (b).

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6.14 1.

2.

3.

Exercises

A gymnastics bar is supported by wires as shown below. (a) If one wire is inclined at an angle of 55c to the horizontal and is 1.4 m out from the base of the bar, find the height of the bar, to the nearest metre. (b) The second wire is inclined at an angle of 68c to the horizontal. How long is the wire (to 1 decimal place)? (c) The third wire is 2.2 m long. What is its angle of elevation?

A pole has two supporting ropes, 2.5 m and 3.1 m long. (a) If the 3.1 m rope makes an angle of elevation of 38c , find the length of the pole, correct to 1 decimal place. (b) What angle of elevation does the other rope make?

A 25 cm #11 cm # 8 cm cardboard box contains an insert (the shaded area) made of foam. (a) Find the area of foam in the insert, to the nearest cm 2 . (b) Find the angle, i, the insert makes at the corner of the box.

4.

A cone has radius 7 cm and a slant height of 13 cm. Find the vertical angle at the top of the cone, in degrees and minutes.

5.

From a point 15 m due north of a tower, the angle of elevation of the tower is 32c (a) Find the height of the tower, to the nearest metre. (b) Find the angle of elevation of the tower at a point 20 m due east of the tower.

6.

A pole is seen from two points A and B. The angle of elevation from A is 58c . If +CAB = 52c and +ABC = 34c , and A and B are 100 m apart, find: (a) how far A is from the foot of the pole, to the nearest metre. (b) the height of the pole, to 1 decimal place.


7.

Two straight paths up to the top of a cliff are inclined at angles of 25c and 22c to the horizontal.

(a) If path 1 is 114 m long, find the height of the cliff, to the nearest metre. (b) Find the length of path 2, to 1 decimal place. (c) If the paths meet at 47c at the base of the cliff, find their distance apart at the top of the cliff, correct to 1 decimal place. 8.

9.

A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

10. A hot air balloon flying at 950 m/h at a constant altitude of 3000 m is observed to have an angle of elevation of 78c . After 20 minutes, the angle of elevation is 73c . Calculate the angle through which the observer has turned during those 20 minutes.

David walks along a straight road. At one point he notices a tower on a bearing of 053c with an angle of elevation of 21c. After walking 230 m, the tower is on a bearing of 342c , with an angle of elevation of 26c . Find the height of the tower correct to the nearest metre.

Sums and Differences of Angles Sums and differences Angles can be expressed as sums or differences of other angles. This enables us to simplify or evaluate some angles that normally would be too hard to simplify.

cos ^ x - y h = cos x cos y + sin x sin y

367

368


Proof

Remember: cos i = x -coordinate and sin i = y -coordinate.

By the distance formula: d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 AB 2 = ^ cos x - cos y h2 + ^ sin x - sin y h2 = cos 2 x - 2 cos x cos y + cos 2 y + sin 2 x - 2 sin x sin y + sin 2 y = (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) - 2 cos x cos y - 2 sin x sin y = 2 - 2 (cos x cos y + sin x sin y)

(1)

By the cosine rule: c 2 = a 2 + b 2 - 2ab cos C AB 2 = 1 2 + 1 2 - 2 (1) (1) cos (x - y) = 2 - 2 cos (x - y) From (1) and (2): 2 - 2 cos ^ x - y h = 2 - 2 ^ cos x cos y + sin x sin y h ` cos ^ x - y h = cos x cos y + sin x sin y

cos ^ x + y h = cos x cos y - sin x sin y

Proof Substitute - y for y. cos (x - y) = cos x cos y + sin x sin y cos (x - (- y )) = cos x cos (- y ) + sin x sin (- y ) cos (x + y) = cos x cos y + sin x (- sin y ) cos (x + y) = cos x cos y - sin x sin y

sin ^ x + y h = sin x cos y + cos x sin y

(2)


369

Proof Substitute 90c - x for x. cos (x - y) = cos x cos y + sin x sin y cos (90c - x - y) = cos (90c - x) cos y + sin (90c- x) sin y cos (90c - (x + y)) = sin x cos y + cos x sin y sin (x + y) = sin x cos y + cos x sin y

sin ^ x - y h = sin x cos y - cos x sin y

Proof Substitute - y for y. sin (x + y) = sin x cos y + cos x sin y sin (x + (- y)) = sin x cos (- y) + cos x sin (- y) sin (x - y) = sin x cos y + cos x (- sin y) sin (x - y) = sin x cos y - cos x sin y

tan ^ x + y h =

tan x + tan y 1 - tan x tan y

Proof tan (x + y) =

sin ^ x + y h

cos ^ x + y h sin x cos y + cos x sin y = cos x cos y - sin x sin y sin x cos y + cos x sin y cos x cos y = cos x cos y - sin x sin y cos x cos y tan x + tan y tan (x + y) = 1 - tan x tan y

tan ^ x - y h =

tan x - tan y 1 + tan x tan y

Divide top and bottom by cos x cos y.

370


Proof Substitute -y for y. tan (x + y) = tan (x + (- y)) = tan (x - y) =

tan x + tan y 1 - tan x tan y tan x + tan ^ - y h 1 - tan x tan ^ - y h tan x - tan y

1 - tan x ^ - tan y h tan x - tan y tan (x - y) = 1 + tan x tan y

EXAMPLES 1. Simplify sin 2i cos i - cos 2i sin i.

Solution sin 2i cos i - cos 2i sin i = sin (2i - i) = sin i 2. Find the exact value of cos 75c .

Solution cos 75c = cos (30c+ 45c) = cos 30c cos 45c - sin 30c sin 45c = =

3 1 1 1 # - # 2 2 2 2 3-1

=

2 2 3-1

# 2 2 6- 2 = 4

2 2

3. Simplify cos ] i + 60c g + sin ] i + 60c g .

Solution cos (i + 60c) + sin (i + 60c) = cos i cos 60c - sin i sin 60c + sin i cos 60c + cos i sin 60c 3 3 1 1 - sin i # + sin i # + cos i # 2 2 2 2 3 3 1 1 n + sin i d = cos i d + + n 2 2 2 2 1+ 3n 1- 3n = cos i d + sin i d 2 2 = cos i #


371

Ratios of double angles By using the sum of angles, we can find the trigonometric ratios for double angles.

sin 2x = 2 sin x cos x

Proof sin 2x = sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x

cos 2x = cos 2 x - sin 2 x = 1 - 2 sin 2 x = 2 cos 2 x - 1

Proof cos 2x = cos (x + x) = cos x cos x - sin x sin x = cos 2 x - sin 2 x = (1 - sin 2 x) - sin 2 x = 1 - 2 sin 2 x = 1 - 2 (1 - cos 2 x) = 2 cos 2 x - 1

tan 2x =

Proof tan 2x = tan (x + x) tan x + tan x = 1 - tan x tan x 2 tan x tan 2x = 1 - tan 2 x

Remember: sin 2 x + cos 2 x = 1.

2 tan x 1 - tan 2 x

372


EXAMPLES 1. Simplify cos 2 2i - sin 2 2i.

Solution cos 2 2i - sin 2 2i = cos 2 (2i) = cos 4i 2. If sin x =

4 , find the exact value of sin 2x. 7

Solution AC 2 = 7 2 - 4 2 = 33 ` AC = 33 sin 2x = 2 sin x cos x 4 # 7 8 33 = 49 =2#

33 7

PROBLEM 1 ] 3 sin i - sin 3i g to 4 draw up a table of sine ratios. Can you prove this relation?

Ulug Beg (1393–1449) used the relation sin 3 i =

6.15 1.

Exercises

Expand (a) sin ] a - b g (b) (c) (d) (e) (f) (g) (h) (i) (j)

cos ^ p + q h tan ^ a + b h sin (x + 20c) tan ] 48c + x g cos ] 2i - a g cos ( x + 75c) tan ^ 5x - 7y h sin ^ 4a - b h tan ^ a - 3b h

2.

Simplify (a) sin a cos b + cos a sin b tan 36c + tan 29c (b) 1 - tan 36c tan 29c (c) cos 28c cos 27c - sin 28c sin 27c (d) sin 2x cos 3y + cos 2x sin 3y tan 3i - tan i (e) 1 + tan 3i tan i (f) sin 74c cos 42c - cos 74c sin 42c (g) sin ] a + b g + sin ] a - b g (h) sin ^ x + y h - sin ^ x - y h (i) cos ^ x - y h - cos ^ x + y h (j) cos ] m + n g + cos ] m - n g


3.

4.

Find the exact value of (a) sin 75c (b) cos 15c (c) tan 75c (d) tan 105c (e) cos 105c (f) sin 15c (g) sin 105c (h) tan 285c (i) sin (x + 30c) + cos (x + 30c) (j) cos ^ 45c - y h + cos ^ 45c + y h Simplify tan ^ x + y h + tan ^ x - y h 1 - tan ^ x + y h tan ^ x - y h

.

3 2 and cos y = , find 4 3 the exact value of (a) sin ^ x + y h (b) cos ^ x - y h (c) tan ^ x + y h

5.

If sin x =

6.

By taking 2i = i + i, find an expression for (a) sin 2i (b) cos 2i (c) tan 2i

7.

By writing 3i as 2i + i, find an expression in terms of i for (a) sin 3i (b) cos 3i (c) tan 3i tan 7i - tan 3i . 1 + tan 7i tan 3i (b) Find an expression for sin 4i in terms of 7i and 3i.

3 5 and cos y = , find 5 13 the value of (a) cos x (b) sin y (c) sin ^ x - y h (d) tan y (e) tan ^ x + y h

11. If sin x =

12. (a) Write an expression for cos ^ x + y h + cos ^ x - y h . (b) Hence write an expression for cos 50c cos 65c. 13. Find an expression for (a) sin ^ x + y h + sin ^ x - y h (b) (c) (d) (e) (f)

cos ^ x + y h - cos ^ x - y h sin ^ x + y h - sin ^ x - y h cos ^ x + y h + sin ^ x - y h tan ^ x + y h + tan ^ x - y h tan ^ x + y h - tan ^ x - y h

14. Expand (a) sin 2b (b) tan 2i (c) cos 2i (d) sin (x + 2y) (e) cos (2a + b ) (f) tan (x + 2y) (g) sin (2i - d ) (h) cos (i - 2c) (i) tan (x - 2z) (j) sin (2x - 2y)

15. Simplify (a) 2 cos 3x sin 3x (b) cos 2 7y - sin 2 7y 2 tan 5i (c) 1 - tan 2 5i 9. Find an expression for cos 9x in (d) 1 - 2 sin 2 y terms of 2x and 7x. (e) sin 6i cos 6i 10. Find the exact value of (f) ] sin x + cos x g2 (a) cos 23c cos 22c - sin 23c sin 22c (g) 2 cos 2 3a - 1 tan 85c - tan 25c (b) (h) 1 - 2 sin 2 40c 1 + tan 85c tan 25c 2 tan b (c) sin 180c cos 60c (i) 1 - tan 2 b + cos 180c sin 60c (j) ] sin 3x - cos 3x g2 (d) cos 290c cos 80c 8.

(a) Simplify

+ sin 290c sin 80c tan 11c + tan 19c (e) 1 - tan 11c tan 19c

373

374


16. Find the exact value of (a) cos 22.5c sin 22.5c (b) cos 2 30c - sin 2 30c 2 tan 15c (c) 1 - tan 2 15c (d) 2 sin 75c cos 75c 2 tan 120c (e) 1 - tan 2 120c (f) 1 - 2 sin 2 165c (g) 2 cos 2 22.5c - 1 2 tan i (h) where i = 112.5c 1 - tan 2 i (i) sin 67.5c cos 67.5c (j) 2 cos 105c sin 105c 5 , find the exact value 8 of cos 2x and sin 2x.

17. If cos x =

3 12 and tan b = , find 5 5 the exact values of (a) sin ^ a + b h (b) cos 2a (c) sin 2b (d) tan ^ a - b h

18. If sin a =

19. Express sin 4i in terms of i. 20. (a) Simplify

sin 2x . 1 + cos 2x

(b) Hence, find the exact value of tan 15c. 1 21. Find the exact value of tan 22 c 2 by using the expression for tan 2x. 22. Prove (a) sin 2 i = (b) tan

1 sin 2i tan i 2

i 1 - cos i = 2 sin i

23. Show that sin 2 7i - sin 2 4i = sin 11i sin 3i. 24. Prove that cos 3i = 4 cos 3 i - 3 cos i. 25. Find an expression for sin 3x in terms of sin x.

Further Trigonometric Equations Some trigonometric equations are difficult to solve. However, there are some expressions that can be used to solve them.

Ratios in terms of tan

If tan

i 2 i 2t = t, then tan i = 2 1 - t2

Proof 2 tan A 1 - tan 2 A i 2 tan 2 ` tan i = where i = 2A i 1 - tan 2 2 2t = 1 - t2 tan 2A =


If tan

375

i 2t = t, then sin i = 2 1 + t2

Proof tan

i t =t= 2 1

The hypotenuse is 1 + t 2 by Pythagoras’ theorem.

sin 2A = 2 sin A cos A i i ` sin i = 2 sin cos where i = 2A 2 2 t 1 = 2e oe o 2 1+t 1 + t2 2t = 1 + t2

If tan

These ratios for sin and cos

i 2

i come from the 2 triangle above.

i 1 - t2 = t, then cos i = 2 1 + t2

Proof cos 2A = cos 2 A - sin 2 A i i where i = 2A ` cos i = cos 2 - sin 2 2 2 2 2 t 1 =e -e o o 1 + t2 1 + t2 2 t 1 = 2 1+t 1 + t2 1 - t2 = 1 + t2

i The ratios for cos and 2 i sin come from the 2 previous triangle.

376


EXAMPLES 1. Find the exact value of

Solution sin i =

So

2t 1 + t2

2 tan 15c . 1 + tan 2 15c

where t = tan

i 2

2 tan 15c = sin 30c 1 + tan 2 15c 1 = 2

2. Prove that cot

i i - 2 cot i = tan . 2 2

Solution i - 2 cot i 2 1 2 = i tan i tan 2 i 1 2 where t = tan = t 2t 2 1 - t2 2 1 2^1 - t h = t 2t 1 1 - t2 = t t 1 - 1 + t2 = t t2 = t =t

LHS = cot

i 2 = RHS = tan

` cot

i i - 2 cot i = tan 2 2

There is also another expression that will help solve some further trigonometric equations.

a sin i + b cos i = r sin ] i + a g where b r = a 2 + b 2 and tan a = a


377

Proof If tan a =

b , then the a

hypotenuse is a 2 + b 2 by Pythagoras’ theorem.

RHS = r sin (i + a ) =

a 2 + b 2 (sin i cos a + cos i sin a ) a = a 2 + b 2 e sin i # + cos i # 2 a + b2 = a sin i + b cos i = LHS ` a sin i + b cos i = r sin ] i + a g

b a + b2 2

o

b a 2 + b 2 and tan a = a

where r =

EXAMPLES 1. Write

3 sin x + cos x in the form r sin ] x + a g.

Solution a sin i + b cos i = r sin ] i + a g where r =


For 3 sin x + cos x: a = 3, b = 1 r=

a2 + b2 2

3 + 12 = = 3+1 = 4 =2 b tan a = a 1 = 3 a = tan - 1 e

1 o 3

= 30c So

3 sin x + cos x = 2 sin(x + 30c) CONTINUED

378


2. Write 3 sin i + 2 cos i in the form r sin ] i + a g.

Solution a sin i + b cos i = r sin (i + a ) where r =


For 3 sin i + 2 cos i: a = 3, b = 2 r= = = =

a2 + b2 32 + 22 9+4 13 b tan a = a 2 = 3 2 a = tan - 1 c m 3 = 33° 41l

So 3 sin i + 2 cos i =

13 sin (i + 33c 41l)

Class Investigation Can you find similar results for these? • a sin i - b cos i • a cos i + b cos i • a cos i - b sin i

6.16 1.

Exercises

Simplify 2t (a) 1 - t2

(e)

1 - t2 (b) 1 + t2 2 tan 10c (c) 1 - tan 2 10c 1 - tan 25c 1 + tan 2 25c 2

(d)

2 tan i 1 + tan 2 i

i 2 (f) i 1 + tan 2 2 1 - tan 2

2.

Find the exact value of 2 tan 30c (a) 1 + tan 2 30c


3.

(b)

1 - tan 2 22.5c 1 + tan 2 22.5c

(c)

1 - tan 2 30c 1 + tan 2 30c

(d)

2 tan 90c 1 - tan 2 90c

Write each expression in terms of i t where t = tan . 2 (a) cosec i (b) sec i (c) cot i (d) sin i + cos i (e) 1 + tan i

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) 7.

Write each expression in the form r sin ] i - a g. (a) sin i - cos i (b) sin i - 2 cos i (c) sin i - 3 cos i (d) 3 sin i - cos i (e) 5 sin i - 2 cos i

8.

Write the expression 3 cos i + sin i in the form r cos ] i - a g.

9.

Write the expression cos i - 3 sin i in the form r cos ] i + a g.

i 2 (g) 3 cos i + 4 sin i (f) 1 + tan i tan

(h)

1 + sin i + cos i 1 + sin i - cos i

2 sin i + cos i sin i + 3 cos i sin i + cos i 5 sin i + 2 cos i 4 sin i + cos i 3 sin i + cos i 2 sin i + 3 cos i 4 sin i + 7 cos i 5 sin i + 4 cos i 3 sin i + 5 cos i

(i) tan i + sec i (j) sin 2i 1 + sin i - cos i = t. 1 + sin i + cos i

4.

Prove

5.

Find an expression for sin 2i - cos 2i in terms of t.

6.

Write each expression in the form r sin ] i + a g.

10. Write the expression 9 sin i + 2 cos i in the form. (a) r sin ] i + a g (b) r cos ] i - a g

We can use these results to help solve some trigonometric equations.

EXAMPLES 1. Solve 2 sin i = cos i for 0c # i # 360c .

Solution 2 sin i = cos i Dividing both sides by cos i:

(check cos i = 0 does not give a solution)

2 sin i cos i = cos i cos i 2 tan i = 1 tan i = 0.5 CONTINUED

379

380


Since tan i is positive in the first and third quadrants: i = 26c 34l, 180c+ 26c 34l = 26c 34l, 206c 34l 2. Solve cos 2i = cos i for 0c # i # 360c.

Solution cos 2i = cos i 2 cos i - 1 = cos i 2 cos i - cos i - 1 = 0 ] 2 cos i + 1 g ] cos i - 1 g = 0 2

2

` 2 cos i + 1 = 0 or cos i - 1 = 0 2 cos i = -1 cos i = 1 1 i = 0c, 360c cos i = 2 i = 120c, 240c ` solutions are i = 0c, 120c, 240c, 360c 3 sin x + cos x = 1 for 0c # x # 360c.

3. Solve

Solution (Method 1) Use the result for a sin x + b cos x. For 3 sin x + cos x, a = 3 and b = 1 r=

a2 + b2

= ^ 3 h + 12 =2 b tan a = a 1 = 3 ` a = 30c 2

`

3 sin x + cos x = 2 sin ] x + 30c g

Solving:

Sine is positive in the first and second quadrants.

3 sin x + cos x = 1 2 sin ] x + 30c g = 1 1 sin ] x + 30c g = 2 `

for 0c # x # 360c for 30c # x + 30c # 390c

x + 30c = 30c , 180c - 30c , 360c + 30c = 30c , 150c , 390c x = 0c , 120c , 360c


Solution (Method 2) Use the results for t = tan

i . 2

3 sin x + cos x = 1 2t 1 - t2 3d + =1 n 1 + t2 1 + t2 2 3 t + 1 - t2 =1 1 + t2 2 3 t + 1 - t2 = 1 + t2 0 = 2t 2 - 2 3 t = 2t ^ t - 3 h ` 2t = 0 or t - 3 t=0 t x x tan = 0 tan 2 2 x x = 0c, 180c 2 2 ` x = 0c, 120c, 360c

=0 = 3 =

3 for 0c #

x # 180c 2

= 60c

Test x = 180c separately: 3 sin 180c + cos 180c = 0 + (- 1) = -1 !1 ` x = 180c is not a solution Solutions are x = 0c , 120c , 360c .

General solutions of trigonometric equations Often the solutions of trigonometric equations are restricted, for example, to 0c # i # 360c . If the solutions are not restricted, then they can be described by a general formula.

EXAMPLE Find all solutions for sin i =

3 . 2

Solution

CONTINUED

381

382


Sin is positive in the 1st and 2nd quadrants.

sin i = sin 60c i = 60c, 180c - 60c, 360c + 60c, 360c + 180c - 60c, 360c + 360c + 60cg = 60c, 180c - 60c, 360c + 60c, 540c - 60c, 720c + 60c, g If i can also be negative, i = - (180c + 60c), - (360c - 60c), - [360c - (180c - 60c)], g = - 180c - 60c, - 360c + 60c, - 540c - 60c, g So the general solution for sin i = sin 60c is i = 180c # n + ] - 1 g n 60c where n is an integer.

In general, the solution for sin i = sin a is given by i = 180n + (-1) n a where n is an integer.

EXAMPLE Find all solutions for cos i =

Solution

Cos is positive in the 1st and 4th quadrants.

1 . 2

cos i = cos 45c i = 45c, 360c - 45c, 360c + 45c, 360c + 360c - 45c, 360c + 360c + 45c, f = 45c, 360c - 45c, 360c + 45c, 720c - 45c, 720c + 45c, f If i can also be negative, i = - 45c, - (360c - 45c), - (360c + 45c), - [360c + (360c - 45c)], f = - 45c, - 360c + 45c, - 360c - 45c, - 720c + 45c, f So the general solution for cos i = cos 45c is i = 360c # n ! 45c where n is an integer.

In general, the solution for cos i = cos a is given by i = 360n ! a where n is an integer.


383

EXAMPLE Find all solutions for tan i = 1.

Solution

tan i = tan 45c i = 45c, 180c + 45c, 360c + 45c, 360c + 180c + 45c, 360c + 360c + 45c, f

Tan is positive in the 1st and 3rd quadrants.

= 45c, 180c + 45c, 360c + 45c, 540c + 45c, 720c + 45c, f If i can also be negative, then i = - (180c - 45c), - (360c - 45c), - [360c + (180c - 45c)], f = -180c + 45c, - 360c + 45c, - 540c + 45c, f The general solution for tan i = tan 45c is i = 180c # n + 45c where n is an integer.

In general, the solution for tan i = tan a is given by i = 180n + a where n is an integer.

6.17 1.

Exercises

Solve for 0c # x # 360c .

2.

Solve for 0c # i # 360c.

(a) sin x = cos x

(a) 3 sin i + 4 cos i = 0

(b) cos x =

(b) 5 cos i - 12 sin i = -3

3 sin x

(c) sin 2x = sin x

(c) sin i - 3 cos i = 0

(d) tan x - tan x = 0

(d) sin i + cos i = -1

(e) 2 sin x - sin x - 1 = 0

(e) 4 sin i - cos i + 3 = 0

(f) 2 sin 2 x + 3 cos x - 3 = 0

(f) sin i - cos i = 1

(g) sin x cot x - sin x = 0

(g)

(h) cos x - 1 = 0 (i) 2 sin x tan x - tan x + 2 sin x - 1 = 0 (j) 3 cos 2 x - 7 cos x + 4 = 0

(h) 2 sin i - cos i =

2

2

2

2 cos i + sin i = 1 5 2

(i) 3 cos i - 5 sin i + 2 = 0 (j)

2 cos i + sin i + 1 = 0

384


3.

Find the general solution for 1 (a) sin i = 2 (b) tan a = 3 3 2 (d) 2 sin x = -1

7.

Find the general solutions of (a) sin i = 0 (b) cos x = 1 (c) tan x = 0 (d) sin i = -1 (e) cos a = 0

8.

For each question (i) solve for 0c # x # 360cand (ii) find the general solutions (a) 2 sin x - 1 = 0 (b) 4 cos x - 3 = 0 (c) sin x = 3 cos x (d) 3 sin x + cos x = 0 (e) sin x + cos x = 2

9.

Find the general solutions of 2 sin 2 x + sin x – 1 = 0.

(c) cos i =

(e) tan i + 1 = 0 (f) 2 cos 2 b = 1 (g) 4 sin 2 c = 3 1 (h) tan i = 3 (i) cos i = 0.245 (j) sin a = 0.399 4.

Solve sin ] 2x - 45c g =

3 for 2

-180c # x # 180c . 5.

Find the general solutions of sin 2x = cos x.

6.

Solve sin 2 x = sin x for -180c # x # 180c .

10. (a) Solve cos 2x = cos x for 0c # x # 360c. (b) Find the general solutions of cos 2x = cos x.


Test Yourself 6 1.

Find the exact value of cos i and sin i if 3 tan i = . 5

2.

Simplify

11. Find the length of AB as a surd.

(a) sin x cot x cos 40c + sin 50c (b) cos 40c 12. Evaluate x, correct to 2 significant figures. (a)

1 + cot 2 A i 1 - t2 (d) where t = tan 2 2 1+t (e) 1 - 2 sin 2 10i (c)

3.

4.

Evaluate to 2 decimal places. (a) sin 39c 54l (b) tan 61c 30l (c) cos 19c 2l Find i to the nearest minute if (a) sin i = 0.72 (b) cos i = 0.286 5 (c) tan i = 7 2 cos i = 2 + 2 sin i. 1 - sin i 2

5.

Prove that

6.

Find the value of b if sin b = cos ] 2b - 30 g c .

7.

Find the exact value of (a) cos 315c (b) sin ] - 60c g (c) tan 120c (d) 2 sin 105c cos 105c (e) sin ^ x - y h when sin x = cos y =

(b)

13. Evaluate i to the nearest minute. (a)

(b)

8 and 17

5 13

8.

Solve 2 cos x = -1 for 0c # x # 360c.

9.

Sketch the graph of y = cos x, and hence solve cos x = 0 for 0c # x # 360c .

10. A ship sails on a bearing of 215c from port until it is 100 km due south of port. How far does it sail, to the nearest km?

(c)

385

386


14. Find the area of triangle MNO.

15. Solve for -180c # x # 180c . 3 (a) sin 2 x = 4 1 (b) tan 2x = 3 (c) 3 tan 2 x = tan x 5 16. If sec i = - and tan i 2 0, find sin i 4 and cot i. 17. Jacquie walks south from home for 3.2 km, then turns and walks west for 1.8 km. What is the bearing, to the nearest degree, of (a) Jacquie from her home? (b) her home from where Jacquie is now? 18. Find the general solution of 6 sin i - 8 cos i = 5.

(a) Find an expression for the length of AD. (b) Find the height of the pole, to 1 decimal place. 20. A plane flies from Orange for 1800 km on a bearing of 300c . It then turns and flies for 2500 km on a bearing of 205c . How far is the plane from Orange, to the nearest km? 21. Find the exact value of (a) sin 75c (b) cos 105c (c) sin 22c 30l cos 22c 30l. 22. Find the general solutions of (a) 2 cos x – 1 = 0 (b) tan x = 1 3 (c) sin x = . 2 23. Solve 3 sin i + cos i = 1 for 0c # i # 360c . 24. Evaluate a in the figure below. a

19. The angle of elevation from point B to the top of a pole is 39c , and the angle of elevation from D, on the other side of the pole, is 42c. B and D are 20 m apart.

10 mm

4 mm 12 mm

25. (a) Simplify cos x cos y – sin x sin y. (b) Show that cos 2x = 1 – sin2 x.



2.

3.

Two cars leave an intersection at the same time, one travelling at 70 km/h along one road and the other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection? A ship sails from port on a bearing of 055c , then turns and sails on a bearing of 153c for 29.1 km, when it is due east of port. How far, to 1 decimal place, is the ship from its starting point? Evaluate x correct to 3 significant figures.

6.

Simplify sin ] 360c - x g $ tan ] 90c- x g .

7.

Find the exact area of D ABC.

8.

Find the exact value of cos (-315c) .

9.

Solve tan 2x - 1 = 0 for 0c # x # 360c .

10. Find i to the nearest minute.

4.

(a) Find an exact expression for the length of AC. (b) Hence, or otherwise, find the value of h correct to 1 decimal place. 11. The angle of depression from the top of a 4.5 m mast of a boat down to a fish is 56c 28l . How far down, to 1 decimal place, does a pelican sitting at the top of the mast need to fly to catch the fish? 12. Solve 2 cos (i + 10c ) = - 1 for 0c # i # 360c.

5.

A man walks 3.8 km on a bearing of 134c from a house. He then walks 2.9 km on a bearing of 029c . How far is he from the house, to 1 decimal place?

13. Two roads meet at an angle of 74c . Find the distance, correct to 3 significant figures, between two cars, one 6.3 km from the intersection along one road and the other 3.9 km along the other road.

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388


14. Find the exact value of cos i, given 5 sin i = and cos i 1 0. 9 15. From the top of a vertical pole the angle of depression to a man standing at the foot of the pole is 43c . On the other side of the pole is another man, and the angle of depression from the top of the pole to this man is 52c . The men are standing 58 m apart. Find the height of the pole, to the nearest metre. 16. Show that cos i ] sin i + cos i g = 1 + tan i. ] 1 + sin i g ] 1 - sin i g 17. If x = 3 sin i and y = 3 cos i - 2, eliminate i to find an equation relating x and y. 18. From point A, 93 m due south of the base of a tower, the angle of elevation is 35c . Point B is 124 m due east of the tower. Find (a) the height of the tower, to the nearest metre (b) the angle of elevation of the tower from point B.

20. A cone has a base diameter of 14 cm and a perpendicular height of 26 cm. Find the vertical angle at the top of the cone. 21. Show that cos 6i cos 4i - sin 6i sin 4i = 2 cos 2 5i - 1. 22. A cable car 100 m above the ground is seen to have an angle of elevation of 65c when it is on a bearing of 345c . After a minute, it has an angle of elevation of 69c and is on a bearing of 025c . Find how far it travels in that minute, and its speed in ms - 1 . 23. Solve cos 2i - sin i = 0 for 0c # i # 360c .

19. ABCD is a triangular pyramid with 24. Find the general solutions of sin i = -1. BC = 7 m, CD = 10 m, BD = 8 m, AB = AC and +ACB = 67c . Calculate 25. Simplify cosec i ] cos i - 1 g by expressing (a) +BCD i it in terms of t c tan m . (b) length AB, to the nearest metre. 2

7

Linear Functions TERMINOLOGY Collinear points: Two or more points that lie on the same straight line

Interval: A section of a straight line including the end points

Concurrent lines: Two or more lines that intersect at a single point

Midpoint: A point lying exactly halfway between two points

Gradient: The slope of a line measured by comparing the vertical rise over the horizontal run. The symbol for gradient is m

Perpendicular distance: The shortest distance between a point and a line. The distance will be at right angles to the line

Chapter 7 Linear Functions

INTRODUCTION IN CHAPTER 5, YOU STUDIED functions and their graphs. This chapter looks at the linear function, or straight-line graph, in more detail. Here you will study the gradient and equation of a straight line, the intersection of two or more lines, parallel and perpendicular lines, the midpoint, distance and the perpendicular distance from a point to a line.

DID YOU KNOW? Pierre de Fermat (1601–65) was a lawyer who dabbled in mathematics. He was a contemporary of Descartes, and showed the relationship between an equation in the form Dx = By, where D and B are constants, and a straight-line graph. Both de Fermat and Descartes only used positive values of x, but de Fermat used the x-axis and y-axis as perpendicular lines as we do today. De Fermat’s notes Introduction to Loci, Method of Finding Maxima and Minima and Varia opera mathematica were only published after his death. This means that in his lifetime de Fermat was not considered a great mathematician. However, now he is said to have contributed as much as Descartes towards the discovery of coordinate geometry. De Fermat also made a great contribution in his discovery of differential calculus.

Class Assignment Find as many examples as you can of straight-line graphs in newspapers and magazines.

Distance The distance between two points (or the length of the interval between two points) is easy to find when the points form a vertical or horizontal line.

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EXAMPLES Find the distance between 1. ^ -1, 4 h and ^ -1, -2 h

Solution

Counting along the y-axis, the distance is 6 units. 2. ^ 3, 2 h and ^ -4, 2 h

Solution

Counting along the x-axis, the distance is 7 units.

When the two points are not lined up horizontally or vertically, we use Pythagoras’ theorem to find the distance.


393

EXAMPLE Find the distance between points ^ 3, -1 h and ^ -2, 5 h.

Solution

BC = 5 and AC = 6 By Pythagoras’ theorem,

You studied Pythagoras’ theorem in Chapter 4.

c =a +b AB 2 = 5 2 + 6 2 = 25 + 36 = 61 2

2

2

` AB = 61 Z 7.81

DID YOU KNOW? Pythagoras made many discoveries about music as well as about mathematics. He found that changing the length of a vibrating string causes the tone of the music to change. For example, when a string is halved, the tone is one octave higher.

The distance between two points _ x 1, y 1 i and _ x 2, y 2 i is given by d=

2 2 _ x2 - x1 i + _ y2 - y1 i

394


Proof

If points A and B were changed around, the formula would be d =

(x 1 - x 2 ) + (y 1 - y 2 ) , 2

2

which would give the same answer.

Let A = _ x 1, y 1 i and B = _ x 2, y 2 i Length AC = x 2 - x 1 and length BC = y 2 - y 1 By Pythagoras’ theorem AB 2 = AC 2 + BC 2 d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 `d=

2 2 _ x2 - x1 i + _ y2 - y1 i

EXAMPLES 1. Find the distance between the points ^ 1, 3 h and ^ -3, 0 h.

Solution Let ^ 1, 3 h be _ x 1, y 1 i and ^ -3, 0 h be _ x 2, y 2 i d=

2 2 _ x2 - x1 i + _ y2 - y1 i

= ] -3 - 1 g2 + ] 0 - 3 g2 = ] -4 g2 + ] -3 g2 = 16 + 9 = 25 =5

So the distance is 5 units. 2. Find the exact length of AB given that A = ^ -2, -4 h and B = ^ -1, 5 h .

Solution Let ^ -2, -4 h be _ x 1, y 1 i and ^ -1, 5 h be _ x 2, y 2 i d=

You would still get 82 if you used (- 2, - 4) as (x 2 , y 2 ) and (-1, 5) as (x 1 , y 1 ).

2 2 _ x2 - x1 i + _ y2 - y1 i

=

6 -1 - ^ -2 h @ 2 + 6 5 - ^ -4 h @ 2

= = =

12 + 92 1 + 81 82


7.1 Exercises 1.

Find the distance between points (a) ^ 0, 2 h and ^ 3, 6 h (b) ^ -2, 3 h and ^ 4, -5 h (c) ^ 2, -5 h and ^ -3, 7 h

2.

Find the exact length of the interval between points (a) ^ 2, 3 h and ^ -1, 1 h (b) ^ -5, 1 h and ^ 3, 0 h (c) ^ - 2, -3 h and ^ - 4, 6 h (d) ^ -1, 3 h and ^ -7, 7 h

3.

4.

Find the distance, correct to 2 decimal places, between points (a) ^ 1, -4 h and ^ 5, 5 h (b) ^ 0, 4 h and ^ 3, -2 h (c) ^ 8, -1 h and ^ -7, 6 h Find the perimeter of D ABC with vertices A ^ 3, 1 h, B ^ -1, 1 h and C ^ -1, -2 h .

5.

Prove that the triangle with vertices ^ 3, 4 h, ^ -2, 7 h and ^ 6, -1 h is isosceles.

6.

Show that AB = BC, where A = ^ -2, 5 h, B = ^ 4, -2 h and C = ^ -3, -8 h .

7.

8.

9.

Show that points ^ 3, -4 h and ^ 8,1 h are equidistant from point ^ 7, -3 h . A circle with centre at the origin O passes through the point _ 2 , 7 i . Find the radius of the circle, and hence its equation. Prove that the points X _ 2 , -3 i, Y _ -1, 10 i and Z _ - 6 , 5 i all lie on a circle with centre at the origin. Find its equation.

10. If the distance between ^ a, -1 h and ^ 3, 4 h is 5, find the value of a. 11. If the distance between ^ 3, -2 h and ^ 4, a h is 7 , find the exact value of a.

12. Prove that A ^ 1, 4 h, B ^ 1, 2 h and C _ 1 + 3 , 3 i are the vertices of an equilateral triangle. 13. If the distance between ^ a, 3 h and ^ 4, 2 h is 37 , find the values of a. 14. The points M ^ -1, -2 h, N (3, 0), P ^ 4, 6 h and Q ^ 0, 4 h form a quadrilateral. Prove that MQ = NP and QP = MN. What type of quadrilateral is MNPQ? 15. Show that the diagonals of a square with vertices A ^ -2, 4 h, B ^ 5, 4 h, C ^ 5, -3 h and D ^ -2, -3 h are equal. 16. (a) Show that the triangle with vertices A ^ 0, 6 h, B ^ 2, 0 h and C ^ -2, 0 h is isosceles. (b) Show that perpendicular OA, where O is the origin, bisects BC. 17. Find the exact length of the diameter of a circle with centre ^ -3, 4 h if the circle passes through the point ^ 7, 5 h . 18. Find the exact length of the radius of the circle with centre (1, 3) if the circle passes through the point ^ -5, -2 h . 19. Show that the triangle with vertices A ^ -2, 1 h, B ^ 3, 3 h and C ^ 7, -7 h is right angled. 20. Show that the points X ^ 3, -3 h, Y ^ 7, 4 h and Z ^ - 4, 1 h form the vertices of an isosceles right-angled triangle.

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Midpoint The midpoint is the point halfway between two other points.

The midpoint of two points _ x 1, y 1 i and _ x 2, y 2 i is given by M=e

x1 + x2 y1 + y2 o , 2 2

Proof

Can you see why these triangles are similar?

Find the midpoint of points A _ x 1, y 1 i and B _ x 2, y 2 i. Let M = ^ x, y h Then D APQ <; D ABR AQ AP = AR AB x - x1 1 ` x -x = 2 2 1 2 _ x - x1 i = x2 - x1 2x - 2x 1 = x 2 - x 1 2x = x 1 + x 2 x1 + x2 ` x= 2 y1 + y2 Similarly, y = 2 `

EXAMPLES 1. Find the midpoint of ^ -1, 4 h and ^ 5, 2 h.

Solution x=

x1 + x2 2


-1 + 5 2 4 = 2 =2 y1 + y2 y= 2 4+2 = 2 6 = 2 =3 So M = (2, 3) . =

2. Find the values of a and b if ^ 2, -3 h is the midpoint between ^ -7, -8 h and ^ a, b h.

Solution x=

x1 + x2

2 -7 + a 2= 2 4 = -7 + a 11 = a y1 + y2 y= 2 -8 + b -3 = 2 -6 = -8 + b 2=b So a = 11 and b = 2.

Note that the x-coordinate of the midpoint is the average of x 1 and x 2 . The same applies to the y-coordinate.

PROBLEM A timekeeper worked out the average time for 8 finalists in a race. The average was 30.55, but the timekeeper lost one of the finalist’s times. The other 7 times were 30.3, 31.1, 30.9, 30.7, 29.9, 31.0 and 30.3. Can you find out the missing time?

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398


7.2 Exercises 1.

2.

The locus is the path that P (x, y) follows.

Find the midpoint of (a) ^ 0, 2 h and ^ 4, 6 h (b) ^ -2, 3 h and ^ 4, -5 h (c) ^ 2, -5 h and ^ -6, 7 h (d) ^ 2, 3 h and ^ -8, 1 h (e) ^ -5, 2 h and ^ 3, 0 h (f) ^ -2, -2 h and ^ -4, 6 h (g) ^ 1, -4 h and ^ 5, 5 h (h) ^ 0, 4 h and ^ 3, -2 h (i) ^ 8, -1 h and ^ -7, 6 h (j) ^ 3, 7 h and ^ -3, 4 h Find the values of a and b if (a) ^ 4, 1 h is the midpoint of ^ a, b h and ^ -1, 5 h (b) ^ -1, 0 h is the midpoint of ^ a, b h and ^ 3, -6 h (c) ^ a, 2 h is the midpoint of (3, b h and ^ -5, 6 h (d) ^ -2, 1 h is the midpoint of ^ a, 4 h and ^ -3, b h (e) ^ 3, b h is the midpoint of ^ a, 2 h and ^ 0, 0 h

3.

Prove that the origin is the midpoint of ^ 3, -4 h and ^ -3, 4 h .

4.

Show that P = Q where P is the midpoint of ^ -2, 3 h and ^ 6, -5 h and Q is the midpoint of ^ -7, -5 h and ^ 11, 3 h .

5.

Find the point that divides the interval between ^ 3, -2 h and ^ 5, 8 h in the ratio of 1:1.

6.

Show that the line x = 3 is the perpendicular bisector of the interval between the points ^ -1, 2 h and ^ 7, 2 h .

7.

8.

9.

The points A ^ -1, 2 h, B ^ 1, 5 h, C ^ 6, 5 h and D ^ 4, 2 h form a parallelogram. Find the midpoints of the diagonals AC and BD. What property of a parallelogram does this show? The points A ^ 3, 5 h, B ^ 9, -3 h, C ^ 5, -6 h and D ^ -1, 2 h form a quadrilateral. Prove that the diagonals are equal and bisect one another. What type of quadrilateral is ABCD? A circle with centre ^ -2, 5 h has one end of a diameter at ^ 4, -3 h . Find the coordinates of the other end of the diameter.

10. A triangle has vertices at A ^ -1, 3 h, B ^ 0, 4 h and C ^ 2, -2 h . (a) Find the midpoints X, Y and Z of sides AB, AC and BC respectively. 1 (b) Show that XY = BC, 2 1 1 XZ = AC and YZ = AB. 2 2 11. Point P ^ x, y h moves so that the midpoint between P and the origin is always a point on the circle x 2 + y 2 = 1. Find the equation of the locus of P. 12. Find the equation of the locus of the point P ^ x, y h that is the midpoint between all points on the circle x 2 + y 2 = 4 and the origin.

Gradient The gradient of a straight line measures its slope. The gradient compares the vertical rise with the horizontal run.


399

rise Gradient = run

On the number plane, this is a measure of the rate of change of y with respect to x.

The rate of change of y with respect to x is a very important measure of their relationship. In later chapters you will use the gradient for many purposes, including sketching curves, finding the velocity and acceleration of objects, and finding maximum and minimum values of formulae.

EXAMPLES Find the gradient of each interval. 1. You will study the gradient at different points on a curve in the next chapter.

Solution rise Gradient = run 2 = 3 CONTINUED

400


2.

Solution In this case, x is - 3 (the run is measured towards the left). rise Gradient = run 2 = -3 2 =3

Positive gradient leans to the right.

Negative gradient leans to the left.

Gradient given 2 points The gradient of the line between _ x 1, y 1 i and _ x 2, y 2 i is given by y2 - y1 m= x -x 2 1

Proof


401

BC = y 2 - y 1 and AC = x 2 - x 1 rise Gradient = run y2 - y1 = x -x 2 1

This formula could also be y1 - y2 written m = x1 - x2

EXAMPLES 1. Find the gradient of the line between points ^ 2, 3 h and ^ -3, 4 h .

Solution y2 - y1 Gradient: m = x - x 2 1 4-3 = -3 - 2 1 = -5 1 =5 2. Prove that points ^ 2, 3 h, ^ -2, -5 h and ^ 0, -1 h are collinear.

Solution To prove points are collinear, we show that they have the same gradient (slope).

CONTINUED

Collinear points lie on the same line, so they have the same gradients.

402


Gradient of the interval between ^ -2, -5 h and ^ 0, -1 h : y2 - y1 m= x -x 2 1 -1 - ] -5 g = 0 - ] -2 g -1 + 5 = 2 4 = 2 =2 Gradient of the interval between ^ 0, -1 h and ^ 2, 3 h : y2 - y1 m= x -x 2 1 3 - ] -1 g = 2-0 3+1 = 2 4 = 2 =2 Since the gradient of both intervals is the same, the points are collinear.

Gradient given the angle at the x-axis The gradient of a straight line is given by m = tan i where i is the angle the line makes with the x-axis in the positive direction

Proof rise m = run opposite = adjacent = tan i


For an acute angle tan i 2 0.

403

For an obtuse angle tan i 1 0.

Class Discussion 1. Which angles give a positive gradient? 2. Which angles give a negative gradient? Why? 3. What is the gradient of a horizontal line? What angle does it make with the x-axis? 4. What angle does a vertical line make with the x-axis? Can you find its gradient?

EXAMPLES 1. Find the gradient of the line that makes an angle of 135c with the x-axis in the positive direction.

Solution

m = tan i = tan 135c = -1

2. Find the angle, in degrees and minutes, that a straight line makes with the x-axis in the positive direction if its gradient is 0.5.

Solution m = tan i ` tan i = 0.5 i = 26c34l

Can you see why the gradient is negative?

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7.3 Exercises 1.

2.

3.

Find the gradient of the line between (a) ^ 3, 2 h and ^ 1, -2 h (b) ^ 0, 2 h and ^ 3, 6 h (c) ^ -2, 3 h and ^ 4, -5 h (d) ^ 2, -5 h and ^ -3, 7 h (e) ^ 2, 3 h and ^ -1, 1 h (f) ^ - 5, 1 h and ^ 3, 0 h (g) ^ -2, -3 h and ^ -4, 6 h (h) ^ -1, 3 h and ^ -7, 7 h (i) ^ 1, -4 h and ^ 5, 5 h (j) ^ 0, 4 h and ^ 3, -2 h If the gradient of _ 8, y 1 i and ^ -1, 3 h is 2, find the value of y 1 . The gradient of ^ 2, -1 h and ^ x, 0 h is –5. Find the value of x.

4.

The gradient of a line is –1 and the line passes through the points ^ 4, 2 h and ^ x, -3 h . Find the value of x. 5. (a) Show that the gradient of the line through ^ -2, 1 h and ^ 3, 4 h is equal to the gradient of the line between the points ^ 2, -1 h and ^ 7,2 h . (b) Draw the two lines on the number plane. What can you say about the lines?

6.

7.

8.

Show that the points A ^ -1, 2 h, B ^ 1, 5 h, C ^ 6, 5 h and D ^ 4, 2 h form a parallelogram. Find the gradients of all sides. The points A ^ 3, 5 h, B ^ 9, -3 h, C ^ 5, -6 h and D ^ -1, 2 h form a rectangle. Find the gradients of all the sides and the diagonals. Find the gradients of the diagonals of the square with vertices A ^ -2, 1 h, B ^ 3, 1 h, C ^ 3, 6 h and D ^ -2, 6 h .

9.

A triangle has vertices A ^ 3, 1 h, B ^ -1, -4 h and C ^ -11, 4 h . (a) By finding the lengths of all sides, prove that it is a rightangled triangle. (b) Find the gradients of sides AB and BC.

10. (a) Find the midpoints F and G of sides AB and AC where ABC is a triangle with vertices A ^ 0, 3 h, B ^ 2, -7 h and C ^ 8, -2 h . (b) Find the gradients of FG and BC. 11. The gradient of the line between a moving point P ^ x, y h and the point A ^ 5, 3 h is equal to the gradient of line PB where B has coordinates ^ 2, -1 h . Find the equation of the locus of P. 12. Prove that the points ^ 3, -1 h, ^ 5, 5 h and ^ 2, -4 h are collinear. 13. Find the gradient of the straight line that makes an angle of 45c with the x-axis in the positive direction. 14. Find the gradient, to 2 significant figures, of the straight line that makes an angle of 42c51l with the x-axis. 15. Find the gradient of the line that makes an angle of 87c14l with the x-axis, to 2 significant figures. 16. Find the angle, in degrees and minutes, that a line with gradient 1.2 makes with the x-axis. 17. What angle, in degrees and minutes does the line with gradient –3 make with the x-axis in the positive direction?


18. Find the exact gradient of the line that makes an angle with the x-axis in the positive direction of (a) 60c (b) 30c (c) 120c. 19. Show that the line passing through ^ 4, -2 h and ^ 7, -5 h

makes an angle of 135c with the x-axis in the positive direction. 20. Find the exact value of x with rational denominator if the line passing through ^ x, 3 h and ^ 2, 1 h makes an angle of 60c with the x-axis.

Gradient given an equation In Chapter 5 you explored and graphed linear functions. You may have noticed a relationship between the graph and the gradient and y-intercept of a straight line.

Investigation 1. (i) Draw the graph of each linear function. (ii) By selecting two points on the line, find its gradient. (a) y = x (b) y = 2x (c) y = 3x (d) y = - x (e) y = - 2x Can you find a pattern for the gradient of each line? Can you predict what the gradient of y = 5x and y = - 9x would be? 2. (i) Draw the graph of each linear function. (ii) Find the y-intercept. (a) y = x (b) y = x + 1 (c) y = x + 2 (d) y = x - 2 (e) y = x - 3 Can you find a pattern for the y-intercept of each line? Can you predict what the y-intercept of y = x + 11 and y = x - 6 would be?

y = mx + b has m = gradient b = y-intercept

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406


EXAMPLES 1. Find the gradient and y-intercept of the linear function y = 7x - 5.

Solution The equation is in the form y = mx + b where m = 7 and b = - 5. Gradient = 7 y-intercept = - 5 2. Find the gradient of the straight line with equation 2x + 3y - 6 = 0.

Solution First, we change the equation into the form y = mx + b. 2x + 3y - 6 = 0 2x + 3y - 6 + 6 = 0 + 6 2x + 3y = 6 2x - 2x + 3y = 6 - 2x 3y = 6 - 2x = - 2x + 6 3y - 2x + 6 = 3 3 - 2x 6 y= + 3 3 2 = - x +2 3 2 m=3 2 So the gradient is - . 3

There is a general formula for finding the gradient of a straight line.

The gradient of the line ax + by + c = 0 is given by m=-

Proof ax + by + c = 0 by = - ax - c ax c y=b b a ` m=b

a b


EXAMPLE Find the gradient of 3x - y = 2.

Solution 3x - y = 2 3x - y - 2 = 0 a = 3, b = - 1 a m=b 3 =-1 =3 ` gradient is 3

7.4 Exercises 1.

Find (i) the gradient and (ii) the y-intercept of each linear function. (a) y = 3x + 5 (b) f ] x g = 2x + 1 (c) y = 6x - 7 (d) y = - x (e) y = - 4x + 3 (f) y = x - 2 (g) f ] x g = 6 - 2x (h) y = 1 - x (i) y = 9x (j) y = 5x - 2

2.

Find (i) the gradient and (ii) the y-intercept of each linear function. (a) 2x + y - 3 = 0 (b) 5x + y + 6 = 0 (c) 6x - y - 1 = 0 (d) x - y + 4 = 0 (e) 4x + 2y - 1 = 0 (f) 6x - 2y + 3 = 0 (g) x + 3y + 6 = 0 (h) 4x + 5y - 10 = 0 (i) 7x - 2y - 1 = 0 (j) 5x - 3y + 2 = 0

3.

Find the gradient of the straight line. (a) y = 4x (b) y = - 2x - 1 (c) y = 2 (d) 2x + y - 5 = 0 (e) x + y + 1 = 0 (f) 3x + y = 8 (g) 2x - y + 5 = 0 (h) x + 4y - 12 = 0 (i) 3x - 2y + 4 = 0 (j) 5x - 4y = 15 2 (k) y = x + 3 3 x (l) y = 2 x (m) y = - 1 5 2x (n) y = +5 7 3x -2 (o) y = 5 x 1 (p) 2y = - + 7 3 y (q) 3x - = 8 5 x y (r) + =1 2 3 2x (s) - 4y - 3 = 0 3 x 2y + +7=0 (t) 4 3

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408


Equation of a Straight Line There are several different ways to write the equation of a straight line.

General form ax + by + c = 0

Gradient form y = mx + b where m = gradient and b = y-intercept

Intercept form

x y a+b =1 where a and b are the x-intercept and y-intercept respectively

Proof b m = - a, b = b `

b y = -ax + b y

`

b y

x = -a + 1

x a+b =1

Point-gradient formula There are two formulae for finding the equation of a straight line. One of these uses a point and the gradient of the line.

The equation of a straight line is given by y - y1 = m _ x - x1 i This is a very useful formula as it is used in many topics in this course.

where _ x 1, y 1 i lies on the line with gradient m


409

Proof Given point _ x 1, y 1 i on the line with gradient m Let P = ^ x, y h Then line AP has gradient y2 - y1 m= x -x 2 1 y - y1 ` m= x-x 1 m _ x - x1 i = y - y1

Two-point formula The equation of a straight line is given by y - y1 y2 - y1 = x - x1 x2 - x1

This formula is optional as you can use the point–gradient formula for any question.

where _ x 1, y 1 i and _ x 2, y 2 i are points on the line

Proof

The gradient is the same anywhere along a straight line.

Let P = ^ x, y h D APQ <; D ABR PQ BR So = AR AQ y - y1 y2 - y1 i.e. x - x = x - x 1 2 1 The two-point formula is not essential. The right-hand side of it is the gradient of the line. Replacing this by m gives the point–gradient formula.

410


EXAMPLES 1. Find the equation of the straight line with gradient -4 and passing through the point ^ -2, 3 h .

Solution m = -4, x 1 = -2 and y 1 = 3 Equation: y - y 1 = m (x - x 1) y - 3 = - 4 [x - (-2)] = - 4 (x + 2) = - 4x - 8 ` y = - 4x - 5 or 4x + y + 5 = 0

(gradient form) (general form)

2. Find the equation of the straight line that passes through the points ^ 2, -3 h and ^ -4, -7 h .

Solution By two-point formula: y - y1 y2 - y1 = x - x1 x2 - x1 y - ] -7 g -3 - ] -7 g = x - ] -4 g 2 - ] -4 g y+7 -3 + 7 = x+4 2+4 y+7 2 = x+4 3 3 ^ y + 7 h = 2 ]x + 4 g 3y + 21 = 2x + 8 -2x + 3y + 13 = 0 or 2x - 3y - 13 = 0 By point-gradient method: y2 - y1 m= x -x 2 1 -3 - ] -7 g = 2 - ] -4 g -3 + 7 = 2+4 2 = 3 Use one of the points, say ^ -4, -7 h . 2 m = , x 1 = -4 and y 1 = -7 3 Equation:

y - y 1 = m ( x - x 1) y - (-7) =

2 6 x - ( - 4) @ 3


2 ( x + 4) 3 = 2 ]x + 4 g = 2x + 8 =0 =0

y+7= 3^ y + 7h 3y + 21 ` -2x + 3y + 13 or 2x - 3y - 13

3. Find the equation of the line with x-intercept 3 and y-intercept 2.

Solution x y Intercept form is a + = 1, where a and b are the x-intercept and b y-intercept respectively. x y ` + =1 3 2 2x + 3y = 6 ` 2x + 3y - 6 = 0 Again, the point-gradient formula can be used. The x-intercept and y-intercept are the points ^ 3, 0 h and ^ 0, 2 h .

7.5 Exercises 1.

Find the equation of the straight line (a) with gradient 4 and y-intercept -1 (b) with gradient -3 and passing through ^ 0, 4 h (c) passing through the origin with gradient 5 (d) with gradient 4 and x-intercept -5 (e) with x-intercept 1 and y-intercept 3 (f) with x-intercept 3, y-intercept -4 (g) with y-intercept -1 and making an angle of 45c with the x-axis in the positive direction (h) with y-intercept 5 and making an angle of 45c with the x-axis in the positive direction.

2.

Find the equation of the straight line that makes an angle of 135c with the x-axis and passes through the point ^ 2, 6 h .

3.

Find the equation of the straight line passing through (a) ^ 2, 5 h and ^ -1, 1 h (b) ^ 0, 1 h and ^ -4, -2 h (c) ^ - 2, 1 h and ^ 3, 5 h (d) ^ 3, 4 h and ^ -1, 7 h (e) ^ -4, -1 h and ^ - 2, 0 h .

4.

What is the equation of the line with x-intercept 2 and passing through ^ 3, -4 h ?

5.

Find the equation of the line (a) parallel to the x-axis and passing through ^ 2, 3 h (b) parallel to the y-axis and passing through ^ -1, 2 h .

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6.

A straight line passing through the origin has a gradient of - 2. Find its equation.

7.

A straight line has x-intercept 4 and passes through ^ 0, -3 h . Find its equation.

8.

Find the equation of the straight line with gradient -2 that passes through the midpoint of ^ 5, -2 h and ^ -3, 4 h .

9.

What is the equation of the straight line through the point ^ -4, 5 h and the midpoint of ^ 1, 2 h and ^ -9, 4 h ?

10. What is the equation of the straight line through the midpoint of ^ 0, 1 h and ^ -6, 5 h and the midpoint of ^ 2, 3 h and ^ 8, -3 h ?

Parallel and Perpendicular Lines Parallel lines

Class Investigation Sketch the following straight lines on the same number plane. 1. y = 2x 2. y = 2x + 1 3. y = 2x - 3 4. y = 2x + 5 What do you notice about these lines?

If two lines are parallel, then they have the same gradient. That is, m1 = m2

Two lines that are parallel have equations ax + by + c 1 = 0 and ax + by + c 2 = 0


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Proof a b a ax + by + c 2 = 0 has gradient m 2 = b Since m 1 = m 2, the two lines are parallel. ax + by + c 1 = 0 has gradient m 1 = -

EXAMPLES 1. Prove that the straight lines 5x - 2y - 1 = 0 and 5x - 2y + 7 = 0 are parallel.

Solution 5x - 2y - 1 = 0 5x - 1 = 2y 5 1 x- =y 2 2 5 ` m1 = 2 5x - 2y + 7 = 0 5x + 7 = 2 y 5 7 x+ =y 2 2 5 ` m2 = 2 5 m1 = m2 = 2 ` the lines are parallel. 2. Find the equation of a straight line parallel to the line 2x - y - 3 = 0 and passing through ^ 1, -5 h .

Solution 2x - y - 3 = 0 2x - 3 = y ` m1 = 2 For parallel lines m 1 = m 2 ` m2 = 2 Equation:

y - y 1 = m (x - x 1) y - (-5) = 2 (x - 1) y + 5 = 2x - 2 0 = 2x - y - 7

Notice that the equations are both in the form 5x - 2y + k = 0.

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DID YOU KNOW? Parallel lines are usually thought of as lines that never meet. However, there is a whole branch of geometry based on the theory that parallel lines meet at infinity. This is called affine geometry. In this geometry there are no perpendicular lines.

Perpendicular lines

Class Investigation Sketch the following pairs of straight lines on the same number plane. 1. (a) 3x - 4y + 12 = 0 2. (a) 2x + y + 4 = 0

(b) 4x + 3y - 8 = 0 (b) x - 2y + 2 = 0

What do you notice about these pairs of lines?

Gradients of perpendicular lines are negative reciprocals of each other.

If two lines with gradients m 1 and m 2 respectively are perpendicular, then m 1 m 2 = -1 1 i.e. m 2 = - m 1

Proof

Let line AB have gradient m 1 = tan a . Let line CD have gradient m 2 = tan b. EB EC +CBE = 180c - a EC tan ] 180c - a g = EB EB ` cot ] 180c - a g = EC tan b =

^ straight angle h


`

So or

tan b = cot ] 180c - a g = - cot a 1 =tan a 1 m2 = - m 1 m 1 m 2 = -1

Perpendicular lines have equations in the form ax + by + c 1 = 0 and bx - ay + c 2 = 0

Proof a b b bx - ay + c 2 = 0 has gradient m 2 = - - a b =a a b m1 m2 = - # a b = -1 ax + by + c 1 = 0 has gradient m 1 = -

Since m 1 m 2 = -1, the two lines are perpendicular.

EXAMPLES 1. Show that the lines 3x + y - 11 = 0 and x - 3y + 1 = 0 are perpendicular.

Solution 3x + y - 11 = 0 y = -3x + 11 m 1 = -3 ` x - 3y + 1 = 0 x + 1 = 3y 1 1 x+ =y 3 3 1 ` m2 = 3 1 m 1 m 2 = - 3# 3 = -1

Notice that the equations are in the form 3x + y + c 1 = 0 and x - 3y + c 2 = 0.

` the lines are perpendicular.

CONTINUED

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2. Find the equation of the straight line through ^ 2, 3 h perpendicular to the line that passes through ^ -1, 7 h and ^ 3, 3 h .

Solution Line through ^ -1, 7 h and ^ 3, 3 h: y2 - y1 m= x -x 2 1 7-3 m1 = -1 - 3 4 = -4 = -1 For perpendicular lines, m 1 m 2 = - 1 i.e. -1m 2 = - 1 m2 = 1 Equation through ^ 2, 3 h: y - y 1 = m (x - x 1) y - 3 = 1 (x - 2 ) =x-2 0=x-y+1

7.6 Exercises 1.

Find the gradient of the straight line (a) parallel to the line 3x + y - 4 = 0 (b) perpendicular to the line 3x + y - 4 = 0 (c) parallel to the line joining ^ 3, 5 h and ^ -1, 2 h (d) perpendicular to the line with x-intercept 3 and y-intercept 2 (e) perpendicular to the line making an angle of 135c with the x-axis in the positive direction (f) perpendicular to the line 6x - 5y - 4 = 0 (g) parallel to the line making an angle of 30c with the x-axis (h) parallel to the line x - 3y - 7 = 0

(i) perpendicular to the line making an angle of 120c with the x-axis in the positive direction (j) perpendicular to the line passing through ^ 4, -2 h and ^ 3, 3 h . 2.

Find the equation of each straight line (a) passing through ^ 2, 3 h and parallel to the line y = x + 6 (b) through ^ -1, 5 h and parallel to the line x - 3y - 7 = 0 (c) with x-intercept 5 and parallel to the line y = 4 - x (d) through ^ 3, -4 h and perpendicular to the line y = 2x (e) through ^ -2, 1 h and perpendicular to the line 2x + y + 3 = 0


(f) through ^ 7, -2 h and perpendicular to the line 3x - y - 5 = 0 (g) through ^ -3, -1 h and perpendicular to the line 4x - 3y + 2 = 0 . 3.

Show that the straight lines y = 3x - 2 and 6x - 2y - 9 = 0 are parallel.

4.

Show that lines x + 5y = 0 and y = 5x + 3 are perpendicular.

5.

Show that lines 6x - 5y + 1 = 0 and 6x - 5y - 3 = 0 are parallel.

6.

Show that lines 7x + 3y + 2 = 0 and 3x - 7y = 0 are perpendicular.

7.

If the lines 3x - 2y + 5 = 0 and y = kx - 1 are perpendicular, find the value of k.

8.

9.

Show that the line joining ^ 3, -1 h and ^ 2, -5 h is parallel to the line 8x - 2y - 3 = 0. Show that the points A ^ -3, -2 h, B ^ -1, 4 h, C ^ 7, -1 h, and D ^ 5, -7 h are the vertices of a parallelogram.

10. The points A ^ -2, 0 h, B ^ 1, 4 h, C ^ 6, 4 h and D ^ 3, 0 h form a rhombus. Show that the diagonals are perpendicular.

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11. Find the equation of the straight line (a) passing through the origin and parallel to the line x+y+3=0 (b) through ^ 3, 7 h and parallel to the line 5x - y - 2 = 0 (c) through ^ 0, - 2 h and perpendicular to the line x - 2y = 9 (d) perpendicular to the line 3x + 2y - 1 = 0 and passing through the point ^ -2, 4 h . 12. Find the equation of the straight line passing through ^ 6, -3 h that is perpendicular to the line joining ^ 2, -1 h and ^ -5, -7 h . 13. Find the equation of the line through ^ 2, 1 h that is parallel to the line that makes an angle of 135c with the x-axis in the positive direction. 14. Find the equation of the perpendicular bisector of the line passing through ^ 6, -3 h and ^ -2, 1 h . 15. Find the equation of the straight line parallel to the line 2x - 3y - 1 = 0 and through the midpoint of ^ 1, 3 h and ^ -1, 9 h .

Intersection of Lines Two straight lines intersect at a single point ^ x, y h . The point satisfies the equations of both lines. We find this point by solving simultaneous equations.

You may need to revise simultaneous equations from Chapter 3.

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Concurrent lines meet at a single point. To show that lines are concurrent, solve two simultaneous equations to find the point of intersection. Then substitute this point of intersection into the third and subsequent lines to show that these lines also pass through the point.

EXAMPLES 1. Find the point of intersection between lines 2x - 3y - 3 = 0 and 5x - 2y - 13 = 0.

Solution Solve simultaneous equations: 2x - 3y - 3 = 0 5x - 2y - 13 = 0 4x - 6y - 6 = 0 ^ 1 h # 2: 15x - 6y - 39 = 0 ^ 2 h # 3: + 33 = 0 ^ 3 h - ^ 4 h: -11x 33 = 11x 3=x

^1h ^2h ^3h ^4h

Substitute x = 3 into ^ 1 h:

You could use a computer spreadsheet to solve these simultaneous equations.

2 ^ 3 h - 3y - 3 = 0 - 3y + 3 = 0 3 = 3y 1=y So the point of intersection is ^ 3, 1 h . 2. Show that the lines 3x - y + 1 = 0, x + 2y + 12 = 0 and 4x - 3y - 7 = 0 are concurrent.

Solution Solve any two simultaneous equations: 3x - y + 1 = 0 x + 2y + 12 = 0 4x - 3y - 7 = 0 6x - 2y + 2 = 0 ^ 1 h # 2: 2 + 4 : 7 x + 14 = 0 ^ h ^ h

^1h ^2h ^3h ^4h


7x = -14 x = -2 Substitute x = -2 into ^ 1 h: 3 ^ -2 h - y + 1 = 0 -y - 5 = 0 -5 = y

So the point of intersection of (1) and (2) is ^ -2, -5 h . Substitute ^ -2, -5 h into (3): 4x - 3y - 7 = 0 LHS = 4 ^ -2 h - 3 ^ - 5 h - 7 = -8 + 15 - 7 =0 = RHS So the point lies on line (3) ` all three lines are concurrent.

Equation of a line through the intersection of 2 other lines To find the equation of a line through the intersection of 2 other lines, find the point of intersection, then use it with the other information to find the equation. Another method uses a formula to find the equation.

If a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are 2 given lines then the equation of a line through their intersection is given by the formula (a 1 x + b 1 y + c 1) + k (a 2 x + b 2 y + c 2) = 0 where k is a constant

Proof Let l 1 have equation a 1 x + b 1 y + c 1 = 0. Let l 2 have equation a 2 x + b 2 y + c 2 = 0. Let the point of intersection of l 1 and l 2 be P ^ x 1, y 1 h . Then P satisfies l 1 i.e. a 1 x 1 + b 1 y 1 + c 1 = 0 P also satisfies l2 i.e. a 2 x 1 + b 2 y 1 + c 2 = 0 Substitute P into (a 1 x + b 1 y + c 1) + k (a 2 x + b 2 y + c 2) = 0 (a 1 x 1 + b 1 y 1 + c 1) + k (a 2 x 1 + b 2 y 1 + c 2) = 0 0 + k ^0h = 0 0=0

` if point P satisfies both equations l 1 and l 2 then it satisfies l 1 + kl 2 = 0.

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EXAMPLE Find the equation of the line through ^ -1, 2 h that passes through the intersection of lines 2x + y - 5 = 0 and x - 3y + 1 = 0.

Solution Using the formula: a 1 = 2, b 1 = 1, c 1 = -5

a 2 = 1, b 2 = -3, c 2 = 1

^ a1 x + b1 y + c1 h + k ^ a2 x + b2 y + c2 h = 0 ^ 2x + y - 5 h + k ^ x - 3y + 1 h = 0 Since this line passes through ^ -1, 2 h, substitute the point into the equation: ^ -2 + 2 - 5 h + k ^ -1 - 6 + 1 h = 0 -5 - 6k = 0 -5 = 6k 5 - =k 6 So the equation becomes: 5 ^ 2x + y - 5 h - ^ x - 3y + 1 h = 0 6 6 ^ 2x + y - 5 h - 5 ^ x - 3 y + 1 h = 0 12x + 6y - 30 - 5x + 15y - 5 = 0 7x + 21y - 35 = 0 x + 3y - 5 = 0 Another way to do this example is to find the point of intersection, then use both points to find the equation.

Substitute the value of k back into the equation.

7.7 Exercises 1.

Find the point of intersection of straight lines (a) 3x + 4y + 10 = 0 and 2x - 3y - 16 = 0 (b) 5x + 2y + 11 = 0 and 3x + y + 6 = 0 (c) 7x - 3y = 16 and 5x - 2y = 12 (d) 2x - 3y = 6 and 4x - 5y = 10 (e) x - 3y - 8 = 0 and 4x + 7y - 13 = 0 (f) y = 5x + 6 and y = - 4x - 3 (g) y = 2x + 1 and 5x - 3y + 6 = 0

(h) 3x + 7y = 12 and 4x - y - 1 6 = 0 (i) 3x - 5y = - 7 and 2x - 3y = 4 (j) 8x - 7y - 3 = 0 and 5x - 2y - 1 = 0 2.

Show that the lines x - 2y - 11 = 0 and 2x - y - 10 = 0 intersect at the point ^ 3, -4 h .

3.

A triangle is formed by 3 straight lines with equations 2x - y + 1 = 0, 2x + y - 9 = 0


and 2x - 5y - 3 = 0. Find the coordinates of its vertices. 4.

Show that the lines x - 5y - 17 = 0, 3x - 2y - 12 = 0 and 5x + y - 7 = 0 are concurrent.

5.

Show that the lines x + 4y + 5 = 0, 3x - 7y + 15 = 0, 2x - y + 10 = 0 and 6x + 5y + 30 = 0 are concurrent.

6.

Find the equation of the straight line through the origin that passes through the intersection of the lines 5x - 2y + 14 = 0 and 3x + 4y - 7 = 0 .

7.

Find the equation of the straight line through ^ 3, 2 h that passes through the intersection of the lines 5x + 2y + 1 = 0 and 3x - y + 16 = 0.

8.

Find the equation of the straight line through ^ -4, -1 h that passes through the intersection of the lines 2x + y - 1 = 0 and 3x + 5y + 16 = 0.

9.

Find the equation of the straight line through ^ -3, 4 h that passes through the intersection of the lines 2x + y - 3 = 0 and 3x - 2y - 8 = 0 .

10. Find the equation of the straight line through ^ 2, -2 h that passes through the intersection of the lines 2x + 3y - 6 = 0 and 3x + 5y - 10 = 0. 11. Find the equation of the straight line through ^ 3, 0 h that passes through the intersection of the lines x - y + 1 = 0 and 4x - y - 2 = 0 .

12. Find the equation of the straight line through ^ -1, -2 h that passes through the intersection of the lines 2x + y - 6 = 0 and 3 x + 7 y - 9 = 0. 13. Find the equation of the straight line through ^ 1, 2 h that passes through the intersection of the lines x + 2y + 10 = 0 and 2x - y + 5 = 0. 14. Find the equation of the straight line through ^ -2, 0 h that passes through the intersection of the lines 3x + 4y - 7 = 0 and 3 x - 2 y - 1 = 0. 15. Find the equation of the straight line through ^ 3, -2 h that passes through the intersection of the lines 5x + 2y - 13 = 0 and x - 3y + 11 = 0. 16. Find the equation of the straight line through ^ -3, -2 h that passes through the intersection of the lines x + y + 1 = 0 and 3x + 2y = 0 . 17. Find the equation of the straight line through ^ 3, 1 h that passes through the intersection of the lines 3x - y + 4 = 0 and 2x - y + 12 = 0. 18. Find the equation of the straight line with gradient 3 that passes through the intersection of the lines 2x + y - 1 = 0 and 3x + 5y + 16 = 0. 19. Find the equation of the straight line with gradient 2 that passes through the intersection of the lines 5x - 2y - 3 = 0 and 7x - 3y - 4 = 0 .

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20. Find the equation of the straight line parallel to the line 3x - y - 7 = 0 that passes through the intersection of the lines 3x - 2y - 10 = 0 and 4x + y - 17 = 0.

21. Find the equation of the straight line perpendicular to the line x + 5y - 1 = 0 that passes through the intersection of lines 3x - 5y - 3 = 0 and 2x + 3y + 17 = 0.

Perpendicular Distance The distance formula d = _ x 2 - x 1 i2 + _ y 2 - y 1 i2 is used to find the distance between two points. Perpendicular distance is used to find the distance between a point and a line. If we look at the distance between a point and a line, there could be many distances.

So we choose the shortest distance, which is the perpendicular distance.

The perpendicular distance from _ x 1, y 1 i to the line ax + by + c = 0 is A distance is always positive, so take the absolute value.

given by d =

Proof

| ax 1 + by 1 + c | a2 + b2


Let d be the perpendicular distance of _ x 1, y 1 i from the line ax + by + c = 0. - ax 1 - c c c o C = c 0, - m R = e x 1, A = b- a , 0 l b b c2 c2 + a2 b2

In D ACO, AC =

c2 b2 + c2 a2 a2 b2

= = PR = y 1 - e =

c a2 + b2 ab

- ax 1 - c

b ax 1 + by 1 + c

o

b Why?

D ACO is similar to D PRQ `

`

To find A and C, substitute y = 0 and x = 0 into ax + by + c = 0.

PQ PR = AO AC AO . PR PQ = AC ax 1 + by 1 + c c a2 + b2 c d=a# ' b ab c _ ax 1 + by 1 + c i ab = # ab c a2 + b2 ax 1 + by 1 + c = a2 + b2

All points on one side of the line ax + by + c = 0 make the numerator of this formula positive. Points on the other side make the numerator negative. Usually we take the absolute value of d. However, if we want to know if points are on the same side of a line or not, we look at the sign of d.

EXAMPLES 1. Find the perpendicular distance of ^ 4, - 3 h from the line 3x - 4y - 1 = 0.

Solution x 1 = 4, y 1 = - 3, a = 3, b = - 4, c = - 1 | ax 1 + by 1 + c | d= a2 + b2 | 3 ] 4 g + ] - 4 g ] -3 g + ] -1 g | = 3 2 + ] -4 g2 CONTINUED

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=

| 12 + 12 - 1 |

25 23 = 5 = 4 .6 So the perpendicular distance is 4.6 units. 2. Prove that the line 6x + 8y + 20 = 0 is a tangent to the circle x 2 + y 2 = 4.

Solution There are three possibilities for the intersection of a circle and a straight line.

The centre of the circle x 2 + y 2 = 4 is ^ 0, 0 h and its radius is 2 units. A tangent is perpendicular to the centre of the circle. So we prove that the perpendicular distance from the line to the point ^ 0, 0 h is 2 units (the radius). | ax 1 + by 1 + c | d= a2 + b2 | 6 (0) + 8 (0) + 20 | = 62 + 82 | 20 | = 100 20 = 10 =2 ` the line is a tangent to the circle. 3. Show that the points ^ -1, 3 h and ^ 2, 7 h lie on the same side of the line 2 x - 3 y + 4 = 0.


Solution To show that points lie on the same side of a line, their perpendicular distance must have the same sign. We use the formula without the absolute value sign. d=

ax 1 + by 1 + c a2 + b2

^ - 1, 3 h : 2 ]-1 g - 3 ]3 g + 4 d= 22 + ] - 3 g 2 -2 - 9 + 4 = 4+9 -7 = 13 ^ 2, 7 h : 2 ]2 g - 3 ]7 g + 4 d= 2 2 + ] -3 g 2 4 - 21 + 4 = 4+9 - 13 = 13

Since the perpendicular distance for both points has the same sign, the points lie on the same side of the line.

7.8 Exercises 1.

Find the perpendicular distance between (a) ^ 1, 2 h and 3x + 4y + 2 = 0 (b) ^ - 3, 2 h and 5x + 12y + 7 = 0 (c) ^ 0, 4 h and 8x - 6y - 1 = 0 (d) ^ - 3, - 2 h and 4x - 3y - 6 = 0 (e) the origin and 12x - 5y + 8 = 0.

2.

Find, correct to 3 significant figures, the perpendicular distance between (a) ^ 1, 3 h and x + 3y + 1 = 0 (b) ^ -1, 1 h and 2x + 5y + 4 = 0 (c) ^ 3, 0 h and 5x - 6y - 12 = 0 (d) ^ 5, - 3 h and 4x - y - 2 = 0 (e) ^ - 6, - 3 h and 2x - 3y + 9 = 0.

3.

Find as a surd with rational denominator the perpendicular distance between (a) the origin and the line 3x - 2y + 7 = 0 (b) ^ -1, 4 h and 2x + y + 3 = 0 (c) ^ 3, -1 h and 3x + 14y + 1 = 0 (d) ^ 2, - 6 h and 5x - y - 6 = 0 (e) ^ - 4, - 1 h and 3 x - 2 y - 4 = 0.

4.

Show that the origin is equidistant from the lines 7x + 24y + 25 = 0, 4x + 3y - 5 = 0 and 12x + 5y - 13 = 0.

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5.

6.

7.

Equidistant means that two or more objects are the same distance away from another object.

8.

9.

Show that points A ^ 3, - 5 h and B ^ -1, 4 h lie on opposite sides of 2x - y + 3 = 0.

14. Find the perpendicular distance between ^ 0, 5 h and the line through ^ - 3, 8 h parallel to 4x - 3y - 1 = 0.

Show that the points ^ 2, - 3 h and ^ 9, 2 h lie on the same side of the line x - 3y + 2 = 0.

15. The perpendicular distance between the point ^ x, -1 h and the line 3x - 4y + 7 = 0 is 8 units. Find two possible values of x.

Show that ^ - 3, 2 h and ^ 4, 1 h lie on opposite sides of the line 4 x - 3 y - 2 = 0. Show that ^ 0, - 2 h is equidistant from the lines 3x + 4y - 2 = 0 and 12x - 5y + 16 = 0.

16. The perpendicular distance between the point ^ 3, b h and the line 5x - 12y - 2 = 0 is 2 units. Find the values of b.

Show that the points ^ 8, - 3 h and ^ 1, 1 h lie on the same side of the line 6x - y + 4 = 0.

17. Find m if the perpendicular distance between ^ m, 7 h and the line 9x + 12y + 6 = 0 is 5 units.

10. Show that ^ - 3, 2 h and ^ 4, 1 h lie on opposite sides of the line 2x + y - 2 = 0.

18. Prove that the line 3x - 4y + 25 = 0 is a tangent to the circle with centre the origin and radius 5 units.

11. Show that the point ^ 3, - 2 h is the same distance from the line 6x - 8y + 6 = 0 as the point ^ - 4, -1 h is from the line 5x + 12y - 20 = 0.

19. Show that the line 3x - 4y + 12 = 0 does not cut the circle x 2 + y 2 = 1. 20. The sides of a triangle are formed by the lines with equations 2x - y - 7 = 0, 3x + 5y - 4 = 0 and x + 3y - 4 = 0. (a) Find the vertices of the triangle. (b) Find the exact length of all the altitudes of the triangle.

12. Find the exact perpendicular distance with rational denominator from the point ^ 4, 5 h to the line with x-intercept 2 and y-intercept -1. 13. Find the perpendicular distance from ^ - 2, 2 h to the line passing through ^ 3, 7 h and ^ -1, 4 h .

Angle Between Two Lines

The acute angle i between two straight lines is given by tan i =

m1 - m2 1 + m1 m2

where m 1 and m 2 are the gradients of the lines


Proof

Let line l 1 have gradient m 1 and line l 2 have gradient m 2 . Then m 1 = tan b and m 2 = tan a b = a + i ^ exterior angle of DABC h ` i=b-a tan i = tan (b - a ) tan b - tan a = 1 + tan b tan a m1 - m2 = 1 + m1 m2

When tan i is positive, i is acute. When tan i is negative, i is obtuse. ` for the acute angle between lines l 1 and l 2, tan i =

Note: the denominator cannot be zero, so m 1 m 2 ! -1. So this formula doesn’t work for perpendicular lines.

m1 - m2 1 + m1 m2

EXAMPLES 1. Find the acute angle between the lines 3x - 2y + 1 = 0 and x - 3 y = 0.

Solution 3x - 2y + 1 = 0 3x + 1 = 2y 3 1 x+ =y 2 2 3 m1 = So 2 x - 3y = 0 x = 3y 1 x=y 3 1 m2 = So 3 m1 - m2 tan i = 1 + m1 m2 CONTINUED

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3 1 2 3 = 3 1 1+ # 2 3 7 = 9 7 i = tan -1 c m 9 = 37c 52l 2. Find the obtuse angle between the lines 5x - 2y + 6 = 0 and 2x + y - 4 = 0.

Solution

9 Notice that tan -1 d - n 8 gives - 48c 22l so we need to find the obtuse angle by subtracting the acute angle from 180c.

5x - 2y + 6 = 0 5x + 6 = 2y 5 x+3=y 2 5 So m 1 = 2 2x + y - 4 = 0 y = - 2x + 4 So m 2 = - 2 m1 - m2 tan i = 1 + m1 m2 5 ] - -2g 2 = 5 1 + # ]-2 g 2 9 = 8 9 = 8 9 i = tan - 1 c m 8 = 48° 22l This gives the acute angle. Obtuse angle = 180c - 48c 22l = 131c 38l 3. If the angle between the lines 2x - y - 7 = 0 and y = mx + 3 is 25c, find two possible values of m, correct to 1 decimal place.

Solution 2x - y - 7 = 0 2x - 7 = y ` m1 = 2

( 1)


y = mx + 3 ` m2 = m m1 - m2 tan i = 1 + m1 m2 2-m tan 25° = 1 + 2m

(2 )

There are two possibilities: (1)

(2)

2-m 1 + 2m tan 25c (1 + 2m) = 2 - m tan 25c + 2m tan 25c = 2 - m 2m tan 25c + m = 2 - tan 25c m (2 tan 25c + 1) = 2 - tan 25c 2 - tan 25c m= 2 tan 25c + 1 Z 0.8 tan 25c =

2-m 1 + 2m - tan 25c (1 + 2m) = 2 - m - tan 25c - 2m tan 25c = 2 - m - 2m tan 25c + m = 2 + tan 25c m (- 2 tan 25c + 1) = 2 + tan 25c 2 + tan 25c m= - 2 tan 25c + 1 Z 36.6 - tan 25c =

7.9 Exercises 1.

Find the acute angle between the lines (a) 2x + y + 1 = 0 and x+y+4=0 (b) 3x - y - 7 = 0 and 5x + y + 3 = 0 (c) x + 2y = 0 and 3x - 2y + 1 = 0 (d) x + 3y + 2 = 0 and 4x + 4y - 1 = 0 (e) 2x - 5y - 3 = 0 and x - 5y = 0 (f) 3x + y + 1 = 0 and 4x + 7y + 2 = 0

(g) 2x - 7y - 1 = 0 and 3x + 2y - 4 = 0 (h) 2x + 2y + 1 = 0 and x + 2y = 4 (i) 3x + 4y + 1 = 0 and 5x - 2y - 2 = 0 (j) x - 2y - 3 = 0 and 6 x - 3 y + 4 = 0. 2.

Find the obtuse angle between the lines (a) 4x + y + 2 = 0 and x+y-1=0 (b) 2x - 3y - 9 = 0 and x + 2y + 4 = 0

429

430


(c) x + 6y = 2 and 2x - 4y + 3 = 0 (d) 5x + 2y + 1 = 0 and 4x + y - 7 = 0 (e) 4x - 2y - 7 = 0 and x - 3 y = 0. 3.

Find the acute angle between the line 2x - 5y + 1 = 0 and the line joining ^ -1, 2 h and ^ 5, 3 h .

4.

Find the acute angle between the line joining ^ 3, 2 h and ^ -1, 4 h and the line joining ^ 0, 5 h and ^ 2, - 7 h .

5.

6.

A ^ 2, -1 h, B ^ - 3, 4 h and C ^ 1, - 5 h form the vertices of a triangle. Find the interior angles of the triangle. Find two possible values of m if the lines 2x + y - 5 = 0 and y = mx + 1 intersect at an angle of 45c.

7.

Lines y = mx + 2 and y = 5x - 9 intersect at an acute angle whose 2 tangent is . Find the possible 5 values of m.

8.

Find the values of k if the angle between the lines 6x - 3y - 4 = 0 and kx - y + 5 = 0 is 58c.

9.

A ^ 0, 0 h, B ^ 1, 2 h, C ^ 5, 2 h and D ^ 4, 0 h form the vertices of a parallelogram. (a) By finding all the interior angles, show that opposite angles are equal. (b) Find the obtuse angle between the diagonals of the parallelogram.

10. By calculating the interior angles, show that D ABC with vertices A ^ 7, 1 h, B ^ -1, -1 h and C ^ 5, -7 h is an isosceles triangle.

Ratios You have a formula for the midpoint which divides an interval in half. Sometimes we may want to divide an interval into a ratio that is not a half. Here is a formula that we can use to divide an interval into any internal or external ratio.

The coordinates of a point P that divides the interval between points _ x 1, y 1 i and _ x 2, y 2 i in the ratio m: n respectively are given by mx 2 + nx 1 my 2 + ny 1 x= and y = m+n m+n

Proof

431


Let P ^ x, y h be the point dividing the interval AB into the ratio m:n. Then

m AP = n PB

Draw ADC parallel to the x-axis. Then AD = x - x 1 and DC = x 2 - x. PD < BC AP AD ` = ^ intercepts have equal ratios h PB DC x - x1 m ` n = x2 - x m _ x2 - x i = n _ x - x1 i mx 2 - mx = nx - nx 1 mx 2 + nx 1 = mx + nx = x ]m + n g mx 2 + nx 1 =x m+n Similarly, by drawing AEF perpendicular to the x-axis, we can show that my 2 + ny 1 y= . m+n If P divides the interval internally in the ratio m: n, then the ratio is positive and P lies on AB.

A ratio of 1:1 gives the midpoint x =

x1 + x2 2

,y =

y1 + y2 2

.

If P divides the interval externally in the ratio m:n, then the ratio is negative and P lies outside AB. m and n are measured in opposite directions so they have opposite signs.

EXAMPLES 1. Divide AB into the ratio 3:4 where A is ^ 6, - 2 h and B is ^ - 7, 5 h .

Solution

CONTINUED

432


mx 2 + nx 1 m+n ] 3 -7 g + 4 ] 6 g = 3+4 3 = 7 my 2 + ny 1 y= m+n 3 ] 5 g + 4 ] -2 g = 3+4 7 = 7 =1

x=

3 ` P = c ,1m 7 2. If A is ^ - 2, -1 h and B is ^ 1, 5 h, find the coordinates of the point P that divides AB externally in the ratio 2:5.

Solution

You could use - 2: 5 instead and would still get the same answer for P.

Let the ratio be 2: -5. mx 2 + nx 1 x= m+n 2 (1) + [- 5 (- 2)] = 2 + ( - 5) 12 = -3 = -4 my 2 + ny 1 y= m+n 2 (5) + [- 5 (-1)] = 2 + ( - 5) 15 = -3 = -5 ` P = ^ - 4, - 5 h


7.10 1.

2.

3.

Exercises

Divide these intervals internally. (a) ^ -1, 5 h and ^ 0, - 4 h in the ratio 2:3 (b) ^ 3, - 2 h and ^ 2, 5 h in the ratio 4:1 (c) ^ - 3, 3 h and ^ - 2, 1 h in the ratio 5:4 (d) ^ 3, -1 h and ^ 7, - 2 h in the ratio 2:5 (e) ^ - 2, 1 h and ^ 5, - 4 h in the ratio 7:3 (f) ^ - 2, 0 h and ^ - 6, 3 h in the ratio 3:1 (g) ^ 4, 9 h and ^ - 4, 1 h in the ratio 1:6 (h) ^ - 3, 0 h and ^ - 5, - 6 h in the ratio 2:9 (i) ^ 2, 5 h and ^ - 3, -1 h in the ratio 4:3 (j) ^ 1, 1 h and ^ 3, - 7 h in the ratio 1:2. Divide these intervals externally. (a) ^ - 2, 3 h and ^ 6, 1 h in the ratio 1:5 (b) ^ 4, 0 h and ^ - 3, - 5 h in the ratio 2:7 (c) ^ - 1, 1 h and ^ 4, 7 h in the ratio 4:3 (d) ^ 0, - 2 h and ^ 8, 3 h in the ratio 3:1 (e) ^ - 5, 2 h and ^ 4, 4 h in the ratio 5:4 (f) ^ 7, -1 h and ^ 0, 1 h in the ratio 2:9 (g) ^ - 2, 2 h and ^ 6, 7 h in the ratio 1:3 (h) ^ 1, 3 h and ^ 7, 2 h in the ratio 4:1 (i) ^ - 4, 0 h and ^ 5, - 5 h in the ratio 6:7 (j) ^ 2, - 3 h and ^ 7, 7 h in the ratio 8:3. A ^ 0, 0 h, B ^ 1, 3 h and C ^ 3, 0 h are the vertices of a triangle.

(a) Find the coordinates of point E, which divides AB internally in the ratio 2:1. (b) Find the coordinates of point F, which divides CB internally in the ratio 2:1. (c) Hence prove that AC = 3EF. 4.

5.

6.

7.

8.

9.

Divide the interval AB where A = ^ 3, 2 h and B = ^ - 1, 6 h into three equal parts. A has coordinates ^ - 2, 5 h and B has coordinates ^ 4, -3 h . Find the length of PQ if P divides AB internally in the ratio 3:2 and Q divides AB externally in the ratio 3:2. An interval AB is divided internally at P in the ratio 5:4. If A is ^ - 1, 2 h and P is ^ 5, - 6 h, find the coordinates of B. The point ^ 5, 5 h divides the interval between ^ - 1, p h and ^ q, 6 h in the ratio 2:5. Find the value of p and q. A triangle is formed with vertices A ^ 5, 6 h, B ^ 0, - 4 h and C ^ - 3, 3 h . (a) Find the point of intersection of its medians. (b) If D, E and F are the midpoints of AB, AC and BC, divide the intervals CD, BE and AF in the ratio 2:1. What property of medians does this show? If ^ 0, 0 h divides the interval AB where A = ^ a, b h and B = ^ 4, 9 h in the external ratio of 2:1, find the value of a and b.

10. P divides the interval between the point ^ 2, 3 h and the intersection of lines 2x - 3y + 19 = 0 and 5x + 2y = 0 in the ratio of 4:5. Find the coordinates of P.

433

434


Test Yourself 7 1.

Find the distance between points ^ - 1, 2 h and ^ 3, 7 h .

2.

What is the midpoint of the origin and the point ^ 5, - 4 h ?

3.

Find the gradient of the straight line (a) passing through ^ 3, -1 h and ^ - 2, 5 h (b) with equation 2x - y + 1 = 0 (c) making an angle of 30c with the x-axis in the positive direction (d) perpendicular to the line 5 x + 3 y - 8 = 0.

4.

5.

6.

Find the equation of the linear function (a) passing through ^ 2, 3 h and with gradient 7 (b) parallel to the line 5x + y - 3 = 0 and passing through ^ 1, 1 h (c) through the origin, and perpendicular to the line 2x - 3y + 6 = 0 (d) through ^ 3, 1 h and ^ - 2, 4 h (e) with x-intercept 3 and y-intercept –1. Find the perpendicular distance between ^ 2, 5 h and the line 2x - y + 7 = 0 in surd form with rational denominator. Prove that the line between ^ -1, 4 h and ^ 3, 3 h is perpendicular to the line 4x - y - 6 = 0.

7.

Find the x- and y-intercepts of 2x - 5y - 10 = 0.

8.

(a) Find the equation of the straight line l that is perpendicular to the line 1 y = x - 3 and passes through ^ 1, -1 h . 2 (b) Find the x-intercept of l. (c) Find the exact distance from ^ 1, -1 h to the x-intercept of l.

9.

Prove that lines y = 5x - 7 and 10x - 2y + 1 = 0 are parallel.

10. Find the equation of the straight line passing through the origin and parallel to the line with equation 3x - 4y + 5 = 0. 11. Find the point of intersection between lines y = 2x + 3 and x - 5y + 6 = 0. 12. The midpoint of ^ a, 3 h and ^ - 4, b h is ^ 1, 2 h . Find the values of a and b. 13. Find the acute angle between the lines 2x - 5y + 1 = 0 and x + y - 7 = 0 to the nearest minute. 14. Show that the lines x - y - 4 = 0, 2x + y + 1 = 0, 5x - 3y - 14 = 0 and 3x - 2y - 9 = 0 are concurrent. 15. Divide the interval between points ^ 3, - 4 h and ^ 2, 2 h in the ratio 4:5. 16. A straight line makes an angle of 153c 29l with the x-axis in the positive direction. What is its gradient, to 3 significant figures? 17. The perpendicular distance from ^ 3, - 2 h to the line 5x - 12y + c = 0 is 2. Find 2 possible values of c. 18. Find the equation of the straight line through ^ 1, 3 h that passes through the intersection of the lines 2x - y + 5 = 0 and x + 2y - 5 = 0. 19. Divide the interval between ^ 0, 5 h and ^ - 2, 4 h in the external ratio of 2:3. 20. The gradient of the line through ^ 3, - 4 h and ^ x, 2 h is −5. Evaluate x. 21. Find the obtuse angle between the lines 3x - y + 3 = 0 and 2x + 5y - 1 = 0.


22. Show that the points ^ - 2, 1 h and ^ 6, 3 h are on opposite sides of the line 2 x - 3 y - 1 = 0.

24. Find the equation of the line with x-intercept 4 that makes an angle of 45c with the x-axis.

23. Find the acute angle between the lines y = 3x - 4 and y = 5 - x.

25. Find the equation of the line with y-intercept - 2 and perpendicular to the line passing through ^ 3, -2 h and ^ 0, 5 h .


If points ^ - 3k, 1 h, ^ k - 1, k - 3 h and ^ k - 4, k - 5 h are collinear, find the value of k.

2.

Find the equation, in exact form, of the line passing through _ 3 , -2 i that makes an angle of 30c with the positive x-axis.

3.

Find the equation of the circle whose centre is at the origin and with tangent x - 3y + 9 = 0.

4.

ABCD is a rhombus where A = ^ - 3, 0 h, B = ^ 0, 4 h, C = ^ 5, 4 h and D = ^ 2, 0 h . Prove that the diagonals are perpendicular bisectors of one another.

5.

6.

7.

8.

and the point of intersection of lines 3x - 7y = 15 and 4x - y = - 5. 9.

Find the magnitude of the angle, in degrees and minutes, that the line joining ^ -1, 3 h and ^ 2, - 4 h makes with the x-axis in the positive direction.

10. Find the equation of the line that passes through the point of intersection of lines 2x + 5y + 19 = 0 and 4x - 3y - 1 = 0 that is perpendicular to the line 3x - 2y + 1 = 0. 11. Prove A ^ 2, 5 h, B ^ - 4, 5 h and C ^ -1, 2 h are the vertices of a right-angled isosceles triangle.

Prove that the points _ -1, 2 2 i, _ 3 , - 6 i and _ - 5 , 2 i all lie on a circle with centre the origin. What are the radius and equation of the circle?

12. Find the coordinates of the centre of a circle that passes through points ^ 7, 2 h, ^ 2, 3 h and ^ -4, -1 h .

Find the exact distance between the parallel lines 3x + 2y - 5 = 0 and 3x + 2y = 1.

13. If ax - y - 2 = 0 and bx - 5y + 11 = 0 intersect at the point ^ 3, 4 h, find the values of a and b.

A straight line has x-intercept A ^ a, 0 h and y-intercept B ^ 0, b h, where a and b are positive integers. The gradient of line AB is -1. Find +OBA where O is the origin and hence prove that a = b.

14. Find the equation of the straight line through ^ 3, -4 h that is perpendicular to the line with x-intercept and y-intercept −2 and 5 respectively.

Find the exact perpendicular distance between the line 2x + 3y + 1 = 0

15. Find the acute angle between the straight lines with equations 3x - y = 5 and 2x - 4y + 1 = 0.

435

436


16. Find the exact equation of the straight line through the midpoint of ^ 0, - 5 h, and ^ 4, -1 h that is perpendicular to the line that makes an angle of 30c with the x-axis. 17. Point P ^ x, y h moves so that it is equidistant from points A ^ 1, 4 h and B ^ - 2, 7 h . By finding the distances AP and BP, find the equation of the locus of P. 18. Find the value of b if the lines 2x - y + 1 = 0 and bx - 7y + 5 = 0 make an angle of 45c at their intersection. 19. Find the coordinates of trisection of the interval between ^ 3, -1 h and ^ 1, - 5 h . 20. Prove that if two lines with gradients m 1 and m 2 meet at an angle of 45c, then m 1 m 2 = m 1 - m 2 - 1 or m 1 m 2 = m 2 - m 1 - 1. 21. A and B have coordinates ^ 1, 3 h and ^ - 4, 7 h respectively. If P divides AB in the external ratio of p:1, find the coordinates of P in terms of p.

22. (a) Show that the point ^ - 7, 7 h lies on the line joining A ^ - 2, 0 h and B ^ 3, - 7 h . (b) Find the ratio in which the point divides AB. 23. The interval AB where A = ^ - 5, 3 h and B = ^ x, y h is divided by point P in the ratio of 3:2. If the point P has coordinates ^ 8, - 9 h, find values for x and y. 24. The angle between straight lines 2x - 3y = 0 and mx + 4y = 9 is 32c 51l. Find the value of m correct to 2 significant figures. 25. Given points A ^ 1, 0 h, B ^ 2, 5 h and C ^ 9, 0 h are the vertices of a triangle, (a) find the coordinates of P that divide AB in the ratio 2:1 (b) find the coordinates of Q that divide CB in the ratio 2:1 (c) prove PQ < AC (d) find the coordinates of R that divide AC in the ratio 2:1 (e) prove PR < BC.

8 Introduction to Calculus TERMINOLOGY Composite function: A function of a function. One function, f (x), is a composite of one function to another function, for example g(x) Continuity: Describing a line or curve that is unbroken over its domain Continuous function: A function is continuous over an interval if it has no break in its graph. For every x value on the graph the limit exists and equals the function value

Differentiation: The process of finding the gradient of a tangent to a curve which is called the derivative Differentiation from first principles: The process of finding the gradient of a tangent to a curve by finding the gradient of the secant between two points and finding the limit as the secant becomes a tangent Gradient of a secant: The gradient (slope) of the line between two points that lies close together on a function

Derivative at a point: This is the gradient of a curve at a particular point

Gradient of a tangent: The gradient (slope) of a line that is a tangent to the curve at a point on a function. It is the derivative of the function

Derivative function: The gradient function of a curve obtained through differentiation

Rate of change: The rate at which the dependent variable changes as the independent variable changes

Differentiable function: A function which is continuous and where the gradient exists at all points on the function

Chapter 8 Introduction to Calculus

439

INTRODUCTION CALCULUS IS A VERY IMPORTANT part of mathematics and involves the

Thousand Barrels per Day

measurement of change. It can be applied to many areas such as science, economics, engineering, astronomy, Crude Oil Production (Mbbl/d) sociology and medicine. We also see articles Iran 7,000 in newspapers every day that involve change: the spread of infectious diseases, population 6,000 growth, inflation, unemployment, filling of 5,000 our water reservoirs. For example, this graph shows the 4,000 change in crude oil production in Iran over 3,000 the years. Notice that while the graph shows that production is increasing over recent 2,000 years, the rate at which it is being produced 1,000 seems to be slowing down. Calculus is used to look at these trends and predict what will 0 73 75 77 79 81 83 85 87 89 91 93 95 97 99 01 03 05 07 happen in the future. 74 76 78 80 82 84 86 88 90 92 94 96 98 00 02 04 06 There are two main branches of January 1973–May 2007 calculus. Differentiation is used to calculate the rate at which two variables change in relation to one another. You will learn about Anti-differentiation, or integration, is the inverse of differentiation and integration in the uses information about rates of change to go back and examine the original HSC Course. variables. Integration can also be used to find areas of curved objects.

DID YOU KNOW? ‘Calculus’ comes from the Latin meaning pebble or small stone. In ancient civilisations, stones were used for counting. However, the mathematics practised by these early people was quite sophisticated. For example, the ancient Greeks used sums of rectangles to estimate areas of curved figures. However, it wasn’t until the 17th century that there was a breakthrough in calculus when scientists were searching for ways of measuring motion of objects such as planets, pendulums and projectiles. Isaac Newton, an Englishman, discovered the main principles of calculus when he was 23 years old. At this time an epidemic of bubonic plague closed Cambridge University where he was studying, so many of his discoveries were made at home. He first wrote about his calculus methods, which he called fluxions, in 1671, but his Method of fluxions was not published until 1704. Gottfried Leibniz (1646–1716), in Germany, was also studying the same methods and there was intense rivalry between the two countries over who was first! Search the Internet for further details on these two famous mathematicians. You can find out about the history of calculus and why it was necessary for mathematicians all those years ago to invent it.

Isaac Newton

440


In this chapter you will learn about differentiation, which measures the rate of change of one variable with respect to another.

Gradient Gradient of a straight line The gradient of a straight line measures its slope. You studied gradient in the last chapter. rise m = run

Class Discussion Remember that an increasing line has a positive gradient and a decreasing line has a negative gradient.

positive

negative

Notice also that a horizontal line has zero gradient. Can you see why?

Can you find the gradient of a vertical line? Why?

Gradient plays an important part, not just in mathematics, but in many areas including science, business, medicine and engineering. It is used everywhere we want to find rates. On a graph, the gradient measures the rate of change of the dependent variable with respect to the change in the independent variable.


EXAMPLES 1. The graph shows the average distance travelled by a car over time. Find the gradient and describe it as a rate. d

km

400

t

5 Hours

Solution The line is increasing so it will have a positive gradient. rise m = run 400 = 5 80 = 1 = 80 This means that the car is travelling at the rate of 80 km/hour. 2. The graph shows the number of cases of flu reported in a town over several weeks. N

Number of cases (100s)

15

Weeks

10

t

Find the gradient and describe it as a rate. CONTINUED

441


Solution The line is decreasing so it will have a negative gradient. rise m = run 1500 =10 150 =1 = - 150 This means that the rate is -150 cases/week, or the number of cases reported is decreasing by 150 cases/week.

When finding the gradient of a straight line in the number plane, we think of a change in y values as x changes. The gradients in the examples above show rates of change. However, in most examples in real life, the rate of change will vary. For example, a car would speed up and slow down depending on where it is in relation to other cars, traffic light signals and changing speed limits.

Gradient of a curve

Class Discussion The two graphs show the distance that a bicycle travels over time. One is a straight line and the other is a curve. d

d

20

20

15

15 km

km

442

10 5

10 5

1

2

3 Hours

4

t

1

2

3 Hours

4

Is the average speed of the bicycle the same in both cases? What is different about the speed in the two graphs? How could you measure the speed in the second graph at any one time? Does it change? If so, how does it change?

t


Here is a more general curve. What could you say about its gradient? How does it change along the curve? y

x

Copy the graph and mark on it where the gradient is positive, negative and zero.

Using what we know about the gradient of a straight line, we can see where the gradient of a curve is positive, negative or zero by drawing tangents to the curve in different places around the curve. y

-

+ x

0

Notice that when the curve increases it has a positive gradient, when it decreases it has a negative gradient and when it turns around the gradient is zero.

Investigation There are some excellent computer programs that will draw tangents to a curve and then sketch the gradient curve. One of these is Geometer Sketchpad. Explore how to sketch gradient functions using this or a similar program as you look at the examples below.

443

444


EXAMPLES Describe the gradient of each curve. 1.

Solution Where the curve increases, the gradient is positive. Where it decreases, it is negative. Where it turns around, it has a zero gradient.

2.

Solution


445

Since we have a formula for finding the gradient of a straight line, we find the gradient of a curve by measuring the gradient of a tangent to the curve.

EXAMPLE (a) Make an accurate sketch of y = x 2 on graph paper. (b) Draw tangents to this curve at the points where x = - 3, x = - 2, x = - 1, x = 0, x = 1, x = 2 and x = 3. (c) Find the gradient of each of these tangents. (d) Draw the graph of the gradients (the gradient function) on a number plane.

Solution (a) and (b) y

9 8 7 6 5 4 3 2 1 -3 -2

1

2

3

x

There are computer programs that will draw these tangents.

(c) At x = - 3, m = - 6 At x = - 2, m = - 4 At x = - 1, m = - 2 At x = 0, m = 0 At x = 1, m = 2 At x = 2, m = 4 At x = 3, m = 6 (d)

Use the ‘m’ values as the ‘y’ values on this graph.

446


Drawing tangents to a curve is difficult. We can do a rough sketch of the gradient function of a curve without knowing the actual values of the gradients of the tangents. To do this, notice in the example above that where m is positive, the gradient function is above the x-axis, where m = 0, the gradient function is on the x-axis and where m is negative, the gradient function is below the x-axis.

EXAMPLES Sketch the gradient function of each curve. 1.

Solution First we mark in where the gradient is positive, negative and zero.

Now on the gradient graph, place the points where m = 0 on the x-axis. These are at x 1, x 2 and x 3 .


To the left of x 1, the gradient is negative, so this part of the graph will be below the x-axis. Between x 1 and x 2, the gradient is positive, so the graph will be above the x-axis. Between x 2 and x 3, the gradient is negative, so the graph will be below the x-axis. To the right of x 3, the gradient is positive, so this part of the graph will be above the x-axis.

2.

Solution First mark in where the gradient is positive, negative and zero.

CONTINUED

447

448


The gradient is zero at x 1 and x 2 . These points will be on the x-axis. To the left of x 1, the gradient is positive, so this part of the graph will be above the x-axis. Between x 1 and x 2, the gradient is negative, so the graph will be below the x-axis. To the right of x 2, the gradient is positive, so this part of the graph will be above the x-axis.

8.1 Exercises Sketch the gradient function for each graph. 1.

4.

2.

5.

3.

6.


9.

7.

8.

10.

Differentiation from First Principles Seeing where the gradient of a curve is positive, negative or zero is a good first step, but there are methods to find a formula for the gradient of a tangent to a curve. The process of finding the gradient of a tangent is called differentiation. The resulting function is called the derivative.

Differentiability A function is called a differentiable function if the gradient of the tangent can be found. There are some graphs that are not differentiable in places. Most functions are continuous, which means that they have a smooth unbroken line or curve. However, some have a gap, or discontinuity, in the graph (e.g. hyperbola). This can be shown by an asymptote or a ‘hole’ in the graph. We cannot find the gradient of a tangent to the curve at a point that doesn’t exist! So the function is not differentiable at the point of discontinuity. y

a

x

This function is not differentiable at a since the curve is discontinuous at this point.

449

450


y

b

This function is not differentiable at b as the curve is discontinuous at this point.

x

A function may be continuous but not smooth. It may have a sharp corner. Can you see why curves are not differentiable at the point where there is a corner? y

c

x

The curve is not differentiable at point c since it is not smooth at that point.

A function y = f (x) is differentiable at the point x = a if the derivative exists at that point. This can only happen if the function is continuous and smooth at x = a.


EXAMPLES 1. Find all points where the function below is not differentiable. y

B

x C

A

Solution The function is not differentiable at points A and B since there are sharp corners and the curve is not smooth at these points. It is not differentiable at point C since the function is discontinuous at this point.

2. Is the function f (x) = )

x2 3x - 2

for x $ 1 differentiable at all points? for x 1 1

Solution The functions f (x) = x 2 and f (x) = 3x - 2 are both differentiable at all points. However, we need to look at where one finishes and the other starts, at f (1). For f (x) = x 2 f ] 1 g = 12 =1 For f (x) = 3x - 2 f ]1 g = 3 ]1 g - 2 =1 This means that both pieces of this function join up (the function is continuous). However, to be differentiable, the curve must be smooth at this point. CONTINUED

451

452


Sketching this function shows that it is not smooth (it has a sharp corner) so it is not differentiable at x = 1. y

y = x2

1 1

x

-2 y = 3x - 2

8.2 Exercises For each function, state whether it has any points at which it is not differentiable. 1.

4.

y

x1

x

x

x1

x

2.

y

3.

y

y

x


5.

y

453

y

10.

5 4 3 2

x1

x2

x

1 -4 -3 -2 -1 -1

1

2

3

4

x

-2 -3

6.

4 f (x) = x

7.

y=-

8. 9.

-4 -5

1 x+3

11. y = tan x for 0c # x # 360c

x3 if x 2 2 f (x) = ) x + 1 if x # 2 Z 2x for x 2 3 ] f (x) = [3 for - 2 # x # 3 ] 2 \1 - x for x 1 - 2

x 12. f (x) = x 13. f (i) = -3 cos 2i 14. g (z) = sin 2 z 15. y =

x-3 x2 - 9

Limits To differentiate from first principles, we need to look more closely at the concept of a limit. A limit is used when we want to move as close as we can to something. Often this is to find out where a function is near a gap or discontinuous point. You saw this in Chapter 5 when looking at discontinuous graphs. In this topic, it is used when we want to move from a gradient of a line between two points to a gradient of a tangent.


x2 - x - 2 . x-2

Solution (x + 1) (x - 2) x2 - x - 2 = lim x "2 x "2 x-2 (x - 2 ) = lim (x + 1)

You did this in Chapter 5.

lim

x "2

=2+1 =3

CONTINUED

454


2. Find an expression in terms of x for lim h "0

2xh - h 2 - 3h . h

Solution h (2 x - h - 3) 2xh - h 2 - 3h = lim h "0 h "0 h h = lim (2x - h - 3)

lim

h "0

= 2x - 3 3. Find an expression in terms of x for lim

dx " 0

3x 2 dx + dx 2 - 5dx . dx

Solution d x ( 3x 2 + d x - 5 ) 3x 2 d x + d x 2 - 5 d x = lim dx " 0 dx " 0 dx dx 2 = lim (3x + dx - 5) lim

dx " 0 2

= 3x - 5

8.3 Exercises 1.

Evaluate

2.

x + 3x x x "0 5x 3 - 2x 2 - 7x lim x x "0 x 2 - 3x lim x "3 x - 3 t 2 - 16 lim t "4 t-4 g2 - 1 lim g "1 g - 1 x2 + x - 2 lim x " -2 x+2 h 5 + 2h lim h "0 h 2 x - 7x + 12 lim x "3 x-3 n 2 - 25 lim n "5 n - 5 x 2 + 4x + 3 lim x " -1 x2 - 1

Find as an expression in terms of x

2

x 2 h - 2xh - 4h h "0 h 2x 3 h + xh - h lim h "0 h 3x 2 h 2 - 7xh + 4h 2 - h lim h "0 h 4x 4 h - x 2 h - 4xh 2 lim h "0 h x 2 h 2 + 3xh 2 - 4xh + 3h lim h "0 h 2x 2 h + 5xh 2 + 6h lim h "0 h x 2 dx 2 - 2xdx lim dx " 0 dx 4 x 2 dx - 2 dx 2 lim dx " 0 dx x 3 dx 2 + 3xdx - dx lim dx " 0 dx x 2 dx - 2xdx + 9dx lim dx " 0 dx

(a) lim

(a) lim

(b)

(b)

(c) (d) (e) (f) (g) (h) (i) (j)

(c) (d) (e) (f) (g) (h) (i) (j)


Differentiation as a limit y2 - y1 The formula m = x - x is used to find the gradient of a straight line when we 2 1 know two points on the line. However, when the line is a tangent to a curve, we only know one point on the line—the point of contact with the curve. To differentiate from first principles, we first use the point of contact and another point close to it on the curve (this line is called a secant) and then we move the second point closer and closer to the point of contact until they overlap and the line is at single point (the tangent). To do this, we use a limit. If you look at a close up of a graph, you can get some idea of this concept. When the curve is magnified, two points appear to be joined by a straight line. We say the curve is locally straight.

Investigation Use a graphics calculator or a computer program to sketch a curve and then zoom in on a section of the curve to see that it is locally straight. For example, here is a parabola. 10

2

y

f 1(x) = x2 2

-20

x 20

-10

Notice how it looks straight when we zoom in on a point on the parabola? 7.99 y

2.99

f 1(x) = x2

x

Use technology to sketch other curves and zoom in to show that they are locally straight.

455

456


Before using limits to find different formulae for differentiating from first principles, here are some examples of how we can calculate an approximate value for the gradient of the tangent to a curve. By taking two points close together, as in the example below, we find the gradient of the secant and then estimate the gradient of the tangent. y

(3.01, f (3.01)) (3, f (3)) x

EXAMPLES 1. For the function f ] x g = x 3, find the gradient of the secant PQ where P is the point on the function where x = 2 and Q is another point on the curve close to P. Choose different values for Q and use these results to estimate the gradient of the curve at P.

y Q P

(2.1, f(2.1)) (2, f(2))

x


457

Solution P = ^ 2, f (2) h Take different values of x for point Q, for example x = 2.1 Using different values of x for point Q gives the results in the table. Point Q _ 2 .1 , f ] 2 .1 g i

_ 2.01, f ] 2.01 g i

_ 2.001, f ] 2.001 g i

_ 1 .9 , f ] 1 .9 g i

_ 1.99, f ] 1.99 g i

_ 1.999, f ] 1.999 g i

Gradient of secant PQ

y2 - y1 to find x2 - x1 the gradient of the secant. Use m =

f ( 2 . 1 ) - f (2 ) m= 2 .1 - 2 2 .1 3 - 2 3 = 2 .1 - 2 = 12.61 f (2.01) - f (2) 2.01 - 2 2.01 3 - 2 3 = 2.01 - 2 = 12.0601

m=

f (2.001) - f (2) 2.001 - 2 2.001 3 - 2 3 = 2.001 - 2 = 12.006001

m=

f ( 1 . 9 ) - f (2 ) 1 .9 - 2 1 .9 3 - 2 3 = 1 .9 - 2 = 11.41

m=

f (1.99) - f (2) 1.99 - 2 1.99 3 - 2 3 = 1.99 - 2 = 11.9401

m=

f (1.999) - f (2) 1.999 - 2 1.999 3 - 2 3 = 1.999 - 2 = 11.994001

m=

From these results, a good estimate for the gradient at P is 12. We can say that as x approaches 2, the gradient approaches 12. f (x) - f (2) We can write lim = 12. x "2 x-2

CONTINUED

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2. For the curve y = x 2, find the gradient of the secant AB where A is the point on the curve where x = 5 and point B is close to A. Find an estimate of the gradient of the curve at A by using three different values for B.

Solution A = ^ 5, f (5) h Take three different values of x for point B, for example x = 4.9, x = 5.1 and x = 5.01. (a) B = ^ 4.9, f (4.9) h y2 - y1 m= x -x 2 1 f ( 4 . 9 ) - f (5 ) = 4 .9 - 5 4 .9 2 - 5 2 = 4 .9 - 5 = 9 .9 (b) B = ^ 5.1, f (5.1) h y2 - y1 m= x -x 2 1 f ( 5 . 1 ) - f (5 ) = 5 .1 - 5 5 .1 2 - 5 2 = 5 .1 - 5 = 10.1 (c) B = ^ 5.01, f (5.01) h y2 - y1 m= x -x 2 1 f (5.01) - f (5) = 5.01 - 5 5.01 2 - 5 2 = 5.01 - 5 = 10.01 From these results, a good estimate for the gradient at A is 10. We can say that as x approaches 5, the gradient approaches 10. We can write lim x "5

f (x) - f (5) = 10. x-5

We can find a general formula for differentiating from first principles by using c rather than any particular number. We use general points P ^ c, f (c) h and Q ^ x, f (x) h where x is close to c. The gradient of the secant PQ is given by y2 - y1 m= x -x 2 1 f (x) - f (c) = x-c


459

The gradient of the tangent at P is found when x approaches c. We call this fl(c).

f l(c) = lim x "c

f (x) - f (c) x-c

There are other versions of this formula. We can call the points P ^ x, f (x) h and Q ^ x + h, f (x + h) h where h is small. y Q

(x + h, f(x + h))

P

(x , f(x)) x

Secant PQ has gradient y2 - y1 m= x -x 2 1 f (x + h) - f (x) = x+h-x f (x + h) - f (x) = h To find the gradient of the secant, we make h smaller as shown, so that Q becomes closer and closer to P. y Q

(x + h, f(x + h)) Q

Q P

Q

(x, f (x)) x Search the Internet using keywords ‘differentiation from first principles’, gradient of secant’ and ‘tangent’ to find mathematical websites that show this working.

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As h approaches 0, the gradient of the tangent becomes lim h "0 We call this fl(x).

fl(x) = lim

h

.

f (x + h) - f (x) h

h "0

The symbol d is a Greek letter called delta.

f (x + h) - f (x)

If we use P ^ x, y h and Q ^ x + dx, y + dy h close to P where dx and dy are small: Gradient of secant PQ y2 - y1 m= x -x 2 1 y + dy - y = x + dx - x dy = dx dy . We As dx approaches 0, the gradient of the tangent becomes lim dx " 0 d x dy call this . dx

dy dx

= lim

dx " 0

dy dx

All of these different notations stand for the derivative, or the gradient of the tangent: dy

d d , (y), ^ f (x) h, f l(x), yl dx dx dx These occur because Newton, Leibniz and other mathematicians over the years have used different notation.

Investigation Leibniz used

dy dx

where d stood for ‘difference’. Can you see why he would

have used this? Use the Internet to explore the different notations used in calculus and where they came from.


461

The three formulae for differentiating from first principles all work in a similar way.

EXAMPLE Differentiate from first principles to find the gradient of the tangent to the curve y = x 2 + 3 at the point where x = 1. Remember that y = x 2 - 3 is the same as f (x) = x 2 - 3.

Solution Method 1: f (x) - f (c) x-c 2 f ]x g = x + 3 f ] 1 g = 12 + 3 =4 f (x) - f (c) f l(c) = lim x "c x-c f (x) - f (1) f l(1) = lim x "1 x-1 (x 2 + 3) - 4 = lim x "1 x-1 x2 - 1 = lim x "1 x - 1 (x + 1) (x - 1) = lim x "1 x-1 = lim (x + 1) f l(c) = lim x "c

x "1

=1+1 =2 Method 2: f l(x) = lim

f (x + h ) - f ( x )

h "0

f ] x g = x2 + 3 f ] 1 g = 12 + 3 =4

h

f ] x + h g = ] x + h g2 + 3 When x = 1 f ] 1 + h g = ] 1 + h g2 + 3 = 1 + 2h + h 2 + 3 = 2h + h 2 + 4 CONTINUED

462


f l(x) = lim h "0

f l(1) = lim

f (x + h) - f (x) h f (1 + h) - f (1)

h (2h + h 2 + 4) - 4 = lim h "0 h 2h + h 2 = lim h "0 h h (2 + h) = lim h "0 h = lim (2 + h) h "0

h "0

=2+0 =2 Method 3: dy

= lim

dy

dx dx y = x2 + 3 dx " 0

When x = 1 y = 12 + 3 =4 So point ^ 1, 4 h lies on the curve. Substitute point (1 + dx, 4 + dy): 4 + dy = (1 + dx) 2 + 3 = 1 + 2d x + d x 2 + 3 = 2d x + d x 2 + 4 d y = 2d x + d x 2 dy 2d x + d x 2 = dx dx dx(2 + dx) = dx = 2 + dx dy dy = lim d x " 0 dx dx = lim (2 + dx) dx " 0

=2+0 =2

We can also use these formulae to find the derivative function generally.


463

EXAMPLE Differentiate f ] x g = 2x 2 + 7x - 3 from first principles.

Solution

Try this example using the other two formulae.

f ] x g = 2x + 7x - 3 f ] x + h g = 2 ] x + h g2 + 7 ] x + h g - 3 2

= 2 ^ x 2 + 2xh + h 2 h + 7x + 7h - 3 = 2x 2 + 4xh + 2h 2 + 7x + 7h - 3 f ] x + h g - f ] x g = ^ 2x 2 + 4xh + 2h 2 + 7x + 7h - 3 h - ^ 2x 2 + 7x - 3 h = 2x 2 + 4xh + 2h 2 + 7x + 7h - 3 - 2x 2 - 7x + 3 = 4xh + 2h 2 + 7h f l(x) = lim

f (x + h) - f (x)

h 4xh + 2h 2 + 7h = lim h "0 h h ( 4 x + 2h + 7 ) = lim h "0 h = lim (4x + 2h + 7) h "0

h "0

= 4x + 0 + 7 = 4x + 7

8.4 Exercises 1.

2.

(a) Find the gradient of the secant between the point ^ 1, 2 h and the point where x = 1.01, on the curve y = x 4 + 1. (b) Find the gradient of the secant between ^ 1, 2 h and the point where x = 0.999 on the curve. (c) Use these results to find the gradient of the tangent to the curve y = x 4 + 1 at the point ^ 1, 2 h . A function f ] x g = x 3 + x has a tangent at the point ^ 2, 10 h . f (x) - f (2) (a) Find the value of x-2 when x = 2.1.

(b) Find the value of

f (x) - f (2) x-2

when x = 2.01. f (x) - f (2) (c) Evaluate when x-2 x = 1.99. (d) Hence find the gradient of the tangent at the point ^ 2, 10 h . 3.

For the function f ] x g = x 2 - 4, find the derivative at point P where x = 3 by selecting points near P and finding the gradient of the secant.

4.

If f (x) = x 2, (a) find f (x + h) (b) show that f (x + h) - f (x) = 2xh + h 2

464


(c) show that f (x + h) - f (x) h

5.

6.

= 2x + h

(d) show that fl(x) = 2x .

(b) show that

A function is given by f (x) = 2x 2 - 7x + 3. (a) Show that f (x + h) = 2x 2 + 4xh + 2h 2 - 7x - 7h + 3. (b) Show that f (x + h) - f (x) = 4xh + 2h 2 - 7h. (c) Show that f (x + h) - f (x) = 4x + 2h - 7 . h (d) Find fl(x) .

(c) find

A function is given by f (x) = x 2 + x + 5. (a) Find f ] 2 g. (b) Find f ] 2 + h g. (c) Find f ] 2 + h g - f ] 2 g. (d) Show that f (2 + h) - f ( 2) = 5 + h. h (e) Find fl(2).

7.

Given the curve f (x) = 4x 3 - 3 (a) find f ] -1 g (b) find f ] -1 + h g - f ] -1 g (c) find the gradient of the tangent to the curve at the point where x = -1.

8.

For the parabola y = x - 1 (a) find f ] 3 g (b) find f ] 3 + h g - f ] 3 g (c) find fl(3).

9.

Remember that 1 -1 = x

x

by substituting the point ^ x + dx, y + dy h

2

For the function f (x) = 4 - 3x - 5x 2 (a) find f l(1) (b) similarly, find the gradient of the tangent at the point ^ -2, -10 h .

10. For the parabola y = x 2 + 2x (a) show that dy = 2xdx + dx 2 + 2dx

dy dx

dy dx

= 2x + dx + 2

.

11. Differentiate from first principles to find the gradient of the tangent to the curve (a) f ] x g = x 2 at the point where x=1 (b) y = x 2 + x at the point ^ 2, 6 h (c) f ] x g = 2x 2 - 5 at the point where x = -3 (d) y = 3x 2 + 3x + 1 at the point where x = 2 (e) f ] x g = x 2 - 7x - 4 at the point ^ -1, 6 h . 12. Find the derivative function for each curve by differentiating from first principles (a) f ] x g = x 2 (b) y = x 2 + 5x (c) f ] x g = 4x 2 - 4x - 3 (d) y = 5x 2 - x - 1 (e) y = x 3 (f) f ] x g = 2x 3 + 5x (g) y = x 3 - 2x 2 + 3x - 1 (h) f (x) = -2x 3. 13. The curve y = x has a tangent drawn at the point ^ 4, 2 h . (a) Evaluate

f (x) - f (4) when x-4

x = 3 .9 . f (x) - f (4) (b) Evaluate when x-4 x = 3.999. f (x) - f (4) when (c) Evaluate x-4 x = 4.01. 14. For the function f (x) = x - 1, f (x) - f (5) (a) evaluate when x-5 x = 4.99.


f (x) - f (5) (b) evaluate when x-5 x = 5.01. (c) Use these results to find the derivative of the function at the point where x = 5.

15. Find the gradient of the tangent 4 to the curve y = 2 at point x P ^ 2, 1 h by finding the gradient of the secant between P and a point close to P.

Short Methods of Differentiation The basic rule Remember that the gradient of a straight line y = mx + b is m. The tangent to the line is the line itself, so the gradient of the tangent is m everywhere along the line. y

y = mx + b

x

So if y = mx,

dy dx

=m

d ] g kx = k dx

For a horizontal line in the form y = k, the gradient is zero. y

y=k

x

So if y = k,

dy dx

=0 d ] g k =0 dx

465

466


Investigation Differentiate from first principles: y = x2 y = x3 y = x4 Can you find a pattern? Could you predict what the result would be for xn? Alternatively, you could find an approximation to the derivative of a f (x + 0.01) - f (x) function at any point by drawing the graph of y = . 0.01 Use a graphics calculator or graphing computer software to sketch the derivative for these functions and find the equation of the derivative.

Mathematicians working with differentiation from first principles discovered this pattern that enabled them to shorten differentiation considerably! For example: When y = x 2, yl = 2x When y = x 3, yl = 3x 2 When y = x 4, yl = 4x 3

d ^ nh x = nx n - 1 dx

Proof You do not need to know this proof.

f (x) = x n f (x + h) = (x + h) n f (x + h) - f (x) = (x + h) n - x n = ^ (x + h) - x h [(x + h) n - 1 + (x + h) n - 2 x + (x + h) n - 3 x 2 + (x + h) n - 4 x 3 + . . . + (x + h) x n - 2 + x n - 1] = h [(x + h) n - 1 + (x + h) n - 2 x + (x + h) n - 3 x 2 + (x + h) n - 4 x 3 + . . . + (x + h) x n - 2 + x n - 1]

f l(x) = lim

f (x + h) - f (x)

h h [(x + h) n - 1 + (x + h) n - 2 x + (x + h) n - 3 x 2 + (x + h) n - 4 x 3 + . . . + (x + h) x n - 2 + x n - 1] = lim h "0 h n-1 n-2 n-3 2 = lim [(x + h) + (x + h ) x + (x + h) x + (x + h) n - 4 x 3 + . . . + (x + h) x n - 2 + x n - 1] h "0

h "0

= (x) n - 1 + (x) n - 2 x + (x) n - 3 x 2 + (x) n - 4 x 3 + . . . + (x) x n - 2 + x n - 1 = nx n - 1


467

EXAMPLE Differentiate f (x) = x 7.

Solution f l(x) = 7x 6

There are some more rules that give us short ways to differentiate functions. The first one says that if there is a constant in front of the x (we call this a coefficient), then it is just multiplied with the derivative.

d ^ nh kx = knx n - 1 dx

A more general way of writing this rule is:

d ^ kf (x) h = kf l(x) dx

Proof kf (x + h) - kf (x) d ^ kf (x) h = lim 0 h " dx h k [f (x + h) - f (x)] = lim h "0 h f (x + h) - f (x) = k lim h "0 h = kf l(x)

EXAMPLE Find the derivative of 3x8.

Solution If y = 3x 8 dy = 3 # 8x 7 dx = 24x 7

You do not need to know this proof.

468


Also, if there are several terms in an expression, we differentiate each one separately. We can write this as a rule:

d ^ f (x) + g (x) h = f l(x) + g ’(x) dx

Proof You do not need to know this proof.

[f (x + h) + g (x + h)] - [f (x) + g (x)] d ^ f (x) + g (x) h = lim h "0 dx h f (x + h) + g (x + h) - f (x) - g (x) = lim h "0 h f (x + h) - f (x) + g (x + h) - g (x) = lim h "0 h f (x + h ) - f ( x ) g ( x + h ) - g ( x) G = lim = + h "0 h h f (x + h ) - f ( x ) g ( x + h ) - g ( x) = lim + lim 0 h "0 h " h h = f l(x) + gl(x)

EXAMPLE Differentiate x 3 + x 4.

Solution d 3 (x + x 4) = 3x 2 + 4x 3 dx

Many functions use a combination of these rules.

EXAMPLES Differentiate 1. 7x

Solution d ] g 7x = 7 dx CONTINUED


2. f (x) = x 4 - x 3 + 5

Solution f l(x) = 4x 3 - 3x 2 + 0 = 4x 3 - 3x 2 3. y = 4x 7

Solution dy dx

= 4 # 7x 6 = 28x 6

4. If f (x) = 2x 5 - 7x 3 + 5x - 4, evaluate f l(-1)

Solution f l(x) = 10x 4 - 21x 2 + 5 f l(-1) = 10(-1) 4 - 21(-1) 2 + 5 = -6 5. Differentiate

3x 2 + 5x 2x

Solution Divide by 2x before differentiating. 3x 2 + 5x 3x 2 5x = + 2x 2x 2x 3 5 = x+ 2 2 dy 3 = 2 dx 1 =1 2 6. Differentiate S = 2rr 2 + 2rrh with respect to r.

Solution We are differentiating with respect to r, so r is the variable and r and h are constants. dS = 2r(2r) + 2rh dr = 4r r + 2r h

469

470


8.5 Exercises 1.

Expand brackets before differentiating.

2.

3.

Simplify by dividing before differentiating.

Differentiate (a) x + 2 (b) 5x - 9 (c) x 2 + 3x + 4 (d) 5x 2 - x - 8 (e) x 3 + 2x 2 - 7x - 3 (f) 2x 3 - 7x 2 + 7x - 1 (g) 3x 4 - 2x 2 + 5x (h) x 6 - 5x 5 - 2x 4 (i) 2x 5 - 4x 3 + x 2 - 2x + 4 (j) 4x 10 - 7x 9 Find the derivative of (a) x ] 2x + 1 g (b) ] 2x - 3 g2 (c) ] x + 4 g ] x - 4 g 2 (d) ^ 2x 2 - 3 h (e) ] 2x + 5 g ^ x 2 - x + 1 h Differentiate (a)

x2 -x 6

(b)

x4 x3 +4 2 3

(c)

1 6 2 x ( x - 3) 3

(d)

2x 3 + 5x x

(e)

x 2 + 2x 4x

(f)

2x 5 - 3x 4 + 6x 3 - 2x 2 3x 2

4.

5.

6.

Find f l(x) when f (x) = 8x 2 -7x + 4. dy If y = x 4 - 2x 3 + 5, find when dx x = - 2. dy Find if dx y = 6x 10 - 5x 8 + 7x 5 - 3x + 8. ds . dt

7.

If s = 5t 2 - 20t , find

8.

Find gl(x) given g (x) = 5x - 4.

dv when v = 15t 2 - 9. dt dh 10. If h = 40t - 2t 2, find . dt 9.

Find

11. Given V =

dV 4 3 rr , find . 3 dr

12. If f (x) = 2x 3 - 3x + 4, evaluate f l(1). 13. Given f (x) = x 2 - x + 5, evaluate (a) f l(3) (b) f l(-2) (c) x when f l(x) = 7 14. If y = x 3 - 7, evaluate dy (a) when x = 2 dx dy (b) x when = 12 dx 15. Evaluate gl(2) when g (t) = 3t 3 - 4t 2 - 2t + 1.


471

Tangents and Normals

DID YOU KNOW? • The word tangent comes from the Latin ‘tangens’, meaning ‘touching’. A tangent to a circle intersects it only once.

• However, a tangent to a curve could intersect the curve more than once.

This line is a tangent to the curve at point P.

• A line may only intersect a curve once but not be a tangent.

• So a tangent to a curve is best described as the limiting position of the secant PQ as Q approaches P.

Remember from earlier in the chapter that the derivative is the gradient of the tangent to a curve.

dy dx

is the gradient of the tangent to a curve

472


u

EXAMPLES 1. Find the gradient of the tangent to the parabola y = x 2 + 1 at the point ^ 1, 2 h .

Solution dy dx At ^ 1, 2 h

dy dx

= 2x + 0 = 2x = 2 (1 ) =2

So the gradient of the tangent at ^ 1, 2 h is 2. 2. Find values of x for which the gradient of the tangent to the curve y = 2x 3 - 6x 2 + 1 is equal to 18.

Solution dy dx dy dx

= 6x 2 - 12x is the gradient of the tangent, so substitute

dy dx

= 18.

18 = 6x 2 - 12x 0 = 6x 2 - 12x - 18 = x 2 - 2x - 3 = ]x - 3 g]x + 1 g x - 3 = 0, x + 1 = 0 ` x = 3, x = -1 3. Find the equation of the tangent to the curve y = x 4 - 3x 3 + 7x - 2 at the point ^ 2, 4 h .

Solution dy At ^ 2, 4 h

dx dy dx

= 4x 3 - 9x 2 + 7 = 4 ] 2 g3 - 9 ] 2 g2 + 7

=3 So the gradient of the tangent at ^ 2, 4 h is 3. Equation of the tangent: y - y1 = m _ x - x1 i y - 4 = 3 ]x - 2 g


473

= 3x - 6 y = 3x - 2 or 0 = 3x - y - 2

The normal is a straight line perpendicular to the tangent at the same point of contact with the curve. y

Tangent

x

Normal

If lines with gradients m1 and m2 are perpendicular, then m 1 m 2 = -1

EXAMPLES 1. Find the gradient of the normal to the curve y = 2x 2 - 3x + 5 at the point where x = 4.

Solution dy dx

is the gradient of the tangent. dy

= 4x - 3 dx When x = 4 dy =4#4-3 dx = 13 So m 1 = 13 The normal is perpendicular to the tangent. So m 1 m 2 = -1 CONTINUED

You used this rule in the previous chapter.

474


13m 2 = -1 1 m2 = 13 1 So the gradient of the normal is - . 13

2. Find the equation of the normal to the curve y = x 3 + 3x 2 - 2x - 1 at the point ^ -1, 3 h .

Solution dy dx

is the gradient of the tangent. dy

= 3x 2 + 6x - 2 dx When x = -1 dy = 3 ] -1 g2 + 6 ] -1 g - 2 dx = -5 So m 1 = - 5 The normal is perpendicular to the tangent. So m 1 m 2 = -1 -5m 2 = -1 1 m2 = 5 1 So the gradient of the normal is . 5 Equation of the normal: y - y1 = m _ x - x1 i 1 y - 3 = ] x - ] -1 g g 5 5y - 15 = x + 1 0 = x - 5y + 16

8.6 Exercises 1.

Find the gradient of the tangent to the curve (a) y = x 3 - 3x at the point where x=5 (b) f ] x g = x 2 + x - 4 at the point ^ -7, 38 h (c) f ] x g = 5x 3 - 4x - 1 at the point where x = -1 (d) y = 5x 2 + 2x + 3 at the point ^ -2, 19 h (e) y = 2x 9 at the point where x=1

(f) f ] x g = x 3 - 7 at the point where x = 3 (g) v = 2t 2 + 3t - 5 at the point where t = 2 (h) Q = 3r 3 - 2r 2 + 8r - 4 at the point where r = 4 (i) h = t 4 - 4t where t = 0 (j) f ] t g = 3t 5 - 8t 3 + 5t at the point where t = 2.


2.

3.

4.

Find the gradient of the normal to the curve (a) f ] x g = 2x 3 + 2x - 1 at the point where x = -2 (b) y = 3x 2 + 5x - 2 at the point ^ -5, 48 h (c) f ] x g = x 2 - 2x - 7 at the point where x = - 9 (d) y = x 3 + x 2 + 3x - 2 at the point ^ -4, - 62 h (e) f ] x g = x 10 at the point where x = -1 (f) y = x 2 + 7x - 5 at the point ^ - 7, - 5 h (g) A = 2x 3 + 3x 2 - x + 1 at the point where x = 3 (h) f ] a g = 3a 2 - 2a - 6 at the point where a = - 3 (i) V = h 3 - 4h + 9 at the point ^ 2, 9 h (j) g ] x g = x 4 - 2x 2 + 5x - 3 at the point where x = -1. Find the gradient of the (i) tangent and (ii) normal to the curve (a) y = x 2 + 1 at the point ^ 3, 10 h (b) f ] x g = 5 - x 2 at the point where x = -4 (c) y = 2x 5 - 7x 2 + 4 at the point where x = -1 (d) p ] x g = x 6 - 3x 4 - 2x + 8 where x = 1 (e) f ] x g = 4 - x - x 2 at the point ^ -6, 26 h . Find the equation of the tangent to the curve (a) y = x 4 - 5x + 1 at the point ^ 2, 7 h (b) f (x) = 5x 3 - 3x 2 - 2x + 6 at the point ^ 1, 6 h (c) y = x 2 + 2x - 8 at the point ^ -3, -5 h (d) y = 3x 3 + 1 at the point where x = 2 (e) v = 4t 4 - 7t 3 - 2 at the point where t = 2

5.

Find the equation of the normal to the curve (a) f ] x g = x 3 - 3x + 5 at the point ^ 3, 23 h (b) y = x 2 - 4x - 5 at the point ^ -2, 7 h (c) f ] x g = 7x - 2x 2 at the point where x = 6 (d) y = 7x 2 - 3x - 2 at the point ^ -3, 70 h (e) y = x 4 - 2x 3 + 4x + 1 at the point where x = 1.

6.

Find the equation of the (i) tangent and (ii) normal to the curve (a) f ] x g = 4x 2 - x + 8 at the point ^ 1, 11 h (b) y = x 3 + 2x 2 - 5x at the point ^ -3, 6 h (c) F ] x g = x 5 - 5x 3 at the point where x = 1 (d) y = x 2 - 8x + 7 at the point ^ 3, - 8 h (e) y = x 4 - 2x 3 + 4x + 1 at the point where x = 1.

7.

For the curve y = x 3 - 27x - 5, dy find values of x for which = 0. dx

8.

Find the coordinates of the point at which the curve y = x 3 + 1 has a tangent with a gradient of 3.

9.

A function f (x) = x 2 + 4x - 12 has a tangent with a gradient of -6 at point P on the curve. Find the coordinates of the point P.

10. The tangent at point P on the curve y = 4x 2 + 1 is parallel to the x-axis. Find the coordinates of P. 11. Find the coordinates of point Q where the tangent to the curve y = 5x 2 - 3x is parallel to the line 7x - y + 3 = 0 .

475

476


12. Find the coordinates of point S where the tangent to the curve y = x 2 + 4x - 1 is perpendicular to the line 4x + 2y + 7 = 0. 13. The curve y = 3x - 4 has a gradient of 6 at point A. (a) Find the coordinates of A. (b) Find the equation of the tangent to the curve at A. 2

14. A function h = 3t 2 - 2t + 5 has a tangent at the point where t = 2. Find the equation of the tangent. 15. A function f ] x g = 2x 2 - 8x + 3 has a tangent parallel to the line 4x - 2y + 1 = 0 at point P. Find the equation of the tangent at P.

Further Differentiation and Indices The basic rule for differentiating xn works for any rational number n.

Investigation 1. (a) Show that

-h 1 1 - = . x+h x x (x + h)

1 (b) Hence differentiate y = x from first principles. (c) Differentiate y = x - 1 using a short method. Do you get the same answer as 1(b)? 2. (a) Show that ( x + h - x ) ( x + h + x ) = h. (b) Hence differentiate y = 1 2

x from first principles.

(c) Differentiate y = x and show that this gives the same answer as 2(b).

We sometimes need to change a function into index form before differentiating.

EXAMPLES 1. Differentiate 7 3 x .

Solution 1

7 3 x = 7x 3 dy 1 1-1 = 7$ x 3 3 dx 2 7 -3 = x 3 7 1 = # 2 3 x3


= =

7 1 # 3 3 x2 7 3 3 x2

4 2. Find the equation of the tangent to the curve y = 2 at the point x where x = 2.

Solution 4 x2 = 4x - 2

y= dy dx

= - 8x - 3 =-

8 x3

When x = 2 4 y= 2 2 =1 Gradient of the tangent at ^ 2, 1 h: dy 8 =- 3 dx 2 = -1 Equation of the tangent: y - y1 = m _ x - x1 i y - 1 = -1 ] x - 2 g = -x + 2 y = -x + 3 or x + y - 3 = 0

8.7 Exercises 1.

Differentiate (a) x - 3 (b) x 1.4 (c) 6x 0.2 1

(d) x 2 1

(e) 2x 2 - 3x - 1 1

(f) 3x 3 3

(g) 8x 4 (h) - 2x

-

1 2

2.

Find the derivative function, writing the answer without negative or fractional indices. 1 (a) x (b) 5 x (c) 6 x 2 (d) 5 x 5 (e) - 3 x 1 (f) x

477

478


Note that

1 2x 6

=

1 1 # . 2 x6

(g)

Use index laws to simplify first.

10. Find the equation of the tangent to f (x) = 6 x at the point where x = 9.

1 2x 6

(h) x x

x 11. (a) Differentiate x .

2 (i) 3x 3 1 (j) + 4x 2 x 4 3.

Expand brackets first.

This rule is also called the function of a function rule or chain rule.

(b) Hence find the gradient of the x tangent to the curve y = x at the point where x = 4.

Find the gradient of the tangent to the curve y = 3 x at the point where x = 27.

12. Find the equation of the tangent 4 to the curve y = x at the point 1 c 8, m. 2

dx 12 , find when t = 2. t dt

4.

If x =

5.

A function is given by f (x) = 4 x . Evaluate f l(16) .

6.

Find the gradient of the tangent 3 to the curve y = at the point 2x 2 1 c 1, 1 m. 2 dy

14. The function f (x) = 3 x has 3 f l(x) = . Evaluate x. 4 2 15. The hyperbola y = x has two 2 tangents with gradient - . Find 25 the coordinates of the points of contact of these tangents.

if y = ^ x + x h . 2

7.

Find

8.

A function f (x) =

9.

Find the equation of the tangent 1 to the curve y = 3 at the point x 1 c 2, m. 8

dx

13. If the gradient of the tangent to 1 y = x is at point A, find the 6 coordinates of A.

x has a 2 tangent at ^ 4, 1 h . Find the gradient of the tangent.

Composite Function Rule A composite function is a function composed of two or more other functions. 5 For example, ^ 3x 2 - 4 h is made up of a function u5 where u = 3x 2 - 4. To differentiate a composite function, we need to use the result..

dy dx

=

dy du

#

du dx


Proof Let dx, dy and du be small changes in x, y and u where dx " 0, dy " 0, du " 0. dy dy du = # Then dx du dx As dx " 0, du " 0 dy dy du = lim # lim So lim dx " 0 d x du " 0 d u dx " 0 d x

479

You do not need to learn this proof.

Using the definition of the derivative from first principles, this gives dy dx

=

dy du

#

du . dx

EXAMPLES Differentiate 1. (5x + 4) 7

Solution u = 5x + 4 du Then =5 dx y = u7 dy ` = 7u 6 du dy dy du = # dx du dx = 7u 6 # 5 = 35 (5x + 4) 6 Let

Can you see a quick way of doing this question?

2. (3x 2 + 2x - 1) 9

Solution u = 3x 2 + 2x - 1 du Then = 6x + 2 dx y = u9 dy ` = 9u 8 du dy dy du = # dx du dx 8 = 9u (6x + 2) = 9(6x + 2) (3x 2 + 2x - 1) 8 Let

CONTINUED

480


3.

3-x

Solution 1

3 - x = (3 - x) 2 Let u = 3 - x du = -1 dx 1

y = u2 dy 1 -1 = u 2 2 du dy dy du = # dx du dx 1 - 12 = u (-1) 2 1 1 = - (3 - x) 2 2 1 =2 3-x

The derivative of a composite function is the product of two derivatives. One is the derivative of the function inside the brackets. The other is the derivative of the whole function.

d [ f (x)] n = f l(x) n [ f (x)] n - 1 dx


Proof u = f (x) du = f l(x) Then dx y = un dy ` = nu n - 1 du dy dy du = # dx du dx = nu n - 1 # f l(x) = f l(x) n [ f (x)] n - 1 Let


EXAMPLES Differentiate 1. (8x 3 - 1) 5

Solution dy dx

= f l(x) $ n [ f (x)] n - 1 = 24x 2 $ 5 (8x 3 - 1) 4 = 120x 2 (8x 3 - 1) 4

2. (3x + 8) 11

Solution yl = f l(x) . n [ f (x)] n - 1 = 3 # 11 (3x + 8) 10 = 33 (3x + 8) 10

3.

1 (6x + 1) 2

Solution 1 = (6x + 1) - 2 (6x + 1) 2 yl = f l(x) $ n [ f (x)] n - 1 = 6 # -2 (6x + 1) - 3 = -12 (6x + 1) - 3 12 =( 6 x + 1) 3

8.8 Exercises 1.

Differentiate (a) (x + 3) 4 (b) (2x - 1) 3 (c) (5x 2 - 4) 7 (d) (8x + 3) 6 (e) (1 - x) 5

(f) (g) (h) (i) (j)

3 (5x + 9) 9 2 (x - 4) 2 (2x 3 + 3x) 4 (x 2 + 5x - 1) 8 (x 6 - 2x 2 + 3) 6 1

(k) (3x - 1) 2

481

482


2.

Find the gradient of the tangent to the curve y = ] 3x - 2 g3 at the point ^ 1, 1 h .

(o) (x 3 - 7x 2 + x) 4

3.

If f (x) = 2 (x 2 - 3) 5, evaluate fl(2).

3x + 4 1 (q) 5x - 2 1 (r) (x 2 + 1) 4

4.

The curve y =

5.

For what values of x does the 1 function f (x) = have 4x - 1 4 f l(x) = ? 49

6.

Find the equation of the tangent to y = (2x + 1) 4 at the point where x = - 1.

(l) (4 - x) - 2 (m) (x 2 - 9) - 3 1

(n) (5x + 4) 3

3

(p)

(s) (t) (u) (v) (w) (x) (y)

(7 - 3x) 2 5 4+x 1 2 3x - 1 3 4 (2x + 7) 9 1 x 4 - 3x 3 + 3x

3

3

(4x + 1) 4 1

4

( 7 - x) 5

x - 3 has a 1 tangent with gradient at point 2 N. Find the coordinates of N.

Product Rule Differentiating the product of two functions y = uv gives the result

dy dx

=u

dv du +v dx dx

Proof y = uv Given that dy, du and dv are small changes in y, u and v. y + dy = (u + du) (v + dv) = uv + udv + vdu + dudv `

dy = udv + vdu + dudv ^ since y = uv h dy dv du dv =u +v + du dx dx dx dx


As dx " 0, du " 0 dy dv du dv F lim = lim < u +v + du dx " 0 d x dx " 0 dx dx dx dv du dv F + lim < v F + lim < du F = lim < u dx " 0 d x " 0 d x " 0 dx dx dx dy dv du =u +v dx dx dx

483


It is easier to remember this rule as y l = uv l + vu l. We can also write this the other way around which helps when learning the quotient rule in the next section. If y = uv, y l = u lv + v lu

EXAMPLES Differentiate 1. ] 3x + 1 g ] x - 5 g

Solution You could expand the brackets and then differentiate: ] 3x + 1 g ] x - 5 g = 3x 2 - 15x + x - 5 = 3x 2 - 14x - 5 dy = 6x - 14 dx Using the product rule: y = uv where u = 3x + 1 and v = x - 5 ul = 3 vl = 1 y l = u lv + v l u = 3 ] x - 5 g + 1 ] 3x + 1 g = 3x - 15 + 3x + 1 = 6x - 14 2. 2x 5 ] 5x + 3 g3

Solution y = uv where u = 2x 5 and v = ] 5x + 3 g 3 u l = 10x 4 v l = 5.3 ] 5x + 3 g 2 CONTINUED

484


y l = u lv + v l u = 10x 4 ] 5x + 3 g3 + 5.3 ] 5x + 3 g2 $ 2x 5 = 10x 4 ] 5x + 3 g3 + 30x 5 ] 5x + 3 g2 = 10x 4 ] 5x + 3 g2 6 ] 5x + 3 g + 3x @ = 10x 4 ] 5x + 3 g2 ] 8x + 3 g

We can simplify this further by factorising.

3. (3x - 4) 5 - 2x

Solution 1

Remember

5 - 2x = ] 5 - 2x g 2

1

y = uv where u = 3x - 4 and v = ] 5 - 2x g 2 1 1 ul = 3 v l = - 2 $ (5 - 2 x ) 2 2 y l = u lv + v lu 1

1 1] 5 - 2x g 2 ] 3x - 4 g 2 1 5 - 2x - (3x - 4) ] 5 - 2x g 2 3x - 4 5 - 2x 1 (5 - 2x) 2 3x - 4 5 - 2x 5 - 2x 5 - 2x $ 5 - 2x - (3x - 4)

= 3 ] 5 - 2x g 2 +- 2 $ =3 =3 =3 = =

3

5 - 2x 3(5 - 2x) - (3x - 4)

5 - 2x 15 - 6x - 3x + 4 = 5 - 2x 19 - 9x = 5 - 2x

8.9 Exercises 1.

Change this into a product before differentiating.

Differentiate (a) (b) (c) (d) (e) (f)

x 3 ] 2x + 3 g ] 3x - 2 g ] 2 x + 1 g 3x ] 5x + 7 g

4x 4 ^ 3x 2 - 1 h 2x ^ 3x 4 - x h x 2 ] x + 1 g3

4x ] 3x - 2 g5 3x 4 ] 4 - x g3 ] x + 1 g ] 2x + 5 g4 ^ x 3 + 5x 2 - 3 h ^ x 2 + 1 h 5 x 2-x 5x + 3 (l) 2x - 1

(g) (h) (i) (j) (k)


2.

Find the gradient of the tangent to the curve y = 2x ] 3x - 2 g4 at the point ^ 1, 2 h .

7.

Find the equation of the tangent to h = (t + 1) 2 (t - 1) 7at the point ^ 2, 9 h .

3.

If f (x) = (2x + 3) (3x - 1) 5, evaluate fl(1) .

8.

4.

Find the exact gradient of the tangent to the curve y = x 2x + 5 at the point where x = 1.

Find exact values of x for which the gradient of the tangent to the curve y = 2x ] x + 3 g2 is 14.

9.

Given f (x) = (4x - 1) (3x + 2) 2, find the equation of the tangent at the point where x = -1.

5.

Find the gradient of the tangent where t = 3, given x = ] 2t - 5 g ] t + 1 g3.

6.

Find the equation of the tangent to the curve y = x 2 ] 2x - 1 g4 at the point ^ 1, 1 h .

Quotient Rule u Differentiating the quotient of two functions y = v gives the result.

dy dx

=

v

du dv -u dx dx v2

Proof u y= v Given that dy, du and dv are small changes in y, u and v. u + du y + dy = v + dv u + du u u ` dy = - v a since y = v k v + dv v (u + du) u (v + dv) = v (v + dv) v (v + dv) v (u + du) - u (v + dv) = v (v + dv) vu + vdu - uv - udv = v (v + dv) vdu - udv = v (v + dv) du dv v -u dy dx dx = v (v + dv) dx As dx " 0, dv " 0

485

486


lim

dx " 0


dy dx dy dx

R S v du - u dv S dx dx = lim S dx " 0 v ( v + d v) T du dv v -u dx dx = v2

V W W W X

It is easier to remember this rule as y l =

u lv - v lu . v2

u u lv - v lu If y = v , y l = v2

EXAMPLES Differentiate 1. 3x - 5 5x + 2

Solution u y = v where u = 3x - 5 and v = 5x + 2 ul = 3 vl = 5 yl = =

u lv - v lu v2 3 (5x + 2) - 5 (3x - 5)

(5x + 2) 2 15x + 6 - 15x + 25 = (5x + 2) 2 31 = (5x + 2) 2 2. 4x 3 - 5x + 2 x3 - 1

Solution u y = v where u = 4x 3 - 5x + 2 and v = x 3 - 1 u l = 12x 2 - 5 v l = 3x 2 u l v - v lu yl = v2 (12x 2 - 5) (x 3 - 1) - 3x 2 (4x 3 - 5x + 2) = (x 3 - 1 ) 2 12x 5 - 12x 2 - 5x 3 + 5 - 12x 5 + 15x 3 - 6x 2 = (x 3 - 1 ) 2 10x 3 - 18x 2 + 5 = (x 3 - 1) 2


8.10 1.

Exercises

Differentiate 1 2x - 1 (b) 3x x+5 x3 (c) 2 x -4 (d) x - 3 5x + 1 (e) x - 7 x2 (f) 5x + 4 x+3 x (g) 2 2x - x (h) x + 4 x-2 (i) 2x + 7 4x - 3 (j) x + 5 3x + 1 (k) x + 1 3x 2 - 7 2x 2 (l) 2x - 3 2 (m) x + 4 x2 - 5 3 (n) x x+4 3 (o) x + 2x - 1 x+3 2 (p) x - 2x - 1 3x + 4 3 x +x (q) 2 x -x-1 2x (r) 1 (x + 5) 2

3 (s) (2x - 9) 5x + 1 x-1 (t) (7x + 2) 4 5 (u) (3x + 4)

(a)

(2x - 5) 3 (v) 3x + 1 x+1 (w)

x-1 2x - 3

(x)

x2 + 1 (x - 9) 2

2.

Find the gradient of the tangent to 2x the curve y = at the point 3x + 1 1 c 1, m. 2

3.

If f (x) =

4.

Find any values of x for which the gradient of the tangent to the 4x - 1 curve y = is equal to - 2. 2x - 1

5.

Given f (x) = f l(x) =

4x + 5 evaluate f l(2). 2x - 1

2x find x if x+3

1 . 6

6.

Find the equation of the tangent x to the curve y = at the x+2 2 point c 4, m. 3

7.

Find the equation of the tangent x2 - 1 to the curve y = at the x+3 point where x = 2.

Angle Between 2 Curves To measure the angle between two curves, measure the angle between the tangents to the curves at that point.

487

488


tan i =

m1 - m2 1 + m1 m2

where m 1 and m 2 are the gradients of the tangents to

the curves at the point of intersection.

EXAMPLE Find the acute angle formed at the intersection of the curves y = x 2 and y = (x - 2) 2 .

Solution

The curves intersect at the point (1,1) . For y = x 2 dy = 2x dx dy At (1, 1), = 2 (1 ) dx m1 = 2 ` For y = (x - 2) 2 dy = 2 (x - 2 ) dx dy At (1, 1), = 2 (1 - 2) dx m2 = - 2 ` tan i =

m1 - m2 1 + m1 m2

2 - (-2) 1 + 2 (-2) 4 = 3 i = 53c 08l =


8.11

Exercises

1.

(a) Sketch the curves y = x 2 - 4 and y = x 2 - 8x + 12 on the same set of axes. (b) Show that the curves intersect at the point Q (2, 0) . (c) Find the gradient of the tangent of each curve at point Q. (d) Find the acute angle at which the curves intersect at Q.

6.

The curves y = 2x 2 - 4x and y = x 2 - x + 4 intersect at two points X and Y. (a) Find the coordinates of X and Y. (b) Find the gradient of the tangent to each curve at X and Y. (c) Find the acute angle between the curves at X and Y.

2.

(a) Sketch the curve y = x 2 and the line y = 6x - 9 on the same set of axes. (b) Find the point P, their point of intersection. (c) Find the gradient of the curve y = x 2 at P. (d) Find the acute angle between the curve and the line at P.

7.

Find the acute angle between the curve f (x) = x 2 - 1 and the line g (x) = 3x - 1 at their 2 points of intersection.

8.

(a) Find the points of intersection between y = x 3 and y = x 2 + 2x. (b) Find the acute angle between the curves at these points.

9.

Show that the acute angle between the curves y = x 2 and y = 4x - x 2 is the same at both the points of intersection.

3.

Find the acute angle between the curves y = x 2 and y = x 3 at point (1,1) .

4.

Find the acute angle between the curves y = x 3 and y = x 2 - 2x + 2 at their point of intersection.

5.

What is the obtuse angle between the curves f (x) = x 2 - 4x and g (x) = x 2 - 12 at the point where they meet?

10. Find the obtuse angles between the curves y = x 3 + 2x and y = 5x - 2x 2 at their points of intersection.

489

490


Test Yourself 8 1.

Sketch the derivative function of each graph (a)

6.

7. 8.

Find the gradient of the tangent to the curve y = x 3 - 3x 2 + x - 5 at the point (-1, -10) . dh If h = 60t - 3t 2, find when t = 3. dt Find all x-values that are not differentiable on the following curves. (a)

(b)

y

(b) 5 4 3 2 1

2.

Differentiate y = 5x 2 - 3x + 2 from first principles.

3.

Differentiate (a) 7x 6 - 3x 3 + x 2 - 8x - 4 (b) 3x - 4 2x + 1

(c)

-4 -3 -2 -1-1

5.

Given f (x) = (4x - 3) 5, find the value of (a) f (1) (b) f 1 (1).

2

3

1 2 3 4

-2 -3 -4 -5

dv if v = 2t 2 - 3t - 4. dt

Find

1

5 4 3 2 1

(c) (x 2 + 4x - 2) 9 (d) 5x(2x - 1) 4 (e) x 2 x 5 (f) 2 x 4.

-44 -3 -2 -1 -1 -2 -3 -4 -5 y

9.

Differentiate (a) f ] x g = 2 ] 4x + 9 g4 5 (b) y = x-3 (c) y = x ] 3x - 1 g2 4 (d) y = x (e) f (x) = 5 x

x

4

x


10. Sketch the derivative function of the following curve.

15. Find the equation of the tangent to the curve y = x 2 + 2x - 5 that is parallel to the line y = 4x - 1.

y

16. Find the gradient of the tangent to the curve y = ] 3x - 1 g3 ] 2x - 1 g2 at the point where x = 2. 17. Find f l(4) when f ] x g = ] x - 3 g9 . x

18. Find the equation of the tangent to the 1 1 curve y = at the point where x = . 3x 6 1 2 at with respect 2 to t and find the value of t for which ds = 5, u = 7 and a = - 10. dt

19. Differentiate s = ut +

11. Find the equation of the tangent to the curve y = x 2 + 5x - 3 at the point ^ 2, 11 h . 12. Find the point on the curve y = x 2 - x + 1 at which the tangent has a gradient of 3. 13. Find

dS if S = 4rr 2. dr

14. At which points on the curve y = 2x 3 - 9x 2 - 60x + 3 are the tangents horizontal?

20. Find the x-intercept of the tangent to 4x - 3 the curve y = at the point where 2x + 1 x = 1. 21. Find the acute angle between the curve y = x 2 and the line y = 2x + 3 at each point of intersection. 22. Find the obtuse angle between the curve y = x 2 and the line y = 6x - 8 at each point of intersection.


If f (x) = 3x 2 (1 - 2x) 5, find the value of f (1) and fl(1) .

2.

If A =

3.

dA 5h + 3 , find when h = 1. 7h - 1 dh

dx Given x = 2t 4 + 100t 3, find and find dt dx values of t when = 0. dt

4.

Find the equations of the tangents to the curve y = x (x - 1) (x + 2) at the points where the curve cuts the x-axis.

5.

Find the points on the curve y = x 3 - 6 where the tangents are parallel to the line y = 12x - 1. Hence find the equations of the normals to the curve at those points.

491

492


6.

Find f l(2) if f (x) =

7.

Differentiate (5x + 1) 3 (x - 9) 5 .

8.

Find the derivative of y =

9.

If f (x) = 2x 3 + 3x 2 + 4, for what exact values of x is fl(x) = 7?

3x - 2 .

2x + 1 . (4x - 9) 4

10. Find the equation of the normal to the curve y = 3 x + 1 at the point where x = 8. 11. The tangent to the curve y = ax 3 + 2 at the point where x = 3 is inclined at 135c to the x-axis. Find the value of a. 12. The normal to the curve y = x 2 + 1 at the point where x = 2, cuts the curve again at point P. Find the coordinates of P.

18. The function f (x) = 3x + 1 has a tangent that makes an angle of 30c with the x-axis. Find the coordinates of the point of contact for this tangent and find its equation in exact form. 19. Find all x values of the function f (x) = (x 2 - 3) (2x - 1) 8 for which f l(x) = 0. 20. (a) Find any points at which the graph below is not differentiable. (b) Sketch the derivative function for the graph. y

90c

180c 270c

360c

x

13. Find the exact values of the x- coordinates of the points on the curve y = (3x 2 - 2x - 4) 3 where the tangent is horizontal. 14. Find the gradient of the normal to the curve y = 2x 5 - x at the point (4, 8) . 15. Find the equation of the tangent to the curve y = x 3 - x 2 + 2x + 6 at point P (1, 8) . Find the coordinates of point Q where this tangent meets the y-axis and calculate the exact length of PQ. 16. (a) Show that the curves y = ] 3x - 2 g5 and 5x - 3 y= intersect at ^ 1, 1 h x+1 (b) Find the acute angle between the curves at this point. 17. The equation of the tangent to the curve y = x 4 - nx 2 + 3x - 2 at the point where x = - 2 is given by 3x - y - 2 = 0. Evaluate n.

21. Find the point of intersection between the tangents to the curve y = x 3 - 2x 2 - 5x + 3 at the points where x = 2 and x = - 1. 22. Find the equation of the tangent to the x2 - 3 parabola y = at the point where 2 the tangent is perpendicular to the line 3x + y - 3 = 0. 23. Differentiate

3x - 2 . 2x 3

24. (a) Find the equations of the tangents to the parabola y = 2x 2 at the points where the line 6x - 8y + 1 = 0 intersects with the parabola. (b) Show that the tangents are perpendicular.


25. Find any x values of the function 2 f (x ) = 3 where it is not x - 8x 2 + 12x differentiable.

33. Find fl(7) as a fraction, given 1 . f (x) = 3 x+1

26. The equation of the tangent to the curve y = x 3 + 7x 2 - 6x - 9 is y = ax + b at the point where x = -4. Evaluate a and b.

34. For the function f (x) = ax 2 + bx + c, f (2) = 4, f l(1) = 0 and f l(x) = 8 when x = -3. Evaluate a, b and c.

27. Find the exact gradient with rational denominator of the tangent to the curve y = x 2 - 3 at the point where x = 5. p 28. The tangent to the curve y = x has a 1 gradient of - at the point where x = 3. 6 Evaluate p. 2r dV when r = and h = 6 given 3 dr 1 V = rr 3 h. 3

29. Find

30. Evaluate k if the function f (x) = 2x 3 - kx 2 + 1 has f l(2) = 8. 31. Find the equation of the chord joining the points of contact of the tangents to the curve y = x 2 - x - 4 with gradients 3 and -1. 32. Find the equation of the straight line passing through ^ 4, 3 h and parallel to the tangent to the curve y = x 4 at the point ^ 1, 1 h .

35. Find the equation of the tangent to the curve S = 2rr 2 + 2rrh at the point where r = 2 (h is a constant). 36. Differentiate (a) 2x 3 - x ] 3x - 5 g4 2x + 1 (b) (x - 3) 3 37. The tangents to the curve y = x 3 - 2x 2 + 3 at points A and B are perpendicular to the tangent at ^ 2, 3 h . Find the exact values of x at A and B. 38. (a) Find the equation of the normal to the curve y = x 2 + x - 1 at the point P where x = 3. (b) Find the coordinates of Q , the point where the normal intersects the parabola again.

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9

Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference of a circle Concentric circles: Circles that have the same centre Concyclic points: Points that lie on the circumference of the same circle Cyclic quadrilateral: A cyclic quadrilateral is a figure whose four vertices are concyclic points. The four vertices lie on the circumference of a circle

Radius: A radius is the distance from the centre of a circle out to the circumference (radii is plural, meaning more than one radius) Subtend: Form an angle at some point (usually the centre or circumference of a circle) Tangent: A straight line external to a curve or circle that just touches the curve or circle at a single point

Chapter 9 Properties of the Circle

INTRODUCTION IN CHAPTER 4, YOU STUDIED the geometry of angles, triangles, quadrilaterals

and other polygons. This chapter shows you some properties of the circle.

DID YOU KNOW? A rainbow is the shape of an arc of a circle. If you could see the whole rainbow, it would form a circle. Research the rainbow on the Internet and find out more about its shape and other properties.

Parts of a Circle An arc is a part of the circumference.

A tangent touches the circle at one point.

499

500


Concentric circles are circles that have the same centre.

Equal circles have the same radius.

DID YOU KNOW? Leonardo da Vinci (1452–1519) was a great artist, scientist and inventor. He studied geometry, and many of his model drawings show this influence. His drawings included designs for flying machines, spring-driven automobiles, bridges and weapons. Leonardo’s designs were revolutionary, and the scientists of his time did not have the knowledge needed to make the models work.

Arcs, Angles and Chords Equal arcs subtend equal angles at the centre of the circle.


501

Proof Let two equal arcs have lengths l 1 and l 2, and subtend angles of a and b at the centre of the circle. Using l = ri , l 1 = ra and l 2 = rb But l 1 = l 2 ` ra = rb ` a=b The converse is also true:

If two arcs subtend equal angles at the centre of the circle, then the arcs are equal.

EXAMPLE

AB is a diameter of the circle with centre O. Arc CB = arc BD. Prove +AOC = +AOD.

Solution Since arc CB = arc BD,+COB = +DOB Let +COB = +DOB = x Then +AOC = 180c - +COB = 180c - x Also +AOD = 180c - +DOB = 180c - x ` +AOC = +AOD

(+AOB is a straight angle) (similarly)

You will study the formula l = r i in Chapter 5 of the HSC Course book.

502


Equal chords subtend equal angles at the centre of the circle.

Proof OA = OC OB = OD AB = CD ` by SSS, DOAB / DOCD ` +AOB = +COD

(equal radii) (similarly) (given) (corresponding +s in congruent Ds)

The converse in also true:

Equal angles subtended at the centre of the circle cut off equal chords.

EXAMPLE

AD is a diameter of the circle with centre O, where AB = CD. Prove that BC < AD.


503

Solution Since AB = CD, +AOB = +COD Let +AOB = +COD = x Then +BOC = 180c - (x + x) = 180c - 2x OB = OC ` D OBC is isosceles with +OBC = +OCB +OBC + +OCB + 180c - 2x = 180c +OBC + +OCB = 2x ` +OBC = +OCB = x ` +OBC = +AOB

(+AOD is a straight +) (equal radii) (+ sum of DOBC)

But these are equal alternate angles ` BC < AD

The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.

These figures show that this property can look quite different in different situations.

Proof

Join AO and produce to D. Let +BAO = x and +CAO = y ` +BAC = x + y OA = OB

(equal radii)

504


` +OBA = x = +BAO OA = OC

(base +s of isosceles D)

` +OCA = y = +CAO

(base +s of isosceles D)

(equal radii)

+BOD = x + x = 2x

(exterior + of DOBA)

+COD = y + y = 2y +BOC = +BOD + +COD = 2x + 2y = 2 (x + y ) = 2+BAC

(exterior + of DOCA)

EXAMPLES 1. Find the values of x and y.

Solution x = 40 (+ at centre is twice the + at the circumference) Reflex +BOC = 360c - 80c (+ of revolution) = 280c ` y = 140 (+ at centre is twice the + at the circumference) 2. Prove +BOC is twice the size of +OCA.


Solution Let +OAC = x +BOC = 2+OAC = 2x ` `

(+ at centre is twice the + at the circumference)

OA = OC ^ equal radii h +OCA = +OAC = x (base +s of isosceles D) +BOC = 2+OCA

Angles in the same segment of a circle are equal.

Proof

Join A and D to centre O +AOD = 2+ABD +AOD = 2+ACD ` +ABD = +ACD

(+ at centre is twice the + at the circumference)

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506


EXAMPLE Prove D ABC and D DEC are similar. Hence find the value of y correct to 1 decimal place.

Solution +ABC = +DEC +BCA = +ECD

(+s in same segment) (vertically opposite +s)

` D ABC ||| D DEC EC ED ` = BA BC y 6 .5 = 1 .7 1.2 1 .2y = 1 . 7 # 6. 5 1 .7 # 6 .5 y= 1 .2 = 9.2 cm correct to 1 decimal place.

The angle in a semicircle is a right angle.

Proof


+AOB = 180c +AOB = 2+ACB ` +ACB = 90c

(straight +) (+ at centre is twice the + at the circumference)

EXAMPLE AB is a diameter of the circle below. If BC = 2 m and AB = 9 m, find the exact length of AC.

Solution +ACB = 90c ` AB 2 = AC 2 + BC 2 9 2 = AC 2 + 2 2 81 = AC 2 + 4 77 = AC 2 ` AC =

77 m

9.1 Exercises 1.

Find values of all pronumerals (O is the centre of each circle). (a)

(b)

507

508


(c)

3.


(d)

(b)

(e) (c)

(f) (d)

2.

The circumference of a circle is 16r cm. Find the length of the arc that subtends an angle of 40c at the centre of the circle.

(e)


(f)

4.


(g)

(b)

(h)

(c)

(i)

(d)

(j)

509

510


(e)

(f)

(j)

5.

(a) Prove D ABC ||| D DEC. (b) Hence find the value of x correct to 1 decimal place.

6.

Find x and y, giving reasons.

7.

Find x and y, giving reasons.

(g)

(h)

(i)


8.

Evaluate x, giving reasons for each step in your calculation.

11. Find x, giving reasons for each step in your calculations.

9.

Prove D STV and DWUV are similar. Hence find x.

12. The circle below has centre O. D

C

O y A

x 38c B

10. AB = 6 cm and BC = 3 cm. O is the centre of the circle. Show that the radius of the circle is 3 5 cm. 2

(a) Evaluate x and y. (b) Show that AD = BC. 13. Show that AD < BC in the circle below. A

D

33c

114c

C

B

511

512


14. A circle has centre O and radius r as shown.

15. The circle below has centre O and +DAB = i. A i

A

B

O

C

B

(a) Show that triangles AOB and ABC are similar. (b) Show that BC = 2 r.

O

C

Show that +DAB and +BCD are supplementary.

Chord Properties A perpendicular line from the centre of a circle to a chord bisects the chord.

Proof

D


+ADO = +BDO = 90c OA = OB

(given) (equal radii)

OD is common ` by RHS DOAD / DOBD ` AD = BD

(corresponding sides in congruent Ds)

So OD bisects AB The converse is also true: A line from the centre of a circle that bisects a chord is perpendicular to the chord.

EXAMPLES 1. Line OC is perpendicular to chord AB. If the radius of the circle is 6 cm and the chord is 11 cm long, find the length of OC, correct to 1 decimal place.

Solution AB = 11 ` AC = 5.5 Also OA = 6

(OC bisects AB) (radius-given)

OA = AC + OC 6 2 = 5.5 2 + OC 2 36 = 30.25 + OC 2 5.75 = OC 2 2

`

2

2

OC = 5.75 = 2.4 cm CONTINUED

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514


2. Given XY = YZ and +OXY = 54c, find +XOY.

Solution +OYX = 90c

(OY bisects XZ)

+XOY = 180c - (90c + 54c) = 36c

(+ sum of DOXY)

Equal chords are equidistant from the centre of the circle.

Proof

Let CD = AB +OEB = +OFD = 90c OB = OD AB = CD

(given) (equal radii) (given)


1 AB 2 1 DF = CD 2 BE = DF BE =

`

by RHS DOEB / DOFD OE = OF

` `

(OE bisects AB) (OF bisects CD)


The converse is also true:

Chords that are equidistant from the centre are equal.

Class Exercise Prove that chords that are equidistant from the centre are equal.

EXAMPLE In the circle below, with centre O, OE = 35 mm, DE = 56 mm and OC = 21 mm. Show that AB = DE.

Solution EF = 28 mm

(OF bisects DE)

OE = EF + OF 35 2 = 28 2 + OF 2 35 2 - 28 2 = OF 2 441 = OF 2 2

` ` So

2

OF = 441 = 21 mm OF = OC AB = DE.

2

(chords equal when equidistant from the centre)

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516


The products of intercepts of intersecting chords are equal.

AE: EB = DE: EC

Proof

+AED = +CEB +DAE = +ECB +ADE = +EBC

(vertically opposite +s) (+s in same segment) (similarly)

D AED <;D CEB AE DE ` = EB EC ` AE: EB = DE: EC `

EXAMPLE

Given chord AB = 12.3 cm, EB = 2.7 cm and DE = 10.6 cm, find the length of EC, correct to 1 decimal place.


Solution AE = AB - EB = 12.3 - 2.7 = 9 .6 AE: EB = DE: EC 9.6 # 2.7 = 10.6 # EC 9 .6 # 2 .7 ` EC = 10.6 = 2.4 cm

9.2 Exercises 1.

Find the values of all pronumerals (O is the centre of each circle).

(d)

(a)

(e)

(b)

(f)

(c)

517

518


(g)

3.

A circle with radius 89 mm has a chord drawn 52 mm from the centre. How long, to the nearest millimetre, is the chord?

4.

O and P are the centres of intersecting circles with radii 20 cm and 8 cm respectively. If AB = 10 cm, find the distance OP, correct to 1 decimal place.

5.

Show AB = CD.

6.

AC = 20 cm and AD = 26 cm. Find OB, correct to 1 decimal place.

(h)

(i)

(j)

2.

Find the exact radius of a circle with a chord that is 8 cm long and 5 cm from the centre.


7.

Evaluate x and y, correct to 1 decimal place.

9.

A circle with centre O has radius r and chord AB = x. D

O

A

C

8.

B

Find the values of all pronumerals.

Show that CD =

2r + 4r 2 - x 2 . 2 A

10.

E C D

B

(a) Prove that triangles ABC and CDE are similar. (b) Show that AC : CD = BC : CE.

Concyclic Points Concyclic points are points that lie on the circumference of a circle.

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520


Any 3 non-collinear points are concyclic. They lie on a unique circle, with centre at the point of intersection of the perpendicular bisectors of the intervals joining these points.

Four or more non-collinear points may not necessarily lie on a circle.

Cyclic quadrilaterals A cyclic quadrilateral is a figure whose 4 vertices are concyclic points.

The opposite angles in a cyclic quadrilateral are supplementary.


Proof

Join B and D to O. Obtuse +DOB = 2+A Reflex +DOB = 2+C Obtuse +DOB + reflex +DOB = 360c ` 2+A + 2+C = 360c ` +A + +C = 180c

(+ at centre is doubl e +at circumference) (+ of revolution)

Similarly, it can be proven that +B + +D = 180c by joining A and C to O. The converse is also true:

If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

The property of opposite angles being supplementary in a cyclic quadrilateral can also be used to prove the following property:

The exterior angle at a vertex of a cyclic quadrilateral is equal to the interior opposite angle.

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522


Proof

Let Then `

+A = x +BCD = 180c - x

(opposite +s supplementary in cyclic quadrilateral) (+DCE straight angle)

+BCD + +BCE = 180c

`

+BCE = 180c - (180c - x) = 180c - 180c + x =x +A = +BCE

`

EXAMPLE Evaluate a, b and c.

Solution OB = OD ` +ODB = 20c a + 20c + 20c = 180c ` a = 180c - 40c = 140c b = 70c c = 180c - 70c = 110c

(equal radii) (base + s of isosceles D equal) (+ sum of D)

(+ at centre double + at circumference) (opposite + s in cyclic quadrilateral)


9.3 Exercises 1.

Find the values of all pronumerals.

(e)

(a)

(f) (b)

(g) (c)

(h)

(d)

(i)

523

524


(e)

(j) 118c

yc xc

O 46c

2.

(f) Find the values of all pronumerals. (a)

(g)

(b)

(h)

(c)

(i)

(d)

(j)

xc 111c


3.

Show that ABCD is a cyclic quadrilateral. (a)

A

i

C

i

58c C C

B

B

D

B 58c

(b)

(c)

525

D

A E

D

A

Tangent Properties

The tangent to a circle is perpendicular to the radius drawn from the point of contact.

The perpendicular distance is the shortest distance — any other distance would be greater than the radius.

The converse is also true: The line perpendicular to the radius at the point where it meets the circle is a tangent to the circle at that point.

526


Here is another property of tangents to a circle: Tangents to a circle from an exterior point are equal.

Proof

Join OB +A = +C = 90c OB is common OA = OC ` by RHS, DOAB / DOCB ` AB = CB

(tangent = radius) (equal radii) (corresponding sides in congruent Ds)

EXAMPLE A circle with centre O and radius 10 cm has a tangent AB drawn to it where OB = 26 cm. Find the length of AB.


527

Solution OA = 10 +OAB = 90c

(radius) (tangent = radius)

` OB 2 = OA 2 + AB 2 26 2 = 10 2 + AB 2 676 = 100 + AB 2 576 = AB 2 ` AB = 576 = 24 cm

When two circles touch, the line through their centres passes through their point of contact.

Proof

You could also prove this result for when circles touch internally.

AB is a tangent to circle with centre O ` +OCB = 90c AB is a tangent to circle with centre P ` +PCB = 90c +OCB + +PCB = 90c + 90c = 180c ` OCP is a straight line.

(tangent = radius) (similarly)

528


EXAMPLE Two circles touch at A and the larger circle has centre O. Prove (a) TABC and TAOD are similar (b) CB < DO (c) BC = 2DO.

Solution (a) OA is a diameter of small circle

(line through centres passes through point of contact) (+ in semicircle)

` +ODA = 90c Since AB is a diameter of the larger circle, +BCA = 90c (similarly) ` +BCA = +ODA +A is common ` D ABC ;;; D AOD (b) +BCA = +ODA These are equal corresponding angles. ` CB < DO (c)

AB AB ` OA AB ` OA ` BC

= 2 OA

[from (a)]

(OA radius)

=2 AC BC = =2 AD DO = 2DO =

(by similar Ds)

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.


Proof

Draw in diameter CF and join EF. Let +ECB = x. +FCB = 90c ` +FCE = 90c - x +FEC = 90c

(tangent = radius) (+ in semicircle)

` +EFC = 180c - (90c + 90c - x) (angle sum of D ) =x +EFC = +EDC (angles in same segment) ` +EDC = +ECB

EXAMPLE Evaluate x and y.

CONTINUED

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530


Solution +ABC = 180c - (95c + 50c) = 35c ` x = 35c y = 35c

(+ sum of D ) (+ s in alternate segment) ( y and +ABC in same segment)

The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point.

PQ 2 = QR:QS where PQ is a tangent to the circle.

Proof

The third pair of angles is equal by angle sum of a triangle.

+QPR = +PSR +Q is common ` PQR ;;;D SPQ PQ QR PR = = ` QS SP PQ PQ QR ` = QS PQ 2 PQ = QR:QS

(angles in alternate segments)


EXAMPLE AB is a tangent to the circle and CD = 1.3 cm, BC = 1.7 cm. Find the length of AB, correct to 1 decimal place.

Solution BD = 1.3 + 1.7 =3 AB 2 = BC: BD = 1 .7 # 3 = 5 .1 AB = 5.1 = 2.3 cm correct to 1 decimal place.

9.4 Exercises 1.


(b)

531

532


(c)

(h)

(d)

(i)

y x

38c O

(e) (j)

(f)

2. (g)

Find the values of all pronumerals (all external lines are tangents to the circles). (a)


(g)

(b)

(h) (c)

(i) (d)

xc

63c

O

yc

(j)

9 cm

15

cm

13 cm

(e)

O

y P

(f)

533

534


3.

4.

Find the values of all pronumerals, giving reasons for each step of your working (O is the centre of circle, AB is a tangent).

AB, BC and AC are tangents, with AB = 24 cm, BC = 27 cm and BM = 15 cm. Find the length of AC.

6.

Find the values of all pronumerals (O is the centre of each circle; all external lines are tangents). (a)

(b)

(c)

5.

AB = 6.5 m, AC = 3.9 m and BC = 5.2 m. Prove A lies on a diameter of the circle, given BC is a tangent to the circle. (d)


(e)

(i)

(f)

(j)

(g)

7.


(h)

(b)

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536


(c)

(h)

(d)

(i)

(e) (j)

(f) 8.

(g)

Find AB, given AD = 4.9 m, BC = 5.1 m and CD = 7.8 m.


Test Yourself 9 1.

O is the centre of the circle. Evaluate i.

2.

Evaluate y to 1 decimal place.

3.

4.

5.

Evaluate x, y and z, giving reasons for each step of your working.

6.

O is the centre of the larger circle. Find the value of x.

7.

AB is a tangent to the circle. Evaluate a, b and c.

8.

O is the centre of the circle, and AB is a tangent. Evaluate a, b, c and d, giving reasons for each step of your working.

AB is a tangent to the circle. Find the value of x to 1 decimal place.

O is the centre of the circle. Find the length of tangents x and y.

537

538


9.

Find the length of the radius of the circle. O is the centre.

13. Calculate the length of AB to 3 significant figures, given that A and B are the centres of the circles.

10. Find the length of tangent AB. 14. Find the value of x to 1 decimal place.

11. Evaluate a and b.

12. O is the centre of the circle. Find the value of a and b.

15. Find the length of AB.

16. Evaluate a and b.


17. Evaluate x and y, giving reasons for your working.

18. Evaluate x, y and z.

19. Prove that DBCD is similar to DABC.

20. O is the centre of the circle. (a) Prove that DOAC and DOBC are congruent. (b) Show that OC bisects AB.


Find the length of the radius, to the nearest centimetre, if AC = 10 cm and BD = 3 cm.

2.

In the circle below with centre O, OD = DC. Prove +AOE = 3+DCB.

539

540


3.

Two circles meet at points A and B. A tangent to each circle is drawn from A to meet the circles at D and C. Prove +DAC = 90c.

5.

The triangles below have +BDE = 2+BAD and +CDE = 2+CAD. Prove that a circle can be drawn through A, B and C with centre D.

4.

Three equal circles touch each other, as in the figure. (a) Prove that the triangle with vertices the centres of the circles is equilateral. (b) Find the perimeter of the curved figure DEF in terms of the radius r of the circles. (c) Find the exact area of the shaded region.

6.

Two chords AB and CD intersect at 90c Prove, for obtuse +AOD,+AOD + +COB = 180c where O is the centre of the circle.

7.

Prove that any kite ABCD with +ADC = +ABC = 90c is a cyclic quadrilateral with diameter AC.


8.

A large circle with radius R is surrounded by 7 smaller circles with radius r. A circle is drawn through the centres of the smaller circles. If R = area in terms of r.

3r , find the shaded 2

9.

Prove that if an interval subtends equal angles at two points on the same side of it, then the endpoints of the interval and the two points are concyclic.

10. Prove that if both pairs of opposite angles in a quadrilateral are supplementary, then the quadrilateral is cyclic.

541

10 The Quadratic Function TERMINOLOGY Axis of symmetry: A line about which two parts of a graph are symmetrical. One half of the graph is a reflection of the other Coefficient: A constant multiplied by a pronumeral in an algebraic term e.g. in ax3 the a is the coefficient Discriminant: Part of the quadratic formula, the algebraic expression b 2 - 4ac is called the discriminant as its value determines the number and nature of the roots of a quadratic equation Equations reducible to quadratics: Equations that can be reduced to the form: ax 2 + bx + c = 0 Indefinite: A quadratic function where f(x) can be both positive and negative for varying values of x

Maximum value: The maximum or greatest y-value of a graph for a given domain Minimum value: The minimum or smallest y-value of a graph for a given domain Negative definite: A quadratic function where f(x) is always negative for all values of x Positive definite: A quadratic function where f(x) is always positive for all values of x Root of an equation: The solution of an equation

Chapter 10 The Quadratic Function

543

INTRODUCTION THE SOLUTION OF QUADRATIC equations is important in many fields,

such as engineering, architecture and astronomy. In this chapter you will study quadratic equations in detail, and look at the relationship between quadratic equations and the graphs of quadratic functions (the parabola). You will study the axis of symmetry and maximum and minimum values of the quadratic function. You will also look at the quadratic formula in detail, and at the relationships between the roots (solutions) of quadratic equations, the formula and the quadratic function.

DID YOU KNOW? Thousands of clay tablets from ancient Babylonia have been discovered by archaeologists. These tablets are from as far back as 2000 BC. They show that the Babylonians had mastered many mathematical skills. Geometry, including Pythagoras’ theorem, was well developed, and geometric problems were often worked out by using algebra. Quadratic equations were used in solving geometry problems. The word ‘quadratic’ comes from the Latin ‘quadratum’, meaning ‘four-sided figure’. Completing the square and the quadratic formula were both used to solve quadratic equations. The Babylonians also had some interesting approximations for square roots. For example, 17 2 = . An approximation for 2 that is very accurate was found on a tablet dating back to 1600 BC: 12 51 10 24 2 =1+ + + = 1.414213 60 60 2 60 3

Graph of a Quadratic Function Axis of symmetry EXAMPLE (a ) Sketch the parabola y = x 2 - 4x on the number plane. (b) Find the equation of the axis of symmetry of the parabola. (c ) Find the minimum value of the parabola.

Solution (a) For the y-intercept, x = 0 i.e. y = 0 2 - 4 (0) =0 For the x-intercept, y = 0

The axis of symmetry lies halfway between x = 0 and x = 4.

i.e. 0 = x 2 - 4x = x (x - 4) ` x = 0 or x - 4 = 0 x=4 CONTINUED

544


(b) The axis of symmetry has equation x = 2. (c) Since the parabola is symmetrical about the line x = 2, the minimum value is on this line. Substitute x = 2 into the equation of the parabola i.e. y = 2 2 - 4 (2) = -4 So the minimum value is - 4.

Class Investigation 1. How would you find the axis of symmetry for a graph with no x-intercepts?

2. How would you find the axis of symmetry of a graph where the x-intercepts are irrational numbers?

The axis of symmetry of the quadratic function y = ax 2 + bx + c has the equation b x=2a


Proof The axis of symmetry lies midway between the x-intercepts. For the x-intercepts, y = 0 i.e. ax 2 + bx + c = 0 x=

- b ! b 2 - 4ac 2a

The x-coordinate of the axis of symmetry is the average of the x-intercepts.

i.e.

- b - b 2 - 4ac - b + b 2 - 4ac + 2a 2a x= 2 - 2b 2a = 2 - 2b = 4a b =2a

The parabola has a minimum value if a 2 0. The shape of the parabola is concave upwards.

Minimum value

The parabola has a maximum value if a 1 0. The shape of the parabola is concave downwards. Maximum value

545

546


The minimum or maximum value is f c -

b m 2a

EXAMPLES 1. Find the equation of the axis of symmetry and the minimum value of the quadratic function y = x 2 - 5x + 1.

Solution The equation of the axis of symmetry is given by b 2a (- 5 ) x=i.e. 2 ( 1) 5 = 2 1 ` Equation is x = 2 2 5 2 5 Minimum value: y = c m - 5 c m + 1 2 2 25 25 = +1 4 2 1 = -5 4 x=-

a 2 0 gives a minimum value.

1 So minimum value is - 5 . 4 2. Find the equation of the axis of symmetry and the maximum value of the quadratic function y = - 3x 2 + x - 5.

Solution The equation of the axis of symmetry is given by b 2a 1 x=i.e. 2 (- 3 ) 1 = 6 1 ` Equation is x = 6 1 2 1 Maximum value: y = - 3 c m + c m - 5 6 6 1 1 =+ -5 12 6 11 = -4 12 x=-

a 1 0 gives a maximum value.

So maximum value is - 4

11 . 12


Class Investigation Examine the graph of y = - 3x 2 + x - 5 from the above example. Are there any solutions for the quadratic equation - 3x 2 + x - 5 = 0?

The minimum or maximum point of the parabola is called the vertex.

EXAMPLE (a) Find the equation of the axis of symmetry and the coordinates of the vertex of the parabola y = 2x 2 - 12x + 7. (b) Find the y-intercept and sketch the graph.

Solution (a) Axis of symmetry: b x=2a - 12 =2#2 =3

y

7

When x = 3

y = 2 ] 3 g 2 - 12 ] 3 g + 7 = - 11

3

x

So the vertex is (3, -11) . (b) For y-intercept, x = 0 y = 2 ] 0 g 2 - 12 ] 0 g + 7 =7

-11

(3, -11)

The vertex is the minimum point of the parabola since a 2 0.

10.1 Exercises 1.

By finding the intercepts on the axes, sketch the parabola y = x 2 + 2x. Find the equation of its axis of symmetry, and the minimum value.

2.

Find the equation of the axis of symmetry and the minimum value of the parabola y = 2x 2 + 6x - 3.

3.

Find the equation of the axis of symmetry and the minimum value of the parabola y = x 2 + 3x + 2.

4.

Find the equation of the axis of symmetry and the minimum value of the parabola y = x 2 - 4.

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5.

Find the equation of the axis of symmetry and the minimum point of the parabola y = 4x 2 - 3x + 1.

6.

Find the equation of the axis of symmetry and the maximum value of the parabola y = - x 2 + 2x - 7.

7.

Find the equation of the axis of symmetry and the maximum point of the parabola y = - 2x 2 - 4x + 5.

8.

Find the minimum value of y = x 2 + 4x + 3. How many solutions does the equation x 2 + 4x + 3 = 0 have?

9.

Find the minimum value of y = x 2 + x + 4. How many solutions does the equation x 2 + x + 4 = 0 have?

10. Find the minimum value of y = x 2 + 4x + 4. How many solutions does the equation x 2 + 4x + 4 = 0 have? 11. Find the equation of the axis of symmetry and the coordinates of the vertex for each parabola. (a) y = x 2 + 6x - 3 (b) y = - x 2 - 8x + 1 (c) y = - 2x 2 + 5x (d) y = 4x 2 + 10x - 7 (e) y = 3x 2 + 18x + 4 12. Find (i) the equation of the axis of symmetry (ii) the minimum or maximum value and (iii) the vertex of the parabola. (a) y = x 2 + 2x - 2 (b) y = - 2x 2 + 4x - 1 13. Find the maximum or minimum point for each function. (a) y = x 2 + 2x + 1 (b) y = x 2 - 8x - 7

(c) (d) (e) (f) (g) (h) (i) (j)

f ] x g = x 2 + 4x - 3 y = x 2 - 2x f ] x g = x 2 - 4x - 7 f ] x g = 2x 2 + x - 3 y = - x 2 - 2x + 5 y = - 2x 2 + 8x + 3 f ] x g = - 3x 2 + 3x + 7 f ] x g = - x 2 + 2x - 4

14. For each quadratic function (i) find any x-intercepts using the quadratic formula. (ii) state whether the function has a maximum or minimum value and find this value. (iii) sketch the function on a number plane. (a) f ] x g = x 2 + 4x + 4 (b) f ] x g = x 2 - 2x - 3 (c) y = x 2 - 6x + 1 (d) f ] x g = x 2 + 2x (e) y = 2x 2 - 18 (f) y = 3x 2 + x - 2 (g) f ] x g = - x 2 - 2x + 6 (h) f ] x g = - x 2 - x + 3 (i) y = - x 2 - 3x + 2 (j) y = - 2x 2 + 4x + 5 15. (a) Find the minimum value of the parabola y = x 2 - 2x + 5. (b) How many solutions does the quadratic equation x 2 - 2x + 5 = 0 have? (c) Sketch the parabola. 16. (a) How many x-intercepts has the quadratic function f ] x g = x 2 - 3x + 9 ? (b) Find the minimum point of the function. (c) Sketch the function. 17. (a) Find the maximum value of the quadratic function f ] x g = - 2x 2 + x - 4 . (b) How many solutions has the quadratic equation - 2x 2 + x - 4 = 0 ? (c) Sketch the graph of the quadratic function.


18. (a) Sketch the parabola y = x 2 - 5x + 6. (b) From the graph, find values of x for which x 2 - 5x + 6 2 0. (c) Find the domain over which x 2 - 5x + 6 # 0 . 19. Sketch y = 3x 2 - 2x + 4 and hence show that 3x 2 - 2x + 4 2 0 for all x.

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20. By sketching f ] x g = x 2 + x + 2, show that x 2 + x + 2 2 0 for all x. 21. Show by a sketch that - x 2 + 2x - 7 1 0 for all x. 22. Sketch y = - 5x 2 + 4x - 1 and show that -5x 2 + 4x - 1 1 0 for all x.

Investigation Could you tell without sketching the function y = x 2 - x + 5 if x 2 - x + 5 2 0 for all x? How could you do this? How could you know that - x 2 + 2x - 7 1 0 for all x without sketching the graph of f ] x g = - x 2 + 2x - 7?

Quadratic Inequalities You looked at solving quadratic inequations in Chapter 3 using the number line. You can also solve them using the graph of a parabola.

For any curve on a number plane y = 0 is on the x-axis (all values of y are zero on the x-axis) y 2 0 is above the x-axis (all positive values of y lie above the x-axis) y 1 0 is below the x-axis (all negative values of y lie below the x-axis)

Substituting ax 2 + bx + c for y in the general parabola y = ax 2 + bx + c gives the following results:

For the parabola y = ax 2 + bx + c ax 2 + bx + c = 0 on the x-axis ax 2 + bx + c 2 0 above the x-axis ax 2 + bx + c 1 0 below the x-axis

You will look at this later on in the chapter.

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y a20

ax2 + bx + c 2 0

x

ax 2 + bx + c = 0

ax 2 + bx + c 1 0

y a10

ax 2 + bx + c 2 0

x

ax 2 + bx + c = 0

ax 2 + bx + c 1 0

EXAMPLES 1. Solve x 2 - 3x + 2 $ 0.

Solution First sketch y = x 2 - 3x + 2 showing x-intercepts (a 2 0 so it is concave upwards). For x-intercepts, y = 0 0 = x 2 - 3x + 2 = ]x - 2g]x - 1g x - 2 = 0, x - 1 = 0 x = 2, x=1


y

1

2

x

y $ 0 on and above the x-axis So x 2 - 3x + 2 $ 0 on and above the x-axis. ` x # 1, x $ 2 2. Solve 4x - x 2 2 0.

Solution First sketch y = 4x - x 2 showing x-intercepts (a 1 0 so it is concave downwards). For x-intercepts, y = 0 0 = 4x - x 2 = x ]4 - x g x = 0, 4-x=0 x = 0, 4=x y

0

4

x

y 2 0 above the x-axis So 4x - x 2 2 0 above the x-axis. ` 0 1 x 1 4.

CONTINUED

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3. Solve x 2 - 25 1 0.

Solution First sketch y = x 2 - 25 showing x-intercepts (a 2 0 so it is concave upwards). For x-intercepts, y = 0 0 = x 2 - 25 = ]x + 5 g]x - 5 g x + 5 = 0, x - 5 = 0 x = - 5, x=5 y

-5

5

x

y 1 0 below the x-axis So x 2 - 25 1 0 below the x-axis. ` -5 1 x 1 5

Further inequations You learned how to solve inequations involving the pronumeral in the denominator by using the number line in Chapter 3. Here we use quadratic inequalities to solve them.


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EXAMPLES 1. Solve

1 $ 2. x+1

Solution x ! -1 We don’t know whether x + 1 is positive or negative, but ] x + 1 g 2 is always positive. We can multiply both sides of the inequation by ] x + 1 g 2 without changing the inequality sign. 1 $2 x+1

1 # ]x + 1 g2 $ 2 # ]x + 1 g 2 x+1 x + 1 $ 2 ]x + 1 g2 0 $ 2 ]x + 1 g2 - ]x + 1 g $ ]x + 1 g62 ]x + 1 g - 1 @ $ ] x + 1 g ] 2x + 2 - 1 g $ ] x + 1 g ] 2x + 1 g

Factorise by taking out x + 1 as a common factor.

We solve this by sketching the parabola y = ] x + 1 g ] 2x + 1 g. For x-intercepts: y = 0 0 = ] x + 1 g ] 2x + 1 g x + 1 = 0, 2x + 1 = 0 x = - 1, 2x = - 1 1 x=2

y

-1

-

x

1 2

0 $ ] x + 1 g ] 2x + 1 g on and below the x-axis. However, x ! -1 1 The solution is - 1 1 x # - . 2

2. Solve

4x 1 5. x-2

Solution x!2 We multiply both sides of the inequation by ] x - 2 g 2. CONTINUED

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4x 15 x-2

4x # ] x - 2 g 2 1 5# ] x - 2 g 2 x-2 4x ] x - 2 g 1 5 ] x - 2 g 2 0 1 5 ] x - 2 g 2 - 4x ] x - 2 g 1 ] x - 2 g 6 5 ] x - 2 g - 4x @ 1 ] x - 2 g ] 5x - 10 - 4x g 1 ] x - 2 g ] x - 10 g

Factorise by taking out x - 2 as a common factor.

We solve this by sketching the parabola y = ] x - 2 g ] x - 10 g . For x-intercepts: y = 0 0 = ] x - 2 g ] x - 10 g x - 2 = 0, x - 10 = 0 x = 2, x = 10 y

2

x

10

0 1 ] x - 2 g ] x - 10 g above the x-axis. The solution is x 1 2, x 2 10.


x2 - 9 2 0

6.

2t - t 2 2 0

2.

n2 + n # 0

7.

x 2 + 2x - 8 2 0

3.

a 2 - 2a $ 0

8.

p 2 + 4p + 3 $ 0

4.

4 - x2 1 0

9.

m 2 - 6m + 8 2 0

5.

y 2 - 6y # 0

10. 6 - x - x 2 # 0


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1 23. x $ 1

11. 2h 2 - 7h + 6 1 0 12. x 2 - x - 20 # 0

1 24. - x $ 2

13. 35 + 9k - 2k 2 $ 0

25.

1 23 x -1

26.

1 #1 x+2

27.

2 $5 x-2

28.

3 2 -1 x+3

20. ] x - 3 g ] x + 1 g $ 5

29.

-1 #3 x-1

1 21. x 1 - 2

30.

14. q 2 - 9q + 18 2 0 15. ] x + 2 g2 $ 0 16. 12 - n - n 2 # 0 17. x 2 - 2x 1 15 18. - t 2 $ 4t - 12 19. 3y 2 2 14y + 5

Solve the inequations in Chapter 3 using these methods for extra practice.

x $4 x+2

1 22. x 2 3

The Discriminant The values of x that satisfy a quadratic equation are called the roots of the equation. The roots of ax 2 + bx + c = 0 are the x-intercepts of the graph y = ax 2 + bx + c

1. If y = ax 2 + bx + c has 2 x-intercepts, then the quadratic equation ax 2 + bx + c = 0 has 2 real roots. y

y

a20

a10 x

Since the graph can be both positive and negative, it is called an indefinite function.

x

556


2. If y = ax 2 + bx + c has 1 x-intercept, then the quadratic equation ax 2 + bx + c = 0 has 1 real root y

y

a20

a10 x

x

3. If y = ax 2 + bx + c has no x-intercepts, then the quadratic equation ax 2 + bx + c = 0 has no real roots y

y

a20

a10 x

x

Since this graph is always positive, it is called a positive definite function.

Since this graph is always negative, it is called a negative definite function.

This information can be found without sketching the graph.

Investigation 1. Solve the following quadratic equations using the quadratic formula (a) x 2 - 3x + 2 = 0 (b) x 2 + 4x - 7 = 0 (c) x 2 + x + 5 = 0 (d) x 2 - 6x + 9 = 0 2. Without solving a quadratic equation, can you predict how many roots it has by looking at the quadratic formula?


b 2 - 4ac , the expression b 2 - 4ac is called 2a the discriminant. It gives us information about the roots of the quadratic equation ax 2 + bx + x = 0. In the quadratic formula x =

-b !

EXAMPLES Use the quadratic formula to find how many real roots each quadratic equation has. 1. x 2 + 5x - 3 = 0

Solution - b ! b 2 - 4ac 2a - 5 ! 5 2 - 4 #1# - 3 = 2 #1 - 5 ! 25 + 12 = 2 - 5 ! 37 = 2 There are 2 real roots: x=

x=

- 5 + 37 - 5 - 37 , 2 2

2. x 2 - x + 4 = 0

Solution - b ! b 2 - 4ac 2a - (-1) ! (-1) 2 - 4 #1# 4 = 2 #1 1 ! -15 = 2

x=

There are no real roots since

-15 has no real value.

3. x 2 - 2x + 1 = 0

Solution - b ! b 2 - 4ac 2a - (- 2) ! (- 2) 2 - 4 #1#1 = 2 #1 2! 0 = 2

x=

CONTINUED

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There are 2 real roots: 2+ 0 2- 0 , 2 2 = 1, 1

x=

However, these are equal roots.

Tis the Greek letter 'delta'.

Notice that when there are 2 real roots, the discriminant b 2 - 4ac 2 0. When there are 2 equal roots (or just 1 real root), b 2 - 4ac = 0. When there are no real roots, b 2 - 4ac 1 0. We often use D = b 2 - 4ac.

1. If T2 0, then the quadratic equation ax 2 + bx + c = 0 has 2 real unequal (different) roots. y

y

a20

a10 x

x

If T is a perfect square, the roots are rational. If T is not a perfect square, the roots are irrational. 2. If T = 0, then the quadratic equation ax 2 + bx + c = 0 has 1 real root or 2 equal roots. y

y

a20

a10 x

x


3. If T1 0, then the quadratic equation ax 2 + bx + c = 0 has no real roots. y

y

a10

a20 x

If T1 0 and a 2 0, it is positive definite and ax 2 + bx + c 2 0 for all x.

x

If T1 0 and a 1 0, it is negative definite and ax 2 + bx + c 1 0 for all x.

We can examine the roots of the quadratic equation by using the discriminant rather than the whole quadratic formula.

EXAMPLES 1. Show that the equation 2x 2 + x + 4 = 0 has no real roots.

Solution T = b 2 - 4ac = 12 - 4 ] 2 g ] 4 g = 1 - 32 = - 31 10 So the equation has no real roots. 2. Find the values of k for which the quadratic equation 5x 2 - 2x + k = 0 has real roots.

Solution For real unequal roots, T 2 0. For real equal roots, T = 0. So for real roots, T $ 0. CONTINUED

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560


T $0 b - 4ac $ 0 2

] -2 g 2 - 4 ] 5 g ] k g $ 0

4 - 20k $ 0 4 $ 20k 1 $k 5 3. Show that x 2 - 2x + 4 2 0 for all x.

Solution If a 2 0 and T1 0, then ax 2 + bx + c 2 0 for all x. y

a20 x

a =1 20 T = b 2 - 4ac = ] -2 g 2 - 4 ] 1 g ] 4 g = 4 - 16 = -12 10 Since a 2 0 and T1 0, x 2 - 2x + 4 2 0 for all x. 4. Show that the line 4x + y + 4 = 0 is a tangent to the parabola y = x 2.

Solution For the line to be a tangent, it must intersect with the curve in only 1 point.


y 5 4 3 2 1 x -4 -3 -2

-11

1

2

3

4

5

-2 -3 -4 -5 -6

It is too hard to tell from the graph if the line is a tangent, so we solve simultaneous equations to find any points of intersection. y = x2 4x + y + 4 = 0

]1 g ]2 g

Substitute (1) into (2): 4x + x 2 + 4 = 0 x 2 + 4x + 4 = 0 We don’t need to find the roots of the equation as the question only asks how many roots there are. We find the discriminant. D = b 2 - 4ac = 42 - 4 ] 1 g ] 4 g = 16 - 16 =0 ` the equation has 1 real root (equal roots) so there is only one point of intersection. So the line is a tangent to the parabola.

10.3 Exercises 1.

Find the discriminant of each quadratic equation. (a) x 2 - 4x - 1 = 0 (b) 2x 2 + 3x + 7 = 0 (c) - 4x 2 + 2x - 1 = 0 (d) 6x 2 - x - 2 = 0 (e) - x 2 - 3x = 0 (f) x 2 + 4 = 0 (g) x 2 - 2x + 1 = 0 (h) -3x 2 - 2x + 5 = 0

(i) - 2x 2 + x + 2 = 0 (j) - x 2 + 4x - 4 = 0 2.

Find the discriminant and state whether the roots of the quadratic equation are real or imaginary (not real), and if they are real, whether they are equal or unequal, rational or irrational.

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(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

x2 - x - 4 = 0 2x 2 + 3x + 6 = 0 x 2 - 9x + 20 = 0 x 2 + 6x + 9 = 0 2x 2 - 5x - 1 = 0 - x 2 + 2x - 5 = 0 - 2x 2 - 5x + 3 = 0 - 5x 2 + 2x - 6 = 0 -x2 + x = 0 - 2x 2 + 8x - 2 = 0

3.

Find the value of p for which the quadratic equation x 2 + 2x + p = 0 has equal roots.

4.

Find any values of k for which the quadratic equation x 2 + kx + 1 = 0 has equal roots.

5.

Find all the values of b for which 2x 2 + x + b + 1 = 0 has real roots.

6.

Evaluate p if px 2 + 4x + 2 = 0 has no real roots.

7.

Find all values of k for which ] k + 2 g x 2 + x - 3 = 0 has 2 real unequal roots.

8.

9.

Prove that 3x 2 - x + 7 2 0 for all real x. Find the values of k for which x 2 + ] k + 1 g x + 4 = 0 has real roots.

10. Find values of k for which the expression kx 2 + 3kx + 9 is positive definite. 11. Find the values of m for which the quadratic equation x 2 - 2mx + 9 = 0 has real and different roots.

12. If x 2 - 2kx + 1 = 0 has real roots, evaluate k. 13. Find exact values of p if px 2 - 2x + 3p = 0 is negative definite. 14. Evaluate b if ] b - 2 g x 2 - 2bx + 5b = 0 has real roots. 15. Find values of p for which the quadratic equation x 2 + px + p + 3 = 0 has real roots. 16. Show that the line y = 2x + 6 cuts the parabola y = x 2 + 3 in 2 points. 17. Show that the line 3x + y - 4 = 0 cuts the parabola y = x 2 + 5x + 3 in 2 points. 18. Show that the line y = - x - 4 does not touch the parabola y = x 2. 19. Show that the line y = 5x - 2 is a tangent to the parabola y = x 2 + 3x - 1 . 20. The line y = 3x - p + 1 is a tangent to the parabola y = x 2. Evaluate p. 21. Which of these lines is a tangent to the circle x 2 + y 2 = 4? (a) 3x - y - 1 = 0 (b) 5x + y - 3 = 0 (c) 4x + 3y - 10 = 0 (d) 5x - 12y + 26 = 0 (e) 2x + y - 7 = 0

Quadratic Identities When you use the quadratic formula to solve an equation, you compare a quadratic, say, 3x 2 - 2x + 5 = 0 with the general quadratic ax 2 + bx + c = 0.


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You are assuming when you do this that 3x 2 - 2x + 5 and ax 2 + bx + c are equivalent expressions. We can state this as a general rule: If two quadratic expressions are equivalent to each other then the corresponding coefficients must be equal.

If a 1 x 2 + b 1 x + c 1 / a 2 x 2 + b 2 x + c 2 for all real x then a 1 = a 2, b 1 = b 2 and c 1 = c 2

Proof If a 1 x 2 + b 1 x + c 1 = a 2 x 2 + b 2 x + c 2 for more than two values of x, then (a 1 - a 2) x 2 + (b 1 - b 2) x + (c 1 - c 2) = 0. That is, a 1 = a 2, b 1 = b 2 and c 1 = c 2 .

EXAMPLES 1. Write 2x 2 - 3x + 5 in the form A (x - 1)2 + B (x - 1) + C.

Solution A ] x - 1 g2 + B (x - 1) + C = A (x 2 - 2x + 1) + Bx - B + C = Ax 2 - 2Ax + A + Bx - B + C = Ax 2 + (- 2A + B) x + A - B + C For 2x 2 - 3x + 5 / Ax 2 + (- 2A + B) x + A - B + C A=2 - 2A + B = - 3 A-B+C=5 Substitute (1) into (2): - 2 ( 2) + B = - 3 - 4 + B = -3 B=1 Substitute A = 2 and B = 1 into (3): 2-1+C=5 1+C=5 C=4

( 1) (2) ( 3)

` 2x 2 - 3x + 5 / 2 (x - 1) 2 + (x - 1) + 4

CONTINUED

You learnt how to solve simultaneous equations with 3 unknowns in Chapter 3.

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2. Find values for a, b and c if x 2 - x / a (x + 3)2 + bx + c - 1.

Solution a ] x + 3 g2 + bx + c - 1 = a (x 2 + 6x + 9) + bx + c - 1 = ax 2 + 6ax + 9a + bx + c - 1 = ax 2 + (6a + b) x + 9a + c - 1 2 For x - x / ax 2 + (6a + b) x + 9a + c - 1 a =1 6a + b = -1 9a + c - 1 = 0 Substitute (1) into (2): 6 (1) + b = -1 6 + b = -1 b = -7 Substitute (1) into (3): 9 (1) + c - 1 = 0 8+c=0 c = -8 ` a = 1, b = -7, c = - 8

( 1) ( 2) ( 3)

3. Find the equation of the parabola that passes through the points (-1, -3), (0, 3) and (2, 21) .

Solution The parabola has equation in the form y = ax 2 + bx + c. Substitute the points into the equation: ^ -1, -3 h: -3 = a ] -1 g 2 + b ] -1 g + c =a-b+c `

a - b + c = -3 ^ 0, 3 h : 3 = a ]0 g2 + b ]0 g + c =c `

^ 2, 21 h:

c =3

21 = a ] 2 g 2 + b ] 2 g + c = 4a + 2b + c

` 4a + 2b + c = 21 Solve simultaneous equations to find a, b and c. Substitute (2) into (1): a - b + 3 = -3 a - b = -6

]1 g

]2 g

]3 g

( 4)


Substitute (2) into (3): 4a + 2b + 3 = 21 4a + 2b = 18 (4) # 2: 2a - 2b = -12 (6) + (5): 2a - 2b = -12 4a + 2b = 18 6a =6 a=1 Substitute a = 1 into (5): 4 (1) + 2b = 18 4 + 2b = 18 2b = 14 b=7 ` a = 1, b = 7, c = 3 Thus the parabola has equation y = x 2 + 7x + 3.

(5 ) (6 )

10.4 Exercises 1.

Find values of a, b and c for which (a) x 2 + 4x - 3 / a ]x + 1 g 2 + b ]x + 1 g + c (b) 2x 2 - 3x + 1 / a ]x + 2 g2 + b ]x + 2 g + c (c) x 2 - x - 2 / a ]x - 1 g2 + b ]x - 1 g + c (d) x 2 + x + 6 / a ]x - 3 g2 + b ]x - 3 g + c (e) 3x 2 - 5x - 2 / a ]x + 1 g2 + b ]x - 1 g + c (f) 4x 2 + x - 7 / a ]x - 2 g2 + b ]x - 2 g + c (g) 2x 2 + 4x - 1 / a ]x + 4 g2 + b ]x + 2 g + c (h) 3x 2 - 2x + 5 / a ] x + 1 g 2 + bx + c (i) - x 2 + 4x - 3 / a ]x + 3 g2 + b ]x + 3 g + c (j) - 2x 2 + 4x - 3 / a ]x - 1 g2 + b ]x + 1 g + c

2.

Find values of m, p and q for which 2x 2 - x - 1 / m ] x + 1 g 2 + p ] x + 1 g + q.

3.

Express x 2 - 4x + 5 in the form Ax ] x - 2 g + B ] x + 1 g + C + 4.

4.

Show that x 2 + 2x + 9 can be written in the form a ]x - 2g]x + 3g + b ]x - 2g + c where a = 1, b = 1 and c = 17.

5.

Find values of A, B and C if x 2 + x - 2 / A ] x - 2 g 2 + Bx + C.

6.

Find values of a, b and c for which 3x 2 + 5x - 1 / ax ] x + 3 g + bx 2 + c ] x + 1 g .

7.

Evaluate K, L and M if x 2 / K ] x - 3 g 2 + L ] x + 1 g - 2M.

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8.

9.

10. Find the equation of the parabola that passes through the points (a) (0, -5), (2, -3) and (-3, 7) Find the values of a, b and c if (b) (1, -2), (3, 0) and (-2, 10) 20x - 17 / a ] x - 4 g 2 - b ] 5x + 1 g + c. (c) (-2, 21), (1, 6) and (-1, 12) (d) (2, 3), (1, -4) and (-1, -12) (e) (0, 1), (-2, 1) and (2, -7)

Express 4x 2 + 2 in the form a ] x + 5 g + b ] 2x - 3 g 2 + c - 2 .

Sum and Product of Roots When you solve a quadratic equation, you may notice a relationship between the roots. You also used this to factorise trinomials in Chapter 2.

EXAMPLE (a) Solve x 2 - 9x + 20 = 0. (b) Find the sum of the roots. (c) Find the product of the roots.

Solution (a)

Notice -9 is the coefficient of x and 20 is the constant term in the equation.

x 2 - 9x + 20 = 0 (x - 4) (x - 5) = 0 x - 4 = 0, x - 5 = 0 ` x = 4, x=5

(b) Sum = 4 + 5 =9 (c) Product = 4 # 5 = 20

This relationship with the sum and product of the roots works for any quadratic equation.

The general quadratic equation can be written in the form x 2 - (a + b) x + ab = 0 where a and b are the roots of the equation.

Proof Suppose the general quadratic equation ax 2 + bx + c = 0 has roots a and b. Then this equation can be written in the form


567

(x - a ) (x - b) = 0 i.e. x - bx - a x + ab = 0 x 2 - (a + b) x + ab = 0 2

EXAMPLES 1. Find the quadratic equation that has roots 6 and - 1.

Solution Method 1: Using the general formula x 2 - (a + b) x + ab = 0 where a = 6 and b = -1 a + b = 6 + -1 =5 ab = 6 # -1 = -6 Substituting into x 2 - (a + b ) x + ab = 0 gives x 2 - 5x - 6 = 0 Method 2: If 6 and -1 are the roots of the equation then it can be written as ]x - 6 g]x + 1 g = 0 x 2 + x - 6x - 6 = 0 x 2 - 5x - 6 = 0 2. Find the quadratic equation that has roots 3 + 2 and 3 - 2 .

Solution Method 1: Using the general formula a+b=3+ 2+3- 2 =6 ab = (3 + 2 ) # (3 - 2 ) = 3 2 - ( 2 )2 =9-2 =7 Substituting into x 2 - (a + b ) x + ab = 0 gives x 2 - 6x + 7 = 0 Method 2: If 3 + 2 and 3 - 2 are the roots of the equation then it can be written as _x - "3 + 2 ,i_x - "3 - 2 ,i = 0 ^x - 3 - 2 h^x - 3 + 2 h = 0 x 2 - 3x + 2 x - 3x + 9 - 3 2 - 2 x + 3 2 - 2 = 0 x 2 - 6x + 7 = 0

It doesn’t matter which way around we name these roots.

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We can find a more general relationship between the sum and product of roots of a quadratic equation.

If a and b are the roots of the quadratic equation ax 2 + bx + c = 0: b Sum of roots: a + b = - a c Product of roots: ab = a

Proof If an equation has roots a and b, it can be written as x 2 - (a + b )x + ab = 0. But we know that a and b are the roots of the quadratic equation ax 2 + bx + c = 0. Using quadratic identities, we can compare the two forms of the equation. ax 2 + bx + c = 0 ax 2 bx c 0 a + a +a=a c b x2 + a x + a = 0

c b For x 2 - (a + b ) x + ab / x 2 + a x + a b - (a + b ) = a b ` a + b = -a c Also ab = a

EXAMPLES 1. Find (a) a + b (b) ab (c) a2 + b2 if a and b are the roots of 2x 2 - 6x + 1 = 0.

Solution b (a) a + b = - a ==3

(-6) 2


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c (b) ab = a 1 = 2 (c)

a 2 + b 2 ! (a + b ) 2

^ a + b h2 = a 2 + 2ab + b2 ^ a + b h2 - 2ab = a 2 + b2 ] 3 g 2 - 2 c 1 m = a2 + b2 2 9 - 1 = a 2 + b2

8 = a 2 + b2 2. Find the value of k if one root of kx 2 -7x + k + 1 = 0 is - 2.

Solution If - 2 is a root of the equation then x = -2 satisfies the equation. Substitute x = - 2 into the equation: k ] -2 g 2 - 7 ] -2 g + k + 1 = 0 4k + 14 + k + 1 = 0 5k + 15 = 0 5k = - 15 k = -3 3. Evaluate p if one root of x 2 + 2x - 5p = 0 is double the other root. You could use b and 2b instead.

Solution If one root is a then the other root is 2a . Sum of roots: b a + b = -a 2 1 3a = - 2 2 a=3

a + 2a = -

CONTINUED

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Product of roots: c ab = a - 5p a # 2a = 1 2 2a = - 5p 2 2 2 c - m = - 5p 3 4 2 c m = - 5p 9 8 = - 5p 9 8 =p 45

10.5 Exercises 1.

2.

Find a + b and ab if a and b are the roots of (a) x 2 + 2x + 1 = 0 (b) 2x 2 - 3x - 6 = 0 (c) 5x 2 - x - 9 = 0 (d) x 2 + 7x + 1 = 0 (e) 3y 2 - 8y + 3 = 0 If a and b are the roots of the quadratic equation x 2 - 3x - 6 = 0, find the value of (a) a + b (b) ab 1 1 + (c) a b

4.

Find the value of m in x 2 + 2mx - 6 = 0 if one of the roots is 2.

5.

If one of the roots of the quadratic equation 2x 2 - 5x + k - 1 = 0 is - 3, find the value of k.

6.

One root of 3x 2 - 2 (3b + 1) x + 4b = 0 is 8. Find the value of b.

7.

In the quadratic equation 2x 2 - 3x + k = 0, one root is double the other. Find the value of k.

8.

In the quadratic equation x 2 - 8x + p - 1 = 0, one root is triple the other. Find the value of p.

9.

In the quadratic equation (k - 2) x 2 + 50x + 2k + 3 = 0, the roots are reciprocals of each other. Find the value of k.

(d) a2 + b2 3.

Reciprocals are n and

1 . n

Find the quadratic equation whose roots are (a) 2 and - 5 (b) - 3 and 7 (c) - 1 and - 4 (d) 4 + 5 and 4 - 5 (e) 1 + 2 7 and 1 - 2 7


10. In the quadratic equation x 2 + mx + 2 = 0, the roots are consecutive. Find the values of m. 11. In the quadratic equation - 3x 2 - (k + 1) x + 5 = 0, the roots are equal in magnitude but opposite in sign. Find the value of k. 12. Find values of n in the equation 2x 2 - 5(n - 1) x + 12 = 0 if the two roots are consecutive. 13. If the sum of the roots of x 2 + px + r = 0 is - 2 and the product of roots is - 7, find the values of p and r. 14. One root of the quadratic equation x 2 + bx + c = 0 is 4 and the product of the roots is 8. Find the values of b and c. 15. The roots of the quadratic equation x 2 + 4x - a = 0 are b + 1 and b - 3. Find the values of a and b. 16. Show that the roots of the quadratic equation 3mx 2 + 2x + 3m = 0 are always reciprocals of one another.

17. Find values of k in the equation k +1 m = 0 if: x 2 + (k + 1 ) x + c 4 (a) roots are equal in magnitude but opposite in sign (b) roots are equal (c) one root is 1 (d) roots are reciprocals of one another (e) roots are real. 18. Find exact values of p in the equation x 2 + px + 3 = 0 if (a) the roots are equal (b) it has real roots (c) one root is double the other. 19. Find values of k in the equation x 2 + kx + k - 1 = 0 if (a) the roots are equal (b) one root is 4 (c) the roots are reciprocals of one another. 20. Find values of m in the equation mx 2 + x + m - 3 = 0 if (a) one root is -2 (b) it has no real roots (c) the product of the roots is 2.

Equations Reducible to Quadratics To solve a quadratic equation such as ] x - 3 g 2 - ] x - 3 g - 2 = 0, you could expand the brackets and then solve the equation. However, in this section you will learn a different way to solve this. There are other equations that do not look like quadratic equations that can also be solved this way.

Consecutive numbers are numbers that follow each other in order, such as 3 and 4.

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EXAMPLES 1. Solve ] x + 2 g 2 - 3 ] x + 2 g - 4 = 0.

Solution u=x+2

Let

Then u - 3u - 4 ]u - 4 g]u + 1 g u - 4 = 0, u + 1 u = 4, u 2

But

=0 =0 =0 = -1

u=x+2

So x + 2 = 4, x = 2,

x + 2 = -1 x = -3

2 2. Solve x + x = 3 where x ! 0.

Solution 2 x+ x =3 2 x#x+x#x =3#x x 2 + 2 = 3x x 2 - 3x + 2 = 0 ]x - 2 g]x - 1 g = 0 x - 2 = 0, x - 1 = 0 x = 2, x=1 3. Solve 9 x - 4.3 x + 3 = 0.

Solution x 2 9x = ^ 32 h = ^ 3x h So 9 x - 4.3 x + 3 = 0 can be written as ^ 3 x h2 - 4.3 x + 3 = 0 Let k = 3x

k 2 - 4k + 3 = 0 ]k - 3 g]k - 1 g = 0 k - 3 = 0, k - 1 = 0 k = 1, k=3 But k = 3 x So 3 x = 1, x = 0,

3x = 3 x=1


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4. Solve 2 sin 2 x + sin x - 1 = 0 for 0c # x # 360c.

Solution sin x = u

Let

Then 2u + u - 1 = 0 ] 2u - 1 g ] u + 1 g = 0 2u - 1 = 0 or u + 1 = 0 u = -1 2u = 1 2

u=

1 2

But u = sin x 1 So sin x = or sin x = -1 2 1 sin x = has solutions in the 1st and 2nd quadrants 2 1 sin 30c = 2 So x = 30c, 180c - 30c = 30c, 150c

30c 3

2

600c

1

For sin x = - 1, we use the graph of y = sin x y

1

90c

180c

270c

360c

x

-1

From the graph: x = 270c So solutions to 2 sin 2 x + sin x - 1 = 0 are x = 30c, 150c, 270c CONTINUED

See Chapter 6 if you have forgotten how to solve a trigonometric equation.

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10.6 Exercises 1.

2.

3.

4.

Solve (a) ] x - 1 g 2 + 7 ] x - 1 g + 10 = 0 (b) ^ y - 3 h 2 - ^ y - 3 h - 2 = 0 (c) ] x + 2 g 2 - 2 ] x + 2 g - 8 = 0 (d) ] n - 5 g 2 + 7 ] n - 5 g + 6 = 0 (e) ] a - 4 g 2 + 6 ] a - 4 g - 7 = 0 (f) ^ p + 1 h 2 - 9 ^ p + 1 h + 20 = 0 (g) ] x + 3 g 2 - 4 ] x + 3 g - 5 = 0 (h) ] k - 8 g 2 - ] k - 8 g - 12 = 0 (i) ] t - 2 g 2 + 2 ] t - 2 g - 24 = 0 (j) ] b + 9 g 2 - 2 ] b + 9 g - 15 = 0 Solve (x ! 0) . 6 (a) x - x = 1 6 (b) x + x = 5 20 (c) x + x - 9 = 0 15 (d) x + x = 8 12 (e) 2x + x = 11

(c) 5 2x - 5 x - 20 = 0 (d) 9 x + 3 x - 12 = 0 (e) 4 x - 10.2 x + 16 = 0 5.

6.

7.

4 = 5 ( x ! 0) . x2

1 2 1 Solve b x + x l + b x + x l - 2 = 0 (x ! 0) . Solve 1 2 1 n - 9 d x 2 + 2 n + 20 = 0 x2 x correct to 2 decimal places (x ! 0) .

d x2 +

8.

Solve (a) x 4 - 7x 2 - 18 = 0 9. (b) y 4 - 6y 2 + 8 = 0, giving exact values 2 (c) ^ x 2 - x h + ^ x 2 - x h - 2 = 0 giving exact values (d) ^ x 2 + 3x -1 h2 - 7 ^ x 2 + 3x -1 h +10 = 0 correct to 2 decimal places 10. 2 (e) ^ a 2 + 4a h + 2 ^ a 2 + 4a h - 8 = 0 giving exact values. Solve (a) 2 2x - 9.2 x + 8 = 0 (b) 3 2p + 3 p - 12 = 0

Solve x 2 +

Solve for 0c # x # 360c. (a) sin 2 x - sin x = 0 (b) cos 2 x + cos x = 0 (c) 2 sin 2 x - sin x - 1 = 0 (d) 2 cos 2 x = cos x (e) sin x = cos 2 x - 1 Solve for 0c # x # 360c. (a) tan 2 x - tan x = 0 (b) cos 2 x - 1 = 0 (c) 2 sin 2 x - sin x = 0 (d) 8 sin 4 x - 10 sin 2 x + 3 = 0 (e) 3 tan 4 x - 10 tan 2 x + 3 = 0 Show that the equation 2 x+3+ = 5 has 2 real x+3 irrational roots (x ! - 3) .


Test Yourself 10 1.

Solve (a) x 2 - 3x # 0 (b) n 2 - 9 2 0 (c) 4 - y 2 $ 0

2.

Evaluate a, b and c if 2x 2 - 5x + 7 = 2a(x + 1) 2 + b (x + 1) + c.

3.

Find (a) the equation of the axis of symmetry and (b) the minimum value of the parabola y = x 2 - 4x + 1.

4.

Show that y = x 2 - 2x + 7 is a positive definite quadratic function.

5.

If a and b are roots of the quadratic equation x 2 - 6x + 3 = 0, find (a) a + b (b) ab 1 1 + (c) a b (d) ab2 + a 2 b (e) a 2 + b2

9.

Find (a) the equation of the axis of symmetry and (b) the maximum value of the quadratic function y = - 2x 2 - x + 6.

10. Write 3x 2 + 7 in the form a (x - 2) 2 + b (x + 3) + c. 11. Solve 2 sin 2 x + sin x - 1 = 0 for 0c # x # 360c. 12. Find the value of k in x 2 + 3x + k - 1 = 0 if the quadratic equation has (a) equal roots (b) one root - 3 (c) one root double the other (d) consecutive roots (e) reciprocal roots. 3 13. Solve 2x = 5 + x

(x ! 0) .

14. Find values of m such that mx 2 + 3x - 4 1 0 for all x.

6.

Solve (3x - 2) 2 - 2 (3x - 2) - 3 = 0.

15. Solve 5 2x - 26.5 x + 25 = 0.

7.

Describe the roots of each quadratic equation as (i) real, different and rational (ii) real, different and irrational (iii) equal or (iv) unreal. (a) 2x 2 - x + 3 = 0 (b) x 2 - 10x - 25 = 0 (c) x 2 - 10x + 25 = 0 (d) 3x 2 + 7x - 2 = 0 (e) 6x 2 - x - 2 = 0

16. For each set of graphs, state whether they have (i) 2 points (ii) 1 point (iii) no points of intersection. (a) xy = 7 and 3x - 5y - 1 = 0 (b) x 2 + y 2 = 9 and y = 3x - 3 (c) x 2 + y 2 = 1 and x - 2y - 3 = 0 2 (d) y = x and y = 3x + 1 (e) y = x 2 and y = 4x - 4

8.

Show that - 4 + 3x - x 2 1 0 for all x.

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17. State if each quadratic function is (i) indefinite (ii) positive definite or (iii) negative definite. (a)

(d)

18. Show that kx 2 - px + k = 0 has reciprocal roots for all x. (b)

19. Find the quadratic equation that has roots (a) 4 and - 7 (b) 5 + 7 and 5 - 7 20. Solve 2 2x - 10.2 x + 16 = 0. 21. Solve

(c)

(a) (b) (c) (d) (e)

3 17 x+1 2n $1 n-3 2 23 5y - 1 3x #2 2x + 5 2x + 1 $5 x-4


Show that the quadratic equation 2x 2 - kx + k - 2 = 0 has real rational roots.

2.

Find the equation of a quadratic function that passes through the points (- 2, 18), (3, - 2) and (1, 0) .

3.

Find the value of a, b and c if x 2 + 5x - 3 / ax (x + 1) + b (x + 1)2 + cx.

25 = 10. x2 + 1

4.

Solve x 2 + 1 +

5.

Find the maximum value of the function f (x) = - 2x 2 - 4x + 9.

6.

Find the value of n for which the equation (n + 2) x 2 + 3x - 5 = 0 has one root triple the other.


7.

Find the values of p for which x 2 - x + 3p - 2 2 0 for all x.

12. Find exact values of k for which x 2 + 2kx + k + 5 = 0 has real roots.

8.

Show that the quadratic equation x 2 - 2px + p 2 = 0 has equal roots.

13. Solve 3 - 2 cos 2 x - 3 sin x = 0 for 0c # x # 360c.

9.

Solve 2 2x + 1 - 5.2 x + 2 = 0.

1 2 1 14. Solve b x + x l - 5 b x + x l + 6 = 0.

10. Find values of A, B and C if 4x 2 - 3x + 7 / (Ax + 4)2 + B (x + 4) + C. 4x + 1 in the form x -x-2 a b + . x-2 x+1

11. Express

2

15. Solve 2 sin 2 x + cos x - 2 = 0 for 0c # x # 360c. 16. If a and b are the roots of the quadratic equation 2x 2 + 4x - 5 = 0, evaluate a 3 + b 3.

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11 Locus and the Parabola TERMINOLOGY Axis: A line around which a curve is reflected e.g. the axis of symmetry of a parabola

Latus rectum: A focal chord that is perpendicular to the axis of the parabola

Cartesian equation: An equation involving two variables x and y

Locus: The path traced out by a point that moves according to a particular pattern or rule. Locus can be described algebraically or geometrically

Chord: An interval joining any two points on a curve. In this chapter, any two points on a parabola Circle: The locus of a point moving so that it is equidistant from a fixed point on a plane surface Directrix: A fixed line from which all points equidistant from this line and a fixed point called the focus form a parabola Focal chord: A chord that passes through the focus Focal length: The distance between the focus and the vertex of a parabola or the shortest distance between the vertex and the directrix Focus: A fixed point from which all points equidistant from this point and the directrix form a parabola

Parabola: The locus of a point moving so that it is equidistant from a fixed point called the focus and a fixed line called the directrix Parametric equations: A set of equations where variables x and y are both written in terms of a third variable, called a parameter, usually p or t Tangent: A straight line that touches a curve at a single point only. Vertex: The turning point (maximum or minimum point) of a parabola. It is the point where the parabola meets the axis of symmetry

Chapter 11 Locus and the Parabola

INTRODUCTION THIS CHAPTER EXPANDS THE work on functions that you have already learned.

It shows a method of finding the equation of a locus. In particular, you will study the circle and the parabola, defined as a locus. A parabola can also be defined as a set of parametric equations, and you will study these in this chapter.

DID YOU KNOW? Locus problems have been studied since very early times. Apollonius of Perga (262–190 BC), a contemporary (and rival) of Archimedes, studied the locus of various figures. In his Plane Loci, he described the locus points whose ratio from two fixed points is constant. This locus is called the ‘Circle of Apollonius’. 2 Apollonius also used the equation y = lx for the parabola. René Descartes (1596–1650) was another mathematician who tried to solve locus problems. His study of these led him to develop analytical (coordinate) geometry.

Locus A relation can be described in two different ways. It can be a set of points that obey certain conditions, or a single point that moves along a path according to certain conditions. A locus is the term used to describe the path of a single moving point that obeys certain conditions.

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EXAMPLES Describe the locus of the following. 1. A pencil on the end of compasses.

Solution The path of the pencil is a circle with centre at the point of the compasses.

2. A person going up an escalator (standing still on one step).

What would the locus be if the person walks up the escalator?

Solution The body travels along a straight line parallel to the escalator.

3. A doorknob on a closing door.


Solution If the door could swing right around it would follow a circle. So a door closing swings through an arc of a circle.

4. A point on the number line that is 3 units from 0.

Solution

The locus is !3. 5. A point in the number plane that moves so that it is always 3 units from the y-axis.

Solution The locus is 2 vertical lines with equations x = !3.

Class Discussion Describe the path of a person abseiling down a cliff.

11.1 Exercises Describe the locus of the following: 1.

a racing car driving around a track

2.

a person climbing a ladder

3.

a child on a swing

4.

a ball’s flight when thrown

5.

a person driving up to the 5th floor of a car park

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6.

7.

a point that moves along the number line such that it is always less than 2 units from 0 a point on the number plane that moves so that it is always 2 units from the origin

8.

a point that moves so that it is always 1 unit from the x-axis

9.

a point that moves so that it is always 5 units from the y-axis

10. a point that moves so that it is always 2 units above the x-axis 11. a point that moves so that it is always 1 unit from the origin 12. a point that moves so that it is always 4 units from the point ^ 1, - 2 h

13. a point that is always 5 units below the x-axis 14. a point that is always 3 units away from the point (1, 1) 15. a point that is always 7 units to the left of the y-axis 16. a point that is always 3 units to the right of the y-axis 17. a point that is always 8 units from the x-axis 18. a point that is always 4 units from the y-axis 19. a point that is always 6 units from the point (- 2, 4) 20. a point that is always 1 unit from the point (- 4, 5).

A locus describes a single point P ^ x, y h that moves along a certain path. The equation of a locus can often be found by using P ^ x, y h together with the information given about the locus.

EXAMPLES 1. Find the equation of the locus of a point P ^ x, y h that moves so that it is always 3 units from the origin.

Solution

You studied this formula in 2 Chapter 7. It is easier to use d than d to find the equation of the locus.

You may recognise this locus as a circle, centre ^ 0, 0 h radius 3 units. Its equation is given by x 2 + y 2 = 9. Alternatively, use the distance formula. d=

2 2 _ x2 - x1 i + _ y2 - y1 i

or d 2 = _ x 2 - x 1 i2 + _ y 2 - y 1 i2


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Place P anywhere on the number plane.

Let P ^ x, y h be a point of the locus. We want PO = 3 i.e. PO 2 = 9 ^x - 0h2+ ^y - 0h2 = 9 x2 + y2 = 9 2. Find the equation of the locus of point P ^ x, y h that moves so that distance PA to distance PB is in the ratio 2:1 where A = ^ - 3, 1 h and B = ^ 2, - 2 h .

Solution

Let P ^ x, y h be a point of the locus. PA : PB = 2:1 i.e. `

2 PA = PB 1 PA = 2PB PA 2 = ] 2PB g 2 = 4PB 2

[x - ^ - 3 h] 2 + ^ y - 1 h2 = 4 $ ^ x - 2 h 2 + [y - ^ - 2 h] 2 . ^ x + 3 h 2 + ^ y - 1 h2 = 4 [^ x - 2 h 2 + ^ y + 2 h 2 ] 2 x + 6x + 9 + y 2 - 2y + 1 = 4 ^ x 2 - 4x + 4 + y 2 + 4x + 4 h = 4x 2 - 16x + 16 + 4y 2 + 16y + 16 0 = 3x 2 - 22x + 3y 2 + 18y + 22 or 3x 2 - 22x + 3y 2 + 18y + 22 = 0

i.e.

3. Find the equation of the locus of a point P ^ x, y h that moves so that the line PA is perpendicular to line PB, where A = ^ 1, 2 h and B = ^ - 3, -1 h . CONTINUED

Use the distance formula as in Example 1.

This is the equation of a circle.

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Solution

These results come from Chapter 7.

The locus is a circle with diameter AB.

Let P ^ x, y h be a point of the locus. For perpendicular lines, m 1 m 2 = -1 y2 - y1 Using m = x - x 2 1 y-2 PA: m 1 = x -1 y - ] -1 g PB: m 2 = x - ]-3g y +1 = x+3 For PA perpendicular to PB y - 2 y +1 # = -1 x -1 x + 3 y2 - y - 2 = -1 x 2 + 2x - 3 y 2 - y - 2 = - ^ x 2 + 2x - 3 h = - x 2 - 2x + 3 i.e. x 2 + 2x + y 2 - y - 5 = 0 4. Find the equation of the locus of point P ^ x, y h that is equidistant from fixed point A ^ 1, - 2 h and fixed line with equation y = 5.

Solution

Let P ^ x, y h be a point of the locus. B has coordinates ^ x, 5 h . We want PA = PB PA 2 = PB 2 ^ x - 1 h + [ y - ^ - 2 h] 2 = ^ x - x h 2 + ^ y - 5 h 2 ^x - 1h2+ ^y + 2h2 = ^y - 5h2 x 2 - 2x + 1 + y 2 + 4y + 4 = y 2 - 10y + 25 x 2 - 2x + 14y - 20 = 0 i.e.

2

This is the equation of a parabola. Can you see where the parabola lies?


11.2 Exercises 1.

Find the equation of the locus of point P ^ x, y h that moves so that it is always 1 unit from the origin.

2.

Find the equation of the locus of point P ^ x, y h that moves so that it is always 9 units from the point ^ -1, -1 h .

3.

4.

5.

6.

Find the equation of the locus of a point that moves so that it is always 2 units from the point ^ 5, - 2 h . Find the equation of the locus of point P ^ x, y h that moves so that it is equidistant from the points ^ 3, 2 h and ^ -1, 5 h . Find the equation of the locus of a point that moves so that it is equidistant from the points ^ - 4, 6 h and ^ 2, -7 h . Find the equation of the locus of point P ^ x, y h that moves so that it is equidistant from the x-axis and the y-axis.

7.

Find the equation of the locus of a point P that moves so that PA is twice the distance of PB where A = ^ 0, 3 h and B = ^ 4, 7 h .

8.

Find the equation of the locus of point P ^ x, y h that moves so that the ratio of PA to PB is 3:2 where A = ^ - 6, 5 h and B = ^ 3, -1 h .

9.

Find the equation of the locus of a point that moves so that it is equidistant from the point ^ 2, - 3 h and the line y = 7.

10. Find the equation of the locus of a point that moves so that it is equidistant from the point ^ 0, 5 h and the line y = - 5.

11. Find the equation of the locus of a point that moves so that it is equidistant from the point ^ 2, 0 h and the line x = 6. 12. Find the equation of the locus of a point that moves so that it is equidistant from the point ^ 1, -1 h and the line y = 3. 13. Find the equation of the locus of a point that moves so that it is equidistant from the point ^ 0, - 3 h and the line y = 3. 14. Find the equation of the locus of a point P ^ x, y h that moves so that the line PA is perpendicular to line PB where A = ^ 1, - 3 h and B = ^ 4, 5 h . 15. Find the equation of the locus of a point P ^ x, y h that moves so that the line PA is perpendicular to line PB, where A = ^ - 4, 0 h and B = ^ 1, 1 h . 16. Find the equation of the locus of a point P ^ x, y h that moves so that the line PA is perpendicular to line PB where A = ^ 1, 5 h and B = ^ - 2, - 3 h . 17. Point P moves so that PA 2 + PB 2 = 4 where A = ^ 3, -1 h and B = ^ - 5, 4 h . Find the equation of the locus of P. 18. Point P moves so that PA 2 + PB 2 = 12 where A = ^ - 2, - 5 h and B = ^ 1, 3 h . Find the equation of the locus of P. 19. Find the equation of the locus of a point that moves so that its distance from the line 3x + 4y + 5 = 0 is always 4 units.

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20. Find the equation of the locus of a point that moves so that its distance from the line 12x - 5y - 1 = 0 is always 1 unit. 21. Find the equation, in exact form, of the locus of a point that moves so that its distance from the line x - 2y - 3 = 0 is always 5 units. 22. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 = 0 and the line 3x + 4y - 7 = 0.

23. Find the equation of the locus of a point that moves so that it is equidistant from the line 3x + 4y - 5 = 0 and the line 5x + 12y - 1 = 0. 24. Given two points A ^ 3, - 2 h and B ^ -1, 7 h, find the equation of the locus of P ^ x, y h if the gradient of PA is twice the gradient of PB. 25. If R is the fixed point ^ 3, 2 h and P is a movable point ^ x, y h, find the equation of the locus of P if the distance PR is twice the distance from P to the line y = -1.

PROBLEM Can you see 2 mistakes in the solution to this question? Find the locus of point P ^ x, y h that moves so that its perpendicular distance from the line 12x + 5y - 1 = 0 is always 3 units.

Solution Let P ^ x, y h be a point of the locus. d= 3= = =

| ax 1 + by 1 + c | a2 + b2 | 5x + 12y - 1| 5 2 + 12 2 | 5x + 12y - 1| 25 + 144 | 5x + 12y - 1|

169 | 5x + 12y - 1| = 13 ` 39 = 5x + 12y - 1 0 = 5x + 12y - 40 Can you find the correct locus?


Circle as a Locus The locus of point P (x, y) that is always a constant distance from a fixed point is a circle.

The circle, centre ^ 0, 0 h and radius r, has the equation x2 + y2 = r2

Proof Find the equation of the locus of point P ^ x, y h that is always r units from the origin.

Let P ^ x, y h be a point of the locus. OP = r i.e. OP 2 = r 2 2 ^ x - 0 h + ^ y - 0 h 2 = r2 x2 + y2 = r2 So x 2 + y 2 = r 2 is the equation of the locus. It describes a circle with radius r and centre ^ 0, 0 h .

The circle, centre ^ a, b h and radius r, has the equation ^ x - a h 2 + ^ y - b h 2 = r2

Proof Find the equation of the locus of point P ^ x, y h that is always r units from point A ^ a, b h .

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Let P ^ x, y h be a point of the locus. AP = r i.e. AP 2 = r 2 ^ x - a h 2 + ^ y - b h 2 = r2 So ] x - a g2 + ^ y - b h2 = r 2 is the equation of the locus. It describes a circle with radius r and centre ^ a, b h .

EXAMPLES 1. Find the equation of the locus of a point that is always 2 units from the point ^ -1, 0 h .

Solution

You could find this equation by using P (x, y) and treating the question as a locus problem.

This is a circle with radius 2 and centre ^ -1, 0 h . Its equation is in the form ^ x - a h 2 + ^ y - b h 2 = r2 i.e. [x - ^ -1 h] 2 + ^ y - 0 h 2 = 2 2 ^ x + 1 h 2 + y2 = 4 x 2 + 2x + 1 + y 2 = 4 x 2 + 2x + y 2 - 3 = 0 2. Find the radius and the coordinates of the centre of the circle x 2 + 2x + y 2 - 6y - 15 = 0.

Solution You learned how to complete the square in Chapter 3.

We put the equation into the form ^ x - a h 2 + ^ y - b h 2 = r 2 . To do this we complete the square. b 2 In general, to complete the square on x 2 + bx, add c m to give: 2 b 2 b 2 x 2 + bx + c m = c x + m 2 2 First we move any constants to the other side of the equation, then complete the square. 2 2 To complete the square on x 2 + 2x, we add c m = 1. 2


6 2 To complete the square on y 2 - 6y, we add c m = 9. 2 x 2 + 2x + y 2 - 6y - 15 = 0 x 2 + 2x + y 2 - 6y = 15 2 x + 2x + 1 + y 2 - 6y + 9 = 15 + 1 + 9 ^ x + 1 h 2 + ^ y - 3 h 2 = 25 ^ x - ] - 1 g h 2 + ^ y - 3 h 2 = 52 The equation is in the form ^ x - a h 2 + ^ y - b h 2 = r 2 . This is a circle, centre ^ -1, 3 h and radius 5.

11.3 Exercises 1.

2.

3.

4.

Find the length of the radius and the coordinates of the centre of each circle. (a) x 2 + y 2 = 100 (b) x 2 + y 2 = 5 (c) ^ x - 4 h 2 + ^ y - 5 h 2 = 16 (d) ^ x - 5 h 2 + ^ y + 6 h 2 = 49 (e) x 2 + ^ y - 3 h2 = 81 Find the equation of each circle in expanded form (without grouping symbols). (a) Centre (0, 0) and radius 4 (b) Centre (3, 2) and radius 5 (c) Centre ^ -1, 5 h and radius 3 (d) Centre (2, 3) and radius 6 (e) Centre ^ - 4, 2 h and radius 5 (f) Centre ^ 0, - 2 h and radius 1 (g) Centre (4, 2) and radius 7 (h) Centre ^ - 3, - 4 h and radius 9 (i) Centre ^ - 2, 0 h and radius 5 (j) Centre ^ - 4, -7 h and radius 3 . Find the equation of the locus of a point moving so that it is 1 unit from the point ^ 9, - 4 h . Find the equation of the locus of a point moving so that it is 4 units from the point ^ - 2, - 2 h .

5.

Find the equation of the locus of a point moving so that it is 7 units from the point ^ 1, 0 h .

6.

Find the equation of the locus of a point moving so that it is 2 units from the point ^ - 3, 8 h .

7.

Find the equation of the locus of a point moving so that it is 2 units from the point ^ 5, - 2 h .

8.

Find the equation of a circle with centre ^ 0, 0 h and radius 3 units.

9.

Find the equation of a circle with centre ^ 1, 5 h and radius 1 unit.

10. Find the equation of a circle with centre ^ - 6, 1 h and radius 6 units. 11. Find the equation of a circle with centre ^ 4, 3 h and radius 3 units. 12. Find the equation of a circle with centre ^ 0, - 3 h and radius 2 2 units. 13. Find the coordinates of the centre and the length of the radius of each circle. (a) x 2 - 4x + y 2 - 2y - 4 = 0 (b) x 2 + 8x + y 2 - 4y - 5 = 0 (c) x 2 + y 2 - 2y = 0

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(d) (e) (f) (g) (h) (i) (j)

x2 x2 + x2 x2 + x2 + x2 x2 +

10x + y 2 + 6y - 2 = 0 2x + y 2 - 2y + 1 = 0 12x + y 2 = 0 6x + y 2 - 8y = 0 20x + y 2 - 4y + 40 = 0 14x + y 2 + 2y + 25 = 0 2x + y 2 + 4y - 5 = 0

14. Find the centre and radius of the circle with equation given by x 2 - 6x + y 2 + 2y - 6 = 0. 15. Find the centre and radius of the circle with equation given by x 2 - 4x + y 2 - 10y + 4 = 0. 16. Find the centre and radius of the circle with equation given by x 2 + 2x + y 2 + 12y - 12 = 0. 17. Find the centre and radius of the circle with equation given by x 2 - 8x + y 2 - 14y + 1 = 0. 18. Find the centre and radius of the circle with equation given by x 2 + 3x + y 2 - 2y - 3 = 0. 19. Sketch the circle whose equation is given by x 2 + 4x + y 2 - 2y + 1 = 0. 20. Prove that the line 3x + 4y + 21 = 0 is a tangent to the circle x 2 - 8x + y 2 + 4y - 5 = 0.

Concentric circles have the same centre.

21. (a) Show that x 2 - 2x + y 2 + 4y + 1 = 0 and x 2 - 2x + y 2 + 4y - 4 = 0 are concentric. (b) Find the difference between their radii. 22. Given two points A ^ 2, - 5 h and B ^ - 4, 3 h, find the equation of the circle with diameter AB.

23. Find the exact length of the tangent from ^ 4, - 5 h to the circle x 2 + 4x + y 2 - 2y - 11 = 0. 24. Find the exact length of AB where A and B are the centres of the circles x 2 - 6x + y 2 = 0 and x 2 + 4x + y 2 + 6y - 3 = 0 respectively. 25. (a) Find the length of XY where X and Y are the centres of the circles x 2 + 6x + y 2 - 2y + 1 = 0 and x 2 - 4x + y 2 - 2y + 1 = 0 respectively. (b) Find the radius of each circle. (c) What conclusion can you draw from the results for (a) and (b)? 26. Show that the circles x 2 + y 2 = 4 and x 2 + 2x + y 2 - 4y - 4 = 0 both have 3x + 4y + 10 = 0 as a tangent. 27. A circle has centre C ^ -1, 3 h and radius 5 units. (a) Find the equation of the circle. (b) The line 3x - y + 1 = 0 meets the circle at two points. Find their coordinates. (c) Let the coordinates be X and Y, where Y is the coordinate directly below the centre C. Find the coordinates of point Z, where YZ is a diameter of the circle. (d) Hence show +ZXY = 90c. 28. (a) Find the perpendicular distance from P ^ 2, - 5 h to the line 5x + 12y - 2 = 0. (b) Hence find the equation of the circle with centre P and tangent 5x + 12y - 2 = 0.


Parabola as a Locus The locus of a point that is equidistant from a fixed point and a fixed line is always a parabola. The fixed point is called the focus and the fixed line is called the directrix.

Work on the parabola as a locus is very important, as the properties of the parabola are useful to us. The parabola is used in lenses of glasses and cameras, in car headlights, and for bridges and radio telescope dishes.

DID YOU KNOW? Any rope or chain supporting a load (e.g. a suspension bridge) is in the shape of a parabola. Find some examples of suspension bridges that have a parabola shaped chain. Other bridges have ropes or chains hanging freely. These are not in the shape of a parabola, but are in a shape called a catenary. Can you find some bridges with this shape? More recent bridges are cable-stayed, where ropes or chains are attached to towers, or pylons, and fan out along the sides of the bridge. An example is the Anzac Bridge in Sydney. There are many different bridge designs. One famous bridge in Australia is the Sydney Harbour Bridge. Research different bridge designs and see if you can find some with parabolic shapes.

Parabola with vertex at the origin Just as the circle has a special equation when its centre is at the origin, the parabola has a special equation when its vertex is at the origin. Both also have a more general formula.

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The locus of a point that is equidistant from a fixed point and a fixed line is always in the shape of a parabola. If the fixed point is (0, a) and the fixed line is y = - a (where a 2 0), then one of the equidistant points is the origin (0, 0). The distance between the points (0, 0) and (0, a) is a units. The point on y = - a directly below the origin is ^ 0, - a h and the distance from (0, 0) to ^ 0, - a h is also a units. y

(0, a) a x a (0, -a)

y =- a

To find the equation of the parabola, we use the general process to find the equation of any locus. The features of the parabola have special names.

A parabola is equidistant from a fixed point and a fixed line. • The fixed point is called the focus. • The fixed line is called the directrix. • The turning point of the parabola is called the vertex. • The axis of symmetry of the parabola is called its axis. • The distance between the vertex and the focus is called the focal length. • An interval joining any two points on the parabola is called a chord. • A chord that passes through the focus is called a focal chord. • The focal chord that is perpendicular to the axis is called the latus rectum. • A tangent is a straight line that touches the parabola at a single point.


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PARABOLA x 2 = 4 ay The locus of point P ^ x, y h moving so that it is equidistant from the point ^ 0, a h and the line y = - a is a parabola with equation x 2 = 4ay

Proof

Let P ^ x, y h be a point of the locus. Taking the perpendicular distance from P to the line y = - a, point B = ^ x, - a h . PA = PB ` PA 2 = PB 2 ^ x - 0 h 2 + ^ y - a h 2 = ^ x - x h 2 + [y - ^ - a h] 2 x2 + ^ y - a h 2 = ^ y + a h 2 x 2 + y 2 - 2ay + a 2 = y 2 + 2ay + a 2 x 2 = 4ay

The parabola x 2 = 4ay has • focus at ^ 0, a h • directrix with equation y = - a • vertex at ^ 0, 0 h • axis with equation x = 0 • focal length the distance from the vertex to the focus with length a • latus rectum that is a horizontal focal chord with length 4a

Class Investigation Find the equation of the locus if point P ^ x, y h is equidistant from ^ 0, - a h and y = a.

Since the focal length is a, a is always a positive number.

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EXAMPLES 1. Find the equation of the parabola whose focus has coordinates ^ 0, 2 h and whose directrix has equation y = - 2.

Solution The focus has coordinates in the form ^ 0, a h and the directrix has equation in the form y = - a, where a = 2. ` the parabola is in the form x 2 = 4ay where a = 2 i.e. x 2 = 4 (2) y x 2 = 8y 2. (a) Find the coordinates of the focus and the equation of the directrix of the parabola x 2 = 20y. (b) Find the points on the parabola at the endpoints of the latus rectum and find its length.

Solution (a) The parabola x 2 = 20y is in the form x 2 = 4ay 4a = 20 ` a=5 The focal length is 5 units. We can find the coordinates of the focus and the equation of the directrix in two ways. Method 1: Draw the graph x 2 = 20y and count 5 units up and down from the origin as shown. y x2 = 20y (0, 5) 5 x 5 (0, -5) y = -5

The focus is (0, 5) and the directrix has equation y = -5.


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Method 2: The focus is in the form (0, a) where a = 5. So the focus is (0, 5). The directrix is in the form y = - a where a = 5. So the directrix is y = - 5. (b) The latus rectum is a focal chord that is perpendicular to the axis of the parabola as shown y x2 = 20 y

(0, 5) x

The endpoints of the latus rectum will be where the line y = 5 and the parabola intersect. Substitute y = 5 into the parabola. x 2 = 20y = 20 ] 5 g = 100 x = ! 100 = !10 So the endpoints are (-10, 5) and (10, 5). y x2 = 20 y

(-10, 5)

(0, 5)

(10, 5) x

From the graph, the length of the latus rectum is 20 units.

CONTINUED

The latus rectum is 4a units long which gives 20 units.

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3. Find the equation of the focal chord to the parabola x 2 = 4y that passes through (- 4, 4).

Solution The parabola x 2 = 4y is in the form x 2 = 4ay. 4a = 4 à=1 The focal length is 1 unit. The focus is 1 unit up from the origin at (0, 1) and the focal chord also passes through (- 4, 4). y (-4, 4)

x2 = 4y

(0, 1) x

You used these formulae in Chapter 7.

We can find the equation of the line between (0, 1) and (- 4, 4) by using either formula y - y1 y2 - y1 y - y 1 = m _ x - x 1 i or x - x = x - x 1 2 1 y - y1 y2 - y1 x - x1 = x2 - x1 y -1 4 -1 = x-0 -4 - 0 y -1 3 x = -4 - 4 ^ y - 1 h = 3x - 4y + 4 = 3x 0 = 3x + 4y - 4

As you saw in the previous chapter, a parabola can be concave downwards. Can you guess what the equation of this parabola might be? PARABOLA x 2 = - 4 ay The locus of a point P(x, y) moving so that it is equidistant from the point ^ 0, - a h and the line y = a is a parabola with equation x 2 = - 4ay


Proof y

B(x, a)

y=a

x P(x, y) A(0, -a)

Let P(x, y) be a point of the locus. Taking the perpendicular distance from P to the line y = a, point B = ^ x, a h . PA = PB `

PA 2 = PB 2

2 ^ x - 0 h 2 + 7 y - ^ - a h A = ^ x - x h 2 + ^ y - a h2 x 2 + ^ y + a h2 = ^ y - a h2 2 x + y 2 + 2ay + a 2 = y 2 - 2ay + a 2 x 2 = - 4ay

The parabola x 2 = - 4ay has • focus at ^ 0, - a h • directrix with equation y = a • vertex at (0, 0) • axis with equation x = 0 • focal length a • latus rectum a horizontal focal chord with length 4a

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EXAMPLES 1. Find the equation of the parabola with focus ^ 0, - 4 h and directrix y = 4.

Solution If we draw this information, the focus is below the directrix as shown. So the parabola will be concave downwards (the parabola always turns away from the directrix). y

y=4 4 x 4 (0, -4)

The focal length is 4 so a = 4. The parabola is in the form x 2 = - 4ay where a = 4. x 2 = - 4ay = -4]4gy = - 16y 2. Find the coordinates of the vertex, the coordinates of the focus and the equation of the directrix of the parabola x 2 = - 12y.

Solution The parabola x 2 = -12y is in the form x 2 = - 4ay. 4a = 12 à=3 The focal length is 3 units. The vertex is (0, 0). We can find the coordinates of the focus and the equation of the directrix in two ways. Method 1: Draw the graph x 2 = -12y and count 3 units up and down from the origin as shown. (The parabola is concave downward.)


y

y=3 3 x 3 (0, -3)

x2 = -12y

Counting down 3 units, the focus is ^ 0, - 3 h . Counting up 3 units, the directrix has equation y = 3. Method 2: The focus is in the form ^ 0, - a h where a = 3. So the focus is ^ 0, - 3 h . The directrix is in the form y = a where a = 3. So the directrix is y = 3. 3. Find the equation of the parabola with focal length 5 and whose vertex is ^ 0, 0 h and equation of the axis is x = 0.

Solution Vertex ^ 0, 0 h and axis given by x = 0 give a parabola in the form x 2 = !4ay, since there is not enough information to tell whether it is concave upwards or downwards. This gives two possible parabolas.

CONTINUED

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Focal length of 5 means a = 5 i.e.

The equation is x 2 = !4 (5) y x 2 = !20y

11.4 Exercises 1.

2.

3.

Find the equation of each parabola. (a) focus (0, 5), directrix y = - 5 (b) focus (0, 9), directrix y = - 9 (c) focus (0, 1), directrix y = -1 (d) focus (0, 4), directrix y = - 4 (e) focus (0, 10), directrix y = -10 (f) focus (0, 3), directrix y = - 3 (g) focus (0, 6), directrix y = - 6 (h) focus (0, 11), directrix y = -11 (i) focus (0, 2), directrix y = - 2 (j) focus (0, 12), directrix y = -12 Find the equation of each parabola. (a) focus (0, - 1), directrix y = 1 (b) focus (0, - 3), directrix y = 3 (c) focus (0, - 4), directrix y = 4 (d) focus (0, - 7), directrix y = 7 (e) focus (0, - 6), directrix y = 6 (f) focus (0, - 9), directrix y = 9 (g) focus (0, - 8), directrix y = 8 (h) focus (0, - 2), directrix y = 2 (i) focus (0, -15), directrix y = 15 (j) focus (0, -13), directrix y = 13 Find (i) the coordinates of the focus and (ii) the equation of the directrix of (a) x 2 = 4y (b) x 2 = 28y (c) x 2 = 16y

(d) (e) (f) (g) (h) (i) (j)

x 2 = 36y x 2 = 40y x 2 = 44y x 2 = 12y x 2 = 6y x 2 = 10y x 2 = 15y

4.

Find (i) the coordinates of the focus and (ii) the equation of the directrix of (a) x 2 = - 4y (b) x 2 = - 24y (c) x 2 = - 8y (d) x 2 = - 48y (e) x 2 = - 20y (f) x 2 = - 16y (g) x 2 = - 32y (h) x 2 = - 40y (i) x 2 = - 2y (j) x 2 = - 22y

5.

Find the equation of the parabola with (a) coordinates of the focus ^ 0, 7 h and equation of the directrix y = -7 (b) coordinates of the focus ^ 0, 11 h and equation of the directrix y = -11 (c) coordinates of the focus ^ 0, - 6 h and equation of the directrix y = 6 (d) coordinates of the focus ^ 0, 2 h and coordinates of the vertex ^ 0, 0 h .


(e) coordinates of the vertex ^ 0, 0 h, equation of the axis x = 0 and focal length 3 (f) coordinates of the vertex ^ 0, 0 h, equation of the axis x = 0 and focal length 8 (g) coordinates of the vertex ^ 0, 0 h and equation of the axis x = 0, and passing through the point ^ - 8, 2 h (h) coordinates of the vertex ^ 0, 0 h and equation of the axis x = 0, and passing through the point ^ -1, 7 h . 6.

7.

Find the coordinates of the focus, the equation of the directrix and the focal length of the parabola (a) x 2 = 8y (b) x 2 = 24y (c) x 2 = -12y (d) x 2 = 2y (e) x 2 = - 7y (f) 2x 2 = y Find the equation of the focal chord that cuts the curve x 2 = 8y at ^ - 4, 2 h .

8.

The tangent with equation 2x - y - 4 = 0 touches the parabola x 2 = 4y at A. Find the coordinates of A.

9.

The focal chord that cuts the parabola x 2 = - 6y at ^ 6, - 6 h cuts the parabola again at X. Find the coordinates of X.

10. Find the coordinates of the endpoints of the latus rectum of the parabola x 2 = - 8y. What is the length of the latus rectum?

11. The equation of the latus rectum of a parabola is given by y = - 3. The axis of the parabola is x = 0, and its vertex is ^ 0, 0 h . (a) Find the equation of the parabola. (b) Find the equation of the directrix. (c) Find the length of the focal chord that meets the parabola at 1 c 2, - m . 3 12. (a) Show that the point ^ - 3, 3 h lies on the parabola with equation x 2 = 3y. (b) Find the equation of the line passing through P and the focus F of the parabola. (c) Find the coordinates of the point R where the line PF meets the directrix. 13. (a) Find the equation of chord 1 PQ where P c -1, m and Q ^ 2, 1 h 4 lie on the parabola x 2 = 4y. (b) Show that PQ is not a focal chord. (c) Find the equation of the circle with centre Q and radius 2 units. (d) Show that this circle passes through the focus of the parabola. 14. (a) Show that Q _ 2aq, aq 2 i lies on the parabola x 2 = 4ay. (b) Find the equation of the focal chord through Q. (c) Prove that the length of the latus rectum is 4a.

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Investigation Sketch the parabola x = y 2. You may like to complete the table below to help you with its sketch. x y

-3

-2

-1

0

1

2

3

Is this parabola a function? What is its axis of symmetry?

The parabola that has y2 rather than x2 in its equation is a sideways parabola. It still has the same properties, but generally the x and y values are swapped around. PARABOLA y 2 = 4 ax The locus of point P ^ x, y h moving so that it is equidistant from the point ^ a, 0 h and the line x = - a is a parabola with equation y 2 = 4ax

Proof Find the equation of the locus of point P ^ x, y h, which moves so that it is equidistant from the point ^ a, 0 h and the line x = - a.

Coordinates of B are ^ - a, y h . We want PA = PB i.e.

PA 2 = PB 2 ^ x - a h 2 + ^ y - 0 h 2 = [x - ^ - a h 2 ] 2 + ^ y - y h 2 ^ x - a h 2 + y2 = ^ x + a h 2 x 2 - 2ax + a 2 + y 2 = x 2 + 2ax + a 2 y 2 = 4ax


The parabola y 2 = 4ax has • focus at ^ a, 0 h • equation of directrix x = - a • vertex at ^ 0, 0 h • axis with equation y = 0 • focal length the distance from the vertex to the focus with length a • latus rectum that is a vertical focal chord with length 4a

EXAMPLES 1. Find the equation of the parabola with focus (7, 0) and directrix x = - 7.

Solution If we draw this information, the focus is to the right of the directrix as shown (the parabola always turns away from the directrix). So the parabola turns to the right. y x=-7

7

7 (7, 0)

x

CONTINUED

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The focal length is 7 so a = 7. The parabola is in the form y 2 = 4ax where a = 7. y 2 = 4ax = 4^7hx = 28x. 2. Find the coordinates of the focus and the equation of the directrix of the parabola y 2 = 32x.

Solution The parabola y 2 = 32x is in the form y 2 = 4ax. 4a = 32 ` a=8 The focal length is 8 units. Method 1: Draw the graph y 2 = 32x and count 8 units to the left and right from the origin as shown. (The parabola turns to the right.) y

x=-8

8

8 (8, 0)

x

y2 = 32x 2

Counting 8 units to the right, the focus is (8, 0). Counting 8 units to the left, the directrix has equation x = - 8. Method 2: The focus is in the form (a, 0) where a = 8. So the focus is (8, 0). The directrix is in the form x = - a where a = 8. So the directrix is x = - 8.

A parabola can also turn to the left.


PARABOLA y 2 = - 4 ax The locus of a point P(x, y) moving so that it is equidistant from the point ^ - a, 0 h and the line x = a is a parabola with equation y 2 = - 4ax

Proof y P(x, y)

B(a, y)

x

A( - a, 0)

x=a

Let P(x, y) be a point of the locus. Taking the perpendicular distance from P to the line x = a, point B = ^ a, y h . PA = PB `

PA 2 = PB 2

7 x - ^ - a h A + ^ y - 0 h2 = ^ x - a h 2 + ^ y - y h 2 ^ x + a h 2 + y2 = ^ x - a h 2 2

x 2 + 2ax + a 2 + y 2 = x 2 - 2ax + a 2 y 2 = - 4ax

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The parabola y 2 = - 4ax has • focus at (- a, 0) • directrix with equation x = a • vertex at (0, 0) • axis with equation y = 0 • focal length a • latus rectum a vertical focal chord with length 4a

EXAMPLES 1. Find the equation of the parabola with focus (- 4, 0) and directrix x = 4.

Solution Drawing this information shows that the parabola turns to the left. y

4

4 x

( - 4, 0)

x=4

The focal length is 4 so a = 4. The parabola is in the form y 2 = - 4ax where a = 4. y 2 = - 4ax = -4^ 4h x = -16x. 2. Find the coordinates of the focus and the equation of the directrix of the parabola y 2 = - 2x.

Solution The parabola y 2 = - 2x is in the form y 2 = - 4ax. 4a = 2 ` a=

1 2

The focal length is

1 unit. 2


Method 1: 1 Draw the graph y 2 = - 2x and count unit to the left and right from the 2 origin as shown. (The parabola turns to the left.) y

1 2

(

- 1, 0 2

1 2 x

) x=

1 2

1 1 units to the left, the focus is c - , 0 m . 2 2 1 1 Counting units to the right, the directrix has equation x = . 2 2

Counting

Method 2: 1 The focus is in the form (- a, 0) where a = . 2 1 So the focus is c - , 0 m . 2 1 The directrix is in the form x = a where a = . 2 1 So the directrix is x = . 2

11.5 Exercises 1.

Find the equation of each parabola. (a) focus (2, 0), directrix x = - 2 (b) focus (5, 0), directrix x = - 5 (c) focus (14, 0), directrix x = -14 (d) focus (9, 0), directrix x = - 9 (e) focus (8, 0), directrix x = - 8 (f) focus (6, 0), directrix x = - 6 (g) focus (7, 0), directrix x = - 7 (h) focus (3, 0), directrix x = - 3 (i) focus (4, 0), directrix x = - 4 (j) focus (1, 0), directrix x = -1

2.

Find the equation of each parabola. (a) focus (- 9, 0), directrix x = 9 (b) focus (- 4, 0), directrix x = 4 (c) focus (-10, 0), directrix x = 10 (d) focus (- 6, 0), directrix x = 6 (e) focus (- 2, 0), directrix x = 2 (f) focus (-12, 0), directrix x = 12 (g) focus (-11, 0), directrix x = 11 (h) focus (- 5, 0), directrix x = 5 (i) focus (- 3, 0), directrix x = 3 (j) focus (- 7, 0), directrix x = 7

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3.

4.

5.

Find (i) the coordinates of the focus and (ii) the equation of the directrix of (a) y 2 = 8x (b) y 2 = 12x (c) y 2 = 16x (d) y 2 = 4x (e) y 2 = 28x (f) y 2 = 32x (g) y 2 = 24x (h) y 2 = 36x (i) y 2 = x (j) y 2 = 18x Find (i) the coordinates of the focus and (ii) the equation of the directrix of (a) y 2 = - 8x (b) y 2 = -12x (c) y 2 = - 28x (d) y 2 = - 4x (e) y 2 = - 24x (f) y 2 = - 52x (g) y 2 = - 60x (h) y 2 = - 2x (i) y 2 = - 26x (j) y 2 = - 5x Find the equation of the parabola with (a) coordinates of the focus ^ 5, 0 h and equation of the directrix x = -5 (b) coordinates of the focus ^ 1, 0 h and equation of the directrix x = -1 (c) coordinates of the focus ^ - 4, 0 h and equation of the directrix x = 4 (d) coordinates of the focus ^ 3, 0 h and coordinates of the vertex ^ 0, 0 h (e) coordinates of the vertex ^ 0, 0 h equation of the axis y = 0 and focal length 9

(f) coordinates of the vertex ^ 0, 0 h, equation of the axis y = 0 and focal length 2 (g) coordinates of the vertex ^ 0, 0 h and equation of the axis y = 0 and passing through the point ^ 3, 6 h (h) coordinates of the vertex ^ 0, 0 h and equation of the axis y = 0 and passing through the point ^ 2, 1 h . 6.

Find the coordinates of the focus, the equation of the directrix and the focal length of the parabola (a) y 2 = 8x (b) y 2 = 4x (c) y 2 = -12x (d) y 2 = 6x (e) y 2 = - 5x (f) 3y 2 = x

7.

Find the equation of the focal chord that cuts the curve y 2 = 16x at ^ 4, 8 h .

8.

Find the length of the latus rectum of the parabola y 2 = 12x. What are the coordinates of its endpoints?

9.

The line with equation x - 3y - 27 = 0 meets the parabola y 2 = 4x at two points. Find their coordinates.

1 10. Let R c , - 2 m be a point on the 5 parabola y 2 = 20x. (a) Find the equation of the focal chord passing through R. (b) Find the coordinates of the point Q where this chord cuts the directrix. (c) Find the area of DOFQ where O is the origin and F is the focus. (d) Find the perpendicular distance from the chord to the point P ^ -1, -7 h . (e) Hence find the area of DPQR.


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Application A parabolic satellite dish receives its signals through the focus. If the dish has height 12 m and a span of 20 m, find where the focus should be placed, to the nearest mm.

SOLUTION

The parabola is of the form x 2 = 4ay and passes through (10, 12) and (-10, 12) Substituting (10, 12) gives 2 10 = 4a (12)

100 = 48a 2.083 = a So the focus should be placed 2.083 m from the vertex.

Here is a summary of the 4 different types of parabola with the vertex at the origin.

1. x 2 = 4ay y

x2 = 4ay Focus (0, a) x

Directrix y = -a

This is 2083 mm to the nearest millimetre.

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2. x 2 = - 4ay y

Directrix y=a x Focus (0, -a) x 2 = -4ay

3. y 2 = 4ax y Directrix x = -a

Focus (a, 0)

x

y 2 = 4a x

4. y 2 = - 4ax y Directrix x=a

Focus (-a, 0)

x

y2 = -4ax

General Parabola When the parabola does not have its vertex at the origin, there is a more general formula. Since we use a to mean the focal length, we cannot use (a, b) as the vertex. We use (h, k) instead.


PARABOLA (x - h)2 = 4a(y - k) The concave upwards parabola with vertex (h, k) and focal length a has equation ^ x - h h 2 = 4a ^ y - k h

Proof Find the equation of the parabola with vertex ^ h, k h and focal length a.

Counting up a units from vertex V gives the focus F = ^ h, k + a h . Counting down a units from V gives the point on the directrix D = ^ h, k - a h . So the equation of the directrix is given by y = k - a. We find the equation of the locus of P ^ x, y h that is equidistant from point F ^ h, k + a h and line y = k - a.

B has coordinates ^ x, k - a h . We want PF = PB i.e. PF 2 = PB 2 ^ x - h h 2 + [ y - ^ k + a h] 2 = ^ x - x h 2 + [ y - ^ k + a h] 2 ^x - hh2 + ^ y - k - ah2 = ^ y - k + ah2 ^x - hh2 = ^ y - k + ah2- ^ y - k - ah2 = [^ y - k + a h + ^ y - k - a h ] # [^ y - k + a h - ^ y - k - a h ] ^ difference of two squares h = ^ 2y - 2k h ^ 2a h = 4ay - 4ak = 4a ^ y - k h

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The parabola ^ x - h h2 = 4a ^ y - k h has • axis parallel to the y-axis • vertex at ^ h, k h • focus at ^ h, k + a h • directrix with equation y = k - a

EXAMPLES 1. Find the equation of the parabola with focus ^ 2, 3 h and directrix with equation y = - 7.

Solution

Draw a diagram to find the vertex and to find a.

Coordinates of B are ^ 2, -7 h . The vertex is the midpoint of ^ 2, 3 h and ^ 2, -7 h . ` vertex = ^ 2, - 2 h Focal length is the distance from the focus to the vertex. ` a=5 From the diagram the parabola is concave upwards. The equation is in the form ^ x - h h 2 = 4a ^ y - k h i.e. ^ x - 2 h 2 = 4 ^ 5 h [ y - ^ - 2 h] = 20 ^ y + 2 h x 2 - 4x + 4 = 20y + 40 x 2 - 4x - 20y - 36 = 0 2. Find the coordinates of the vertex and the focus, and the equation of the directrix, of the parabola with equation x 2 + 6x - 12y - 3 = 0.


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Solution Complete the square on x. x 2 + 6x - 12y - 3 = 0 x 2 + 6x = 12y + 3 x 2 + 6x + 9 = 12y + 3 + 9 ^ x + 3 h2 = 12y + 12 = 12 (y + 1) So the parabola has equation ^ x + 3 h2 = 12 ^ y + 1 h . Its vertex has coordinates ^ - 3, -1 h . 4a = 12 ` a=3 The parabola is concave upwards as it is in the form ^ x - h h 2 = 4a ^ y - k h .

Count up 3 units to the focus ` focus = ^ - 3, 2 h Count down 3 units to the directrix ` directrix has equation y = - 4.

PARABOLA (x - h)2 = - 4a(y - k) The concave downwards parabola with vertex (h, k) and focal length a has equation ^ x - h h 2 = - 4a ^ y - k h

Proof Find the equation of the concave downwards parabola with vertex (h, k) and focal length a.

It is easy to find the focus and the directrix by counting along the y-axis.

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Counting down a units from the vertex V gives the focus F = ^ h, k - a h . Counting up a units from the vertex V gives the point on the directrix D = ^ h, k + a h . So the equation of the directrix is given by y = k + a. We find the equation of the locus of P(x, y) that is equidistant from point F ^ h, k - a h and line y = k + a. y B

y=k+a

x

P (x, y)

F (h , k - a)

B has coordinates ^ x, k + a h . We want

PF = PB PF 2 = PB 2

2 2 ^x - hh2 + 7y - ^k - ahA = ^x - xh2 + 7y - ^k + ahA 2 2 2 ^x - hh + ^y - k + ah = ^y - k - ah ^x - hh2 = ^y - k - ah2- ^y - k + ah2 = 7^y - k - ah + ^y - k + ahA7^y - k - ah - ^y - k + ahA (difference of two squares) = ^ 2y - 2k h ^ - 2a h = - 4ay + 4ak = - 4a ^ y - k h


The parabola ^ x - h h 2 = - 4a ^ y - k h has • axis parallel to the y-axis • vertex at (h, k) • focus at ^ h, k - a h • directrix with equation y = k + a

EXAMPLES 1. Find the equation of the parabola with focus (- 2, 1) and directrix y = 3.

Solution y

B

1 1 (-2, 1)

3

y=3

2 1

-2 -1

x

Coordinates of B are (- 2, 3). The vertex is the midpoint of (- 2, 1) and (- 2, 3). ` vertex = (- 2, 2) Focal length a = 1. From the diagram the curve is concave downwards. The equation is in the form ^ x - h h 2 = - 4a ^ y - k h i.e.

2 7x - ^ -2 h A = -4 ]1 g^ y - 2 h ^ x + 2h 2 = -4^ y - 2h

x 2 + 4x + 4 = - 4y + 8 x 2 + 4x + 4y - 4 = 0. 2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola x 2 - 8x + 8y - 16 = 0. CONTINUED

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Solution Complete the square on x. x 2 - 8x + 8y - 16 = 0 x 2 - 8x = - 8y + 16 x 2 - 8x + 16 = - 8y + 16 + 16 ^ x - 4 h2 = - 8y + 32 = -8^ y - 4h So the parabola has equation ^ x - 4 h 2 = - 8 ^ y - 4 h . Its vertex has coordinates (4, 4). 4a = 8 à=2 The parabola is concave downwards as it is in the form ^ x - h h 2 = - 4a ^ y - k h . y y=6

5

(4, 4)

4 3

2

2 (4, 2)

2 1 1

2

3

4

Count down 2 units to the focus ` focus = ^ 4, 2 h Count up 2 units to the directrix ` directrix has equation y = 6.

PARABOLA ( y - k)2 = 4a(x - h) The parabola with vertex (h, k) and focal length a that turns to the right has equation ^ y - k h 2 = 4a ^ x - h h

Proof Find the equation of the parabola that turns to the right with vertex (h, k) and focal length a.


Counting a units to the right from the vertex V gives the focus F = ^ h + a, k h . Counting a units to the left from the vertex V gives the point on the directrix D = ^ h - a, k h . So the equation of the directrix is given by x = h - a. We find the equation of the locus of P(x, y) that is equidistant from point F ^ h + a, k h and line x = h - a.

x=h-a

y

B

P (x, y)

x

F (h +a, k)

B has coordinates ^ h - a, y h . We want

PF = PB PF 2 = PB 2

2 2 7x - ^h + ahA + ^y - kh2 = 7x - ^h - ahA + ^y - y h2 ^x - h - ah2+ ^y - kh2 = ^x - h - ah2 ^y - kh2 = ^x - h + ah2- ]x - h - ag2

= 7^x - h + ah + ^x - h - ahA7^x - h + ah - ^x - h - ahA (difference of two squares) = ^ 2x - 2h h ^ 2a h = 4ax - 4ah = 4a ^ x - h h

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The parabola ^ y - k h 2 = 4a ^ x - h h has • axis parallel to the x-axis • vertex at ^ h, k h • focus at ^ h + a, k h • directrix with equation x = h - a

EXAMPLES 1. Find the equation of the parabola with focus (1, -1) and directrix x = - 5.

Solution y x = -5

5 4 3 2 1

1 2 3 4 5 (1, -1)

-5 -4 -3 -2 -1 B

3

3

-2 -3

Coordinates of B are (- 5, -1). The vertex is the midpoint of (- 5, -1) and (1, -1). ` vertex = ^ - 2, -1 h Focal length a = 3 From the diagram the parabola curves to the right. The equation is in the form ^ y - k h 2 = 4a ] x - h g i.e.

2 7 y - ^ -1 h A = 4 ] 3 g 7 x - ^ - 2 h A ^ y + 1 h 2 = 12 ] x + 2 g

y 2 + 2y + 1 = 12x + 24 y 2 + 2y - 12x - 23 = 0

x


2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola y 2 + 12y - 4x - 8 = 0.

Solution Complete the square on y. y 2 + 12y - 4x - 8 = 0 y 2 + 12y = 4x + 8 y 2 + 12y + 36 = 4x + 8 + 36 ^ y + 6 h 2 = 4x + 44 = 4 ^ x + 11 h So the parabola has equation ^ y + 6 h2 = 4 ^ x + 11 h or 7 y - ] - 6 g A 2 = 4 6 x - ] -11 g @ . Its vertex has coordinates (-11, - 6). 4a = 4 ` a=1 The parabola turns to the right as it is in the form ^ y - k h 2 = 4a ^ x - h h . y x = -12 x (-11, -6) 1

(-10, -6)

1

Count 1 unit to the right for the focus ` focus = ^ -10, - 6 h . Count 1 unit to the left for the directrix ` directrix has equation x = -12.

PARABOLA (y – k)2 = – 4a(x – h) The parabola with vertex (h, k) and focal length a that turns to the left has equation ^ y - k h 2 = - 4a ^ x - h h

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Proof Find the equation of the parabola that turns to the left with vertex (h, k) and focal length a.

Counting a units to the left from the vertex V gives the focus F = ^ h - a, k h. Counting a units to the right from the vertex V gives the point on the directrix D = ^ h + a, k h. So the equation of the directrix is given by x = h + a. We find the equation of the locus of P(x, y) that is equidistant from point F ^ h - a, k h and line x = h + a. y x=h+a B

P (x, y)

x

F (h -a, k)

B has coordinates ^ h + a, y h . We want PF = PB PF 2 = PB 2 2 7x - ^h - ahA + ^y - kh2 = 7x - ^h + ahA + ^y - y h2 ^x - h + ah2+ ^y - kh2 = ^x - h - ah2 2

^y - kh2 = ^x - h - ah2- ^x - h + ah2 = 7^x - h - ah + ^x - h + ahA7^x - h - ah - ^x - h + ahA (difference of two squares) = ^ 2x - 2h h ^ - 2a h = - 4ax + 4ah = - 4a ^ x - h h = - 4a ^ y - k h


The parabola ^ y - k h 2 = - 4a ] x - h g has • axis parallel to the x-axis • vertex at (h, k) • focus at ^ h - a, k h • directrix with equation x = h + a

EXAMPLES 1. Find the equation of the parabola with focus (2, 1) and directrix x = 3.

Solution y x=3 1 1 2 2 1

(2, 1) 1

2

B

1

(2 2x, 1)

Coordinates of B are (3, 1). The vertex is the midpoint of (3, 1) and (2, 1). 1 ` vertex = c 2 , 1 m 2 1 Focal length a = 2 From the diagram the parabola curves to the left. The equation is in the form

^ y - k h 2 = - 4a ^ x - h h 1 1 i.e. ^y - 1h2 = -4c m cx - 2 m 2 2 1 2 ^y - 1h = -2cx - 2 m 2 y 2 - 2y + 1 = - 2x + 5 y 2 - 2y + 2x - 4 = 0 2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola y 2 + 4y + 8x - 4 = 0. CONTINUED

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Solution Complete the square on y. y 2 + 4y + 8x - 4 = 0 y 2 + 4y = - 8 x + 4 y 2 + 4y + 4 = - 8x + 4 + 4 ^ y + 2 h 2 = - 8x + 8 = -8 ]x - 1 g So the parabola has equation ^ y + 2 h2 = - 8 ] x - 1 g or 7 y - ] - 2 g A 2 = - 8 ] x - 1 g . Its vertex has coordinates ^ 1, - 2 h . 4a = 8 ` a=2 The parabola turns to the left as it is in the form ^ y - k h 2 = - 4a ^ x - h h y x=3

1 -3 -2 -1

1 -1

(-1, -2)

-22

2

3

x

(1, -2)

2

2

Count 2 units to the left for the focus ` focus = ^ -1, - 2 h . Count 2 units to the right for the directrix ` directrix has equation x = 3.

11.6 Exercises 1.

Complete the square on x to write each equation in the form ] x - h g2 = !4a ^ y - k h . (a) x 2 - 6x - 8y - 15 = 0 (b) x 2 - 10x - 4y + 1 = 0 (c) x 2 - 2x - 4y - 11 = 0 (d) x 2 - 8x + 12y - 20 = 0

(e) (f) (g) (h) (i) (j)

x 2 - 12x - 8y - 20 = 0 x 2 + 14x + 16y + 1 = 0 x 2 - 4x + 4y - 16 = 0 x 2 + 18x - 12y + 9 = 0 x 2 + 2x - 8y - 7 = 0 x 2 - 6 x + 4y + 1 = 0


2.

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) 3.

4.

(g) (h) (i) (j)

Complete the square on y to write each equation in the form ^ y - k h2 = !4a ] x - h g y 2 - 8y - 4x = 0 y 2 - 2y - 8x - 15 = 0 y 2 + 4y - 12x - 8 = 0 y 2 - 20y + 4x - 16 = 0 y 2 + 6y + 16x - 7 = 0 y 2 - 12y - 8x + 4 = 0 y 2 + 10y + 24x - 23 = 0 y 2 + 24y - 4x = 0 y 2 - 4y + 20x - 16 = 0 y 2 + 8y + 8x = 0

Find the equation of each parabola (a) focus ^ -1, 3 h, directrix y = - 1 (b) focus ^ - 4, 1 h, directrix y = -1 (c) focus (2, 0), directrix y = - 4 (d) focus (3, 6), directrix y = 2 (e) focus ^ - 2, 5 h, directrix y = -3 (f) focus ^ -1, - 4 h, directrix y = 4 (g) focus ( 4, - 3), directrix y = 7 (h) focus ^ - 5, 1 h, directrix y = 5 (i) focus ^ - 3, - 6 h, directrix y = 0 (j) focus ^ 0, -7 h, directrix y = - 5 (k) focus (2, 3), directrix x = - 4 (l) focus ^ -1, 4 h, directrix x = - 3 (m) focus (6, 0), directrix x = 2 (n) focus ( 3, - 2 ), directrix x = -5 (o) focus ^ 1, -1 h, directrix x = - 3 (p) focus ^ - 2, - 4 h, directrix x = 4 (q) focus (2, 1), directrix x = 4 (r) focus ^ - 5, 3 h, directrix x = 3 (s) focus ^ - 1, 2 h, directrix x = 0 (t) focus (3, 1), directrix x = 4 Find (i) the coordinates of the focus and (ii) the equation of the directrix of (a) x 2 - 6x - 4y - 3 = 0 (b) x 2 - 2x - 8y - 7 = 0 (c) x 2 + 4x - 4y = 0 (d) x 2 - 8x - 12y + 4 = 0 (e) x 2 + 10x - 8y + 1 = 0 (f) x 2 - 6x + 4y + 1 = 0

x2 x2 x2 x2

+ 2x + 8y - 15 = 0 - 4 x + 4y = 0 - 8x + 12y + 4 = 0 + 4x + 16y - 12 = 0

5.

Find (i) the coordinates of the focus and (ii) the equation of the directrix of (a) y 2 + 2y - 4x - 3 = 0 (b) y 2 - 8y - 12x + 4 = 0 (c) y 2 - 6y - 8x - 7 = 0 (d) y 2 + 4y - 16x - 12 = 0 (e) y 2 - 2y - 24x + 25 = 0 (f) y 2 + 10y + 8x + 1 = 0 (g) y 2 + 14y + 4x + 1 = 0 (h) y 2 - 12y + 20x - 4 = 0 (i) y 2 - 4y + 32x - 28 = 0 (j) y 2 + 6y + 40x + 29 = 0

6.

Find the equation of the parabola with vertex ^ 0, 3 h if it is concave upwards and a = 3.

7.

Find the equation of the parabola with vertex ^ - 2, -1 h, focal length 2, and axis parallel to the y-axis.

8.

A parabola has its vertex at ^ 1, - 5 h and its focal length as 1. If the parabola is concave upwards, find its equation.

9.

A parabola has its axis parallel to the x-axis. If its vertex has coordinates ^ 2, 6 h and a = 3, find its equation if it turns to the left.

10. Find the equation of the parabola with vertex at ^ 1, 0 h and focus at ^ 1, 4 h . 11. Find the equation of the parabola that has vertex ^ 1, 1 h and focus ^ 1, 8 h . 12. A parabola has its vertex at ^ 2, - 2 h and focus at ^ - 4, - 2 h . Find its equation.

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13. Find the equation of the parabola with vertex ^ 0, 3 h and focus ^ 8, 3 h . 14. Find the equation of the parabola with vertex ^ 3, 3 h and equation of directrix y = 5.

21. The latus rectum of a parabola has endpoints ^ - 2, 3 h and ^ 6, 3 h . Find two possible equations for the parabola. 22.

15. Find the equation of the parabola with vertex ^ 3, -1 h and directrix x = -1. 16. A parabola has directrix y = 5 and focus ^ - 3, 3 h . Find its equation. 17. Find the equation of the locus of a point moving so that it is equidistant from the point ^ 2, 2 h and the line y = - 4. 18. Find the equation of the parabola with focus ^ 2, -1 h and directrix x = 10. 19. Find the coordinates of the vertex and focus and the equation of the directrix for the parabola (a) x 2 + 4x - 8y + 12 = 0 (b) x 2 - 6x - 12y + 33 = 0 (c) x 2 - 2x + 4y + 5 = 0 (d) y 2 - 8y - 16x + 64 = 0 (e) y 2 + 4y - 24x + 4 = 0 (f) y 2 + 8x + 40 = 0. 20. For the parabola x 2 + 2x + 28y - 111 = 0, find the coordinates of its vertex and focus, and the equations of its directrix and axis. What is its maximum value?

(a) Find the equation of the arch above. (b) Find the coordinates of its focus and the equation of its directrix. 23. (a) Sketch y = x 2 + 2x - 8, showing intercepts and the minimum point. (b) Find the coordinates of the focus and the equation of the directrix of the parabola. 24. Find the equation of the parabola with vertex ^ - 2, 3 h that also passes through ^ 2, 1 h and is concave downwards. 25. A parabolic satellite dish has a diameter of 4 m at a depth of 0.4 m. Find the depth at which its diameter is 3.5 m, correct to 1 decimal place.

DID YOU KNOW? The word ‘directrix’ is due to the Dutch mathematician Jan De Witt (1629–72). He published a work called Elementa curvarum, in which he defined the properties of the parabola, ellipse, circle and hyperbola. These curves are all called conic sections.


De Witt was well known as the ‘Grand Pensionary of Holland’. He took part in the politics and wars of his time, opposing Louis XIV. When the French invaded Holland in 1672, De Witt was seized and killed.

Tangents and Normals Remember that the gradient of the tangent to a curve is given by the derivative. The normal to the curve is perpendicular to its tangent at that point. That is, m 1 m 2 = - 1 for perpendicular lines.

EXAMPLES 1. Find the gradient of the tangent to the parabola x 2 = 8y at the point ^ 4, 2 h .

Solution

`

x 2 = 8y x2 y= 8 dy 2x = 8 dx x = 4 CONTINUED

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dy

4 4 =1 So the gradient of the tangent at ^ 4, 2 h is 1. At ^ 4, 2 h,

dx

=

2. Find the equation of the normal to the parabola x 2 = 4y at the point ^ - 8, 16 h .

Solution x 2 = 4y

dy dx

At (- 8, 16):

x2 4 2x = 4 x = 2 -8 = 2 = -4 = - 4.

y=

So

dy dx

So the gradient of the tangent m 1

The normal is perpendicular to the tangent. ` m 1 m 2 = -1 So `

] - 4 g m 2 = -1

m2 =

1 4

The equation of the normal is given by y - y 1 = m (x - x 1) i.e.

1 [x - (- 8)] 4 1 = (x + 8 ) 4 4y - 64 = x + 8 0 = x - 4y + 72. y - 16 =

11.7 Exercises 1.

Find the gradient of the tangent to the parabola x 2 = 12y at the point where x = 2.

3.

Find the gradient of the normal to the parabola x 2 = 4y at the point where x = 2.

2.

Find the gradient of the tangent to the parabola x 2 = - 3y at the point ^ 6, -12 h .

4.

Find the gradient of the tangent to the parabola x 2 = 16y at the point ^ 4, 1 h .


5.

Show that the gradient of the tangent to the curve x 2 = 2y at any point is its x-coordinate.

6.

Find the equation of the tangent to the curve x 2 = 8y at the point ^ 4, 2 h .

7.

Find the equation of the normal to the curve x 2 = 4y at the point where x = - 4.

8.

Find the equations of the tangent and normal to the parabola x 2 = - 24y at the point ^ 12, - 6 h .

9.

Find the equations of the tangent and normal to the parabola x 2 = 16y at the point where x = 4.

10. Find the equation of the tangent to the curve x 2 = - 2y at the point ^ 4, - 8 h . This tangent meets the directrix at point M. Find the coordinates of M. 11. Find the equation of the normal to the curve x 2 = 12y at the point ^ 6, 3 h . This normal meets the parabola again at point P. Find the coordinates of P.

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12. The normal of the parabola x 2 = 18y at ^ - 6, 2 h cuts the parabola again at Q. Find the coordinates of Q. 13. Find the equations of the normals to the curve x 2 = - 8y at the 1 points ^ -16, - 32 h and c - 2, - m . 2 Find their point of intersection and show that this point lies on the parabola. 14. Find the equation of the tangent at ^ 8, 4 h on the parabola x 2 = 16y. This tangent meets the tangent at the vertex of the parabola at point R. Find the coordinates of R. 15. (a) Show that the point P _ 2p, p 2 i lies on the parabola x 2 = 4y. (b) Find the equation of the normal to the parabola at P. (c) Show that p 2 + 1 = 0 if the normal passes through the focus of the parabola ^ p ! 0 h .

Parametric Equations of the Parabola An equation involving x and y, for example x 2 = 4ay, is called a Cartesian equation. Equations can also be written in parametric form. In this form, x and y are both written in terms of a third variable called a parameter. An example of a Cartesian equation is y = x 2 - 1. An example of parametric equations is x = 2t + 3, y = t - 2. Any Cartesian equation can be written in parametric form.

The word ‘Cartesian’ comes from the name of the mathematician Descartes.

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EXAMPLE Write y = 3x + 1 in parametric form.

Solution There are many different ways this can be done. For example: Given parameter p (a) Let x = p Then y = 3x + 1 = 3p + 1 So parametric equations are x = p, y = 3p + 1. (b) Let x = p - 5 Then y = 3x + 1 = 3^ p - 5h + 1 = 3p - 15 + 1 = 3p - 14 So parametric equations are x = p - 5, y = 3p - 14.

There are many different ways to write parametric equations. Can you find some more for the example above? We can also change parametric equations back into Cartesian form.

EXAMPLES 1. Find the Cartesian equation of x = 3t + 1, y = 2t - 3.

Solution We use the process for solving simultaneous equations to eliminate the parameter. ]1 g x = 3t + 1 ]2 g y = 2t - 3 From (1) x - 1 = 3t x -1 =t 3 Substitute in (2) y = 2t - 3 x -1 m-3 3 3y = 2 ] x - 1 g - 9 = 2x - 2 - 9 = 2x - 11 0 = 2x - 3y - 11 = 2c


2. Find the Cartesian equation of x = 2q, y = q 2 - 3.

Solution x = 2q

]1 g

y=q -3

]2 g

2

From (1) x = 2q x =q 2 Substitute in (2) y = q2 - 3 x 2 =c m -3 2 x2 = -3 4 4y = x 2 - 12 0 = x 2 - 4y - 12

The equation of a parabola can be written as a set of parametric equations.

The parabola x 2 = 4ay can be written as x = 2at y = at 2 where t is a parameter.

Proof Substitute x = 2at into x 2 = 4ay

^ 2at h2 = 4ay 4a 2 t 2 = 4ay at 2 = y ` x = 2at and y = at 2 satisfy the equation x 2 = 4ay

Class Investigation 1. How would you write x 2 = - 4ay in parametric form? 2. How would you write y 2 = 4ax in parametric form? 3. How would you write y 2 = - 4ax in parametric form?

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The parabola x 2 = - 4ay can be written as x = 2at y = - at 2

Proof Substitute x = 2at into x 2 = - 4ay ^ 2at h2 = - 4ay 4a 2 t 2 = - 4ay at 2 = - y - at 2 = y ` x = 2at and y = - at 2 satisfy the equation x 2 = - 4ay.

The parabola y 2 = 4ax can be written as x = at 2 y = 2at

Proof Substitute y = 2at into y 2 ] 2at g2 4a 2 t 2 at 2

= 4ax = 4ax = 4ax =x

` x = at 2 and y = 2at satisfy the equation y 2 = 4ax.

The parabola y 2 = - 4ax can be written as x = - at 2 y = 2at

Proof Substitute y = 2at into y 2 ] 2at g2 4a 2 t 2 at 2 - at 2

= - 4ax = - 4ax = - 4ax = -x =x

` x = - at 2 and y = 2at satisfy the equation y 2 = - 4ax.


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EXAMPLES 1. Given the parabola x = 4t and y = 2t 2, find (a) its Cartesian equation (b) the points on the parabola when t = !2.

Solution (a) x = 4t x ` 4 =t Substitute into y = 2t 2: x 2 y = 2c m 4 2x 2 = 16 x2 = 8 8y = x 2 (b) When t = 2 x = 4^2h =8 y = 2 ^ 2 h2 =8 When t = - 2 x = 4^-2h = -8 y = 2 ^ - 2 h2 =8

When t = !k, the points are symmetrical about the y-axis.

` points are ^ 8, 8 h and ^ - 8, 8 h . 2. Find the coordinates of the focus and the equation of the directrix of the parabola x = -12t, y = - 6t 2 .

Solution Method 1: We can find the Cartesian equation. x = -12t

]1g ]2g

y = - 6t From ] 1 g x = -12t x =t -12 2

CONTINUED

632


Substitute in (2) y = - 6t 2 x 2 m -12 x2 = -6c m 144 x2 =24 - 24y = x 2 = -6c

This is in the form x 2 = - 4ay (concave downwards parabola with vertex at the origin). 4a = 24 a=6 So focal length is 6 units. y

y=6 6 x 6 (0, - 6)

Method 2: The equations x = -12t, y = - 6t 2 are in the form x = - 2at, y = - at 2 . ` a=6 The equations satisfy x 2 = - 4ay x2 = - 4 ] 6 g y = - 24y This is a concave downward parabola with focus ^ 0, - a h and directrix y = a. So focus = ^ 0, - 6 h and directrix has equation y = 6. 3. Write x 2 = 32y as a set of parametric equations.

Solution 4a = 32 So a = 8 Equations are in the form x = 2at, y = at 2 . So x = 2 ] 8 g t, y = 8t 2 x = 16t, y = 8t 2


4. Write y 2 = 12x in parametric form.

Solution 4a = 12 a=3 Equations are in the form x = at 2, y = 2at So x = 3t 2, y = 2 ] 3 g t x = 3t 2, y = 6t

11.8 Exercises 1.

Sketch the graph of (a) x = t - 2, y = t 2 (b) x = t - 2, y = 3t - 1 (c) x = 2t, y = 4t - 3 (d) x = t + 1, y = t 2 (e) x = 2t, y = 2t 2 - 3 (f) x = 6t, y = 3t 2

2.

Find the Cartesian equation of (a) x = 4t, y = 2t - 1 (b) x = t + 3, y = 2t - 5 (c) x = t - 1, y = t 2 + t t (d) x = , y = 4t 2 - 1 2 1 (e) x = , y = 2t t

3.

Write as a set of parametric equations (a) x 2 = 4y (b) x 2 = 12y (c) x 2 = - 8y (d) x 2 = 16y (e) x 2 = - 36y (f) x 2 = 20y (g) x 2 = - 6y (h) x 2 = y (i) 2x 2 = y (j) x 2 = -10y

4.

Find the Cartesian equation for each parabola (a) x = 8t, y = 4t 2 (b) x = 10t, y = 5t 2 (c) x = 2t, y = t 2 (d) x = -14t, y = - 7t 2 (e) x = 4t, y = - 2t 2 (f) x = 2at, y = at 2 (g) x = 2m, y = - m 2 (h) x = 12p, y = 6p 2 1 (i) x = - t, y = - t 2 2 (j) x = 2aq, y = aq 2

5.

(a) Show that _ 6t, - 3t 2 i lies on the parabola x 2 = -12y for all values of t. (b) Find the coordinates of point P where t = - 2. (c) Find the equation of the tangent to the parabola at P.

6.

(a) Find the coordinates of Q on the parabola x = 8t, y = 4t 2 at the point where t = - 1. (b) Find the equation of the normal to the parabola at Q.

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634


7.

A parabola has equations x = 4t 2, y = 8t. Find the coordinates of its focus and the equation of its directrix.

8.

Find the coordinates of point P on the parabola x = t 2, y = - 2t where t = 2. Find the equation of line PS where S is the focus of the parabola.

9.

(a) Find the Cartesian equation of the parabola x = 12t, y = 6t 2 . 3 m lies on the 8 parabola. What is the value of t at this point?

(b) The point c 3,

10. Find the equation of the tangent to the parabola x = 4t, y = 2t 2 at the point where t = 3.

Chords, Tangents and Normals If P(2ap, ap2) and Q(2aq, aq2) are any two points on the parabola x 2 = 4ay, p+q then the chord PQ has gradient 2 and equation y -

1 ^ p + q h x + apq = 0 2

Proof y2 - y1 m= x -x 2 1 = = =

ap 2 - aq 2 2ap - 2aq a _ p2 - q2 i 2a ^ p - q h a ^ p + qh^ p - qh

2a ^ p - q h p+q = 2

Learn to derive these equations rather than memorise them.

The equation formula is y - y 1 = m (x - x 1) p+q ` y - aq 2 = (x - 2aq) 2 1 = (p + q) x - aq (p + q) 2 1 = (p + q) x - apq - aq 2 2 1 ` y - (p + q) x + apq = 0 2


If PQ is a focal chord, then pq = - 1

Proof x 2 = 4ay has focus (0, a). 1 PQ has equation y - ^ p + q h x + apq = 0. 2 For PQ to be a focal chord, it passes through (0, a). 1 i.e. a - (p + q) $ 0 + apq = 0 2 a + apq = 0 apq = - a pq = -1 The tangent to the parabola x 2 = 4ay at the point P(2ap, ap2) has gradient p and equation given by y - px + ap 2 = 0

Proof x 2 = 4ay x2 ` y= 4a dy 2x = 4 a dx x = 2a At P ^ 2ap, ap 2 h dy 2ap = 2a dx =p The equation formula is y - y1 = m _ x - x1 i `

y - ap 2 = p ^ x - 2ap h = px - 2ap 2 ` y - px + ap 2 = 0

The tangents to the parabola x 2 = 4ay at points P _ 2ap, ap 2 i and Q _ 2aq, aq 2 i intersect at the point [a ^ p + q h , apq @

Proof Equation of tangent at P is Equation of tangent at Q is

y - px + ap 2 = 0 y - qx + aq 2 = 0

(1) ( 2)

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636


(1) – (2). - px + qx + ap 2 - aq 2 = 0 x ( q - p ) = a (q 2 - p 2 ) = a (q + p) (q - p) x = a (q + p) Substitute in (1): y - p $ a (q + p) + ap 2 = 0 y - apq - ap 2 + ap 2 = 0 y = apq ` point of intersection is [a ^ p + q h , apq @ 1 The normal to the curve x 2 = 4ay at point P _ 2ap, ap 2 i has gradient p and equation given by x + py = ap 3 + 2ap

Proof Tangent at P has gradient p. For perpendicular lines, m 1 m 2 = - 1 1 ` normal has gradient - . p The equation formula is y - y1 = m _ x - x1 i 1 ` y - ap 2 = - ^ x - 2ap h p p (y - ap 2) = - (x - 2ap) py - ap 3 = - x + 2ap x + py = ap 3 + 2ap

The normals to the parabola x 2 = 4ay at P _ 2ap, ap 2 i and Q _ 2aq, aq 2 i intersect at [- apq ^ p + q h, a _ p 2 + pq + q 2 + 2 i]

Proof Equation of normal at P is Equation of normal at Q is

x + py = ap 3 + 2ap x + qy = aq 3 + 2aq

(1) - (2):

py - qy = ap 3 - aq 3 + 2ap - 2aq

( 1) (2 )

y ^ p - q h = a _ p 3 - q 3 i + 2a ^ p - q h = a ^ p - q h _ p 2 + pq + q 2 i + 2a ^ p - q h y = a _ p 2 + pq + q 2 i + 2a = a _ p 2 + pq + q 2 + 2 i


Substitute in (1): x + p $ a (p 2 + pq + q 2 + 2) = ap 3 + 2ap x + ap 3 + ap 2 q + apq 2 + 2ap = ap 3 + 2ap x = - ap 2 q - apq 2 = - apq (p + q)

` point of intersection is [- apq ^ p + q h, a _ p 2 + pq + q 2 + 2 i] EXAMPLES 1. Find the equation of the chord joining points where t = 3 and t = –2 on the parabola x = 2at, y = at 2 .

Solution When t = 3 x = 2a ^ 3 h = 6a When t = –2 x = 2a ^ - 2 h x = - 4a

y = a ^ 3 h2 = 9a y = a ^ - 2 h2 = 4a

` points are (6a, 9a) and ^ –4a, 4a h y2 - y1 Gradient m = x - x 2 1 4a - 9a = - 4a - 6a - 5a = -10a 1 = 2 The equation formula is y - y1 = m _ x - x1 i 1 ` y - 4a = ^ x + 4a h 2 2y - 8a = x + 4a 0 = x - 2y + 12a 2. Find the equation of the tangent to the parabola x 2 = 8y at the point _ 4t, 2t 2 i .

Solution x 2 = 8y x2 ` y= 8

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638


dy dx

2x 8 x = 4 =

At (4t, 2t2) dy 4t = 4 dx =t The equation formula is y - y1 = m _ x - x1 i ` y - 2t 2 = t ^ x - 4 t h = tx - 4t 2 ` 0 = tx - y - 2t 2

The equations of the tangent, normal and chord can also be derived from points in Cartesian form rather than parametric form.

If point A(x1, y1) lies on the parabola x 2 = 4ay, then the equation of the tangent at A is given by xx 1 = 2a _ y + y 1 i

Proof x2 4a dy 2x = 4a dx x = 2a y=

At (x1, y1) dy dx

=

x1 2a

The equation formula is y - y1 = m ^ x - x1 h x1 = ^ x - x1 h 2a 2a (y - y 1) = x 1 ^ x - x 1 h 2ay - 2ay 1 = xx 1 - x 12 = xx 1 - 4ay 1 2ay + 2ay 1 = xx 1 2a (y + y 1) = xx 1

(since x 12 = 4ay 1)


If point A (x1, y1) lies on the parabola x 2 = 4ay, then the equation of the normal at A is given by 2a y - y1 = - x _ x - x1 i 1

Proof x2 4a dy 2x = 4a dx x = 2a y=

At (x1, y1) dy dx

=

x1 2a

For normal, m 1 m 2 = –1 2a m2 = - x ` 1 The equation formula is y - y1 = m _ x - x1 i 2a ` y - y1 = - x ^ x - x1 h 1

The equation of the chord of contact XY of tangents drawn from external point (x1, y1) to the parabola x 2 = 4ay is given by xx 1 = 2a _ y + y 1 i

Proof

Let X = _ 2ap, ap 2 i and Y = _ 2aq, aq 2 i . Then chord XY has equation y -

1 ^ p + q h x + apq = 0. 2

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640


Now the tangents at X and Y intersect at P _ x 1, y 1 i . But the intersection of tangents is [a ^ p + q h , apq @ ` x1 = a ^ p + q h and y 1 = apq From (1): x1 p+q= a

(1) (2)

(3)

Substituting (2) and (3) into the equation of chord XY gives 1 d x1 n x + y1 = 0 2 a 2ay - x 1 x + 2ay 1 = 0 2a (y + y 1) - x 1 x = 0 2a (y + y 1) = x 1 x y-

EXAMPLE 1 1 Tangents are drawn from the point c , - m to the points P and Q on the 2 2 parabola x 2 = 4y. Find the equation of the chord of contact PQ and the coordinates of P and Q.

Solution x 2 = 4y ` 4a = 4 a=1 PQ has equation xx 1 = 2a _ y + y 1 i where x 1 =

1 1 and y 1 = - . 2 2

1 1 x = 2cy - m 2 2 = 2y - 1 x = 4y - 2 ` x - 4y + 2 = 0 is the equation of the chord of contact. To find P and Q, solve simultaneous equations. x 2 = 4y x - 4y + 2 = 0 From (2): x + 2 = 4y Substitute into (1): x2 = x + 2 x -x-2=0 (x - 2) (x + 1) = 0 ` x = 2, -1 2

(1) (2) ( 3)


Substitute x = 2 into (3): 2 - 4y + 2 = 0 4 = 4y 1=y Substitute x = –1 into (3): -1 - 4y + 2 = 0 1 = 4y 1 =y 4 So P and Q are points (2, 1) and c -1,

1 m. 4

11.9 Exercises 1.

Find the (i) gradient and (ii) equation of chord AB on the parabola (a) x 2 = 16y where A = _ 8t, 4t 2 i and B = _ 8n , 4n 2 i (b) x 2 = 8y where A = _ 4p, 2p 2 i and B = _ 4q , 2q 2 i (c) x 2 = 12y where A = _ 6m, 3m 2 i and B = _ 6n , 3n 2 i (d) x 2 = 20y where A = _ 10p, 5p 2 i and B = _ 10q, 5q 2 i (e) x 2 = 4y where A = _ 2a, a 2 i and B = _ 2b, b 2 i (f) x 2 = - 8y where A = _ 4p, - 2p 2 i and B = _ 4 q , - 2q 2 i (g) x 2 = - 24y where A = _ 12a, - 6a 2 i and B = _ 12b, - 6b 2 i (h) x 2 = -16y where A = _ - 8p, - 4p 2 i and B = _ - 8q, - 4 q 2 i (i) x 2 = - 4y where A = _ 2s, - s 2 i and B = _ 2t , - t 2 i

(j) x 2 = - 28y where A = _ - 14p, -7p 2 i and B = _ -14q, -7q 2 i 2.

Find (i) the gradient of the tangent, (ii) the gradient of the normal, (iii) the equation of the tangent and (iv) the equation of the normal to the curve (a) x 2 = 4y at the point (2p, p2) (b) x 2 = 12y at the point (6q, 3q2) (c) x 2 = 8y at the point (4t, 2t2) (d) x 2 = 20y at the point (10n, 5n2) (e) x 2 = 24y at the point (12p, 6p2) (f) x 2 = -16y at the point (8k, - 4k2) (g) x 2 = - 4y at the point (- 2q, - q 2) (h) x 2 = - 8y at the point (4t, - 2t 2) (i) x 2 = - 12y at the point (- 6m, - 3m 2) (j) x 2 = - 32y at the point (16a, - 8a2)

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642


3.

4.

5.

Find the point of intersection between the (i) tangents and (ii) normals to the curve (a) x 2 = 4y at the points (2p, p2) and (2q, q2) (b) x 2 = 16y at the points (8p, 4p2) and (8q, 4q2) (c) x 2 = 8y at the points (4a, 2a2) and (4b, 2b2) (d) x 2 = 12y at the points (6s, 3s2) and (6t, 3t2) (e) x 2 = 20y at the points (10t, 5t2) and (10w, 5w2) (f) x 2 = - 24y at the points (12p, - 6p2) and (12q, - 6q2) (g) x 2 = - 16y at the points (8m, - 4m2) and (8n, - 4n2) (h) x 2 = - 40y at the points (20p, -10p2) and (20q, -10q2) (i) x 2 = - 20y at the points (10h, - 5h2) and (10k, - 5k2) (j) x 2 = - 12y at the points (- 6p, - 3p2) and (- 6q, - 3q2) Find the equation of the (i) tangent and (ii) normal at the point (x1, y1) to the parabola (a) x 2 = 8y (b) x 2 = 12y (c) x 2 = 16y (d) x 2 = 4y (e) x 2 = 20y (f) x 2 = - 4y (g) x 2 = - 8y (h) x 2 = - 24y (i) x 2 = - 44y (j) x 2 = - 28y Find the equation of the chord of contact AB of tangents drawn from an external point (x1, y1) to the parabola (a) x 2 = 16y (b) x 2 = 4y (c) x 2 = 8y

(d) (e) (f) (g) (h) (i) (j)

x 2 = 12y x 2 = 20y x 2 = - 4y x 2 = - 24y x 2 = - 8y x 2 = -16y x 2 = - 36y

6.

Derive the equation of the tangent to the curve x 2 = 4ay at the point (a) _ 2ap, ap 2 i (b) _ x 0, y 0 i

7.

Find the equation of chord XY on the parabola x 2 = 8y where X = _ 4t, 2t 2 i and Y = _ 4r, 2 r 2 i .

8.

Find the equation of chord PQ on the parabola x = 6t, y = 3t 2, given that t = 2 at P and t = –3 at Q.

9.

Show that the equation of the normal to the parabola x 2 = -18y at the point d - 9t, -

9t 2 n is given 2

by 2x + 2ty + 9t 3 + 18t = 0. 10. Derive the equation of the normal to the parabola x 2 = 4ay at the point _ 2at, at 2 i . 11. Find the equation of the chord of contact of tangents drawn from the external point ^ 3, -1 h to the parabola x 2 = 8y. 12. Show that 3x + 4y + 4 = 0 is a focal chord of the parabola x 2 = - 4 y. 13. Show that if PQ is a focal chord of x 2 = 4ay where P is the point _ 2ap, ap 2 i and Q is the point _ 2aq, aq 2 i then pq = -1. 14. Find the point of intersection of the tangents to the curve 1 x 2 = 12y at ^ - 6, 3 h and c 2, m . 3


15. Show that the tangents to the curve x 2 = 4ay at P _ 2ap, ap 2 i and Q _ 2aq, aq 2 i intersect at the point [a ^ p + q h , apq @ . 16. (a) Find the equation of the chord joining P ^ –8, 8 h and 1 Q c 2, m where P and Q are points 2 on the parabola x 2 = 8y. (b) Show that PQ is a focal chord. 17. Points P _ 2ap, ap 2 i and Q _ 2aq, aq 2 i lie on the parabola x 2 = 4ay. (a) Show that the normal at P is given by x + py = ap 3 + 2ap. (b) Find the point N where this normal meets the axis of the parabola. 18. Point M ^ 4, - 8 h lies on the parabola x 2 = - 2y. (a) Find the equation of the focal chord through M. (b) Find point N where this chord cuts the parabola again. 19. Tangents are drawn from an external point P ^ - 2, -1 h to the parabola x 2 = 12y. (a) Find the equation of the chord of contact of the tangents. (b) Find the coordinates of the points where the tangents meet the parabola. 20. (a) Find the coordinates of the focus F of the parabola x = 12t, y = 6t 2 .

(b) Find the equation of the focal chord PF where P is the point 1 c 6, 1 m on the parabola. 2 (c) Find Q where this chord cuts the parabola again. (d) Find the equations of the tangents to the parabola at P and Q. (e) Prove that the tangents are perpendicular. (f) Find the point of intersection R of these tangents. (g) Show that R lies on the directrix. 21. The chord of contact of two tangents drawn from an external point P to the parabola x 2 = 8y has equation x + 2y - 3 = 0. Find the coordinates of P. 22. Find the equation of the normal to the parabola x = 6t, y = 3t 2 at the point where t = - 1. 23. Prove that the tangents at the end of a focal chord are perpendicular. 24. Show that the tangents at the ends of a focal chord intersect on the directrix. 25. Show that the equation of the tangent at the point P _ x 0, y 0 i on the parabola x 2 = 4ay is given by xx 0 = 2a _ y + y 0 i .

Properties of the Parabola

The tangents at the end of a focal chord intersect at right angles on the directrix.

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644


Proof Let PQ be a focal chord of x 2 = 4ay where P = _ 2ap, ap 2 i and Q = _ 2aq, aq 2 i . Then pq = -1 Tangent at P has gradient m 1 = p Tangent at Q has gradient m 2 = q pq = -1 i.e. m 1 m 2 = -1 ` the tangents are perpendicular Tangents intersect at [a ^ p + q h , apq @ i.e. y = apq But pq = - 1 ` y = -a This is the equation of the directrix. ` tangents intersect on the directrix

EXAMPLE 1 m and Q ^ - 8, - 8 h lie on the parabola x 2 = - 8y. 2 (a) Find the equation of line PQ. (b) Show that PQ is a focal chord. (c) Prove that the tangents at P and Q intersect at right angles on the directrix.

Points P c 2, -

Solution (a) Equation of PQ y - y1 y2 - y1 x - x1 = x2 - x1 1 - +8 y+8 2 = x+8 2+8 3 = 4 4y + 32 = 3x + 24 0 = 3x - 4y - 8

(1)


(b)

x 2 = - 8y ` 4a = 8 a=2 Focus = (0, - 2)

Substitute (0, - 2) into (1) RHS = 3 (0) - 4 (- 2) - 8 =0 = LHS ` PQ is a focal chord (c)

x2 8 dy x =4 dx dy 2 =At P, 4 dx 1 =2 y=-

The equation formula is y - y 1 = m (x - x 1) 1 1 y + = - ( x - 2) 2 2 2y + 1 = - x + 2 ]1 g

x + 2y - 1 = 0 At Q ,

dy dx

=-

]-8g 4

=2

The equation formula is y - y 1 = m (x - x 1) y + 8 = 2 (x + 8 ) = 2x + 16 ]2 g

0 = 2x - y + 8 P has gradient m 1 = -

1 2

Q has gradient m 2 = 2 1 #2 2 = -1

m1 m2 = -

` the tangents are perpendicular Solve simultaneous equations to find the point of intersection. x + 2y - 1 = 0 2x - y + 8 = 0 (1) # 2: 2x + 4y - 2 = 0 (2) - (3): - 5y + 10 = 0 CONTINUED

]1 g ]2 g

]3 g

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646


10 = 5y 2=y Substitute in (1)

x+4-1=0 x = -3

` point of intersection is ^ - 3, 2 h The directrix has equation y = a i.e. y=2 The point ^ - 3, 2 h lies on the line y = 2 ` the tangents meet on the directrix.

The tangent at point P on a parabola is equally inclined to the axis of the parabola and the focal chord through P.

Proof

FP = PR ^ definition of a parabola h PR = PS + SR = ap 2 + a ` FP = ap 2 + a Tangent PQ has equation y - px + ap 2 = 0


At Q , x = 0 y - 0 + ap 2 = 0 y = - ap 2 `

Q = _ 0, - ap 2 i FQ = FO + OQ = a + ap 2 = FP

` +FQP = +FPQ ] base+s of isosceles D g ` tangent is equally inclined to the axis and the focal chord.

Application This property of the parabola—that is, that the tangent at P is equally inclined to the axis of the parabola and the focal chord through P—is used in many practical applications, including telescopes, headlights and radar.

Class Investigation Explore the use of the parabola in everyday life. You could go on an excursion to the Observatory, the physics section of a university, an optics manufacturer, an engineering company or a camera manufacturer. Write about the use of the parabola in any of the above applications, or any others you can think of.

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Locus Problems

EXAMPLES 1. Find the locus of the midpoints of the chords in the parabola x 2 = 4ay that pass through (0, 2).

Solution Equation of chord PQ where P = _ 2ap, ap 2 i and Q = _ 2aq, aq 2 i is given by 1 y - ^ p + q h x + apq = 0. 2 If PQ passes through (0, 2): 2-

1 (p + q) 0 + apq = 0 2 apq = - 2

]1 g

For midpoint M(x, y) 2ap + 2aq 2 = a ^ p + qh x `p + q = a ap 2 + aq 2 y= 2 1 = a _ p2 + q2 i 2 1 = a 7 ^ p + q h2 - 2pq A 2 Substitute (2) into (3) x=

]2 g

]3 g

1 ; b x l2 a a - 2pq E 2 x2 2y = a - 2apq x2 = a +4 2ay = x 2 + 4a x 2 = 2ay - 4a = 2a ( y - 2) y=

` locus is a parabola with vertex (0, 2) and focal length

6 from ] 1 g @

a . 2

2. Points P(2ap, ap2) and Q(2aq, aq2) lie on the parabola x 2 = 4ay and chord PQ passes through ^ 0, - 4a h. Find the locus of the intersection of the normals drawn from P and Q.


649

Solution 1 ^ p + q h x + apq = 0. 2 PQ passes through ^ 0, - 4a h 1 ` - 4a - (p + q) 0 + apq = 0 2 apq = 4a pq = 4 PQ has equation y -

( 1)

Normals intersect at [- apq ^ p + q h, a ^ p 2 + pq + q 2 + 2 h] . i.e.

x = - apq ^ p + q h = - 4a ^ p + q h

[from (1)]

x =p+q - 4a y = a ^ p 2 + pq + q 2 + 2 h = a ^ p2 + 4 + q2 + 2 h

(2)

= a _ p2 + q2 + 6 i y 2 2 a =p +q +6 = ^ p + q h2 - 2pq + 6 = ^ p + q h2 - 8 + 6 = ^ p + q h2 - 2 x2 = -2 16a 2 y x2 a + 2 = 16a 2 16ay + 32a 2 = x 2

[from (2)]

This is the equation of a parabola with vertex (0, - 2a) and focal length 4a.

16a ^ y + 2a h = x 2

11.10 Exercises 1.

2.

(a) Find the equation of the focal chord PF on the parabola x 2 = 8y where P = ^ - 8, 8 h and F is the focus. (b) Find the coordinates of Q where the focal chord intersects the parabola again. (c) Find the point of intersection of the tangents at P and Q. (d) Show that the tangents at P and Q are perpendicular. (a) Find the equation of the tangent to the parabola x 2 = 4y at the point P(2p, p2).

(b) Find the length of PF where F is the focus. (c) Show that PF = FR where R is the y-intercept of the tangent. 3.

(a) Find the equation of the tangent to the parabola x 2 = 12y at the point T(6t, 3t 2). (b) Find the coordinates of Y, the y-intercept of the tangent. (c) Show that TF = FY where F is the focus.

4.

(a) Find the equation of the tangent to the parabola x 2 = - 20y at the point Q(10q, - 5q2).

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(b) Find the coordinates of R, the y-intercept of the tangent. (c) Show that +FQR = +FRQ where F is the focus. 5.

6.

7.

8.

9.

(a) Find the equation of chord AB on the parabola x 2 = 12y where 1 A = c 2, m and B = ^ - 18, 27 h . 3 (b) Show that AB is a focal chord. (c) Show that the tangents at A and B are perpendicular. (d) Show that the tangents intersect on the directrix. Find the equation of the locus of the midpoint M of all chords PQ where P(2ap, ap2) and Q(2aq, aq2) lie on the parabola x 2 = 4ay and PQ passes through (0, 2a). Find the equations of the tangents to the curve x 2 = 8y at the points P(4p, 2p2) and Q(4q, 2q2). Find the equation of the locus of their point of intersection if PQ is a focal chord. Find the equation of the locus of point R that is the intersection of the normals at P(2p, p2) and Q(2q, q2) on the parabola x 2 = 4y, given that pq = - 4. The chord PQ is a focal chord of the parabola x 2 = 4ay where P = ^ 2ap, ap 2 h and Q = ^ 2aq, aq 2 h . Find the equation of the locus of the midpoint of PQ.

10. Tangents to the parabola x 2 = 4ay drawn from points P(2ap, ap2) and Q(2aq, aq2) intersect at right angles at point R. Find the equation of the locus of (a) point R (b) the midpoint of PQ.

11. The normal at any point P _ - 8p, - 4p 2 i on the parabola x 2 = -16y cuts the y-axis at point M. Find the equation of the locus of the midpoint of PM. 12. Given that P(2ap, ap2) and Q(2aq, aq2) lie on the parabola x 2 = 4ay, chord PQ subtends a right angle at the origin.

(a) Show pq = –4. (b) Find the equation of the locus of the midpoint of PQ. (c) Show that this locus is a parabola, and find its vertex and focal length. 13. Find the locus of the midpoint of PF where P is the point (2ap, ap2) on the parabola x 2 = 4ay and F is its focus. 14. (a) Find the point of intersection T of the tangents at P(2ap, ap2) and Q(2aq, aq2) on the parabola x 2 = 4ay. (b) Given that PQ passes through (0, 6a), find the equation of the locus of T. 15. Normals to the parabola x = 2at, y = at 2 from points P(2ap, ap2) and Q(2aq, aq2) intersect at N. Find the equation of the locus of N if PQ passes through the point (0, 3a).


Class Investigation Can you spot 6 mistakes in the solution to this question? Find the equation of the normal to the parabola x 2 = 4ay at the point P(2ap, ap2). SOLUTION x 2 = 4ay x2 y= ` 4a dy x =4a dx dy ap 2 At P, = 4a dx p2 m1 = ` 4 For normal, m 1 m 2 = - 1 p2 m = -1 4 2 4 m2 = 2 p if y - y 1 = m _ x - x 1 i 4 y - 2ap = 2 _ x - ap 2 i p p 2 y - 2ap 2 = 4 _ x - ap 2 i = 4x - 4ap 2 ` p 2 y = 4x + 2ap 2 i.e.

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Test Yourself 11 1.

Find the equation of the locus of a point moving so that it is equidistant from A ^ - 1, 2 h and B ^ 3, 5 h.

2.

Find the equation of the parabola with focus ^ 2, 1 h and directrix y = - 3.

3.

Find the radius and centre of the circle x 2 - 6x + y 2 - 2y - 6 = 0.

4.

Find the coordinates of (a) the vertex and (b) the focus of the parabola (y + 3) 2 = 12 ] x - 1 g .

5.

(a) Find the coordinates of P on the parabola x = 4t, y = 2t 2, where t = 2. (b) Find the equation of the tangent at P.

6.

Find the equation of the locus of a point that is always 5 units from the origin.

7.

Find (a) the equation of the directrix and (b) the coordinates of the focus of the parabola x 2 = - 8y.

8.

A point P ^ x, y h moves so that AP and BP are perpendicular, given A = ^ 3, 2 h and B = ^ - 4, 1 h . Find the equation of the locus of P.

9.

Point P ^ x, y h is equidistant from the point A ^ 4, - 2 h and the line y = 6. Find the equation of the locus.

10. Find (a) the coordinates of the (i) vertex and (ii) focus and (b) the equation of the directrix of the parabola x 2 - 2x - 4y + 5 = 0. 11. Find the equation of the tangent to the parabola x 2 = 18y at the point ^ - 6, 2 h .

12. Find the length of the diameter of the circle x 2 + 8x + y 2 - 12y + 3 = 0. 13. Find the equation of the parabola with directrix x = 6 and focus ^ - 6, 0 h . 14. A parabola has a focus at ^ 0, 4 h and its vertex is at ^ 0, 2 h. Find the equation of the parabola. 15. Find the equation of the locus of a point that is always 3 units from the line 4x - 3y - 1 = 0 . 16. A point is equidistant from the x- and y-axis. Find the equation of its locus. 17. Find the equation of the parabola with vertex at the origin, axis y = 0 and 1 passing through the point c 1 , 5 m . 4 18. Find the gradient of (a) the tangent and (b) the normal to the parabola x 2 = - 12y at the point where x = 3. 19. Find the Cartesian equation of (a) x = 6t, y = 3t 2 (b) x = - 8t 2, y = - 16t. 20. (a) Find the equation of the normal to the parabola x 2 = 4y at the point ^ - 8, 16 h . (b) This normal cuts the parabola again at Q. Find the coordinates of Q. 21. Show that 7x - 3y + 12 = 0 is a focal chord of the parabola x 2 = 16y. 22. Find the point of intersection of the normals to the parabola x 2 = -12y at the 1 1 points c 4, -1 m and c - 2, - m . 3 3


Q(2aq, aq2) are points on the parabola x 2 = 4ay. (b) If PQ is a focal chord show that pq = - 1.

23. Find the equation of the chord PQ on the parabola x = 4t, y = 2t 2 if t = 5 at P and t = –2 at Q. 24. Points P (10p, 5p2) and Q (10q, 5q2) lie on the parabola x 2 = 20y. Find the equation of the locus of the midpoint of PQ if pq = - 2. 25. Find the equation of the tangent to the parabola x = 2at, y = at 2 the point where t = 3. 26. (a) Find the equation of the tangent to the parabola x 2 = 12y at the point P (6, 3). (b) Find R, the y-intercept of the tangent. (c) Show that FP = FR where F is the focus. 27. (a) Find the equation of the chord PQ given that P(2ap, ap2) and

28. Find the equation of the normal to the parabola x = 8t, y = 4t 2 at the point where t = - 2. 29. Tangents are drawn from an external point P ^ 2, - 3 h to the parabola x 2 = 4y. (a) Find the equation of the chord of contact of the tangents. (b) Find the coordinates of the points at which each tangent meets the parabola. 30. Chord PQ is a focal chord of x 2 = 4ay where P = ^ 2ap, ap 2 h and Q = ^ 2aq, aq 2 h . Find the equation of the locus of the points of intersection of the tangents at P and Q.


(a) Find the equation of the locus of point P, which is equidistant from fixed points A ^ 3, 5 h and B ^ -1, 2 h . (b) Show that this locus is the perpendicular bisector of line AB.

2.

(a) Find the equation of the circle with centre ^ 1, 3 h and radius 5 units. (b) Show that the circle cuts the x-axis at the points ^ 5, 0 h and ^ - 3, 0 h .

3.

4.

5.

(a) Find the equation of the normals to the parabola x 2 = 8y at the points 1 M c - 2, m and N ^ 8, 8 h . 2 (b) Show that these normals are perpendicular. (c) Find the point of intersection X of the normals. (d) Find the equation of line MN and show that it is a focal chord.

Write in Cartesian form the equation x = sin i, y = cos 2i.

6.

From which point on the parabola x 2 = 4ay does the normal pass through the focus?

The line with equation 5x - 12y + 36 = 0 is a chord of the parabola x 2 = 12y. Find the point of intersection of the tangents to the parabola from the endpoints of the chord.

7.

(a) Find the equation of the tangents to the parabola x 2 = 4y at the points 1 A c 1, m and B ^ - 4, 4 h . 4 (b) Show that the point of intersection of these tangents lies on the directrix.

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8.

Find the equation of the parabola with axis parallel to the y-axis and passing through points ^ 0, - 2 h, ^ 1, 0 h and ^ 3, - 8 h .

9.

Find the equation of the straight line through the centres of the circles with equations x 2 + 4x + y 2 - 8y - 5 = 0 and x 2 - 2x + y 2 + 10y + 10 = 0.

10. Sketch the region x 2 + 2x + y 2 - 4y - 4 # 0. 11. (a) Find the equation of the locus of a point P moving so that PA is perpendicular to PB where A = ^ - 4, 3 h and B = ^ 0, 7 h . (b) Show that this locus is a circle with centre ^ - 2, 5 h and radius 2 2 . 12. Find the exact gradient, with rational denominator, of the normal to the parabola y 2 = 12x at the point where x = 4 in the first quadrant. 13. (a) Find the equation of the parabola with vertex ^ 3, - 2 h and focus ^ 7, - 2 h . (b) Find the equation of the tangent to the parabola at the point where x = 4 in the first quadrant. 14. Find the exact length of the line from ^ 2, 7 h to the centre of the circle x 2 + 4x + y 2 - 6y - 3 = 0 . 15. Find the equation of the locus of midpoints of all chords of length 2 units in the circle with equation x 2 + y 2 - 2 y - 3 = 0.

16. A satellite dish is to be 3.5 m wide and 1.1 m deep. Find the position of the focus in millimetres, correct to the nearest millimetre. 1.1 m 3.5 m

17. Find the equation of the locus of point P that moves such that the distance from P to the lines 3x - 4y + 1 = 0 and 12x + 5y + 3 = 0 is in the ratio 3:1. 18. PQ is a chord of x 2 = 4ay where P = (2ap, ap 2) and Q = _ 2aq, aq 2 i . (a) Find the coordinates of point N that divides PQ in the ratio 2:3. (b) Find the locus of the midpoint of PQ if pq = 2. 19. The chord of contact of the tangents to the parabola x 2 = 4ay from an external point R(x1, y1) passes through the point N(0, 2a). Find the equation of the locus of the midpoint of RN. 20. (a) Find the coordinates of T where T is the point of intersection of the tangents at the points t = - 2 and t = 5 on the parabola x = 4t, y = 2t 2. (b) Find the coordinates of P where P is the point of intersection of the tangents at the points X(2at, at2) and Y(2as, as2) on the parabola x 2 = 4ay. (c) The tangents from X and Y meet at s-1 s+1 45c. Show that t = or t = . s+1 1-s

12

Polynomials 1 TERMINOLOGY Coefficient: A constant multiplied by a pronumeral in an algebraic term e.g. in ax3 the a is the coefficient

Monic polynomial: A polynomial where the leading coefficient is 1

Degree: The value of the highest power of x in a polynomial

Polynomial: A sum or difference of terms involving integral powers of a variable, usually x. A function of the form P (x) = a0 + a1 x + a2 x 2 + f + an x n where a0, a1, ... are real numbers and n is a positive integer or zero

Dividend: The number, algebraic expression or polynomial that is being divided by another of the same type Divisor: A number, algebraic expression or polynomial that divides another of the same type Factor theorem: If P(x) is divided by x - a and P (a) = 0 then x - a is a factor of P(x) Leading term: The term with the highest power of x. e.g. 5x 3 - 2x 2 + 3 has a leading term of 5x 3 Long division: A division of one polynomial into another polynomial using a method similar to long division of numbers

Quotient: The result when two numbers, algebraic expressions or polynomials are divided Remainder theorem: If P(x) is divided by x - a then the remainder is given by P(a) Root of a polynomial equation: The solution of polynomial equation P (x) = 0. Graphically it is where the polynomial crosses the x-axis. Zeros: The zeros of a polynomial are the roots of the polynomial equation P (x) = 0. They are the values that make P(x) zero.

Chapter 12 Polynomials 1

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INTRODUCTION POLYNOMIALS ARE AN IMPORTANT part of algebra and are used in many branches of mathematics. Some examples of polynomials that you have already studied are linear and quadratic functions. In this chapter you will study some properties of polynomials in general, and relate polynomial expressions to equations and graphs.

DID YOU KNOW? The word ‘polynomial’ means an expression with many terms. (A binomial has 2 terms and a trinomial has 3 terms). ‘Poly’ means ‘many’, and is used in many words, for example, polyanthus, polygamy, polyglot, polygon, polyhedron, polymer, polyphonic, polypod and polytechnic. Do you know what all these words mean? Do you know any others with ‘poly-’?

Definition of a Polynomial A polynomial is a function defined for all real x involving positive powers of x in the form:

P ] x g = p 0 + p 1 x + p 2 x 2 + f + p n - 1 x n - 1 + p n x n where n is a positive integer or zero.

P(x) is a continuous and differentiable function. Although the definition has the term pnxn last, we generally write polynomials from the highest order down to the lowest. e.g. f ] x g = x 2 - 5x + 4. We can describe various aspects of polynomial as follows:

p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 is called a polynomial expression P ] x g = p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 has degree n where p n ! 0 p n, p n - 1, p n - 2, f p 0 are called coefficients pnxn is called the leading term and pn is the leading coefficient p0 is called the constant term If p n = 1, P ] x g is called a monic polynomial If p 0 = p 1 = p 2 = f = p n = 0 then P(x) is the zero polynomial

The degree of a polynomial is the highest power of x with non-zero coefficient.

Coefficients can be any real number but we generally use integers in this course for convenience.

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EXAMPLES 1. Which of the following are polynomial expressions? (a) 4 - x + 3x 2 (b) 3x 4 - x 2 + 5x - 1 (c) x 2 - 3x + x -1

Solution (a) and (b) are polynomials but (c) is not, since it has a term x -1 that is not a positive power of x. 2. For the polynomial P ] x g = x 6 - 2x 4 + 3x 3 + x 2 - 7x - 3 (a) Find the degree. (b) Is the polynomial monic? (c) State the leading term. (d) What is the constant term? (e) Find the coefficient of x4.

Solution (a) (b) (c) (d) (e)

Degree is 6 since x6 is the highest power. Yes, the polynomial is monic since the coefficient of x6 is 1. The leading term is x6. The constant term is -3. The coefficient of x4 is -2.

Polynomial equation P ] x g = 0 is a polynomial equation of degree n The real values of x that satisfy the equation are called the real roots of the equation or the real zeros of the polynomial.

EXAMPLES 1. Find the zeros of the polynomial P ] x g = x 2 - 5x.

Solution To find the zeros of the polynomial, we solve P ] x g = 0. x 2 - 5x = 0 x ]x - 5 g = 0


x = 0, x - 5 = 0 x=5 So the zeros are 0, 5. 2. Find the roots of the polynomial equation x 3 - 2x 2 - 3x = 0.

Solution x 3 - 2x 2 - 3x = 0 x ^ x 2 - 2x - 3 h = 0 x ]x - 3 g]x + 1 g = 0 x = 0, x - 3 = 0, x + 1 = 0 x = 3, x = -1 The roots are x = 0, 3, -1. 3. Show that the polynomial p ] x g = x 2 - x + 4 has no real zeros.

Solution We look at the polynomial equation p ] x g = 0. x2 - x + 4 = 0 The discriminant will show whether the polynomial has real zeros. b 2 - 4ac = ] -1 g2 - 4 ] 1 g ] 4 g = 1 - 16 = -15 10 So the polynomial has no real zeros. 4. For the polynomial P ] x g = ax 5 - 3x 4 + x 3 - 7x + 1 (a) Evaluate a if the polynomial is monic. (b) Find the degree of the derivative Pl(x) .

Solution (a) For a monic polynomial, a = 1 (b) P l^ x h = 5ax 4 - 12x 3 + 3x 2 - 7 Pl(x) has degree 4 (highest power).

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12.1 1.

2.

3.

4.

Exercises

Write down the degree of each polynomial. (a) 5x 7 - 3x 5 + 2x 3 - 3x + 1 (b) 3 + x + x 2 - x 3 + 2x 4 (c) 3x + 5 (d) x 11 - 5x 8 + 4 (e) 2 - x - 5x 2 + 3x 3 (f) 3 (g) 2x 4 - x For the polynomial P(x) = x 3 - 7x 2 + x - 1, find (a) P(2) (b) P(-1) (c) P(0) Given P (x) = x + 5 and Q (x) = 2x - 1, find (a) P (-11) (b) Q (3) (c) P (2) + Q (-2) (d) the degree of P (x) + Q (x) (e) the degree of P (x) $ Q (x) For the polynomial P (x) = x 5 - 3x 4 - 5x + 4, find (a) the degree of P (x) (b) the constant term (c) the coefficient of x 4 (d) the coefficient of x 2

5.

Find the zeros of the following polynomials. (a) P (x) = x 2 - 9 (b) p (x) = x + 5 (c) f (x) = x 2 + x - 2 (d) P (x) = x2 - 8x + 16 (e) g (x) = x 3 - 2x 2 + 5x

6.

Find the derivative of each polynomial P (x) and state the degree of Pl(x) (a) P (x) = 3x 4 - 2x 3 - x 2 + 4x - 5 (b) P (x) = 5x 2 + 3 (c) P (x) = 9x 12 - 7x 5 + 8x (d) P (x) = x 7 - 3x 3 + x 2 - 7x - 3 (e) P (x) = 8x + 5

7.

Which of the following are not polynomials? 1 (a) 5x 4 - 3x 2 + x + x x2 + 3x x 2 + 3x - 7 3x + 5 0 1 3x 2 - x + 1 2 3 (g) 4x + 7x -2 + 5

(b) (c) (d) (e) (f)

8.

For the polynomial P (x) = (a + 1) x 3 + (b - 7) x 2 + c + 5, find values for a, b or c if (a) P (x) is monic (b) the coefficient of x 2 is 3 (c) the constant term is -1 (d) P (x) has degree 2 (e) the leading term has a coefficient of 5

9.

Given P (x) = 2x + 5, Q (x) = x 2 - x - 2 and R (x) = x 3 + 9x, find (a) any zeros of P (x) (b) the roots of Q (x) = 0 (c) the degree of P (x) + R (x) (d) the degree of P (x) $ Q (x) (e) the leading term of Q (x) $ R (x)

10. Given f (x) = 3x 2 - 2x + 1 and g (x) = 3x - 3, (a) show f (x) has no zeros (b) find the leading term of f (x) $ g (x) (c) find the constant term of f (x) + g (x) (d) find the coefficient of x in f (x) $ g (x) (e) find the roots of f (x) + g (x) = 0 11. State how many real roots there are for each polynomial equation P ] x g = 0. (a) P ] x g = x 2 - 9 (b) P ] x g = x 2 + 4 (c) P ] x g = x 2 - 3x - 7


(d) (e) (f) (g)

P ] x g = 2x 2 + x + 3 P ] x g = 3x 2 - 5x - 2 P ]x g = x ]x - 1 g]x + 4 g]x + 6 g P ]x g = ]x + 1 g]x - 2 g]x - 5 g

12. For the polynomial P ] x g = 2x 3 + 3x 2 - 36x + 17, find the roots of the derivative polynomial equation Pl(x) = 0.

13. If P ] x g = 3x 4 - 4x 3 - 1, find the zeros of Pl(x) . 14. Show that Pl(x) = 0 has no real roots if P ] x g = x 3 - x 2 + 9x. 15. Show that Ql(x) = 0 has equal roots given Q ] x g = x 3 - 3x 2 + 3x + 5.

Division of Polynomials You would have learned how to do long division in primary school, but have probably forgotten how to do it! We use this method to divide polynomials.

Class Investigation Here are two examples of long division. 1. Divide 5715 by 48. 119 r3 48 g 5715 48 91 48 435 432 3 This means

3 5715 = 119 + 48 48

3 5715 # 48 = 119 # 48 + # 48 48 48 So

5715 = 48 # 119 + 3

(check this on your calculator)

The number 5715 is called the dividend, the 48 is the divisor, 119 is the quotient and 3 is the remainder. 2. Divide 4871 by 35. 139 r6 35 g 4871 35 137 105 321 315 6

CONTINUED

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This means

4871 6 = 139 + 35 35

or

4871 = 35 #139 + 6

(check this on your calculator)

The number 4871 is called the dividend, the 35 is the divisor, 139 is the quotient and 6 is the remainder. Use long division to divide other numbers and write them in the form above. For example: 1. 2. 3. 4. 5.

2048 ' 5876 ' 3546 ' 2992 ' 8914 '

15 17 21 33 19

A polynomial P(x) can be written as P ] x g = A ] x g $ Q ] x g + R ] x g where P(x) is the dividend, A(x) is the divisor, Q(x) is the quotient and R(x) is the remainder.

Proof If we divide a polynomial P(x) by A(x), we can write P(x) in the form of P (x) R (x) = Q (x) + where Q(x) is the quotient and R(x) is the remainder. A (x) A (x) P (x) R (x ) # A (x ) = Q ( x ) # A ( x ) + # A ( x) A (x ) A (x ) P ]x g = A ]x g $ Q ]x g + R ]x g The division continues until the remainder can no longer be broken down further by division. The degree of remainder R(x) is always less than the degree of the divisor A(x).

Proof Suppose the degree of R(x) is higher than the degree of A(x). This means that R(x) can be divided by A(x). R 1(x) R (x ) = Q 1 (x) + A (x ) A (x ) R ] x g = A ] x g $ Q1] x g + R1] x g So = R2] x g This gives P ] x g = A ] x g $ Q ] x g + R 2 ] x g .


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EXAMPLES 1. (a) Divide P (x) = 3x 4 - x 3 + 7x 2 - 2x + 3 by x - 2. (b) Hence write P (x) in the form P (x) = A (x) Q (x) + R (x) . (c) Show that P (2) is equal to the remainder.

Solution (a) Step 1: Divide the leading term by x. i.e. 3x 4 ' x = 3x 3 3x 3 x - 2 g 3 x 4 - x 3 + 7x 2 - 2x + 3

x - 2 is called the divisor.

Step 2: Multiply 3x 3 by (x - 2) and find the remainder by subtraction. i.e. 3x 3 (x - 2) = 3x 4 - 6x 3 3x 3 x - 2 g 3x 4 - x 3 + 7x 2 - 2x + 3 3x 4 - 6x 3 5x 3 Step 3: Bring down the 7x 2 and divide 5x 3 by x. 3x 3 + 5x 2 x - 2 g 3x 4 - x 3 + 7x 2 - 2 x + 3 3x 4 - 6x 3 5x 3 + 7x 2 Step 4: Multiply 5x 2 by (x - 2) and find the remainder by subtraction. i.e. 5x 2 ] x - 2 g = 5x 3 - 10x 2 3x 3 + 5x 2 x - 2 g 3x 4 - x 3 + 7 x 2 - 2 x + 3 3x 4 - 6x 3 5x 3 + 7x 2 5x 3 - 10x 2 17x 2 CONTINUED

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The quotient is 3 2 3x + 5x + 17x + 32.

Continuing this way until we have finished, we will have 3x 3 + 5x 2 + 17x + 32 x - 2 g 3x 4 - x 3 + 7 x 2 - 2 x + 3 3x 4 - 6x 3 5x 3 + 7x 2 5x 3 - 10x 2 17x 2 - 2x 17x 2 - 34x 32x + 3 32x - 64 67

The remainder is 67.

(b) This means that (3x 4 - x 3 + 7x 2 - 2x + 3) ' (x - 2) = (3x 3 + 5x 2 + 17x + 32), remainder 67 3x 4 - x 3 + 7x 2 - 2x + 3 67 = 3x 3 + 5x 2 + 17x + 32 + x-2 x-2 or 3x 4 - x 3 + 7x 2 - 2x + 3 = (x - 2) (3x 3 + 5x 2 + 17x + 32) + 67

i.e.

i.e. P (x) = A (x) Q (x) + R (x) where A (x) is the divisor, Q (x) is the quotient and R (x) is the remainder. (c) P (2) = 3 (2) 4 - 2 3 + 7 (2) 2 - 2 (2) + 3 = 48 - 8 + 28 - 4 + 3 = 67 ` P (2) is equal to the remainder. 2. Divide x 3 - 3x 2 + x + 4 by x 2 - x.

Solution

Check this is true by expanding and simplifying.

x-2 x 2 - x g x 3 - 3x 2 + x + 4 x3 - x2 - 2x 2 + x - 2x 2 + 2x -x + 4 This means that (x 3 - 3x 2 + x + 4) ' (x 2 - x) = (x - 2), remainder - x + 4 x 3 - 3x 2 + x + 4 -x + 4 =x-2+ 2 i.e. 2 x -x x -x or x 3 - 3x 2 + x + 4 = (x - 2) (x 2 - x) + (- x + 4)


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3. Divide x 5 + x 3 + 5x 2 - 6x + 15 by x 2 + 3.

Solution x 3 - 2x + 5 x + 3 g x 5 + x 3 + 5x 2 - 6 x + x 5 + 3x 3 - 2x 3 + 5x 2 - 6x - 2x 3 - 6x 2 5x + 2 5x + 2

15

15 15 0

R (x) = 0, so there is no remainder.

This means that (x 5 + x 3 + 5x 2 - 6x + 15) ' (x 2 + 3) = (x 3 - 2x + 5) x 5 + x 3 + 5x 2 - 6x + 15 i.e. = x 3 - 2x + 5 x2 + 3 or x 5 + x 3 + 5x 2 - 6x + 15 = (x 3 - 2x + 5) (x 2 + 3)

12.2

Exercises

Divide the following polynomials and put them in the form P (x) = A (x) Q (x) + R (x) .

11. (6x 2 - 3x + 1) ' (3x - 2) 12. (x 4 - 2x 3 - x 2 - 2) ' (x 2 - x) 13. (3x 5 - 2x 4 - 3x 3 + x 2 - x - 1) ' (x + 2 )

1.

(3x 2 + 2x + 5) ' (x + 4)

2.

(x 2 - 7x + 4) ' (x - 1)

3.

(x + x + 2 x - 1 ) ' (x - 3 )

4.

(4x + 2x - 3) ' (2x + 3)

5.

(x - 5x + x + 2) ' (x + 3x)

6.

(x + x - x - 3) ' (x - 2)

7.

(5x - 2x + 3x + 1) ' (x + x)

8.

(x - x - 2x + x - 3) ' (x + 4)

19. (x 4 - 2x 3 + 4x 2 + 2x + 5) ' (x 2 + 2x - 1)

9.

(2x 4 - 5x 3 + 2x 2 + 2x - 5) ' (x 2 - 2x)

20. (3x 5 - 2x 3 + x - 1) ' (x + 1)

3

2

2

3

2

3

2

3

4

2

2

3

2

2

10. (4x 3 - 2x 2 + 6x - 1) ' (2x + 1)

14. (x 2 + 5x - 2) ' (x + 1) 15. (x 4 - 2x 2 + 5x + 4) ' (x - 3) 16. (2x 4 - x 3 + 5) ' (x 2 - 2x) 17. (x 3 - 3x 2 + 3x - 1) ' (x 2 + 5) 18. (2x 3 + 4x 2 - x + 8) ' (x 2 + 3x + 2)

Check this by expanding and simplifying.

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Remainder and Factor Theorems Dividing polynomials helps us to factorise them, which in turn makes sketching their graphs easier. There are two theorems that will also help us to work with polynomials.

Remainder theorem

If a polynomial P(x) is divided by x - a, then the remainder is P(a)

Proof The degree of R(x) is less than the degree of A(x).

P ] x g = A ] x g $ Q ] x g + R ] x g where A ] x g = x - a P ]x g = ]x - a gQ ]x g + R ]x g The degree of A(x) is 1, so the degree of R(x) must be 0. So R ] x g = k where k is a constant ` P ]x g = ]x - a gQ ]x g + k Substituting

x = a: P ]a g = ]a - a gQ ]a g + k = 0 $ Q ]x g + k =k

So P ] a g is the remainder.

EXAMPLES 1. Find the remainder when 3x 4 - 2x 2 + 5x + 1 is divided by x - 2.

Solution The remainder when P(x) is divided by x - a is P(a). The remainder when P(x) is divided by x - 2 is P(2). P ] 2 g = 3 ] 2 g4 - 2 ] 2 g2 + 5 ] 2 g + 1 = 51 So the remainder is 51. 2. Evaluate m if the remainder is 4 when dividing 2x 4 + mx + 5 by x + 3.

Solution x + 3 = x - (- 3) .

The remainder when P(x) is divided by x + 3 is P ^ -3 h . So P ] -3 g = 4 4 2 ] - 3 g + m ] -3 g + 5 = 4 162 - 3m + 5 = 4


167 - 3m = 4 167 = 3m + 4 163 = 3m 54

1 = m. 3

Factor theorem The factor theorem is a direct result of the remainder theorem.

For a polynomial P(x), if P ] a g = 0 then x - a is a factor of the polynomial.

Proof P ] x g = A ] x g $ Q ] x g + R ] x g where A ] x g = x - a P ]x g = ]x - a gQ ]x g + R ]x g The remainder when P(x) is divided by x - a is P(a). So P ] x g = ] x - a g Q ] x g + P ] a g But if P ] a g = 0: P ]x g = ]x - a gQ ]x g + 0 = ]x - a gQ ]x g So x - a is a factor of P(x). The converse is also true: For a polynomial P(x), if x - a is a factor of the polynomial, then P ] a g = 0

Proof If x - a is a factor of P(x), then we can write: P ]x g = ]x - a gQ ]x g This means that when P(x) is divided by x - a, the quotient is Q(x) and there is no remainder. So P ] a g = 0

673

674


EXAMPLE (a) Show that x - 1 is a factor of P ] x g = x 3 - 7x 2 + 8x - 2. (b) Divide P(x) by x - 1 and write P(x) in the form P ] x g = ] x - 1 g Q ] x g.

Solution (a) The remainder when dividing the polynomial by x - 1 is P(1) P ] 1 g = 1 3 - 7 ] 1 g2 + 8 ] 1 g - 2 =0 So x - 1 is a factor of P(x). (b) x 2 - 6x + x - 1 g x 3 - 7x 2 + x3 - x2 - 6x 2 + - 6x 2 +

Notice that x 2 - 6x + 2 won’t factorise.

2 8x - 2

8x 6x 2x - 2 2x - 2 0 3 2 ] g So P x = x - 7x + 8x - 2 = ] x - 1 g ^ x 2 - 6x + 2 h

Further properties of a polynomial Some properties of polynomials come from the remainder and factor theorems.

If polynomial P(x) has k distinct zeros a 1, a 2, a 3, ... a k, then (x - a 1) (x - a 2) (x - a 3) ... (x - a k) is a factor of P(x)

Proof If a1 is a zero of P(x) then (x - a 1) is a factor of P(x). If a2 is a zero of P(x) then (x - a 2) is a factor of P(x). If a3 is a zero of P(x) then (x - a 3) is a factor of P(x). Similarly, if a k is a zero of P(x) then (x - a k) is a factor of P(x). ` P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a k) g ] x g So (x - a 1) (x - a 2) (x - a 3) ... (x - a k) is a factor of P ] x g .

If polynomial P(x) has degree n and n distinct zeros a 1, a 2, a 3, ... a n, then P ] x g = p n (x - a 1) (x - a 2) (x - a 3) ... (x - a n)


Proof Since a 1, a 2, a 3, ... a n are zeros of P(x), (x - a 1) (x - a 2) (x - a 3) ... (x - a n) is a factor of the polynomial. So P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a n)Q(x) But (x - a 1) (x - a 2) (x - a 3) ... (x - a n) has degree n and P(x) has degree n so Q(x) must be a constant. ` P ] x g = p n (x - a 1) (x - a 2) (x - a 3) ... (x - a n)

A polynomial of degree n cannot have more than n distinct real zeros.

Proof P(x) has degree n So P ] x g = p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 where p n ! 0 Suppose P(x) has more than n distinct zeros, say n + 1 Then (x - a 1) (x - a 2) (x - a 3) ... (x - a n + 1) is a factor of P(x). So P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a n + 1) Q ] x g. But this gives P(x) at least degree n + 1, and P(x) only has degree n. So the polynomial cannot have more than n distinct real zeros. This also means that the polynomial equation cannot have more than n real roots.

EXAMPLE If a polynomial has degree 2, show that it cannot have 3 zeros.

Solution Let P ] x g = p 2 x 2 + p 1 x + p 0 where p 2 ! 0 Assume P(x) has 3 zeros, a1, a2 and a3 Then _ x - a 1 i _ x - a 2 i _ x - a 3 i is a factor of the polynomial. ` P (x) = (x - a 1) (x - a 2) (x - a 3) Q (x) But this polynomial has degree 3 and P(x) only has degree 2. So P(x) cannot have 3 zeros.

A polynomial of degree n with more than n distinct real zeros is the zero polynomial P ] x g = 0 ( p 0 = p 1 = p 2 = ... = p n = 0)

Proof Let P(x) be a polynomial of degree n with zeros a 1, a 2, a 3, ... a n Then P ] x g = (x - a 1) (x - a 2) (x - a 3) ... (x - a n) k

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Suppose P(x) has another distinct zero a n + 1 Then P _ a n + 1 i = 0 `(a n + 1 - a 1) (a n + 1 - a 2) (a n + 1 - a 3) ... (a n + 1 - a n) k = 0 But a n + 1 ! a 1, a 2, a 3, ... a n since all zeros are distinct. So k = 0 `P ] x g = 0

If two polynomials of degree n are equal for more than n distinct values of x, then the coefficients of like powers of x are equal. That is, if a 0 + a 1 x + a 2 x 2 + ... + a n x n / b 0 + b 1 x + b 2 x 2 + ... + b n x n then a 0 = b 0, a 1 = b 1, a 2 = b 2, ... a n = b n

Proof Let A ] x g = a 0 + a 1 x + a 2 x 2 + ... + a n x n and B ] x g = b 0 + b 1 x + b 2 x 2 + ... + b n x n where A ] x g = B ] x g for more than n distinct x values. Let P ] x g = A ] x g - B ] x g Then P ] x g = (a 0 - b 0) + (a 1 - b 1) x + (a 2 - b 2) x 2 + ... + (a n - b n) x n and P(x) has degree n. If A ] x g = B ] x g for more than n distinct x values, then A ] x g - B ] x g = 0 for more

You learned a special case of this result in Chapter 10 under quadratic identities. This is a more general result for all polynomials.

than n distinct x values. This means P ] x g = 0 for more than n distinct x values. This means that P(x) has more than n zeros. ` P(x) is the zero polynomial P ]xg = 0 (a 0 - b 0) + (a 1 - b 1) x + (a 2 - b 2) x 2 + ... + (a n - b n) x n = 0 So a 0 - b 0 = 0, a 1 - b 1 = 0, a 2 - b 2 = 0, ..., a n - b n = 0 ` a 0 = b 0, a 1 = b 1, a 2 = b 2, ..., a n = b n

EXAMPLE Write x 3 - 2x 2 + 5 in the form ax 3 + b ] x + 3 g2 + c ] x + 3 g + d.

Solution ax 3 + b ] x + 3 g2 + c (x + 3) + d = ax 3 + b ^ x 2 + 6x + 9 h + c ] x + 3 g + d = ax 3 + bx 2 + 6bx + 9b + cx + 3c + d = ax 3 + bx 2 + ] 6b + c g x + 9b + 3c + d For

x 3 - 2x 2 + 5 / ax 3 + b ] x + 3 g2 + c ] x + 3 g + d


677

]1 g

a=1 b = -2 6b + c = 0 9b + 3c + d = 5

]2 g

]3 g ]4 g

Substitute (2) into (3): 6 ] -2 g + c = 0 - 12 + c = 0 c = 12 Substitute b = - 2 and c = 12 into (4): 9 ] - 2 g + 3 ] 12 g + d -18 + 36 + d 18 + d d

=5 =5 =5 = -13

` x 3 - 2x 2 + 5 / x 3 - 2 ] x + 3 g2 + 12 ] x + 3 g - 13.

If x - a is a factor of polynomial P(x), then a is a factor of the constant term of the polynomial.

Proof Let P ] x g = p n x n + p n - 1 x n - 1 + p n - 2 x n - 2 + f + p 2 x 2 + p 1 x + p 0 where p n ! 0 If x - a is a factor of P(x) we can write P ] x g = ] x - a g Q ] x g where Q(x) has degree n - 1. P ] x g = ] x - a g _ q n - 1 x n - 1 + q n - 2 x n - 2 + g + q 2 x 2 + q 1 x + q 0 i where q n - 1 ! 0 = xq n - 1 x n - 1 + xq n - 2 x n - 2 + g + xq 1 x + xq 0 - aq n - 1 x n - 1 - aq n - 2 x n - 2 - g - aq 2 x 2 - aq 1 x - aq 0 = q n - 1 x n + q n - 2 x n - 1 + g + q 1 x 2 + q 0 x - aq n - 1 x n - 1 - aq n - 2 x n - 2 - g - aq 2 x 2 - aq 1 x - aq 0 = q n - 1 x n + _ q n - 2 - a i x n - 1 + g + _ q 1 - a i x 2 + _ q 0 - a i x - aq 0 ` p 0 = - aq 0 So a is a factor of p0.

We already use this when factorising a trinomial. This is a more general result for all polynomials.

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EXAMPLE Factorise x 2 + 2x - 15.

Solution Factors of -15 are - 3# 5, 3# - 5, -1#15, 1# -15. We choose - 3# 5 since - 3x + 5x = 2x, the middle term. So x 2 + 2x - 15 = ] x - 3 g ] x + 5 g.

To factorise polynomials in general, we also look for factors of the constant term.

Class Investigation Why are factors of the polynomial factors of the constant term? Use the knowledge you have of trinomials to help you in your discussion.

EXAMPLES 1. Find all factors of f (x) = x 3 + 3x 2 - 4x - 12.

Solution Try factors of - 12 (i.e. !1, !2, !3, !4, !6, !12) . e.g. f (1) = 1 3 + 3 (1) 2 - 4 (1) - 12 = -12 !0 ` x - 1 is not a factor of f (x)

x - 2 is called a linear factor as it has degree 1.

f (2) = 2 3 + 3 (2) 2 - 4 (2) - 12 =0 Since f (2) = 0, the remainder when f (x) is divided by x - 2 is 0. ` x - 2 is a factor of f (x) . We divide f (x) by x - 2 to find other factors: x 2 + 5x + 6 x - 2 g x 3 + 3x 2 - 4x - 12 x 3 - 2x 2 5x 2 - 4x 5x 2 - 10x 6x - 12 6x - 12 0


679

` f (x) = (x - 2) (x 2 + 5x + 6) = (x - 2) (x + 2) (x + 3) 2. Find all factors of P (x) = x 3 + 3x 2 + 5x + 15.

Solution Try factors of 15 (i.e.!1, !3, !5, !15) . e.g. P (- 3) = (- 3) 3 + 3 (- 3) 2 + 5 (- 3) + 15 =0 ` x + 3 is a factor of f (x) We divide P (x) by x + 3 to find other factors: +5

x2

x + 3 g x + 3x + 5x + 15 x 3 + 3x 2 0 + 5x + 15 5x + 15 3

2

0

x 2 + 5 will not factorise for any real x.

` P (x) = (x + 3) (x + 5) 2

12.3 Exercises 1.

Use the remainder theorem to find the remainder in each question. (a) (x 3 - 2x 2 + x + 5) ' (x - 4)

(c) the remainder is 0 when 2x 5 + 7x 2 + 1 + k is divided by x + 6. (d) 2x 4 - kx 3 + 3x 2 + x is divisible by x - 3. (e) the remainder is 25 when 2x 4 - 3x 2 + 5 is divided by x - k.

(b) (x 2 + 5x + 3) ' (x + 2) (c) (2x 3 - 4x - 1) ' (x + 3) (d) (3x 5 + 2x 2 - x + 4) ' (x - 5) (e) (5x 3 + 2x 2 + 2x - 9) ' (x - 1)

3.

(a) Find the remainder when f (x) = x 3 - 4x 2 + x + 6 is divided by x - 2. (b) Is x - 2 a factor of f (x) ? (c) Divide x 3 - 4x 2 + x + 6 by x - 2. (d) Factorise f (x) fully and write f (x) as a product of its factors.

4.

(a) Show that x + 3 is a factor of P (x) = x 4 + 3x 3 - 9x 2 - 27x. (b) Divide P (x) by x + 3 and write P (x) as a product of its factors.

(f) (x - x + 3x - x - 1) ' (x + 2) 4

3

2

(g) (2x 2 + 7x - 2) ' (x + 7) (h) (x 7 + 5x 3 - 1) ' (x - 3) (i) (2x 6 - 3x 2 + x + 4) ' (x + 5) (j) (3x 4 - x 3 - x 2 - x - 7) ' (x + 1) 2.

Find the value of k if (a) the remainder is 3 when 5x 2 - 10x + k is divided by x - 1. (b) the remainder is - 4 when x 3 - (k - 1) x 2 + 5kx + 4 is divided by x + 2.

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5.

The remainder is 5 when P (x) = ax 3 - 4bx 2 + x - 4 is divided by x - 3 and the remainder is 2 when P (x) is divided by x + 1. Find the values of a and b.

6.

When f (x) = ax 2 - 3x + 1 and g (x) = x 3 - 3x 2 + 2 are divided by x + 1 they leave the same remainder. Find the value of a.

7.

(a) Show that x - 3 is not a factor of P (x) = x 5 - 2x 4 + 7x 2 - 3x + 5. (b) Find a value of k such that x - 3 is a factor of Q (x) = 2x 3 - 5x + k.

Linear factors are in the form x - a.

8.

9.

The polynomial P (x) = x 3 + ax 2 + bx + 2 has factors x + 1 and x - 2. Find the values of a and b. (a) The remainder, when f (x) = ax 4 + bx 3 + 15x 2 + 9x + 2 is divided by x - 2, is 216, and x + 1 is a factor of f (x) . Find a and b. (b) Divide f (x) by x + 1 and write the polynomial in the form f (x) = (x + 1) g (x) . (c) Show that x + 1 is a factor of g (x) . (d) Write f (x) as a product of its factors.

10. Write each polynomial as a product of its factors. (a) x 2 - 2x - 8 (b) x 3 + x 2 - 2x (c) x 3 + x 2 - 10x + 8 (d) x 3 + 4x 2 - 11x - 30 (e) x 3 - 11x 2 + 31x - 21 (f) x - 12x + 17x + 90 3

2

(g) x 3 - 7x 2 + 16x - 12 (h) x 4 + 6x 3 + 9x 2 + 4x (i) x 3 + 3x 2 - 4 (j) x 3 - 7x - 6

11. (a) Write P (x) = x 3 - 7x + 6 as a product of its factors. (b) What are the zeros of P (x) ? (c) Is (x - 2) (x + 3) a factor of P (x) ? 12. If f (x) = x 4 + 10x 3 + 23x 2 - 34x - 120 has zeros - 5 and 2 (a) show (x + 5) (x - 2) is a factor of f (x) (b) write f (x) as a product of its linear factors. 13. If P (x) = x 4 + 3x 3 - 13x 2 - 51x - 36 has zeros - 3 and 4, write P (x) as a product of its linear factors. 14. (a) Show that P (x) = x 3 - 3x 2 - 34x + 120 has zeros - 6 and 5. (b) Write P (x) as a product of its linear factors. 15. (a) Write the polynomial P ] u g = u 3 - 4u 2 + 5u - 2 as a product of its factors. (b) Hence or otherwise, solve ] x - 1 g3 - 4 ] x - 1 g2 + 5 ] x - 1 g - 2 = 0. 16. (a) Write the polynomial f ^ p h = p 3 - 2p 2 - 5p + 6 as a product of its factors. (b) Hence or otherwise, solve ] 2x + 1 g3 - 2 ] 2x + 1 g2 - 5 ] 2x + 1 g + 6 = 0. 17. (a) Write P ] k g = 2k 3 + 3k 2 - 1 as a product of its factors. (b) Hence or otherwise, solve 2 sin 3 x + 3 sin 2 x - 1 = 0 for 0c # x # 360c. 18. (a) Write f ] u g = u 3 - 13u 2 + 39u - 27 as a product of its factors. (b) Hence or otherwise, solve 3 3x - 13.3 2x + 39.3 x - 27 = 0. 19. Solve ] x + 4 g4 - ] x + 4 g3 - 2 ] x + 4 g2 = 0.


20. Solve 2 cos 3 i - cos 2 i - cos i = 0 for 0c# x # 360c. 21. Evaluate a, b, c and d if (a) x 3 + 3x 2 - 2x + 1 / ax 3 + b ] x - 1 g2 + cx + d (b) x 3 - x 2 + 4x / ax 3 + b ] x + 2 g2 + c ] x + 2 g + d (c) 2x 3 - x + 7 / ax 3 + b ] x + 1 g2 + c ] x + 1 g + d + 2 (d) x 3 + x 2 + 5x - 3 / ax 3 + b ] x - 3 g2 + cx + d (e) 4x 3 - x + 3 / ] a + 1 g x 3 + b ] x + 4 g2 + c ]x + 4 g + d - 1 (f) x 3 + x 2 - 8x - 6 / ax 3 + b ] x - 2 g2 + cx + d - 3 (g) 3x 3 - 2x 2 + x / ] a - 2 g x 3 + b ] x - 5 g2 + c ]x - 5g + d - 2 (h) - x 3 + x 2 - 4x - 2 / a ] x + 1 g3 + bx 2 + cx + d

(i) - 2x 3 + 3x 2 - 1 / 2ax 3 + b ] x - 1 g2 + cx + d (j) - x 3 - 4x 2 + x + 3 / a ] x - 2 g3 + b ] x - 2 g2 + c ]x - 2 g + d + 1 22. A monic polynomial of degree 3 has zeros - 3, 0 and 4. Find the polynomial. 23. Polynomial P ] x g = ax 3 - bx 2 + cx - 8 has zeros 2 and - 1 and P ] 3 g = 28. Evaluate a, b and c. 24. A polynomial with leading term 2x4 has zeros - 2, 0, 1 and 3. Find the polynomial. 25. Show that a polynomial of degree 3 cannot have 4 zeros.

Graph of a Polynomial We can use the graphing techniques that you have learned to sketch the graph of a polynomial.

Using intercepts Finding the zeros of a polynomial or the roots of the polynomial equation helps us to sketch its graph.

EXAMPLES 1. (a) Write the polynomial P ] x g = x 3 + x 2 - 6x as a product of its factors. (b) Sketch the graph of the polynomial.

Solution (a) P ] x g = x 3 + x 2 - 6x = x ^ x2 + x - 6 h = x ]x + 3 g]x - 2 g (b) For the graph of P ] x g = x 3 + x 2 - 6x For x-intercepts: y = 0 CONTINUED

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0 = x 3 + x 2 - 6x = x ]x + 3 g]x - 2 g x = 0, x + 3 = 0, x - 2 = 0 x = - 3, x=2 So x-intercepts are 0, - 3 and 2. For y-intercepts: x = 0 P ] 0 g = 0 3 + ] 0 g2 - 6 ] 0 g =0 So y-intercept is 0. y 4 3 2 1 -4

-3

-2

-1 0 -1

x 1

2

3

4

-2 -3 -4

We look at which parts of the graphs are above and which are below the x-axis between the x-intercepts. Test x 1 - 3, say x = - 4: P ] x g = x 3 + x 2 - 6x = x ]x + 3 g]x - 2 g P ]-4 g = -4 ]-4 + 3 g]-4 - 2 g = -4 ]-1 g]-6 g = - 24 10 So the curve is below the x-axis. Test - 3 1 x 1 0, say x = - 1: P ] - 1 g = - 1 ] - 1 + 3 g ] -1 - 2 g = -1 ]2 g]-3 g =6 20 So the curve is above the x-axis.


683

Test 0 1 x1 2, say x = 1: P ]1g = 1]1 + 3g]1 - 2g = 1]4g]-1g = -4 10 So the curve is below the x-axis. Test x 2 2, say x = 3: P ]3g = 3]3 + 3g]3 - 2g = 3]6g]1g = 18 20 So the curve is above the x-axis. We can sketch the polynomial as shown. y 4 3 2 1 -4

-3

-2

-1 0 -1

x 1

2

3

4

-2 -3 -4

2. (a) Write the polynomial P ] x g = x 3 - x 2 - 5x - 3 as a product of its factors. (b) Sketch the graph of the polynomial.

Solution (a) Factors of - 3 are ! 1 and ! 3. P ] -1 g = ] -1 g3 - ] -1 g2 - 5 ] -1 g - 3 =0 CONTINUED

Later on, in a class investigation in this chapter you will learn how to make the graph more accurate by finding the maximum and minimum points. This is a topic in the HSC Course.

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So x + 1 is a factor of the polynomial. x 2 - 2x - 3 x + 1 g x 3 - x 2 - 5x - 3 x3 + x2 -2x2 - 5x -2x2 - 2x -3x - 3 -3x - 3 0 P ]x g = ]x + = ]x + = ]x +

1 g ^ x 2 - 2x - 3 h 1 g]x - 3 g]x + 1 g 1 g2 ] x - 3 g

(b) For the graph of P ] x g = x 3 - x 2 - 5x - 3 For x-intercepts: y = 0 0 = x 3 - x 2 - 5x - 3 = ] x + 1 g2 ] x - 3 g ] x + 1 g2 = 0, x-3=0 x+1=0 x=3 x = -1 So x-intercepts are - 1 and 3. For y-intercepts: x = 0

y 4 3 2 1

-4

-3

-2

-1 0 -1

P ] 0 g = 0 3 - ] 0 g2 - 5 ] 0 g - 3 = -3

-2

So y-intercept is - 3.

-4

x 1

2

3

4

-3

We look at which parts of the graphs are above and which are below the x-axis between the x-intercepts. Test x 1 -1, say x = - 2: P ] x g = x 3 - 3x 2 - x + 3 = ] x + 1 g2 ] x - 3 g P ] - 2 g = ] - 2 + 1 g2 ] - 2 - 3 g = ] - 1 g2 ] - 5 g = -5 10 So the curve is below the x-axis. Test -11 x 1 3, say x = 0: P ] 0 g = ] 0 + 1 g2 ] 0 - 3 g = ] 1 g2 ] - 3 g = -3 10


So the curve is below the x-axis. Test x 2 3, say x = 4: P ] 4 g = ] 4 + 1 g2 ] 4 - 3 g = ] 5 g2 ] 1 g = 25 20 So the curve is above the x-axis. We can sketch the polynomial as shown. y 4 3 2 1 -4

-3

-2

x

-1 0 -1

1

2

3

4

-2 -3 -4 -5 -6 -7

12.4 Exercises 1.

Sketch the graph of each polynomial by finding its zeros and showing the x- and y-intercepts. (a) f ] x g = ] x + 1 g ] x - 2 g ] x - 3 g (b) P ] x g = x ] x + 4 g ] x - 2 g

2.

(i) Write each polynomial as a product of its factors (ii) Sketch the graph of the polynomial (a) P ] x g = x 3 - 2x 2 - 8x

(c) p ] x g = - x ] x - 1 g ] x - 3 g (d) f ] x g = x ] x + 2 g2

(b) f ] x g = - x 3 - 4x 2 + 5x (c) P ] x g = x 4 + 3x 3 + 2x 2 (d) A ] x g = 2x 3 + x 2 - 15x

(e) g ] x g = ] 5 - x g ] x + 2 g ] x + 5 g

(e) P ] x g = - x 4 + 2x 3 + 3x 2

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3.

4.

5.

(a) Find the x-intercepts of the polynomial P ] x g = x 4 + 3x 3 - 4x. (b) Sketch the graph of the polynomial. (a) Show that x - 2 is a factor of P ] x g = x 3 - 3x 2 - 4x + 12. (b) Write P(x) as a product of its factors. (c) Sketch the graph of the polynomial.

(a) P ] x g = x 3 + 3x 2 - 10x - 24 (b) P ] x g = x 3 + x 2 - 9x - 9 (c) P ] x g = 12 - 19x + 8x 2 - x 3 (d) P ] x g = x 3 - 13x + 12 (e) P ] x g = - x 3 + 2x 2 + 9x - 18 (f) P ] x g = x 3 + 2x 2 - 4x - 8 (g) P ] x g = x 3 - 5x 2 + 8x - 4 (h) P ] x g = x 3 + x 2 - 5x + 3 (i) f (x) = 16x + 12x 2 - x 4 (j) P ] x g = x 4 - 2x 2 + 1

Sketch the graph of each polynomial, showing all x- and y-intercepts.

Class Investigation The graphs in the examples above are not very accurate, as we don’t know where they turn around. We can use calculus to help find these points. You will look at the applications of calculus in sketching graphs in the HSC Course.

You used the axis of symmetry to find the minimum and maximum values of quadratic functions in Chapter 10. You can also use calculus to find the minimum or maximum turning points of functions.

You looked at the gradient of tangents to a curve in Chapter 8.

Notice that the graph below has both a maximum and minimum turning point. We can find these by looking at the gradient of the tangents dy around the curve, or . dx y

Maximum turning point

x

Minimum turning point


Notice that at both these turning points,

dy dx

= 0.

We can also examine each type of turning point more closely. Maximum turning point:

The maximum turning point has a zero gradient at the point itself but notice that it has a positive gradient on the left-hand side and a negative gradient on the right-hand side. dy dy So 2 0 on the LHS and 1 0 on the RHS. dx dx Minimum turning point:

The minimum turning point has a zero gradient at the point itself but it has a negative gradient on the left-hand side and a positive gradient on the right-hand side. dy dy So 1 0 on the LHS and 2 0 on the RHS. dx dx CONTINUED

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There is also another type of point that you see in graphs such as f ] x g = x 3.

This is called a point of inflexion and has

dy dx

= 0.

However, the gradient has the same sign on both the LHS and RHS. These three types of points are called stationary points. We can use them to sketch the graph of a polynomial. Here is an example. Sketch the polynomial P ] x g = 2x 3 + 3x 2 - 12x - 7 showing any stationary points. dy dx

= 6x 2 + 6x - 12

For stationary points

dy dx

= 0:

6x 2 + 6x - 12 = 0 6 ^ x2 + x - 2 h = 0 6 ]x - 1 g]x + 2 g = 0 x - 1 = 0, x + 2 = 0 x = 1, x = -2 So there are two stationary points when x = 1, - 2. ] 1 g When x = 1

P ] 1 g = 2 ] 1 g3 + 3 ] 1 g2 - 12 ] 1 g - 7 = -14 So there is a stationary point at ^ 1, -14 h. We can check the gradient on the LHS and RHS of this point to determine if it is a maximum or minimum turning point. When x = 0 dy = 6 (0) 2 + 6 (0) - 12 dx = -12 When x = 2 dy = 6 (2) 2 + 6 (2) - 12 dx = 24


x dy dx Since

dy

0

1

2

-12

0

24

1 0 on the LHS and

dx turning point.

dy dx

2 0 on the RHS, ^ 1, -14 h is a minimum

] 2 g When x = - 2

P ] - 2 g = 2 ] - 2 g3 + 3 ] - 2 g2 - 12 ] - 2 g - 7 = 13 So there is a stationary point at ^ - 2, 13 h. Check the gradient on the LHS and RHS of this point. When x = - 3 dy = 6 (- 3) 2 + 6 (- 3) - 12 dx = 24 When x = -1 dy = 6 (-1) 2 + 6 (-1) - 12 dx = -12 x dy dx Since

dy dx

-3

-2

-1

24

0

-2

2 0 on the LHS and

dy dx

1 0 on the RHS, ^ - 2, 13 h is a maximum

turning point. Now we look for intercepts. For x-intercepts: y = 0 0 = 2x 3 + 3x 2 - 12x - 7 The expression 2x 3 + 3x 2 - 12x - 7 will not factorise so we cannot find the x-intercepts. For y-intercept: x = 0 P ] x g = 2 ] 0 g3 + 3 ] 0 g2 - 12 ] 0 g - 7 = -7 So the y-intercept is - 7. CONTINUED

Factors of - 7 are !1 and !7 and none of these factors will satisfy the polynomial equation.

689

690


We sketch the polynomial using the stationary points and y-intercept. y (-2, 13)

x -7

(1, -14)

Can you sketch the following polynomials using calculus to find their stationary points? 1. P ] x g = x 2 + 6x - 3 2. P ] x g = - x 2 + 4x + 1 3. p ] x g = x 3 - 5 4. f ] x g = x 4 + 2 5. g ] x g = 2x 3 + 3x 2 - 1 6. P ] x g = 2x 3 - 21x 2 + 72x - 12 7. f ] x g = - 2x 3 + 9x 2 - 12x + 4 8. P ] x g = x 3 - 3x 2 + 3x - 5 9. A ] x g = x 4 + 8x 3 - 18x 2 - 7 10. Q ] x g = - 3x 4 + 20x 3 - 48x 2 + 48x - 3

You may have noticed some of these properties while sketching the graphs of polynomials.

Limiting behaviour of polynomials The limiting behaviour of a function describes what happens to the function as x " !3.

For very large x , P (x) . p n x n


Investigation Use a graphics calculator or graphing computer software to explore the behaviour of polynomials as x becomes large (both negative and positive values). For example, sketch f ] x g = 2x 5 + 3x 2 - 7x - 1 and f ] x g = 2x 5 together. What do you notice at both ends of the graphs where x is large? Zoom out on these graphs and watch the graph of the polynomial and the graph of the leading term come together. Try sketching other polynomials along with their leading term as different graphs. Do you find the same results?

So the leading term shows us what its limiting behaviour will be. If the degree of a polynomial P(x) is even and the leading coefficient is positive, then the polynomial will be positive as x becomes large. This means that for any polynomial with a positive leading coefficient and even degree, P ] x g " 3 as x " !3. On the graph, both ends of the graph will go up as shown by the examples below. y

x

y

x

All positive or negative values of x to an even power will always be positive.

691

692


y

x

If the degree of a polynomial P(x) is even and the leading coefficient is negative, then the polynomial will be negative as x becomes large. This means that for any polynomial with a negative leading coefficient and even degree, P ] x g " - 3 as x " ! 3. On the graph, both ends of the graph will go down as shown by the examples below. y

x

y

x


y

x

If P(x) is an odd degree polynomial with positive leading coefficient, then as x becomes a very large positive value, P(x) will also be positive. As x becomes a very large negative value, P(x) will also be negative. This means that P ] x g " - 3 as x " - 3 and P ] x g " 3 as x " 3. On the graph, the end of the graph on the LHS will go down and the end on the RHS will go up as shown in the examples. y

x

y

x

693

694


y

x

y

x

If P(x) is an odd degree polynomial with negative leading coefficient, then as x becomes a very large positive value, P(x) will be negative. As x becomes a very large negative value, P(x) will be positive. This means that P ] x g " 3 as x " - 3 and P ] x g " - 3 as x " 3. On the graph, the end of the graph on the LHS will go up and the end on the RHS will go down as shown in the examples. y

x


y

x

y

x

y

x

695

696


If P(x) has even degree, the ends of the graph both go the same way. y

Leading coefficient 2 0

y Leading coefficient 1 0

x

x

If P(x) has odd degree, the ends of the graph both go different ways.

y

Leading coefficient 2 0

y Leading coefficient 1 0

x

x

A polynomial of odd degree always has at least one real zero.

This comes from the results above. A polynomial with odd degree will go up at one end and down the other as x becomes large. This means that it must cross the x-axis at least once. ` the polynomial must have at least one real zero.

At least one maximum or minimum value of P(x) occurs between any two distinct real zeros.

You can see this on a graph. If there are two distinct real zeros of a polynomial, then they will show up on the graph as two x-intercepts since the zeros make P ] x g = 0.


When the graph passes through one x-intercept, say x1, it must turn around again to pass through the other x-intercept x2 as shown in the examples below. So there must be at least one maximum or minimum value between the zeros. y

x1

x

x2

y

x1

x2

x

y

x1

x2

x

Multiple roots In quadratic functions, you saw that if a quadratic expression is a perfect square, it has equal roots (and the discriminant is zero).

697

698


EXAMPLE Solve x 2 - 2x + 1 = 0.

Solution x 2 - 2x + 1 = 0 ]x - 1g]x - 1g = 0 ] x - 1 g2 = 0 x - 1 = 0, x = 1,

x -1= 0 x =1

The solution is x = 1 but we say that there are two equal roots.

If P(x) has two equal roots at x = a then we can write P ] x g = ] x - a g2 Q ] x g We say that the polynomial has a double root at x = a. If P ] x g = ] x - a g3 Q ] x g, the polynomial has a triple root at x = a. There are three equal roots at x = a.

If P ] x g = ] x - a gn Q ] x g, the polynomial has a multiple root at x=a It has n equal roots at x = a

EXAMPLES 1. Sketch the graph of f ] x g = ] x + 2 g2.

Solution This graph is f ] x g = x 2 translated 2 units to the left. y 5 4 3 2 1 -4 -3 -2 -1 0 -1

x 1

2

-2 -3 -4 See class investigations on pages 686–690.

Notice that there is a minimum turning point at the root x = - 2.


699

2. Sketch the graph of F ] x g = ] x - 1 g3.

Solution This is the graph of F ] x g = x 3 translated 1 unit to the right. y 8 6 4 2 -4

-3

-2

-1 0 -2

x 1

2

3

4

-4 -6 -8 See class investigations on pages 686–690.

Notice that there is a point of inflexion at the root x = 1.

Generally, a graph cuts the x-axis at a single root but touches the x-axis at a multiple root in a special way.

EXAMPLE (a) Examine the polynomial P ] x g = ] x + 2 g2 ] x - 1 g close to the roots. (b) Describe the behaviour of the polynomial as x becomes very large. (c) Draw a sketch of the polynomial showing its roots.

Solution (a) P ] x g = ] x + 2 g2 ] x - 1 g has roots when P ] x g = 0. ] x + 2 g2 ] x - 1 g = 0 x + 2 = 0, x - 1 = 0

x = - 2,

x=1

Notice that there is a double root at x = - 2. Look at the sign of P(x) close to x = 1: When x = 0.9 P ] 0.9 g = ] 0.9 + 2 g2 ] 0.9 - 1 g = +#=So the curve is below the x-axis on the LHS. CONTINUED

700


When x = 1.1 P ] 1.1 g = ] 1.1 + 2 g2 ] 1.1 - 1 g = +#+ =+ So the curve is above the x-axis on the RHS. Look at the sign of P(x) close to x = - 2: When x = - 2.1 P ] - 2.1 g = ] - 2.1 + 2 g2 ] - 2.1 - 1 g = +#=So the curve is below the x-axis on the LHS. When x = - 1.9 P ] - 1.9 g = ] - 1.9 + 2 g2 ] - 1.9 - 1 g = +#=So the curve is below the x-axis on the RHS. At the single root x = 1, the curve passes through the root from below the x-axis to above the x-axis. At the double root x = - 2, the curve touches the x-axis from below and turns around and continues to be below the x-axis. (b) Expanding P ] x g = ] x + 2 g2 ] x - 1 g gives x3 as the leading term. P ] x g = ] x + 2 g2 ] x - 1 g = ^ x 2 + 4x + 4 h ] x - 1 g There is no need to expand the brackets fully as we only need the leading term.

= x 3 - x 2 + 4x 2 - 4x + 4 x - 4 So the polynomial has degree 3 since the highest power is x3. Also the leading coefficient is 1. Since P(x) has odd degree and a positive leading coefficient, as x becomes a larger positive number, P ] x g "3 and as x becomes a larger negative number, P ] x g " -3. (c)

y

-2

1

x


701

Investigation Use a graphics calculator or graphing computer software to draw graphs with multiple roots. (a) Examine values close to the roots. (b) Look at the relationship between the degree of the polynomial, the leading coefficient and its graph. Here are some examples of polynomials but you could choose others to examine. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

P ]xg = ]x + 1g]x - 3g P ]xg = ]x + 1g2]x - 3g P ]xg = -]x + 1g3]x - 3g P ]xg = -]x + 1g4]x - 3g P ]xg = ]x + 1g]x - 3g2 P ]xg = ]x + 1g]x - 3g3 P ]xg = -]x + 1g]x - 3g4 P ]xg = -]x + 1g2]x - 3g2 P ]xg = -]x + 1g2]x - 3g3 P ]xg = ]x + 1g3]x - 3g2

Where there is a multiple root, there is always a stationary point (maximum, minimum or point of inflexion). This means that

dy dx

= 0 at that point.

If the root is at x = a, then we can write this as Pl(a) = 0 .

If P ] x g = ] x - a gn Q ] x g has a multiple root at x = a then P (a) = P l(a) = 0 There is a stationary point at x = a: If n is even, there is a maximum or minimum turning point at x = a If n is odd, there is a point of inflexion at x = a

Proof P ] x g = ] x - a gn Q ] x g P ] a g = ] a - a gn Q ] a g = 0n $ Q ] x g =0

See class investigation on pages 686–690.

702


P l (x) = u lv + v lu = n (x - a) n - 1 $ 1 $ Q (x) + Ql(x) $ (x - a) n = n (x - a) n - 1 Q (x) + (x - a) n Ql(x) P l (a) = n (a - a) n - 1 Q (a) + (a - a) n Ql(a) = n $ 0 n - 1 $ Q (a) + 0 n $ Ql(a) =0 ` P (a) = P l (a) = 0

EXAMPLES 1. Draw a sketch of P ] x g = - x ] x - 3 g3 .

Solution Roots of the polynomial equation P ] x g = 0: - x ] x - 3 g3 = 0 x = 0, x - 3 = 0 x=3 There is a single root at x = 0 so the curve crosses the x-axis at this point. There is a triple root at x = 3. Since n is odd, there is a point of inflexion at x = 3. P ] x g = - x ] x - 3 g3 There is no need to fully expand the polynomial as we only want to find the leading term.

= - x ^ x 3 - 9x 2 + 27x - 27 h = - x4 f Since –x 4 is the leading term, P(x) has degree 4 and the leading coefficient is negative. So as x becomes large (both negative and positive) the value of P ] x g " -3. y

0

3

x


703

2. A polynomial has a double root at x = 5. (a) Write an expression for the polynomial. (b) Prove that P ] 5 g = P l (5) = 0.

Solution (a) If P(x) has a double root at x = 5, then ] x - 5 g 2 is a factor So P ] x g = ] x - 5 g 2Q ] x g (b) P ] x g = ] x - 5 g 2Q ] x g P ] 5 g = ] 5 - 5 g 2Q ] 5 g = 02 #Q ] 5 g =0 To find P l (5), first we differentiate P(x) using the product rule. P l (x) = u lv + v lu = 2 (x - 5) 1 $ 1 $ Q (x) + Q l (x) $ (x - 5) 2 = 2 ( x - 5 ) Q ( x ) + ( x - 5 ) 2 Q l (x ) 2 P l ( 5 ) = 2 (5 - 5 ) Q ( 5 ) + (5 - 5 ) Q l ( 5 ) 2 = 2 # 0 # Q (5 ) + 0 # Q l ( 5 ) =0 3. A monic polynomial has degree 5 and has a double root at a1 and a triple root at a2. Draw a sketch of the polynomial where a 1 1 a 2 .

Solution Since P(x) is monic and has degree 5, the leading term is x5. We could write P ] x g = _ x - a 1 i 2 _ x - a 2 i 3. Since the polynomial has odd degree and a positive leading coefficient, as x becomes a positive large value, P ] x g " 3 and as x becomes a negative large value, P ] x g " - 3. The double root at x = a 1 gives a maximum or minimum turning point and the triple root at x = a 2 means a point of inflexion. Putting all this information together gives the graph below. y

a1

a2

x

You learned this rule in Chapter 4.

704


12.5 Exercises 1.

Find the roots of each polynomial equation P (x) = 0 and state if they are multiple roots. (a) P ] x g = x 2 - 6x + 9 (b) P ] x g = x 3 - 9x 2 + 14x (c) P ] x g = x 3 - 3x 2 (d) f ] x g = x 3 - 2x 2 - 4x + 8 (e) P ] x g = x 3 - 6x 2 + 12x - 8 (f) A ] x g = x 4 - 4x 3 + 5x 2 - 2x (g) P ] x g = x 4 - 4x 3 - 2x 2 + 12x + 9 (h) Q ] x g = x 5 - 8x 4 + 16x 3 (i) P ] x g = x 4 + 2x 3 - 12x 2 + 14x - 5 (j) f ] x g = 8x 3 - 36x 2 + 54x - 27

(c)

y

x

(d)

y

x

2.

For each graph, state if (i) the leading coefficient is positive or negative and (ii) the degree of the polynomial is even or odd. (a)

y

(e)

y

x

(b)

x

y

(f)

y

x x


(g)

y

(b) If P ] 2 g = 5, write the expression for the polynomial. 5.

Polynomial P ] x g = x 3 - 7x 2 + 8x + 16 has a double root at x = 4. (a) Show that ] x - 4 g 2 is a factor of P(x). (b) Write P(x) as a product of its factors. (c) Prove P ] 4 g = P l(4) = 0.

6.

Polynomial f ] x g = x 4 + 7x 3 + 9x 2 - 27x - 54 has a triple root at x = - 3. (a) Show that ] x + 3 g3 is a factor of f(x). (b) Write f (x) as a product of its factors. (c) Prove f ] - 3 g = f l(- 3) = 0 .

7.

A polynomial has a triple root at x = k and degree n. (a) Write an expression for the polynomial. (b) Prove that P (k) = P l (k) = 0.

8.

Draw an example of a polynomial with leading term (a) x3 (b) - 2x 5 (c) 3x2 (d) - x 4 (e) - 2x 3

9.

Draw an example of a polynomial with a double root at x = 2 and a leading term of 2x3.

x

(h)

y

x

(i)

y

x

(j)

y

x

3.

4.

A monic polynomial of degree 2 has a double root at x = - 4. Write down an expression for the polynomial P(x). Is this a unique expression? A polynomial of degree 3 has a triple root at x = 1. (a) Write down an expression for the polynomial. Is this unique?

10. Draw an example of a polynomial with a double root at x = -1 and leading term - x 3. 11. Sketch an example of a polynomial with a double root at x = 2 and a leading term of x4. 12. Draw an example of a polynomial with a double root at x = - 3 and leading term x 6 .

705

706


13. A polynomial has a triple root at x = 1 and it has a leading term of x3. Draw an example of a graph showing this information. 14. Given a polynomial with a triple root at x = 0 and leading term - x 4, sketch a polynomial on a number plane that fits this information. 15. If a polynomial has a triple root at x = - 2 and a leading term of x8 sketch a polynomial fitting this information. 16. A polynomial has a triple root at x = 4 and its leading term is - 4x 3 . Show this on a number plane.

18. A polynomial with leading term - x 8 has a triple root at x = - 2. Show by a sketch that the polynomial has at least one other root in the domain x 2 - 2. 19. A polynomial has a double root at x = 2 and a double root at x = - 3. Its leading term is 2x5. By drawing a sketch, show that the polynomial has another root in the domain x 2 2. 20. Show that a polynomial with leading term - x 3 and a double root at x = 1 has another root at a point where x 21.

17. A monic polynomial has degree 3 and a double root at x = -1. Show on a sketch that the polynomial has another root in the domain x 2 -1.

Roots and Coefficients of Polynomial Equations In Chapter 10, you studied the relationship between the roots and coefficients of the quadratic equation. In this section you will revise this and also study this relationship for cubic and quartic equations.

Quadratic equation The quadratic equation ax 2 + bx + c = 0 can be written in monic form as c b x2 + a x + a = 0 If the quadratic equation has roots a and b, then the equation can be written in monic form as ( x - a ) (x - b ) = 0 x 2 - bx - a x + ab = 0 x 2 - (a + b ) x + ab = 0 i.e.

c b x 2 + a x + a / x 2 - (a + b) x + ab


This gives the results below:

For the quadratic equation ax 2 + bx + c = 0: • Sum of roots: b a + b = -a • Product of roots: c ab = a

Cubic equation The cubic equation ax 3 + bx 2 + cx + d = 0 can be written in monic form as d c b x 3 + a x 2 + a x + a = 0. If the cubic equation has roots a, b and c then the equation can be written in monic form as (x - a ) (x - b ) (x - c) = 0 (x 2 - bx - a x + ab ) (x - c) = 0 x 3 - cx 2 - bx 2 + bcx - a x 2 + acx + abx - abc = 0 x 3 - (a + b + c) x 2 + (ab + bc + ac) x - abc = 0 d c b x 3 + a x 2 + a x + a / x 3 - (a + b + c) x 2 + (ab + bc + ac) x - abc This gives the results below:

For the cubic equation ax 3 + bx 2 + cx + d = 0: • Sum of roots 1 at a time: b a + b + c = -a • Sum of roots 2 at a time: c ab + ac + bc = a • Product of roots (sum of roots 3 at a time) d abc = - a

Quartic equation The quartic equation ax 4 + bx 3 + cx 2 + dx + e = 0 can be written in monic form d c e b as x 4 + a x 3 + a x 2 + a x + a = 0.

707

708


If the quartic equation has roots a, b, c and d then the equation can be written in monic form as (x - a ) (x - b) (x - c) (x - d) = 0 [x - (a + b + c) x + (ab + bc + ac) x - abc] (x - d) = 0 x - dx - (a + b + c) x 3 + d (a + b + c) x 2 + (ab + bc + ac) x 2 - d (ab + bc + ac) x - abcx + abcd = 0 x 4 - (a + b + c + d) x 3 + (ad + db + dc + ab + bc + ac) x 2 (abd + bdc + adc + abc) x + abcd = 0 e b 3 c 2 d 4 ` x + ax + ax + ax + a 3

4

2

3

/ x 4 - (a + b + c + d) x 3 + (ad + db + dc + ab + bc + ac) x 2 - (abd + bdc + adc + abc) x + abcd This gives the results below:

For the quartic equation ax 4 + bx 3 + cx 2 + dx + e = 0: • Sum of roots 1 at a time: b a + b + c + d = -a • Sum of roots 2 at a time: c ab + ac + ad + bc + bd + cd = a • Sum of roots 3 at a time: d abc + abd + acd + bcd = - a • Product of roots (sum of roots 4 at a time): e abcd = a

This pattern extends to polynomials of any degree.

Class Investigation Can you find results for sums and products of roots for equations of degree 5, 6 and so on?


EXAMPLES 1. If a, b, c are the roots of 2x 3 - 5x 2 + x - 1 = 0, find (a ) (a + b + c) 2 (b) (a + 1) ( b + 1) (c + 1) (c)

1 1 1 + + . a b c

Solution b a + b + c = -a =-

(- 5) 2

5 2 c ab + ac + bc = a 1 = 2 =

d abc = - a ==

(-1) 2

1 2

5 2 1 (a) (a + b + c) 2 = c m = 6 4 2 (b) (a + 1) ( b + 1) (c + 1) = (a + 1) ( bc + b + c + 1) = abc + ab + ac + a + bc + b + c + 1 = abc + (ab + ac + bc) + (a + b + c) + 1 5 1 1 = + + +1 2 2 2 1 =4 2 bc + ac + ab (c) 1 1 1 + + = a c b abc 1 2 = 1 2 =1

CONTINUED

709

710


2. If one root of x 3 - x 2 + 2x - 3 = 0 is 4, find the sum and product of the other two roots.

Solution Roots are a, b, c where, say, c = 4. b a + b + c = -a `

`

a+b+4 =1 a + b = -3 d abc = - a ab (4) = 3 3 ab = 4

3. Solve 12x 3 + 32x 3 + 15x - 9 = 0, given that 2 roots are equal.

Solution Let the roots be a, a and b. b a + a + b = -a `

2a + b = -

32 12

(1)

c aa + ab + ab = a 15 ` a 2 + 2ab = 12 d aab = - a 9 ` a2 b = 12

(2)

(3)

From (1): b=-

32 - 2a 12

Substitute in (2): 32 15 a 2 + 2a c - 2a m = 12 12 32 2 12a + 24a c - 2a m = 15 12 12a 2 - 64a - 48a 2 = 15 0 = 36a 2 + 64a + 15 = (2a + 3) (18a + 5) 2a + 3 = 0 18a + 5 = 0 2a = - 3 18a = - 5 -5 1 a = -1 a= 2 18

(4)


Substitute in (4): 32 1 1 a = -1 : b = - 2 c -1 m 12 2 2 1 = 3 32 5 5 a=- :b=- 2cm 12 18 18 1 = -2 9 Substitute in (3): 9 1 1 1 2 1 a = - 1 , b = : c- 1 m c m = 2 3 2 3 12 3 3 = 4 4 5 5 2 9 1 1 a = - , b = - 2 : cm c- 2 m = 18 9 18 9 12 This is impossible as LHS is negative and RHS is positive. ` the roots are -1

1 1 and 2 3

12.6 Exercises 1.

Given that a and b are the roots of the equation, find (i) a + b and (ii) ab for the following quadratic equations. (a) x 2 - 2x + 8 = 0 (b) 3x 2 + 6x - 2 = 0 (c) x 2 + 7x + 1 = 0 (d) 4x 2 - 9x - 12 = 0 (e) 5x 2 + 15x = 0

2.

Find (i) a + b + c, (ii) ab + ac + bc, and (iii) abc, where a, b and c are the roots of the equation, for the following cubic equations. (a) x 3 + x 2 - 2x + 8 = 0 (b) x 3 - 3x 2 + 5x - 2 = 0 (c) 2x 3 - x 2 + 6x + 2 = 0 (d) - x 3 - 3x 2 - 11 = 0 (e) x 3 + 7x - 3 = 0

3.

For the following quartic equations, where a, b, c and d are the roots of the equation, find (i) a + b + c + d, (ii) ab + ac + ad + bc + bd + cd, (iii) abc + abd + acd + bcd and (iv) abcd (a) x 4 + 2x 3 - x 2 - x + 5 = 0 (b) x 4 - x 3 - 3x 2 + 2x - 7 = 0 (c) - x 4 + x 3 + 3x 2 - 2x + 4 = 0 (d) 2x 4 - 2x 3 - 4x 2 + 3x - 2 = 0 (e) 2x 4 - 12x 3 + 7 = 0

4.

If a and b are the roots of x 2 - 5x - 5 = 0, find (a) a + b (b) ab 1 1 (c) + a b (d) a2 + b2 (e) a3 + b3

711

712


5.

6.

If a , b and c are the roots of 2x 3 + 5x 2 - x - 3 = 0, find (a) abc (b) ab + ac + bc (c) a + b + c 1 1 1 (d) + + a b c (e) (a + 1) ( b + 1) (c + 1) If a, b, c and d are the roots of x 4 - 2x 3 + 5x - 3 = 0, find (a) abcd (b) abc + abd + acd + bcd 1 1 1 1 (c) + + + a b c d

7.

One root of x 2 - 3x + k - 2 = 0 is - 4. Find the value of k.

8.

One root of x 3 - 5x 2 - x + 21 = 0 is 3. Find the sum a + b and the product ab of the other two roots.

9.

Given P (x) = 2x 3 - 7x 2 + 4x + 1, if the equation P (x) = 0 has a root at x = 1, find the sum and product of its other roots.

10. Find the value(s) of k if the quadratic equation x 2 - (k + 2) x + k + 1 = 0 has (a) equal roots (b) one root equal to 5 (c) consecutive roots (d) one root double the other (e) reciprocal roots 11. Two roots of x 3 + mx 2 + 15x - 7 = 0 are equal and rational. Find m.

12. Two roots of x 3 + ax 2 + bx - 5 = 0 are equal to 4 and - 2. Find the values of a and b. 13. (a) Show that 1 is a zero of the polynomial P (x) = x 4 - 2x 3 + 7x - 6. (b) If a, b and c are the other 3 zeros, find the value of a + b + c and abc. 14. If x = 2 is a double root of ax 4 - 2x 3 - 8x + 16 = 0, find the value of a and the sum of the other two roots. 15. Two of the roots of x 3 - px 2 - qx - 4 = 0 are 3 and 5. (a) Find the other root. (b) Find p and q. 16. The product of two of the roots of x 4 + 2x 3 - 18x - 5 = 0 is - 5. Find the product of the other two roots. 17. The sum of two of the roots of x 4 + x 3 + 7x 2 + 14x - 1 = 0 is 4. Find the sum of the other two roots. 18. Find the roots of 8x 3 - 20x 2 + 6x + 9 = 0, given that two of the roots are equal. 19. Solve 12x 3 - 4x 3 - 3x + 1 = 0 if the sum of two of its roots is 0. 20. Solve 6x 4 + 5x 3 - 24x 2 - 15x + 18 = 0 if the sum of two of its roots is zero.


Test Yourself 12 1.

Write p (x) = x 4 + 4x 3 - 14x 2 - 36x + 45 as a product of its factors.

11. Find the x- and y-intercepts of the curve y = x 3 - 3x 2 - 10x + 24.

2.

If a, b and c are the roots of x 3 - 3x 2 + x - 9 = 0, find (a) a + b + c (b) abc (c) ab + ac + bc 1 1 1 (d) + + a b c

12. Divide p (x) = 3x 5 - 7x 3 + 8x 2 - 5 by x - 2, and write p (x) in the form p (x) = (x - 2) a (x) + b (x) .

3.

4.

5.

A monic polynomial P (x) of degree 3 has zeros - 2,1 and 6. Write down the polynomial. (a) Divide P (x) = x 4 + x 3 - 19x 2 - 49x - 30 by x 2 - 2x - 15. (b) Hence, write P (x) as a product of its factors. For the polynomial P (x) = x 3 + 2x 2 - 3x, find (a) the degree (b) the coefficient of x (c) the zeros (d) the leading term.

6.

Sketch f (x) = (x - 2) (x + 3)2 showing the intercepts.

7.

If ax 4 + 3x 3 - 48x 2 + 60x = 0 has a double root at x = 2, find (a) the value of a (b) the sum of the other two roots.

8.

Show that x + 7 is not a factor of x 3 - 7x 2 + 5x - 4.

9.

If the sum of two roots of x 4 + 2x 3 - 8x 2 - 18x - 9 = 0 is 0, find the roots of the equation.

10. The polynomial f (x) = ax 2 + bx + c has zeros 4 and 5, and f (-1) = 60. Evaluate a, b and c.

13. Solve 2 cos 3 x + cos 2 x - cos x = 0 for 0c# x # 360c. 14. When 8x 3 - 5kx + 9 is divided by x - 2, the remainder is -1. Evaluate k. 15. Find the zeros of g (x) = - x 2 + 9x - 20. 16. Sketch P (x) = 2x (x - 3) (x + 5), showing intercepts. 17. Find the value of k if the remainder is - 4 when x 3 + 2x 2 - 3x + k is divided by x - 2. 18. The sum of 2 roots of x 4 - 7x 3 + 5x 2 - x + 3 = 0 is 3. Find the sum of the other 2 roots. 19. A polynomial is given by P (x) = A (x) (x - a) 3 . Show that P (a) = Pl(a) = 0. 20. Show that x - 5 is a factor of f (x) = x 3 - 6x 2 + 12x - 35. 21. (a) Show that x - 5 is a factor of f ] x g = x 3 - 7x 2 - 5x + 75. (b) Show that f ] 5 g = f l(5) = 0 . (c) What can you say about the root at x = 5? (d) Write f (x) as a product of its factors. 22. The leading term of a polynomial is 3x 3 and there is a double root at x = 3. Draw an example of a graph of the polynomial.

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23. A polynomial P(x) has a triple root at x = - 6. (a) Write an expression for P(x). (b) If P(x) has leading coefficient 3 and degree 4, draw a sketch showing this information.

24. Draw an example of a polynomial with leading term 3x 5 . 25. If P ] x g = ax 3 + bx 2 + cx + d has a remainder of 8 when divided by x - 1, P ] 2 g = 17, P ] -1 g = - 4 and P ] 0 g = 5, evaluate a, b, c and d.


Write P (x) = x 5 + 2x 4 + x 3 - x 2 - 2x - 1 as a product of its factors.

8.

Find the value of a if (x + 1) (x - 2) is a factor of 2x 3 - x 2 + ax - 2.

2.

A polynomial P (x) = (x - b) 7 Q (x) . (a) Show that P (b) = P l(b) = 0. (b) Hence find a and b, if (x - 1) 7 is a factor of P (x) = x 7 + 3x 6 + ax 5 + x 4 + 3x 3 + bx 2 x +1.

9.

Prove that if x - a is a factor of polynomial P (x), then P (a) = 0.

Solve tan 4 i - tan 3 i - 3 tan 2 i + 3 tan i = 0 for 0c# i # 360c.

11. Write down an example of a polynomial with the graph below.

3.

4.

(a) Find the equation of the tangent to the curve y = x 3 at the point where x = 1. (b) Find the point where this tangent cuts the curve again.

5.

(a) Find the remainder when p (x) = 2x 4 - 7x 3 + ax 2 + 3x - 9 is divided by 2x - 1. (b) If the remainder, when p (x) is divided by x + 2, is 17, find the value of a.

6.

If a, b and c are roots of the cubic equation 2x 3 + 8x 2 - x + 6 = 0, find (a) abc (b) a2 + b2 + c2

7.

Solve 4 sin 3 i - 3 sin i - 1 = 0 for 0c# i # 360c.

10. Find the points of intersection between the polynomial P ] x g = x 3 + 5x 2 + 4x - 1 and the line 3x + y + 4 = 0.

y

-1

2

x

12. Sketch an example of a polynomial with a double root at x = a 1 and a double root at x = a 2, if the polynomial is monic and has even degree _ a 2 2 a 1 i .

13 Permutations and Combinations TERMINOLOGY Arrangements: Different ways of organising objects Combinations: Arrangements of objects without replacement or repetition when order is not important. The notation used is nC r for selecting r objects from n where order doesn’t matter Factorial: A factorial is the product of n consecutive positive integers from n down to one. For example 6! = 6 x 5 x 4 x 3 x 2 x 1 Fundamental counting principle: If one event can occur in p ways and a second independent event can occur in q ways, then the two successive events can occur in p x q different ways Ordered selections: Selections that are taken in a particular position

Permutations: The arrangement of objects without replacement or repetition when order is important. The notation used is nPr for selecting r objects from n where order matters Random experiments: Experiments that are made with no pattern or order where each outcome is equally likely to occur Sample space: The set of all possible outcomes in an event or series of events Unordered selections: Selections that are made when the order of arrangements is not important or relevant

Chapter 13 Permutations and Combinations

INTRODUCTION THIS CHAPTER IS AN introduction to some of the concepts you will meet in probability in the HSC Course. Probability is the study of how likely it is that something will happen. It is used to make predictions in different areas, ranging from games of chance to business decision-making. In this chapter you will study general counting techniques based on the fundamental counting principle. These will lead on to the study of permutations and combinations. These have applications in finding the size of the sample space, or the ways that certain events can happen in probability. It can also tell us the number of postcodes a state can have, the number of telephone numbers that is possible in a city and the number of combinations making up serial numbers for appliances.

Fundamental Counting Principle Simple probability You have studied probability in earlier stages of mathematics. We can measure probability in theory. However, probability only gives us an approximate idea of the likelihood of certain events happening. For example, in Lotto draws, there is a machine that draws out the balls at random and a panel of supervisors checks that this happens properly. Each ball is independent of the others and is equally likely to be drawn out. In a horse race, it is difficult to measure probability as the horses are not all equally likely to win. Other factors such as ability, training, experience and weight of the jockey all affect it. The likelihood of any one horse winning is not random. The probability of an event E happening, P(E), is given by the number of ways the event can occur, n(E), compared with the total number of outcomes possible n(S) (the size of the sample space). n (E ) P ]Eg = n (S) If P ] E g = 0 the event is impossible. If P ] E g = 1 the event is certain (it has to happen). 0 # P ]Eg # 1

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The sum of all probabilities is 1. Complementary events: P ] not E g = 1 – P ] E g or P^L E h = 1 - P (E) where L E is the complement of E P ]Eg + P ^ L Eh = 1

EXAMPLES 1. Alison buys 5 raffle tickets and 100 are sold altogether. What is the probability that Alison (a) wins (b) doesn’t win first prize in the raffle?

Solution (a) The size of the sample space, or total number of outcomes is 100, since there are 100 tickets altogether. Alison has 5 tickets so has 5 different ways of winning the raffle. 5 100 1 = 20 (b) There are 100 - 5 or 95 other tickets that could win if Alison loses. P (Win) =

95 100 19 = 20 Or, if we know that the sum of all probabilities is 1, we could say P (Loss) =

P (Loss) = 1 - P (Win) =1=

1 20

19 20

2. There are 56 books on music at the school library and there are 2000 books altogether. If Anthony selects a book at random, find the probability that it will be a book on music.

Solution The size of the sample space is 2000 and there are 56 ways that Anthony could select a music book. 56 P (Music book) = 2000 7 = 250


13.1 Exercises 1.

A lottery is held in which 20 000 tickets are sold. If I buy 2 tickets, what is the probability of my winning the prize in the lottery?

2.

The probability of a bus arriving 17 on time is estimated at . What 33 is the probability that the bus will not arrive on time?

3.

7.

A shoe shop orders in 20 pairs of black, 14 pairs of navy and 3 pairs of brown school shoes. If the boxes are all mixed up, find the probability that one box selected at random will contain brown shoes.

8.

A biased coin is weighted so that heads comes up twice as often as tails. Find the probability of tossing a tail.

9.

A die has the centre dot painted white on the 5 so that it appears as a 4. Find the probability of rolling (a) a 2 (b) a 4 (c) a number less than 5.

The probability of a seed 7 . 9 Find the probability that the seed will produce a different coloured flower.

producing a pink flower is

4.

In a lottery, 200 000 tickets are sold. If Lucia buys 10 tickets, what is the probability of her winning first prize?

5.

A machine has a 1.5% chance of breaking down at any given time. What is the probability of the machine not breaking down?

6.

A bag contains 6 red balls and 8 white balls. If one ball is drawn out at random, find the probability that it will be (a) white (b) red.

10. A book has 124 pages. If the book is opened at any page at random, find the probability of the page number being (a) either 80 or 90 (b) a multiple of 10 (c) an odd number (d) less than 100. 11. In the game of pool, there are 15 balls, each with the number 1 to 15 on it. In Kelly Pool, each person chooses a number at random from a container and has to try and sink the ball with the corresponding number. If Tracey chooses a number, find the probability that her ball will be (a) the eight ball (b) an odd number (c) a number less than 10.

A multiple of 10 is a number that is divisible by 10.

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12. A box containing a light globe 1 has a probability of holding 20 a defective globe. If 160 boxes are checked, how many globes would be expected to be defective? 13. There are 29 red, 17 blue, 21 yellow and 19 green jelly beans in a packet. If Kate chooses one at random, find the probability that it will be (a) red (b) blue or yellow (c) not green. 14. The probability of breeding a 2 white budgie is . If Mr Seed 15 breeds 240 budgies over the year, how many would be expected to be white? 15. A die is rolled. Calculate the probability of rolling (a) a 6 (b) an even number (c) a number less than 3 (d) 4 or more (e) a multiple of 2. 16. The probability that an arrow will 13 . hit a target is 18 (a) Find the probability that it will miss the target. (b) If 126 arrows are fired, how many would be expected to hit a target? 17. A dog can catch a ball in its mouth 8 times out of 10. (a) What is the probability of the dog catching a ball? (b) If the ball is thrown to the dog 45 times, how many times would the dog be expected to catch it?

18. In a bag there are 21 lollies with pink wrappers and 23 with blue wrappers. If Leila chooses a lolly at random from the bag, find the probability that she selects a lolly with (a) a blue wrapper (b) a pink wrapper. 19. In a survey, it was found that 18 people preferred Brand A of soft drink while 39 people preferred Brand B. What is the probability that a person chosen at random prefers Brand A? 20. A school has 653 junior and 329 senior students. If a student is chosen at random, what is the probability that it will be a senior student? 21. A class has 12 girls and 19 boys. Eight of the girls and 12 of the boys play a sport. If a student is chosen at random, find the probability that the person chosen (a) is a boy (b) plays a sport (c) is a girl who doesn’t play a sport (d) doesn’t play a sport. 22. Amie’s CD collection includes 21 R&B, 14 rock and 24 jazz albums. If she selects one CD to play at random, find the probability that the CD is (a) rock (b) R&B or rock (c) jazz (d) not R&B.


23. The probability of winning a 3 game of chance is and the 8 5 probability of losing is . What 12 is the probability of a draw?

What percentage would vote for Greens? 25. An arrow has a 0.37 probability of hitting a target outside the bullseye zone and 0.12 probability of missing the target altogether. What is the probability of the arrow hitting the bullseye area?

24. In a poll, 39% said they would vote Labor, 34% said they would vote Liberal and 8% said they would vote for independents or small parties.

Counting techniques In the HSC Course you will learn about multi-stage events (events where there are two or more events such as rolling two dice or tossing two coins). The counting can become quite difficult, which is why we introduce counting techniques in the Preliminary Course.

Class Discussion Break up into pairs and try these experiments with one doing the activity and one recording the results. 1. Toss two coins as many times as you can in a 5-minute period and record the results in the table: Result

Two heads

One head and one tail

Two tails

Tally Compare your results with others in the class. What do you notice? Is this surprising? 2. Roll two dice as many times as you can in a 5-minute period, find the total of the two uppermost numbers on the dice and record the results in the table: Total

2

3

4

5

6

7

8

9

10

11

12

Tally Compare your results with others in the class. What do you notice? Is this surprising? Why don’t these results appear to be equally likely?

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There are many examples of where counting techniques are useful, in probability and in areas such as manufacturing, business, biology and economics. For example, in genetics, the number of molecules on DNA strands can be difficult to find.

Investigation 1. To travel to work, Cassie needs to catch a bus and a train. She lives near a bus stop and there are three different buses she could catch into town. When she arrives in town, she needs to catch one of four trains to work. If there are three buses and four trains possible for Cassie to catch, in how many ways is it possible for her to travel to work? Buses

Cassie’s house

Trains

A B C

1 2 3 4

2. At a restaurant, there are three entrees, four main meals and two desserts. Every time Rick eats at the restaurant he chooses to eat a different combination of courses. How many times would he need to go to the restaurant to cover all possible combinations?

FUNDAMENTAL COUNTING PRINCIPLE If one event can happen in a different ways, a second event can happen in b different ways, a third in c different ways and so on, then these successive events can happen in abc … different ways.

EXAMPLES 1. A personal identification number (PIN) has 4 letters followed by 3 numbers. How many different PINs of this type are possible?

Solution There are 26 letters and 10 numbers (0 – 9) possible for the positions in the PIN.


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Total number = 26 # 26 # 26 # 26 #10 #10 #10 = 26 4 #10 3 = 456 976 000 So 456 976 000 PINs are possible. 2. A restaurant serves 5 different types of entree, 12 main courses and 6 desserts. (a) If I order any combination of entree, main course and dessert at random, how many different combinations are possible? (b) If my friend makes 3 guesses at which combination I will order, what is the probability that she will guess correctly?

Solution (a) Total number of combinations = 5#12 # 6 = 360 3 360 1 = 120

(b) P ^ correct guess h =

Here are some examples of counting when there is no repetition or replacement.

EXAMPLES 1. To win a trifecta in a race, a person has to pick the horses that come first, second and third in the race. If a race has 9 horses, how many different combinations could be a trifecta?

Solution The probabilities will be different for where each horse will come in the race, but the number of possible different trifecta combinations will be the same.

Any of the 9 horses could come first. Any of the remaining 8 could come second. Any of the remaining 7 horses could come third. Total ways = 9 # 8 # 7 = 504 CONTINUED

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2. A group of 15 people attend a concert and 3 of them are randomly given a free backstage pass. The first person receives a gold pass, the second one a silver pass and the third one a bronze pass. In how many different ways can the passes be given out?

Solution Any of the 15 people can receive the first pass. There are 14 people left who could receive the second pass. Similarly there are 13 people that could receive the third pass. Total number of possibilities = 15#14 #13 = 2 730 3. In Lotto, a machine contains 45 balls, each with a different number from 1 to 45. (a) In how many ways can 6 balls be randomly drawn? (b) To win first prize in Lotto, a person must choose all 6 numbers correctly. Lisa has 3 tickets in the same draw of Lotto. What is the probability that she will win first prize?

Solution (a) The first ball could be any of the 45 balls. The second could be any of the remaining 44 balls and so on. The number of ways = 45# 44 # 43# 42 # 41# 40 = 5 864 443 200 3 (b) P ^ first prize h = 5 864 443 200 1 = 1 954 814 400

13.2 Exercises 1.

A password has 4 letters. How many combinations are possible?

2.

A motorcycle numberplate is made up of 2 letters followed by 2 numbers. How many numberplates of this type are available?

3.

A password can have up to 5 letters followed by 4 numbers on it. If I could use any letter of the alphabet or number, how many different passwords could be formed? Leave your answer in index form.


4.

A witness saw most of the numberplate on a getaway car except for the first letter and the last number. How many different cars do the police need to check in order to find this car?

5.

A certain brand of computer has a serial number made up of 10 letters then 15 numbers. How many computers with this type of serial number can be made? Leave your answer in index form.

6.

Victoria has postcodes starting with 3. How many different postcodes are available in Victoria?

7.

8.

9.

A country town has telephone numbers starting with 63 followed by any 6 other numbers from 0 to 9. How many telephone numbers are possible in this town? Yasmin has 12 tops, 5 pairs of jeans and 5 pairs of shoes in her wardrobe. If she randomly chooses a top, pair of jeans and shoes, how many combinations are possible? A car manufacturer produces cars in 8 different colours, with either manual or automatic gear transmission, and 4 different types of wheels. How many different combinations can it produce?

10. A PIN has 4 numbers. If I forget my PIN I am allowed 3 tries to get it right. Find the probability that I get it within the 3 tries. 11. A restaurant offers 7 main courses and 4 desserts, as well as 3 different types of coffee. (a) How many different combinations of main

course, dessert and coffee are possible? (b) Find the probability that I randomly pick the combination voted most favourite. 12. A telephone number in a capital city can start with a 9 and has 8 digits altogether. (a) How many telephone numbers are possible? (b) If I forget the last 3 digits of my friend’s telephone number, how many numbers would I have to try for the correct number? 13. A company manufactures 20 000 000 computer chips. If it uses a serial number on each one consisting of 10 letters, will there be enough combinations for all these chips? 14. A password consists of 2 letters followed by 5 numbers. What is the probability that I randomly guess the correct password? 15. A city has a population of 3 500 000. How many digits should its telephone numbers have so that every person can have one? 16. A manufacturer of computer parts puts a serial number on each part, consisting of 3 letters, 4 numbers then 4 letters. The number of parts sold is estimated as 5 million. Will there be enough combinations on this serial number to cope with these sales? 17. A bridal shop carries 12 different types of bridal dresses, 18 types of veils and 24 different types of shoes. If Kate chooses a combination of dress, veil and

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shoes at random, what is the probability that she chooses the same combination as her friend Jane? 18. Kate chooses a different coloured dress for each of her 3 bridesmaids. If the colours are randomly given to each bridesmaid, how many different possibilities are there?

3 free haircuts randomly given away, in how many ways could this be done? 23. A flock of 28 pelicans is fed 6 fish carcasses. If each carcass is given to a different pelican, in how many ways can this happen?

19. In a computer car race game, the cars that come first, second and third are randomly awarded. If there are 20 cars, how many possible combinations of first, second and third are there? 20. Jacquie only has 4 chocolates left and decides to randomly choose which of her 6 friends will receive one each. How many possible ways are there in which can she give the chocolates away? 21. Three prizes are given away at a concert by taping them underneath random seats. If there are 200 people in the audience, in how many ways can these prizes be won? 22. There are 7 clients at a hairdressing salon. If there are

24. A set of cards is numbered 1 to 100 and 2 chosen at random. (a) How many different arrangements of ordered pairs are possible? (b) What is the probability that a particular ordered pair is chosen? 25. Each of 10 cards has a letter written on it from A to J. If 3 cards are selected in order at random, find the probability that they spell out CAB.

Factorial notation Counting outcomes when repetition or replacement is allowed is quite straightforward, even when the numbers become very large.

EXAMPLE A card is drawn randomly from a set of 25 cards numbered 1 to 25 in turn and then replaced before the next is selected. How many possible outcomes are there if 25 cards are chosen this way? Answer in scientific notation, correct to 3 significant figures.


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Solution Each time there is a card drawn, there are 25 possibilities. Total number = 25# 25# 25#. . .# 25

] 25 times g

= 25 = 8.88 #10 34 25

When there is no repetition or replacement, the calculations can be quite long.

EXAMPLE A card is drawn randomly from a set of 25 cards numbered 1 to 25 in turn without replacing it before the next is drawn. How many possible outcomes are there if all 25 cards are drawn out? Answer in scientific notation, correct to 3 significant figures.

Solution First card: there are 25 possibilities. Second card: there are only 24 possibilities since one card has already been drawn out. Third card: there are 23 possibilities and so on. Total number = 25# 24 # 23#…# 3# 2 #1 = 1.55#10 25

This calculation is quite tedious!

Factorial notation allows us to easily calculate the number of possible outcomes when selecting all objects in order with no replacement or repetition. You can find a x!

n! = n ] n - 1 g ] n - 2 g ] n - 3 g ] n - 4 g . . . 3# 2 #1

key on most scientific calculators.

Since the sequence of numbers multiplied doesn’t go further than 1, then by convention we say that

0! = 1

Check 0! on your calculator.

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EXAMPLES 1. Evaluate (a) 4! (b) 7! (c) 25! (answer in scientific notation correct to 3 significant figures.)

Solution 4! = 4 # 3 # 2 #1 = 24 (b) 7! = 7 # 6 # 5 # 4 # 3 # 2 # 1 = 5040 (c) 25! = 1.55#10 25 (a)

It is much easier to use the x! key on a calculator to find this.

2. A group of 9 teenagers is waiting to be served in a café. They are each randomly assigned a number from 1 to 9. (a) In how many ways is it possible for the numbers to be assigned? (b) One of the group needs to be served quickly as he has to leave. If he is given the first number, in how many ways is it possible for the numbers to be assigned?

Solution (a) The first number could be assigned 9 ways. The second number could be assigned 8 ways and so on. Total ways = 9! = 362 880 (b) One of the group is given the first ticket (this can only happen in one way) The second number could be assigned 8 ways and so on. Total ways = 1# 8! = 40 320

13.3 Exercises 1.

Evaluate (a) 6! (b) 10! (c) 0! (d) 8! - 7! (e) 5# 4! 7! (f) 4!

12! 5! 13! (h) 4!9! 8! (i) 3!5! 11! (j) 4!7! (g)


2.

A group of 9 jockeys are each given a set of riding colours to wear. If these are given out in order randomly, how many different arrangements are possible?

8.

A group of 7 people line up to do karaoke. If they are each randomly given a song to sing, how many possible outcomes are there?

9.

A kindergarten class has a rabbit, a mouse and a parrot. Three children are selected to take these pets home for the holidays. If the pets are randomly given out to these children, how many different ways are possible?

10. A PIN consists of 6 numbers, with no repetition of any numbers allowed. How many different PINs are possible? 3.

Each of 6 people at a restaurant is given a different coloured glass. How many possible combinations are there?

4.

A mountain trail only has room for one person at a time. If 12 people are waiting at the bottom of the trail and are randomly picked to start out, in how many ways can this happen?

5.

6.

7.

A dog walker has 5 dogs and 5 leashes. In how many different ways is it possible to put a leash on each dog? There are 11 people in a choir and each receives a musical score. (a) In how many different ways could the scores be handed out? (b) Russell, the musical director, also needs a musical score. In how many ways could the scores be handed out for the choir and the musical director? A row of seats in a theatre seats 8 people. In how many ways could a group of 8 friends be randomly seated in this row?

11. In a chorus for a school musical, 7 students each wear a different mask. In how many different ways can the masks be worn by these students? 12. If 15 people play a game of Kelly pool, each person in turn chooses a number at random between 1 and 15. In how many different ways can this occur? Answer in scientific notation, correct to one decimal place. 13. (a) A school talent quest has 11 performers and each one is randomly given the order in which to perform. In how many ways can the order of performances be arranged? (b) If one performer is chosen to perform first, in how many ways can the others be arranged? 14. A group of 6 friends sit in the same row at a concert. (a) In how many different ways can they arrange themselves? (b) If one friend must sit on the aisle, in how many ways can they be arranged?

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15. A group of 8 friends go to a restaurant and sit at a round table. If the first person can sit anywhere, in how many ways can the others be arranged around the table? 16. In a pack of cards, the 4 aces are taken out and shuffled. (a) What is the probability of picking out the Ace of Hearts at random? (b) If all the aces are arranged in order, what is the probability of guessing the correct order?

(b) What is the probability that the bride will have a pink rose at her table? 18. In a maths exam, a student has to arrange 5 decimals in the correct order. If he has no idea how to do this and arranges them randomly, what is the probability that he makes the right guess for all the decimals? 19. In a car race, the fastest car is given pole position and the other cars are randomly given their starting positions. If there are 14 cars altogether, in how many ways can this be arranged? 20. Show that 8! (a) = 8 #7 # 6 # 5 4! (b)

11! = 11#10 # 9 # 8 #7 6!

n! r! = n ]n - 1 g]n - 2 g]n - 3 g

(c)

17. At a wedding, each of the 12 tables is to have a centrepiece with a different coloured rose. (a) In how many different ways can the roses be arranged at random?

... ] r + 1 g where n 2 r n! (n - r) ! = n ]n - 1 g]n - 2 g]n - 3 g ... ] n - r + 1 g where n 2 r

(d)

Permutations Factorial notation is useful for finding the number of possible outcomes when arranging all objects in order without replacement. However, sometimes we need to find the number of possible outcomes when arranging only some of the objects in order without replacement. It is easy to arrange objects with replacement.


EXAMPLE In how many ways can 5 cards be selected from a total of 20 cards if each one is replaced before selecting the next one?

Solution Each selection can be made in 20 possible ways. Total = 20 # 20 # 20 # 20 # 20 = 20 5 = 3 200 000 ways.

For r selections from n objects (with repetitions), the number of possible outcomes is n # n # n # n f (r times) or nr

However, when arranging r objects from n objects in order without replacement, it is not so easy.

EXAMPLE In how many ways can 5 cards be selected from a total of 20 cards if there is no replacement?

Solution The 1st card can be selected in 20 different ways. The 2nd card can be selected in 19 different ways as the first card is no longer being used. The 3rd card can be selected in 18 different ways, and so on. Total = 20 #19 #18 #17 #16 = 1 860 480 ways

The calculations can become tedious if we select a larger number of objects.

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EXAMPLE If there are 20 cards and 13 cards are chosen in order at random without replacement, find the possible number of ways the cards can be chosen in scientific notation correct to 1 decimal place.

Solution The first card can be any of the 20 numbers. The second card can be any of the remaining 19 numbers. The third can be any of the remaining 18 numbers. The number of ways the cards can be chosen = 20 #19 #18 #17 #f# 8 = 4.8 #1014

For r ordered selections from n objects without replacement, the number of possible outcomes is n # ] n - 1 g # ] n - 2 g # ] n - 3 g f ] r times g or n ]n - 1 g]n - 2 g]n - 3 g f ]n - r + 1 g

A permutation describes an arrangement of r objects from a total of n objects in a certain order without replacement or repetition.

Permutation n Pr is the number of ways of making ordered selections of r objects from a total of n objects. n! n Pr = ]n - r g!

You can find a n Pr key on most scientific calculators.

Proof n

Pr = n ] n - 1 g ] n - 2 g ] n - 3 g f ] n - r + 1 g

]n - r g]n - r - 1g]n - r - 2gf 3 $ 2 $ 1 ]n - r g]n - r - 1g]n - r - 2g f 3 $ 2 $ 1 n ]n - 1g]n - 2g]n - 3g f ]n - r + 1g]n - r g]n - r - 1g]n - r - 2g f 3 $ 2 $ 1 = ]n - r g]n - r - 1g]n - r - 2g f 3 $ 2 $ 1

= n ]n - 1g]n - 2g]n - 3g f ]n - r + 1g #

=

n!

]n - r g!

A special case of this result is:

n

Pn = n!


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Proof n! ]n - r g! n! ` n Pn = ]n - n g! n! = 0! n! = 1 = n! n

Pr =

EXAMPLES You can evaluate this on a calculator.

1. Evaluate 9 P4

Solution 9

9! ]9 - 4 g! 9! = 5! 9$8$7$6$5$4$3$2$1 = 5$4$3$2$1 =9$8$7$6

P4 =

= 3024 2. (a) Find the number of arrangements of 3 digits that can be formed using the digits 0 to 9 if each digit can only be used once. (b) How many 3 digit numbers greater than 700 can be formed?

Solution (a) There are 10 digits from 0 to 9. The 1st digit can be any of the 10 digits. The 2nd digit can be any of the remaining 9 digits. The 3rd digit can be any of the remaining 8 digits. Total permutations = 10 # 9 # 8 = 720 10! or 10P3 = ] 10 - 3 g ! 10! = 7! = 720 CONTINUED

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(b) The 1st digit must be 7 or 8 or 9 (3 possible digits). The 2nd digit can be any of the remaining 9 digits. The 3rd digit can be any of the remaining 8 digits. Total arrangements = 3# 9#8 = 216 Another method: There are 3 ways to get the 1st digit. The possible arrangements of the remaining 2 digits is 9 P2 Total arrangements = 3# 9 P2 = 3#72 = 216

There are some special examples that need very careful counting, such as arrangements around a circle. Others involve counting when there are identical objects.

EXAMPLES 1. (a) In how many ways can 6 people sit around a circular table? (b) If seating is random, find the probability that 3 particular people will sit together.

Solution (a) The 1st person can sit anywhere around the table so we only need to arrange the other 5 people.

The 2nd person can sit in any of the 5 remaining seats. The 3rd person can sit in any of the remaining 4 seats and so on. Total arrangements = 5! = 120


(b) The 3 people can sit anywhere around the table together in 3# 2 #1 or 3! ways. The remaining 3 people can sit together in 3! ways. Total arrangements = 3!# 3! = 36 36 P (3 sit together) = 120 3 . = 10 2. In how many ways can the letters of the word EXCEPTIONAL be arranged?

Solution EXCEPTIONAL has 11 letters with the letter E repeated. If each E was different, i.e. E1 and E2, then there would be 11! arrangements. However, we cannot tell the difference between the 2 Es. Since there are 2! ways of arranging the Es, then there are 2! arrangements of the word EXCEPTIONAL that are identical. We need to divide by 2! to eliminate these identical arrangements. 11! Total arrangements = 2! = 19 958 400.

The number of different ways of arranging n objects in which a of the objects are of one kind, b objects are of another n! kind, c of another kind and so on, is given by where a!b!c!f a + b + c +f# n

EXAMPLE Find the number of ways that the word ANAETHEMA can be arranged.

Solution There are 9 letters, including 3 As and 2 Es. There are 9! ways of arranging the letters, with 3! ways of arranging the As and 2! ways of arranging the Es. 9! Total arrangements = 3!2! = 30 240

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Some questions involving counting need different approaches and sometimes it is just a matter of logically working it out.

EXAMPLES A bag contains 5 balls of different colours—red, yellow, blue, green and white. In how many ways can these 5 balls be arranged (a) with no restrictions (b) if the yellow ball must be first (c) if the first ball must not be red or white (d) if blue and green must be together (e) if red, blue and green must be together?

Solution (a) The 1st can be any of the 5 balls. The 2nd can be any of the remaining 4 balls and so on. Total arrangements = 5! = 120 (b) The 1st ball must be yellow, so there is only 1 way of arranging this. The 2nd ball can be any of the remaining 4 balls. The 3rd ball can be any of the remaining 3 balls and so on. Total arrangements = 4! = 24 (c) The 1st ball could be yellow, blue or green so there are 3 possible arrangements. The 2nd ball could be any of the remaining 4 balls and so on. Total arrangements = 3# 4! = 72 (d) When two objects must be together, we treat them as a single object with 2! possible arrangements. So we arrange 4 balls in 4! ways: R, Y, BG and W. But there are 2! ways in which to arrange the blue and green balls. Total arrangements = 4!# 2! = 48 (e) When three objects are together, we treat them as a single object with 3! possible arrangements. We are then arranging 3 balls in 3! ways: RBG, Y, W. But there are 3! ways in which to arrange the red, blue and green balls. Total arrangements = 3!# 3! = 36


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13.4 Exercises 1.

Write each permutation in factorial notation and then evaluate. (a) 6 P3 (b) 5 P2

(c) of odd numbers (d) of even numbers. 5.

(a) How many arrangements of the letters A, B, C and D are possible if no letter can be used twice? (b) How many arrangements of any 3 of these letters are possible?

6.

A 4 digit number is to be selected at random from the numbers 0 to 9 with no repetition of digits. (a) How many arrangements can there be? (b) How many arrangements of numbers over 6000 are there? (c) How many arrangements of numbers less than 8000 are there?

(c) 8 P3 (d)

10

P7

(e) P6 9

(f)

7

P5

(g) P6 8

2.

3.

4.

P8

(h)

11

(i)

9

P1

(j)

6

P6

A set of 26 cards, each with a different letter of the alphabet, is placed into a hat and cards drawn out at random. Find the number of ‘words’ possible if selecting (a) 2 cards (b) 3 cards (c) 4 cards (d) 5 cards. A 3 digit number is randomly made from cards containing the numbers 0 to 9. (a) In how many ways can this be done if the cards cannot be used more than once and zero cannot be the first number? (b) How many numbers over 400 can be made? (c) How many numbers less than 300 can be made? A set of 5 cards, each with a number from 1 to 5 on it, is placed in a box and 2 drawn out at random. Find the possible number of combinations (a) altogether (b) of numbers greater than 50 possible

7.

The numbers 1, 2, 3, 4 and 5 are arranged in a line. How many arrangements are possible if (a) there is no restriction (b) the number is less than 30 000 (c) the number is greater than 20 000 (d) the number is odd (e) any 3 numbers are selected at random?

8.

There are 12 swimmers in a race. (a) In how many ways could they finish? (b) In how many ways could they come in first, second and third?

9.

How many different ordered arrangements can be made from the word COMPUTER with (a) 2 letters (b) 3 letters (c) 4 letters?

The first number cannot be zero.

The first number is not zero.

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10. How many different ordered arrangements can be made from these words? (a) CENTIPEDE (b) ALGEBRA (c) TELEVISION (d) ANTARCTICA (e) DONOR (f) BASKETBALL (g) GREEDY (h) DUTIFUL (i) MANUFACTURER (j) AEROPLANE 11. A group of friends queue outside a restaurant in a straight line. Find how many ways the friends can be arranged if there are (a) 4 friends (b) 7 friends (c) 8 friends (d) 10 friends (e) 11 friends. 12. A group of friends go into a restaurant and are seated around a circular table. Find how many arrangements are possible if there are (a) 4 friends (b) 7 friends (c) 8 friends (d) 10 friends (e) 11 friends.

All beads are different from each other.

13. A string of beads looks the same if turned over. Find the number of different arrangements possible with (a) 10 beads (b) 12 beads (c) 9 beads (d) 11 beads (e) 13 beads. 14. In how many ways can a group of 6 people be arranged (a) in a line (b) in a circle?

15. Find how many different ways a group of 9 people can be arranged in (a) a line (b) a circle. 16. In how many ways can a set of 10 beads be arranged (a) in a line (b) in a circle around the edge of a poster (c) on a bracelet? 17. (a) How many different arrangements can be made from the playing cards Jack, Queen, King and Ace? (b) If I choose 2 of these cards randomly, how many different arrangements could I make? (c) If I choose 3 of these cards randomly, how many different arrangements could I make? 18. A group of 7 people sit around a table. In how many ways can they be arranged (a) with no restrictions (b) if 2 people want to sit together (c) if 2 people cannot sit together (d) if 3 people sit together? 19. A group of 5 boys and 5 girls line up outside a cinema. In how many ways can they be arranged (a) with no restriction? (b) If a particular girl stands in line first? (c) If they alternate between boys and girls (with either a girl or boy in first place)? 20. Find the probability that if 10 people sit around a table, 2 particular people will be seated together at random.


21. A bookshelf is to hold 5 mathematics books, 8 novels and 7 cookbooks. (a) In how many different ways could they be arranged? (Leave answer in factorial notation.) (b) If the books are grouped in categories, in how many ways can they be arranged? (Answer in factorial notation.) (c) If one book is chosen at random, find the probability that it is a cookbook.

25. A minbus has 6 forward facing and 2 backward facing seats. If 8 people use the bus, in how many ways can they be seated (a) with no restrictions (b) if one person must sit in a forward facing seat (c) if 2 people must sit in a forward facing seat? 26. If 3 letters of the word VALUED are selected at random, find the number of possible arrangements if (a) the first letter is D. (b) the first letter is a vowel. 27. The letters of the word THEORY are arranged randomly. Find the number of arrangements. (a) with no restrictions. (b) if the E is at the beginning. (c) if the first letter is a consonant and the last letter is a vowel.

22. (a) How many different arrangements can be made from the numbers 3, 4, 4, 5 and 6? (b) How many arrangements form numbers greater than 4000? (c) How many form numbers less than 5000? (d) If an arrangement is made at random, find the probability that it is less than 4000.

28. Find the number of arrangements possible if x people are (a) in a straight line (b) in a circle (c) in a circle with 2 people together (d) in a straight line with 3 people together (e) in a circle with 2 people not together.

23. Find the probability that an arrangement of the word LAPTOP will start with T.

29. (a) Use factorial notation to 8 8 P3 P5 show that = 3! 5! n n Pn - r Pr (b) Prove that = ] r! n - r g!

24. What is the probability that if a 3 letter ‘word’ is formed randomly from the letters of PHYSICAL, it will be CAL?

30. Prove that n + 1 Pr = n Pr + r nPr - 1

Vowels are letters a, e, i, o and u while consonants are all other letters.

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Combinations The permutation n Pr is the number of arrangements possible for an ordered selection of r objects from a total of n objects. When the order is not important, for example when AB is the same as BA, the number of arrangements is called a combination.

EXAMPLES 1. A committee of 2 is chosen from Scott, Rachel and Kate. In how many ways can this be done?

Solution Number of ordered arrangements = 3 P2 =6 However, a committee of Scott and Rachel is the same as a committee of Rachel and Scott. This is the same for all other arrangements of the committee. There are 2! ways of arranging each committee of two people. To get the number of unordered arrangements, we divide the number of ordered arrangements by 2! 3 P2 Total arrangements = 2! =3 2. There are 3 vacancies on a school council and 8 people who are available. If the vacancies are filled randomly, in how many ways can this happen?

Solution Number of orderedarrangements = 8 P3 However, order is not necessary here, since the 3 vacancies filled by, say, Hamish, Amie and Marcus, would be the same in any order. There are 3! different ways of arranging Hamish, Amie and Marcus. 8 P3 So total arrangements = 3! = 56

The number of ways of making unordered selections of r n

objects from n is

Pr

r!

which is the same as

n! ] n - r g ! r!


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Proof n

Pr is the ordered selection of r objects from n objects. There are r! ways of arranging r objects. If order is unimportant, the unordered selection of r objects from n is given n Pr by . r! n! n Pr ]n - r g! = r! r! n! 1 # = ] n - r g! r! n! = ] n - r g !r !

n Combination n Cr or a r k is the number of ways of making unordered selections of r objects from a total of n objects. n! n Cr = ] n - r! g r !

EXAMPLES 1. A bag contains 3 white and 2 black counters labelled W1, W2, W3 and B1, B2. If two counters are drawn out of the bag, in how many ways can this happen if order is not important?

Solution Possible arrangements (unordered) are: W1 W2 W1 W3 W1 B1 W1 B2

W2 W3 W2 B1 W2 B2

W3 B1 W3 B2

B1 B2

There are 10 different combinations. Using combinations, the number of different arrangements of choosing 2 counters from 5 is 5 C 2 . 5! 5 C2 = (5 - 2)!2! 5! = 3!2! = 10

CONTINUED

We can call this ‘choose’ notation.

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2. If 12 coins are tossed, find the number of ways of tossing 7 tails.

Solution The order is not important. There are 12 C7 ways of tossing 7 tails from 12 coins 12

12! (12 - 7)!7! 12! = 5!7! = 792

C7 =

3. (a) A committee of 5 people is formed randomly from a group of 15 students. In how many different ways can the committee be formed? (b) If the group consists of 9 senior and 6 junior students, in how many ways can the committee be formed if it is to have 3 senior and 2 junior students in it?

Solution (a) The order of the committee is not important. l Number of arrangements = b 15 5 = 3003 l (b) 3 senior students can be chosen in b 9 3 or 84 ways. l 2 junior students can be chosen in b 6 2 or 15 ways. 9 6 Total number of arrangements = c m # c m 3 2 = 84 #15 = 1260 4. A team of 6 men and 5 women is chosen at random from a group of 10 men and 9 women. If Kaye and Greg both hope to be chosen in the team, find the probability that (a) both will be chosen (b) neither will be chosen.

Solution The number of possible teams = 10 C 6 # 9 C 5 = 210 #126 = 26 460 (a) For Kaye to be chosen, then 4 out of the other 8 women will be chosen i.e. 8 C 4 For Greg to be chosen, 5 out of the other 9 men will be chosen i.e. 9 C5


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Number of combinations = 8 C4 # 9 C5 = 70 # 126 = 8820 8820 26 460 1 = 3

Probability =

(b) For Kaye and Greg not to be included, then 5 out of the other 8 women and 6 out of the other 9 men will be chosen. Number of combinations = 8 C5 # 9 C6 = 56 #84 = 4704 4704 Probability = 26 460 8 = 45

13.5 Exercises 1.

2.

Write in factorial notation and evaluate. l (a) b 9 5 l (b) b 12 7 l (c) b 8 3 (d)

10

C4

(e)

11

C5

people can be made randomly from a group of (a) 8 people (b) 9 people (c) 11 people (d) 15 people (e) 20 people. 4.

(a) Evaluate (i) 10 C 0 (ii)

7

C0

l (iii) b 14 0

(iv)

9

C9

11 m 11 (b) Hence complete (i) n C 0 = (v)

(ii) 3.

c

n

Cn =

Find the number of different ways that a committee of 6

5.

(a) A set of 3 red cards and 3 blue cards are placed in a box. By naming the red cards R1, R2 and R3 and the blue cards B1, B2 and B3, list the number of different arrangements possible when 2 cards are drawn out at random, with order not important. How many arrangements are possible? (b) If there are 10 red and 10 blue cards and 7 are drawn out at random, how many different combinations are possible? A coin is tossed 20 times. How many different arrangements are there for tossing 5 heads?

The cards are all distinct from each other.

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6.

A set of 10 marbles are placed in a bag and 6 selected at random. In how many different ways can this happen?

7.

In poker, 5 cards are dealt from a pack of 52 playing cards. How many different arrangements are possible?

8.

Three cards are drawn randomly from a set of 10 cards with the numbers 0 to 9 on them. How many different arrangements are possible if order is (a) important (b) unimportant?

9.

A debating team of 3 is chosen from a class of 14 students. In how many ways can the team be selected if order is (a) important (b) unimportant?

10. A bag contains 12 different lollies with blue wrappers and 15 different lollies with red wrappers. If I take 6 lollies out of the bag, how many different combinations are possible? 11. A team of 4 players is chosen at random from a group of 20 tennis players to play an exhibition match. In how many ways could the team be chosen? 12. A group of 3 students to go on a student representative council is chosen at random from a class of 27. In how many different ways could this be done? 13. A board of 8 people is chosen from a membership of 35. How many different combinations are possible? 14. A basketball team of 5 players is selected at random from a group of 12 PE students. (a) In how many ways can the team be selected?

(b) Find the probability that Erik is selected as one of the team members. (c) Find the probability that Erik and Jens are both selected. 15. A committee of 6 people is to be selected randomly from a group of 11 men and 12 women. Find the number of possible committees if (a) there is no restriction on who is on the committee (b) all committee members are to be male (c) all members are to be female (d) there are to be 3 men and 3 women (e) a particular woman is included (f) a particular man is not included (g) there are to be 4 women and 2 men. 16. A horse race has 15 horses competing and at the TAB, a quinella pays out on the horses that come in first and second, in either order. Ryan decides to bet on all possible combinations of quinellas. If it costs him $1 a bet, how much does he pay? 17. A group of 25 students consist of 11 who play a musical instrument and 14 who don’t. Find the number of different arrangements possible if a group of 9 students is selected at random (a) with no restriction (b) who all play musical instruments (c) where 5 play musical instruments (d) where 2 don’t play musical instruments.


18. A set of cards consists of 8 yellow and 7 red cards. (a) If 10 cards are selected at random, find the number of different arrangements possible. (b) If 8 cards are selected, find the number of arrangements of selecting (i) 4 yellow cards (ii) 6 yellow cards (iii) 7 yellow cards (iv) 5 red cards. 19. Ten cards are selected randomly from a set of 52 playing cards. Find the number of combinations selected if (a) there are no restrictions (answer in scientific notation correct to 3 significant figures) (b) they are all hearts (c) there are 7 hearts (d) they are all red cards (e) there are 4 aces. 20. An animal refuge has 17 dogs and 21 cats. If a nursing home orders 12 animals at random, find the number of ways that the order would have (a) 7 dogs (b) 9 dogs (c) 10 dogs (d) 4 cats (e) 6 cats. 21. There are 8 white, 9 red and 5 blue marbles in a bag and 7 are drawn out at random. Find the number of arrangements possible (a) with no restriction (b) if all marbles are red (c) if there are 3 white and 2 red marbles (d) if there are 4 red and 1 blue marbles (e) if there are 4 white and 2 blue marbles.

22. Out of a group of 25 students, 7 walk to school, 12 catch a train and 6 catch a bus. If 6 students are selected, find the number of combinations if (a) all walk to school (b) none catch a bus (c) 3 walk to school and 1 catches a bus (d) 1 walks to school and 4 catch a train (e) 3 catch a train and 1 catches a bus.

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The cards are all distinct from each other.

23. At a karaoke night, a group of 14 friends decide that 4 of them will sing a song together. Of the friends, 5 have previously sung this song before. In how many ways can they do this if they select (a) friends who have all sung the song previously (b) 2 of the friends who sang the song previously (c) none of the friends who sang the song previously? 24. (a) Evaluate 12 C 5 (b) Evaluate 12 C 7 (c) By using factorial notation, show why 12 C 5 = 12 C 7 25. By evaluating both sides, show that 9 C 6 = 8 C 6 + 8 C 5

k a 13 k 26. Show that a 13 7 = 6 l b9 l b9 l 27. Show that b 10 4 = 4 + 3

n n l 28. Prove that b l = b n r r 29. Prove that n Pr = r! nC r n - 1 l bn - 1 l 30. Prove that b l = b n k -1 + k k

All marbles are distinct.

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Test yourself 13 1.

Find the number of ways of arranging 7 people (a) in a straight line (b) in a circle.

8.

A set of cards numbered from 1 to 20 is arranged randomly. In how many ways can this be done? (Answer in scientific notation, to 2 significant figures.)

2.

A bag contains 8 red, 5 green and 9 yellow marbles. If a marble is chosen at random, find the probability that it is (a) red (b) green or red (c) not green.

9.

The probability of a missile hitting a 8 target is . What is the probability of the 9 missile missing the target?

3.

A carriage has 2 seats facing forwards and 2 seats facing backwards. Find the number of ways of seating 4 people in the carriage if (a) there are no restrictions (b) one person must sit facing forwards.

4.

A set of 10 cards, numbered 1 to 10, is placed into a box and 3 drawn out at random. Find the number of arrangements possible if order is (a) important (b) unimportant.

5.

A group of 10 boys and 16 girls are on a school excursion. Five of them are chosen at random to help the teacher check the rolls. Find the number of ways these students could be selected if (a) there is no restriction (b) 3 are girls (c) all are boys.

6.

The probability of winning a game of chance is 59% and the probability of a draw is 12%. What is the probability of losing?

7.

In how many ways can 3 letter ‘words’ be selected at random from the word RANDOM?

10. A 5 person committee is selected from a class of 30 students. In how many ways can the committee be selected? 11. In a horse race, a person bets at the TAB on a trifecta. To win, the person must pick the first 3 horses in order. In how many ways is it possible to win if there are 11 horses in the race? 12. (a) A group of 9 friends go out to dinner and sit at a round table. In how many ways can this be done? (b) The 9 friends then go to a nightclub and randomly queue up to get in. How many arrangements are there? (c) If Jack and Jill queue up together, in how many ways can the friends line up? (d) Once they get into the nightclub, there is only a table for 3 available, and the others will have to stand up. If the friends randomly assign who sits at the table, in how many different ways can they be seated? 13. How many different arrangements are there of the word (a) PERMUTATION (b) COMBINATION (c) FACTORIAL (d) PROBABILITY (e) SELECTION


14. A set of n coins are tossed. Find the number of ways if tossing k tails. 15. A set of 20 cards is numbered 1 to 20 and 6 selected at random. Find the number of arrangements of selecting (a) all odd numbers (b) the last 2 numbers less than 5. 16. In Australian Idyll, there are 12 singers who must choose a song to sing from a list of 32 songs. Each singer takes turns in order to randomly choose a song. In how many ways could these choices be made? Answer in scientific notation correct to 2 decimal places.

ballet exam. If 8 students are chosen at random, find the number of ways that (a) 5 are practising for the exam (b) all are practising for the exam (c) 3 are practising for the exam. 18. Evaluate k (a) a 6 4 (b) 9 P7 19. A serial number is made up of 4 letters and 2 numbers. If zero is not allowed, find how many serial numbers are possible. 20. (a) Evaluate 0! n n (b) Show that a k = a k n 0

17. A ballet class has 30 students in it. Of these students, 21 are practising for a


2.

3.

4.

Numbers are formed from the digits 1, 2, 3, 3, and 7 at random. (a) In how many ways can they be arranged with no restrictions? (b) In how many ways can they be arranged to form a number greater than 30 000? A charm bracelet has 6 charms on it. In how many ways can the charms be arranged if the bracelet (a) has a clasp (b) has no clasp? n n - 1l bn - 1 l Show that b l = b + for k k k 1 1 # k # n - 1. A group of n people sit around a circular table. (a) In how many ways can they be arranged?

(b) How many arrangements are possible if k people sit together? 5.

(a) How many different arrangements of the word CHALLENGE are there? (b) How many different arrangements are possible if 3 letters are randomly selected from the word CHALLENGE and arranged into ‘words’?

6.

A subcommittee of 5 people is formed from the 12 members of a board. (a) If this is a random selection, in how many different ways can the committee be formed? (b) If there are 4 NSW members and 3 Queensland members on the board, what is the probability that 2 NSW and 2 Queensland members will be on the committee?

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748


7.

Prove that n Pr = r! nC r

8.

A management committee is made up of 5 athletes and 3 managers. If the committee is formed randomly from a group of 20 athletes and 10 managers, find (a) the number of different ways in which the committee could be formed (b) the probability that Marcus, an athlete, is included (c) the probability that both Marcus and his girlfriend, Rachel who is a manager, are included (d) the probability that Marcus and Rachel are excluded from the committee.

9.

A set of 100 counters, numbered from 1 to 100, is placed in a bag and 4 drawn

out at random in order. Find the number of different possible arrangements if (a) there is no restriction (b) all the numbers are 90 or more (c) all numbers are even (d) all numbers are less than 20 (e) the first number is greater than 60 (f) the first 2 numbers are odd. 10. In a group of 35 students, 18 play soccer and 21 play basketball. All students play at least one of these sports. If one of these students is selected at random, find the probability that this student (a) plays both soccer and basketball (b) plays basketball but not soccer.

PRACTICE ASSESSMENT TASK 1

Practice Assessment Task SET 1 p-3 p+1 = 1. 5 2

1.

Solve for p:

2.

Factorise fully: 10x + 2xy - 10y - 2y 2.

3.

Write in index form 1 (a) x (b)

3

2x + 6 . 2

17. Solve 2x 2 - 3x - 1 = 0 correct to 3 significant figures. 18. The radius r of a circle with area A is A given by r = . Find r, correct to r 2 decimal places, if A = 7.59.

x4

4.

Simplify the expression 8y - 2 ^ y + 5 h .

5.

5 Rationalise the denominator of . 5- 2 Expand and simplify ] x - 3 g ^ x 2 + 5x - 1 h .

6.

16. Simplify

7.

3x 2 Solve the equation = . ^ x ! -1h x+1 3

8.

Simplify

9.

Show that TABC and TEDC are congruent triangles. Hence, or otherwise, show that TACE is an isosceles triangle.

x 2 - 2x - 3 x+1 . ' 5 10

19. Solve 5 - 2x 1 3 and sketch the solution on a number line. 20. Evaluate

3 5 1 2 + + . 20 15 3 12

21. Solve the equation x 2 - 4x + 1 = 0, giving exact solutions in simplest surd form. 22. Write 7 - 2 as a rational number. 23. Solve simultaneous equations y = 3x - 1 and y = x 2 - 5. 24. Find integers x and y such that 3 = x + y 3. 2 3+3

10. Evaluate ] 3.9 g4, correct to 1 decimal place.

25. Evaluate |-2 | 2 - | -1 | + | 4 |. 26. Find the value of x.

11. Simplify 2 3 - 27 . 12. Find the size of each interior angle in a regular octagon. 13. Evaluate 0.72 ' 9.82 in scientific notation, correct to 3 significant figures. 14. Expand and simplify

2 ^3 5 - 2 2 h.

15. Find, correct to 2 decimal places, the ] 2.14 g3 value of . 6.94 - 3.72

27. Factorise 8x 3 - 1. 28. Rationalise the denominator of 2 3 . 3 5- 2

199

200


29. Simplify 2 | -4 | - | 3 | + | -2 |.

o as a fraction. 42. Write 0.16

30. Find the sale price if a discount of 8% is given on a DVD player that usually sells for $699.

43. Prove that the diagonals in any rhombus bisect the angles they make with the sides of the rhombus.

5.21 + 4.71 correct to 3.94 # 1.76 2 significant figures.

31. Evaluate 3

32. Rationalise the denominator of

5 +1

. 2 2+3 33. The price of roller skates has increased by 6.5% to $89. Find the price before the increase.

34. Find the values of all pronumerals, giving reasons for each step of your calculations.

1 in index form. x+3 45. Expand and simplify ] x + 2 g3.

44. Write

46. Find the value of a 3 b - 2 in index form if 4 2 1 3 a = c m and b = c m . 5 2 47. Find the value of x, giving reasons for each step of your working out.

48. Find values of x and y. 35. Find the area of this figure.

36. Simplify

^ a - 4 h3 # b 6

a9 # ^ b-1 h

. 4

37. Solve 5x - 9 2 21. 38. Evaluate 4

-

3 2

as a rational number.

39. Simplify 2 ] x - 5 g -3 ] x - 1 g . 40. Solve 4 2x + 1 = 8. 41. Solve x 2 $ 9.

5 49. Solve x 2 1. -

50. Write ] 3x + 2 g

1 2

without an index.

51. Simplify (a) 8x - 7y - y + 4x (b)

124

(c)

x2 - 9 x 3 + 27


(d) (e)

1 + 2+1

56. ABCD is a parallelogram with CD produced to E so that ED = AD. Prove that +ABC = 2+DEA.

2 2-1

3 2 4 + x + 1 x2 - 1 x - 1

1 (f) x - x when x = 2 3 (g)

^ x - 2 h5 y 4 z - 3

x4 _ y3 i

-1

^ z - 4 h- 2

a+b a + 2ab + b ' 3 - 6b 5a - 20ab 2 2

(h)

2

(i) 8 5 - 3 20 + 2 45 a3 b2 ^ c4 h 1 2 2 3 c m c m , if a = , b = 2 3 ^ a 2 h2 bc 5 4 -1 and c = c m 9 2

(j)

52. Find the values of x and y, correct to 1 decimal place.

53. Evaluate x.

2 1 5 3 57. Find the exact value of . 5 16 1 1 58. Tran spent of her salary on rent, 4 3 1 1 on food, on bus and taxi fares, and 5 6 on going out. If she puts the rest of her salary into savings, what percentage of her salary is savings? 59. The speed of light is about 2.99 # 10 8 ms - 1 . If a rocket travels at one-fifth the speed of light, find its speed in kmh - 1 (in scientific notation correct to 2 significant figures). 60. Find the value of k if ] 2x + 5 g2 = 4x 2 + kx + 25. 61. Simplify

54. The volume of a sphere is given by the 4 formula V = rr 3. Find the exact radius r, 3 2 if the volume V is 10 cm 3. 3 55. Find the perimeter of the figure below, correct to 3 significant figures.

81x 2 y 3 .

62. The sum of the interior angles in a regular polygon is 1620c. Find the size of each interior angle, to the nearest minute. 63. Find the area of the shaded region in this figure, correct to 2 decimal places.

201

202


64. Factorise (a) 5 ] a - 2 g3 + 40b 3 (b) ] 2a - b + c g2 - ] a + 5b - c g2 65. Solve -2 #

8x - 1 1 9. 5

20 m

66. ABCD and BCEF are parallelograms. Show that AFED is a parallelogram. 25 m

72. In the figure, BD is the perpendicular bisector of AC. Prove that triangle ABC is isosceles. B

67. Find the value of b correct to 2 decimal places.

A

D

C

73. The diagonals of a rhombus are x and y. Find the length of its side. 68. The diagonals of a rhombus are 6 cm and 10 cm long. Find the (a) exact length of the sides of the rhombus (b) area of the rhombus. 69. Write as a single fraction with a rational 2 1 denominator . 3 3- 2 2+ 5 70. The exterior angles of a regular polygon are 18c . How many sides has the polygon? 71. A cable is used to support a 20 m tower as shown. If the cable is placed 25 m away from the base of the tower, how long must it be, to the nearest metre?

74. Write

1 in index form. 3 ] x - 2 g5 -

5 3

-

5 2

(a) ] x - 2 g

]x - 2 g (b) 3

-

(c) 3 ] x - 2 g 1 (d) 5 ] x - 2 g3

5 2

75. Write the number 54 000 000 in scientific notation. (a) 5.4 # 108 (b) 54 # 106 (c) 5.4 # 107 (d) 54 # 10−8


76. Simplify

^ 2a 3 b h 3

] ab g

2

.

(a) 8a b (b) 8a8b (c) 2a7b (d) 2a8b 77. A computer costs $1850. If it has increased in cost by 4% since last week, how much did it cost last week? (a) $1924.00 (b) $1778.85 (c) $1867.80 (d) $1776.00

o to a fraction. 79. Convert 0.36

7

78. Evaluate 4 (a) - 8 (b)

1 8

(c)

1 6

(d) −6

-

3 2

.

(a)

9 25

(b)

12 33

(c) 3 (d) 80.

1 3

11 30

A E C

B

D

The triangles ABC and CDE can be proven congruent by using the test (a) SSS (b) SAS (c) RHS (d) AAS.

203

494


Practice Assessment Task SET 2 3 . 2x - 1

1.

Evaluate sin 309c 41l to 2 decimal places.

13. Find the domain and range of y =

2.

Simplify sin 2 38c + sin 2 52c.

3.

Find the equation of the straight line through ^ -1, 3 h that is perpendicular to the line 2x - 5y = 9.

14. (a) On a number plane, draw the line 2x - y = 4. (b) On your diagram, shade the region given by y $ 0, 2x - y $ 4.

4.

Calculate the point of intersection of the lines 2x - 3y - 10 = 0 and 5 x + 4 y - 2 = 0.

5.

Evaluate cot 107c 9l to 3 decimal places.

6.

ABCD is a parallelogram in which BC = 2 cm, +ABC = 150c and AB = 3 cm. Find (a) the exact area of the parallelogram (b) the exact lengths of both diagonals.

7.

The lines AB and AC have equations 3x - 4y + 9 = 0 and 8x + 6y - 1 = 0 respectively. Show that the lines are perpendicular and find the coordinates of A.

8.

If sin ^ x + 5 hc = cos 70c, find a possible value of x.

9.

Simplify as a fraction the expression sin 2 60c - cos 2 45c + tan 2 30c without using a calculator.

15. Sketch y = x 2 - 5x + 4. 16. Simplify sin 2 i - sin 2 i cos 2 i. 17. Find the perpendicular distance from ^ -2, 3 h to the line 5x - 12y + 20 = 0. 18. Find the equation of the straight line through ^ 1, 2 h parallel to the line through ^ -3, 4 h and ^ 5, 5 h . 19. Solve cos i = -

1 for 0c # i # 360c. 2

20. Find the gradient of the line through the origin and ^ -3, 5 h . 21. Simplify

tan 20c + cot 70c . tan 20c

22. Find a in degrees and minutes.

23. Find the value of y correct to 3 significant figures.

10. Sketch 5x - 2y - 10 = 0. 11. On the number plane, shade in the region given by x 2 + y 2 # 9 and x - y $ 2. 12. A soccer goal is 8 m wide. A man shoots for goal when he is 9 m from one post and 11 m from the other. Within what angle must a shot be made in order to score a goal?

24. Find the equation of the line passing through ^ 1, 2 h that passes through the intersection of lines x - 2y - 11 = 0 and 5x - y - 19 = 0.


25. Solve 3x - 7 1 2. 26. If f ] x g = 9 - 2x 2, find the value of f ] -1 g . 27. Find the value of x if f ] x g = 7 where f ] x g = 2 x - 1. 28. Find the exact value of tan 300c. 29. Show that 3x - 4y + 10 = 0 is a tangent to the circle x 2 + y 2 = 4. 3-x if x 2 1 30. If g ] x g = ) 2x if x # 1 (a) find g ] 2 g and g ] -3 g (b) sketch the graph of g ] x g. 3x 2 h - 4xh + 2h 2 . h "0 h

31. Find lim

32. Find the exact value of cos 135c. 33. A bird at the top of a 10 m tree sees a mouse on the ground. If the angle of depression is 34c51l, how far, to 1 decimal place, does the bird need to fly to reach the mouse? 34. If the point ^ 3, -7 h is the midpoint of ^ x, 3 h and ^ 8, y h, find the values of x and y. 35.

(c) Find the area of TOPQ where O is the origin. 38. Two points A and B are 100 m apart on the same side of a tower. The angle of elevation of A to the top of the tower is 20c and the angle of elevation from B is 27c. Find the height of the tower, to the nearest metre. 39. The lines 3x - y = 6, 2x + y = 14 and y = 0 enclose a triangle. Find the area of the triangle. 40. Show that f ] x g = x 6 - x 2 - 3 is an even function. 41. Differentiate x ^ 2x 2 + 1 h . 4

42. Solve 3y - 7 # 20. 3 43. Differentiate x . 44. (a) Find the equation of the tangent to the curve y = x 3 - 2 at the point P ^ 1, -1 h . (b) The curve y = x 3 - 2 meets the y-axis at Q. Find the equation of PQ. (c) Find the equation of the normal to y = x 3 - 2 at the point ^ -1, - 3 h . (d) Find the point R where this normal cuts the x-axis. 45. Find the distance between the parallel lines 5x - 12y - 3 = 0 and 5x - 12y + 5 = 0.

(a) Find AB, correct to 1 decimal place. (b) Find the area of TABC, to 3 significant figures. 36. Simplify

9 - 9 sin 2 i .

37. (a) Find the equation of the straight line l through ^ -1, 2 h that is perpendicular to the line 3x + 6y - 7 = 0. (b) Line l cuts the x-axis at P and the y-axis at Q. Find the coordinates of P and Q.

46. What is the domain and range of the 3 function f ] x g = ? x+4 47. Differentiate

2x - 7 .

48. Justin walks for 3.1 km due west, then turns and walks for 2.7 km on a bearing of 205c. How far is he from his starting point? Give your answer correct to 1 decimal place.

495

496


49. Find the derivative of 4x 2 - 7x + 3 + 5x - 2 . 50. Find the derivative of

5x . x+1

51. Differentiate from first principles f ] x g = x 2 - 3x + 2. 52. Write as a single fraction 3 4 2 . - + x 2 + 5x x x + 5 53. Find the equation of the normal to the curve y = 3x 2 - 6x + 9 at the point where x = 2. 54. If f ] x g = 2x - 5x + 4x - 1, find f ] -2 g and f l ] -2 g . 3

2

55. At the point ^ 2, -3 h on the curve y = ax 2 + bx + 7, the tangent is inclined at 135c to the x-axis. Find the values of a and b. 56. Find the equation of the straight line passing through ^ 3, 6 h that also passes through the intersection of the lines x - 2y = 0 and 3x + y + 7 = 0. 57. Find the equation of the tangent to the curve f ] x g = 2x 3 - 7x + 1 at the point where x = -3. 58. Solve for x: 4 2x - 1 =

1 . 8

59. Divide the interval AB into the external ratio 5:4, given A^ -2, 3 h and B^ 5, -1 h . 60. Find the exact value of sin 75c. 61. Find the acute angle between the lines 5x - y - 1 = 0 and 2x + 3y - 3 = 0. 62. Solve 2 cos x - sin x = 0 for 0c # x # 360c . 63. Sketch y =

x . x-2

65. Simplify i 1 - t2 where t = tan 2 2 1+t (b) cos i cos b - sin i sin b (a)

(c)

2 tan 7a 1 - tan 2 7a

65. Find xlim "∞ 66. Solve

3x 3 + x . x3 - 2

3 1 5. x-4

67. Find the acute angle between the curves y = x 2 and y = 2x 2 - 1 at their points of intersection. 68. Find the domain and range of (a) y = 2x - 1 5 (b) y = x+7 (c) y = - 4 - x 2 69. The point P ^ -3, 2 h divides the interval AB into the ratio 2:3 where A = ^ 5, 4 h and B = ^ a, b h . Find the values of a and b. 70. Simplify 1 - 2 sin 2 i. 71. (a) Find all points of intersection between the curves y = 2x 3 + x 2 and y = x 4 + 2x. (b) Find the acute angle between the two curves at each of these points of intersection. 72. Find the general solutions of 1 (a) cos x = 2 (b) tan x =

1 3

(c) sin x =

3 . 2


73. For the interval AB where A = ^ 3, -2 h and B = ^ -1, 4 h find (a) the midpoint (b) the exact distance (c) the gradient (d) the equation of line AB. 74. Differentiate ] 3x - 2 g ] 2x + 1 g6 (a) 12 ] 3x - 2 g ] 2x + 1 g5 + 3 ] 2x + 1 g6 (b) 36 ] 2x + 1 g5 (c) 6 ] 3x - 2 g ] 2x + 1 g5 + 3 ] 2x + 1 g6 (d) 18 ] 2x + 1 g5. 75. Find an expression involving i for the triangle below (there may be more than one answer).

i

a 7

52 + 42 - 72 2#5#4 sin i sin a (b) = 4 5 sin i sin a (c) = 5 4 2 5 + 72 - 42 (d) cos i = 2#5#7 (a) cos i =

8x if x 2 3 76. If f ] x g = * 3x 2 - 2 if 0 # x # 3 9 if x 1 0 evaluate f ] 3 g + f ] 1 g + f ] -1 g (a) 35 (b) 226 (c) 233 (d) 53 3

78. The linear function with equation 4x - 2y + 3 = 0 has 1 (a) gradient -2, y-intercept -1 2 3 1 (b) gradient , y-intercept 4 2 1 (c) gradient 2, y-intercept 1 2 (d) gradient 4, y-intercept 3. 79. Write an expression for cos 2x (there may be more than one answer)

4

5

77. The equation of the normal to the parabola y = 3x 2 - 5x + 1 at the point ^ 2, 3 h is (a) 7x - y - 11 = 0 (b) 7x - y - 17 = 0 (c) x + 7y - 23 = 0 (d) x + 7y - 19 = 0

(a) (b) (c) (d)

sin 2 x - cos 2 x cos 2 x - sin 2 x 2 sin 2 x - 1 2 cos 2 x - 1

80. Find the point if the interval AB where A = ^ 4, 2 h and B = ^ -3, 5 h is divided in an external ratio of 2:3. 1 1 (a) c 1 , 3 m 5 5 (b) ^ -17, 11 h (c) ^ 18, -4 h 1 4 (d) c - , 3 m 5 5

497


Practice Assessment Task SET 3 1.

Solve m 2 - 5m + 6 $ 0.

2.

Find the locus of point P that moves so that it is equidistant from the points A^ -3, 1 h and B ^ 5, 7 h .

3.

Write x = 4t, y = 2t 2 as an equation in Cartesian form.

10.

4.

Show that AF < CD given AC and FD are straight lines. 11. Find the equation of the locus of a point whose distance from the line 3x - 4y + 1 = 0 is 3 units.

AB, AC and CB are tangents with CZ = 3 cm, ZB = 7 cm and AY = 2 cm. Find the perimeter of TABC. 5.

Find the centre and radius of the circle with equation x 2 + 6x + y 2 - 10y - 15 = 0.

6.

If a and b are the roots of the quadratic equation 3x 2 - 2x - 1 = 0, find the value of (a) a + b (b) ab (c) a 2 + b2

7.

Find the coordinates of the focus and the equation of the directrix of the parabola x 2 = - 8 y.

8.

Solve ] x + 3 g + 5(x + 3) + 6 = 0.

9.

Find the value of k in the equation x 2 - ] k - 4 g x + 3k = 0 if the sum of the roots is -5.

2

12. Find the coordinates of the vertex and focus of the parabola y = x 2 + 8x - 1. 13. Solve 2 2x - 9.2 x + 8 = 0. 14.

Find the value of k correct to 1 decimal place. 15. Find the equation of the tangent to the parabola x 2 = 16y at the point ^ -4, 1h . 16. For what values of b does the equation x 2 + 4x - 2b = 0 have real roots?

655

656


17.

25. i 0

32c

Find i. O is the centre of the circle. 18. A and B are the points ^ -4, 0 h and ^ 4, 0 h respectively. Point P ^ x, y h moves so that PA 2 + PB 2 = 64. Find the equation of the locus of P and describe it geometrically. 19. Find the equation of the circle with centre ^ -2, -3 h and radius 5 units. 20. The lines PA and PB are perpendicular, where A is ^ -2, 7 h, B is ^ 5, -1 h and P is ^ x, y h . Find the equation of the locus of P. 21.

Find x and y. O is the centre of the circle. 26. Differentiate

9 - x2 .

27. The point P _ 2ap, ap 2 i lies on the parabola x 2 = 4ay. (a) Find the equation of the tangent to the curve at P. (b) Find the point R where this tangent meets the directrix. (c) Find the equation for FR where F is the focus. 28. Find the locus of the point that is equidistant from the point ^ 2, 5 h and the line y = -3. 29. Show that D ABC is similar to DCDE and hence find y, correct to 1 decimal place.

O is the centre of the circle. Show +DAE = 90c - +BDC. 22. Find the gradient of the normal to the curve x 2 = - 6y at the point where x = - 4. 23. Find the locus of a point moving so that the ratio of PA to PB is 2:3 where A is ^ 3, 2 h and B is ^ 0, 7 h . 24. If 2x 2 - 3x + 1 / a(x - 1) 2 + b(x - 1) + c, find the values of a, b and c.

30. Find the equation of the tangent to the curve ] x - 2 g2 = 8y at the point where x = 6. 31. Find the equation of the locus of point P ^ x, y h that moves so that it is always equidistant from the point ^ -1, 3 h and the line y = - 5. 32. Solve 2 2x - 5.2 x + 4 = 0.


33. Show that - x 2 + x - 9 1 0 for all x. 34. Differentiate ^ 3x - 1 h ^ 2x + 5 h4. 35. Simplify cot x + tan x. 36. Prove that the opposite angles are supplementary in any cyclic quadrilateral. 37. Find the centre and radius of the circle whose equation is x 2 + 10x + y 2 - 6y + 30 = 0.

45. (a) Change the set of parametric equations x = 2t, y = 4t 2 - 1 into Cartesian form. (b) Find the coordinates of the point where t = -2. (c) Find the equation of the normal to the curve at the point where t = -2. 46. Find the value of i in degrees and minutes.

38. 47.

AC is a tangent and AC < DE. Prove FGED is a cyclic quadrilateral.

O is the centre of the circle. Find x.

39. Show that x 2 - x + 3 2 0 for all x.

48. Show that the quadratic equation 6x 2 + x - 15 = 0 has 2 real, rational roots.

40. Find the value of k in the quadratic equation x 2 - 3x + k + 1 = 0 if the roots are consecutive numbers.

49. Find the equation of the normal to the curve y = 2x 4 - 5x 2 - 1 at the point ^ -1, -4 h.

41. Find the equation of the locus of the point that is equidistant from ^ -2, 1 h and ^ 4, 5 h .

50. Find values of k for which the quadratic equation x 2 - 2x + k - 2 = 0 has real roots.

42. A ship sails from port due east for 150 km, then turns and sails on a bearing of 195c for 200 km. (a) How far from port is the ship, to the nearest kilometre? (b) On what bearing, to the nearest degree, is the ship from port?

51. Sketch y =

43. Find the values of a, b and c if 3x 2 - 7 / a ] x + 3 g2 + bx + c.

54. Find the exact value of tan 75c.

44. Solve 2x - 7 2 1.

x . 2x + 1

52. Find the equation of the straight line through ^ 5, -4 h , that is parallel to the line through ^ 7, 4 h and ^ 3, -1 h. 53. Divide the interval AB where A = ^ 1, -4 h and B = ^ 7, 0 h in the ratio 2:3.

55. Solve

y $ 5. y+1

657

658


56. Rationalise the denominator of 2+1 3 3+ 5

.

57. Find the values of x and y correct to 1 decimal place.

AB is a diameter of the larger circle and DB is a straight line. Show AD is a diameter of the smaller circle. 67. Solve 2 cos 2 x = 1 for 0c # x # 360c . 68. Solve equations x 2 + xy + 1 = 0 and 3x - y + 5 = 0 simultaneously. 69. Factorise a 3 - 8b 3 . 70. Solve

x+1 x+2 = 7. 2 3

58. Given f ] x g = 8x - 3, find the value of x for which f ] x g = 5.

71. Find the gradient of the normal to the curve y = 2x 3 + 7x + 1 at the point where x = - 2.

59. Find the distance between ^ 0, 7 h and ^ -2, -1 h correct to 3 significant figures.

72. Find the perpendicular distance from ^ 3, -2 h to the line 4x - 3y - 9 = 0.

60. Find the value of p correct to 1 decimal place.

73. Simplify

] sec i + 1 g ] sec i - 1 g .

74. Differentiate ] 2x + 5 g (x 2 - 1) 4. x-2 . x2 - 4 76. Find the equation of the locus of point P(x, y) if PA is perpendicular to PB, given A = ^ 3, -2 h and B = ^ -5, 5 h .

75. Find lim x "2

a3 ^ b2 h 2 4 if a = and b = . 3 9 ^ a - 1 h2 b 7 4

61. Simplify

1 62. Solve cos 2x = - for 0c # x # 360c . 2 63. Find the equation of the straight line through ^ 3, -1 h perpendicular to the line 3 x - 2 y - 7 = 0. 64. Solve 5y - 3 = 5 - y. 65. Find the size of each internal angle in a regular 20-sided polygon. 66.

77. Find the coordinates of the focus and the equation of the directrix of the parabola x 2 - 4x + 8y - 20 = 0. 78. Find the equation of the tangent to the parabola x 2 = 36y at the point P(18p, 9p2). 79. Find the equation of the normal to the parabola x 2 = -12y at the point where x = 12. 80. If points P(2ap, ap2) and Q(2aq, aq2) lie on the parabola x 2 = 4ay, find (a) the equation of chord PQ (b) the equation of the locus of the midpoint of PQ if PQ passes through (0, 2a) (c) Describe the shape of this locus.


81. The equation of the locus of point P(x, y) that moves so that it is always 4 units from ^ -1, 3 h is (a) ^ x - 1 h2 + ^ y + 3 h2 = 4 (b) ^ x + 1 h + ^ y - 3 h = 4 (c) ] x + 1 g2 + ^ y - 3 h2 = 16 2

2

(d) ^ x - 1 h + ^ y + 3 h = 16 2

2

85. Find the centre and radius of the circle x 2 + 2x + y 2 - 8y + 13 = 0. (a) Centre ^ -1, 4 h, radius 4 (b) Centre ^ 1, -4 h, radius 2 (c) Centre ^ -1, 4 h, radius 2 (d) Centre ^ 1, -4 h, radius 4 86. In the circle, O is the centre. Evaluate x.

82. If a and b are the roots of the quadratic equation x 2 - 5x + 2 = 0, a b evaluate + b a (a) (b) (c) (d)

1 11 2 1 12 2 1 2 2 1 10 2

83. The equation of the locus of point P(x, y) moving so that it is equidistant from (3, 2) and the line x = -1 is given by (a) x 2 - 2x + 8y - 15 = 0 (b) y 2 - 4y - 8x + 12 = 0 (c) x - 2x - 8y + 17 = 0 2

(d) y - 4y + 8x - 4 = 0 2

84. The quadratic equation x 2 + ] k - 3 g x + k = 0 has real roots. Evaluate k (a) k # 1, k $ 9 (b) k = 1, 9 (c) 1 # k # 9 (d) k 1 1, k 2 9

O

x

84⬚

(a) x = 42c (b) x = 168c (c) x = 84c (d) x = 96c 87. Find the Cartesian equation for x = 8t, y = 4t 2 . (a) x 2 = 32y (b) x 2 = 4y (c) x 2 = 16y (d) x 2 = 8y 88. The equation of the normal to the curve x 2 = 20y at the point (10p, 5p2) is (a) x + py = 5p 3 + 10p (b) x - py = 5p 3 - 10p (c) px + y = 15p 2 (d) px - y + 15p 2 = 0

659

660


89. AB is a tangent to the circle. Which statement is true (there may be more than one answer)?

90. For the quadratic function y = ax 2 + bx + c to be positive definite (a) a 2 0, b 2 - 4ac 2 0 (b) a 1 0, b 2 - 4ac 2 0 (c) a 2 0, b 2 - 4ac 1 0

A D

C B

AB BD = AB BC CD AB (b) = AB BC (a)

(c) AB 2 = BC $ CD (d) AB 2 = BC $ BD

(d) a 1 0, b 2 - 4ac 1 0


Practice Assessment Task SET 4 1.

Find the zeros of f ^ x h = ^ 2x - 1 h5 .

2.

Write P ^ x h = x 3 + 4x 2 - x - 4 as a product of its factors.

3.

For the polynomial f ^ x h = 3x 4 - 2x 3 - x + 8, what graph does the polynomial approach as x becomes very large?

4.

Find the number of ways of seating 10 people around a table (a) randomly (b) if three people are to sit together (c) if two people must not sit together.

5.

Find the zeros of the polynomial f (x) = x 4 - x 3 + x 2 - 3x - 6.

6.

Find the coordinates of the point that divides the interval between (3, 4) and (-5, 1) in the external ratio of 2:3.

7.

8.

9.

If a , b and c are the roots of x 3 + 2x 2 - 3x + 4 = 0, find (a) abc (b) a + b + c 1 1 1 (c) + + a b c (d) a2 + b2 + c2 A and B are the points (-4, 0) and (4, 0) respectively. Point P (x, y) moves so that PA 2 + PB 2 = 36. Find the equation of the locus of P and describe it geometrically. Find the sum of the interior angles of a regular polygon with 19 sides. How large is each angle (in degrees and minutes)?

10. Prove that the line 6x - 8y + 40 = 0 is a tangent to the circle with centre the origin and radius 4 units. 11. In the quadratic equation (k -1) x 2 - 5x + 3k + 4 = 0, the roots are reciprocals of each other. Find the value of k. 12. Find x and y, giving reasons.

13. Evaluate 10 C 3. 14. Solve x 2 - 4 2 0. 15. A polynomial P (x) has a double root at x = 2. Show that P (2) = Pl(2) = 0. 16. If a , b, c and d are the roots of the equation x 4 - 3x 3 + 2x + 1 = 0, find (a) a bcd (b) a + b + c + d (c) ab and a + b if c = -2 and d = 5. 17. In how many different ways can a committee of 4 people be selected from a group of 9 people? 18. Find the remainder if P ^ x h = 3x 5 - 4x 3 + 2x 2 - 7x - 3 is divided by x + 1.

749

750


19.

29. The angle of elevation from a point A to the top of a tower BC is 38c 54l . A is 10 m due south of the tower.

CD is a tangent to the circle and AB = 5.6 m, BC = 4.8 m. Find the length of CD, correct to 1 decimal place. 20. Find the obtuse angle between the lines 2x - y = 0 and 5x + 2y - 3 = 0. 21. (a) Write P ] x g = x 3 - 7x 2 + 15x - 9 as a product of its factors. (b) Sketch the graph of the polynomial on a number plane. x . x2 - 4 23. Show that the remainder, on dividing P (x) = 2x 3 - 7x 2 + x - 9 by (x + 2), is P ( - 2) . 22. Sketch y =

24. If a , b, c and d are the roots of x4 - x3 + x2 - 1 = 0 (a) show that abc + acd + bcd + abd = 0 (b) find (a + b + c + d) 2 (c) find a 2 + b2 + c2 + d2.

(a) Find the height of the tower, to 1 decimal place. (b) If point D is 11.2 m due east of the tower, find the angle of elevation from D to the tower. 30. If P (x) = (x 2 - 1)3 (x 2 + 5) (a) find all zeros of P (x) (b) show P (1) = P l(1) = 0. 31. What is the domain and range of y = x 2 - 3? 32. Prove that TABC is congruent to TCDE.

33. Find the area of the figure below.

25. Show that if the polynomial P(x) has a double root at x = 3, P ] 3 g = Pl(3) = 0. 26. A team of 3 boys and 5 girls is chosen at random from a class of 12 boys and 18 girls. In how many ways can this be done? 27. A circle with centre at the origin O passes through the point (2, 5 ). Find the radius of the circle, and hence its equation. 28. Find values of a, b and c for which 3x 2 - 2x - 7 / a (x + 2) 2 + b (x + 2) + c.

34. Find the equation of the straight line through the midpoint of (-5, 7) and (1, 3) and making an angle of 135c with the x-axis.


35. Complete the square on x 2 - 12x. 36. Find the acute angle between the lines 3x - y - 4 = 0 and 7x + 3y + 1 = 0. 37. Solve 2y + 4 $ 9. 3 38. Sketch y = 2 . x -1 39. Find the point P that divides the interval AB in the ratio 4:3, given that A has coordinates (3, 6) and B has coordinates (-5, 8) . 40. (a) Find the equation of the tangent to the curve y = x 3 - 3x at the point P (-2, -2). (b) Find the equation of the normal to y = x 3 - 3x at P. (c) Find the point Q where this normal cuts the x-axis. (d) The curve y = x 3 - 3x meets the line 3x - 2y + 2 = 0 at P. Find the angle between the line and the curve at P. 41. What is the domain and range of the x function f (x) = 2 ? x -4 42. Simplify (a) sin a cos b - cos a sin b (b) cos 15c cos 30c - sin 15c sin 30c (c) 1 - sin 2 15c 43. Jason measures the angle of elevation of a tower as 38c 52l at a point A, 185 m due west of the tower. He then walks to a point B, 140 m due south of the tower. (a) What is the height of the tower, correct to 1 decimal place? (b) What is the angle of elevation of the tower from point B? 44. Point X divides AB externally in the ratio 3:5, given A (3, 4) and B (0, -5). Find the coordinates of X.

45. Find the exact value of tan 75c - tan 15c . 1 + tan 75c tan 15c 3 46. Solve 1 5. x-1 47. Solve sin x - 3 cos x = 2 for 0c # x # 360c. 48. (a) Find the equation of the normal to the curve y = x 2 - 6x + 9 at the point where x = -1. (b) This normal cuts the curve again at point R. Find the coordinates of R. 49. Write sin i + cos 2i in terms of t where i t = tan . 2 50. Show that x - 3 is a factor of f ] x g = 3x 3 - 7x 2 - 5x - 3. 51. The function f (x) = ax 2 + bx + c has a tangent at (1, -3) with a gradient of -1. It also passes through (4, 3) . Find the values of a, b and c. 52. Find two possible equations for the straight line passing through (1, 2) that intersects the line x - 2y + 5 = 0 at an angle of 45c . 53. In the cone below, the vertical angle is 54c and the perpendicular height is 20 cm. Find the volume and surface area of the cone, correct to the nearest whole number.

54. Solve y 1

-2 y+3

751

752


3 12 and tan y = , express 5 5 cos (x - y) as a fraction.

55. If sin x =

64. How many committees of 5 people could be formed randomly from a meeting of 20 people?

56. Solve for 0c # x # 360c (a) sin 2 x = 3 cos 2 x (b) sin x + cos x - 1 = 0 (c) cos 2 x + sin x + 1 = 0

65. Write the polynomial P ] x g = - x 3 + 3x 2 + 9x + 5 as a product of its factors.

x-1 . x2 - 5 x-1 (b) Sketch y = 2 . x -5

66. A string of beads is made up of 5 orange and 5 black beads. If they are randomly arranged around the necklace, how many possible combinations are there?

57. (a) Find xlim "3

58. Find the general solution of sin 2i = -1. 59. Find the angle between the curves y = x and y = x 2 at (1, 1) .

3

67. An example of a polynomial with leading coefficient -3 and degree 4 is (a) y

60. The point P (1, 2) divides the interval A (x, 3) and B ^ -5, y h in the external ratio of 5:4. Find the values of x and y. 61. For the polynomial f ] x g = - x 4 + x 3 - 2x 2 + x - 2 , what graph does the polynomial approach as x becomes very large? 62. If a , b, and c are the roots of the cubic equation x 3 - 4x 2 - 3x + 2 , evaluate (a) a + b + c (b) abc (c) ab + bc + ac 1 1 1 (d) + + a b c

x

(b)

y

(e) a 2 + b2 + c2 63. By dividing the polynomial P ] x g = x 3 - 2x 2 + x + 3 by x - 2 , write P(x) in the form P ]x g = ]x - 2 gQ ]x g + R ]x g.

x


(c)

69. The graph of the polynomial P ^ x h = x 3 ^ x + 3 h 2 is

y

(a)

y

x

x

3

(d)

y

x

(b)

68. What is the number of possible outcomes when arranging the letters of the word LITERATURE? 10! (a) 2! 2! 10! (b) 2! 2! 2! (c) 10! 10! (d) 3!

y

3

x

753

754


72. The number of possible seating positions for 12 people sitting at a round table is

y

(c)

(a)

12

C 11

(b) 11! (c) 12! x

-3

(d)

12

P11

73. The graph below is of the polynomial (a) P ^ x h = x ^ x - 2 h 2 (b) P ^ x h = x ^ x + 2 h 2 (c) P ^ x h = x 2 ^ x + 2 h (d) P ^ x h = x 2 ^ x - 2 h

(d)

y

y

x

2

70. The polynomial P ^ x h = x 3 - 5x 2 + 3x - 8 (a) is monic (b) has degree 3 (c) has leading coefficient - 8 (d) has constant term - 8. (there may be more than one answer)

74. Combination nCr is equal to (a) ] n - r g ! n Pr

-3

71. The number of possible different PINs with a combination of 4 numbers and 2 letters is (a) 4 435 236 (b) 6 760 000 (c) 1 000 000 (d) 10 676

n

(b)

Pr

(n - r) !

(c) r! n Pr n

(d)

Pr

r!

x


75. The graph below could have a leading term of (a) x 6 (b) x 5 (c) - x 4 (d) - x 3 y

x

755

756


Answers Chapter 1: Basic arithmetic

8.

o (b) 0.07 oo (c) 0.13 oo (d) 0.16 o (a) 0.83 o oo o (g) 0.142857 or 0. 142857 (h) 1.18

9.

(a)

8 9

(h)

13 60

Problem 5

Exercises 1.1 1.

2.

(a) Rational (b) Rational (e) Rational (f) Irrational (i) Rational (j) Irrational

(e) - 4.3

(a) 18 (b) 11 (c) 6 (d) 11 (h) 1

3.

(c) Rational (g) Irrational

19 20

(i) 2

(j) 3

(d) Irrational (h) Rational

(f) −1

(g) 2

7 15

1 3

(f) 0.17 (g) 0.36 (h) 1.20 (i) - 4.27 1300

8.

600

5. 950

16. 1.7

6. 3000

(j) 8.16

1.

7. 11 000

8.

17. 79 cents 18. 2.73 19. 1.1 20. 3.6 m

21. $281.93

22. 1.8 g

(b) 2

3 20 (d)

12. 0.73 13. 33 14. 3.248 15. 4.21

23. $3.20

24. (a) 7.95 (b) 30.03 (c) 0.37 (d) 5.74 (e) 0.52 25. 0.2

1.

1

6.

- 1.2

10. - 2 15. 5

3. - 56

4. 10

(a)

7. - 7.51

8. - 35.52

9. 6.57

11. - 7

12. −23

13. 10

16. 3

16 25

(b)

17. 1

51 1000

(c) 5

1 20

(d) 11

14. 1

4 5

7 20 3 (e) 5

3.

(a)

4.

(a) 0.27 (b) 1.09 (c) 0.003 (d) 0.0623

5.

1 (a) 35% (b) 33 % 3

6.

(a) 124% (b) 70%

7.

(a) 0.52;

13 25

(d) 1.09; 1

oo (d) 0.68

8 11

7 18

(c)

67 99

(f)

6 11

(g)

7 45

(d) 2

oo (e) 1.72 4 45

(e)

14. 17.5%

15. 41.7%

1 20 7 4. $547.56 5. 714.3 g 6. 24

2. 3 28

17 20

3. (a)

(b)

7 10

(c) 1

7. $65

179 cm 9. (a) 11.9 (b) 5.3 (c) 19 (d) 3.2 (e) 3.5 (f) 0.24 (g) 0.000 18 (h) 5720 (i) 0.0874 (j) 0.376

14. 5.9%

15. 402.5 g 19. 573

12. 1152.125 g

16. 41.175 m

13. $10.71

17. $30.92

20. $2898

3 8

(c)

1 1000

Exercises 1.5

(d) 1

(a) 500

(b) 145

(c)

2.

(a) 13.7

(b) 1.1

(c) 0.8

3.

(a) a 17

(e) 0.434;

(c) 0.168; 217 500

(h) x 21

4.

21 125

(f) 0.1225;

(h) p - 1 5.

49 400

(a) x14

(a) p5q15 (f) x4y10

y

(d) 3 (d) 2.7

(c) a - 4

(i) 4x 10

(d) w

(j) 81y - 8

(o) x -3

(n) p 5

(q) x - 5 y 2 or

(c) 40.5% (d) 127.94% 7 100

(b) y 0 = 1

(m) w 10

97 1000

(d) 0.1%

1 64

1.

oo (d) 0.63

2 (c) 226 % 3

(b) 0.07;

9 100

5 minutes after 1 o’clock. 11

(g) y 6

o (a) 0.4 (b) 1.875 (c) 0.416 (b)

o (c) 0.73

11. 54.925 mL

5

18. 60 19. −20 20. 9

2.

1 50

(e)

37 495

10. $52.50

5. - 4

Exercises 1.3 1.

(j) 1

7 9

Problem

2. - 11

4 15

1 8

(d) 3

13. 77.5%

18. 3.2 m

Exercises 1.2

217 990

5 9

Exercises 1.4

9. $8 000 000 10. $34 600 000

11. 844 km

5 8

(c) 1

(b) 7.4

12. 74%

(a) 16.36 (b) 21.87 (c) 8.80 (d) 22.71 (e) - 13.20

4.

(i)

o 10. (a) 0.5 11. (a)

2 9

(b)

oo (f) 0.15

o (e) 0.6

2

(e) 2 (e) - 2.6 (e) x 5 (k) a

(p) a - 2 b 3 or

(f) 0.5

(f) p 10 (l)

x 10

b3

y 45

a2

x5

(b) a -7

(c) m4 (d) k10 (e) a -8

(f) x

(g) mn2

(i) 9x22 (j) x21 (b)

a8 8

b 2k 23 (g) 27

(c)

64a 3 b 12

(d) 49a10b2 (e) 8m17

(h) 16y47 (i) a3 (j) 125x - 21 y 18

ANSWERS

6.

4

1 2

7. 324

8. 2

10 27

9. (a) a3b

3

1 25

(b)

-

1 2

5.

(a) x 2

6.

(a) x + x 2 + 2x 2

(b) x

2

5

5

(c) x 3

(d) x 3

(e) x 4

3

7 (b) 32

10. (a) pq r

2 2

14.

1 81

4 11. 9

1 108

15.

1 12. 18

1 12

16.

4 13. 27

5 22

17.

49 3888

18.

2 58

(d) x + x - 1 + 2 7.

Exercises 1.6 1.

2.

3.

(d)

1 1 1 1 1 (b) (c) (d) (e) (f) 1 4 27 343 10 000 256 1 1 1 1 1 1 (g) (h) (i) (j) (k) (l) (m) 1 7 64 9 32 81 81 1 1 1 1 (n) (o) (p) (q) (r) 1 36 125 100 000 128 1 1 (s) (t) 64 64

1.

(e) x

1

1 2

- 3x

^ y - 3 h2

3

(e)

3 4 ^ x + y h5

-

1

(b)

a - 2b

-

3 2

+x

(c)

-

5 2

4

7

] 6a + 1 g4

6

7 9 ] 3x + 8 g2

(a) m - 3

(b) x - 1

-4

(h) 3y

(c) p - 7 −2

2.

(d) d −9 (e) d −5 (f) x - 2 3t - 8 (j) 5

1 z- 6 (i) z - 6 or 2 2

2x - 1 (k) 7

2y - 7 5m (m) (n) ] 3x + 4 g- 2 (o) ] a + b g- 8 2 3 (p) ] x - 2 g- 1 (q) ^ 5p + 1 h- 3 (r) 2 ] 4t - 9 g- 5 ]x + 1g 4

(a) (h)

1 5

t 5 x7

(m)

(t) 1

(b)

(c)

6

x 1

(i)

5 ] a + 3b g 9 1 y

3

1 n

8

(e)

1 w

(k)

2 x

(f)

10

1

(g)

1 (l) 8y + z

] x + 1 g6

1

(n)

]k - 3g

(d)

1 (j) 4n

8x 3

1

(o) x5 (p) y10 (q)

^ 3x + 2y h x-y 3x + y 7 o (s) (t) e x+y 2w - z

2

(r) ] a + b g2

9

3 m

4

(a) 2.19 (b) 2.60 (c) 1.53 (d) 0.60 (e) 0.90 (f) 0.29

3.

(a) 3 y

(b) 3 y 2 or _ 3 y i

(c)

(f) 3 6q + r

(g)

2

1

1

3

(a) t 2

(b) y 5

(c) x 2

(f) ] 2t + 3 g

-

(i) ] x - 2 g

-

2 3

1 2

2a 2 2 y - 1k 3

x

-

3 2

(d) 1

5

] x + 7 g2 1

1

(j)

1

(d) ] 9 - x g 3

(g) ^ 5x - y h

1

(l)

(e) 8.67 # 10 9

(f) 4.16 # 10 5

(h) 1.376 # 10

2

(a) 5.7 # 10 - 2 -4

-6

4

(i) 2 # 10 7

(b) 5.5 # 10 - 5 (e) 2 #10

-6

(h) 2.3#10

(j) 8 #10 4

(c) 4 # 10 - 3

(f) 8#10 - 8 -1

(i) 8.5#10 - 3

(j) 7#10 - 11

3.

(a) 36 000 (b) 27 800 000 (c) 9 250 (d) 6 330 000 (e) 400 000 (f) 0.072 3 (g) 0.000 097 (h) 0.000 000 038 (i) 0.000 007 (j) 0.000 5

4.

(a) 240 000 (b) 9 200 000 (c) 11 000 (d) 0.36 (e) 1.3 (f) 9.0 (g) 16 (h) 320 (i) 2900 (j) 9.1

5.

(a) 6.61

6.

1.305 # 10 10

(b) 0.686

(c) 8.25

(d) 1.30

7. 6.51 # 10 - 10

Exercises 1.9 1.

(a) 7 (b) 5 (c) 6 (d) 0 (e) 2 (f) 11 (g) 6 (h) 24 (i) 25 (j) 125 2. (a) 5 (b) −1 (c) 2 (d) 14 (e) 4 (f) −67 (g) 7 (h) 12 (i) −6 (j) 10 3. (a) 3 (b) 3 (c) 1 (d) 3 (e) 1 4. (a) a (b) - a (c) 0 (d) 3a (e) −3a (f) 0 (g) a + 1 (h) -a - 1 (i) x - 2 (j) 2 - x

5.

(a) | a + b | = 6 (b) | a + b | = 3 (c) | a + b | = 1 (d) | a + b | = 1 (e) | a + b | = 10

6.

(a)

x2 = | x | = 5

(b)

x2 = | x | = 2

(d)

x2 = | x | = 4

(e)

x2 = | x | = 9

2

2.

1

(d) 1.2 #10 7

(g) 7.6#10

p

(a) 9 (b) 3 (c) 4 (d) 2 (e) 7 (f) 10 (g) 2 (h) 8 (i) 4 (j) 1 (k) 3 (l) 2 (m) 0 (n) 5 (o) 7 (p) 2 1 1 (q) 4 (r) 27 (s) (t) 2 16

3x - 1

(c) 6.19 # 10 4

(d) 6.2 #10

Exercises 1.7

(e)

(b) 1.23#10 6

-7

- 11

(s)

(a) 3.8 # 10 3 (g) 9 #10

(l)

4.

1

(c) p 2 + p - 1 + 2p 2

Exercises 1.8

1 11 1 (a) 1 (b) 16 (c) 1 (d) 1 (e) 1 (f) 125 (g) 1 2 25 3 3 13 19 1 (h) 49 (i) 3 (j) 32 (k) 2 (l) 1 (m) 1 (n) 1 8 3 36 81 5 16 7 (o) 1 (p) 16 (q) - 15 (r) (s) 1 (t) 8 23 25

-6

1.

1 3

2

(a)

(g) 2x

4.

(a)

2

(b) a 3 - b 3

2x + 5 or

|a | + | b |= 6 ` | a + b | # | a | + | b | |a | + | b |= 3 ` | a + b | # | a | + | b | |a | + | b |= 5 ` | a + b | # | a | + | b | |a | + | b |= 9 ` | a + b | # | a | + | b | | a | + | b | = 10 ` | a + b | # | a | + | b | (c)

x2 = | x | = 3

7.

(a) x + 5 for x 2 - 5 and - x - 5 for x 1 - 5 (b) b - 3 for b 2 3 and 3 - b for x 1 3 (c) a + 4 for a 2 - 4 and - a - 4 for a 1 - 4 (d) 2y - 6 for y 2 3 and 6 - 2y for y 1 3 (e) 3x + 9 for x 2 - 3 and - 3x - 9 for x 1 - 3 (f) 4 - x for x 1 4 and x - 4 for x 2 4 1 1 (g) 2k + 1 for k 2 - and - 2k - 1 for k 1 2 2 2 2 (h) 5x - 2 for x 2 and - 5x + 2 for x 1 5 5 (i) a + b for a 2 - b and - a - b for a 1 - b (j) p - q for p 2 q and q - p for p 1 q

8.

x = !3

1 ^ 5 x + 7 h2 1

(e) ] 4s + 1 g 2 5

(h) ] 3x + 1 g 2 1

1 ^ y + 7 h 2 (k) 5 ] x + 4 g 3 2 3 3 4 (m) _ x 2 + 2 i 5

9. !1

10. !1, x ! 2

757

758


Test yourself 1 1.

(a)

9 20

(b) 0.14 (c) 0.625 2. (a)

(f) 73.3% 3.

Chapter 2: Algebra and surds

1 49

(b)

157 200 1 (c) 3

(d)

1 5

Exercises 2.1

(e) 1.2%

(a) 8.83 (b) 1.55 (c) 1.12 (d) 342 (e) 0.303 4. (a) 1 (e) - 10 (f) - 1 (g) 4 5. (a) x 9 8x 18 29 (b) 25y 6 (c) a 11 b 6 (d) (e) 1 6. (a) 27 40 1 1 1 (b) 3 (c) 12 (d) 2 (e) 12 7. (a) 4 (b) 6 (c) 19 7 2 2 1 1 (d) (e) 4 (f) 3 (g) (h) 2 (i) 1 (j) 4 7 64 (b) 1 (c) 39 (d) 2

8.

5

30 18

(a) a

(b) x y

(c) p

(g)

1 -3 x 2

36

11

(d) 16b

(d) ] x + 1 g

(b) x - 5 (c) ^ x + y h- 1 (f) 2x - 1

9

4

(e) 8x y

1 4

9. (a) n

9

-

3 4

1 ] 4t - 7 g4 1 1 (f) 5 a + b (g) (h) 4 b 3 (i) 3 ] 2x + 3 g4 (j) 3 x x3 11. | a + b | = 2 | a | +| b | = 8 ` | a + b | # | a | + | b | 10. (a)

1

a5

(b) 4 n

1 13. 192

12. 1

x+1

(d)

1

(c) ] x + 3 g 6

(b) y - 1

(d) ] 2x - 3 g- 11

20. (a) 1.3 # 10 - 5

(b) 1.23 # 10 11

1

b 5 (c) c m a

x

3

(e)

7 14. 689 mL 15. (a) 6 h (b) 12

1

22. (a)

1 (c) 8

(b)

1 2a + 5

21. (a)

7 9

7

(e) y 3 (b)

41 330

23. 14 500

24. LHS = | -2 + - 5 | = 7, RHS = | -2 | + | -5 | = 7. So | a + b | # | a | +| b | since 7 # 7.

1.

4

4.

1 1 53 % 5. 3 16

9.

18 h

11.

2. 1

11 18

3. 0.502, 51%,

6. 3.04 # 10

14

51 o , 0. 5 99

3271 7. 83% 8. 1 9990

10. 1.98

LHS = 2 ^ 2 k - 1 h + 2 k + 1 = 2k+1 - 2 + 2k+1 = 2:2 k + 1 - 2 = 2 ^ 2k+1 - 1 h = RHS ` 2 ^ 2k - 1 h + 2k+1 = 2 ^ 2k+1 - 1 h

.

o , 0, 12. −24 35 13. - 0.34, 2, 1. 5

3 7

2 14. 6 % 3

1 1 15. when x 2 - 1, when x 1 - 1 16. 0.73 x-1 1-x 17. 0.6%

7.

-y

3. z

8. −5x

13. - m

9. 0

14. - x

19. 6x - 6y

4. 6a

10. 3k

15. 0

20. a - 3b

23. m 2 - 6m + 12 26. - 2ab + 10b

5. 3b

6. −3r

11. 9t

16. 5b

12. 10w

17. 11b

21. 4xy + 2y

24. p 2 - 2p - 6 27. 2bc - ac

29. x 3 - 2xy 2 + 3x 2 y + 2y 3

18. - 10x

22. - 6ab 2

25. 8x + 3y

28. 2a 5 - 9x 3 + 1

30. 3x 3 + x 2 - 7x - 6

1.

10b

2. 8xy

3. 10p 2

5.

15ab 6. 14xyz 7. 48abc 8. 12d 2

9.

12a3

10. - 27y3

12. 6a 2 b 3 19. - 14m

11. 32x10

13. - 10a 3 b 2

15. 5a 3 b 3

4. - 6wz

14. 21p 3 q 4

16. - 8n 10 17. k 3 p 3 20. 24x y

11

6

18. 81t 12

3

Exercises 2.3 1.

2. 2

6x

3. 4a 2

4. 8a

5. 4a

y

6.

2

7. 3p

ab 4 1 -2 9. 10. - 3x 3 11. 3a 12. 13. qs 3y 2 3ab 2 4 7 6 2 a b 2 z b 14. 15. 16. 6p 4 q 17. 18. 4c 2a 3c 2 d 2x 2

8.

19. -

x3 z3 3y

a 13

20.

2b 6

Exercises 2.4

Challenge exercise 1 278 303

2. 3a

16. $38 640 17. 70% 18. 6.3 # 10 23

(d) 33.3% 19. (a) x 2

(c)

1 x-y

(j) m

7x

Exercises 2.2

1

(e) ] a + b g 7

(i) ] 5x + 3 g 7

(h) x 3

1 2

1.

18 4.54 19. 4.14 # 10 - 20

20. | a + b | = | a | + | b | when a 2 0, b 2 0 or a 1 0, b 1 0; | a + b | 1 | a | + | b | when a 2 0, b 1 0 or a 1 0, b 2 0; ` | a + b | # | a + b | for all a, b

1.

2x - 8

5.

x 2 - 2x 6. 6a 2 - 16ab 7. 2a 2 b + ab 2 8. 5n 2 - 20n

2. 6h + 9

3. - 5a + 10

9.

3x3 y2 + 6x2 y3

10. 4k + 7

11. 2t - 17

12. 4y + 11y

13. - 5b - 6

15. - 3m + 1

16. 8h - 19 17. d - 6

2

19. 3x - 9x - 5 2

22. - 7y + 4

14. 8 - 2x

20. 2ab - 2a b + b 2

23. 2 b

4. 2xy + 3x

24. 5t - 6

18. a 2 - 2a + 4 21. 4x - 1

25. 2a + 26

Exercises 2.5 1.

a 2 + 7a + 10

2. x 2 + 2x - 3

4.

m 2 - 6m + 8

5. x 2 + 7x + 12

7.

2x 2 + x - 6

8. h 2 - 10h + 21

3. 2y 2 + 7y - 15 6. y 2 - 3y - 10 9. x 2 - 25

10. 15a 2 - 17a + 4 11. 8y 2 + 6y - 9 12. xy + 7x - 4y - 28 13. x 3 - 2x 2 + 3x - 6 16. 16 - 49y 2 20. y 2 - 36

14. n 2 - 4

17. a 2 - 4b 2

21. 9a 2 - 1

15. 4x 2 - 9

18. 9x 2 - 16y 2

22. 4z 2 - 49

19. x 2 - 9

ANSWERS

23. x 2 - 2xy + 11x - 18y + 18 24. 2ab + 2b 2 - 7b - 6a + 3

Exercises 2.8

25. x + 8

1.

]x + 4g]2 + bg

4.

]m - 2g]m + 3g

5. ] d - c g ] a + b g 6. ] x + 1 g ^ x 2 + 3 h

7.

] 5a - 3 g ] b + 2 g

8. ^ 2y - x h ^ x + y h

3

26. a - 27

27. a + 18a + 81

3

28. k - 8k + 16

29. x + 4x + 4

2

33. 9a + 24ab + 16b 35. 4a + 4ab + b 2

2

32. 4t 2 - 4t + 1

2

38. a - 2ab + b

30. y - 14y + 49

2

31. 4x 2 + 12x + 9

2

2

34. x - 10xy + 25y

2

2

36. a - b

2

2

39. a + b

2

3

3

10. ] x + 5 g ] x - 1 g

2

37. a + 2ab + b

2

2

40. a - b 3

1.

t + 8t + 16

4.

y 2 + 16y + 64

7.

n 2 + 2n + 1

5. q 2 + 6q + 9

6. k 2 - 14k + 49

25. ^ y + 7 h ] x - 4 g

8. 4b 2 + 20b + 25

9. 9 - 6x + x 2

10. 9y - 6y + 1

11. x + 2xy + y 2

13. 16d + 40de + 25e 2

21. 16a 2 - 1

24. x + 10x + 25 4

27. a 2 -

2

2

2

28. x 2 - ^ y - 2 h2 = x 2 - y 2 + 4y - 4

1 a2

28. 3 (a + 2b) (a + 3)

29. 5 (y - 3) (1 + 2x)

30. ] r + 2 g ] rr - 3 g

23. x 4 - 4 4 26. x + 4 + 2 x

2

26. (x - 4) (x 3 - 5)

19. 4a 2 - 9

22. 49 - 9x 2

25. 9a b - 16c

2

2

15. x 2 - 9

18. x 2 - 100

24. ] a - 3b g ] 4 + c g

27. (2x - 3) (2x + 4) = 2 (2x - 3) (x 2 + 2)

12. 9a - 6ab + b

2

17. ] x - 3 g ^ 7 - y h

20. ] a + 3 g ] 2 - b g

2

2

14. t - 16

2

17. r 2 - 36

20. x 2 - 25y 2

2

2

14. ^ a + b h ] ab - 4 g

22. ^ q - 3 h ^ p + q h

23. ] x - 2 g ^ 3x 2 - 5 h

2

12. (m - 2) (1 - 2y)

2

18. ] d + 3 g ] 4 - e g 19. ] x - 4 g ^ 3 + y h

3. x - 2x + 1

2. z - 12z + 36 2

9. ^ y + 1 h ] a + 1 g

15. ] 5 - x g ] x + 3 g 16. (x + 7) (x 3 - 4)

3

21. (x - 3) (x 2 + 6)

2

3. ] x + 5 g ] x + 2 g

11. (y + 3) (1 + a)

13. ^ x + 5y h ^ 2x - 3y h

2

Exercises 2.6

16. p 2 - 1

2. ^ y - 3 h ] a + b g

29. ] a + b g2 + 2 ] a + b g c + c 2 = a 2 + 2ab + b 2 + 2ac + 2bc + c 2

Exercises 2.9 1.

]x + 3g]x + 1g

4.

] t + 4 g2

7.

]v - 3g]v - 5g

2. ^ y + 4 h ^ y + 3 h

5. ] z + 3 g ] z - 2 g 8. ] t - 3 g

2

3. ] m + 1 g2

6. ] x + 1 g ] x - 6 g 9. ] x + 10 g ] x - 1 g

30. ] x + 1 g2 - 2 ] x + 1 g y + y 2 = x 2 + 2x + 1 - 2xy - 2y + y 2

10. ^ y - 7 h ^ y - 3 h

11. ] m - 6 g ] m - 3 g

12. ^ y + 12 h ^ y - 3 h

13. ] x - 8 g ] x + 3 g

31. 12a

14. ] a - 2 g

32. 32 - z

2

34. x 2 + 3xy + y 2 - 2x

33. 9x + 8x - 3 2

35. 14n 2 - 4

36. x - 12x + 48x - 64 3

2

37. x

2

38. x - 2x y + y 4

2

2

4

2

15. ] x - 2 g ] x + 16 g

16. ^ y + 4 h ^ y - 9 h

17. ] n - 6 g ] n - 4 g 18. ] x - 5 g 2

19. ^ p + 9 h ^ p - 1 h

20. ] k - 2 g ] k - 5 g 21. ] x + 4 g ] x - 3 g

39. 8a + 60a + 150a + 125

22. ] m - 7 g ] m + 1 g 23. ^ q + 10 h ^ q + 2 h

40. 4x + 16x + 15x - 4x - 4

24. ] d - 5 g ] d + 1 g 25. ] l - 9 g ] l - 2 g

Problem

Exercises 2.10

a = 2, b = 7, c = 9, d = 4, e = 3, f = 8, g = 0, h = 6, i = 1

1.

(2a + 1) (a + 5) 2. ^ 5y + 2 h ^ y + 1 h

3.

(3x + 7) (x + 1) 4. (3x + 2) (x + 2) 5. (2b - 3) (b - 1)

6.

(7x - 2) (x - 1) 7. ^ 3y - 1 h ^ y + 2 h

9.

^ 5p - 2 h ^ p + 3 h 10. ] 3x + 5 g ] 2x + 1 g

3

2

4

3

2

Exercises 2.7 2. 5 ] x - 2 g 3. 3 ] m - 3 g 4. 2 ] 4x + 1 g

1.

2^y + 3h

5.

6 ^ 4 - 3y h

9.

3a ] 5 - a g 10. ab ] b + 1 g 11. 2xy ] 2x - 1 g

6. x ] x + 2 g 7. m ] m - 3 g 8. 2y ^ y + 2 h

12. 3mn ^ n 2 + 3 h

13. 2xy ] 4x - z g 14. a ] 6b + 3 - 2a g

15. x ^ 5x - 2 + y h

17. 5b 2 ] b + 3 g

16. q 2 _ 3q 3 - 2 i

18. 3a b ] 2b - a g 19. (m + 5) (x + 7) 2

20. ^ y - 1 h ^ 2 - y h

2

21. (7 + y) (4 - 3x)

22. ] a - 2 g ] 6x + 5 g

23. ] 2t + 1 g ^ x - y h

24. ] 3x - 2 g ] a + 2b - 3c g

25. 3x ] 2x + 3 g 2

28. 4x 2 ] x - 6 g

26. 3q _ pq 2 - 2 i 3

29. 5m 2 n ^ 7mn 3 - 5 h

31. 2rr ] r + h g 32. ] x - 3 g ] x + 2 g 34. - ] a + 1 g

27. 3ab ^ 5a 3 b 2 + 1 h

35. (a 2 + 1) (4ab - 3)

30. 4ab 2 ^ 6ab 3 + 4 h

33. (x + 4) (y 2 + 2)

8. ] 2x + 3 g ] x + 4 g

11. (2y + 1) (y - 6)

12. ] 5x - 1 g ] 2x + 1 g

13. (4t - 1) (2t - 3)

14. (3x + 4) (2x - 3)

15. ^ 6y - 1 h ^ y + 8 h

16. ] 4n - 3 g ] n - 2 g

17. ] 4t - 1 g ] 2t + 5 g 18. ^ 3q + 2 h ^ 4q + 5 h 19. ] 4r - 1 g ] 2r + 6 g = 2 ] 4r - 1 g ] r + 3 g 20. ] 2x - 5 g ] 2x + 3 g

21. ^ 6y - 1 h ^ y - 2 h

22. ^ 2p - 3 h ^ 3p + 2 h

23. (8x + 7) (x + 3)

24. ] 3b - 4 g ] 4b - 9 g

25. (6x + 1) (x - 9)

26. ] 3x + 5 g2

27. ^ 4y + 3 h2

29. ] 6a - 1 g2

30. ] 7m + 6 g2

28. ] 5k - 2 g2

759

760


Exercises 2.11 1.

^y - 1h

5.

(x - 6)

9.

] 5x - 4 g2

Exercises 2.14

2. (x + 3)

2

3. (m + 5)

2

6. ] 2x + 3 g

2

2

8. ] 3a + 2 g

7. ] 4b - 1 g

2

2

2

10. ^ 7y + 1 h2 2

13. ] 5x + 1 g

14. ] 9a - 2 g

2

2

2

12. ] 4k - 3 g

11. ^ 3y - 5 h

2

15. ] 7m + 6 g2 1 2 19. c x + m x

4. (t - 2)

2

16. d t +

1 n 2

2

17. d x -

2

18. d 3y +

2 n 3

1 n 5

2 2 20. d 5k - n k

(a + 2) (a - 2)

2. (x + 3) (x - 3)

4.

]x + 5g]x - 5g

5. (2x + 7) (2x - 7)

7.

(1 + 2z) (1 - 2z) 8. ] 5t + 1 g ] 5t - 1 g 9. ] 3t + 2 g ] 3t - 2 g

3. (y + 1) (y - 1) 6. (4y + 3) (4y - 3)

10. ] 3 + 4x g ] 3 - 4x g

11. (x + 2y) (x - 2y)

12. ^ 6x + y h ^ 6x - y h

13. ] 2a + 3b g ] 2a - 3b g

18. ] z + w + 1 g ] z - w - 1 g

1 1 19. d x + n d x - n 2 2

3

+ 1oe

3

5 ] a - 1 g2 6. - ] 2x - 3 g ] x - 4 g 7. 3z ] z + 5 g ] z + 4 g

8.

ab ] 3 + 2ab g ] 3 - 2ab g 9. x ] x + 1 g ] x - 1 g

10. 2 ] 3x - 2 g ] x + 2 g 11. ] m - 5 g ] 3 + n g

12. - 7 ] 2x + 1 g

14. ] x - 1 g ] x + 2 g ^ x 2 - 2x + 4 h

13. ^ y + 5 h ^ y + 4 h ^ y - 4 h

19. 3 ] 2 - b g ^ 4 + 2b + b 2 h

18. y (2xy + 1) (2xy - 1)

20. 3 ] 3x - 2 g ] 2x + 5 g 21. 3 ] x - 1 g2 23. z ] z + 3 g2

22. (x + 2) (x + 5) (x - 5)

24. ] x + 1 g ] x - 1 g ] 2x + 3 g ] 2x - 3 g 27. 5x ] 2 - x g ^ 4 + 2x + x 2 h

28. (a + 2) (a - 2) (a + 3) (a - 3)

29. 4k (k + 5) 2

30. 3 (x + 1) (x - 1) (x + 3)

Exercises 2.15

- 1 o 21. ^ x + 2y + 3 h ^ x - 2y + 1 h

22. (x 2 + 1) (x 2 - 1) = (x 2 + 1) (x + 1) (x - 1) 23. _ 3x 3 + 2y i _ 3x 3 - 2y i 24. _ x 2 + 4y 2 i ^ x + 2y h ^ x - 2y h 25. (a 4 + 1) (a 2 + 1) (a + 1) (a - 1)

1.

x 2 + 4x + 4 = ] x + 2 g2

2. b 2 - 6b + 9 = ] b - 3 g2

3.

x 2 - 10x + 25 = ] x - 5 g2

5.

m - 14m + 49 = ] m - 7 g

7.

x 2 + 2x + 1 = ] x + 1 g2

9.

x 2 - 20x + 100 = ] x - 10 g2

4. y 2 + 8y + 16 = ^ y + 4 h2

2

6. q 2 + 18x + 81 = ^ q + 9 h2

2

8. t 2 - 16t + 64 = ] t - 8 g2

10. w 2 + 44w + 484 = ] w + 22 g2

Exercises 2.13 2. ] x + 3 g ^ x 2 - 3x + 9 h

1.

(b - 2) (b 2 + 2b + 4)

3.

]t + 1g^t - t + 1h

5.

(1 - x) (1 + x + x )

7.

(y + 2z) (y 2 - 2yz + 4z 2)

9.

^ 2x + 3y h _ 4x 2 - 6xy + 9y 2 i 10. ] ab - 1 g ^ a 2 b 2 + ab + 1 h

11. x 2 - 32x + 256 = ] x - 16 g2

2

2

6. ^ 2 + 3y h _ 4 - 6y + 9y 2 i

12. d

x x 2 3x - 3ne + + 9o 4 2 2

10 1 100 10 1 13. d + ne 2 + o a b ab b 2 a

15. ^ 5xy + 6z h _ 25x y - 30xyz + 36z i 2

14. a 2 + a +

15. x 2 + 9x +

81 9 2 = dx + n 4 2

16. y 2 -

17. k 2 -

11k 121 11 n + = dk 4 2 16

5y 2

1 1 2 = da + n 4 2

+

25 5 2 = dy - n 4 16

2

16. - 9 ^ a - a + 1 h

20. p 2 - 8pq + 16q 2 = ^ p - 4q h2

2

Exercises 2.16

2

x x x ne1 + + o 9 3 3

18. ^ x + y + 3 h _ y 2 - 3y - xy + 9 + 6x + x 2 i 19. ^ x + y - 1 h _ x 2 + 4x - xy + y 2 - 5y + 7 i 20. (2a + 6 - b) (4a + 24a + 2ab + 6b + b + 36) 2

49 7 2 = dx - n 4 2

3 2 9 = dy + n 4 2

18. x 2 + 6xy + 9y 2 = ^ x + 3y h2 19. a 2 - 4ab + 4b 2 = ] a - 2b g2

14. ^ x + 1 - y h _ x 2 + 2x + 1 + xy + y + y 2 i 2

13. x 2 - 7x +

8. (x - 5y) (x 2 + 5xy + 25y 2)

11. (10 + 2t) (100 - 20t + 4t 2)

2

12. y 2 + 3y +

4. (a - 4) (a + 4a + 16)

2

17. d 1 -

5.

26. 4a (a + 3) (a - 3)

15. ] 2a + 9b g ] 2a - 9b g 17. (a + b - 3) (a - b + 1)

20. e

5 ^ y - 1 h _ y 2 + y + 1 i 4. 2ab ^ a + 2b) (2a - 1 h

25. 2 ] x + 2 g ] x - 2 g ^ x + y h _ x 2 - xy + y 2 i

16. ^ x + 2 + y h ^ x + 2 - y h

y

3.

16. x ] x + 2 g ] x - 5 g 17. ] x + 3 g (x - 3) 2

1.

y

2 ] x + 3 g ] x - 3 g 2. 3 ^ p + 3 h ^ p - 4 h

15. ] x + 1 g ^ x 2 - x + 1 h ] x - 1 g ^ x 2 + x + 1 h

Exercises 2.12

14. ^ x + 10y h ^ x - 10y h

1.

2

1.

a+2

2. 2t - 1

6.

1 y-4

7.

10. 14.

p+5 3

2 ] b - 2a g a-3

11.

a+1 a+3

p-2 4p - 2p + 1 2

3.

15.

4y + 1 3

s-1 s+3

8.

12.

4.

4 2d - 1 9.

3+y x + 2x + 4

a+b 2a - b

2

5.

x 5x - 2

b2 + b + 1 b+1 13. x - 3

ANSWERS

Exercises 2.17 1.

2.

(a)

(a)

(d) 3.

5x 4

(b)

Exercises 2.20

13y + 3

b 2a - 1

(b)

a+8 12

(d)

^ p - 2 h _ q2 - q + 1 i

ab

a+b+3 (c) a+b

-x + 2 (b) x ]x - 1g

(a)

x - 13 6

b 2 ^ x + 2y h

10 ] 2b - 1 g

2 _ 3y 2 + 14y + 13 i

^y + 2h^y + 3h^y - 1h x2 ] x + 2 g

(b)

8 _ y 2 - 3y + 9 i

1. 3 5

(e)

3p 2 + 5pq - 2q 2

pq ^ p + q h ^ p - q h

9. - 4 2

10. 4 5

13. - 3

14.

15. 5 7

16.

- ] 5x + 22 g (j) ]x + 4g]x - 4g]x + 3g

1.

5.5 7. 377

2.

15y

2. 47

8. 14

14.

3 4

9. 60

14. - 84

16. 28

18. - 2 105

17.

30

25. 2 3

26.

3

31.

2 2

3 10 3

32.

2

(a)

4. 375

5. - 196

17. 2 3

18. 23.987 19. 352.47 20. 93 21. 4

10 + 6

3 4

2.

2.

(n) 9 3

(p) 6 3

(q) 3 11

(r) 5 5

(a) 6 3

(b) 20 5

(c) 28 2

(f) 8 14 3.

4.

(a)

18

(f)

160

(g) 72 5 (b) (g)

20 117

(d) 4 7

(h) 30 2

(c)

176

(h)

98

2 2

2 3

35.

5 7

(c) 12 + 8 15

(e) 16 5 (j) 24 5

(d)

128

(e)

75

(i)

363

(j)

1008

(a) x = 45 (b) x = 12 (c) x = 63 (d) x = 50 (e) x = 44 (f) x = 147 (g) x = 304 (h) x = 828 (i) x = 775 (j) x = 960

(h) 5 - 5 15

(l) 210 - 14 15

(n) - 10 - 2 2

(o) 4 3 - 12

(h) - 1

(i) - 12

(n) 7 + 2 10

(j) 43

(k) 3

(o) 11 - 4 6

(l) - 241 (p) 25 + 6 14

(r) 27 - 4 35

(s) 77 - 12 40 = 77 - 24 10

(t) 53 + 12 10

3.

(a) 18

(d) 19 + 6 2

4.

(a) a = 21, b = 80

5.

(a) a - 1

6.

k = 25

9.

a = 107, b = - 42

(o) 7 5

(i) 14 10

34.

(f) 15 - 15 + 18 10 - 6 6

(j) 3 6

(i) 4 2

5

(c) 2 10 - 6 + 10 15 - 15 6 (d) 12 20 + 18 60 - 8 10 - 12 30 = 24 5 + 36 15 - 8 10 - 12 30

(q) 57 + 12 15

(h) 5 3

2 5

33.

3 5

6

1 2

(a) 10 + 3 6 + 3 5 + 9 3 (b) 10 - 35 - 2 + 14

(e) 6 2

(m) 8 2

9

29.

(g) - 6 - 12 6

1.

(g) 4 3

2 5

1

(j) 2 54 + 6 = 6 6 + 6

(m) - 6

(l) 10 3

28.

8

24.

(f) 5 33 + 3 21

(g) 4

(k) 4 7

1

23. 1

(e) - 6 + 4 18 = - 6 + 12 2

Exercises 2.19 (f) 10 2

22. 4 3

(b) 2 6 - 15

(e) 52 - 13 10

(d) 5 2

19. 18

(d) 5 14 - 2 21

(m) 10 6 - 120

15. 15 16. 10

(c) 2 6

27.

6. 30

15. 2

21. 2 6

1

5. - 6 6

12 = 2 3

12. 15 28 = 30 7

(e) 0.6

13. 1838.8

(b) 3 7

10.

11. 2 48 = 8 3

1.

12. 22.4

(a) 2 3

7-5 2

Exercises 2.22

10. 51.935 11. - 1

8. 284 9. - 40

4. 10 14

(k) - 8 + 12 12 = - 8 + 24 3

(d) - 37.7

3. - 7

21.

24. - 2 - 2 3

13. 2 20 = 4 5

30.

a 2 - 2ab - b 2 + 1 ]a + bg]a - bg

(c) 48.1

3. 3 6

15

(i) 6 + 30

(f) 2.3 (g) - 5.3 6.

21

7. - 12 55

^x + yh^x - yh

(b) - 6.9

12. 5 3

2

17. 13 6

2

20. 5 2 - 2 3

23. 7 6 + 3 5

20. 30 50 = 150 2

y ^x + y + 1h

(a) - 7.1

2

19. 47 3

11.

6. 3 6

Exercise 2.21

2]x - 1g (f) ]x + 1g]x - 3g

Exercises 2.18 1.

5. - 3 5

25. - 17 5 + 10 2 2x (d) x+2

^y + 2h^y + 1h

(d)

4. 3 3

8. 8 5

22. - 2 3 - 4 5

x 2 + 10x - 24 3b 2 - 5b - 10 (d) (e) x 2 ]x - 3g]x - 4g 2b ] b + 1 g 3x - 13 3 - 5x (a) (b) ]x - 5g]x - 2g]x + 3g ]x + 2g]x - 2g (c)

3. 6 3

2

7. - 7 2

(c)

5.

2.

18. - 9 10

a+2 (h) ] a + 1 g2

- 3x + 8 (g) ]x + 2g]x - 2g

4.

(e)

6

]x - 3g]x - 1g (e) ]x - 5g]x - 2g

^ p + qh^ p - qh + 1 p2 - q2 + 1 = (e) p+q p+q

(i)

4p + 3

(c)

q+1

x 2 - xy + y 2

5 (a) x

(c)

15

(b) 108 2

(c) 432 2

(e) 9

(b) a = 19, b = - 7

(b) 2p - 1 - 2 p ^ p - 1 h 7. 2x - 3y - 5 xy

8. a = 17, b = 240

10. 9 + 5 units 2

761

762


Exercises 2.23 1.

(a) (e) (h)

2.

7 7

8. 6 4

(b)

3+ 6 3

(f)

3 14 - 4 7 14

(c)

2 15 5

(d)

12 - 5 2 2 (i)

6 14 3 14 = 5 10

(g)

(a) 4 3 - 4 2 = 4 ^ 3 - 2 h

(b)

(j)

4 15 - 2 10 35

6 15 - 9 6 + 2 10 - 6 2

3.

So rational 9.

1.

2.

(j) (l) 4.

(i)

2-1

=

28 - 2 6 - 7 3 13

(b) a = 1, b = 8

8 5 (d) a = - 1 , b = 9 9

=

(k)

2 15 + 2 10 - 2 6 - 3 - 5 2

(a) a = 45, b = 10

5.

4 6+9 3 21

15 30 - 30 5 - 4 3 30

+ 2+1 2-1

2 2-1

2+1 2-1 ^ 2 - 1h^ 2 - 1h

+ +

4 2

(a) 4 (b) 14

7.

3 5 - 2 - 15 - 3 3

#

4 2 2

^ 2 h2 - 1 2 2- 2- 2+1 = +2 2 2-1 3-2 2 = +2 2 1 =3-2 2+2 2 =3 So rational 6.

1 1 (c) a = - , b = 2 2

(e) a = 5, b = 32

4

#

x = -^ 3 + 2h

10.

2 2

(a) - 2y

(b) a + 4

b+4 b+4 b-4

(f) 6 2

(g) 4 5

(c) - 6k 5

(d)

5x + 3y

(e) 3a - 8b

15

(b) ] a + 3 g ] a - 1 g (c) 4ab ] b - 2 g

(a) ] x + 6 g ] x - 6 g

(e) 2 ^ 2n - p + 3 h

(d) (y - 3) (5 + x)

(f) (2 - x) (4 + 2x + x 2) 3.

20 12 + 19 6 + 25 3 - 6 19 6 + 65 3 - 6 (g) = 15 15 6+9 2+2 3 6

2

Test yourself 2

(a) 2 2 (b) - ^ 2 + 6 - 3 2 + 3 3 h = - 2 - 6 + 3 2 - 3 3 22 5 + 14 2 (c) 39 ^ (d) - 6 6 - 16 - 3 84 + 8 14 h 10 - 3 6 + 8 + 3 21 - 4 14 = 5 (e) - 4 (f) 4 2

(h)

2

6-4 2 +4 2 9-4#2 6-4 2 = +4 2 1 =6-4 2+4 2 =6

(c)

(f)

2 3-2 2 8 = # + # 3+2 2 3-2 2 2 2^3 - 2 2 h 8 2 + = 2 2 32 - ^ 2 2 h =

-^ 6 + 7 3 h 47

- ^ 2 15 - 4 18 h - 2 ^ 15 - 6 2 h = 19 19 - ^ 19 - 8 3 h 8 3 - 19 = (d) (e) 6 + 2 + 5 3 + 5 2 13 13

8

+

2

5 + 2 10 5

8 5 + 3 10 20

2 3+2 2

(b) 2x 2 + 5x - 3

(a) 4b - 6

(d) 16x - 24x + 9 (g) 2 6 - 5 3 (a)

5.

V = 157.464

(f) - 1 - 7a

(h) 3 3 - 6 + 21 - 2 7

8

4.

(c) 4m + 17

(e) p 2 - 25

2

(b)

b 2 ^ a 2 + 3a + 9 h

6. (a) 17

15 ] m - 2 g2 (b)

6 15 - 9 17

4x + 5 8. (a) 36 (b) - 2 ]x + 3g]x - 2g 1 9. (a) (b) 8 10. d = 11.25 5 2 3 2+ 6 11. (a) (b) 15 2 7.

12. (a) 3 6 - 6 - 4 3 + 4 2

(c) 2

(d) 216

(b) 11 + 4 7

(b) 6 ] x - 3 g ] x + 1 g

13. (a) 3 (x - 3) (x + 3)

(c) 5 ^ y + 2 h _ y 2 - 2y + 4 i

14. (a)

x3

(b)

3y 4

15. (a) 99

1 3x - 1

(b) 24 3

16. (a) a 2 - b 2

(b) a 2 + 2ab + b 2

17. (a) ] a - b g2

(b) ] a - b g ^ a 2 + ab + b 2 h

18.

3 3+1 2

20.

21 5 - 46 - 2 7

(c) 16

19. (a)

4b + 3a ab

(c) a 2 - 2ab + b 2

(b)

3x - 11 10

(e) 2

ANSWERS

21. (a) 6 2 (f)

(b) - 8 6

m

24. (a)

(d)

3 7 7

6 15

5+1 2

(c)

(e)

x + 10 10

17a - 15 21

1 k-1

(b)

(e)

71 121

20 + 3 15 + 4 10 + 3 6 53 (c)

3 - 2x (x + 1) (x - 1)

15 - 6 - 15 3 - 15 2 3

(b) n = 175

(d) n = 5547

27. (b), (c)

(c) n = 392

28. (d)

33. (a)

29. (a), (d)

34. (d)

30. (c)

35. (b)

(b) y 4 - 4

2. 4.

x2 +

2 3. or 4 2 2

b b2 b 2 n x + 2 = dx + a a 2 4a

4x 2 + 12x + 9 = ] 2x + 3 g2 ]a + 1g

a2 - a + 1

10. d

11. w = 13

7.

(d) ] b - 2 g ] a + 2 g ] a - 2 g y+1

2]x - 1g

8. 2 5

13. x = 14

16. p = 3

5. k = 5

5 8

24. y = 1

14. x = - 1

17. t = 8.2

20. x = - 3

1.

t = 8.5

6.

r = 6.68

21. b = 0.8 2 25. t = - 1 3

18. x = - 9.5 22. a = - 0.375

2x 1 1 2 + = dx + n 9 3 3

20. r =

21. s = 2 + 6 3

4. a = 41

3 4 r

=

71 121 3 r 4r

(b) a =

8. n = 15

11. (a) BMI = 25.39 12. r = 0.072

19. x = 5.5

20. r = 3.3

5. y = 4

9. y 1 = 3

2 3

(b) w = 69.66 13. x 1 = - 9

17. r = 10.46

14. t = 2.14

18. x = 1.19

Exercises 3.4 1.

2.

16. x = 2

(a) 3

7. x = 6.44

16. r = 2.12

(a) x 2 3

-4

- 66 6 + 4 2 - 15 + 4 5 - 65 3 13

18.

3. b = 8

-3

-2

-1

0

1

2

3

-2

-1

0

1

2

3

4

(b) y # 4

3x + 4 (b) ] 2x - 1 g2

2

400 - 59 5 10

2. l = 122

15. x = ! 2

-4

13. x 3 - 7x 2 + 15x - 9

1 2

12. t = 30

15. x = - 0.4

2 a 2 a + nd - n x b x b

12. (a) 8x - 12x + 6x - 1

19. i = 1

9 35

x = 36 7. t = 0.6 8. x = - 3 9. y = - 1.2 10. x = 69

]x - 3g]x + 3g]x - 2g 3

17.

4. x = 1

6.

3x 3 - 6x 2 + 3x + 4xy - 6y

15. x 2 +

2. x = 35 3. y = 4

4 9

b =3

(c) h = 1.94

(a) ] x + 4 g ] x + 9 g

2

14.

1 3

1.

10. h = 3.7

(c) ] 5x + 7 g ^ 25x 2 - 35x + 49 h

11.

30. x Z 4.41

1

(b) _ x 2 - 3y i _ x 2 + 2y i = (x + 3 y) (x - 3 y) _ x 2 + 2y i

9.

29. p = 5

2

17 3 + 2 5 + 20 17

6.

16. x = 20 17. m = 20 18. x = 4 19. a = - 7 20. y = 3 2 21. b = - 4 22. x = 3 23. a = - 1 24. t = - 4 3 1 25. x = 1.2 26. a = 1.6 27. b = 28. t = 39 8

Exercises 3.3

(c) 8x - 60x + 150x - 125

5.

2. z = - 5.6 3. y = 1 4. w = 6.7 5. x = 12 1 8. b = 35 9. n = - 16 10. r = 4 6. x = 4 7. y = 15 11. y = 9 12. k = 6 13. d = 2 14. x = 5 15. y = 15

23. x = 3

(a) 2a 2 b - 8ab 2 + 6a 3 3

t = -5

19. q = 22

Challenge exercise 2 1.

1.

Exercises 3.2

(e) n = 1445

32. (b)

Chapter 3: Equations

Exercises 3.1

12 - 2 6 15

31. (c)

(e) 30a 2 b

3

(b) 10 14 - 5 21 - 6 10 + 3 15

(b)

25. (a) n = 48

26. 3

4

(d) 43 (e) 65 - 6 14

(c) 7

(d)

(d)

(g) 2x - 3y

3n 4

22. (a) 2 6 + 4

23. (a)

(c) 2 3

17 14 , b=23 23

-3

4

(a) t 2 7 (b) x $ 3 (c) p 2 - 1 (d) x $ - 2 (e) y 2 - 9 1 2

(f) a $ - 1

(g) y $ - 2

(j) y 1 12

(k) b 1 - 18

(l) x 2 30

(n) m 2 14

2 3

1 4

(o) b $ 16

(h) x 1 - 2

(i) a # - 6

(m) x # 3

(p) r # - 9

3 4

(q) z 2 8

763

764


(r) w 1 2

4 5

(s) x $ 35 2 3

(v) x 2 - 1 3.

(t) t $ - 9

(w) b # - 11

(u) q 2 - 6

(e) y = 1.89 (f) d = ! 2.55 (g) k = ! 4.47 (h) x = 2.22

2 5

(i) y = ! 3.81

1 4

3.

(a) 1 1 x 1 7 1

0

2

3

4

5

6

7

8

0

1

2

3

4

5

4.

(b) - 2 # p 1 5 -3

-2

-1

(c) 1 1 x 1 4 -3

-2

0

-1

1

2

3

4

-3

-2 -1 1 2 (e) 1 y 1 1 6 3 -1

-2

0

1

2

3

0

1

2

3

2.

(a) x = ! 5

(b) y = ! 8

5

(c) - 4 1 a 1 4

(d) k $ 1, k # - 1

(e) x 2 6, x 1 - 6

(f) - 10 # p # 10

(g) x = 0

(i) - 12 1 y 1 12

(j) b $ 20, b # - 20

(a) x = 5, - 9

(g) - 3

1 1y 12 2

(a) x = 1

1 4

(b) a = 3, -

3 1 (h) d = 2 , -1 4 2 (a) x = 2, -

1 2

(d) x = 4, -7 5.

1 3

2 (a) t = 3, -1 5 -3

-2

-1

2.

(a) x = 3

(f) a = 2 (g) x = ! 2 (h) b = 9 2 3

1 3

(c) b = 2 2 7

(c) m = 9 (h) x = 2

(a) x = 2

(b) x = 1 (c) x = - 2 (d) n = 2 (e) x = 0 1 (g) y = (h) x = 2 (i) x = 2 (j) a = 0 3

(i) x = ! 12

(b) y = 3, 2

1 3

1 3 2 3

(c) a = - 10, 1

(b) y = ! 8

(e) p = 10

(f) x = ! 5

(i) n = ! 4

(j) q = - 2

(f) n =

(i) k = -

1 6

(j) x = 1

(a) x = - 1

(c) n = ! 2 (g) y = ! 3

(c) x =

3 4

2 3

(d) x = 5 (i) x = 1

1 3

(g) x = 1

(d) k = -

1 2

(h) n =

2 3

1 2

(b) x = - 1

1 3

(c) k = - 4

(f) x = -

2 3

(g) x = - 4

1 2

10. (a) m =

1 4

(b) k = - 2

(e) n =

1 18

(f) n = 1 7

(e) m = 0 (j) k = 2

2 3

(j) x = 18

(i) x = - 1

2

1 3

4 5

3 5

2 (b) - 1 1 t 1 3 5 1

(b) x =

(e) k = -

(i) x = 1

(e) d = 4, -5

0

1 2

(a) m =

(e) x = - 2

(j) No solutions

1 2

(b) y = 5

8.

(f) x = 7 (g) m = 5, 1

4 , -2 5

(i) y =

(j) b = ! 1

1 7

(a) x =

(f) x = 6

3 4

1 2

(c) x = 2 (g) x =

4 5

3 8

(d) n = 3 1 2

(h) x = - 1

(d) k = 1

1 2

(h) b = - 3

1 6

7 11

(j) m = 5

Puzzle 3

4

1.

All months have 28 days. Some months have more days as well. 2. 10 3. Bottle $1.05; cork 5 cents

4.

16 each time

5

Exercises 3.6 1.

2 3

(d) x = !

(g) x = 2

9.

(d) No solutions (e) y = - 2

4.

(e) n =

(c) y =

(f) x = 3 7.

(j) 2 # a # 10 3.

(b) a =

(a) n = 4

5 (f) x = 5, -4 7

(h) x $ 9, x # - 6

1 2

1 5

6.

(c) a 2 2, a 1 - 2

(e) x = 3, -6

1 2

(h) y = 27

(a) x =

(h) a 2 14, a 1 - 14

(b) n = 4, -2

(d) 4 # x # 6

(j) t = 81

(g) b = 216

1 1 1 1 (b) x = 6 (c) a = (d) k = 4 512 81 625 19 1 (e) x = (f) x = 4 (g) y = 8 (h) n = 7 8 32 127 (i) b = 8 (j) m = 1 216

Exercises 3.5 1.

(i) a = 128

(d) t = 8

5. 5

4

(c) x = 32

(f) m = 625

5

4

(b) t = 16

(e) p = 243

(i) x = !

(d) - 3 # y # 5

-3

(a) n = 27

(j) y = 3.01

(d) x = ! 2 5 (h) w = 2

(a) p = ! 6.71 (b) x = 4.64 (c) n = 2.99 (d) x = ! 5.92

5. Friday

Exercises 3.7 1.

y = 0, -1

5.

x = -2, -7

2. b = 2, -1 6. q = !3

3. p = 3, -5 7. x = !1

4. t = 0, 5

8. a = 0, -3

ANSWERS

9.

x = 0, - 4

12. y = 1, -1 16. x = 1, 2

10. x = ! 1 2

20. x = 3, 4

11. x = -1, -1

3 1 , 4 2

13. b =

1 2

1 2

17. x = 0, 5

10. x 1 2, x 2 2

14. x = 5, -2 15. x = 0, 18. y = - 1, 2

21. m = - 6, 1

23. y = 1, -5, -2

1 3 2 3

19. n = 3, 5

22. x = 0, -1, -2

24. x = 5, -7

25. m = 8, -1

(b) a = ! 7 + 3

(d) x = ! 13 - 1

(c) y = ! 23 + 4

(e) p = ! 44 - 7 = ! 2 11 - 7

(f) x = ! 28 + 5 = ! 2 7 + 5

(g) y = ! 88 - 10 = ! 2 22 - 10 = 2 ^ ! 22 - 5 h (h) x = ! 2 + 1

2.

(i) n = ! 137 - 12

! 5+3 2

(a) x = 3.45, -1.45

(b) x = - 4.59, -7.41

(c) q = 0.0554, -18.1

(d) x = 4.45, - 0.449

(e) b = - 4.26, -11.7

(f) x = 17.7, 6.34

(g) r = 22.3, - 0.314

(h) x = - 0.683, -7.32

(i) a = 0.162, - 6.16

(j) y = 40.1, - 0.0749

(a) y = - 0.354, - 5.65 (c) b = 3.54, - 2.54

(f) n = 0.243, -8.24

- 1 ! 17 2

(c) q =

4 ! 28 = 2! 7 2

(b) x =

(e) s =

8 ! 40 4 ! 10 = 6 3

(f) x =

- 11 ! 133 2

(j) x =

(i) x = 1, - 6

5 ! 13 6

(g) d =

2 ! 32 =1!2 2 2

4.

01m #

7.

11x 11

2 7 1 4

1.

-3 1 x 1 0

- 5 ! 73 12

1 2

4.

x # - 2, x $ 2

7.

c 1 - 1, c 2 2

3 5. x 1 - , x > 0 5 8. - 3

2 4 1x 17 3

1 2

9. 2 1 x # 2

3 4

3. n # 0, n $ 1

5. n 1 - 1, n 2 1

6. - 5 # n # 3

8. - 4 # x # - 2 1 2

2 5

9. 4 1 x 1 5

11. a 1 - 1, a 2 13. x # 15. - 1

1 3

2 ,x $1 3

1 1 #x #2 3

17. x 1 - 4, x 2 4

18. - 1 # a # 1

19. - 2 1 x 1 3

20. x # - 1, x $ 3

21. 0 1 x 1 2

1 2

23. y # - 2, y $

1 2

4 5

4 5

25. 1 # x # 1

27. x 1 0, x $

28. y 1 - 1, y 2 -

6. - 2 # b 1 0

1 1 z 1 -3 5

29. x 1 - 4, -

16. - 4 # y # 3

26. 0 1 x 1

1! 5 2

3. x 1 0, x 2 1

27. x $ 3, - 1 # x 1 2

2. 0 1 y 1 4

30. x # - 8, x 2 - 5

2. 0 1 x 1

25. x # - 1, 3 1 x # 4

1 ,11x #7 2

2 1 24. m 1 - 1 , m 2 1 3 2

Exercises 3.10 y 2 1, y 1 0

1 2 #x 1 2 3

22. 1 # a # 1

(i) t =

23. x 2 5, - 3 1 x 1 0

24. m # - 2, - 1 1 m # 6

14. b 1 - 3, b 2

7 ! 41 4

1.

22. 0 1 n # 2, n $ 4

10. b # - 2, b $ -

- 12 ! 128 -3 ! 2 2 (d) h = = 8 2

(h) x =

1 2 2 19. t # , t 2 2 5 5 3

1 12. y 1 - 1 , y 2 2 2

(a) x =

1 1 p 1 26 2

Exercises 3.11

(b) x = 1, 1.5

(h) x = 0, 7

17. 4

8 1 m 1 0 21. x 1 - 5, 0 1 x 1 1 9

30. x #

(j) y = 2.62, 0.382 2.

15. y 1 - 2, y 2 - 1

28. n 1 - 1, 3 1 n 1 5

(d) x = 1, - 0.5

(e) x = - 0.553, 0.678 (g) m = - 2, - 5

20. -

26. x # - 2,

Exercises 3.9 1.

5 1 1x 1 9 2

5 # x 1 -4 9

1 1 13. a 1 - 3 , a 2 - 2 4 2

1 7 1x 11 3 15

18. x # - 1, x 2 -

(a) x = ! 5 - 2

(j) y =

14.

11. - 4

7 16. x # - , x 2 4 8

Exercises 3.8 1.

12. 1

1 6

1 2

29. 3 1 n # 3 31. x 1

1 3

1 2

3 2 ,x 2 5 7

1 32. x # 4 , x 2 5 5

1 33. x # - 1 , x 2 - 1 4

34. x 1 - 3, x 2 2

35. -

3 3 #x 14 5

765

766


Exercises 3.12 1.

a = 1, b = 3

4.

x = 6, y = 17

7.

x = - 3, y = 2

10. m = 2, n = 3

2. x = 2, y = 1

3. p = 2, q = - 1

5. x = - 10, y = 2 11. w 1 = - 1, w 2 = 5

9. x = 3, y = - 4

15. x = - 1, y = - 4 16. s = 2, t = - 1 18. k = - 4, h = 1

19. v 1 = - 2, v 2 = 4

4.

(a) x = - 2, y = 5

5.

(a) x = 2

6.

(a) b = 2, -1

7.

(a) A = 36

20. x = 2, y Z 1.41

9.

-1 1 y # 3

1 3

1 4

(b) g = 2,

(b) b = 12

10. (a) x = - 0.298, -6.70

Problem

1 4

(c) x $ 4, x # 3

8. x =

1 ,1 2

(b) y = 4.16, -2.16

(c) n = 0.869, -1.54

23 adults and 16 children.

Exercises 3.13

11. (a) V = 764.5

(b) r = 2.9

13. x 1 2, x 2 9

14. x = 2.4, y = 3.2

(b) r = 3.9

1.

x = 0, y = 0 and x = 1, y = 1

2.

x = 0, y = 0 and x = - 2, y = 4

3.

x = 0, y = 3 and x = 3, y = 0

4.

x = 4, y = - 3 and x = 3, y = - 4

6.

x = 3, y = 9

8.

m = - 4, n = 0 and m = 0, n = - 4

9.

x = 1, y = 2 and x = - 1, y = - 2

13. x = 1, y = 5 and x = 4, y = 11 1 14. x = , y = 4 and x = - 1, y = - 1 4

16. (a) ii (b) i

18. n 2 0, n 1 - 3 5. x = - 1, y = - 3

7. t = - 2, x = 4 and t = 1, x = 1

11. x = 2, y = 1 and x = - 1, y = - 2

19. x = - 4

20. x = - 2

(c) x = 2

15. (a) V = 2100 (d) iii

(d) x = 2

(g) - 4 # x # 2

21. (a) y 2 3 (e) x = 3, -1

(h) x = - 3

12. x = 0, y = 1 (k) x =

(j) x # - 1, x $ 1 1 1 15. t = - , h = 4 2

(m) No solutions

(e) iii

18. x = 0, y = 0 and x = 1, y = 1 and x = - 1, y = 1

(s) 2 1 n # 2

5 12 ,y =13 13

2 5

2 5

(b) - 3 # n # 0 (f) t $ 1, t # - 2

(i) y 2 2, y 1 - 2

5 6

(l) -

1 3 (n) t = 2 , 3 5

(p) m # - 3, m $ 2

17. x = 0, y = 0 and x = - 2, y = - 8 and x = 3, y = 27

20. x = -

(c) ii

1 4

1 3

16. x = 2, y = 0

3 1 ,y =2 4 2

12. x 2 71

17. a = 3, b = 2, c = - 4

10. x = 0, y = 0 and x = 1, y = 1

19. x =

(b) k 2 + 4k + 4 = ] k + 2 g2

1 (b) x = 4, y = 1 and x = - , y = - 8 2

(b) y =

12. a = 0, b = 4

14. x 1 = 1, x 2 = - 1

17. a = - 2, b = 0

(a) x 2 - 8x + 16 = ] x - 4 g2

6. t = 3, v = 1

8. x = - 64, y = - 39

13. p = - 4, q = 1

3.

1 #b #2 2

(o) - 1 1 x 1 3

(q) t 1 - 1, t 2 0

(t) -

(r) 1 1 y 1 3

1 1 1x # 5 2

Challenge exercise 3

Exercises 3.14 2. a = - 2, b = - 1, c = 2

1.

y =1

4.

x = 2.56, -1.56

2. x 1 - a, x 2 a

3. a = 3, b = !2

1.

x = - 2, y = - 8, z = - 1

3.

a = - 4, b = 2, c = 7

4. a = 1, b = 2, c = - 3

6.

5.

x = 5, y = 0, z = - 2

6. x = 0, y = - 5, z = 4

] x + 3 g ] x - 3 g ] x - 2 g ^ x 2 + 2x + 4 h; x = ! 3, 2

7.

7.

p = - 3, q = 7, r = 4

8. x = 1, y = - 1, z = 2

x = 1, y = 2 and x = - 1, y = 0

8.

9.

h = - 3, j = 2, k = - 4

b = 4; x = ! 17 + 4 Z 8.12, - 0.123

10. a = 3, b = - 1, c = - 2

10. - 1 1 t 1 1

Test yourself 3

13. r = 2.31

1.

16. y # - 2,

2.

(a) b = 10

(b) a = - 116 (c) x = - 7 1 (d) x # - 4 , x 2 - 3 (e) p # 4 3 (a) A = 1262.48

(b) P = 8558.59

18. x =

5. y # - 2, 0 1 y # 3

11. - 3 # x # 8

14. No solutions 1 2 #y 1 2 3

2 ^ 4 ! 10 h 3

20. y 1 -1, y 2

3 5

9. x = ! 1 1 12. x = 4

15. x = ! b + a 2 + a

17. P = 2247.36

19. x 1 - 4, - 2.2 1 x 1 0.7

ANSWERS


10. +AEB + +BEC + +CED = 50 - 8y + 5y - 20 + 3y + 60 = 90c

Exercises 4.1 1.

4.

7.

So +AED is a right angle.

(a) y = 47c (b) x = 39c (c) m = 145c (d) y = 60c (e) b = 101c (f) x = 36c (g) a = 60c (h) x = 45c (i) y = 40c (j) x = 80c 2. (a) 121c (b) 72c 29l (c) 134c 48l 3. (a) 42c (b) 55c 37l (c) 73c 3l (a) (i) 47c (ii) 137c (b) (i) 9c (ii) 99c (c) (i) 63c (ii) 153c (d) (i) 35c (ii) 125c (e) (i) 52c (ii)142c (f) (i) 15c7l (ii) 105c7l (g) (i) 47c36l (ii) 137c36l (h) (i) 72c21l (ii)162c21l (i) (i) 26c11l (ii) 116c11l (j) (i) 38c51l (ii) 128c51l 5. (a) x = 49c (b) 41c (c) 131c 6. (a) y = 15c, x = z = 165c (b) x = 142c, y = 48c, z = 28c (c) a = 43c, b = 137c, c = 101c (d) a = 97c, b = d = 41c, c = 42c (e) a = 68c, b = 152c, c = 28c (f) a = 10c, b = 150c 8x - 10 + 2x - 10 + x + 10 + 7x + 10 = 360 (angle of revolution)

18x = 360 x = 20 +ABE = 8x - 10 = 8 (20) - 10 = 150c +EBC = 2x - 10 = 2 (20) - 10 = 30c +ABE + +EBC = 150c + 30c = 180c ` +ABC is a straight angle +DBC = 7x + 10 = 7 (20) + 10 = 150c +DBC + +EBC = 150c + 30c = 180c ` +DBE is a straight angle ` AC and DE are straight lines 8.

+DFB = 180c - (180 - x) c (+AFB is a straight angle)

=x ` +AFC = x


+CFE = 180c - (x + 180c - 2x) (+AFB is a straight angle)

=x ` +AFC = +CFE ` CD bisects +AFE 9.

+ABD + +DBC = 110 - 3x + 3x + 70 = 180c So +ABC is a straight angle. AC is a straight line.

Exercises 4.2 1.

(a) a = b = e = f = 148c , c = d = g = 32c (b) x = z = 70c , y = 110c (c) x = 55c , y = 36c , z = 89c

(d) y = 125c , x = z = 55c

(e) n = e = g = a = c = z = x = 98c, o = m = h = f = b = d = y = w = 82c (f) a = 95c , b = 85c , c = 32c (g) a = 27c , b = 72c , c = 81c (h) x = 56c , y = 124c , z = a = 116c , b = 64c (i) x = 61c 2.

(a)

(j) y = 37c

+CGF = 180c - 121c

(FGH is a straight angle)

= 59c ` +BFG = +CGF = 59c These are equal alternate angles. ` AB < CD (b) +BAC = 360c - 292c = 68c (angle of revolution)

` +BAC + +DCA = 68c + 112c = 180c These are supplementary cointerior angles. ` AB < CD (c) +BCD = 180 - 76 (+BCE is a straight angle) = 104c +ABC = +BCD = 104c These are equal alternate angles. ` AB ; CD (d) +CEF = 180 - 128 (+CED is a straight angle) = 52c +CEF = +ABE = 52c These are equal corresponding angles. ÀB ; CD (e) +CFH = 180 - ] 23 + 115 g (+EFG is a straight angle) = 42c `+BFD = 42c (vertically opposite angles) +ABF + +BFD = 138c + 42c = 180c These are supplementary cointerior angles. ` AB ; CD

767

768


Exercises 4.3 1.

(a) x = 60c

(b) y = 36c

(c) m = 71c

(e) x = 30c

(f) x = 20c

(g) x = 67c

(d) x = 37c (h) a = 73c

10. +OQP = 180 - ] 75 + 73 g (angle sum of triangle) = 32c ` +MNO = +OQP = 32c These are equal alternate angles.

(i) a = 75c , b = 27c , c = 46c

` MN ; QP

(j) a = 36c , b = 126c , c = 23c (k) x = 67c , y = z = 59c , w = 121c 2.

3. 4.

5.

Exercises 4.4

All angles are equal. Let them be x. (angle sum of D) Then x + x + x = 180 3x = 180 x = 60

AB = EF = 5cm

(given) (given)

So all angles in an equilateral triangle are 60c.

AC = DE = 8 cm

(given)

] 90 - x g c

` D ABC / DDEF

(SSS)

(b)Yes

(vertically opposite angles) +ACB = 50c +ABC = 180c - (50c + 45c) (angle sum of D) = 85c ` +DEC = +ABC = 85c These are equal alternate angles.

XY = BC = 4.7 m

(given)

+XYZ = +BCA = 110c (given) YZ = AC = 2.3 m

(given)

` AB < DE

` D XYZ / DABC

(SAS)

+ACB = 180c - 124c = 56c +CBA + 68c = 124c +CBA = 124c - 68c = 56c ` +CBA = +ACB = 56c ` D ABC is isosceles

(c) No

y = 38c

7.

(a) x = 64c

(DCB is a straight angle)

(b) x = 64c , y = 57c

+HJI = 180c - (35c + 25c) = 120c +IJL = 180c - 120c = 60c +JIL = 180c - (90c + 30c) = 60c +ILJ = 180 - (60c + 60c) = 60c

(d) Yes

(exterior angle of D)

2. (angle sum of D HJI)

(angle sum of D IKL) (angle sum of D JIL)

(given) (given)

`DPQR / DSTU

(AAS)

(given) (a) AB = KL = 4 (given) +B = +L = 38c (given) BC = JL = 5 ` by SAS, D ABC / D JKL

(given) (d) +Y = +T = 90c (given) +Z = +S = 35c (given) XY = TR = 1.3 ` by AAS, D XYZ / D STR

(angle sum of D JKL)

` +JLK = +JKL = 30° ` D JKL is isosceles

(given) (e) BC = DE = 4 (given) +C = +E = 90c (given) AC = EF = 7 ` by SAS, D ABC / D DEF

BC = BD

` AB ; ED

+PRQ = +STU = 52c QR = TU = 8 cm

(given) (c) MN = QR = 8 (given) NO = PR = 8 (given) MO = PQ = 5 ` by SSS, D MNO / D PQR

(KJI is a straight angle)

`+BDC = 46c (base angles of isosceles triangle) +CBD = 180 - 2 # 46 = 88c `+CBD = +BDE = 88c These are equal alternate angles.

(given)

(given) (b) +Z = +B = 90c (given) XY = AC = 7 (given) YZ = BC = 2 ` by RHS, D XYZ / D ABC

(HJL is a straight angle)

Since +IJL = +JIL = +ILJ = 60c, D IJL is equilateral +KJL = 180c - 60c = 120c +JLK = 180c - (30c + 120c) = 30c

+PQR = +SUT = 49c

(e) No

(c) x = 63c

(d) a = 29c , b = 70c

9.

(a) Yes

BC = DF = 6 cm

6.

8.

1.

3.

(a)

+B = +C (base angles of isosceles D) +BDA = +CDA = 90c (given) AD is common ` by AAS, D ABD / D ACD

ANSWERS

4.

(b) ` BD = DC (corresponding sides in congruent Ds) ` AD bisects BC

But +OBA + +OBC = 180c

+ABD = +BDC

OB is perpendicular to AC.


5.

(a)

OA = OC

(equal radii)

OB = OD

(similarly)

+AOB = +COD

So +OBA = +OBC = 90c

(alternate angles, AB < CD)

(alternate angles, AD < BC) +ADB = +DBC BD is common ` by AAS, D ABD / D CDB ` AD = BC

10. (a) AD = BC +ADC = +BCD = 90c DC is common `DADC / DBCD (b) AC = BD

`DAOB / DCOD

(SAS)

(b) AB = CD

(corresponding sides in congruent

(given)

BC = DC

(given)

1.

`DABC / DADC

(SSS)

(b) +ABC = +ADC

(corresponding angles in congruent

`DOAB / DOBC

(SAS)

(b) +OCB = +OBC

(base angles of OBC, an isosceles

4.

right angled triangle) (angle sum of triangle)

5.

So +OCB = +OBC = 45c Similarly +OBA = 45c ` +OBA + +OBC = 45c + 45c = 90c So +ABC is right angled (a) +AEF = +BDC = 90c

(given)

AF = BC

(given)

FE = CD

(given)

`DAFE / DBCD

(RHS)

(b) +AFE = +BCD

(corresponding angles in

(e) b = 4.5 (g) p = 9.7

(alternate angles, AB < ED) (similarly) (vertically opposite angles)

(given) +GFE = +EFD 1.5 GF o = = 0.5 EF 2.7 2.7 EF o = = 0.5 DF 4.86 GF EF ` = EF DF Since two pairs of sides are in proportion and their included angles are equal, then DDEF ||| DFGE

1.3 AB = = 0.714 DE 1.82 4.2 AC = = 0.714 DF 5.88 4.9 BC = = 0.714 EF 6.86 AC BC AB = = ` DE DF EF Since three pairs of sides are in proportion, D ABC ||| D DEF y = 41c

6.

congruent triangles)

(a) OA = OC

(c) m = 6.6

(f) a = 115c , x = 19c , y = 3.2

+BAC = +EDC +ABC = +DEC +ACB = +ECD

+AOB = +COB = 90c (given)

9.

(corresponding sides in congruent

(d) a = 76c , i = 23c , b = 81c

3.

OB is common

8.

(b) x = 4.4

a = 1.81, b = 5.83

(equal radii)

But +OCB + +OBC = 90c

(SAS)

` since 3 pairs of angles are equal, DABC ||| DCDE

triangles)

(a) OA = OC

(a) x = 15.1

2.

AC is common

7.

(given)

Exercises 4.5


(a) AB = AD

(given)

triangles)

triangles)

6.

(ABC is a straight angle)

(equal radii)

OB is common

(a) OA = OB OC = OD OA OB ` = OD OC +AOB = +COD

(equal radii) (similarly)


AB = BC

(given)

Since two pairs of sides are in proportion and their included angles are equal, 3 OAB ||| 3 OCD

`DOAB / DOBC

(SSS)

(b) AB = 5.21 cm

(b) +OBA = +OBC

(corresponding angles in congruent triangles)

7.

(a) +A is common +ABC = +ADE +ACB = +AED

(corresponding angles, BC < DE) (similarly)

769

770


6.

` since 3 pairs of angles are equal, D ABC ||| D ADE (b) x = 2.17, y = 2.25 8.

+ABF = +BEC +CBE = +BFA ` +C = +A

YZ 2 = XY 2 = 1, XZ 2 = 2 YZ 2 + XY 2 = 1 + 1 =2 = XZ 2 ` D XYZ is right angled

(alternate angles, AB z CD) (similarly, BC z AD) (angle sum of Ds)

` since 3 pairs of angles are equal, D ABF ||| DCEB 9.

+A is common 1.2 AD = = 0.4 AB 3 0.8 AE = = 0.4 AC 2 AD AE ` = AB AC Since two pairs of sides are in proportion and their included angles are equal, D AED ||| D ABC, m = 4.25

AB 10. CD BC AC AC AD AB ` CD

= = = =

XY = YZ = 1 ` D XYZ is isosceles

7.

AC 2 = AB 2 + BC 2 2 2 2 = ^ 3 h + BC 2 4 1 `1 AC

8.

2 = 0.769 2.6 3 = 0.769 3.9 3.9 = 0.769 5.07 BC AC = AD AC

= 3 + BC 2 = BC 2 = BC =2 =2#1 = 2BC

(a) AC = 5 (b) AC 2 = 25, CD 2 = 144, AD 2 = 169 AC 2 + CD 2 = 25 + 144 = 169 = AD 2 ` D ACD has a right angle at +ACD ` AC is perpendicular to DC

9.

D ABC ||| D ACD, x = 109c, y = 47c

11. d 2 = ] 20 - 3t g 2 + ] 15 - 2t g 2 = 400 - 120t + 9t 2 + 225 - 60t + 4t 2 = 13t 2 - 180t + 625

11. (a) x = 7.8

(b) m = 4.0, p = 7.2

(d) x = 6.2, y = 4.4 AB AD 12. (a) = DE BC AD AF Also = DE FG AB AF ` = BC FG (c)

(c) x = 6.5

(e) x = 1.4, y = 9.2

Also `

16. 4.3 m

14. y = 0.98

(a) x = 6.4

2.

(a) p =

3.

s = 6.2 m

5.

AB 2 = 81, CB 2 = 144, CA 2 = 225 AB 2 + CB 2 = 81 + 144 = 225 = CA 2 ` D ABC is right angled

61

(c) b = 5.7

(b) t =

(c) x =

58

15. 134.6 cm

17. 42.7 cm

20. (a) BC 2 = 6 2 - 4 2 = 20 BC = 20 AO = 6 cm (equal radii) So AC 2 = 6 2 - 4 2 = 20 AC = 20 Since BC = AC, OC bisects AB (b) +OCA = +OCB = 90c (given) OA = OB (equal radii) OC is common ` DOAC / DOBC (RHS) So AC = BC (corresponding sides in congruent triangles) ` OC bisects AB

(d) m = 6.6 65

14. 12.6 m

19. No. The diagonal of the boot is the longest available space and it is only 1.4 m.

15. x = 3.19, y = 1.64

(b) y = 6.6

13. 683 m

18. 1.3 + 1.1 2 = 2.9 and 1.5 2 = 2.25 1.3 2 + 1.1 2 ! 1.5 2 so the triangle is not right angled ` the property is not a rectangle

Exercises 4.6 1.

10.

2

BD AD = AE CE AD DF Also = AE EG BD DF ` = CE EG

13. a = 4.8, b = 6.9

3b

12. 1471 mm

AB AD = AE AC AD AF = AE AG AB AF = AC AG

(b)

AB =

x2 + y2 x

Since three pairs of sides are in proportion,

(d) y =

33

4. CE = 15.3 cm

Exercises 4.7 1.

(a) x = 94c (b) y = 104c (c) x = 111c (d) x = 60c (e) y = 72c (f) x = 102°, y = 51° (g) x = 43°, y = 47°

ANSWERS

2.

9.

D ABE is isosceles. ` +B = +E = 76c (base +s equal) +CBE = +DEB = 180c - 76c = 104c (straight +s) +D + 62c + 104c + 104c = 360c (angle sum of quadrilateral) +D + 270c = 360c +D = 90c ` CD is perpendicular to AD`

3.

(a)

+D = 180c - x (+A and +D cointerior angles, AB < DC)

+C = 180c - (180c - x)

(+C and +D cointerior angles, AD < BC)

= 180c - 180c + x =x `+A = +C = x +B = 180c - x (+B and +C cointerior angles, AB < DC) `+B = +D = 180c - x (b) Angle sum = x + x + 180c - x + 180c - x = 360c 4.

a = 150c , b = 74c

5.

(a) a = 5 m, b = 3 m, x = z = 108c, y = 72c (b) x = 53c, y = 56c, z = 71c (c) x = y = 5 cm, a = b = 68c (d) a = 121c, b = 52c, i = 77c (e) x = 60c (f) x = 3, y = 7

6.

+ADB = +CDB +CDB = +ABD +ADB = +DBC ` +ABD = +DBC ` BD bisects +ABC

7.

(a) AD = BC = 3.8 cm AB = DC = 5.3 cm

(BD bisects +ADC) (alternate angles, AB < DC) (alternate angles, AD < BC)

(given) (given)

Since two pairs of opposite sides are equal, ABCD is a parallelogram. (b) AB = DC = 7cm (given) AB < DC (given) Since one pair of opposite sides is both equal and parallel, ABCD is a parallelogram. (c) +X + +M = 54c + 126c = 180c These are supplementary cointerior angles. ` XY < MN Also, XM < YN

(given)

` XMNY is a parallelogram (d) AE = EC = 5 cm DE = EB = 6 cm

(given) (given)

Since the diagonals bisect each other, ABCD is a parallelogram. 8.

(a) x = 5 cm, i = 66c (b) a = 90c, b = 25c, c = 65c (c) x = 3 cm, y = 5 cm (d) x = 58c, y = 39c (e) x = 12 cm

6.4 cm

11. 4 2 cm

10. +ECB = 59c, +EDC = 31c, +ADE = 59c 12. x = y = 57c

Exercises 4.8 1.

(a) 540c (b) 720c (c) 1080c (d) 1440c (e) 1800c (f) 2880c 2. (a) 108c (b) 135c (c) 150c (d) 162c (e) 156c 3. (a) 60c (b) 36c (c) 45c (d) 24c

4.

128c34l 5. (a) 13

8.

2340c

(b) 152c18l 6. 16

7. 3240c

9. 168c23l

10. Sum = 145n = (n - 2) # 180c 145n = 180n - 360 = 35n 10.3 = n But n must be a positive integer. ` no polygon has interior angles of 145c. 11. (a) 9

(b) 12

(c) 8

(d) 10

(e) 30

12. (a) ABCDEF is a regular hexagon. AF = BC (equal sides) FE = CD (equal sides) +AFE = +BCD (equal interior angles) ` D AFE / D BCD (SAS)

S = ] n - 2 g # 180c = (6 - 2) # 180c = 720c 720c +AFE = 6 = 120c Since AF = FE, triangle AFE is isosceles. So +FEA = +FAE (base angles in isosceles triangle) 180 - 120c ` +FEA = (angle sum of triangle) 2 = 30c +AED = 120 - 30c = 90c Similarly, +BDE = 90c (b)

So +AED + +BDE = 180c These are supplementary cointerior angles ÀE < BD 13. A regular octagon has equal sides and angles. AH = AB (equal sides) GH = BC (equal sides) +AHG = +ABC (equal interior angles) ` D AHG / D ABC (SAS) So AG = AC (corresponding sides in congruent triangles)

S = ] n - 2 g # 180c = (8 - 2) # 180c = 1080c 1080c ` +AHG = 8 = 135c +HGA = +HAG

(base angles in isosceles triangle)

771

772


180 - 135c (angle sum of triangle) 2 = 22c30l +GAC = 135 - 2 # 22c30l = 90c We can similarly prove all interior angles are 90c and adjacent sides equal. So ACEG is a square. `+HAG =

2.

3.

118.28 cm2

4.

(common) (a) +DAE = +BAC (corresponding angles, DE < BC) +ADE = +ABC (similarly) +AED = +ACB ` D ABC and D ADE are similar (AAA)

] 5 - 2 g # 180c 5 = 108c

(b) x = 3.1 cm, y = 5.2 cm 162c

8.

(a) AB = AD BC = DC

(adjacent sides in kite) (similarly)

AC is common ` Δ ABC and Δ ADC are congruent (SSS) (b)

AO = CO BO = DO +AOB = +COD

(equal radii) (similarly) (vertically opposite angles)

` Δ AOB and Δ COD are congruent (SAS) 9.

11.

73.5 cm2

AF AD = AE AG AD AB = AE AC AF AB ` = AG AC

12. (a) AB = AC +B = +C BD = DC

Exercises 4.9 1.

(a) 26.35 m 2 (b) 21.855 cm 2 (c) 18.75 mm 2 (d) 45 m 2 (e) 57 cm 2 (f) 81 m 2 (g) 28.27 cm 2 2. 4.83 m 2

3.

(a) 42.88 cm 2 (b) 29.5 m 2 (c) 32.5 cm 2 (d) 14.32 m 2 (e) 100.53 cm 2 4. (a) 25 m 2 (b) 101.85 cm 2 (c) 29.4 m 2 (d) 10.39 cm 2 (e) 45 cm 2

5.

7 51 + 98 = 7 ^ 51 + 14 h cm 2

7.

$621.08

9.

(a) 48 cm (b) 27 cm 10. 12w units 2

6. 22.97 cm 2

(b) 89 m 2

(similarly)

(given) (base +s of isosceles D) (AD bisects BC, given)

(b) +ADB = +ADC (corresponding +s in congruent Ds) (straight +) But + ADB + +ADC = 180c So +ADB = +ADC = 90c So AD and BC are perpendicular. 13.

+ACB = 68c +CAD = 68c - 34c = 34c ` ˚+CAD = +ADC = 34c So Δ ACD is isosceles

14.

(a) x = 43c, y = 137c, z = 147c (b) x = 36c (c) a = 79c, b = 101c, c = 48c (d) x = 120c (e) r = 7.2 cm (f) x = 5.6 cm, y = 8.5 cm (g) i = 45c

(equal ratios on intercepts)

` D ABD / D ACD ] SAS g

(c) 10.5 m

Test yourself 4 1.

6. 1020.7 cm3 7. 36 m

5.

2 10. 6 2 + ^ 2 7 h = 36 + 28 = 64 = 8 2 ` ΔABC is right angled (Pythagoras)

360 p (b) Each interior angle: 360 180 p 180p 360 = p p 180p - 360 = p 180 ^ p - 2 h = p

15. (a)

8. (a) 161.665 m 2

(vertically opposite +HGB)

So +AGF = +CFE = i These are equal corresponding +s. ` AB < CD

14. +EDC =

ED = CD (equal sides in regular pentagon) So EDC is an isosceles triangle. `+DEC = +ECD (base angles in isosceles triangle) 180 - 108c +DEC = (angle sum of triangle) 2 = 36c +AEC = 108 - 36c = 72c Similarly, using triangle ABC, we can prove that +EAC = 72c So EAC is an isosceles triangle. (Alternatively you could prove EDC and ABC congruent triangles and then AC = EC are corresponding sides in congruent triangles.)

+AGF = i

(base +s of isosceles D) (exterior + of D)

^ base +s equal h

ANSWERS

+DAC = +ACB +BAC = +ACD

(alternate +s, AD < BC) (alternate +s, AB < DC)

7.

AC is common ` D ABC / D ADC (AAS) (corresponding sides in congruent Ds) ` AB = DC Similarly, AD = BC ` opposite sides are equal 15. (a) 24 cm2 (b) 5 cm

Let ABCD be a square with diagonals AC and BD and +D = 90c (adjacent sides of square) AD = DC ` D ADC is isosceles (base angles of isosceles D) `+DAC = +DCA (angle sum of D) +DAC + +DCA = 90° ` +DAC = +DCA = 45° Similarly, +BAC = +BCA = 45°

16. 9

17. +BFG + +FGD = 109c - 3x + 3x + 71c = 180c These are supplementary cointerior +s. ` AB < CD 18. 57 cm2

(other angles can be proved similarly)

19. +ACB = 180c - ] +A + +B g = 180c - x - y +ACD = 180c - +ACB z = 180c - (180c - x - y) = 180c - 180c + x + y =x+y 20. (a)

+A AC EF AB DE AC ` EF

= +E 2.97 = = 1.1 2.7 3.96 = = 1.1 3.6 AB = DE

(+sum of D) (straight +)

8.

^ given h

Let ABCD be a kite (given) AD = AB (given) DC = BC

So Δ ABC and Δ DEF are similar (two sides in proportion, included +s equal). (b) x = 4.3 cm

AC is common ` by SSS, D ADC / D ABC ` +DAC = +BAC


94c

2. x = 75c , y = 46c , z = 29c

4.

+BAD = +DBC +ABD = +BDC ` +ADB = +DCB

(corresponding angles in congruent Ds) (given) (found)

3. 1620c, 32c 44l

AD = AB +DAE = +BAE

(given) (alternate angles, AB < DC) (angle sum of D)

AE is common ` by SAS, D ADE / D ABE ` +DEA = +BEA (corresponding angles in congruent Ds) But +DEA + +BEA = 180c (DEB is a straight angle) ` +DEA = +BEA = 90c ` the diagonals are perpendicular

` since 3 pairs of angles are equal, D ABD <; D BCD d = 6.74 cm 5.

AB = DC (given) +A + +D = 131c + 49c = 180c +A and +D are supplementary cointerior angles ` AB < DC Since one pair of opposite sides are both parallel and equal, ABCD is a parallelogram.

6.

27.36 m 2

9.

(exterior angle of D MNZ) +MNY + 84c = (15c + 112c) ` +MNY = 43c (exterior angle of D XYZ) +XYZ + 69c = 112c ` +XYZ = 43c ` +MNY = +XYZ = 43c These are equal corresponding angles. ` MN < XY

10. x = 2.12 m

11. (a) 6 m 2

12. x = 28.7 cm, y = 3.8 cm

(b) 10 + 2 5 = 2 ^ 5 + 5 h m 13. x = 7.40 m, y = 4.19 m

773

774


14. (a)

(adjacent sides in square) AB = BC +ABE = +CBE = 45 (diagonals in square make 45c with sides)

34. x = 22c, y = 29c, w = z = 90c 35. 56.7 cm 2 36. a - 21 b 10 =

EB is common.

b 10 a 21

37. x 2 6, x 1 -2

2 5

` by SAS, D ABE / DCBE ` AE = CE (corresponding sides in congruent Ds)

39. - x - 7

Since AB = BC and AE = CE, ABCE is a kite.

43. Given diagonal AC in rhombus ABCD:

40. x =

1 4

1 8

38.

41. x # -3, x $ 3

42.

1 6

(adjacent sides in rhombus) AB = BC (alternate +s, AD < BC) +DAC = +ACB (base +s of isosceles D ABC) +BAC = +ACB ` +DAC = +BAC ` diagonal AC bisects the angle it meets. Similarly, diagonal BD bisects the angle it meets.

44. ] x + 3 g-1 45. x 3 + 6x 2 + 12x + 8 46. (b) BD =

x2 + x2

= =

2x 2 2x

48. x = 98c, y = 41c

1 BD 2 2x = units 2

DE =

51. (a) 12x - 8y

2. 2 ^ 5 + y h ^ x - y h

p =9

25 + 5 2 5. 23

(b) x 3

7.

x=

9.

(given) +ABC = +EDC = 90° (vertically opposite angles) +ACB = +ECD (given) AB = ED ` by AAS D ABC / DEDC (corresponding sides in congruent triangles) ` AC = EC ` D ACE is isosceles

10. 231.3

14. 3 10 - 4

6. x 3 + 2x 2 - 16x + 3

2 x-3

11. - 3

12. 135c

13. 7.33 # 10 - 2

15. 3.04 16. x + 3

17. x = 1.78, -0.281

52. x = 2.7, y = 3.1

21. x =

4 ! 12 =2! 3 2

28.

26. 7.02 cm

6 15 + 2 6 43

22.

1 49

24. x = 2, y = -1

27. ] 2x - 1 g ^ 4x 2 + 2x + 1 h 29. 7

(d) 3 2 + 1

53. x = 25

54. r =

2 3

r

cm

57.

2 5

58. 5%

61. 9xy y

59. 2.2 # 10 8 kmh -1

60. k = 20

62. 147c 16l 63. 5.57 m 2

64. (a) 5 ] a + 2b - 2 g ^ a 2 - 4a - 2ab + 4b 2 + 4b + 4 h (b) ] 3a + 4b g ] a - 6b + 2c g

23. x = 4, y = 11 or x = -1, y = - 4 25. 7

x-3 x 2 - 3x + 9

56. Let +DEA = x (base +s of isosceles D) Then +EAD = x +CDA = x + x (exterior + of DEAD) = 2x ` +ABC = 2x (opposite +s of < gram are equal) ` +ABC = 2+DEA

65. - 1 7 15

3x + 2

55. 17.3 cm

18. r = 1.55 19. x 2 1

20.

1

50.

4

3. (a) x - 1

6y - 10

8.

(c)

47. x = 53c

y7 - ]x + 5g 11 3 (f) (g) x - 14 y 7 z -11 = 14 11 ]x + 1g]x - 1g 6 x z 3 1 (h) (i) 8 5 (j) 13 2 5a ] a + b g ] 1 + 2b g

4.

2 7

(b) 2 31

2 17

(e)

Practice assessment task set 1 1.

49. 0 1 x 1 5

54

30. $643.08 31. 1.1

32. -2 10 + 3 5 - 2 2 + 3

66.

3 1 #x 15 4 8

BC < AD (ABCD is a < gram) BC < FE (BCEF is a < gram) ` AD < FE Also BC = AD ^ opposite sides of < gram h BC = FE ^ similarly h ` AD = FE Since AD and FE are both parallel and equal, AFED is a parallelogram.

67. b = 11.95 m 33. $83.57

68. (a)

34 cm

(b) 30 cm2

ANSWERS

69.

18 3 + 31 2 - 25 5 75

70. 20 71. 32 m

(g) x-intercepts 3, 5, y-intercept 15

72. BD bisects AC So AD = DC +BDC = +BDA = 90c (given) BD is common ` DBAD / DBCD (SAS) ` AB = CB (corresponding sides in congruent

(h) x-intercept - 3 5 , y-intercept 5 (i) x-intercept -3, no y-intercept (j) x-intercept !3, y-intercept 9 2.

triangles)

3.

So triangle ABC is isosceles 73.

x2 + y2 2

79. (d)

74. (b)

75. (c)

76. (a) 77. (b)

78. (b)

4.

Chapter 5: Functions and graphs 5.

Exercises 5.1 Yes

2. No

3. No

8.

Yes

9. Yes

10. No

4. Yes

5. Yes

11. Yes

6. Yes

12. No

7. No

6.

13. Yes

14. No 15. Yes

7.

Exercises 5.2 1.

f ] 1 g = 4, f ] -3 g = 0

3.

f ] 5 g = -25, f ] -1 g = -1, f ] 3 g = -9, f ] -2 g = -4

5.

-35

6. x = 9

2. h ] 0 g = -2, h ] 2 g = 2, h ] -4 g = 14

7. x = !5

8. x = -3

4. 14

9. z = 1, -4

8.

10. f ^ p h = 2p - 9, f ] x + h g = 2x + 2h - 9 11. g ] x - 1 g = x 2 + 2

12. f ] k g = ] k - 1 g ^ k 2 + k + 1 h

13. t = -1; t = 2, -4

14. 0

9.

15. f ] 5 g = 125, f ] 1 g = 1, f ] -1 g = -1

18. 7

25. (a) 2 (b) 0

2 , y-intercept -2 3 (b) x-intercept -10, y-intercept 4 (c) x-intercept 12, y-intercept 4 (d) x-intercepts 0, -3, y-intercept 0 (e) x-intercepts !2, y-intercept -4 (f) x-intercepts -2, -3, y-intercept 6 (a) x-intercept

f ] - x g = - x = -f ] x g ` odd function f ] - x g = ] - x g2 - 1 = x2 - 1 = f (x) ` even function f ] -x g = 4 ] -x g - ] -x g 3 = - 4x + x 3 = - ^ 4x - x 3 h = - f ]xg ` odd function f ] -x g = ] -x g 4 + ] -x g 2 = x4 + x2 = f ]xg ` even function f ]xg - f ]- xg = 0 (a) Odd (b) Neither (c) Even (d) Neither (e) Neither

(c) n 4 + n 2 + 2

(b) Yes, when n is odd (1, 3, 5, …)

12. (a) (i) x 2 0

(ii) x 1 0

(iii) Even

(b) (i) x 1 2

(ii) x 2 2

(iii) Neither

(c) (i) -2 1 x 1 2 (d) (i) All real x ! 0

22. 4x + 2h + 1

Exercises 5.3 1.

g ] - x g = ] - x g + 3 ] - x g4 - 2 ] - x g 2 = x 8 + 3x 4 - 2 x 2 = g (x) ` even function

11. (a) No value of n

20. (a) 3 (b) x - 3 = 3 - 3 = 0 Denominator cannot be 0 so the function doesn’t exist for x = 3. (c) 4

23. 5] x - c g 24. 3k 2 + 5

(d) Neither odd nor even

8

(b) Odd values i.e. n = 1, 3, 5, f

19. -28

21. f ] x + h g - f ] x g = 2xh + h 2 - 5h

(b) 7 f ] x g A 2 = x 6 + 2x 3 + 1

10. (a) Even values i.e. n = 2, 4, 6, f

16. f ] 2 g - f ] -2 g + f ] -1 g = 0 - 4 + 1 = -3 17. 10

(a) f ^ x 2 h = x 6 + 1

(c) f ] - x g = - x 3 + 1

80. (d)

1.

f ] -x g = ] -x g 2- 2 = x2 - 2 = f (x) ` even function

(e) (i) None

(ii) x 1 -2, x 2 2 (ii) None

(ii) All real x

(iii) Odd

(iii) Neither

Exercises 5.4 1.

(iii) Neither

(a) x-intercept 2, y-intercept -2 1 (b) x-intercept -1 , y-intercept 3 2 1 (c) x-intercept , y-intercept 1 2 (d) x-intercept -3, y-intercept 3 2 1 (e) x-intercept , y-intercept 3 3

775

776


2.

(a)

y

(e) 5

5 4 3 2 1

4 3 2 1 1

-4 -3 -2 -1 -1

2

3 4

-2 -3 -4 -5

2 3 4

x

y

(f)

y

(b)

1 21

-4 -3 -2 -1 -1 -2 -3 -4 -5

5 4 3 2 1

5 4 3 2 1 -4 -3 -2 -1 -1

1

2 3

-4 -3 -2 -1 -1 -2 -3 -4 -5

x

4

-2 -3

x

1 2 3 4

-4 -5

5 4 3 2 2 - 1

y

(c) 5 4 3

3

2

-4 -3 -2 -1 -1 -2 -3 -4 -5

1 -4 -3 -2 -1 -1

1

2

3

x

4

-2 -3 -4

x

5 4 3 2 1

y 5 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 -5

1 2 3 4

y

(h)

-5 (d)

y

(g)

1 2 3 4

x

-4 -3 -2 -1 -1 -2 -3 -4 -5

1 2 3 4

x

ANSWERS

Exercises 5.5

y

(i) 5

1.

(a) x-intercepts 0, -2, y-intercept 0 (b) x-intercepts 0, 3, y-intercept 0 (c) x-intercepts !1, y-intercept -1 (d) x-intercepts -1, 2, y-intercept -2 (e) x-intercepts 1, 8, y-intercept 8

2.

(a)

4 3 2 1 1

-4 -3 -2 -1 -1

2

3

4

x

6 5 4 3 2 1

-2 -3 -4 -5 y

(j)

y

-4 -3 -2 -1 -1 -2 -3 -4 -5

5 4 3 2 1 111 2

-4 -3 -2 -1 -1

3.

4.

2 3

4

6

-2

5

-3

4

-4

3

-5

2 1

(a) " all real x ,, " all real y , (b) " all real x ,, " y: y = 2 , (c) ! x: x = -4 +, " all real y , (d) ! x: x = 2 +, " all real y , (e) ! all real x +, " y: y = 3 , (a) Neither

(b) Even

5.

(c) Neither

(d) Odd

-4 -3 -2 -1 -1

2 1

-4 -5

4

x

5

-3

y

(c)

3

-3

3

-4

4

-2

2

-5

5

-4 -3 -2 -1 -1

1

-2

(e) Odd

y

(3, -1)

y

(b)

x

x

1 2 3 4 5

111 2 2

3

4

x

6 5 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 -5

1 2 3 4

5

x

777

778


y

(d) 6 5 4 3 2 1

5 4 3 2 1

1 2 3 4

-4 -3 -2 -1 -1

5

-4 -3 -2 -1 -1 -2 -3 -4 -5 -6

x

-2 -3 -4 -5 y

(e)

1 2 3 4

x

5

y

(i) 5 4 3 2 1

6 5 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 -5

1 2 3 4

5

-4 -3 -2 -1 -1 -2

x

1112 2 3 4

5

1 2 3 4

5

x

-3 -4 -5 -6 y

(f)

y

(h)

y

(j)

12 10 8 6 4 2

5 4 3 2 1

1 2

-4 -3 -2 -1 -2 -4 -6 -8

3 4

5

-4 -3 -2 -1 -1 -2 -3 -4

x

x

-5 -6

-10 3.

5 4 3

(a) (i) x-intercepts 3, 4, y-intercept 12 (ii) {all real x}, 1 ( y: y $ - 2 4 (b) (i) x-intercepts 0, -4, y-intercept 0 (ii) {all real x}, " y: y $ -4 ,

2 1

(c) (i) x-intercepts -2, 4, y-intercept -8 " y: y $ - 9 ,

y

(g)

-4 -3 -2 -1 -1 -2 -3 -4 -5 -6

1 2 3 4

5

(ii) {all real x},

(d) (i) x-intercept 3, y-intercept 9 (ii) {all real x}, " y: y $ 0 ,

x

(e) (i) x-intercepts ! 2, y-intercept 4 " y: y # 4 , 4.

(a) {all real x}, " y: y $ -5 ,

(ii) {all real x},

(b) {all real x}, " y: y $ - 9 ,

ANSWERS

1 (c) {all real x}, ( y: y $ -2 2 4

(d) {all real x}, " y: y # 0 ,

(e) {all real x}, " y: y $ 0 , 5.

(a) 0 # y # 9

6.

5 4

(b) 0 # y # 4

(c) -1 # y # 24 1 (e) -18 # y # 2 4

(d) -4 # y # 21

(a) (i) x 2 0 (ii) x 1 0

3 2 1

(b) (i) x 1 0 (ii) x 2 0

(c) (i) x 2 0 (ii) x 1 0 (d) (i) x 1 2 (ii) x 2 2 (e) (i) x 2 -5 (ii) x 1 -5 7.

8.

y

(c)

-4 -3 -2 -1 -1

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

-2

f ] -x g = - ] -x g 2 = -x2 = f (x) ` even

-3 -4 -5

(a) Even (b) Even (c) Even (d) Neither (e) Neither (f) Even (g) Neither (h) Neither (i) Neither (j) Neither

y

(d) 5 4 3

Exercises 5.6 1.

2.

2

(a) x-intercept 0, y-intercept 0 (b) No x-intercepts, y-intercept 7 (c) x-intercepts ! 2, y-intercept -2 (d) x-intercept 0, y-intercept 0 (e) x-intercepts ! 3, y-intercept 3 (f) x-intercept -6, y-intercept 6 2 (g) x-intercept , y-intercept 2 3 4 (h) x-intercept - , y-intercept 4 5 1 (i) x-intercept , y-intercept 1 7 (j) No x-intercepts, y-intercept 9

1

-4 -3 -2 -1 -1 -2 -3 -4 -5

5 4 3

5

2

4

1

3

-4 -3 -2 -1 -1 -2

2 1 -4 -3 -2 -1 -1

1

2

3

4

5

-4 -5

-3

(f)

-4

y 5

-5

4

y

3 2

5 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 -5

x

-3

x

-2

(b)

y

(e)

y

(a)

x

1 -4 -3 -2 -1 -1

1 2 3 4 5

x

-2 -3 -4 -5

x

779

780


y

(g)

3.

5 4 3 2 1 1

-4 -3 -2 -1 -1 -2

2

3

4

4.

(a) (i) x 2 2 (ii) x 1 2 (b) (i) x 2 0 (ii) x 1 0 1 1 (d) (i) x 2 0 (ii) x 1 0 (c) (i) x 2 1 (ii) x 1 1 2 2 (e) (i) x 1 0 (ii) x 2 0

5.

(a) 0 # y # 2

x

5

-3

(a) {all real x}, " y: y $ 0 , (b) {all real x}, " y: y $ -8 , (c) {all real x}, " y: y $ 0 , (d) {all real x}, " y: y $ -3 , (e) {all real x}, " y: y # 0 ,

(b) - 8 # y # -4

(d) 0 # y # 11

-4 -5

6.

(a) x 2 -3 (e) x 1 -2

(b) x 1 0

7.

(a) x = !3

(b) x 2 1, x 1 -1

y

(h) 5

(c) 0 # y # 6

(e) -1 # y # 0 (c) x 2 9

(d) x 2 2

(c) -2 # x # 2

(d) x = -1, -3 (e) x = 3 (f) x = 1, 2 (g) -3 1 x 1 5

4

(h) - 4 # x # 2

(j) x # 2, x $ 4 1 (k) - 4 # x # 1 (l) x # 0, x $ 1 (m) x = 2, 2 (n) No solutions (o) x = 0 (p) x = 1 (q) x = 0, -2 1 (t) x = 0, 6 (r) No solutions (s) x = 3

3 2 1 1

-4 -3 -2 -1 -1

2

3

4

x

5

-2 -3

Exercises 5.7

-4

1.

-5

(i) x 2 4, x 1 0

(a) (i) {all real x: x ! 0}, {all real y: y ! 0} (ii) no y-intercept y

(iii) y

(i)

5

5

4

4

3

3

2

2

1

1 1

-4 -3 -2 -1 -1

2

3

4

x

5

-2

2

3

x

4

-2

-3 -4

-3

-5

-4 -5

y

(j)

1

-4 -3 -2 -1 -1

(b) (i) {all real x: x ! 0}, {all real y: y ! 0} (ii) no y-intercept

5 4

y

(iii)

3 2

2

1 -4 -3 -2 -1 -1

1

2

3

4

5

x

1

-2 -3 -4 -5

-2

1

-1 -1

-2

2

x

ANSWERS

(c) (i) {all real x: x ! -1}, {all real y: y ! 0} (ii) 1 y

(iii)

y

(iii) 5 4

2

3 2

1

1

-2

-1

1

x

2

-4 -3 -2 -1 -1

1

2 3

4

5

x

-2

-1

-3 -4

-2

-5

1 (d) (i) {all real x: x ! 2}, {all real y: y ! 0} (ii) -1 2

(g) (i) {all real x: x ! 1}, {all real y: y ! 0} (ii) -4

y

(iii)

y

(iii)

5 4

5

3

4

2

3

1

2

-4 -3 -2 -1 -1

1 2

3

4

1

x

5

-2

-4 -3 -2 -1 -1

-3

-2

-4

-3

-5

-4

(e) (i) {all real x: x ! -2}, {all real y: y ! 0} (ii)

2

3

4

5

x

-5

1 6

(h) (i) {all real x: x ! -1}, {all real y: y ! 0} (ii) -2

y

(iii)

1

y

(iii) 2

5 4

1

3

-2

1

-1

2

-1

2 1

-4 -3 -2 -1 -1

-2 (f) (i) {all real x: x ! 3}, {all real y: y ! 0} (ii)

x

-2 2 3

-3 -4 -5

1

2

3

4

5

x

781

782


(i) (i) ' all real x: x !

1 2 1, {all real y: y ! 0} (ii) 2 3

Exercises 5.8 1.

y

(iii)

y

(a) (i)

2

3

1

1 2

-1

-2

-

-1

1

2

x

3

-3

x

2 3

-3 -2 (ii) ! x: -3 # x # 3 +, " y: -3 # y # 3 , (j) (i) {all real x: x ! -2}, {all real y: y ! 0} (ii) -3

y

(b) (i) y

(iii)

4

5 4 3 2 1

-4 -3 -2 -1 -1

1

2

3

4

5

x

4

-4 x

-2 -4

-3 -4 -5

2.

3.

4.

(ii) ! x: -4 # x # 4 +, " y: -4 # y # 4 ,

f ] -x g =

2 -x 2 =x = - f (x) ` odd function

y

(c) (i) 5 4 3

(a)

1 #y #1 9

(b)

1 #y #1 3

(d)

3 #y #3 7

(e) - 2 # y # -

(a) 1 # x # 3

(b) 1 # x # 4

(d) 1 # x # 4

(e) 1 # x # 2

(c) -2

1 1 #y #2 2

1 8

(c) - 6 # x # 0

2

(2, 1)

1 -4 -3 -2 -1 -1 -2 -3 -4 -5

1

2

3

4

x

ANSWERS

(ii) ! x: 0 # x # 4 +, " y: -1 # y # 3 ,

(iii) ! x: -5 # x # 5 +, " y: -5 # y # 0 ,

y

(d) (i)

(b) (i) Above x-axis y

(ii) 5 4 3 1

2 1 -4 -3 -2 -1 -1

1

2

3

x

4

-2

x

1

-1

-3 (iii) ! x: -1 # x # 1 +, " y: 0 # y # 1 ,

-4 -5

(c) (i) Above x-axis

(ii) ! x: -4 # x # 2 +, " y: -3 # y # 3 ,

y

(ii)

y

(e) (i)

6

5 4 3 2 (-2, 1)

6

-6

x

1 1

-4 -3 -2 -1 -1

2

3

x

4

(iii) ! x: -6 # x # 6 +, " y: 0 # y # 6 , (d) (i) Below x-axis

-2

y

(ii)

(ii) ! x: -3 # x # -1 +, " y : 0 # y # 2 , 2.

(a) (i) Below x-axis (ii)

y

8

-8

-8

5

-5

-5

x

(iii) ! x: -8 # x # 8 +, " y: -8 # y # 0 ,

x

783

784


6.

(e) (i) Below x-axis y

(ii)

(a) {y: - 9 # y # 3} (b) {y: 0 # y # 9} (c) {y: -8 # y # 1} 1 (d) ' y: # y # 1 1 (e) {y: 0 # y # 4} 5 (f) {y: -1 # y # 15} (g) {y: -1 # y # 0} (h) " y: - 1 # y # 8 , (i) {y: - 4 # y # 21} 1 (j) ' y: - 6 # y # 6 1 4

7

- 7

x

7.

(a) {all real x: x ! -1} (b) x-intercept: y = 0 3 0= x+1 0=3 This is impossible so there is no x-intercept (c) {all real y: y ! 0}

8.

(a) {all real x: x ! 0}

(b) {all real y: y ! !1}

9.

(a)

y

- 7 (iii) " x: - 7 # x # 3.

7 , , # y: - 7 # y # 0 -

(a) Radius 10, centre (0, 0) (b) Radius

5 , centre (0, 0)

25

(c) Radius 4, centre (4, 5) (d) Radius 7, centre (5, -6) (e) Radius 9, centre (0, 3) 4.

20 15 10

(a) x 2 + y 2 = 16 (b) x - 6x + y - 4y - 12 = 0 (c) x 2 + 2x + y 2 - 10y + 17 = 0 (d) x 2 - 4x + y 2 - 6y - 23 = 0 (e) x 2 + 8x + y 2 - 4y - 5 = 0 (f) x 2 + y 2 + 4y + 3 = 0 (g) x 2 - 8x + y 2 - 4y - 29 = 0 (h) x 2 + 6x + y 2 + 8y - 56 = 0 (i) x 2 + 4x + y 2 - 1 = 0 (j) x 2 + 8x + y 2 + 14y + 62 = 0 2

2

5

-4 -3 -2 -1 -5

5.

(a) {all real x}, {all real y} (b) {all real x}, {y: y = -4} (c) {x: x = 3}, {all real y} (d) {all real x}, {y: y $ -1} 1 (e) {all real x}, {all real y} (f) {all real x}, ' y: y # 12 1 4 (g) {x: -8 # x # 8}, {y: -8 # y # 8} (h) {all real t: t ! 4}, {all real f (t): f (t) ! 0} (i) {all real z: z ! 0}, {all real g ^ z h: g ^ z h ! 5} (j) {all real x}, {y: y $ 0}

(a) {x: x # - 1, x $ 2}

(b) {t: t # - 6, t $ 0}

1

2

3

4

x

5

8

1.

(a) x = 0, 5 (b) x = -3, 1, 2 (c) x = 0, 2, 4 (d) x = 0, ! 4 (e) x = !7 4. (a) -1 # x # 1 (b) {x: -1 # x # 1}

4

y

(b)

6

3.

3

-15

4

(a) {x: x $ 0}, {y: y $ 0} (b) {x: x $ 2}, {y: y $ 0} (c) {all real x}, {y: y $ 0} (d) {all real x}, {y: y $ -2} 1 (e) ' x: x $ -2 1, {y: y # 0} 2 (f) {all real x}, {y: y # 5} (g) {all real x}, {y: y 2 0} (h) {all real x}, {y: y 1 0} (i) {all real x: x ! 0}, {all real y: y ! 1} (j) {all real x: x ! 0}, {all real y: y ! 2}

2

-10

Exercises 5.9

2.

1

2

-4 -3 -2 -1 -2

x

-4 -6 -8 y

(c) 25 20 15 10 5

-4 -3 -2 -1 -5 -10 -15

1

2

3

4

5

x

ANSWERS

y

(d)

(g)

y

8 6

3

4

2

2 1

-4 -3 -2 -1 -2

2

3

4

1

x

-4

x

1

-1 -1

-6 -8

10. (a) " x: x $ 1 , " y: y $ 0 , y

(b) y

(e)

2 1

8 6

-1

4

1

2

2 1

-4 -3 -2 -1 -2

2

3

4

x

x

3

y

11. 6

-4

5

-6

4

-8

3 2 1

-1

y

(f)

1

x

12. (a) (i) {all real x}, {all real y} (ii) All x (iii) None (b) (i) {all real x}, " y: y 2 -2 , (ii) x 2 0 (iii) x 1 0 (c) (i) {all real x: x ! 0}, {all real y: y ! 0} (ii) None (iii) All x ! 0 (d) (i) {all real x}, {all real y} (ii) All x (iii) None (e) (i) {all real x}, " y: y 2 0 , (ii) All x (iii) None 13. (a) - 2 # x # 2 (b) (i) {x: - 2 # x # 2}, { y: 0 # y # 2} (ii) {x: - 2 # x # 2}, { y: - 2 # y # 0}

10

-10

-1

10

x

Exercises 5.10 1.

-10

(b) -10 (c) 8 (d) 3 (e) 3 (f) 75 1 (j) 1 (k) - 7 (l) x 2 - 3x (h) - 6 (i) 4 (a) 21

(m) 2x 3 + 3x - 5 2.

(g) 0

(n) 3c 2

(a) Continuous (b) Discontinuous at x = - 1 (c) Continuous (d) Continuous (e) Discontinuous at x = !2

785

786


3.

5.

(a)

(a)

(b)

(b)

(c)

(c)

(d)

Exercises 5.11 1.

(a) 0 (b) 0 (c) 0 (d) 2 (e) 1 (h) 0 (i) 5x (j) 3

2.

(a) RHS = 1 + =

(f) 6 (g)

3 1 + 2 x x x2 + x + 3

x2 = LHS (b) 1 from above (c) 1 from below 3.

(a) 2 from below (b) 2 from above

4.

(a)

x 3

(b)

5x 2 4

2 3

(e)

ANSWERS

(f)

7.

21x #2

9.

x#

2 5

2 ,x 21 3

8. x 1 - 6, x 2 - 3 10. - 2

2 # x 1 -2 3

Exercises 5.13 1.

y

(a) 6

(g)

5 4 3 2 1 1

-4 -3 -2 -1 -1

2

3

4

x

-2

(h)

-3 -4

y

(b) 6 5

(i)

4 3 2 1

-4 -3 -2 -1 -1

1

2

3

4

1

2

3

4

x

-2 (j)

-3 -4 y

(c) 6 5 4 3 2 1

Exercises 5.12 1.

-

1 1x 10 2

2. 0 1 x 1

1 3

4.

-

1 #x 10 2

5. 1 1 x 1 1

3. 0 1 x # 1 1 3

6. x $ -1, x 1 - 2

-4 -3 -2 -1 -1 -2 -3 -4

x

787

788


y

(d)

y

(g)

6

6

5

5

4

4

3

x+y = 1

2

3 2 1

1

-4 -3 -2 -1 -1

2

1

3

x

4

-4 -3 -2 -1 -1

-2

-2

-3

-3

-4

-4 y

(e)

4

3

3

2

2

1

1 1

2 3

x

4

-4 -3 -2 -1 -1

-2

-2

-3

-3

-4

-4

1

2

3

4

x

x

3x - y - 6 = 0

-5

y

(f)

4

5

4

-4 -3 -2 -1 -1

3

6

y = x +1

5

2

y

(h)

6

1

-6

6

y

(i)

5

y = 2x -3

4

6

3

5

2

4

1

-4 -3 -2 -1 -1 -2 -3

1

2

3

4

3

x x + 2y - 2 = 0

2 1

-4 -3 -2 -1 -1 -2 -3 -4 -5 -6

1

2

3

4

x

ANSWERS

y

(j)

y

(c)

6 1

5 4 3 2 1

-4 -3 -2 -1 -1

1

2

3

x

1

-1 x

4

-2

-1

-3 -4 -5

x=

-6 2.

(a) x 2 -3 (e) y $ 2

3.

(b) y $ -2

1 2

y

(d) 5

(d) y 2 x 2 - 4

(c) y $ x + 1

y=x2

4

x

3

y

(a)

2 1

5 4

-4 -3 -2 -1

y = x2 - 1

3

-1

2

3

4

5

3

4

x

-2

2

-3

1

-4 -3 -2 -1 -1

1

1

2

3

4

5

-4

x

-5

-2

y

(e)

-3 -4

8

-5

6 4

y

(b)

2

-4 -3 -2 -1 -2

3

-4 y = x3

-3

3

-3

x 4.

(a) y 1 3x - 2 (b) y 2 x 2 + 2 (c) x 2 + y 2 1 49 (d) x 2 + y 2 2 81 (e) x 1 5, y 2 2

-6 -8

1

2

x

789

790


5.

y

(a)

y

(b)

5

6

4

5

3

4

2

3 2

1 1

-4 -3 -2 -1 -1

2

3

4

x

y=x -3

1

-4

-2

1

-3 -2 -1 -1

2

3

x

4

-2 -3

y

(b)

-4

5

-5

4

-6

3 2

y

(c)

1

-4 -3 -2 -1 -1

1

2

3

4

6

x

5

y = 3x – 5

4

-2

3 2

y

(c)

1

5 4

-4 -3

3

1

-1 -1

2

3

4

x

-2

2

-3

1

-5 -4 -3 -2 -1 -1

1

2

3

4

-4

x

-5 -6

-2 6.

-2

y

(d)

y

(a)

6

6

5

5 4

3

3

2

2

1

1

-4 -3 -2 -1 -1 -2 -3 -4

y=x+1

4

1

2

3

4

x

-4

-3

-2 -1 -1 -2 -3 -4 -5 -6

1

2

3

4

y=3–x

x

ANSWERS

y

(e)

y

(h) x = -2

3

8

y = x3

6

y=3

4 2

y=1

-3

-4 -3 -2 -1 -2

x

3

1

2

3

x

4

-4 -6 -8

-3

y

(f)

y

(i)

2

1

1 1

-2

-1

x

2

-2

x

1

-1

-1

x=–1 y

(g)

y

(j) y=4

5

6

4

5

3

4

2

y = x2

3

1

-4 -3 -2 -1 -1

x - y = -1

2 1

2

3

4

5

x

x-y=2

1

-2

-4 -3 -2 -1 -1

-3

-2

-4

-3

-5

-4 -5 -6

1

2

3

4

x

791

792


7.

y

(a)

y

(d)

y = x2 5 4

2

3 2

1

1

-4 -3 -2 -1 -1

1

2

3

4

x

5

1

-1

2

3

x

4

2 y= x

-2

-2

-3 -4 -5 (e)

y

y

(b) 8

y = x3

2

6 4

y=1

1

2 1

-4 -3 -2 -1 -2

2

3

x

4

-4 y=

-4

-3 -2

-1

1

2

3

x

4

-1

1 x+2

-2

-6 -8 8. y

(c)

(a)

y y = x2

y=5

5

2

4 3

2

1

-2

2

1

x -4 -3 -2 -1

-1

1

2

3

-2 -3

-2

-4 x=1

-5

x=2

4

5

x

ANSWERS

(e)

y

(b)

y

6

6

5

y = |x|

5

4

4

3

3

2

2

1

1

y=3

x -4 - 3 - 2 - 1 -1 -2

1

2

3

x -4 -3 -2 -1 -1

4

y = -1

-3

2

-3

-5

3

4

x=2

Test yourself 5

-6

y

(c)

1

-2

x=3

-4 y=x-2

793

1.

(a) f ] - 2 g = 6

2.

(a)

(b) f ] a g = a 2 - 3a - 4

(c) x = 4, -1

y = 2x + 1

6 5 4 3 2

2x - 3y = 6

1

(b)

x

-4 -3 -2 -1 -1 -2

1

2

3

4

-3 -4 -5 -6 (c) y

(d)

3

y=2

(d)

-3

3

x

(e) x = -3

Answer S1-S5.indd 793

-3

8/11/09 11:31:52 AM

794


8.

(f)

9. (g)

(h) 10.

3.

4.

1 4 (b) Domain: all real x; range: all real y (c) Domain: - 1 # x # 1; range: - 1 # y # 1 (d) Domain: - 1 # x # 1; range: 0 # y # 1 (e) Domain: - 1 # x # 1; range: - 1 # y # 0 (f) Domain: all real x ! 0; range: all real y ! 0 (g) Domain: all real x; range: all real y (h) Domain: all real x; range: y $ 0 (a) Domain: all real x; range: y $ - 6

15

5. (a) 4 (b) 5

(c) 9

(d) 3

11. (a) y # 3

(b) y 2 x + 2

(c) y $ - x 2, y # 0

12. (a) Domain: all real x ! 3, range: all real y ! 0 (b)

(e) 2

6.

13. (a)

7.

(b) (i) x = 2, -4 (ii) - 4 1 x 1 2 (iii) x 2 2, x 1 - 4

ANSWERS

14. (a) 2

(b) x = 3

2 3

1 3

(c) 1

y

(b)

15. (a) x-intercept -10, y-intercept 4 (b) x-intercepts - 2, 7, y-intercept y -14 16. (a) i (b) iii (c) ii 17. (a) 4

(b)

2 5

(c) - 1

(d) i 1 2

(e) iii

(d) 3 -1

18.

x

1

y

(c)

19. (a) Domain: x $ 2, range: y $ 0

2

(b)

y

(d) f (x) = x 4 3x 2 1 f ( x) = ] - x g4 3 ] x g2 = x 4 3x 3x 2 1 = f (x) So f ] x g is even.

x

4

-4

20. (a)

1 1 x

f (x) = x 3 x f ( x) = ] - x g3 ( ) = - x3 + x = -( 3 ) = - f (x) So f ] x g is odd. (b)

y

(e)

21. (a a)

y

2

1

1 x

x -4 -3 -2 -1 -1 4

-2

1

2

3

4

5

795

796



6.

f ] 3 g = 9, f ] -4 g = 16, f ] 0 g = 1

7.

Domain: all real x ! ! 1; range: y # - 1, y 2 0

2 b=- ,3 3

2.

y

3.

8.

-2

2

x

9.

Domain: x $ 0; range: y $ 0

10. x = 0, 3, - 2

11.

4.

12. h ] 2 g + h ] -1 g - h ] 0 g = - 3 + 0 - ] -1 g = - 2 5.

ANSWERS

13.

18.

14.

19.

15. f ^ (-a) h = 2 (-a ) - 1 = 2a 2 - 1 = f (a 2) 2

16. x =

17. (a)

20. Domain: x $ 3; range: y $ 0 21. Domain: - 2 # x # 2

2

22.

1 ! 41 4

1 x+3 2]x + 3g 1 = + x+3 x+3 2x + 6 + 1 = x+3 2x + 7 = x+3 = LHS 2x + 7 1 =2+ ` x+3 x+3 RHS = 2 +

23. (a) 0 (b)

(b) Domain: all real x ! - 3; range: all real y ! 2 (c) 24.

797

798



Exercises 6.1 cos i =

2.

3 5 4 sin b = , cot b = , sec b = 5 4 3

3.

sin b =

4.

cos x =

5 , tan x = 9

5.

cos i =

3 4 , sin i = 5 5

6.

5 5 3 tan i = , sec i = , sin i = 2 2 3

7.

cos i =

35 , tan i = 6

8.

tan i =

51 51 , sin i = 7 10

9.

(a)

2

10. (a)

3

7 74

7 , cos b = 5

, tan b =

(a) 17c 20l (b) 34c 20l (c) 34c 12l

1.6 m

5.

(a) 18.4 cm

7.

47.4 mm 8. 20.3 m (c) 9.0 cm

9 56

3. 20.3 cm

4. 13.9 m

(b) 13.8 cm

6. 10 cm and 10.5 cm

9. (a) 7.4 cm

(b) 6.6 cm

1 10. (a) 6.8 cm

35

(b) 6.5 cm

1.

(a) x = 39c 48l (b) a = 35c 06l (c) i = 37c 59l (d) a = 50c 37l (e) a = 38c 54l (f) b = 50c 42l (g) x = 44c 50l (h) i = 30c 51l (i) a = 29c 43l (j) i = 45c 37l (k) a = 57c 43l (l) i = 43c 22l (m) i = 37c 38l (n) i = 64c 37l (o) b = 66c 16l (p) a = 29c 56l (q) i = 54c 37l (r) a = 35c 58l (s) i = 59° 2l (t) c = 56c 59l

2.

37c 57l 3. 22c 14l

3 1 1 , cos 30c = , tan 30c = 2 2 3 3

6.

(a) 11.4 cm

13. tan 48c = cot 42c = 1.11

14. (a) 2 cos 61c or 2 sin 29c

8.

(a) 13 m

(e) 2

15. x = 80c

16. y = 22c

19. t = 20c

20. k = 15c

11. 38 cm

Exercises 6.4

3 1 , cos 60c = , tan 60c = 2 2

(d) 1

(e) 1.393

2.

74

12. sec 82c = cosec 8c = 7.19

(c) 0

(d) 0.928

(a) x = 6.3 (b) y = 5.6 (c) b = 3.9 (d) x = 5.6 (e) m = 2.9 (f) x = 13.5 (g) y = 10.0 (h) p = 3.3 (i) x = 5.1 (j) t = 28.3 (k) x = 3.3 cm (l) x = 2.9 cm (m) x = 20.7 cm (n) x = 20.5 mm (o) y = 4.4 m (p) k = 20.6 cm (q) h = 17.3 m (r) d = 1.2 m (s) x = 17.4 cm (t) b = 163.2 m

11. sin 67c = cos 23c = 0.92

(b) 0

(c) 0.339

1. 5

(b) 45c 1 1 (c) sin 45c = , cos 45c = , tan 45c = 1 2 2

(c) sin 60c =

6.

(b) 0.697

Exercises 6.3

56 , cosec x = 5

(b) sin 30c =

(a) 0.635

(d) 46c 34l (e) 79c 10l

5 12 12 , sin i = , tan i = 5 13 13

1.

5.

4. 36c52l 5. 50c

(b) 37c 52l 7. a = 31c 58l, b = 45c 44l

(b) 65c 17l 9. (a) 11c 19l (b) 26 cm

10. 4.96 cm and 17.3 cm

17. p = 31c

18. b = 25c

11. (a) 12.9 m

(b) 56c 34l

Exercises 6.5 1.

(a)

North

Exercises 6.2 1.

(a) 47c

2.

(a) 47c 13l (b) 81c 46l (c) 19c 26l

(b) 82c

(c) 19c

(d) 77c

(e) 52c

(d) 76c 37l (e) 52c 30l 3.

4.

(a) 77.75c

(b) 65.5c

(c) 24.85c

(d) 68.35c

(e) 82.517c

(a) 59c 32l (b) 72c 14l (c) 85c 53l (d) 46c 54l (e) 73c 13l

Beach house

100c

Boat

ANSWERS

North

(b)

(f) North

Farmhouse

Jamie

12c

Campsite

Dam (g)

320c

North

North

(c)

House 160c

Jetty

200c

Mohammed (h)

North

Seagull (d) North

Alistair Mine shaft 80c Town 50c

(i)

Bus stop

Yvonne

North

North

(e)

Plane

349c

B Hill 285c

School

799

800


North

(j)

4.

(a) 2nd

6.

(a) 1st (e) -

8.

Boat ramp Island

280c

2.

(a) 248c

3.

080c

7.

4. 210c 5. 160c 6. 10.4 m

12. 1.8 km

(g)

3 2

1 2

(b)

13. 12 m 14. 242c

15. 035c

25. 198 m 26. 4.8 km 27. 9.2 m 28. 217c

3+1 2

(g) 1

1 4

(b) 1 (c)

2

6+ 2 = 4

(h)

(k) 0

(d) 4

2 ^ 3 + 1h 4

(l) 1

(q) 2 3

3 2 2

9 3 2

(a) x =

3.

60c

7.

(a) 6 2 m

4.

(b) y =

2m

5.

3m

4 3 3 (i)

(f)

(r) -

1 2

(s) 6

2 3 3

(n) 2 3

(t)

6 2- 3 2

(c) p = 2 3 6.

10 3 m 3

(b) 4 m 8. 0.9 m

9.

5^3 + 3 h m 3

10. 100 3 m

(a) 1st, 4th (b) 1st, 3rd (c) 1st, 2nd (e) 3rd, 4th (f) 2nd, 3rd (g) 3rd (h) 3rd (i) 2nd (j) 4th

2.

(a) 3rd

(b) -

1 2

3. (a) 4th (b) -

1 2

3

1

3 2

(c)

(d) 1

(i)

3

(i) −1

5 21

(d) 2nd, 4th

65

(d)

(j)

2 1 2

(c) - 3 3 2

(i) 3 2

1 2

(j) -

(e) -

(j) -

(e) -

(d)

1 2

3 2

1 2

1 2

(f)

3

1 2

, cot x = -

, sec i =

3 10

20. (a) sin i

2 21

21 2

, tan x = -

9 65

(b) cos x = -

5 18. cot a = - , sec a = 6 19. sin i =

89 5

55 8 8 , sec x = - , cosec x = 3 3 55

17. (a) sin x =

91 3 , tan x = 10 91

61 61 , cosec a = 5 6

51 7 , cot i = 10 51 (b) cos x

(f) - sin i (g) cos a

(c) tan b

(d) - sin a

(e) - tan i

(h) - tan x

Exercises 6.8 1.

Exercises 6.7 1.

(c)

, cosec x = -

4

15. tan i = 16. tan x =

3

(m) 2 ^ 2 - 1 h

(p) 3 - 2 2

2.

(e)

3 2

3

2

7 74 5 74 , sin x = 74 74

14. cos x = -

(j) - ^ 2 + 3 h 1 3

89

(b) 7.2 km 30. (a) 13.1 m (b) 50c26l

(a)

(o) 1

8

13. cosec x = -

Exercises 6.6 1.

(h) -

2 1

1

33 4 , tan i = 7 33

11. cos i = 12. cos x =

22. 1931.9 km 23. 34.6 m 24. 149c

29. (a) 1.2 km

3 2

3

2

1

(b)

(g)

(h) -

1

(h)

16. 9.2 m 17. 171 m 18. 9.8 km 19. 51c 41l 20. 2.6 m 21. 9c21l

1 2

(b)

3 4 10. sin i = - , cos i = 5 5

10. (a) 1056.5 km (b) 2265.8 km (c) 245c 11. 83.1 m

(b) -

2

5. (a) 2nd

7. (a) 1

1 2

(f) -

1

(a) (g)

8. 126.9 m 9. 72c48l

21 m

1 2

(a) -

3 2

(b)

(f) - 3

9.

(b) 145c (c) 080c (d) 337c (e) 180c

(b) - 3

(a) i = 20c 29l, 159c 31l (b) i = 120c, 240c (c) i = 135c, 315c (d) i = 60c, 120c (e) i = 150c, 330c (f) i = 30c, 330c (g) i = 30c, 120c, 210c, 300c ] 0c # 2i # 720c g (h) i = 70c, 110c, 190c, 230c, 310c, 350c ] 0c # 3i # 1080c g

ANSWERS

(i) i = 30c, 150c, 210c, 330c (j) i = 15c, 45c, 75c, 105c, 135c, 165c, 195c, 225c, 255c, 285c, 315c, 345c 2.

(a) i = !79c 13l (b) i = 30c, 150c (d) i = - 60c, -120c

16.

(c) i = 45c, -135c

(e) i = 150c, -30c

(f) i = !30c, !150c (g) i = 22c 30l, 112c 30l, -67c 30l, -157c 30l (h) i = !15c, !45c, !75c, !105c, !135c, !165c (i) i = 135c, -45c

(j) i = !30c, !60c, !120c, !150c

3. 17.

4.

-1

5.

Exercises 6.9 1.

(a) cos i (b) - tan i (c) cos i (d) tan i (e) - sec a

2.

(a) sin i (b) sec i (c) cosec x (f) cosec x 2

(j) sin2 x 3.

6.

x = 0c, 180c, 360c

9.

x = 0c, 360c

7. - 1

8. 1

(g) sec x

(k) 1

2

(l) sin i cos i

(a) LHS = cos x - 1 = 1 - sin 2 x - 1 = - sin 2 x = RHS So cos 2 x - 1 = -sin 2 x 2

(b) LHS = sec i + tan i sin i 1 = + cos i cos i 1 + sin i = cos i = RHS 1 + sin i So sec i + tan i = cos i

10.

(c) LHS = 3 + 3 tan 2 a

11. 0

12. x = 270c

14. x = 0c, 180c, 360c

13. x = 0c, 180c, 360c 15. x = -270c, 90c

= 3 (1 + tan 2 a ) = 3 sec 2 a 3 = cos 2 a 3 = 1 - sin 2 a = RHS So 3 + 3 tan 2 a =

(d) cos2 x

(e) sin a

(h) tan i (i) 5 cosec 2 i

2

3 1 - sin 2 a

801

802


(d) LHS = sec 2 x - tan 2 x = tan 2 x + 1 - tan 2 x =1 = cosec 2 x - cot 2 x = RHS

(j) LHS = =

=

(e) LHS = ] sin x - cos x g 3 = ] sin x - cos x g ] sin x - cos x g 2 = ] sin x - cos x g ^ sin 2 x - 2 sin x cos x + cos 2 x h = ] sin x - cos x g ] 1 - 2 sin x cos x g = sin x - 2 sin 2 x cos x - cos x + 2 sin x cos 2 x = RHS So ] sin x - cos x g 3 = sin x - 2 sin 2 x cos x - cos x + 2 sin x cos 2 x

=

RHS =

1 - sin 2 i + 2 sin i sin i cos i cos 2 i + 2 sin i = sin i cos i 2 sin i cos 2 i = + sin i cos i sin i cos i cos i 2 = + sin i cos i = cot i + 2 sec i = LHS

=

=

(g) LHS = cos 2 ] 90c - i g cot i = sin 2 i cot i cos i = sin 2 i # sin i = sin i cos i = RHS So cos 2 ] 90c - i g cot i = sin i cos i

So 4.

So

cos 2 i

= tan 2 i + cos 2 i

sin b cos b + cos b sin b sec b sin 2 b + cos 2 b sin b cos b sec b

1 + cot b cosec b

- cos b = sin b

LHS = x 2 + y 2 = ] 2 cos i g 2 + ] 2 sin i g 2 = 4 cos 2 i + 4 sin 2 i = 4 (cos 2 i + sin 2 i) = 4 ]1g =4 = RHS So x 2 + y 2 = 4

1 - sin 2 i cos 2 i

1 - sin 2 i cos 2 i

tan b + cot b sec b

LHS = RHS

So ] cosec x + cot x g ] cosec x - cot x g = 1 cos 2 i sin 2 i cos 2 i = cos 2 i cos 2 i = sec 2 i - sin 2 i = tan 2 i + 1 - (1 - cos 2 i) = tan 2 i + 1 - 1 + cos 2 i = tan 2 i + cos 2 i = RHS

sec b

1 sin b cos b cos b sin b = sec b # 1 cos b sin b 1 = # 1 cos b = sin b

1 - sin i + 2 sin i sin i cos i

1

cosec b 1 + cot b - cot b

=

2

(i) LHS =

cosec b

cosec b 1 = cosec b = sin b

(f) RHS =

(h) LHS = ] cosec x + cot x g ] cosec x - cot x g = cosec 2 x - cot 2 x = 1 + cot 2 x - cot 2 x =1 = RHS

- cos b cosec b 1 + cot b - cos b cosec b 1 + cot b - cos b #

So sec 2 x - tan 2 x = cosec 2 x - cot 2 x

So cot i + 2 sec i =

1 + cot b

5.

LHS = x 2 + y 2 = ] 9 cos i g 2 + ] 9 sin i g 2 = 81 cos 2 i + 81 sin 2 i = 81 (cos 2 i + sin 2 i) = 81 ] 1 g = 81 = RHS So x 2 + y 2 = 81

1 sin b

ANSWERS

Exercises 6.10 1.

(a) x = 8.9

Exercises 6.13 (b) y = 9.4 cm

(d) b = 10.7 m 2.

(c) a = 10.0

1.

(a) 7.5 cm 2 (b) 32.3 units 2 (c) 9.9 mm 2 (d) 30.2 units 2 (e) 6.3 cm 2

2.

15 3 2 m 2

6.

1.2 m 2

9.

(a) 7.8 cm

(e) d = 8.0

(a) i = 54c 57l (b) a = 61c 23l (c) x = 43c 03l (d) a = 87c 04l (e) i = 150c 56l

3.

126c 56l 4. (a) 13.5 mm

5.

(a) 1.8 m

7.

(a) 10.3 m

9.

(a) 14.1 cm (b) 15.6 cm

(b) 2.7 m 6. 5.7 cm (b) 9.4 m 8. (a) 60c 22l (b) 57c 9l

(b) b = 10.4 m

(d) n = 16.4 2.

7. 42 cm 2

(c) h = 7.4 cm

(e) y = 9.3

(a) i = 51c 50l (b) i = 60c 27l (c) x = 57c 42l (d) b = 131c 31l (e) i = 73c 49l

10. (a) 5.6 cm

(c) 19.1 cm 2

1.

(a) 2 m (b) 2.2 m (c) 65c 21l 2. (a) 1.9 m (b) 49c 46l

3.

(a) 109 cm 2

5.

(a) 9 m

7.

(a) 48 m

(b) 128.6 m

9.

16c 50l

10. 11c 10l

(b) 16c 20l 4. 65c 9l

(b) 25c 7l 6. (a) 56 m

Exercises 6.15

5.

(a) 11.9 cm (b) 44c 11l (c) 82c 13l

1.

9.

8. 247.7 mm 2

(b) 18.5 cm 2

32.94 mm 4. 11.2 cm and 12.9 cm

7.

+XYZ = +XZY = 66c 10l, +YXZ = 47c 40l (a) 18.1 mm (b) 80c49l 8. (a) 6.2 cm

(c)

(b) 12.7 cm

(b) 30c

tan a + tan b 1 - tan a tan b

(b) 89.7 m

(c) 97.7 m 8. 84 m

(a) sin a cos b - cos a sin b

(e)

12.9 cm 10. (a) 11 cm

(b) cos p cos q - sin p sin q

(d) sin x cos 20c + cos x sin 20c

tan 48 + tan x 1 - tan 48c tan x

(f) cos 2i cos a + sin 2i sin a

(g) cos x cos 75c - sin x sin 75c

(h)

Exercises 6.12 1.

12.5 cm and 4.7 cm 2. (a) 040c (b) 305c 3. 16.4 m

4.

103c

7.

(a) 1.21 km

5. 1.97 m 6. 11c

(i) sin 4a cos b - cos 4a sin b (j) 2.

(b) 1 minute 8. 32 m 9. 107 m

12. 7.7 km

13. 5.7 km and 5.4 km

14. 1841 km

15. 35.8 m 16. 89c 52l 17. 9.9 km

18. 163.5 km

19. 64.1 m 20. 3269 km

21. (a) 11.3 cm (b) 44c 40l 22. 141c 23. (a) 11.6 cm (b) 73c 14l

(a) sin ] a + b g (b) tan 65c (e) tan 2i (f) sin 32c (i) 2 sin x sin y

5.8 sin 42c 29l 10. (a) AC = (b) i = 74c 50l sin 101c 36l 11. h = 8.5

3.

(a)

(c)

(d)

(e)

24. (a) 265.5 km (b) 346c 33l 25. (a) 35c 5l (b) (i) 4.5 m

S6.indd 803

5. 34.8 cm 2

(b) 180.8 cm 2

3.

6.

4. 15.5 cm 2

Exercises 6.14

Exercises 6.11 (a) m = 5.8

3. 7.5 cm 2

(b) 25 mm

10. (a) 54.7 mm (b) 35.1 mm

1.

803

(ii) 0.55 m

(g)

1+ 3

=

2 2 1+ 3

2 2

(c) cos 55c

(d) sin ^ 2x + 3y h

(g) 2 sin a cos b

(h) 2 cos x sin y

(j) 2 cos m cos n 2+ 6 4

(b)

1+ 3

2+ 6 4

=

2 2

=

-^2 3 + 4h = -^ 3 + 2h 2

=

2 2 1+ 3

1 + tan a tan 3b

2 3+4 = 2

1- 3 1- 3

tan a - tan 3b

=

3-1 1+ 3

tan 5x - tan 7y 1 + tan 5x tan 7y

=

2- 6 4

(f)

3+2

3-1 2 2

=

6- 2 4

2+ 6 4

8/11/09 11:38:03 AM

804


(h)

1+ 3 1- 3

=

15. (a) sin 6x

-^4 + 2 3 h = -^2 + 3 h 2

(e)

3-1 1+ 3 o + cos x e o (i) sin x e 2 2 (j)

4.

2 2

cos y =

(c)

2 cos y

5. (a)

6 + 35 3 5-2 7

6 + 35 12

=

3 5+2 7 12

(b)

6.

(a) 2 sin i cos i (b) cos 2 i - sin 2 i (c)

7.

(a) 3 sin i cos 2 i - sin 3 i

8. 9.

(g)

32 5 + 27 7 17

(b) cos 3 i - 3 sin 2 i cos i (c)

2 tan i 1 - tan 2 i

3 tan i - tan 3 i

3 (c) 2 11. (a)

4 5

(b)

3 (d) 2 12 13

(c) -

(e) 33 65

10. (a)

1 2

18. (a)

(b)

(d)

12 5

(e) - 3

(b) -2 sin x sin y

21.

15 16

(c) 2 cos x sin y

(f)

2 tan y ^ tan 2 x + 1 h 1 - tan 2 x tan 2 y

2 tan i 1 - tan 2 i

(c) cos 2 i - sin 2 i

(d) sin x cos 2y + cos x sin 2y = sin x _ cos 2 y - sin 2 y i + 2 cos x sin y cos y (e) cos 2a cos b - sin 2a sin b = ^ cos 2 a - sin 2 a h cos b - 2 sin a cos a sin b (f)

tan x + tan 2y

1 - tan x tan 2y tan x - tan x tan 2 y + 2 tan y = 1 - tan 2 y - 2 tan x tan y (g) sin 2i cos d - cos 2i sin d = 2 sin i cos i cos d - cos 2 i sin d + sin 2 i sin d (h) cos i cos 2c + sin i sin 2c = cos i _ cos 2 c - sin 2 c i + 2 sin i sin c cos c tan x - tan 2z tan x - tan x tan 2 z - 2 tan z = 1 + tan x tan 2z 1 - tan 2 z + 2 tan x tan z (j) sin 2x cos 2y - cos 2x sin 2y = 2 sin x cos x _ cos 2 y - sin 2 y i - 2 sin y cos y ^ cos 2 x - sin 2 x h

(i)

(b)

(h) 1

2

63 65

(b)

1 2

(c) 1

(i)

2 2

1

(d)

3 (j) -

1 2

5 39 7 , sin 2x = 32 32 7 25

(c)

120 169

(d) -

33 56

(b)

1 2+ 3

=2- 3

2-1

(d) cos x cos y - sin x sin y + sin x cos y - cos x sin y

(b)

1

2 4

3

22. (a)

13. (a) 2 sin x cos y

14. (a) 2 sin b cos b

2 2

=

17. cos 2x = -

1 [cos 115c + cos ] - 15c g] 2

1 - tan 2 x tan 2 y

1

20. (a) tan x

3

(b)

2 tan x _ 1 + tan 2 y i

(g) cos 6a

1

12. (a) 2 cos x cos y

(e)

1 sin 12i (f) 1 + sin 2x 2

19. 4 sin i cos i ^ cos 2 i - sin 2 i h = 4 sin i cos 3 i - 4 sin 3 i cos i

1 - 3 tan 2 i

(a) tan 4i (b) sin 7i cos 3i - cos 7i sin 3i cos 2x cos 7x - sin 2x sin 7x

(c) tan 10i (d) cos 2y (h) cos 80c

(i) tan 2b (j) 1 - sin 6x 16. (a)

tan 2x

(b) cos 14y

(b)

1 sin 2i tan i 2 1 = (2 sin i cos i) tan i 2 sin i = sin i cos i cos i = sin 2 i = LHS 1 2 ` sin i = sin 2i tan i 2 RHS =

RHS =

1 - cos i sin i

i i - sin 2 n 2 2 = i i cos 2 sin 2 2 i i 1 - cos 2 + sin 2 2 2 = i i cos 2 sin 2 2 i i sin 2 + sin 2 2 2 = i i 2 sin cos 2 2 i 2 sin 2 2 = i i 2 sin cos 2 2 i sin 2 = i cos 2 i = tan 2 = LHS i 1 - cos i ` tan = 2 sin i 1 - d cos 2

1 2

(e)

3

(f)

3 2

ANSWERS

23. RHS = sin 11i sin 3i = sin (7i + 4i) sin (7i - 4i) = (sin 7i cos 4i + cos 7i sin 4i) (sin 7i cos 4i - cos 7i sin 4i) = sin 2 7i cos 2 4i - cos 2 7i sin 2 4i = sin 2 7i (1 - sin 2 4i) - (1 - sin 2 7i) sin 2 4i = sin 2 7i - sin 2 7i sin 2 4i - sin 2 4i + sin 2 7i sin 2 4i = sin 2 7i - sin 2 4i = LHS ` sin 2 7i - sin 2 4i = sin 11i sin 3i

5. 6.

7.

24. LHS = cos 3i = cos (2i + i) = cos 2i cos i - sin 2i sin i = (cos 2 i - sin 2 i) cos i - 2 sin 2 i cos i = cos 3 i - sin 2 i cos i - 2 sin 2 i cos i = cos 3 i - 3 sin 2 i cos i = cos 3 i - 3 (1 - cos 2 i) cos i = cos 3 i - 3 cos i + 3 cos 3 i = 4 cos 3 i - 3 cos i = RHS

8.

(a)

1 + t2 2t

(i)

1

4.

LHS =

=

1 2

1.

1 - t2 (f)

(c)

1+t

2

1-t

2

2.

(j)

(a)

2 sin ] i - 45cg (b)

34 sin ] i + 59c 2lg

5 sin ] i - 63c 26lg

29 sin ] i - 21c 48lg

10 cos ] i - 18c 26lg 9. 2 cos ] i + 60cg 85 sin ] i + 12c 32lg

(b)

85 cos ] i - 77c 28lg

(a) x = 45c, 225c

(b) x = 30c, 210c

(f) x = 0c, 60c, 300c, 360c

(d)

2

+

1 + t2

3 - 3t + 8t 2

1+t

2

(h)

3. 1 t

(b) a = 180n + 60c

(c) i = 360n ! 30c

(d) x = 180n - ] -1 gn # 30c

(e) i = 180n - 45c

(f) b = 360n ! 45c (h) i = 180n + 30c

(i) i = 360n ! 75c 49l (j) a = 180n + ] -1 gn # 23c 31l

2

1 + t2 1 + t 2 + 2t + 1 - t 2

1 + sin i - cos i =t 1 + sin i + cos i

(a) i = 180n + ] - 1 gn # 30c

(g) c = 180n ! 60c

1 - t2

2

(d) i = 180c, 270c

(i) i = 51c 2l, 190c 54l (j) i = 160c 32l, 270c

2t + 1 - t 2

^1 + t h

2t

(a) i = 126c 52l, 306c52l (b) i = 35c58l, 189c16l

(g) i = 90c, 340c 32l (h) i = 56c 34l, 176c 34l

1 - t2 2t (g)

(h) x = 0c, 180c, 360c

(j) x = 0c, 360c

(e) i = 240c 43l, 327c 21l (f) i = 90c, 180c

2 2

1+t 2t 2 + 2t = 2 + 2t 2t ] t + 1 g = 2]1 + t g =t = RHS `

41 sin ] i + 38c 40lg

65 sin ] i + 60c 15lg

(d) cos 50c

4t ^ 1 - t 2 h

1+t 1+t 1 + t 2 + 2t - 1 + t 2

=

(i)

10 sin ] i + 18c 26lg

(f)

(i) x = 30c, 135c, 150c, 315c

(d) 0

1 + sin i - cos i 1 + sin i + cos i 2t 1 - t2 1+ 2 1+t 1 + t2 1+

(h)

(c) i = 60c, 240c

1 + t2

2

(j)

13 sin ] i + 56c 19lg

(e) x = 90c, 210c, 330c

(c)

2

1 - t + 2t 1+t 1-t

(g)

29 sin ] i + 21c 48lg

(g) x = 0c, 45c, 180c, 225c, 360c

(b)

1-t

(e)

17 sin ] i + 14c 2lg

(d) x = 0c, 45c, 180c, 225c, 360c

2

(e)

2 sin ] i + 45cg (d)

Exercises 6.17

(e) sin 2i (f) cos i

3.

(c)

(b) 2 sin ] i + 60cg

(c) x = 0c, 60c, 180c, 300c, 360c

(a) tan i (b) cos i (c) tan 20c

(b)

5 sin ] i + 26c 34lg

10. (a)

Exercises 6.16

2.

(a)

(e)

25. 3 sin x - 4 sin 3 x

3 (a) 2

^ 1 + t 2 h2

(c) 2 sin ] i - 60cg (d) 2 sin ] i - 30cg

` cos 3i = 4 cos 3 i - 3 cos i

1.

4t - 4t 3 - 1 + 6t 2 - t 4

4.

x = 52c 30l, 82c 30l, -97c 30l, -127c 30l

5.

x = 180n + ] - 1 gn # 30c, 360n ! 90c

6.

x = -180c, 0c, 90c, 180c

7.

(a) i = 180n

(b) x = 360n

(d) i = 180n + (-1) 270c n

8.

(a) (i) x = 30c, 150c

(c) x = 180n (e) 360n ! 90c

(ii) x = 180n + ] - 1 gn ! 30c

(b) (i) x = 41c 25l, 318c 35l (ii) x = 360n ! 41c 25l (c) (i) x = 71c 34’, 251c 34l (ii) x = 180n + 71c 34l (d) (i) x = 161c 34l, 341c 34l (ii) x = 180n - 18c 26l (e) (i) x = 45c

(ii) x = 180n + (-1) n 90c - 45c

805

806


9.

x = 180n + ] -1 g n # 30c, 180n + (-1) n 270c

10. (a) x = 0c, 120c, 240c, 360c

cos i =

5

2.

(a) cos x

(b) 2

34

, sin i =

34

(c) cosec A

(d) cos i

(e) cos 20i

(a) i = 46c 3l (b) i = 73c 23l (c) i = 35c 32l

5.

LHS =

8.

(b) -

2 2

2 ^ 3 + 1h 4

=

2 ^1 - 3 h 4

(c)

1 2 2

=

2 4

(b) x = 180n + 45c

24. a = 51c 40l

23. i = 0c, 120c, 360c

(b) cos ] x + x g = cos x cos x - sin x sin x = cos 2 x - sin 2 x = ^ 1 - sin 2 x h - sin 2 x = 1 - 2 sin 2 x


2 cos 2 i So = 2 + 2 sin i 1 - sin i

2

=

25. (a) cos ^ x + y h

2 cos 2 i 1 - sin i 2 ^ 1 - sin 2 i h = 1 - sin i 2 ] 1 + sin i g ] 1 - sin i g = 1 - sin i = 2 (1 + sin i) = 2 + 2 sin i = RHS

1

2 2

3+1

22. (a) x = 360n ! 60c

(b) 1.84 (c) 0.95

7. (a)

1- 3

(b) 8.5 m

(c) x = 180n + ] - 1 gn # 60c

4.

1 (d) 2

21. (a)

(b)

(a) 0.64

b = 40c

20. 2951 km

3

3.

6.

20 sin 39c sin 99c

(b) 360n ! 120c, 360n

Test yourself 6 1.

19. (a) AD =

3 2

1.

92c 58l 2. 50.2 km

4.

(a) AC =

25.3 sin 39c 53l (b) h = 25.2 cm sin 41c 21l

6.

- cos x

7. 16 3 cm 2

9.

x = 22c 30l, 112c 30l, 202c 30l, 292c 30l 10. i = 75c 45l

3. x = 12.7 cm

8.

(c) - 3

140 (e) 221

11. 5.4 m

x = 120c, 240c

14. -

56 9

12. i = 110c, 230c 15. 31 m

1 2

13. 6.43 km

16. LHS =

9.

5. 4.1 km

cos i ] sin i + cos i g

= cos i

1 - sin 2 i ] sin i + cos i g

cos 2 i sin i + cos i = cos i = tan i + 1 = RHS 17. x 2 + y 2 + 4y - 5 = 0 x = 90c, 270c 10. 122 km

11. 5 3

19. (a) 52c 37l (b) 9 m 12. (a) 6.3 cm

(b) 8.7 m

13. (a) i = 65c 5l (b) i = 84c 16l (c) i = 39c 47l 14. 65.3 cm 2

15. (a) x = !60c, !120c

(b) x = 15c, 105c, -75c, -165c (c) x = 0c, !180c, 30c, -150c 3 4 16. sin i = - , cot i = 5 3

17. (a) 209c

18. i = 180n + ] -1 gn 30c + 53c 8l

(b) 029c

18. (a) 65 m

(b) 27c 42l

20. 30c 8l

21. LHS = cos 6i cos 4i - sin 6i sin 4i = cos (6i + 4i) = cos 10i = cos 2 5i - sin 2 5i = cos 2 5i - (1 - cos 2 5i) = 2 cos 2 5i - 1 = RHS ` cos 6i cos 4i - sin 6i sin 4i = 2 cos 2 5i - 1 22. 30.1 m, 0.5 ms - 1

23. i = 30°, 150°, 270°

24. i = 180n + (-1) n 270c

25. - t

ANSWERS


7.

1 1 Midpoint of AC = midpoint of BD = d 2 , 3 n . 2 2 Diagonals bisect each other

8.

AC = BD =

Exercises 7.1 1.

2.

(a) 5 (b) 10

(c) 13

(c)

52 = 2 13

85

(d)

(a)

13

(b)

65

3.

(a) 9.85 (b) 6.71 (c) 16.55 4. 12 units

5.

Two sides =

6.

Show AB = BC =

7.

Show points are

8.

Radius = 3 units, equation x 2 + y 2 = 9

9.

Distance of all points from ^ 0, 0 h is x 2 + y 2 = 11

34 , 1 side =

AC =

11 , equation

11. a = ! 6 - 2

37 , QP = MN =

15. BD = AC =

98

19. AB =

29 , BC =

(b) XY =

17 units from ^ 7, -3 h

14. MQ = NP =

20 , so parallelogram

16. (a) AB = AC = 17. 2 101

18.

116 , AC =

145

65 , YZ =

130 , XZ =

1.

40 , BC = 4 61 units 3.

12. x 2 + y 2 = 1

(b) 1

(g) - 4

1 2

x = 1.8

1 3

(c) - 1

(h) -

2 3

4. x = 9

(-2, 1)

7.

^ 4, 3 h 6. x = 3 is the vertical line through midpoint ^ 3, 2 h.

(f) -

2. y 1 = 21

(3, 4) (7, 2)

1 1

2 3 4 (2, -1)

6

5

Gradient of AB = gradient of CD = 1

1 2

Gradient of AB = gradient of CD = -1 Gradient of BC = gradient of AD =

1 3

3 4

1 Gradient of AC = - 5 , 2 1 gradient of BD = 2

(c) a = -1, b = - 2 (d) a = -1, b = - 2

5.

(j) - 2

Gradient of BC = gradient of AD = 0

(b) a = - 5, b = 6

4. P = Q = ^ 2, -1 h

2 3

(e)

2

-3 -2 -1 -1

1 1 (e) ^ -1, 1 h (f) ^ - 3, 2 h (g) d 3, n (h) d 1 , 1 n 2 2 1 1 1 (i) d , 2 n (j) d 0, 5 n 2 2 2

3 + ]-3g -4 + 4 = 0, =0 2 2

1 4

2 5

3

Exercises 7.2

3.

(d) - 2

5. (a) Show m 1 = m 2 =

-2

(e) a = 6, b = 1

1 3

(i) 2

4

(Pythagoras’ theorem)

(a) ^ 2, 4 h (b) ^ 1, -1 h (c) ^ - 2, 1 h (d) ^ - 3, 2 h

2

y

6.

(a) a = 9, b = - 3

(a) 2

65

30.2

2.

2 , AB = 2

(b) Lines are parallel.

Problem

1.

34 ; YZ =

34 , 2

40 = 2 10 ; XZ =

Exercises 7.3

Since XY = YZ, triangle XYZ is isosceles. XY 2 + XZ 2 = 65 + 65 = 130 = YZ 2 So triangle XYZ is right angled.

10 , BC =

11. x 2 + y 2 = 4

AB 2 + BC 2 = 29 + 116 = 145 = AC 2 So triangle ABC is right angled (Pythagoras’ theorem) 20. XY =

9. ^ - 8, 13 h

1 1 1 1 10. (a) X = d - , 3 n , Y = d , n , Z = ^ 1, 1 h 2 2 2 2

128

12. All 3 sides are 2 units. 13. a = 10, - 2

(b) OC = OB = 2

1 BD = d 4, - n ; rectangle 2

85

10. a = 3

125 , midpoint AC = midpoint

8.

Gradient of AC = 1, gradient of BD = -1

9.

(a) Show AB 2 + BC 2 = AC 2 (b) Gradient of AB = gradient of BC = -

4 5

5 , 4

7

3 5

1 8

807

808


10. (a) F = ^ 1, - 2 h, G = d 4,

Exercises 7.6

1 n 2

(b) Gradient of FG = gradient of BC =

1.

5 6

11. 4x - 3y - 11 = 0 12. Gradient of ^ 2, - 4 h and ^ 3, -1 h = gradient of ^ 3, -1 h and ^ 5, 5 h = 3

18. (a) 19.

(h)

14. 0.93 15. 21 16. 50c 12l 17. 108c 26l

13. 1

3

1

(b)

3

-5 - ] -2 g 7-4 -3 = 3 = -1 m = tan i -1 = tan i ` i = 180c - 45c ^ 2nd quadrant h = 135c

(b) (i) 2

(ii) -7

(ii) 1 (c) (i) 6

(d) (i) -1

(ii) 0

(e) (i) - 4

(ii) 3

(f) (i) 1

(ii) - 2

(g) (i) - 2

(ii) 6 (h) (i) -1

(ii) 1

(i) (i) 9 (ii) 0

(j) (i) 5 (ii) - 2 2. (a) (i) - 2 (ii) 3 (b) (i) - 5 (ii) - 6 1 (c) (i) 6 (ii) -1 (d) (i) 1 (ii) 4 (e) (i) - 2 (ii) 2 1 1 4 (f) (i) 3 (ii) 1 (g) (i) (ii) - 2 (h) (i) (ii) 2 5 2 3 (ii) -

1 2

(j) (i) 1

2 3

(ii)

2 3

3. (a) 4

1 (c) 0 (d) - 2 (e) -1 (f) - 3 (g) 2 (h) 4 3 1 2 1 1 2 (j) 1 (k) (l) (m) (n) (o) 4 5 7 5 3 2 3 1 1 1 (p) (q) 15 (r) - 1 (s) (t) 6 14 2 8

(b) - 2

3.

(j)

1 5

1

(i)

3

(d) 1

1 2

(d) x + 2y + 5 = 0

(e) x - 2y + 4 = 0

(f) x + 3y - 1 = 0

(g) 3x + 4y + 13 = 0

4.

m1 m2 = -

(g)

1 3

1 # 5 = - 1 so perpendicular 5 1 5

m1 # m2 = -

3 7 # = -1 7 3

7. k = -

2 3

8. m 1 = m 2 = 4

5 AB < CD _ m 1 = m 2 = 3 i and BC < AD d m 1 = m 2 = - n 8 1 10. Gradient of AC: m 1 = , gradient of BD: m 2 = - 2, 2 1 m 1 # m 2 = # - 2 = -1 2 11. (a) y = - x

(b) 5x - y - 8 = 0

(d) 2x - 3y + 16 = 0

(c) 2x + y + 2 = 0

12. 7x + 6y - 24 = 0

13. x + y - 3 = 0 14. 2x - y - 5 = 0 15. 2x - 3y + 18 = 0

Exercises 7.7 1.

(a) ^ 2, - 4 h

(b) ^ -1, - 3 h

(e) ^ 5, -1 h

(f) ^ -1, 1 h

(j) d 3.

(c) ^ 4, 4 h

(g) ^ 3, 7 h

^ 2, 5 h, ^ 4, 1 h and ^ -1, -1 h

at ^ 2, -3 h

5x + 6y - 27 = 0

(d) y = 4x + 20 (e) 3x + y - 3 = 0 (f) 4x - 3y - 12 = 0

x+y-1=0

10. 2x + y - 2 = 0

(g) y = x - 1

11. x + y - 3 = 0

12. x - 2y - 3 = 0

13. x - y + 1 = 0

14. x - 3y + 2 = 0

(h) y = x + 5

(a) 4x - 3y + 7 = 0 (e) x - 2y + 2 = 0

2. x + y - 8 = 0

(b) 3x - 4y + 4 = 0 (d) 3x + 4y - 25 = 0

4. 4x + y - 8 = 0

6. y = - 2x

5. (a) y = 3

7. 3x - 4y - 12 = 0

9. x = - 4

10. 3x + 8y - 15 = 0

(i) ^ 41, 26 h

4. All lines intersect

9.

(c) y = 5x

(d) ^ 0, - 2 h

(h) ^ 4, 0 h

1 7 n 2. Substitute ^ 3, - 4 h into both lines ,19 19

7.

(a) y = 4x - 1 (b) y = - 3x + 4

2x + y - 3 = 0

5 6

(a) x - y + 1 = 0 (b) x - 3y + 16 = 0 (c) x + y - 5 = 0

All lines meet at ^ - 5, 0 h

(b) x = -1

(f) -

(e) 1

5.

(c) 4x - 5y + 13 = 0

8.

3 4

m 1 = m 2 = 3 so parallel

1 (i) 1 2

Exercises 7.5 1.

(c)

9.

(a) (i) 3 (ii) 5

1 2

1 3

3.

6.

2^ 3 + 3h 3

(i) (i) 3

1 3

(b)

5. m 1 = m 2 = 1

Exercises 7.4 1.

2.

(c) - 3

m=

20. x =

(a) - 3

15. 3x + y - 7 = 0

21. 5x - y + 17 = 0

8. 4x + 7y + 23 = 0

16. x + 5y + 13 = 0

17. 27x - 5y - 76 = 0 19. 2x - y - 1 = 0

6. 11x + 6y = 0

18. 3x - y - 14 = 0

20. 3x - y - 11 = 0

ANSWERS

Exercises 7.8

Exercises 7.10 3 13

8 13

1.

(a) 2.6 (b) 1

2.

(a) 3.48 (b) 1.30 (c) 0.384 (d) 5.09 (e) 1.66

3.

(a)

4.

d1 = d2 = d3 = 1

5.

7 13 13

A: d =

14 5

(b)

(c) 2.5 (d) 2.4

5

, B: d =

(c)

4 205 205

(e)

(d)

5 26 13

1.

(e)

9 1 1 2 1 (d) d 4 , -1 n (e) d 2 , - 2 n (f) d - 5, 2 n 7 7 4 2 10 14 13 13

^ 2, - 3 h: d =

13 10

6 6 6 4 4 1 n (i) d - , 1 n (g) d 2 , 7 n (h) d - 3 , -1 7 7 7 7 11 11 2 2 (j) d 1 , -1 n 3 3

-3 5

Opposite signs so points lie on opposite sides of the line 6.

2.

5

, ^ 9, 2 h: d =

^ - 3, 2 h : d = - 4 , ^ 4 , 1 h : d = 2

8.

d 1 = d 2 = 2 so the point is equidistant from both lines

9.

^ 8, - 3 h: d =

37

9

, ^ 1, 1 h: d =

10. ^ - 3, 2 h: d =

5

, ^ 4, 1 h: d =

A (3, 2)

5.

3 1 P = d 1 , n , Q = ^ 16, -19 h, PQ = 24 units 5 5

6.

3 4 2 B = d 9 , -12 n 7. p = 4 , q = 20 5 5 5

8.

2 2 2 2 (a) d , 1 n (b) Each ratio gives d , 1 n . This means 3 3 3 3 that the intersection of the medians divides each median in the ratio 2:1.

9.

a = 8, b = 18

Opposite signs so points lie on opposite sides of the line 8 5 units 5

11. d 1 = d 2 = 4 so same distance 12. 13. 1

14. 4.2 15. x = 9

17. m = - 1

2 or -17 3

16. b = 3

1 1 or -1 4 12

2 1 or -18 3 3

2 2 (a) E = d , 2 n (b) F = d 1 , 2 n 3 3

4.

7 5

(1 23 , 3 13 )

18. Show distance between ^ 0, 0 h and the line is 5

Test yourself 7

19. Show distance between ^ 0, 0 h and the line is greater than 1

1.

6.4 units

4 1 20. (a) ^ 3, -1 h, d 3 , n, ^ - 2, 2 h 7 7

3.

(a) - 1

(b)

2 10 13 5 26 34 , , 5 5 119

4.

Exercises 7.9 1.

(a) 149c 2l (b) 119c 45l (c) 143c 58l (d) 172c 14l (e) 135c

3. 12c 20l

1 3

21c 2l, 120c 58l, 38c

7.

m = - 5.4, 1.53

9.

(a) +A = +C = 63c 26l, +B = +D = 116c 34l

6. m = 3, -

8. k Z -1.64, 0.095

(b) 124c 31l 10. +A = 61c 56l, +B = +C = 59c 2l

1 2. d 2 , - 2 n 2 (b) 2

1

(c)

(d)

3

(a) 7x - y - 11 = 0

3 5

(b) 5x + y - 6 = 0

6.

1 m 1 = - , m 2 = 4 so m 1 m 2 = -1 4 ` lines are perpendicular.

7.

x-intercept 5, y-intercept - 2

8.

(a) 2x + y - 1 = 0

9.

m 1 = m 2 = 5, so lines are parallel

(b)

(c) 3x + 2y = 0

(e) x - 3y - 3 = 0

6 5 units 5

4. 53c 58l

5.

2 8 10. P = d , 3 n 9 9

5.

(i) 74c 56l (j) 36c 52l 2.

1 5

B (-1, 6)

( 13 , 4 23 )

(d) 3x + 5y - 14 = 0

(a) 18c 26l (b) 29c 45l (c) 82c 52l (d) 26c 34l (e) 10c 29l (f) 41c 49l (g) 72c 15l (h) 18c 26l

(j) ^ 10, 13 h

(c) EF = 1, AC = 3 ` AC = 3EF

37

Same signs so points lie on same side of the line -6

3.

1 (d) d 12, 5 n 2

4 1 2 (f) d 9, -1 n (g) d - 6, - n (h) d 9, 1 n 7 2 3

(i) ^ - 58, 30 h

1 5

Opposite signs so points lie on opposite sides of the line

55

1 4 (a) d - 4, 3 n (b) d 6 , 2 n (c) ^ 19, 25 h 5 2 (e) ^ 40, 12 h

10

Same signs so points lie on the same side of the line 7.

3 2 1 3 4 8 (a) d - , 1 n (b) d 2 , 3 n (c) d - 2 , 1 n 5 5 5 5 9 9

1 2

(c)

5 units 2 10. 3x - 4y = 0

809

810


11. ^ -1, 1 h

12. a = 6, b = 1

22. (a) AB: 7x + 5y + 14 = 0

13. 66c 48l

^ -7, 7 h lies on the line (show by substitution)

14. Solving simultaneously, x - y - 4 = 0 and

(b) -1:2 or 1:- 2

2x + y + 1 = 0 have point of intersection ^ 1, - 3 h . Substitute ^ 1, - 3 h in 5x - 3y - 14 = 0:

2 23. x = 16 , y = -17 3

LHS = 5 # 1 - 3 # - 3 - 14 = 0 = RHS

2 1 1 1 25. (a) P = d 1 , 3 n (b) Q = d 4 , 3 n 3 3 3 3

` point lies on 5x - 3y - 14 = 0: Substitute ^ 1, - 3 h in 3x - 2y - 9 = 0:

(c) PQ has gradient m 1 = 0

LHS = 3 # 1 - 2 # - 3 - 9 = 0 = RHS

AC has gradient m 2 = 0

` point lies on 3x - 2y - 9 = 0:

Since m 1 = m 2, PQ < AC

` lines are concurrent 5 1 15. d 2 , -1 n 16. - 0.499 9 3 19. ^ 4, 7 h

18. y = 3

22. ^ - 2, 1 h: d =

-8 13

1 (d) R = d 6 , 0 n 3

17. c = -13, - 65 4 5 2

(e) PR has gradient m 1 = -

21. 93c22l

20. x = 1

, ^ 6, 3 h: d =

BC has gradient m 2 = -

13

24. x - y - 4 = 0

Chapter 8: Introduction to calculus

2. x - 3 y - 3 3 = 0

3. 10x 2 + 10y 2 = 81

1.

k = -2

4.

Show AC and BD have the same midpoint ^ 1, 2 h and m AC # m BD = -1

5.

Show distance of all points from ^ 0, 0 h is 3; radius 3; equation x 2 + y 2 = 9

6.

4 13 13

8.

12 13 13

1 18. b = 2 , - 21 3 m1 - m2

`

1 + m1 m2 m1 - m2

14. 2x + 5y + 14 = 0

12. ^ 3, - 5 h

2.

15. 45c

17. x - y + 6 = 0

1 1 2 2 19. d 2 , - 2 n, d 1 , - 3 n 3 3 3 3 =1

=1 1 + m1 m2 m1 m2 + 1 = m1 - m2 m1 m2 = m1 - m2 - 1 m1 - m2 or = -1 1 + m1 m2 m 1 - m 2 = -1 - m 1 m 2 m1 m2 = m2 - m1 - 1

21. P = f

1.

9. 113c12l 10. 2x + 3y + 13 = 0

3x +y + 3 - 2 3 =0

20.

Exercises 8.1

7. +OBA = 45c; a = b (sides of isosceles D)

11. BC = AC = 18 , AB = 6, so D is isosceles; m BC # m AC = -1, so D is right angled.

16.

- 4p - 1 7p - 3 p , p -1 p -1

5 7

25. 3x - 7y - 14 = 0


13. a = 2, b = 3

5 7

Since m 1 = m 2, PR < BC

Opposite signs so points lie on opposite sides of the line 23. 63c 26l

24. m = - 0.059, - 9.2

3.

ANSWERS

4.

5.

10.

Exercises 8.2 2. Yes, x = x 1

1.

Yes, x = 0

5.

Yes, x = x 1, x = x 2

8.

Yes, x = 2

3. No

6. Yes, x = 0

9. Yes, x = - 2, 3

11. Yes, x = 90c, 270c

4. Yes, x = 0 7. Yes, x = - 3

10. Yes, -1 # x 1 0

12. Yes, x = 0

13. No

14. No

15. Yes, x = !3 6.

Exercises 8.3 1.

(a) 3 (b) -7 (c) 3 (d) 8 (h) -1 (i) 10 (j) -1

2.

(a) x 2 - 2x - 4 (b) 2x 3 + x - 1 (c) - 7x - 1 (d) 4x 4 - x 2 (e) - 4x + 3 (f) 2x 2 + 6 (g) - 2x (h) 4x 2 (i) 3x - 1 (j) x 2 - 2x + 9

(e) 2

(f) - 3

(g) 2

Exercises 8.4 7.

1.

(a) 4.06

2.

(a) 13.61

4.

(a) f ] x + h g = x 2 + 2xh + h 2

(b) 3.994

(c) 4

(b) 13.0601

(c) 12.9401

(d) 13

3. 6

2 2 2 (b) f (x + h) - f (x) = x + 2xh + h - x 2 = 2xh + h

(c)

f ]x + hg - f ]xg h

8. (d) f l(x) = lim

2xh + h 2 h h ] 2x + h g = h = 2x + h f ]x + hg - f ]xg =

h "0 h = lim (2x + h) h "0

= 2x 5.

9.

(a) f (x + h) = 2 ] x + h g2 - 7 (x + h) + 3 = 2 (x 2 + 2xh + h 2) - 7x - 7h + 3 = 2x 2 + 4xh + 2h 2 - 7x - 7h + 3 2 2 (b) f (x + h) - f (x) = (2x + 4xh + 2h - 7x - 7h + 3) - (2x 2 - 7x + 3) = 2x 2 + 4xh + 2h 2 - 7x - 7h + 3 - 2x 2 + 7x - 3 = 4xh + 2h 2 - 7h

811

812


(c) f ] x + h g - f ] x g h

(d) f l] x g = 4x - 7 6.

4xh + 2h 2 - 7h h h ] 4x + 2h - 7 g = h = 4x + 2h - 7 =

(a) 4x + 1 (b) 8x - 12 (e) 6x 2 + 6x - 3

3.

(a)

4.

(c) f ] 2 + h g - f ] 2 g = h + 5h 2

(d) f ] 2 + h g - f ] 2 g h

(e) f l] 2 g = 5 7.

6.

h 2 + 5h = h h ]h + 5 g = h =h+5

(a) f ] -1 g = -7 (c) 12

8.

(a) f ] 3 g = 8

(c) f l] 3 g = 6

9.

(a) f l] 1 g = - 13

(b) f ] 3 + h g - f ] 3 g = 6h + h 2

11.

dV = 4rr 2 dr

(c) -12 (b)

(c) f l] x g = 8x - 4 (e) (g)

dy dx dy dx

= 3x 2

dy dx

(d) 15

3.

dy dx

13. (a) 5

(b) - 5

(c) x = 4

15. 18

(b) -13

(c) 11

(d) -18

(h) 136

(i) - 4

(j) 149

(a) 1 7

1 26

(b)

(g) -

1 25

1 71

(ii) -

(a) (i) 6

(c) (h)

1 6

1 20

1 20

(d) (i) -

(b) (i) 8

1 24 1 (ii) 11

(ii) -

(d) (i) - 8

1 8

(e) 18

1 43

(e)

(j) -

(f) 27

1 10

1 5

1 8 1 (ii) 8

(ii) -

(a) x + 24y - 555 = 0 (c) x - 17y - 516 = 0 (e) x + 2y - 9 = 0

6.

(a) (i) 7x - y + 4 = 0 (ii) x + 7y - 78 = 0 (b) (i) 10x - y + 36 = 0 (ii) x + 10y - 57 = 0 (c) (i) 10x + y - 6 = 0 (ii) x - 10y - 41 = 0 (d) (i) 2x + y + 2 = 0 (ii) x - 2y - 19 = 0 (e) (i) 2x - y + 2 = 0 (ii) x + 2y - 9 = 0

7.

x = !3

10. (0, 1)

(b) x - 8y + 58 = 0 (d) x - 45y + 3153 = 0

8. (1, 2) and (-1, 0)

15. -1 11. (1, 2)

Exercises 8.5 (a) 1 (b) 5 (c) 2x + 3 (d) 10x - 1 (e) 3x 2 + 4x - 7 (f) 6x 2 - 14x + 7 (g) 12x 3 - 4x + 5 (h) 6x 5 - 25x 4 - 8x 3 (i) 10x 4 - 12x 2 + 2x - 2 (j) 40x 9 - 63x 8

dh = 40 - 4t dt

5.

(h) f l] x g = - 6x 2

(c) - 0.04

10.

(a) 27x - y - 47 = 0 (b) 7x - y - 1 = 0 (c) 4x + y + 17 = 0 (d) 36x - y - 47 = 0 (e) 44t - v - 82 = 0

= 10x - 1

(b) - 0.03992

12. 3

dv = 30t dt

4. (e) - 9

13. (a) 0.252 (b) 0.25 (c) 0.2498

1.

(e)

ds = 10t - 20 dt

7.

(a) 72

(e) (i) 11

(f) f l] x g = 6x 2 + 5

9.

(b) x = ! 2

(c) (i) 24

= 2x + 5

(d)

= 3x 2 - 4 x + 3

14. (a) - 0.04008

(d) 4x

5. - 56

(g) 11

(f)

= 2x + 2

12. (a) f l] x g = 2x

8x 7 - 6x 5 3

Exercises 8.6

2.

y + dy = ] x + dx g2 + 2 (x + dx) = x 2 + 2xdx + dx 2 + 2x + 2dx Since y = x 2 + 2x dy = 2xdx + dx 2 + 2dx dy 2 x d x + d x 2 + 2d x (b) = dx dx d x ] 2x + d x + 2 g = dx = 2x + dx + 2

11. (a) 2 (b) 5

(c)

= 60x 9 - 40x 7 + 35x 4 - 3

gl] x g = - 20x - 5

1.

(b) 17

Substitute _ x + dx, y + dy i:

dy

dy dx

8.

10. (a) y = x 2 + 2x

dx

(b) 2x 3 - x 2

f l] x g = 16x - 7

14. (a) 12

(b) f ] -1 + h g - f ] -1 g = 4h 3 - 12h 2 + 12h

(c)

x -1 3

(d) 16x 3 - 24x

(c) 2x

(f) 2x 2 - 2x + 2

(b) f ] 2 + h g = h 2 + 5h + 11

(a) f ] 2 g = 11

2.

13. (a) (1, -1)

9. (- 5, -7)

3 15 n 12. d - 1 , - 4 4 16

(b) 6x - y - 7 = 0

14. 10t - h - 7 = 0

15. 4x - 2y - 19 = 0

1 4

ANSWERS

Exercises 8.7

(k) 1

1.

(a) - 3x - 4 (e) x

2.

-

(a) (f) (j) -

3. 8.

1 27 1 8

1 2

+ 3x - 2

1

(b)

x2 1

2x

3

(f) x 5

-

2 3

(g) 6x

(c)

2 x (g) -

2 x3 1

(c) 1.2x - 0.8

(b) 1.4x 0.4

3

1 3 x 2

-

3 2

10

(i) -

x

1

x4

2

1 32

(b) -

15. d 5,

2 ] 4 + x g3

(w) 2.

9

5

1 16

10. x - y + 9 = 0

10x - y - 9 = 0

8.

x=

5 11 =2x - 1 ] 2x - 1 g2 1

7+

7

8

=

5. 176

7

7. 69x - y - 129 = 0

- 6 ! 30 3

12. x + 16y - 16 = 0

13. (9, 3)

2 2 n, d - 5, - n 5 5

(u) -

^ x 4 - 3x 3 + 3x h

2

3

(v) -

16 3 4x + 1 3

4. (4, 1) 5. x = 2, -1

16 ] 5x + 1 g2 - 2x 2

(g)

^ 2x 2 - x h2

9. 34x - y + 29 = 0

1 2

4

6. 8x + y + 7 = 0

(a) 8x 3 + 9x 2 (b) 12x - 1 (c) 30x + 21 (d) 72x 5 - 16x 3 (e) 30x 4 - 4x (f) x ] 5x + 2 g ] x + 1 g2 (g) 8 ] 9x - 1 g ] 3x - 2 g4 (h) 3x 3 ] 16 - 7x g ] 4 - x g 2 (i) ] 10x + 13 g ] 2x + 5 g3 4 5 (j) 10x ^ x 3 + 5x 2 - 3 h ^ x 2 + 1 h + ^ 3x 2 + 10x h ^ x 2 + 1 h 4 3 2 2 ^ h ^ h = x 13x + 60x + 3x - 20 x + 1

- x + 14x 2

(e) (h)

2x 3 + 9x 2 + 7 ] x + 3 g2

(o)

x4 -6 ] x - 2 g2

= (i)

x 4 - 12x 2 ^x - 4h

2

2

- x + 14 x3 - 34 ] 4x - 3 g2

=

x 2 ^ x 2 - 12 h ^ x 2 - 4 h2

(f)

11 ] x + 3 g2

(j)

-14 ] 3x + 1 g2

(q)

(p)

3x 2 + 8x - 5 ] 3x + 4 g2

^ x 2 - x - 1 h2

(s) (t)

(l)

x 4 - 2x 3 - 4x 2 - 1

] 7x + 2 g4 - 28 ] x - 1 g ] 7x + 2 g3

(u)

1 2

-

2]x + 5g - x ]x + 5g (r) x+5

1 2

] 2x - 9 g2 ] 20x + 51 g 6 ] 5x + 1 g ] 2x - 9 g2 - 5 ] 2x - 9 g3 = 2 ] 5x + 1 g ] 5x + 1 g2 ] 7x + 2 g8

=

- 21x + 30 ] 7x + 2 g5

15 ] 2x - 5 g3 ] 3x + 4 g4 - 6 ] 3x + 4 g5 ] 2x - 5 g2 ] 2x - 5 g6 4 ] ] g 3 3x + 4 4x - 33 g = ] 2x - 5 g4

5

] 7 - x g9

(c)

4x ] x - 3 g 4x 2 - 12x = ] 2x - 3 g2 ] 2x - 3 g2 ^ 3x 2 - 7 h 2x 2 ] x + 6 g - 18x 2x 3 + 12x 2 = (m) (n) 2 2 2 ] g ] x + 4 g2 ^x - 5h x+4

27

4

15 ] x + 5 g2

2

2 ] 2x + 7 g10

(y)

(b)

- 3x 2 - 6x - 7

(k)

4 ] 3x - 1 g3 (x)

-2 ] 2x - 1 g2

(a) (d)

Exercises 8.9 1.

6.

1.

^ 4x 3 - 9x 2 + 3 h

3. 40

4.

26

6. −3 7. 2x + 3 x + 1

6 (a) 4 ] x + 3 g3 (b) 6 ] 2x - 1 g2 (c) 70x ^ 5x 2 - 4 h 5 4 (d) 48 ] 8x + 3 g (e) - 5 ] 1 - x g (f) 135 ] 5x + 9 g8 3 (g) 4 ] x - 4 g (h) 4 ^ 6x 2 + 3 h ^ 2x 3 + 3x h 7 2 ^ h (i) 8 ] 2x + 5 g x + 5x - 1 1 3 5 (j) 6 ^ 6x 5 - 4x h ^ x 6 - 2x 2 + 3 h (k) ] 3x - 1 g 2 2 2 5 -4 (l) 2 ] 4 - x g- 3 (m) - 6x ^ x 2 - 9 h (n) ] 5x + 4 g 3 3 1 3^ 2 3 4 3x - 14x + 1 h ^ x 3 - 7x 2 + x h (o) (p) 4 2 3x + 4 8x 5 2 (q) (r) (s) 3 ] 5x - 2 g2 ^ x 2 + 1 h5 7 - 3x

(t) -

3. 1264

2.

3x 2

Exercises 8.8 1.

+

4 - 3x 2 2-x

Exercises 8.10 5.

2 x3

] 2x - 1 g2

15

(e)

x6

+ 2-x =

5

9. 3x + 16y - 8 = 0

14. x = 4

(h) x

(l)

12

-

4. −3

11. (a) -

1 4

(d) -

6 6 x5

(h)

x7

-

1 -2 x 2

(d)

x

2 2-x - 2 ] 5x + 3 g

3x + 1 3 x+1 -2 x+1 3x + 5 (v) = x+1 2 ] x + 1 g3 2x - 3 (w)

2 x-1 -2 x-1 - 2x + 1 = ] 2x - 3 g2 2 x - 1 ] 2x - 3 g2

x ] x - 9 g2 (x)

x2 + 1

- 2 ] x - 9 g x2 + 1 ] x - 9 g4

=

- x 2 - 9x - 2

x2 + 1 ] x - 9 g3

813

814


5 9

2.

1 8

6.

x - 18y + 8 = 0

3. - 1

4. x = 0, 1

(b)

5. x = - 9, 3

7. 17x - 25y - 19 = 0

Exercises 8.11 1.

(a)

2.

dy dx (b)

(b) Substitute Q into both equations. (c) y = x 2 - 4 has m 1 = 4 y = x 2 - 8x + 12 has m 2 = - 4 (d) 28c 4l 2.

(a)

(d) (e)

dy dx dy dx dy dx

=

11 ] 2x + 1 g2

=

5 x3 2

4. 71c 34l

5. 162c 54l

(d) 0c

3. 8c 8l

6. (a) X = ^ 4, 16 h, Y = ^ -1, 6 h

(f)

dy dx

5. (a) 1

8.

(a) x = - 2

(b) x = 1

9.

(a) f l] x g = 32 ] 4x + 9 g3 dy

dy

(c)

dv = 4t - 3 dt

dx

= 42x 5 - 9x 2 + 2x - 8

dx

= 9 (2x + 4) (x 2 + 4x - 2)8

= 40x ] 2x - 1 g3 + 5 ] 2x - 1 g4 = 5 ] 2x - 1 g3 (10x - 1) =-

10 x3

(b) 20

6. 10

7. 42

(c) x = 2 (b)

dy dx

= ] 9x - 1 g ] 3x - 1 g (d)

(e) f l] x g = (c) m = 6

dx

4.

(c)

(b) P = ^ 3, 9 h

dy

3. (a)

= 10x - 3

=dy dx

5 ] x - 3 g2

=-

4 x2

1 5 5 x4

10.

y

(b) At X: m 1 = 12, m 2 = 7 At Y : m 1 = - 8, m 2 = - 3 (c) At X: 3c 22l At Y : 11c19l 7.

71c 34l, 8c 58l 8. (a) (0, 0), (2, 8), (-1, -1) (b) 63c 26l at (0, 0), 4c 42l at (2, 8), 71c 34l at (-1, -1)

9.

At (0, 0), m 1 = 0 and m 2 = 4 At (2, 4), m 1 = 4 and m 2 = 0 Angle at both is 75c 58l

10. 164c 45l at (0, 0), 178c 37l at (- 3, - 33), 146c 19l at (1, 3)

Test yourself 8 1.

(a)

11. 9x - y - 7 = 0

12. (2, 3)

14. (- 2, 71), (5, - 272) 17. 9 20.

7 10

13.

dS = 8rr dr

15. 4x - y - 6 = 0

18. 12x + y - 4 = 0

19.

16. 3525

ds 1 = u + at, t = 5 dt

21. 17c6l at (3, 9), 53c8l at (-1, 1)

22. 175c 26l at (2, 4), 177c 40l at (4, 16)

ANSWERS


f ] 1 g = - 3, f l] 1 g = - 36

3.

dx = 8t 3 + 300t 2; t = 0, - 37.5 dt

4.

2x + y = 0, 3x - y - 3 = 0, 6x - y + 12 = 0

5.

^ 2, 2 h, ^ - 2, -14 h, x + 12y - 26 = 0, x + 12y + 170 = 0

6.

3 4

7.

5 ] 5x + 1 g3 ] x - 9 g4 + 15 ] x - 9 g5 ] 5x + 1 g2 = 10 ] 5x + 1 g2 ] x - 9 g4 (4x - 13)

8.

9.

2. -

13 18

2 ] 4x - 9 g4 - 16 ] 2x + 1 g ] 4x - 9 g3 ] 4x - 9 g8 - 2 ] 12x + 17 g = ] 4x - 9 g5 x=

- 6 ! 204 - 3 ! 51 = 6 12

11. a = 1 14. 2

1 27

y

(b)

1 1 1 1 ! 13 n 13. x = , 12. P = d - 2 , 6 4 16 3 3 10

16. (a) Substitute (1, 1) into both curves: y = ] 3x - 2 g5: LHS = 1 RHS = ] 3 # 1 - 2 g5 = 15 =1 = LHS So (1, 1) lies on the curve y = ] 3x - 2 g5 5x - 3 : x+1 LHS = 1 5#1-3 RHS = 1+1 2 = 2 =1 = LHS

90c

21. ^ - 4, -73 h

10. 2x + y - 25 = 0

15. 3x - y + 5 = 0, Q = ^ 0, 5 h, PQ =

1

25. x = 0, 2, 6 28. p = 1

1 2

3 ] 4 - 5x g

23.

4x 4 3x - 2

29.

8r 3 dV = 3 dr 33. -

27.

30. k = 4 1 48

5 22 22 31. x - y - 4 = 0

34. a = -1, b = 2, c = 4

35. S = 8rr - 8r + 2rrh 36. (a) 6x 2 - 5 ] 3x - 1 g ] 3x - 5 g3

(b)

- ] 5x + 6 g

] x - 3 g4 2x + 1

4 ! 13 6

1 1 n (b) Q = d - 4 , 12 7 49

5x - 3 x+1

Practice assessment task set 2 - 0.77

(b) 22c 45l

5.

- 0.309

3 3 o, 12x - 12 3 y + 31 = 0 2

7.

m1 m2 =

20. (a) x = 90c, 270c

9.

7 12

1 1 3 , -1 , 1 2 2 5

26. a = - 14, b = 7

32. 4x - y - 13 = 0

1.

19. x =

x

So perpendicular

` (1, 1) is a point of intersection

11 18. e 1 , 12

22. 3x - 9y - 14 = 0

360c

38. (a) x + 7y - 80 = 0

So (1, 1) lies on the curve y =

17. n = 8

270c

24. (a) 16x + 32y + 1 = 0, 4x - 2y - 1 = 0 1 (b) m 1 $ m 2 = - # 2 2 = -1

37. x =

y=

180c

2. 1

3. 5x + 2y - 1 = 0

6. (a)

3 cm 2

(b) AC =

4. ^ 2, - 2 h 13 cm, BD = 1 cm

3 8 1 # - = -1; A = d -1, 1 n 4 6 2

8. x = 15c

815

816


10.

19. i = 120c, 240c 23. y = 16.5 26. 7

20. - 1

2 3

21. 2

24. 3x + y - 5 = 0

27. x = 3

22. a = 115c 56l

25. 1

2 1x 13 3

28. - 3

29. Show perpendicular distance from ^ 0, 0 h to the line is 2 units, or solving simultaneous equations gives only one solution. 30. (a) g ] 2 g = 1, g ] - 3 g = - 6 (b) 11.

31. 3x 2 - 4x

32. -

34. x = - 2, y = -17 12. 45c 49’ 13. Domain: all real x ! real y ! 0

1 ; range: all 2

14.

1

33. 17.5 m

2

35. (a) AB = 7.0 m

36. 3 cos i 37. (a) 2x - y + 4 = 0

(b) 27.8 m2

(b) P ^ - 2, 0 h, Q ^ 0, 4 h

2

(c) 4 units

39. 15 units2 40. f (- x) = ] - x g6 - ] - x g2 - 3 = x6 - x2 - 3 = f (x)

38. 127 m

41. 16x 2 ^ 2x 2 + 1 h + ^ 2x 2 + 1 h = ^ 18x 2 + 1 h ^ 2x 2 + 1 h 3

42. - 4

1 # y #9 3

4

43. -

44. (a) 3x - y - 4 = 0

15.

47. 50.

8 units 13 1 2x - 7 5 ] x + 1 g2

3 x2

(b) x - y - 2 = 0

(c) x + 3y + 10 = 0 45.

3

(d) R = ^ -10, 0 h

46. Domain: all x ! - 4; range: all y ! 0 48. 4.9 km 49. 8x - 7 - 10x - 3 51. 2x - 3

53. x + 6y - 56 = 0 55. a = 2, b = - 9

52.

-17 - 2x x 2 + 5x

=

- ] 17 + 2x g x 2 + 5x

54. f ] - 2 g = - 45, f l] - 2 g = 48 56. 7x - 5y + 9 = 0

57. 47x - y + 109 = 0 58. x = - 0.25 59. ^ 33, -17 h 16. sin 4 i 17. 2 units 18. x - 8y + 15 = 0

60.

3+1 2 2

=

6+ 2 4

62. x = 63c 26l, 243c 26l

61. 67c 37l

ANSWERS

63.

4.

(a) x = 112c, y = 56c, z = 34c (c) x = 55c, y = 43c

(b) x = 49c

(d) x = 166c, y = 7c

(e) x = 62c, b = 31c (f) x = y = 32c, z = 58c, v = 32c, w = 17c (h) y = 102c

(g) x = 5c

(i) x = 57c 30l, y = 32c 30l

(j) x = 75c, y = 77c, z = 13c 5. 64. (a) cos i (b) cos ^ i + b h

(c) tan 14a

65. 3

D ABC <; D DEC

67. 12c 32l at both points

66. x 1 4, x 2 4.6

1 range: y $ 0 2 (b) domain: all real x ! -7 range: all real y ! 0

(b) x = 5.5 cm

68. (a) domain: x $

(c) domain: - 2 # x # 2 69. a = -15, b = -1

6.

x = 30° (angle at centre is double the + at the circumference) y = (180° - 30°) ' 2 (+ sum of isosceles D) = 75c

7.

360° - x = 2 # 110° (+ at the centre is double the + at the circumference) ` x = 140° y = 70° (similarly)

8.

+ABC = 90c (+ in semicircle) ` +BAC = 90c - 29c (+ sum of D) = 61c ` x = 61c (+ in same segment)

9.

+STV = +WUV (+ in same segment) +TSV = +UWV (similarly) +TVS = +UVW (vertically opposite +s)

range: - 2 # y # 0

70. cos 2i

71. (a) (0, 0), (1, 3), (-1, -1), (2, 20) (b) 63c 26l at (0, 0), 2c 20l at (1, 3), 40c 36l at (-1, -1), 0c 22l at (2, 20) 72. (a) x = 360n ! 45c

(b) x = 180n + 30c

(c) x = 180n + ] -1 gn # 60c 73. (a) (1, 1) (b) 2 13 units

(c) - 1

1 2

(d) 3x + 2y - 5 = 0 74. (a)

75. (b), (d) 76. (a)

79. (b), (d)

77. (c)

` Since all pairs of angles are equal,

78. (c)

D STV ||| DWUV x = 2.4 cm

80. (c) 10.

+B = 90c (+ in semicircle) AC 2 = AB 2 + BC 2 = 62 + 32 = 36 + 9 = 45 AC = 45 =3 5 1 Radius = AC 2 3 5 = cm 2

11.

+OAC = 30c (Base +s of isosceles D) +BAO = 25c (similarly) ` +CAB = 30c + 25c = 55c x = 2 +CAB (+ at the centre is double the + at the circumference) = 2 # 55c = 110c

Chapter 9: Properties of the circle The proofs given as answers to this chapter are informal. Also, they may not be the only way to answer the question.

Exercises 9.1 1.

(a) i = 32c (b) x = 8 cm (c) i = a = 68c 30l (d) i = 31c (e) x = 9 mm

2.

16r cm 9

(f) i = 22c30l

3. (a) i = 29c

(c) a = 83c , b = 42c (f) y = 97c

(b) x = 18c

(d) x = 68c

(e) x = 10 cm

(g) x = 15c, y = 150c, z = 75c

(h) x = 47c, y = 43c, z = 94c (j) x = y = 39c

(a) +DCE = +ACB (vertically opposite +s) +EDC = +BAC (+s in the same segment) +DEC = +ABC (similarly) ` Since all pairs of +s are equal,

(i) b = 40c

817

818


12. (a) x = 52c, y = 76c (b) AC = BD (equal diameters) Diagonals are equal so ABCD is a rectangle. (opposite sides of a rectangle) ` AD = BC

6.

OB = 8.3 cm

8.

x Z 4.4 m, a = 78c, b = 38c, i = 64c

9.

OA = r

13. +ECB = 33c (angles in same segment) +EBC = 180 - ] 114 + 33 g (angle sum of triangle) = 33° `+ECB = +ADE These are equal alternate angles. ` AD < BC

AC = OC = = =

14. (a) +AOB = 90c (given) +ABC = 90c (angle in semi-circle) `+AOB = +ABC `+A is common `D AOB ||| D ABC (AAA) (Note 2 pairs of angles equal means 3 pairs will be equal by angle sum of triangle.) (b) AO = BO AB =

r +r 2

= =

(equal radii)

=

2 r2

= =

2 # r2 2r

=

x2 4 4r 2 x 2 4 4 2 4r - x 2 4 4r 2 - x 2 2 r2 -

4r 2 - x 2 2

2r + 4r 2 - x 2 2

So BC =

AC. CD = BC.CE

2r (angle at centre double

Reflex +BOD = 360 - 2i

(angle of revolution)

1 +BCD = +BOD 2

(angle at centre double

(angles in same segment)

1.

(a) x = 107c, y = 94c (b) i = 134c, c = 90c (c) x = 112c, y = 112c, z = 68c (d) x = 92c, y = 114c (e) b = 73c, a = 107c, c = 107c (f) x = 141c, y = 63c (g) x = 65c, y = 43c (h) w = 89c, x = 86c, y = 54c, z = 35c (i) w = 69c, x = 111c, y = 82c, z = 98c (j) x = 118c

2.

(a) x = 62c, y = 31c (b) x = 75c, y = 105c (c) x = 88c, y = 65c (d) x = 62c, y = 82c, z = 36c (e) x = 90c, y = 113c (f) x = 38c, y = 71c (g) x = 85c, y = 95c (h) x = 48c, y = 78c (i) x = 107c, y = 73c (j) a = 81c, b = 55c, c = 83c, d = 16c, e = 28c

3.

(a) +A = 180c - 58c

(+A and +B cointerior angles, AD ; BC)

+D = 180c - 58c

(+C and +D cointerior angles, AD ; BC)

angle at circumference)

So +BCD and +DAB are supplementary (add to180c)

Exercises 9.2 (a) x = 5 cm (b) y = 15 cm (c) x = 2.4 m (d) x = 42c (e) z = 90c (f) x Z 10.3 m (g) x = 6 m, y = 3 m (h) m Z 13.4 cm (i) y Z 5 cm (j) x = 5 mm 4. 25.6 cm

CE = 11.5 2 - 6.9 2 = 9.2 CD = 2 # 9.2 (perpendicular from O bisects chord) = 18.4 = AB

(AAA)

Exercises 9.3

1 = (360 - 2i) 2 = 180 - i

3. 144 mm


`DABC ||| DCDE (b) By similar triangles AC BC = CE CD

41 cm

(Pythagoras’ theorem)

+A = +E

angle at circumference)

5.

centre bisects a chord)

x 2 r -d n 2 2

But AO = BO so AB = BC

15. Obtuse +BOD = 2i

2.

(perpendicular from the

10. (a) +ECD = +ACB

By similar triangles AO BO = AB BC

1.

x 2

CD = r +

2

7. x = 4.7 m, y = 1.8 m

So +A = 180c - +C and +D = 180c - +B Since opposite angles are supplementary, ABCD is a cyclic quadrilateral. (b) +B = +D = 90c

(given)

` +B = 180c - +D Let +A = x +C = 360 - ] 90 + 90 + x g

(angle sum of quadrilateral)

ANSWERS

= 360 - 180 - x = 180 - x = 180 - +A Since opposite angles are supplementary, ABCD is a cyclic quadrilateral. (c)

+CDA = 180 - i

(straight angle)

` +B = 180c - +CDA

6.

(a) x = 67c (b) y Z 7.5 cm (c) x = 72c, y = 121c (d) x = 63c, y = 126c (e) x = 8.9 m, y Z 5.1 m (f) x = 63c, y = 63c (g) x = 98c, y = 65c, z = 17c (h) x = 57c, y = 57c (i) x = 72c, y = 15c (j) x = 61c, y = 70c, z = 52c

7.

(a) x = 26c, y = 74c, z = 48c (b) x = 68c, y = 44c, z = 68c (c) x = y = z = 45c (d) x = 70c, y = 31c (e) x = 20c, y = 57c, z = 103c (f) x Z 5.4 cm (g) x Z 7.7 cm (h) x = 77c, y = 13c (i) x Z 1.2 cm, y Z 2.1 cm (j) x = 55c, y = 112c, z = 57c

8.

AB Z 13 m

Let +A = x +C = 360 - ] 90 + 90 + x g (angle sum of quadrilateral) = 360 - 180 - x = 180 - x = 180 - +A Since opposite angles are supplementary, ABCD is a cyclic quadrilateral.

Exercises 9.4 1.

2.

3.

(a) i = 47c (b) x = 5 m (c) y = 11.3 cm (d) x = y = 26c (e) a = 64c, b = 32c (f) i = 57c (g) p = 145 Z 12 cm (h) y = 10 mm (i) x Z 5.79 cm (j) x = 33c, y = 33c (a) x = 10 cm (b) x = 64c, y = 26c (c) x = 13 cm (d) x = 27c, y = 54c (e) y = 5 cm (f) x = 32c, y = 7c (g) x = 72c, y = 42c (h) x = 35c, y = 90c (i) m = 23c, n = 67c, p = 67c, q = 23c (j) x = 71c, y = 62c +OAB = 90c (tangent = to radius) z = 90c - 48c (+ sum of D AOB) = 42c (equal radii) OA = OC ` +OAC = +OCA = y

Test yourself 9 1.

i = 56c

4.

x = y = 12 cm

5.

z = 19c (+s in same segment) y = 180c - (131c + 19c ) (+ sum of D ) = 30c x = 30c (+s in same segment)

6.

x = 10 cm

7.

a = 3c , b = 44c , c = 136c

(+ at centre twice + at circumference)

= 50c +OCA = 90c (tangent perpendicular to radius) b = 90c - 83c ` = 7c OC = OE (equal radii) ` D OCE is isosceles

(base +s of isosceles D )

y = (180c - 48c) ' 2 (+ sum of D OAC)

= 66c +ACD = 180c - +AED

`

+OCE = + OEC = c 2c + 100c = 180c (+ sum of D) 2c = 80c c = 40c Reflex +COE = 360c - 100c (+ of revolution) = 260c d = 360c - (260c + 50c + 7c)

(opposite +s of cyclic quad.)

= 180c - 62c = 118c = 52c = +OAB - +OAC = 90c - 66c = 24c 1 v = +AOC 2

y+u 66c + u u +BAC ` x

(+ at centre twice + at circumference)

1 # 48c 2 = 24c =

4.

21 cm

5.

AC 2 + BC 2 = 3.9 2 + 5.2 2 = 42.25 AB 2 = 6.5 2 = 42.25 ` AB 2 = AC 2 + BC 2 ` +ACB = 90c (by Pythagoras’ theorem) ` A lies on a diameter of the circle (tangent ⊥ radius)

3. x = 7.2 m

1 # 100c 2

a=

8.

`

`

2. y = 2.3 mm

(+ sum of quadrilateral)

= 43c

9.

17 cm

10. 5.3 m

12. a = 61c, b = 29c 15. 18 cm

11. a = 101c, b = 98c 13. 14.9 cm

14. x = 4.9 m

16. a = 127c, b = 53c

17. +D = 180c - (80c + 53c) (+ sum of T) = 47c ` y = 47c (+s in same segment) x = 47c (+s in alternate segment) 18. x = 55c, y = 56c, z = 54c

819

820


6.

19. +C is common +A = +CBD (+s in alternate segment)

Let +ODC = x and +OAB = y. Then you can find all these angles

(giving reasons).

` D BCD ||| D ABC ] AAA g 20. (a) +OCB = +OCA = 90c (given) OA = OB (equal radii) OC is common ` DOAC / DOBC ] RHS g (b) AC = BC (corresponding sides in / D s) ∴ OC bisects AB


6 cm

2.

Let +DOB Then +EDO EO ` +OED

= +DCB = x (base +s of isosceles D ODC) = 2x (ext. + of D) = DO (equal radii) = +EDO = 2x (base of+s of isosceles D EOD)

+AOC + +COB + +BOD + +AOD = 360c (+ of revolution)

90c - y + x + +COB + y + 90c - x + +AOD = 360c +COB ++AOD + 180c = 360c ` +COB ++AOD = 180c

+EOD = 180c - (+OED + +EDO) (+ sum of D EOD) = 180c - 4x +AOE = 180c - (+EOD + +DOB) (+AOC straight +)

= 180c - (180c - 4x + x) = 3x ` +AOE = 3+DCB 3.

Let+DAB Then+DAC +ACB +ADB +DBA +CBA +DBA + +CBA

7.

= x and +CAB = y =x+y = +DAB = x (+s in alternate segment) = +CAB = y (similarly) = 180c - (x + y) (+ sum of D ADB) = 180c - (x + y) (+ sum of D ACB) = 180c (DBC is straight +)

D Let ABCD be a kite with AB = AD and BC = DC, and +ADC = +ABC = 90°. AC is common.

(a) AD = DB = BE = EC = CF = FA (equal radii)

∴ by SSS (or RHS) D ABC / D ADC

` AB = BC = CA `D ABC is equilateral

`+BAC = +DAC and +BCA = +DCA (corresponding 1s in congruent D s)

(b) rr units (c) 5.

3 r2 -

Let +BAC = +DAC = a Then +BAD = 2a +BCA = +DCA = 90c - a (+sum of D ) +BCD = 180c - 2a ` Opposite angles are supplementary.

2 3-r 1 2 o units 2 rr = r 2 e 2 2

+BDE = +ABD + +BAD (ext. + of D BAD) ` 2a = +ABD + a a = +ABD ` D BAD is isosceles with AD = BD +CDE = +ACD + +CAD (ext. +of D CAD ) ` 2b = +ACD + b b = +ACD ` DCAD is isosceles with AD = CD ` AD = BD = CD So a circle can be drawn through A, B and C with centre D.

C

A

` 180c - (x + y) + 180c - (x + y) = 180c 180c = 2 (x + y) ` 90c = x + y ` ` +DAC = 90c 4.

B

∴ ABCD is a cyclic quadrilateral, and A, B, C and D are concyclic points Since +ABC = 90c, AC is a diameter. (+ in semicircle) 8.

25rr 2 units 2 28

ANSWERS

9.

Now ABCE is a cyclic quadrilateral, so

Let interval AB subtend angles of x at +ADB and +ACB.

+AEC + +B = 180c (opposite +s supplementary) Also, +D + +B = 180c (given) +D = +AEC These are equal corresponding angles, so DA < EA (this is impossible!) ∴ A, B, C and D must be concyclic ∴ ABCD is a cyclic quadrilateral.

Chapter 10: The quadratic function Assume A, B, C and D are not concyclic. Draw a circle through A, B and C that cuts AD at E.

Then +AEB = +BCA = x (+s in same segment) But +AEB and +EDB are equal corresponding angles. ` EB || DB (this is impossible!) ∴ A, B, C, D must be concyclic 10. Let ABCD be a quadrilateral with opposite angles supplementary. i.e. +A + +C = 180c and +B + +D = 180c Assume the points are not concyclic. Draw a circle through A, B and C, cutting CD at E.

Exercises 10.1 1.

Axis of symmetry x = - 1, minimum value - 1

2.

Axis of symmetry x = - 1.5, minimum value - 7.5

3.

Axis of symmetry x = - 1.5, minimum value - 0.25

4.

Axis of symmetry x = 0, minimum value - 4

5.

Axis of symmetry x =

6.

Axis of symmetry x = 1, maximum value -6

7.

Axis of symmetry x = - 1, maximum point ^ - 1, 7 h

8.

Minimum value -1, 2 solutions

9.

Minimum value 3.75, no solutions

3 7 3 n , minimum point d , 8 16 8

10. Minimum value 0, 1 solution 11. (a) x = -3; (-3, -12) 1 1 1 (c) x = 1 ; d 1 , 3 n 4 4 8

(b) x = -4; (-4, 17) 1 1 1 (d) x = -1 ; d -1 , -13 n 4 4 4

(e) x = -3; ^ -3, -23 h 12. (a) (i) x = -1 (b) (i) x = 1

(ii) -3

(iii) (-1, -3)

(ii) 1

(iii) (1, 1)

821

822


13. (a) Minimum (-1, 0) (b) Minimum (4, -23) (c) Minimum (-2, -7) (d) Minimum (1, -1) (e) Minimum (2, -11) 1 1 (f) Minimum d - , -3 n 4 8

(c) (i) 5.83, 0.17

(ii) Minimum -8 y

(iii) 10 8 6

(g) Maximum (-1, 6)

4

(h) Maximum (2, 11) 1 3 (i) Maximum d , 7 n 2 4

2 1

-4 -3 -2 -1 -2

(j) Maximum (1, -3)

2

3

4

5

x

6

-4 14. (a) (i) -2 (iii)

-6

(ii) Minimum 0 y

-8 -10

5 4 3

(d) (i) -2, 0

2

y

(iii)

1 -4 -3 -2 -1 -1

5

x

2

1

4

-2

3

-3

2 1

(b) (i) -1, 3 (ii) Minimum -4

-2

5

-3

4

(e) (i) ! 3

3 2

-2 -3

(ii) Minimum -18 y

(iii)

1

-4 -3 -2 -1 -1

x

2

1

-4 -3 -2 -1 -1

y

(iii)

(ii) Minimum -1

1

2

3

4

2

x

1 -4 -3 -2 -1 -2

1

-4

-4

-6

-5

-8 -10 -12 -14 -16 -18 (f) (i) -1,

2 3

(ii) Minimum - 2

1 12

2

3

4

5

x

ANSWERS

y

(iii)

(i) (i) 0.56, -3.56

5

(ii) Minimum 4

1 4

y

(iii)

4

41 4

5

3

4

2

3

1 1

-4 -3 -2 -1 -1

3

2

4

2

x

5

1

2 3

-2

-4 -3 -2 -1 -1

-2 1 -3 12 -4

1

2

3

x

5

4

-2 -3

-5 -6 (j) (i) 2.87, -0.87 (g) (i) 1.65, -3.65 (ii) Maximum 7

y

(iii)

y

(iii)

(ii) Maximum 7

7 6

7

5

6

4

5

3

4

2

3

1

2

1

-4 -3 -2 -1 -1

1 2

1

-4 -3 -2 -1 -1

3

4

5

x

2

3

4

5

x

-2 -3

-2 -3 15. (a) 4

(h) (i) 1.3, -2.3 (ii) Maximum 3

(c)

1 4

y 7

y

(iii)

(b) None

6

5

5

4

3

1 4

4

3

3

2

2 1

1

-4 -3 -2 -1 -1 -2 -3

1

2

3

4

5

x

-4 -3 -2 -1 -1 -2 -3

1

2

3

4

5

x

823

824


16. (a) None

(b) 6

y

19.

3 4

8

y

(c)

6

14

4

12

2

10 8

-4

6

1

-3 -2 -1 -2

3

2

4

x

5

-4

4 2

-6 2

1

-4 -3 -2 -1 -1

3

4

x

5

Graph is always above the x-axis so y 2 0 for all x ` 3x 2 - 2x + 4 2 0 for all x

-2

y

20.

-3

8

7 17. (a) - 3 8

6

(b) None

4

y

(c)

2

2

1

-4 -3 -2 -1 -2

1 1

-4 -3 -2 -1 -2

2

3

4

x

5

2

3

4

5

x

-4 -6

-4 -6

Graph is always above the x-axis so y 2 0 for all x ` x 2 + x + 2 2 0 for all x

-8 -10

4

-14

2

-16 -18

-4 -3 -2 -1 -2

y

18. (a)

y

21.

-12

2

3

4

5

-4

8

-6

6

-8

4

-10 -12

2

-4 -3 -2 -1 -1

1

1

-2

2

3

4

5

x

-14 -16 -18

-3 (b) x 1 2, x 2 3

(c) 2 # x # 3

Graph is always below the x-axis so y 1 0 for all x ` - x 2 + 2x - 7 1 0 for all x

x

ANSWERS

22.

y

8.

2 1

-4 -3 -2 -1 -1

3

2

1

4

5

9.

-6

1.

x 1 -3, x 2 3

2. - 1 # n # 0

4.

x 1 - 2, x 2 2

5. 0 # y # 6

7.

x 1 - 4, x 2 2

8. p # - 3, p $ - 1

3. a # 0, a $ 2 6. 0 1 t 1 2

1 11. 1 1 h 1 2 2 14. q 1 3, q 2 6

15. All real x

17. - 3 1 x 1 5

1 19. y 1 - , y 2 5 3

20. x # - 2, x $ 4

1 3

30. - 2

24. -

18. - 6 # t # 2

29. x #

21. -

1 1x 10 2

1 #x 10 2

26. x $ - 1, x 1 - 2

28. x 1 - 6, x 2 - 3

9. m 1 2, m 2 4

12. - 4 # x # 5

16. n # - 4, n $ 3

25. 1 1 x 1 1

1 2

(1) (2)

x 2 + 3 = 2x + 6 x 2 - 2x - 3 = 0 b 2 - 4ac = ] - 2 g2 - 4 ] 1 g ] - 3 g = 16 20 So there are 2 points of intersection

Exercises 10.2

23. 0 1 x # 1

14. 0 # b # 2

Substitute (2) in (1):

Graph is always below the x-axis so y 1 0 for all x ` - 5x 2 + 4x -1 1 0 for all x

1 3

3

11. m 1 - 3, m 2 3

y = x2 + 3

-7

22. 0 1 x 1

13. p 1 -

1

16. Solving simultaneously: y = 2x + 6

-5

1 #k #7 2

10. 0 1 k 1 4

15. p # - 2, p $ 6

-4

13. - 2

k # - 5, k $ 3

12. k # - 1, k $ 1

-3

10. x # - 3, x $ 2

b 2 - 4ac = ] - 1 g2 - 4 ] 3 g ] 7 g = - 83 10 So 3x 2 - x + 7 2 0 for all x

x

-2

a =320

27. 2 1 x # 2

2 5

2 ,x 21 3

2 # x 1 -2 2

Exercises 10.3 1.

(a) 20 (b) -47 (c) -12 (d) 49 (e) 9 (h) 64 (i) 17 (j) 0

(f) -16 (g) 0

2.

(a) 17 unequal real irrational roots (b) -39 no real roots (c) 1 unequal real rational roots (d) 0 equal real rational roots (e) 33 unequal real irrational roots (f) -16 no real roots (g) 49 unequal real rational roots (h) -116 no real roots (i) 1 unequal real rational roots (j) 48 unequal real irrational roots

3.

1 7 6. p 2 2 7. k 2 - 2 p = 1 4. k = ! 2 5. b # 12 8

17. 3x + y - 4 = 0 y = x 2 + 5x + 3 From (1): y = - 3x + 4 Substitute (2) in (3): x 2 + 5x + 3 = - 3x + 4 x 2 + 8x - 1 = 0 b 2 - 4ac = 8 2 - 4 ] 1 g ] - 1 g = 68 20 So there are 2 points of intersection

(1) (2)

18. y = - x - 4 y = x2 Substitute (2) in (1): x2 = - x - 4 2 x +x+4=0 b 2 - 4ac = 1 2 - 4 ] 1 g ] 4 g = - 15 10 So there are no points of intersection

(1) (2)

19. y = 5x - 2 y = x 2 + 3x - 1 Substitute (2) in (1): x 2 + 3x - 1 = 5 x - 2 x 2 - 2x + 1 = 0 b 2 - 4ac = ] - 2 g2 - 4 ] 1 g ] 1 g =0 So there is 1 point of intersection ` the line is a tangent to the parabola

(1) (2)

20. p = 3

1 4

21. (c) and (d)

(3)

825

826


Exercises 10.4 1.

2.

(a) a = 1, b = 2, c = -6 (b) a = 2, b = -11, c = 15 (c) a = 1, b = 1, c = - 2 (d) a = 1, b = 7, c = 18 (e) a = 3, b = -11, c = -16 (f) a = 4, b = 17, c = 11 (g) a = 2, b = -12, c = -9 (h) a = 3, b = - 8, c = 2 (i) a = - 1, b = 10, c = - 24 (j) a = - 2, b = 0, c = - 1

20. (a) m = 1

(b) m 1

x 2 - 4x + 5 = x ] x - 2 g - 2 ] x + 1 g + 3 + 4

4.

RHS = a ] x - 2 g ] x + 3 g + b ] x - 2 g + c = 1 ] x - 2 g ] x + 3 g + 1 ] x - 2 g + 17 = x 2 + 3x - 2x - 6 + x - 2 + 17 = x 2 + 2x + 9 = RHS ` true

1.

(a) x = -1, - 4 (b) y = 2, 5 (c) x = - 4, 2 (d) n = - 1, 4 (e) a = - 3, 5 (f) p = 3, 4 (g) x = 2, - 4 (h) k = 5, 12 (i) t = 6, - 4 (j) b = -12, - 4

2.

(a) x = - 2, 3

3.

6. a = 2, b = 1, c = - 1

7.

K = 1, L = 6, M = 7.5 8. 12 ] x + 5 g + ] 2x - 3 g - 65 - 2

9.

a = 0, b = - 4, c = - 21

2

(b) y = x 2 - 3x (d) y = x 2 + 4x - 9

(a) a + b = - 2, ab = 1 (b) a + b = 1.5, ab = - 3 (c) a + b = 0.2, ab = - 1.8 (d) a + b = - 7, ab = 1 2 (e) a + b = 2 , ab = 1 3

m = 0.5

5. k = - 32

9.

k = -5

10. m = ! 3

x = ! 1, !2

7.

x = ! 2.19, !0.46, !1.93, !0.52

8.

(a) x = 0c , 90c , 180c , 360c (b) x = 90c , 180c , 270c (c) x = 90c , 210c , 330c (d) x = 60c , 90c , 270c , 300c (e) x = 0c , 180c , 270c , 360c

9.

(a) x = 0c , 45c , 180c , 225c , 360c (b) x = 0c , 180c , 360c (c) x = 0c , 30c , 150c , 180c , 360c (d) x = 45c , 60c ,135c , 120c , 225c , 240c , 315c , 300c (e) x = 30c , 60c , 120c , 150c , 210c , 240c , 300c , 330c

10.

x+3+

6. b = 4

7. k = 1

11. k = - 1

2

2 =5 x+3

14. b = - 6, c = 8

15. a = 0, b = - 1

(c) k = - 1.8

(b) p # - 2 3 , p $ 2 3

So u has 2 real irrational roots. ` x + 3 and so x has 2 real irrational roots

8. p = 13

12. n = - 1, 3

(e) k # - 1, k $ 0

3 3

6. x = - 1

u 2 - 5u + 2 = 0 b 2 - 4ac = ] - 5 g2 - 4 ] 1 g ] 2 g = 17 20

1 16. ab = 1 ` b = a

(c) p = !

(e) a = - 2, - 2 ! 6

Let u = x + 3

4.

18. (a) p = ! 2 3

1! 5 2

5.

(d) 21

(b) k = - 1, 0

(c) x =

2 # (x + 3) = 5 # (x + 3) ]x + 3g ] x + 3 g2 + 2 = 5 ] x + 3 g ] x + 3 g2 - 5 ] x + 3 g + 2 = 0

(a) x 2 + 3x - 10 = 0 (b) x 2 - 4x - 21 = 0 (c) x 2 + 5x + 4 = 0 (d) x 2 - 8x + 11 = 0 (e) x 2 - 2x - 27 = 0

17. (a) k = - 1

(b) y = ! 2, ! 2

(x + 3) # (x + 3) +

3.

13. p = 2, r = - 7

(a) x = ! 3

(d) x = 3, 5

(a) x = 0, 3 (b) p = 1 (c) x = 1 (d) x = 1 (e) x = 1, 3

Exercises 10.5

(c) - 0.5

(c) x = 4, 5

4.

(e) y = - x 2 - 2x + 1

(a) 3 (b) - 6

(b) x = 2, 3

(d) x = 1.37, - 4.37, 0.79, - 3.79

A = 1, B = 5 , C = - 6

2.

3 - 10 3 + 10 ,m2 2 2

1 (e) x = 1 , 4 2

5.

1.

(c) k = 2

Exercises 10.6

3.

(c) y = 2x 2 - 3x + 7

(b) k = - 3

(c) m = - 3

m = 2, p = - 5, q = 2

10. (a) y = x 2 - x - 5

19. (a) k = 2

(d) k = 3

Test yourself 10 1.

(a) 0 # x # 3

(b) n 1 - 3, n 2 3

2.

a = 1, b = - 9, c = 14

4.

a =120 D = b 2 - 4ac = ] -2 g 2 -4 # 1 # 7 = - 24 10 ` positive definite

3. (a) x = 2

(c) - 2 # y # 2 (b) - 3

ANSWERS

2 1 (d) 18 (e) 30 6. x = 1 , 3 3

5.

(a) 6 (b) 3

7.

(a) iv (b) ii (c) iii

8.

a = -1 1 0 D = b 2 - 4ac = 3 2 - 4 # (-1) # (- 4) = -7 10

(c) 2

(d) ii

(e) i

` - 4 + 3x - x 2 1 0 for all x 9.

(a) x = -

1 4

(b) 6

1 8

10. 3 ] x - 2 g2 + 12 ] x + 3 g - 41 12. (a) k = 3

1 4

1 13. x = - , 3 2

(b) k = 1

(c) k = 3

14. m 1 -

16. (a) i (b) i (c) iii

(d) i

17. (a) iii (b) i (c) i

(d) ii

18. For reciprocal roots b ab 1 a LHS a

11. x = 30c , 150c , 270c

9 16

(d) k = 3

(e) k = 2

19. (a) x + 3x - 28 = 0

1 1 (c) 1 y 1 5 3

(e) ii

y = x 2 - 5x + 4

5.

11

9.

x = !1

(b) x - 10x + 18 = 0

(b) n # - 3, n 2 3

6.

The straight line - 2 1 x 1 2 or | x | 1 2

7.

A circle, centre the origin, radius 2 (equation x2 + y2 = 4 i

8.

lines y = !1

4. A (parabolic) arc

5. A spiral

9. lines x = !5

10. line y = 2

11. Circle x 2 + y 2 = 1 (centre origin, radius 1) 12. Circle, centre ^ 1, -2 h, radius 4

13. y = -5 15. x = -7

16. x = 3

18. x = !4

1.

x2 + y2 = 1

2. x 2 + 2x + y 2 + 2y - 79 = 0

3.

x 2 - 10x + y 2 + 4y + 25 = 0

5.

12x - 26y - 1 = 0

7.

3x 2 - 32x + 3y 2 - 50y + 251 = 0

8.

5x 2 - 102x + 5y 2 + 58y - 154 = 0

9.

x 2 - 4x + 20y - 36 = 0

11. y 2 + 8x - 32 = 0

1 (d) x # - 10, x 2 - 2 2

4. 8x - 6y + 13 = 0

6. y = ! x

(e) 4 1 x # 7

13. x 2 + 12y = 0

10. x 2 - 20y = 0

12. x 2 - 2x + 8y - 7 = 0

14. x 2 - 5x + y 2 - 2y - 11 = 0

15. x 2 + 3x + y 2 - y - 4 = 0

3. a = 4, b = - 3, c = 7 7. p 2 0.75

4. x = ! 2

8. Show D = 0

16. x 2 + x + y 2 - 2y - 17 = 0 17. 2x 2 + 4x + 2y 2 - 6y + 47 = 0 18. 2x 2 + 2x + 2y 2 + 4y + 27 = 0 19. 3x + 4y + 25 = 0, 3x + 4y - 15 = 0 20. 12x - 5y - 14 = 0, 12x - 5y + 12 = 0

4x + 1

3 1 = + 11. 2 x-2 x+1 x -x-2 1 - 21 1 + 21 ,k$ 2 2

13. x = 30c , 90c , 150c

An arc

Exercises 11.2

10. A = 2, B = - 19, C = 67 or A = - 2, B = 13, C = - 61

12. k #

3.

2

4 7

6. n = - 2.3375

2. A straight line parallel to the ladder.

20. Circle, centre ^ -4, 5 h, radius 1

1 = a c = a k = k = RHS = 1

D = ] k - 4 g2 $ 0 and a perfect square ∴ real, rational roots

2.

A circle

19. Circle, centre ^ -2, 4 h, radius 6


1.

17. y = !8

20. x = 1, 3 21. (a) x 1 - 1, x 2 -

Exercises 11.1

14. Circle, centre (1, 1), radius 3

15. x = 0, 2

∴ roots are reciprocals for all x. 2

Chapter 11: Locus and the parabola

3! 5 14. x = 1, 2

15. x = 60c , 90c , 270c , 300c

16. - 23

21. x - 2y - 3 ! 5 5 = 0 22. x - 7y + 9 = 0, 7x + y - 5 = 0 23. 7x - 4y - 30 = 0, 32x + 56y - 35 = 0 24. xy - 16x - 7y + 40 = 0 25. x 2 - 6x - 3y 2 - 12y + 9 = 0

827

828


Problem

23.

12x + 5y - 40 = 0, 12x + 5y + 38 = 0

Exercises 11.3 1.

(a) Radius 10, centre (0, 0) (b) Radius 5 , centre (0, 0) (c) Radius 4, centre (4, 5) (d) Radius 7, centre (5, −6) (e) Radius 9, centre (0, 3)

2.

(a) x 2 + y 2 = 16 (b) x 2 - 6x + y 2 - 4y - 12 = 0 (c) x 2 + 2x + y 2 - 10y + 17 = 0 (d) x 2 - 4x + y 2 - 6y - 23 = 0 (e) x 2 + 8x + y 2 - 4y - 5 = 0 (f) x 2 + y 2 + 4y + 3 = 0 (g) x 2 - 8x + y 2 - 4y - 29 = 0 (h) x 2 + 6x + y 2 + 8y - 56 = 0 (i) x 2 + 4x + y 2 - 1 = 0 (j) x 2 + 8x + y 2 + 14y + 62 = 0

3.

x 2 - 18x + y 2 + 8y + 96 = 0

4.

x 2 + 4x + y 2 + 4y - 8 = 0

6.

x 2 + 6x + y 2 - 16y + 69 = 0

7.

x 2 - 10x + y 2 + 4y + 27 = 0

9.

x 2 - 2x + y 2 - 10y + 25 = 0

26. Perpendicular distance from centre ^ 0, 0 h to the line is equal to the radius 2 units; perpendicular distance from centre ^ -1, 2 h to the line is equal to the radius 3 units. 27. (a) x 2 + 2x + y 2 - 6y - 15 = 0 (b) ^ 2, 7 h, ^ -1, -2 h (c) Z = ^ -1, 8 h 1 (d) m zx # m yx = - # 3 3 = -1 ` +ZXY = 90c

Exercises 11.4 1.

8. x 2 + y 2 - 9 = 0

(b) x 2 = 36y

(c) x 2 = 4y

(e) x 2 = 40y

(f) x 2 = 12y

(g) x 2 = 24y

(i) x = 8y 2.

12. x 2 + y 2 + 6y + 1 = 0

13. (a) Radius 3, centre (2, 1) (b) Radius 5, centre (−4, 2) (c) Radius 1, centre (0, 1) (d) Radius 6, centre (5, −3) (e) Radius 1, centre (−1, 1) (f) Radius 6, centre (6, 0) (g) Radius 5, centre (−3, 4) (h) Radius 8, centre (−10, 2) (i) Radius 5, centre (7, −1) (j) Radius 10 , centre (−1, −2)

(a) x 2 = -4y

(b) x 2 = -12y

(d) x = -28y

(e) x = -24y

(g) x 2 = -32y

(h) x 2 = -8y

2

4.

(a) (i) (0, −1) (ii) y = 1 (b) (i) (0, −6) (ii) y = 6 (c) (i) (0, −2) (ii) y = 2 (d) (i) (0, −12) (ii) y = 12 (e) (i) (0, −5) (ii) y = 5 (f) (i) (0, −4) (ii) y = 4 (g) (i) (0, −8) (ii) y = 8 (h) (i) (0, −10) (ii) y = 10

6.

22. x 2 + 2x + y 2 + 2y - 23 = 0

(i) x 2 = -60y

(a) (i) (0, 1) (ii) y = -1 (b) (i) (0, 7) (ii) y = -7 (c) (i) (0, 4) (ii) y = -4 (d) (i) (0, 9) (ii) y = -9 (e) (i) (0, 10) (ii) y = -10 (f) (i) (0, 11) (ii) y = -11 1 1 (g) (i) (0, 3) (ii) y = -3 (h) (i) c (0, 1 m (ii) y = -1 2 2 3 1 1 (i) (i) c 0, 2 m (ii) y = -2 (j) (i) c 0, 3 m 4 2 2 3 (ii) y = -3 4

5.

21. (a) Both circles have centre ^ 1, -2 h (b) 1 unit

(f) x 2 = -36y

3.

1 1 (i) (i) c 0, - m (ii) y = 2 2

20. Show perpendicular distance from the line to ^ 4, -2 h is 5 units, or solve simultaneous equations.

(c) x 2 = -16y

(j) x = -52y

17. Centre (4, 7), radius 8

19.

(h) x 2 = 44y

2

2

15. Centre ^ 2, 5 h , radius 5

1 1 18. Centre d - 1 , 1 n , radius 2 2 2

(d) x 2 = 16y

(j) x = 48y

2

11. x 2 - 8x + y 2 - 6y + 22 = 0

16. Centre ^ - 1, -6 h , radius 7

(a) x 2 = 20y 2

2

14. Centre ^ 3, -1 h , radius 4

(b) x 2 - 4x + y 2 + 10y + 13 = 0

28. (a) 4 units

5. x 2 - 2x + y 2 - 48 = 0

34 units

25. (a) 5 units (b) 3 units and 2 units (c) XY is the sum of the radii. The circles touch each other at a single point, ^ 0, 1 h .

10. x + 12x + y - 2y + 1 = 0 2

24.

56 units

1 1 (j) (i) c 0, -5 m (ii) y = 5 2 2

(a) x 2 = 28y (b) x 2 = 44y (c) x 2 = -24y (d) x 2 = 8y (e) x 2 = !12y (f) x 2 = !32y 1 (g) x 2 = 32y (h) x 2 = y 7 (a) Focus ^ 0, 2 h, directrix y = -2, focal length 2 (b) Focus ^ 0, 6 h, directrix y = -6, focal length 6

(c) Focus ^ 0, -3 h, directrix y = 3, focal length 3 1 1 1 (d) Focus d 0, n, directrix y = - , focal length 2 2 2 3 3 3 (e) Focus d 0, -1 n, directrix y = 1 , focal length 1 4 4 4 1 1 1 (f) Focus d 0, n, directrix y = - , focal length 8 8 8

ANSWERS

7.

y =2

8. ^ 4, 4 h

9.

3 1 X = d -1 , - n 2 8

10. ^ 4, -2 h and ^ -4, -2 h ; 8 units 11. (a) x 2 = - 12y

(b) y = 3

(c) 33

7.

x = 4 (latus rectum)

9.

^ 9, - 6 h, ^ 81, 18 h

(d) 4

2 units 13

(e) 11.7 units2

Exercises 11.6

13. (a) x - 4y + 2 = 0 (b) ^ 0, 1 h does not lie on the line (c) x 2 - 4x + y 2 - 2y + 1 = 0 (d) Substitute ^ 0, 1 h into the equation of the circle. 14. (a) Substitute Q into the equation of the parabola. (b) _ q 2 - 1 i x - 2qy + 2aq = 0 (c) Equation of latus rectum is y = a. Solving with x 2 = 4ay gives two endpoints A ^ -2a, a h, B ^ 2a, a h . Length of AB = 4a.

1.

(a) ] x - 3 g2 = 8 ^ y + 3 h

(b) ] x - 5 g2 = 4 ^ y + 6 h

(c) ] x - 1 g = 4 ^ y + 3 h (d) ] x - 4 g2 = -12 ^ y - 3 h (e) ] x - 6 g2 = 8 ^ y + 7 h (f) ] x + 7 g2 = -16 ^ y - 3 h (g) ] x - 2 g2 = -4 ^ y - 5 h (h) ] x + 9 g2 = 12 ^ y + 6 h 2

(i) ] x + 1 g2 = - 4 ^ y - 2 h

2.

(j) ] x - 3 g2 = 8 ^ y + 1 h

(a) ^ y - 4 h2 = 4 ] x + 4 g (b) ^ y - 1 h2 = 8 ] x + 2 g (c) ^ y + 2 h2 = 12 ] x + 1 g (d) ^ y - 10 h2 = - 4 ] x - 29 g (e) ^ y + 3 h2 = - 16 ] x - 1 g

(f) ^ y - 6 h2 = 8 ] x + 4 g

(g) ^ y + 5 h2 = - 24 ] x - 2 g (h) ^ y + 12 h2 = 4 ] x + 36 g

Exercises 11.5

2.

5 1 units2 (b) d - 5, - 4 n (c) 10 6 12

10. (a) 5x - 12y - 25 = 0

1 units 3

12. (a) Substitute the point into the equation. 3 (b) 3x + 4y - 3 = 0 (c) d 2, - n 4

1.

8. 12, ^ 3, 6 h, ^ 3, -6 h

(a) y 2 = 8x (b) y 2 = 20x (c) y 2 = 56x (d) y 2 = 36x (e) y 2 = 32x (f) y 2 = 24x (g) y 2 = 28x (h) y 2 = 12x (i) y 2 = 16x (j) y 2 = 4x

(i) ^ y - 2 h2 = - 20 ] x - 1 g (j) ^ y + 4 h2 = - 8 ] x - 2 g 3.

2

(b) x 2 + 8x - 4y + 16 = 0

(c) x - 4x - 8y - 12 = 0

(d) x 2 - 6x - 8y + 41 = 0

2

(a) y = -36x (b) y = - 16x (c) y = -40x (d) y 2 = -24x (e) y 2 = - 8x (f) y 2 = -48x (g) y 2 = - 44x (h) y 2 = -20x (i) y 2 = -12x (j) y 2 = -28x 2

(a) x 2 + 2x - 8y + 9 = 0

2

(e) x 2 + 4x - 16y + 20 = 0

(f) x 2 + 2x + 16y + 1 = 0

(g) x - 8x + 20y - 24 = 0

(h) x 2 + 10x + 8y + 1 = 0

(i) x 2 + 6x + 12y + 45 = 0

(j) x 2 + 4y + 24 = 0

2

(k) y - 6y - 12x - 3 = 0

(l) y 2 - 8y - 4x + 8 = 0

2

3.

4.

5.

(a) (i) (2, 0) (c) (i) (4, 0) (e) (i) (7, 0) (g) (i) (6, 0) 1 (i) (i) c , 0 m 4

= -2 = -4 = -7 = -6 1 (ii) x = 4

(ii) x (ii) x (ii) x (ii) x

(a) (i) (−2, 0) (ii) x = 2 (b) (i) (−3, 0) (ii) x = 3 (c) (i) (−7, 0) (ii) x = 7 (d) (i) (−1, 0) (ii) x = 1 (e) (i) (−6, 0) (ii) x = 6 (f) (i) (−13, 0) (ii) x = 13 1 1 (g) (i) (−15, 0) (ii) x = 15 (h) (i) c - , 0 m (ii) x = 2 2 1 1 1 1 (i) (i) c - 6 , 0 m (ii) x = 6 (j) (i) c - 1 , 0 m (ii) x = 1 4 4 2 2 (a) y 2 = 20x

(b) y 2 = 4x

(e) y 2 = !36x 6.

(b) (i) (3, 0) (ii) x = -3 (d) (i) (1, 0) (ii) x = -1 (f) (i) (8, 0) (ii) x = -8 (h) (i) (9, 0) (ii) x = -9 1 1 (j) (i) c 4 , 0 m (ii) x = -4 2 2

(c) y 2 = -16x

(f) y 2 = !8x

(m) y 2 - 8x + 32 = 0

(o) y + 2y - 8x - 7 = 0

(h) y 2 =

(s) y - 4y + 2x + 5 = 0 4.

(a) (i) (3, −2)

(ii) y = -4

(b) (i) (1, 1)

(ii) y = - 3

(c) (i) (−2, 0)

(ii) y = -2

(d) (i) (4, 2)

(ii) y = - 4

(e) (i) (−5, −1)

(ii) y = -5

(f) (i) (3, 1)

(g) (i) (−1, 0)

(ii) y = 4

(h) (i) (2, 0)

(i) (i) (4, −2)

(ii) y = 4

(j) (i) (−2, −3)

(ii) y = 3

(ii) y = 2 (ii) y = 5

5.

(a) (i) (0, −1) (ii) x = -2 (b) (i) (2, 4) (ii) x = - 4 (c) (i) (0, 3) (ii) x = -4 (d) (i) (3, −2) (ii) x = -5 (e) (i) (7, 1) (ii) x = -5 (f) (i) (1, −5) (ii) x = 5 (g) (i) (11, −7) (ii) x = 13 (h) (i) (−3, 6) (ii) x = 7 1 1 (i) (i) (−7, 2) (ii) x = 9 (j) (i) c -10 , -3 m (ii) x = 9 2 2

6.

x 2 - 12y + 36 = 0

(a) Focus ^ 2, 0 h, directrix x = - 2, focal length 2

(c) Focus ^ -3, 0 h, directrix x = 3, focal length 3

(r) y 2 - 6y + 16x + 25 = 0 (t) y 2 - 2y + 2x - 6 = 0

2

1 x 2

(b) Focus ^ 1, 0 h, directrix x = - 1, focal length 1

(p) y 2 + 8y + 12x + 4 = 0

(q) y 2 - 2y + 4x - 11 = 0

(d) y 2 = 12x

(g) y 2 = 12x

(n) y 2 + 4y - 16x - 12 = 0

2

1 1 1 (d) Focus d 1 , 0 n, directrix x = - 1 , focal length 1 2 2 2

7.

x 2 + 4x - 8y - 4 = 0, x 2 + 4x + 8y + 12 = 0

8.

x 2 - 2x - 4y - 19 = 0

9. y 2 - 12y + 12x + 12 = 0

1 1 1 (e) Focus d -1 , 0 n, directrix x = 1 , focal length 1 4 4 4

10. x 2 - 2x - 12y + 1 = 0

11. x 2 - 2x - 28y + 29 = 0

(f) Focus d

12. y 2 + 4y + 24x - 44 = 0

13. y 2 - 6y - 32x + 9 = 0

14. x 2 - 6x + 8y - 15 = 0

15. y 2 + 2y - 16x + 49 = 0

1 1 1 , 0 n, directrix x = - , focal length 12 12 12

16. x 2 + 6x + 4y - 7 = 0

17. x 2 - 4x - 12y - 8 = 0

829

830


18. y 2 + 2y + 16x - 95 = 0

Exercises 11.8

19. (a) Vertex ^ - 2, 1 h, focus ^ - 2, 3 h, directrix y = -1

1.

(a)

(b) Vertex ^ 3, 2 h, focus ^ 3, 5 h, directrix y = -1

(c) Vertex ^ 1, -1 h, focus ^ 1, - 2 h, directrix y = 0 (d) Vertex ^ 3, 4 h, focus ^ 7, 4 h, directrix x = -1

(e) Vertex ^ 0, - 2 h, focus ^ 6, - 2 h, directrix x = -6

(f) Vertex ^ - 5, 0 h, focus ^ - 7, 0 h, directrix x = -3 20. Vertex ^ - 1, 4 h, focus ^ -1, -3 h , directrix y = 11, axis x = - 1, maximum value 4 21. x 2 - 4x - 8y + 12 = 0 or x 2 - 4x + 8y - 36 = 0 (b) d 0, 7

22. (a) 8x 2 + 9y - 72 = 0

23 9 n, y = 8 32 32

(b)

23. (a)

(c) 3 1 (b) d -1, -8 n, y = -9 4 4 24. x 2 + 4x + 8y - 20 = 0

25. 0.3 m

Exercises 11.7 1. 5.

m= dy dx

1 3

=x

2. m = -4

3. m = -1

6. x - y - 2 = 0

4. m =

1 2

7. x - 2y + 12 = 0

8.

x + y - 6 = 0, x - y - 18 = 0

9.

x - 2y - 2 = 0, 2x + y - 9 = 0

(d)

7 1 10. 4x + y - 8 = 0, M = d 1 , n 8 2 11. x + y - 9 = 0, P = ^ - 18, 27 h 12. Q = ^ 33, 60.5 h 13. x + 4y + 144 = 0, 4x + 2y + 9 = 0, ^ 18, -40.5 h ; show the point lies on the parabola by substituting it into the equation of the parabola 14. x - y - 4 = 0, R = ^ 4, 0 h 15. (a) Substitute P into the equation of the parabola (b) x + py - 2p - p 3 = 0 (c) Substitute ^ 0, 1 h into the equation of the normal. 0 + p - 2p - p 3 = 0 0 = p3 + p = p (p 2 + 1) 2 Since p ! 0, p + 1 = 0

(e)

ANSWERS

(f)

2.

(a) (i) p

(ii) -

1 p

(iii) y - px + p 2 = 0

(iv) x + py = p 3 + 2p 1 (iii) y - qx + 3p 2 = 0 (b) (i) q (ii) q (iv) x + qy = 3q 3 + 6q 1 (iii) y - tx + 2t 2 = 0 (c) (i) t (ii) t

2.

(a) x - 2y - 2 = 0 (b) 2x - y - 11 = 0 (c) y = x 2 + 3x + 2 (d) y = 16x 2 - 1 (e) xy = 2

3.

(a) x = 2t, y = t 2

(iv) x + ny = 5n 3 + 10n 1 (iii) y - px + 6p 2 = 0 (e) (i) p (ii) p (iv) x + py = 6p 3 + 12p 1 (iii) y + kx - 4k 2 = 0 (f) (i) −k (ii) k

(b) x = 6t, y = 3t 2

(c) x = - 4t, y = -2t 2

(d) x = 8t, y = 4t 2

(e) x = - 18t, y = - 9t

2

(f) x = 10t, y = 5t 2

t t2 ,y= 4 2 5t 2 (j) x = - 5t, y = 2

(g) x = -3t, y = (i) x =

(iv) x + ty = 2t 3 + 4t 1 (iii) y - nx + 5n 2 = 0 (d) (i) n (ii) n

t t2 ,y= 4 8

3t 2

2

(h) x =

4.

(a) x 2 = 16y (b) x 2 = 20y (c) x 2 = 4y (d) x 2 = -28y (e) x 2 = - 8y (f) x 2 = 4ay (g) x 2 = -4y (h) x 2 = 24y (i) x 2 = -2y (j) x 2 = 4ay

5.

(a) Substitute _ 6t, - 3t i into the equation (b) P = ^ -12, -12 h (c) 2x - y + 12 = 0

(iv) x - ky = 4k 3 + 8k 1 (iii) y - qx - q 2 = 0 (g) (i) q (ii) q (iv) x + qy = -q 3 - 2q 1 (iii) y + tx - 2t 2 = 0 (h) (i) −t (ii) t (iv) x - ty = 2t 3 + 4t 1 (iii) y - mx - 3m 2 = 0 (i) (i) m (ii) m

2

6.

(a) Q = ^ - 8, 4 h

7.

^ 4, 0 h, x = - 4

9.

(a) x 2 = 24y

(iv) x + my = -3m 3 - 6m 1 (iii) y + ax - 8a 2 = 0 (j) (i) −a (ii) a

(b) x - y + 12 = 0

(iv) x - ay = 8a 3 + 16a

8. P = ^ 4, -4 h; 4x + 3y - 4 = 0

(b)

1 4

10. 3x - y - 18 = 0

3.

(ii) 8 -4pq ^ p + q h, 4 _ p 2 + pq + q 2 + 2 i B

Exercises 11.9 1.

t+n (a) (i) 2 (b) (i) (c) (i) (d) (i) (e) (i)

p+q 2 m+n 2 p+q 2 a+b 2

(f) (i) -

p+q 2

a+b (g) (i) 2 (h) (i)

p+q 2

(i) (i) (j) (i)

s+t 2

p+q 2

(a) (i) ^ p + q, pq h (ii) 7 - pq ^ p + q h, p 2 + pq + q 2 + 2 A (b) (i) 7 4 ^ p + q h, 4pq A (c) (i) 6 2 ] a + b g, 2ab @

(ii) 7 -2ab ] a + b g, 2 ^ a 2 + ab + b 2 + 2 h A

1 (ii) y - ] t + n g x + 4tn = 0 2

(d) (i) 6 3 ] s + t g, 3st @

(ii) y -

1 ^ p + q h x + 2pq = 0 2

(ii) y -

1 ] m + n g x + 3mn = 0 2

(ii) 7 -5tw ] t + w g, 5 ^ t 2 + tw + w 2 + 2 h A

(ii) y -

1 ^ p + q h x + 5pq = 0 2

(ii) 8 -6pq ^ p + q h, -6 _ p 2 + pq + q 2 + 2 i B

(ii) y -

1 ] a + b g x + ab = 0 2

(ii) y +

(ii) 7 -3st ] s + t g, 3 ^ s 2 + st + t 2 + 2 h A

(e) (i) 6 5 ] t + w g, 5tw @

(f) (i) 7 6 ^ p + q h, -6pq A

(g) (i) 6 4 ] m + n g, -4mn @

(ii) 7 -4mn ] m + n g, - 4 ^ m 2 + mn + n 2 + 2 h A

(h) (i) 7 10 ^ p + q h, -10pq A

1 ^ p + q h x - 2pq = 0 2

(ii) 8 -10pq ^ p + q h, -10 _ p 2 + pq + q 2 + 2 i B (i) (i) 6 5 ] h + k g, - 5hk @

1 (ii) y + ] a + b g x - 6ab = 0 2

(ii) 7 -5hk ] h + k g, - 5 ^ h 2 + hk + k 2 + 2 h A

1 (ii) y - ^ p + q h x - 4pq = 0 2 (ii) y +

1 ] s + t g x - st = 0 2

1 (ii) y - ^ p + q h x - 7pq = 0 2

(j) (i) 7 -3 ^ p + q h, - 3pq A

(ii) 8 3pq ^ p + q h, - 3 _ p 2 + pq + q 2 + 2 i B

4.

(a) (i) xx 1 = 4 _ y + y 1 i (ii) y - y 1 = -

4 x - x1 i x1 _

(b) (i) xx 1 = 6 _ y + y 1 i (ii) y - y 1 = -

6 x - x1 i x1 _

831

832


(c) (i) xx 1 = 8 _ y + y 1 i (ii) y - y 1 = -

8 x - x1 i x1 _

(d) (i) xx 1 = 2 _ y + y 1 i (ii) y - y 1 = -

2 x - x1 i x1 _

10 (e) (i) xx 1 = 10 _ y + y 1 i (ii) y - y 1 = - _ x - x 1 i x1

5.

(f) (i) xx 1 = -2 _ y + y 1 i

(ii) y - y 1 =

2 x - x1 i x1 _

(g) (i) xx 1 = -4 _ y + y 1 i

(ii) y - y 1 =

4 x - x1 i x1 _

(h) (i) xx 1 = -12 _ y + y 1 i

(ii) y - y 1 =

12 x - x1 i x1 _

(i) (i) xx 1 = -22 _ y + y 1 i

(ii) y - y 1 =

22 x - x1 i x1 _

(j) (i) xx 1 = -14 _ y + y 1 i

(ii) y - y 1 =

14 x - x1 i x1 _

(a) xx 1 = 8 _ y + y 1 i (b) xx 1 = 2 _ y + y 1 i (c) xx 1 = 4 _ y + y 1 i (d) xx 1 = 6 _ y + y 1 i

(e) xx 1 = 10 _ y + y 1 i (f) xx 1 = -2 _ y + y 1 i

(g) xx 1 = -12 _ y + y 1 i

6. 7. 9.

(h) xx 1 = -4 _ y + y 1 i

14. ^ - 2, -1 h 15. Equation of tangent at P: y - px + ap 2 = 0 Equation of tangent at Q: y - qx + aq 2 = 0 - px + qx + ap 2 - aq 2 = 0 x (q - p) - a (q 2 - p 2) = 0 x (q - p) - a (q + p) (q - p) = 0 x - a (q + p) = 0 x = a (q + p) Substitute in (1): y - pa (q + p) + ap 2 = 0 y - apq - ap 2 + ap 2 = 0 y - apq = 0 y = apq (b) Substitute ^ 0, 2 h into equation.

16. (a) 3x + 4y - 8 = 0

17. (a) For proof, see no. 9 above

(j) xx 1 = -18 _ y + y 1 i

18. (a) 15x + 8y + 4 = 0

(a) y - px + ap 2 = 0

(b) xx 0 = 2a _ y + y 0 i

19. (a) x + 3y - 3 = 0

1 ] t + r g x + 2tr = 0 2 x2 y=18 dy x =9 dx 9t 2 o At e -9t, 2 dy -9t n = -d 9 dx =t

8. x + 2y - 36 = 0

For normal, m 1 m 2 = -1 1 t The equation is given by y - y 1 = m (x - x 1) ` m2 = -

9t 2 1 = - (x + 9t) t 2 2ty + 9t 3 = -2 (x + 9t) = -2x - 18t 2x + 2ty + 9t 3 + 18t = 0 `

y+

10. x + ty = at 3 + 2at

11. 3x - 4y + 4 = 0

12. Substitute focus ^ 0, -1 h into equation 3x + 4y + 4 = 0. 13. Equation of chord 1 y - ^ p + q h x + apq = 0 2 Substitute ^ 0, a h into equation 1 a - ( p + q) 0 + apq = 0 2 a + apq = 0 apq = - a pq = - 1

(2)

] 1 g - ] 2 g:

(i) xx 1 = -8 _ y + y 1 i

y-

(1)

(b) N = _ 0, ap 2 + 2a i

1 1 (b) N = c - , - m 4 32

(b) ^ - 6, 3 h c 2,

1 m 3

20. (a) F = ^ 0, 6 h (b) 3x + 4y - 24 = 0 (c) Q = ^ - 24, 24 h

(d) P : x - 2y - 3 = 0; Q: 2x + y + 24 = 0 1 (e) m 1 m 2 = # -2 = -1, ` tangents at P, Q are 2 perpendicular (f) R = ^ -9, -6 h (g) directrix: y = - a = - 6, ` R lies on directrix 21. P = ^ -2, -1.5 h 22. x - y + 9 = 0 23. m 1 m 2 = pq = - 1 (since pq = - 1 for focal chord) ` tangents are perpendicular 24. Tangents intersect at 6 a ^ p + q h, apq @ i.e. y = apq = - a (since pq = - 1 for focal chord) Directrix: y = - a ` tangents meet on the directrix 25.

y=

x2 4a

dy x = 2a dx At P _ x 0, y 0 i, dy dx

=

x0 2a

ANSWERS

(c) Directrix y = -3 Point of intersection = ^ - 8, -3 h So the point lies on the directrix.

The equation is given by y - y 1 = m(x - x 1) x0 ` y - y0 = (x - x 0) 2a 2ay - 2ay 0 = x 0 (x - x 0) = xx 0 - x 02 = xx 0 - 4ay 0 (since x = 4ay 0) 2ay + 2ay 0 = xx 0 2a(y + y 0) = xx 0 2 0

(a) 3x + 4y - 8 = 0 (b) Q = d 2,

x 2 = 2 a ^ y - 2a h

7.

y - px + 2p 2 = 0; y - qx + 2q 2 = 0; y = - 2

8.

x 2 = 16 ^ y - 6 h

dy dx

=

12. (a) PO has gradient

1 n 2

dy

x 4

=

1 2

(a) y - px + p 2 = 0 (b) p 2 + 1 (c) R = _ 0, - p 2 i and F = ^ 0, 1 h FR = p + 1 = PF 2

4.

5.

q 2

#

(b) y = - 6a

15. (a) x 2 = 9a ^ y - 5a h

= -1 So the tangents are perpendicular.

3.

p

2

14. (a) T = 6 a ^ p + q h, apq @

2 4 dx 1 ` m2 = 2

2.

; QO has gradient

(c) x 2 = 2a ^ y - 4a h is a parabola in the form (x - h) 2 = 4a 0 ^ y - k h where ^ h, k h is the vertex and a 0 is the focal length a ` vertex is ^ 0, 4a h and focal length is 2 a 13. x 2 = 2ay - a 2 or x 2 = 2a d y - n 2

1 m: 2

m 1 m 2 = -2 #

p 2

q = -1 2 pq = - 4 ` (b) x 2 = 2a ^ y - 4a h m1 m2 =

At P (−8, 8): dy -8 = 4 dx ` m 1 = -2 At q c 2,

(b) x 2 = 2a ^ y - a h

10. (a) y = -a

(c) (−3, −2) (d)

9. x 2 = 2a ^ y - a h

11. x 2 = - 4 ^ y + 4 h

Exercises 11.10 1.

6.

(a) y - tx + 3t 2 = 0 (b) Y = _ 0, - 3t 2 i (c) F = ^ 0, 3 h TF = FY = 3 ^ t 2 + 1 h (a) y + qx - 5q = 0 (b) R = _ 0, 5q 2 i (c) F = ^ 0, - 5 h FR = FQ = 5 _ q 2 + 1 i So triangle FQR is isosceles. ` +FQR = +FRQ (base angles of isosceles triangle) 2

(a) 4x + 3y - 9 = 0 (b) Focus (0, 3) Substitute into equation: LHS = 4 ] 0 g + 3 ] 3 g -9 =0 = RHS So it is a focal chord.

Test yourself 11 2. x 2 - 4x - 8y - 4 = 0

1.

8x + 6y - 29 = 0

3.

Centre ^ 3, 1 h, radius 4

5.

(a) ^ 8, 8 h

6.

x 2 + y 2 = 25

8.

x 2 + x + y 2 - 3y - 10 = 0

4. (a) ^ 1, - 3 h

(b) ^ 4, - 3 h

(b) 2x - y - 8 = 0

10. (a) (i) ^ 1, 1 h

(b) ^ 0, - 2 h

7. (a) y = 2

(ii) ^ 1, 2 h

9. x 2 - 8x + 16y - 16 = 0

(b) y = 0

11. 2x + 3y + 6 = 0

12. 14 units

13. y = - 24x

2

2

14. x - 8y + 16 = 0

15. 4x - 3y - 16 = 0, 4x - 3y + 14 = 0 16. y = x, y = - x 19. (a) x 2 = 12y

17. y 2 = 20x

18. (a) -

1 2

(b) 2

(b) y 2 = - 32x

20. (a) x - 4y + 72 = 0

1 (b) d 9, 20 n 4

21. Sub ^ 0, 4 h : LHS = 7 # 0 - 3 # 4 + 12 = 0 = RHS 2 22. d , -7 n 9

23. 3x - 2y + 40 = 0

24. x 2 - 10y + 100 = 0

25. y - 3x + 9a = 0

26. (a) x - y - 3 = 0 (b) R = ^ 0, -3 h (c) F = ^ 0, 3 h FP = FR = 6

833

834


27. (a) y a-

1 ^ p + q h x + apq = 0 2

1 (p + q) # 0 + apq 2 a + apq apq pq

(b) Sub ^ 0, a h:

=0

6ap + 4aq 3ap 2 + 2aq 2 p , 5 5

(b) x 2 = 2a ^ y + 2a h

=0 = -a = -1

19. y = 0 20. (a) T = ^ 6, - 20 h

28. x - 2y + 48 = 0 (b) ^ 6, 9 h and ^ - 2, 1 h

29. (a) x - y + 3 = 0

18. (a) N = f

m x = t and m y = s m1 - m2 tan i = 1 + m1 m2 t-s tan 45c = 1 + ts t-s 1= 1 + ts t-s 1= 1 + ts 1 + ts = t - s 1 + s = t - ts = t (1 - s) 1+s =t 1-s t-s -1 = 1 + ts -1 - ts = t - s s - 1 = t + ts = t (1 + s) s-1 =t 1+s

(c)

30. y = - a


(a) 8x + 6y - 29 = 0 (b) Midpoint of AB lies on line; m 1 m 2 = -1

2.

(a) x 2 - 2x + y 2 - 6y - 15 = 0 (b) Put y = 0 into equation

3.

y = 1 - 2x 2

5.

(a) 4x - 2y + 9 = 0; x + 2y - 24 = 0 (b) m 1 m 2 = - 1 (c) X = ^ 3, 10.5 h (d) 3x - 4y + 8 = 0; focus ^ 0, 2 h lies on the line

6.

^ 0, 0 h

7.

(a) 2x - 4y - 1 = 0; 2x + y + 4 = 0 (b) Point lies on line y = - 1

8.

y = - 2 x 2 + 4x - 2

1 4. d 2 , - 3 n 2

or

9. 3x + y + 2 = 0

10.

(b) P = 6 a ] t + s g, ats @

Practice assessment task set 3 1.

m ≤ 2, m ≥ 3

4.

24 cm

2 6. (a) 3

2. 4x + 3y - 16 = 0

Centre ^ - 3, 5 h, radius 7

5.

1 (b) 3

(c) 1

1 9

7.

Focus ^ 0, -2 h, directrix y = 2

8.

x = - 5 or - 6

10.

3. x 2 = 8y

9. k = - 1

+AFE = +CBE (ext.+ equal to opp. interior + in cyclic quadrilateral)

+CBE = 180c - +EDC (opp.+s supplementary in cyclic quadrilateral)

11. (a) x 2 + 4x + y 2 - 10y + 21 = 0 (b) ] x + 2 g2 + ^ y - 5 h2 = 8; centre ^ -2, 5 h; radius = 12. -

8 =2 2

11. 3x - 4y - 14 = 0, 3x - 4y + 16 = 0

2 3 3

12. Vertex ^ - 4, -17 h , focus ^ - 4, -16.75 h

13. (a) y + 4y - 16x + 52 = 0 2

14. 4 2 units

` +AFE = 180c - +EDC These are supplementary cointerior angles. ` AF || CD

(b) 2x - y - 6 = 0

15. x + y - 2y - 2 = 0 2

2

16. 696 mm from the vertex 17. 141x + 127y + 32 = 0; 219x + 23y + 58 = 0

13. x = 0, 3

14. k = 7.2 cm

16. b $ -2 17. i = 16c ^ 0, 0 h and radius 4

15. x + 2y + 2 = 0

18. x 2 + y 2 = 16, circle centre

19. x 2 + 4x + y 2 + 6y - 12 = 0

ANSWERS

Obtuse +AOC = 2+ADC (+at centre double + at circumference) Reflex +AOC = 2+ABC (similarly) Obtuse +AOC + reflex +AOC = 360c (+ of revolution) ` 2+ADC + 2+ABC = 360c +ADC + +ABC = 180c It can be proved similarly that +BAD + +BCD = 180c by drawing BO and DO. ` opposite angles in a cyclic quadrilateral are supplementary

20. x 2 - 3x + y 2 - 6y - 17 = 0 21.

+BCD +DAB +DBC +BDC `+BDC ` +DAE

22. - 0.75

= 90c (+in semicircle) = 90c (similarly) = +DAE (+s in same segment) = 90c - +DBC (+ sum of D BDC) = 90c - +DAE = 90c - +BDC

23. 5x 2 - 54x + 5y 2 + 20y - 79 = 0

24. a = 2, b = 1, c = 0 26. -

25. x = 33°, y = 57°

x

37. Centre ^ -5, 3 h, radius 2 38.

9 - x2

27. (a) y - px + ap 2 = 0

(b) R = f

a _ p2 - 1 i p

, - ap

(c) 2px + _ p 2 - 1 i y + a - ap 2 = 0 28. x 2 - 4x - 16y + 20 = 0 29. `

AC = BC and CD = CE (given) AC BC = CD CE +ACB = +ECD (vertically opposite angles)

` since two sides are in proportion and their included angles are equal, ΔABC is similar to ΔCDE y = 5.3 cm 30. x - y - 4 = 0

39. a 2 0 D = b 2 - 4ac = ] -1 g2 - 4 (1) (3) = -11 10 Since a 2 0 and D 1 0, x 2 - x + 3 2 0 for all x

31. x 2 + 2x - 16y - 15 = 0 33.

Let +DBA = x and +EBC = y Then +EDB = x and +DEB = y (alternate +s, DE < AC) +FDE = 180c - x (+FDB straight +) +GED = 180c - y (+GEB straight +) +FGB = +DBA = x (+s in alternate segment) +GFB = +EBC = y (similarly) ` +FDE = 180c - +FGB and +GED = 180c - +GFB Since opposite angles are supplementary, FGED is a cyclic quadrilateral.

32. x = 0, 2

a 10 D = b 2 - 4ac = 1 2 - 4 (- 1) (- 9) = - 35 10 Since a 1 0 and D 1 0, - x 2 + x - 9 1 0 for all x

34. 8 (3x - 1) (2x + 5) + 3 (2x + 5) = ] 30x + 7 g (2x + 5) 3 3

4

35. sec x cosec x 36.

40. k = 1

41. 3x + 2y - 9 = 0

42. (a) 217 km

(b) 153c

43. a = 3, b = - 18, c = - 34 45. (a) y = x 2 - 1

44. x 2 4, x 1 3

(b) ^ - 4, 15 h

(c) x - 8y + 124 = 0

46. i = 95c 44’ 47. x = 11c 48. T = 361 ^ 2 0 and a perfect square h 49. x + 2y + 9 = 0

50. k # 3

51.

Let ABCD be a cyclic quadrilateral of circle, centre O. Join AO and CO.

52. 5x - 4y - 41 = 0

55. - 1

1 # y 1 -1 4

1+ 3 2 2 53. d 3 , -2 n 54. 5 5 3-1 56.

3 6 - 10 + 3 3 - 5 22

835

836


57. x = 4.9 cm, y = 11.1 cm 60. 4.5 m

128 2187

61.

59. 8.25 units

1 1 64. y = 1 , 3 2

65. 162c

b 2 - 4ac = -104 1 0 So Pl(x) has no real roots 15. Ql] x g = 3x 2 - 6x + 3 b 2 - 4ac = 0 So Ql] x g has equal roots

+ACB = 90c (+ in semicircle) `+DCA = 90c (+DCB straight +) ` AD is a diameter of the circle

Exercises 12.2

67. x = 45°, 135°, 225°, 315° 1 1 68. x = - 1, y = 2 or x = - , y = 4 4 4

69. ] a - 2b g ^ a 2 + 2ab + 4b 2 h

70. x = 43

71. -

1 31

73. tan i

72. 1.8 units

14. Pl] x g = 3x 2 - 2x + 9

62. x = 60°, 120°, 240°, 300°

63. 2x + 3y - 3 = 0 66.

58. x = 1

1.

3x 2 + 2x + 5 = ] x + 4 g ] 3x - 10 g + 45

2.

x 2 - 7x + 4 = ] x - 1 g ] x - 6 g - 2

3.

x 3 + x 2 + 2x - 1 = ] x - 3 g ^ x 2 + 4x + 14 h + 41

74. 8x ] 2x + 5 g (x 2 - 1) 3 + 2 (x 2 - 1) 4 = 2 (x 2 - 1) 3 (9x 2 + 20x - 1)

4.

4x 2 + 2x - 3 = ] 2x + 3 g ] 2x - 2 g + 3

5.

x 3 - 5x 2 + x + 2 = ^ x 2 + 3x h ] x - 8 g + ] 25x + 2 g

1 75. 4

6.

x 3 + x 2 - x - 3 = ] x - 2 g ^ x 2 + 3x + 5 h + 7

77. Focus (2, 1), directrix y = 5

7.

5x 3 - 2x 2 + 3x + 1 = ^ x 2 + x h ] 5x - 7 g + ] 10x + 1 g

78. px - y - 9p 2 = 0

8.

76. x + 2x + y - 3y - 25 = 0 2

2

79. x - 2y - 36 = 0

1 ^ p + q h x + apq = 0 (b) x 2 = 2a ^ y - 2a h 2 (c) Concave upward parabola, vertex (0, 2a)

80. (a) y -

81. (c)

82. (d)

83. (b)

88. (a)

87. (c)

84. (a)

89. (a), (d)

85. (c)

86. (a)

9.

x 4 - x 3 - 2x 2 + x - 3 = ] x + 4 g ^ x 3 - 5x 2 + 18x - 71 h + 281 2x 4 - 5x 3 + 2x 2 + 2x - 5 = ^ x 2 - 2x h ^ 2x 2 - x h + ] 2x - 5 g

10. 4x 3 - 2x 2 + 6x - 1 = ] 2x + 1 g ^ 2x 2 - 2x + 4 h - 5 11. 6x 2 - 3x + 1 = ] 3x - 2 g d 2x +

90. (c)

1 2 n+ 1 3 3

12. x 4 - 2x 3 - x 2 - 2 = ^ x 2 - x h ^ x 2 - x - 2 h + ] -2x - 2 g Chapter 12: Polynomials 1

5 4 3 2 13. 3x - 2x - 3x + x - x - 1 4 3 = ] x + 2 g ^ 3x - 8x + 13x 2 - 25x + 49 h - 99

Exercises 12.1

14. x 2 + 5x - 2 = ] x + 1 g ] x + 4 g - 6

1.

(a) 7 (b) 4 (c) 1

2.

(a) -19 (b) -10 (c) -1 3. (a) -6 (b) 5 (c) 2 (d) 1 (e) 2 4. (a) 5 (b) 4 (c) -3 (d) 0

5.

(a) !3

6.

7.

(b) -5

(d) 11 (e) 3 (f) 0 (g) 4

(c) -2, 1

(d) 4 (e) 0

(a) P l ] x g = 12x 3 - 6x 2 - 2x + 4; 3 (b) Pl] x g = 10x; 1 (c) Pl] x g = 108x 11 - 35x 4 + 8; 11 (d) P l ] x g = 7x 6 - 9x 2 + 2x - 7; 6 (e) Pl] x g = 8; 0 (a), (b), (g) 8. (a) a = 0 (d) a = -1 (c) 3

10. (a)

(d) 3

(e) a = 4

(b) b = 10 1 9. (a) -2 2

(c) c = -6 (b) x = 2, -1

(e) x 5

D = b 2 - 4ac = -8 -8 1 0

` f ] x g has no zeros (b) 9x 3

(c) - 2

(d) 9

11. (a) 2 (b) 0 (c) 2 12. x = -3, 2

(e) x =

(d) 0

13. x = 0, 1

2 , -1 3

(e) 2 (f) 4

(g) 3

15. x 4 - 2x 2 + 5x + 4 = ] x - 3 g ^ x 3 + 3x 2 + 7x + 26 h + 82 16. 2x 4 - x 3 + 5 = ^ x 2 - 2x h ^ 2x 2 + 3x + 6 h + ] 12x + 5 g 17. x 3 - 3x 2 + 3x - 1 = ^ x 2 + 5 h ] x - 3 g + ] - 2x + 14 g 18. 2x 3 + 4x 2 - x + 8 = ^ x 2 + 3x + 2 h ] 2x - 2 g + ] x + 12 g 4 3 2 19. x - 2x + 4x + 2x + 5 = ^ x 2 + 2x - 1 h ^ x 2 - 4x + 13 h + ] - 28x + 18 g

20. 3x 5 - 2x 3 + x - 1 = ] x + 1 g ^ 3x 4 - 3x 3 + x 2 - x + 2 h - 3

Exercises 12.3 1.

(a) 41 (b) -3 (c) -43 (d) 9424 (e) 0 (g) 47 (h) 2321 (i) 31 174 (j) - 3

2.

2 (a) k = 8 (b) k = 7 (e) k = !2

3.

(a) 0 (b) Yes (c) x 3 - 4x 2 + x + 6 = ] x - 2 g ^ x 2 - 2x - 3 h (d) f ] x g = ] x - 2 g ] x - 3 g ] x + 1 g

(c) k = 15 299

(f) 37

(d) k = 6

8 9

ANSWERS

4.

5. 7.

9.

(a) P ] - 3 g = 81 - 81 - 81 + 81 = 0 ` x + 3 is a factor (b) P ] x g = x ] x + 3 g 2 ] x - 3 g 7 17 a = -1 , b = -1 12 48

Exercises 12.4 1.

y

(a)

6. a = - 6

(a) P ] 3 g = 140 ! 0 ` x - 3 is not a factor of P ] x g (b) k = - 39 8. a = -2, b = -1

6

-1

(a) a = 3, b = 11 (b) f ] x g = ] x + 1 g ^ 3x 3 + 8x 2 + 7x + 2 h (c) g ] -1 g = 0 (d) f ] x g = ] 3x + 2 g ] x + 1 g3

10. (a) ] x + 2 g ] x - 4 g (b) x ] x + 2 g ] x - 1 g (c) ] x - 1 g ] x + 4 g ] x - 2 g (d) ] x + 5 g ] x - 3 g ] x + 2 g (e) ] x - 3 g ] x - 1 g ] x - 7 g (f) ] x + 2 g ] x - 9 g ] x - 5 g (g) ] x - 3 g ] x - 2 g2 (h) x ] x + 4 g ] x + 1 g 2 (i) ] x - 1 g ] x + 2 g2 (j) ] x + 1 g ] x - 3 g ] x + 2 g 11. (a) P ] x g = ] x - 1 g ] x + 3 g ] x - 2 g (b) -3, 1, 2

2

x

3

y

(b)

(c) Yes

12. (a) Dividing f ] x g by ] x + 5 g ] x - 2 g gives f ] x g = ] x + 5 g ] x - 2 g ^ x 2 + 7x + 12 h ` ] x + 5 g ] x - 2 g is a factor of f ] x g (b) f ] x g = ] x + 5 g ] x - 2 g ] x + 3 g ] x + 4 g

-4

x

2

13. P ] x g = ] x + 1 g ] x - 4 g ] x + 3 g2 14. (a) P ] -6 g = P ] 5 g = 0

(b) P ] x g = ] x - 4 g ] x + 6 g ] x - 5 g

15. (a) P ] u g = ] u - 2 g ] u - 1 g 2

(b) x = 2, 3

16. (a) f ^ p h = ^ p - 1 h ^ p + 2 h ^ p - 3 h 17. (a) P ] k g = ] 2k - 1 g ] k + 1 g2

y

(c)

1 (b) x = 0, -1 , 1 2

(b) x = 30c , 150c , 270c

18. (a) f ] u g = ] u - 1 g ] u - 3 g ] u - 9 g (b) x = 0, 1, 2 19. x = -5, -4, -2

1

3

x

20. i = 0c, 90c, 120c, 240c, 270c, 360c 21. (a) a = 1, b = 3, c = 4, d = - 2 (b) a = 1, b = -1, c = 8, d = -12 (c) a = 2, b = 0, c = -1, d = 6 (d) a = 1, b = 1, c = 11, d = -12 (e) a = 3, b = 0, c = -1, d = 8 (f) a = 1, b = 1, c = -4, d = -7 (g) a = 5, b = -2, c = -19, d = -43 (h) a = -1, b = 4, c = -1, d = -1 (i) a = -1, b = 3, c = 6, d = -4 (j) a = -1, b = -10, c = -27, d = -20 22. P ] x g = x 3 - x 2 - 12x

23. a = 1, b = -3, c = -6

24. P ] x g = 2x 4 - 4x 3 - 10x 2 + 12x 25. P(x) has degree 3. Suppose P(x) has 4 zeros, a1, a2, a3 and a4. Then _ x - a 1 i _ x - a 2 i _ x - a 3 i _ x - a 4 i is a factor of P(x). So P ] x g = _ x - a 1 i _ x - a 2 i _ x - a 3 i _ x - a 4 i Q ] x g. ` P(x) has at least degree 4 But P(x) only has degree 3. So it cannot have 4 zeros.

y

(d)

-2

x

837

838


(d) (i) A ] x g = x ] 2x - 5 g ] x + 3 g

y

(e)

-5

y

(ii)

50

-2

x

5

2.

(a) (i) P ] x g = x ] x - 4 g ] x + 2 g

(e) (i) P ] x g = - x 2 ] x - 3 g ] x + 1 g

y

(ii)

x

1 22

-3

y

(ii)

-2

x

4

x

3

-1

(b) (i) f ] x g = - x ] x - 1 g ] x + 5 g (ii)

3.

y

(a) x = 0, 1, -2 y

(b)

-5

x

1

-2

(c) (i) P ] x g = x 2 ] x + 1 g ] x + 2 g

4.

y

(ii)

x

1

(a) P ] 2 g = 8 - 12 - 8 + 12 =0 (b) P ] x g = ] x - 2 g ] x - 3 g ] x + 2 g y

(c) 12

-2

-1

x -2

2

3

x

ANSWERS

5.

y

(a)

-4

(e)

-2

y

x

3

-24

-3

x

-18 y

(f)

y

(b)

3

2

-3

-1

3

x

2

-2

x

-8 -9 y

(c)

y

(g)

12

1

3

4

1

x

2

x

-4

y

(d)

y

(h)

12 3

-4

1

3

x -3

1

x

839

840


(i)

(c) P ] 4 g = ] 4 + 1 g ] 4 - 4 g2 =0 Pl(x) = 3x 2 - 14x + 8 Pl(4) = 3 ] 4 g2 - 14 (4) + 8 =0

y

6.

4

-2

x

7. y

(j)

1

1

-1

x

8.

(a) ] x + 3 g 3 = x 3 + 9x 2 + 27x + 27 Dividing by x 3 + 9x 2 + 27x + 27 gives x 4 + 7x 3 + 9x 2 - 27x - 54 = ^ x 3 + 9x 2 + 27x + 27 h ] x - 2 g so ] x + 3 g 3 is a factor (b) f ] x g = ] x - 2 g ] x + 3 g 3 (c) f ] -3 g = ] -3 - 2 g ] -3 + 3 g 3 =0 f l(x) = 4x 3 + 21x 2 + 18x - 27 f l(-3) = 4 ] -3 g3 + 21 ] -3 g2 + 18 (-3) - 27 =0 (a) P ] x g = ] x - k g3 Q ] x g where Q(x) has degree n - 3 (b) P ] k g = ] k - k g3 Q ] k g =0 Pl(x) = ulv + vlu = Ql(x) ] x - k g3 + 3 ] x - k g2 Q (x) Pl(k) = Ql(k) ] k - k g3 + 3 ] k - k g2 Q (k) =0

(a)

y

Exercises 12.5 1.

2.

(a) x = 3, double root (b) x = 0, 2, 7, single roots (c) x = 0, double root, x = 3, single root (d) x = - 2, single root, x = 2, double root (e) x = - 2, triple root (f) x = 0, 2, single roots, x = 1, double root (g) x = - 1, 3, double roots (h) x = 0, triple root, x = 4, double root (i) x = 1, triple root, x = - 5, 1 single root (j) x = 1 , triple root 2

(b)

y

(a) (i) Positive (ii) Even (b) (i) Negative (ii) Odd (c) (i) Negative (ii) Even (d) (i) Negative (ii) Odd (e) (i) Positive (ii) Odd (f) (i) Positive (ii) Even (g) (i) Positive (ii) Odd (h) (i) Negative (ii) Even (i) (i) Positive (ii) Odd (j) (i) Positive (ii)Even

3.

P ] x g = ] x + 4 g 2 Yes, unique

4.

(a) P ] x g = k ] x - 1 g Not unique

5.

x

3

x (b) P ] x g = 5 ] x - 1 g3

(a) ] x - 4 g2 = x 2 - 8x + 16 Dividing by x 2 - 8x + 16 gives x 3 - 7x 2 + 8x + 16 = ^ x 2 - 8x + 16 h ] x + 1 g so ] x - 4 g 2 is a factor (b) P ] x g = ] x + 1 g ] x - 4 g2

ANSWERS

(c)

10.

y

x

x

-1

11.

y

(d)

y

y

x

2

x

y

12. y

(e)

x

-3

x

13. 9.

y

y

1

2

x

x

841

842


17. Odd function with positive leading coefficient starts negative and turns around at the double root. It then becomes positive as x becomes very large so it must cross the x-axis again. So there is another root at k 2 -1

y

14.

y

x

15.

x

k

-1

y

x

-2

18. Even function with negative leading coefficient is negative at both ends. The triple root has a point of inflexion so the curve must cross the x-axis to turn negative again. So there is another root at k 2 -2 y

16.

y

4

x

k

-2

x 19. Odd function with positive leading coefficient starts negative and turns around at both the double roots. It then becomes positive as x becomes very large so it must cross the x-axis again. So there is another root at k 22 y

-3

2

k

x

ANSWERS

20. Odd function with negative leading coefficient starts positive and turns around at the double root. It then becomes negative as x becomes very large so it must cross the x-axis again. So there is another root at k 2 1 y

Test yourself 12 1.

p ]xg = ]x + 3g]x - 3g]x + 5g]x - 1g

2.

(a) 3

3.

P (x) = (x - 6) (x - 1) (x + 2) = x 3 - 5x 2 - 8x + 12

4.

(a) x 2 + 3x + 2

5.

(a) 3 (b) - 3

(b) 9

(c) 1

(d)

1 9

(b) p ] x g = ] x - 5 g ] x + 3 g ] x + 1 g ] x + 2 g (c) - 3, 0, 1

(d) x 3

6. k

1

x

Exercises 12.6 1.

(d) (i) 2

3.

(b) (i) -2 (ii) -

(a) (i) 2 (ii) 8

1 (ii) -3 4

2 3

(c) (i) -7 (ii) 1

(e) (i) -3 (ii) 0

2. (a) (i) -1 (ii) -2

1 (c) (i) (ii) 3 2

(iii) -8

(b) (i) 3 (ii) 5 (iii) 2

(iii) -1

(d) (i) -3 (ii) 0 (iii) -11 (e) (i) 0 (ii) 7 (iii) 3

(a) (i) -2 (ii) -1 (iii) 1 (iv) 5

(b) (i) 1 (ii) -3

(iii) -2 (iv) -7

(c) (i) 1 (ii) -3 (iii) -2 (iv) -4 1 (d) (i) 1 (ii) -2 (iii) -1 (iv) -1 (e) (i) 6 (ii) 0 2 1 (iii) 0 (iv) 3 4. (a) 5 (b) -5 (c) -1 (d) 35 2 3 5 1 1 (e) 200 5. (a) (b) (c) (d) 2 2 2 3 1 (e) 2 2 6. (a) - 3 (b) - 5 (c) 1 7. k = - 26 3 1 1 8. a + b = 2, ab = -7 9. a + b = 2 , ab = 2 2 1 10. (a) k = 0 (b) k = 4 (c) k = ±1 (d) k = - , 1 2 3 1 (e) k = 0 11. m = -9 12. a = -1 , b = -9 4 8 13. (a) P ] 1 g = 0

15. (a)

4 2 , q = -17 15 15

1 1 18. x = - , 1 2 2

(a) a = 3

(b) - 5

8.

p (-7) = ] -7 g3 - 7 ] -7 g2 + 5 (-7) - 4 = - 725 ! 0

9.

x = - 1, !3

10. a = 2, b = - 18, c = 40

11. x-intercepts -3, 2, 4; y-intercept 24 12. 3x 5 - 7x 3 + 8x 2 - 5 = (x - 2) (3x 4 + 6x 3 + 5x 2 + 18x + 36) + 67 13. x = 60c, 90c, 180c, 270c, 300c 15. 4, 5 16.

(b) a + b + c = 1, abc = -6

14. a = 1; a + b = -2 (b) p = 8

7.

19. x =

4 15

16. 1 1 1 ,! 3 2

17. -5 1 2 20. x = ! 3 , -1 , 2 3

17. k = - 14

18. 4

14. k = 7.4

843

844


19. P (a) = A (a) ] a - a g3 =0 Pl(x) = A (x) 3 ] x - a g2 + Al(x) ] x - a g3 Pl(a) = A (a) 3 ] a - a g2 + Al(a) ] a - a g3 =0

y

24.

20. f ] 5 g = 5 3 - 6 ] 5 g 2 + 12 ] 5 g - 35 = 0 x

21. (a) f ] 5 g = 5 3 - 7 ] 5 g 2 - 5 ] 5 g + 75 =0 (b) f l] x g = 3x 2 - 14x - 5 f l(5) = 3 ] 5 g2 - 14 (5) - 5 =0 (c) Double root at x = 5 (d) f ] x g = ] x + 3 g ] x - 5 g 2 y

22.

25. (a) a = 2, b = - 3, c = 4, d = 5

Challenge exercise 12 1. 3

x

2.

(a) P (b) = ] b - b g7 Q (b) =0 Pl(x) = ] x - b g7 Ql(x) + Q (x) 7 ] x - b g6 Pl(b) = ] b - b g7 Ql(b) + Q (b) 7 ] b - b g6 =0 (b) a = -7, b = - 1

23. (a) P ] x g = ] x + 6 g 3Q ] x g (b)

P ] x g = ] x - 1 g ] x + 1 g2 ^ x 2 + x + 1 h

y

3.

i = 0c , 45c , 60c , 120c , 180c , 225c , 240c , 300c , 360c

4.

(a) 3x - y + 2 = 0

(b) ^ 2, 8 h

5. (a)

a - 33 4

(b) a = -14 6.

(a) - 3

(b) 17

7. i = 90c, 210c, 330c

8. a = - 5

If x - a is a factor of P ] x g Then P (x) = (x - a) Q (x) P (a) = (a - a) Q (a) ` =0 10. ^ - 1, -1 h, ^ -3, 5 h 11. P ] x g = - ] x + 1 g 2 ] x - 2 g 3 9.

-6

x

12.

y

a1

a2

x

ANSWERS

Chapter 13: Permutations and combinations

Exercises 13.1 16 33

1.

1 10 000

2.

5.

98.5%

6. (a)

8.

1 3

1 9. (a) 6

2 9

3.

4 7

3 7

7.

1 (b) 3

5 (c) 6

(b)

1 20 000

4. 3 20

6.

(a) 39 916 800

8.

5040

9. 6

12. 1.3 # 10 12 14. (a) 720 16. (a)

1 4

10. 720

13. (a) 39 916 800

(b) 120 (b)

1 62

(b)

3 31

(c)

1 2

11. (a)

1 15

(b)

8 15

(c)

3 5

13. (a)

29 86

(b)

19 43

(c)

67 86

15. (a)

1 6

(b)

4 5

(b) 36

17. (a)

20.

329 982

22. (a)

14 59

1 2

1 3

(c)

(b)

24. 19%

35 59

19 31

(c)

20. (a)

14. 32

(d)

1 2

24 59

(b)

20 31

(b)

(d)

(b)

1 2

(e)

23 44

16. (a) 21 44

(c)

38 59

4 31

23.

19.

(d)

5 18

(b) 91 (c)

6 19

11 31

5 24

(d)

1 24

25. 0.51

1.

5.

2. 67 600

456 976 26 10 # 10 15

8. 300

3. 26 5 # 10 4

6. 1 000

1 11. (a) 84 (b) 84

1 67 600 000

14.

1 17. 5184

18. 6

21. 7 880 400 24. (a) 9900 (b)

1 9900

16. Yes 20. 360

23. 271 252 800 25.

1 720

Exercises 13.3 1.

2.

(a) 720 (b) 3 628 800 (c) 1 (d) 35 280 (e) 120 (f) 210 (g) 3 991 680 (h) 715 (i) 56 (j) 330 362 880

3. 720

1 120

4. 479 001 600

8! 8 # 7 # 6 #...# 2 # 1 = 4#3#2#1 4! =8#7#6#5 11! 11 # 10 # 9 #...# 2 # 1 = 6#5#4#3#2#1 6! = 11 # 10 # 9 # 8 # 7

n ] n - 1 g ] n - 2 g ... ] r + 1 g r ] r - 1 g ... 3.2.1 n! = r! r ] r - 1 g ] r - 2 g ... 3.2.1 = n # ] n - 1 g # ] n - 2 g #...# ] r + 1 g = n (n - 1) (n - 2) #...# (r + 1)

n ] n - 1 g ] n - 2 g ... ] n - r + 1 g ] n - r g ... 3.2.1 n! = ]n - r g! ] n - r g ] n - r - 1 g ] n - r - 2 g ... 3.2.1 = n ] n - 1 g ] n - 2 g ... ] n - r + 1 g

5. 120

5! = 20 ]5 - 2 g!

(d)

10! = 640 800 ] 10 - 7 g !

(f)

7! = 2520 ]7 - 5 g!

(h)

11! = 6 652 800 ] 11 - 8 g !

(j)

15. 7

19. 6840

22. 210

18.

6! = 120 ]6 - 3 g!

7. 1 000 000

12. (a) 10 000 000 (b) 1000

13. Yes

1 12

(a)

4. 260

3 10. 10 000

9. 64

(b)

Exercises 13.4

Exercises 13.2 1.

(b) 3 628 800

19. 6 227 020 800

12. 8

18. (a)

21. (a)

99 124

(d)

11. 5040

15. 5040

(c) 48

17. (a) 479 001 600 10. (a)

7. 40 320

(b) 479 001 600

(b)

(g)

(e)

8! = 336 ]8 - 3 g!

(c)

9! = 60 480 ]9 - 6g!

8! = 20 160 ]8 - 6g! (i)

9! =9 ]9 - 1 g!

6! = 720 ]6 - 6 g!

2.

(a) 650

3.

(a) 648 (b) 432 (c) 144 4. (a) 20 (b) 4 (c) 12 (d) 8

5.

(a) 24

7.

(a) 120

8.

(a) 479 001 600 (b) 1320

9.

(a) 56 (b) 336 (c) 1680

(b) 15 600

(c) 358 800

6. (a) 4536

(b) 24 (b) 48

(c) 96

(d) 7 893 600

(b) 2016

(d) 72

(c) 3528

(e) 60

10. (a) 60 480 (b) 2520 (c) 907 200 (d) 151 200 (f) 453 600 (g) 360 (h) 2520 (i) 59 875 200 (j) 90 720 11. (a) 24 (b) 5040 (e) 39 916 800 12. (a) 6

(b) 720

(c) 40 320

(c) 5040

(e) 60

(d) 3 628 800

(d) 362 880

(e) 3 628 800

845

846


13. (a) 181 440 (b) 19 958 400 (c) 20 160 (d) 1 814 400 (e) 239 500 800

n! ' (n - r) ! ]n - n + r g! n! 1 = # ]n - r g! r! n! = ] n - r g !r! n n Pn - r Pr = ]n - r g! r! =

14. (a) 720 (b) 120 15. (a) 362 880 (b) 40 320 16. (a) 3 628 800 (b) 362 880 (c) 181 440 17. (a) 24 (b) 12 (c) 24

`

18. (a) 720 (b) 240 (c) 480 (d) 144

20.

2 9

21. (a) 20! (b) 5!8!7!3! (c)

22. (a) 60 (b) 48 (c) 36 (d) 24.

1 336

P3

= = = =

P5

5!

= = = =

8

`

=

1 6

= =

3!

8

Pr + r n Pr - 1 =

27. (a) 720 (b) 120 (c) 192

28. (a) x! (b) ] x - 1 g ! (e) ] x - 3 g ] x - 2 g !

29. (a)

23.

n

7 20

Pr =

25. (a) 40 320 (b) 30 240 (c) 21 600

26. (a) 20 (b) 60

8

1 5

n+1

30.

19. (a) 3 628 800 (b) 362 880 (c) 28 800

P3

3!

3! 8! ' 3! 5! 8! 1 # 5! 3! 8! 5!3! 8! ]8 - 5 g! 5! 8! ' 5! 3! 8! 1 # 3! 5! 8! 5!3! 8

=

= = = = `

1.

P5

Pr

= = = =

Pn - r

]n - r g!

=

n+1

n

n

Pr = Pr + r Pr - 1

Exercises 13.5

n! ]n - r g! r! n! ' r! ]n - r g! n! 1 # ]n - r g! r! n! ] n - r g !r! n! (n - 5 n - r ? ) ! ]n - r g!

(a)

9! = 126 ] 9 - 5 g !5!

(c)

8! = 56 ] 8 - 3 g !3 !

(e)

11! = 462 ] 11 - 5 g !5!

(b) (d)

12! ] 12 - 7 g !7!

(a) (i) 1 (ii) 1 (iii) 1 (iv) 1 n n (b) (i) C 0 = 1 (ii) C n = 1

3.

(a) 28

4.

(a) Number of arrangements = 15

(b) 84

(c) 462

= 792

10! = 210 ] 10 - 4 g !4!

2.

5!

r!

n

=

(d) 3! ] x - 2 g !

8! ]8 - 3g!

n

(b)

(c) 2! ] x - 2 g !

]n + 1g! ]n + 1 - r g! n! n! +r ]n - r g! ^n - 5r - 1?h ! n! n! +r ]n - r g! ^n - 5r - 1?h ! n! rn! + ]n - r g! ]n - r + 1g! ] n + 1 - r g n! rn! + ]n + 1 - r g]n - r g! ]n + 1 - r g! ] n + 1 - r g n! rn! + ]n + 1 - r g! ]n + 1 - r g! n $ n! + n! - rn! + rn! ]n + 1 - r g! nn! + n! ]n + 1 - r g! ] n + 1 g n! ]n + 1 - r g! ]n + 1g! ]n + 1 - r g!

(v) 1

(d) 5005

(e) 38 760

R1R2

R2R3

R3B1

B1B2

R1R3

R2B1

R3B2

B1B3

R1B1

R2B2

R3B3

R1B2

R2B3

R1B3 (b) 77 520

B2B3

ANSWERS

5.

15 504

6. 210

8.

(a) 720 (b) 120

10. 296 010

7. 2 598 960 9. (a) 2184 (b) 364

11. 4845

13. 23 535 820

27.

12. 2925

14. (a) 792 (b)

5 12

(c)

5 33

15. (a) 100 947 (b) 462 (c) 924 (d) 36 300 (e) 26 334 (f) 74 613 (g) 27 225 16. $105 17. (a) 2 042 975 (b) 55 (c) 462 462 (d) 30 030 18. (a) 3003 (b) (i) 2450 (ii) 588 (iii) 56 (iv) 1176 19. (a) 1.58 # 10 10 (e) 12 271 512

(b) 286 (c) 15 682 524 (d) 5 311 735

20. (a) 395 747 352 (b) 32 332 300 (c) 4 084 080 (d) 145 495 350 (e) 671 571 264 21. (a) 170 544 (b) 36 (c) 20 160 (d) 17 640 (e) 6300

28.

22. (a) 7 (b) 27 132 (c) 13 860 (d) 20 790 (e) 27 720 23. (a) 5 (b) 360 (c) 126 24. (a) 792 (b) 792 (c)

25.

12! ] 12 - 5 g !5! 12! = 7!5! 12! 12 C7 = ] 12 - 7 g !7! 12! = 5!7! 12 12 ` C5 = C7 12

9 8

C5 =

C 6 = 84

8

C 6 + C 5 = 28 + 56 = 84 9 8 8 ` C6 = C6 + C5 26.

13! b 13 l = 7 ] 13 - 7 g !7! 13! = 6!7! 13! b 13 l = 6 ] 13 - 6 g !6! 13! = 7!6! 13 13 `b 7 l=b 6 l

29.

10! b 10 l = 4 ] 10 - 4 g !4! 10! = 6!4! 9! 9! b9 l + b9 l = + 4 3 ] 9 - 4 g !4! ] 9 - 3 g !3! 9! 9! = + 5!4! 6!3! 6 # 9! 4 # 9! = + 6 # 5!4! 4 # 6!3! 6 # 9! 4 # 9! = + 6!4! 6!4! 6 # 9! + 4 # 9! = 6!4! 10 # 9! = 6!4! 10! = 6!4! 10 l b 9 l b 9 l b ` 4 = 4 + 3 n! bn l = r ] n - r g !r! n! b n l= n-r (n - 5 n - r ? ) ! ] n - r g ! n! = ]n - n + r g! ]n - r g! n! = r! ] n - r g ! n n ` br l = bn - r l n

Pr =

n! ]n - r g!

n! ] n - r g !r! n! = ]n - r g!

r! n C r = r! #

` n Pr = r! n C r n! n 30. b l = k ] n - k g !k! ]n - 1 g! ]n - 1 g! bn - 1 l + bn - 1 l = + k-1 k ] n - 1 - ] k - 1 g g ! ] k - 1 g ! ] n - 1 - k g !k! ]n - 1 g! ]n - 1 g! = + ] n - k g ! ] k - 1 g ! ] n - 1 - k g !k! ]n - k g]n - 1 g! k ]n - 1 g! = + k ] n - k g ! ] k - 1 g ! ] n - k g ] n - 1 - k g !k! k ]n - 1 g! ]n - k g]n - 1 g! = + ] n - k g !k! ] n - k g !k! ]k + n - k g]n - 1 g! = ] n - k g !k! n ]n - 1 g! = ] n - k g !k! n! = ] n - k g !k! n =b l k

847

848


Test yourself 13 1.

13 (b) 22

4 2. (a) 11

(a) 5040 (b) 720

3.

(a) 24 (b) 12

5.

(a) 65 780 (b) 25 200 (c) 252

7.

120

17 (c) 22

4.

(a) ] n - 1 g !

(b)

6.

(a) 792

5 44

4. (a) 720 (b) 120

8. 2.4 # 10 18

9.

7.

6. 29%

1 9

10. 142 506

11. 990 12. (a) 40 320 (b) 362 880 (c) 80 640 (d) 168 13. (a) 19 958 400 (b) 4 989 600 (c) 181 440 (d) 9 979 200 (e) 181 440 n! n 14. b l = k ] n - k g ! k!

15. (a) 151 200 (b) 881 280

16. 1.08 # 10 17

20. (a) 1 n n! c m= ] n - 0 g ! 0! 0 n! = n! 0 ! =1 n! n c m= n ] n - n g ! n! n! = 0!n! =1 n n ` c m=c m 0 n

(a) 60 (b) 72 (c) 30

3.

n! bn l = k ] n - k g ! k!

Pr =

1 4

3 40

21 40

8.

(a) 1 860 480

9.

(a) 94 109 400 (b) 7920 (c) 5 527 200 (e) 37 643 760 (f) 23 289 700 4 35

(b)

(b)

(c)

(d)

(d) 93 024

17 35

Practice assessment task set 4 2. P ] x g = ] x - 1 g ] x + 1 g ] x + 4 g 3. y = 3x 4

1.

1 2

4.

(a) 362 880

(b) 4320

5. -1, 2

(c) 282 240 3 (c) 4

6.

^ 19, 10 h

8.

x 2 + y 2 = 2; circle centre ^ 0, 0 h radius

9.

3060c; 161c3l

d=

7. (a) - 4

(b) −2

(d) 10 2

| ax 1 + by 1 + c | a2 + b2

40 10 =4 = radius ` line is tangent =

2. (a) 360 (b) 60

]n - 1 g! ]n - 1 g! bn - 1 l + bn - 1 l = + k-1 k (n - 1 - 5 k - 1 ? ) ! ] k - 1 g ! ] n - 1 - k g ! k! ]n - 1 g! ]n - 1 g! = + ] n - k g ! ] k - 1 g ! ] n - k - 1 g ! k! ]n - k g]n - 1g! k ]n - 1g! = + k ] n - k g ! ] k - 1 g ! ] n - k g ] n - k - 1 g ! k! k ]n - 1 g! ]n - k g]n - 1 g! = + ] n - k g ! k! ] n - k g ! k! k ] n - 1 g !+ ] n - k g ] n - 1 g ! = ] n - k g ! k! ] n - 1 g !5 k + n - k ? = ] n - k g ! k! ] n - 1 g !n = ] n - k g ! k! n! = ] n - k g ! k!

n n-1 n-1 ` bk l = bk - 1 l + b k l

(b) 246

10. Distance from centre ^ 0, 0 h to line is


5. (a) 90 720

` n Pr = r! n C r

10. (a)

18. (a) 15 (b) 181 440 19. 37 015 056

k!

n! ]n - r g! n! r! n C r = r! ] n - r g !r! n! = ]n - r g! n

17. (a) 1 709 316 (b) 203 490 (c) 167 580

(b)

(b)

]n - k + 1g!

11. k = -2

1 2

12. x = 74c (+s in alternate segment) y = (180c - 74c) ' 2 = 53c (+ sum in isosceles D) 13. 120

14. x 1 - 2, x 2 2

15. P ] x g = ] x - 2 g2 Q ] x g Pl(x) = ] x - 2 g2 Ql(x) + 2 (x - 2) Q (x) P (2) = ] 2 - 2 g 2Q (2) =0 Pl(2) = ] 2 - 2 g2 Ql(2) + 2 (2 - 2) Q (2) =0 16. (a) 1 17. 126

(b) 3

(c) ab = -

18. 7

1 ,a+b=0 10

19. 7.1 m

20. 131c 38l

ANSWERS

31. Domain: all real x; range: y $ - 3

21. (a) P ] x g = ] x - 1 g ] x - 3 g2 (b) y

32.

+ACB +ABC AC ` by AAS, DABC

33. 46 m2 1

x

3

= +ECD ^ vertically opposite angles h = +CED (alternate angles AB||ED) = CD ^ given h / DCDE

34. x + y - 3 = 0

35. x 2 - 12x + 36 = ] x - 6 g2

36. 41c 38l

37. y $ 2.5, y # - 6.5 38.

-9

22. 4 1 39. d - 1 , 7 n 7 7 40. (a) 9x - y + 16 = 0 (c) Q = ^ - 20, 0 h

(b) x + 9y + 20 = 0 (d) 27c 21l

41. Domain: all real x ! ! 2; range: all real y 42. (a) sin ^ a - b h 43. (a) 149.1 m 23. P ] -2 g = - 55 P ] x g = ] x + 2 g ^ 2x 2 - 11x + 23 h - 55 on division ` P ] - 2 g is the remainder 24. (a) abc + acd + bcd + abd = -

d =0 a

(b) 1 (c) -1

25. P ] x g = ] x - 3 g 2 Q ] x g P ]3 g = ]3 - 3 g 2 Q ]3 g = 0Q ] 3 g =0 Pl] x g = 2 ] x - 3 g Q ] x g + ] x - 3 g 2 Ql] x g Pl(3) = 2 (3 - 3) Q (3) + ] 3 - 3 g 2 Ql(3) = 2 (0) Q (3) + 0Ql(3) =0

45.

3

(b) cos 45c =

(b) 46c 48l

2

(c)

^ 3 + 1 h2 8

44. X = ^ 7.5,17.5 h

46. x 11, x 21.6

48. (a) x - 8y + 129 = 0

1

47. x = 150c

1 1 n (b) R = d 7 , 17 64 8

t 4 + 2t 3 - 6t 2 + 2t + 1 ^ 1 + t2 h 2 50. f ] x g = 3x 3 - 7x 2 - 5x - 3 f ]3g = 3]3g3 - 7]3g2 - 5]3g - 3 = 81 - 63 - 15 - 3 =0 So x - 3 is a factor of f ] x g = 3x 3 - 7x 2 - 5x - 3 49.

51. a = 1, b = - 3, c = -1 52. 3x - y - 1 = 0, x + 3y - 7 = 0

26. 1 884 960 27. Radius 3; x 2 + y 2 = 9

53. 2175 cm3; 1045 cm2

28. a = 3, b = -14, c = 9

54. y 1 - 3, - 2 1 y 1 -1

29. (a) 8.1 m (b) 35c 46l

55.

56. (a) x = 60c, 120c, 240c, 300c

30. (a) !1 (b) P (1) = ^ 1 2 - 1 h (1 2 + 5) =0 2 3 Pl(x) = 6x ^ x 2 - 1 h (x 2 + 5) + ^ x 2 - 1 h $ 2x 3

Pl(1) = 6 (1) ^ 1 2 - 1 h (1 2 + 5) + ^ 1 2 - 1 h $ 2 (1) =0 2

3

(c) x = 270c

56 65 (b) x = 0c, 90c, 360c

849

850


57. (a) 0 (b)

60. x = - 6.5, y = 2.8

61. y = - x 4

62. (a) 4

(d) 1

(b) −2

(c) −3

63. P ] x g = ] x - 2 g ^ x 2 + 1 h + 5 65. P ] x g = - ] x - 5 g ] x + 1 g 2

58. i = 90c # n + ] - 1 g n # 135c

59. 8c 8l

1 2

(e) 22 64. 15 504

66. 63

68. (b)

69. (c)

70. (a), (b), (d)

71. (b)

72. (b)

73. (a)

75. (d)

67. (a)

74. (d)

Maths In Focus Preliminary 3 Unit

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