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CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2
MATHEMATICS
9740/01
Paper 1
3 September 2009 3 hours
Additional Materials: Materials:
Answer Paper Graph Paper List of Formulae (MF 15)
READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the all the questions. q uestions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this cover sheet in front and fasten all your work securely together. The number of marks is given in brackets brackets [ ] at the end of each question or part question. question.
Question No.
Marks
Question No.
Marks
1
/4
7
/ 10
2
/6
8
/ 11
3
/6
9
/ 11
4
/7
10
/ 14
5
/7
11
/ 14
6
/ 10
TOTAL
/ 100
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1 Solve the inequality x 2 3x 4 0 . Hence find the exact value of
3 2
1
[1]
x 2 3 x 4 dx .
[3]
2 Referred to the origin O, the position vectors of the points A and B are 3k respectively. respectively. – 3i + 9k and 3i + 3 j – 3 (i)
Find the position vector of the point M on on the line segment AB such that [2] AM : AB = 2 : 3. (ii) Show that OM is perpendicular to AB. [2] (iii) The point C is is the reflection of the origin O in the line AB. Calculate the area of triangle OAC . [2] 3 Find
x sin x d x , 1 cos x 1 x d x , where 0 x 1 , using the substitution x cos
(i) (ii)
[2] 2
u.
[4]
4 The r th th term of a sequence is given by ur = r (3 (3r + + 1), r = 1, 2, 3, …. n
(i)
Write down the values of
u
r
[1]
for n = 1, 2, 3, and 4.
r 1 n
(ii) Make a conjecture for a formula for
u
r
, giving your answer in the form
r 1
[1]
nf(n), where f(n) is a function of n. n
(iii) Prove by induction a formula for
u
r
.
[5]
r 1
5 Use
the
substitution
y
u
to
show
that
the
differential
equation
x
d y d x
2 xy 2 y 1 x 2 x 2
reduces to the differential equation
the general solution for y in terms of x.
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du d x
3u 1 x 2
. Hence find [7]
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2 6 Given that y e 1 x , prove that
d y 2 y x . d x d x 2
d 2 y
Hence expand e 4 x .
1 x in ascending powers of x up to and including the term in
[2]
2
[6]
Use this series expansion to estimate the value of
1 2 0
e
4 x 2 4
d x , correct to 3 decimal places.
[2]
7 A candy maker is interested in using containers in the shape shown below to package his candies. The container is made up of an open cylinder of height h cm and radius r cm, with a hollow hemispherical lid of radius r cm. In order to minimise production cost in this difficult time, the candy maker wants to use containers with the least surface area while maintaining the volume of each container at 500 cm3. If the material used to construct the container cost $0.015 per cm2, find, using differentiation, how much a container with minimum surface area costs to the candy maker. Leave your answer to 2 decimal places. [10] 4 [Volume of sphere, V r 3 ; Surface area of sphere, S 4 r 2 ] 3
r
h
8(a) One of the roots of the equation z 4 z 3 4 z 2 3 z p 0 is 1 – 2i, where p is a constant. A student says that “One of the other roots must be 1 + 2i.” Explain why this statement is not entirely correct.
[1]
(i) Determine the value of p. (ii) Find the exact values of all the other roots of the equation.
[1] [3]
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8(b) The fixed complex number a has modulus R and argument , and a* denotes a3 [3] the conjugate of a. Given that q , express q in trigonometric form. a* 1
If
q 6 is
[3]
purely imaginary, determine the least positive value of .
2 9 A curve is defined by the parametric equations x 4t , y
2t t a 2
, where a is
a positive constant. [4] Using differentiation, find the turning points of this curve in terms of a. 16 x (ii) Show that the cartesian equation of the curve is y 2 . [2] ( x 4a) 2 16 x (iii) Sketch y 2 , indicating clearly the coordinates of the points 2 ( x 4a) where the graph crosses the axes, turning points and the equations of any [3] asymptotes. (i)
(iv) Describe a sequence of geometrical transformations that map the graph of 16(3 x 4a) 16 x 2 [2] onto the graph of . y 2 y ( x 4a) 2 9 x 2
10(a) Susan baked 2500 cookies. She decided to give part of them to her friends and sell the remainder for charity. (i)
To pack the cookies meant for her friends, Susan placed 5 cookies in the first bag. Each subsequent bag she packed contained double the number of cookies in the previous bag. How many complete bags of cookies did [3] Susan have to offer her friends?
(ii) To pack the remaining cookies meant for sale, Susan placed 4 cookies in the first bag. Each subsequent bag she packed contained 3 cookies more than the previous bag. If she charged $0.50 for each cookie, how much [5] would the last complete bag of cookies cost? (b) Given that T r 2 2rb , where b is a constant, n
(i)
show that the terms of the series
ln T form an arithmetic progression, r
[2]
r 1
(ii) express S n , sum of the first n terms of this arithmetic progression, in terms [2] of n and b, (iii) hence find an inequality satisfied by the constant b such that the difference [2] between S 13 and S 14 is not more than 0.5.
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11 The planes p1 and p2, which meet in the line l , have vector equations 2 1 0
r = 4 1 0 1 1 , 6 1 1
2 2 1 r = 4 2 3 2 0 6 0 1 respectively, where 1 , 2 , 1 and 2 are real constants. (i) Show that l is parallel to the vector 5i + 6 j + k . [3] (ii) Calculate the acute angle between p1 and p2. [2] (iii) Find, in exact form, the perpendicular distance from the point with coordinates (4, 2, 2) to p2. [2] The plane p3 has equation ax – 2 y + 2 z = b, where a, b . (iv) Find b in terms of a such that all three planes meet at the single common
2 point with position vector 4 . 6
[4]
(v) If given instead that a = 2, find the values of b, such that the distance between the planes p1 and p3 is
1 3
units.
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JC 2 H2 Mathematics Preliminary Examination 2009 Paper 1 -- Marking Scheme 1 x 2 3 x 4 0
x 1 x 4 0
[B1]
1 x 4
3 2
1
2 x 3 x 4 dx
4
2 x 3 x 4dx
1
3 2 4
2 x 3 x 4dx
[M1]
[M ark awarded for spli ttin g cor r ectly in to 2 sections eit her using modulu s or negative si gn] 4
3 2
1
3 2
x 3 3 x 2 x 3 3 x 2 2 4 x 4 x x 3 x 4 dx 2 2 3 1 3 4
[M1]
[M ark awarded for corr ect integration]
3 2
1
x 2 3 x 4 dx 12
1 6 2 2
OR
2(i)
3 2
1
2 x 3 x 4 dx
OM
[A1]
25 12 2 2
2OB OA 3 3
3 2 3 0 3 9
[M1]
3
1 2 1
[A1]
(ii) If OM is perpendicular to AB, OM AB = 0 6 AB 3 12
1 6 OM AB 2 3 = 0 1 12
[M1]
[M1]
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(iii) Area of triange OAC = OA OM
OR Area of triangle
3 1 = 0 2 9 1 18 = 12 6
1 4 = 2 2 1 8
[M1]
= 6 84 =22.4 units2
= 22.4 units2 (or 6 14 units2)
[A1]
3
x sin xdx
x cos x cos xdx
[M1] [A1]
sin x x cos x c dx 2 sin u cosudu
cos
1 2
2
1 cos 1
u
2
2 sin u cos udu
[M1]
u
du
[M1]
cos u
2 sec udu 2 ln sec u tan u c
[M1]
1 1 x c 2 ln x x
[A1]
4 u1 = 4, u2 = 14, u3 = 30, u4 = 52 1
(i)
u r 1
r
n
(ii)
u
r
2
u
4,
r 1
r
18 ,
3
u
r
4
u
48 ,
r 1
r 1
r
100
[B1]
nn 12
[B1]
r 1
n
(iii) Let Pn be the statement
u r 1
1
LHS of P1 =
u r 1
r
r
2 nn 1 , n Z
[B1/2]
u1 4
RHS of P1 = 1(1 + 1)2 = 4 P1 is true and forms the basis for induction. Assume Pk is true for some k Z , i.e.
k
u r 1
k 1
Required to show Pk +1 is true, i.e.
u
r
r
[M1]
k k 1
k 1k 2
2
[M1]
2
r 1
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LHS k 1
u r r 1
k
u r u k 1 r 1
k k 12 k 13k 4 k 1k 2 k 3k 4 k 1k 2 4k 4
[M1] [M1]
k 1k 2
2
=RHS Since P1 is true and Pk is true Pk+1 is true. Therefore, by mathematical induction, Pn is true n Z . 5 dy dx
x
du
u
[B1]
dx x 2
x
du
u
2u 2
u
1
x Substituting, dx 2 x( x 2) x 2ux 2u x du u x ( x 2) dx du 3ux x x dx ( x 2) du 3u 1 dx
[B1/2]
[M1] [M1]
( x 2)
1 1 du dx 3u 1 x 2
[M1]
1 ln 3u 1 ln x 2 ln c 3
[M1]
ln 3 xy 1 ln c( x 2)
3
3 xy 1 k ( x 2)3 y
[B1]
k ( x 2)3 1
[A1]
3 x
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6
2 y e 1 x 2 dy e 1 x (2 x) 2 xy dx
2
d y dx
2
2 x
dy dx
[M1]
[M1]
y 2
dy 2 y x (shown) dx 2 dy d y dy 2 x 2 dx dx 3 dx dx dy d 2 y 2 2 x 2 dx dx 3
d y
d 4 y dx 4
[M1]
d dy d 2 y 4 2 x 2 dx dx dx
d 3 y d 2 y 4 2 2 x 3 2 dx dx dx d 2 y
6
d 2 y dx
2
2 x
d 3 y dx
[M1]
3
y (0) e , y ' (0) 0 , y' ' (0) 2e , y' ' ' (0) 0 and y ( 4) (0) 12e
[B2]
[deduct B½for ever y mi stake]
e
1 x 2 e 0 x 2e x 2 0 x 3 12e x 4 2! 4! e
[M1] [A1]
e ex 2 x 4 2
1 2 0
e
4 x 2 4
dx
1 2 0 1 2
0
x 2
2
1
e
dx 2
4
x e x e e dx 2 2 2
=1.388 (3dp) [use GC]
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7
1 4
3 r 500 2 3 2 2 r h r 500 3
r 2 h
h
500 2
r
[M1]
2
r
[M1]
3
1 4 r 2 2 500 2 3 r 2 2 r 2 r r 3 S r 2 2 rh
[M1]
[M1]
5 2 1000 r 3 r
dS
dr
dS dr
10 3
r
1000
[M1]
2
r
10 r 3 3000 3r 2
0 10 r 3 3000 0
r 3
300
[M1]
or 4.5708 (4 dp)
[A1]
2
d S 2
dr
10 2000 3 3 r
0 when r 3
300
[M1]
2
5 300 3 1000 3 = 328.1715 (4 dp)
Minimum surface area
3
300
Cost of box with minimum surface area 328.17150.015 4.9226 $4.92 (2 dp) 8a The statement is only true if p is real.
[M1]
[A1] [B1]
(i) Using GC, p = 5.
[B1]
(ii) We have z 4 z 3 4 z 2 3 z 5 ( z (1 2i))( z (1 2i))( z 2 az b) , where a and b are real. = ( z 2 2 z 5)( z 2 az b)
[M1]
Comparing coefficients of similar terms, we have a = b = 1
[B1]
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1
For z 2 z 1 0 , we have z (b)
a3 a
a
2
3 2
i
[A1]
3
a
R 2
[B1]
arg(q) = arg (a 3 ) - arg (a ) = 3 arg(a) + arg(a) = 4 Thus, q R 2 cos( 4 ) i sin(4 ) 1
[M1] [A1]
1
2 2 q 6 = R 3 cos i sin 3 3
[M1] - arg
2 0 , 3
Given that cos
[M1]
2 3 or 2.36 radians 3 2 4
9 (i)
2t 2 2a 2 , 2 dt (t a)
[M1/2]
8t
[M1/2]
dy
dx dt dy
[A1]
dy dt a t 2 =0 . dx dt dx 4t (t 2 a ) 2
[M1/2] – d y/d x [M1/2] – “=0”
t=
[A1]
a
x 4a, y
(2 a ) 2a
Turning points: (4a,
a
a
a a
) , (4a,
a a
).
[A1]
(ii)
y
2t t 2 a
2t y 2 t a
2
2
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y
x
2
y
x a 4 16 x
2
(iii) y
2
[M1]
2
(shown)
x 4a
2
16 x
x 4a
2
Shape – B1 2
[-1/2 if shar p at origin]
y
( 4a, 1
(0, 0)
a a
) x
10
–1
(4a, –2
20
30
a a
40
50
60
70
80
y=0
)
(iv) Method 1: A translation of 4a unit in the direction of positive x-axis followed by a scaling of 1/3 units along the x-axis. Method 2: Scaling of 1/3 units along the x-axis followed by a translation of 4a/3 unit in the direction of positive x-axis. “
Turning pts – B1/2 each (0, 0) – B1/2 “ y = 0” – B1/2
Each transf – B1 [No ½mar k]
”
(A ccept use of the word s hift )
10a GP : a = 5, r = 2 (i) S n 52 n 1 2500
[M1] - Sn [M1] - ineq
n 8.97 (3 sf) No. of complete bags to offer her friends = 8 (ii) No. of cookies left for sale = 2500 – 5(2 -1) = 1225
[A1] [M1]
AP : a = 4, d = 3 S n
n
8 3n 1 1225 2 29.4 n 27.8 (3 sf)
[M1]
Last complete bag = 27th bag
[B1]
T 27 4 26(3) 82
[M1]
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Cost of last complete bag = 82(0.5) = $41 [A1] b(i)
(ii)
ln T r ln T r 1 2 rbln 2 2 r 1bln 2 = b ln 2, a constant (shown) S n
n
r 1
ln T r
n
2
(iii)
22 bln 2 n 1b ln 2 n ln 2
2
n 1b 4
S 14 S 13 0.5 7 ln 215b 4
0.194 b 0.0913
[M1] [B1] – b ln 2 [M1]
[A1]
13 ln 2 14b 4 0.5 2
(3 sf)
[M1] [A1]
[N o marks if modulus sign i s left out]
11i Let n1 and n2 be the normals of p1 and p2 respectively.
1 0 n1 = 0 1 1 1
1 1 1 2 1 3 n2 = 3 0 2 0 1 3
[M1]
[M1]
1 3 5 1 2 6 1 3 1 Therefore l is parallel to 5i + 6 j + k.
(ii) Acute angle between p1 and p2 = cos 1
[M1]
1 3 1 2 1 3 3 22
= 75.7o
(iii) Perpendicular distance =
[M1]
4 2 3 2 4 2 2 6 3 22
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2 3 2 2 4 3
=
22
22
=
22
= (iv)
22
[A1]
The point (2, 4, 6) must lie on p3 So 2a – 8 + 12 = b b = 2a + 4
[M1] [A1]
Since the three planes meet at a single point, we must exclude the case where they meet along a line. If p3 meets in the line l , the normal of p3 is perpendicular to l .
a 5 So 2 6 = 0 2 1 a = 2
[M1]
Therefore b = 2a + 4, where a ≠ 2 (v)
1 p1: r . - 1 4 1
2 p3: r . - 2 b 2
and
1
Distance between p1 and p2 = 4
3 4 3
b
b
3
1
2 3
[A1]
[M1]
3
2 3
1 3
or
4
3
b
2 3
1
[M1]
3
b = 6 or 10
[A1]
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CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2
MATHEMATICS
9740/02
Paper 2
16 September 2009 3 hours
Additional Materials:
Answer Paper Graph Paper List of Formulae (MF 15)
READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this cover sheet in front and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
Name :
Class :
Question No.
Marks
Question No.
Marks
1
/9
7
/8
2
/9
8
/8
3
/ 11
9
/9
4
/ 11
10
/ 11
5
/4
11
/ 13
6
/7
TOTAL
/ 100
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Section A: Pure Mathematics [40 marks]
1 (i)
Express
4 r r (r 1)(r 2) n
(ii)
Hence find
[2]
in partial fractions.
4 r
r (r 1)(r 2) .
[4]
r 3
n
(iii) Use your answer in (ii) to find
2 r
r (r 1)(r 2) .
[3]
r 1
2 (a)
(b)
Calculate the area of the region bounded by the curves y e x 4 1 and y lnx 1.
[3]
The region S is bounded by the curve y x 2 1 and the lines y 1 and x b , where b 0 . y
y x 2 1
S
y 1 0
b
x
V x is the volume of the solid of revolution formed when S is rotated completely
about y 1 . V y is the volume of the solid of revolution when S is rotated completely about the y-axis. Find the value of b such that V x V y .
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3 (a)
Find the eighth roots of the complex number 8 3 8 i, giving your answers in the form r e i , where r is a positive real constant and . [4]
(b) The fixed complex number w is such that 0 < arg (w) <
. In an Argand diagram, 2 w is represented by the point W , and the complex number – i w is represented by V . The variable z is represented by P . Sketch in a single diagram showing W , V and the locus P in each of the following cases: (i) | z – w | = | z + iw| [4]
(ii) arg ( z – w) = arg ( – iw)
Find in terms of w, the complex number representing the point at which the loci [2] intersect. [1] If Q is the point of intersection, describe the geometric shape of OWQV .
4 The functions f and g are defined by 1 f : x → x , x > 0 x g : x → sin x, x [0 , 2] (i) With the aid of a diagram, show that f – 1 exists. (ii) Define f – 1 in a similar form. (iii) Sketch the graphs of f and f – 1 on the same axes. Your sketch should show clearly the axial intercepts, and the geometrical relation between the two graphs and the line y = x. (iv) Show that the composite function fg does not exist. (v) Determine the largest possible domain of g for which the composite function fg exists. Hence, define fg in a similar form.
[1] [4]
[2] [1] [3]
Section B: Statistics [60 marks]
5 The Student Council of Catholic Junior College is organizing Rockella, a music concert, to raise funds for the needy students in school. The Council intends to survey a sample of 300 students to find out the music genre preferences of students in the school. (i)
Given that the school has 1500 students, describe how the sample could be chosen using systematic sampling. [2]
(ii) Suggest why it would be possible to use stratified sampling instead.
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6 In July 2009, Company J introduced the new jPhone 3GS, the fastest, most powerful jPhone, packed with improved performance up to twice as fast as the previous model with longer battery life. The battery life in a randomly chosen jPhone 3GS has a normal distribution and the battery life of the phone is supposed to be 120 hours. A random sample of 90 jPhones 3GS is taken, and the battery life of each phone, x hours, is recorded. The data are summarised by
( x 120) 29 ,
( x 120)
2
419 .
Test, at the 5% significance level, whether the mean battery life of a jPhone 3GS is less than 120 hours.
[6]
Explain, in the context of the question, the meaning of “at the 5% significance level”.
[1]
7 (a)
(b)
Events A and B are such that P ( A)
1
, P ( B)
1
and P ( B | A' )
1
, where A' is 4 4 3 the complement of A. Investigate whether A and B are mutually exclusive, justifying your answer. [3] A game is played using a fair die as follows: If a six is thrown, the game ends and the score is 6. If a three is thrown, the player throws the die one more time and the score is the total of the two numbers thrown. If any other number is thrown, the player throws the die two more times unless his second throw is 6, in which case the game ends. The score is the total of the numbers thrown. For example, if the first throw is 4 and the second throw is 6, the game ends and the score is 10. If the first throw is 5, second throw is 2 and third throw is 1, the score is 8. (i) Find the probability that the score is 5. (ii) Given that the score is at most 5, find the probability that the score is 4.
[2] [3]
8 A group of 10 people consists of 3 married couples and 4 single men. (a) A committee of 4 is to be formed from the 10 people. (i) How many different committees can be formed? (ii) How many different committees can be formed if the committee can consist of at most 1 married couple? (b) The group sits at a round table with 10 seats each of a different colour. (i) If each man sits next to his wife, how many ways can they be seated? (ii) If one man is absent and the rest are allowed to sit without any restrictions, how many ways can they be seated?
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9 After the Monetary Authority of Singapore’s report on the structured financial products sold by financial institutions in Singapore, a survey found that 95% of the respondents would favour greater consumer protection. (i) Find the probability that out of 30 respondents, 25 of them would favour greater consumer protection. [2] (ii) Find the least number of respondents surveyed such that the probability that at least 40 of them would favour greater consumer protection exceeds 0.980. [3] (iii) Using a suitable approximation, find the probability that out of 60 respondents, at most 58 of them would favour greater consumer protection. [4] 10 The ages, x years, and heights, y cm, of 11 boys are as follows: x
6.6
6.8
6.9
7.5
7.8
8.2
9.0
y
112 116 119 123 125 130 135
10.1 139
11.4 140
12.8
13.5
141
141
(i)
Sketch the scatter diagram for the data and comment on the suitability of a linear model between x and y. [3] (ii) State, with a reason, which of the following models is more appropriate to fit the data points: (a) y2 = a + bx2 , where a > 0, b > 0 (b) y = axb where a > 0 , 0 < b < 1 (c) y
ax b x
where a > 0, b < 0
[2]
(iii) For the appropriate model chosen, state the product moment correlation [4] coefficient and estimate the values of a and b, for the transformed data. (iv) Estimate the age of a boy when his height is found to be 110cm. Comment on the [2] reliability of this estimate. 11 The weight of a bar of Brand A chocolate is normally distributed with mean 180g and standard deviation 10g. The weight of a bar of Brand B chocolate is normally distributed with mean 240g and standard deviation 20g. (a) Find the probability that the weight of 3 randomly chosen bars of Brand A chocolate is more than twice the weight of a randomly chosen bar of Brand B chocolate. (b) A sample of 60 bars of Brand A chocolate is sent for inspection. (i) Find the probability that the sample mean exceeds 179g. (ii) Explain whether you need to use Central Limit Theorem in your working. (c) Brand A chocolates are sold at $2 per 100g. (i) Find the probability that a bar of Brand A chocolate costs more than $3.80. (ii) 100 bars of Brand A chocolate are packed in a box. Using a suitable approximation, find the probability that, in one box, more than 5 bars cost more than $3.80 each.
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[5]
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JC 2 H2 Mathematics Preliminary Examination 2009 Paper 2 -- Marking Scheme Section A: Pure Mathematics [40 marks]
1
(i)
4 r r (r 1)(r 2)
A r
B r 1
C
r 2 [B2]
By Cover Up rule or any other methods, A = 2, B = – 3 and C = 1 [ Subtr act 1 mar k f or ever y mistake] n
(ii)
4 r
n
2
1
3
r (r 1)(r 2) r r 1 r 2 r 3
[M1]
r 3
3 1 2 3 3 1 3 2 3 1 2 4 4 1 4 2 3 1 2 5 5 1 5 2 3 1 2 6 6 1 6 2
3 1 2 n 2 n 3 n 4 3 1 2 n 1 n 2 n 3 3 1 2 1 2 n n n
=
3
2
=
2
1
1
2
2 n 1
[M1] 3 n 1
2
[M1]
n
1
[A1]
n n 1 n2 = nn 1 n
(iii)
2 r
r (r 1)(r 2) r 1
n 2
k r 2 k 3
n 2
k 3 n2
r 3
2 (k 2)
[M1]
(k 2)(k 2 1)(k 2 2) 4 k
(k 2)(k 1)(k ) 4 r (r 2)(r 1)(r )
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(n 2) 2 (n 2)(n 2) 1 n
[M1]
(n 2)(n 1)
[A1]
[ Accept answer i n parti al fr actions]
2 (i) By G.C Intersection point (1.05395, -0.947453), (4.3919, 0.47976) Area
4.3919
ln x e x 4 dx [M 1 - corr ect li mits ; M 1 - corr ect for m]
[M1+M1]
1.05395
Area 1.68
[A1]
(ii) V x
b
x dx
[M1]
2 2
0
b
x 5 b5 V x 5 5 0 V y b
2
b
[M1] b2
2
0
ydy
[M 1 - Vol of cylin der ; M 1 - Vol of revoluti on abt y-axi s] [M1+M1]
b2
y 2 b4 4 V y b 2 2 0
b5
5 4 b
[M1]
b4 2
1 0 5 2 b 0 (rejected ) b
b
5
[A1]
2
Alternative Solution
V x
b
x dx 2 2
[M1]
0
x V 5 5
b
x
5
[M1]
0
V y b b 2
b5
2
b 2 1
1
y 1dy
[M 1 - vol of cylin der , M 1 - vol of r evolu tion abt y-
axis] b 2 1
y V y b 2 1 4
y 2
b 2 12 1 b b 2 1 1 2 2 4
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b 2 1
y 2
y V y b 2 1 4
b 0 (rejected ) or b
b 4
b4 2
b4
[M1]
2
5
[A1]
2
5 3 i 4 6 (a) 8 3 8 i = 2 e
[M½ - mod, M½ - arg]
z 2 e 8
z
4
2e
i
5 6
e
5 i ( 2 k ) 8 6
i 2 k
4
= 2 e
5 i ( 2 k ) 6
[M1]
, k = 0, 1,2,3,4
[M1 – taking 8th root, A1]
(b) M1/2, M1/2 - Correct marking of V and W (i) B1 - correct locus B ½ if the perpendicular bisector does not pass through origin (ii) B1 – correct locus B1 - showing awareness that the halfline is parallel to – iw Let Q be the point of intersection. Using vector addition, OQ OW WQ w (iw) It is a square.
[M1,A1] [B1]
y
W w arg(-iw) arg (w)
(ii)
O
x
Q
-iw V
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4 (i)
y
Since any horizontal line y = k , k R will cut the graph of f at most once, hence f is one-one. Thus, f -1 exists. [M1] x
1
(ii) Let y x x
1
x y 1 2
But x > 0, x
2 y
Hence, f -1 : x
(iii)
[M1] y2 4
1
2 2 x 1 2
2
[M1]
y2 4
[M1]
x 2 4 , x (- , )
[A½ - rule, A½ - domain]
[B1] – correct graph of f -1 with y = x [B1] - intercepts y
1 1
x
(iv) Df : (0,) and R g : [-1 , 1]. Since R g Df , therefore fg does not exist.
[B1] [B1]
(v) Largest Dg : (0 , ) fg( x) = f(sin x) = sin x
1
[M1]
sin x
Hence, fg : x sin x
1 sin x
, x (0 , ).
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Section B: Statistics [60 marks]
5(i) Mention of sampling frame/list. Random starting point (i) from first 5 on the list (either explicitly or by giving an example), (ii) otherwise student must mention that once they reach the end of the list, they have to go back to the beginning of the list to pick. Select every 5th student
[B½ ] [B1]
[B½]
E.g. From the school’s registration list [B½] , randomly select a student from the first th 5 on the list as a startin g poin t [ B1] . Select ever y 5 student [B½] down the list from that starting point, until 300 names are obtained.
(ii) A sampling frame is easily accessible. The population can be divided into non-overlapping subgroups (e.g. boys and girls) [demonstrate knowledge of subgroups being non-overlapping by explicitly mentioning or giving examples ]
6 Let X be the battery life of a jPhone 3GS in hours. ( x 120) 29 120 120 119.6777778 Unbiased est. of popn mean, x n 90
Unbiased est. of popn var,
X ~ N 120,
2
ˆ
1
29
89
90
419
[B1] [B1]
[M1]
4.602871411
[M1]
4.602871411 90
H 0 : 120
[B1]
H 1 : 120 Test statistic, z
119.6777778 120
[M1]
4.602871411
90 Use GC, p-value = 0.0771035664 = 0.0771
[A1]
[accept 0.0771 – 0.0785]
Since p-value = 0.0771 > 0.05, we do not reject H 0 . [B½] Hence, there is insufficient evidence at 5% level to indicate that the mean battery life [A½] of a jPhone is less than 120 hours. Probability of concluding the mean battery life of a jPhone is less than 120 hours when it is actually 120 hours is 0.05.
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7
3 P ( B A ' ) (a) P ( B A ' ) 1 4 16 1 4 1 3 7 P ( B A) P ( B) P ( B A ' ) 0 3 16 48 Hence A and B are not mutually exclusive. 1
[M1] – 1st stmt
[M1] – “ 0” [A1]
6 3
1,2,3,4,5,6 6
1,2,4,5
1,2,3,4,5
1,2,3,4,5,6
(bi) P (score is 5) = P[(3,2), (1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1)] =
1 36
5 216
(bii) P ( score 3) P (1,1,1, )
[M1]
11
[A1]
216 1
B1 for finding P(score 3, 4)
216
P ( score 4) = P [(3,1), (1,1,2),(1,2,1), (2,1,1) ] =
9 216
score 4 score 5 score 5
P score 4 score 5 P
score 4 P 5 score =
9 1 9 11
[M1]
3
[A1]
7
8 (a)(i)10C4=210 (ii) 10C4 - 3C2 = 207
[B1] [M1, A1]
(b)(i) 222(7-1)! 10 =57600 [ M 1 - consideri ng permutati on of wif e and hu sband, M 1 - (7-1)!
[M1,M1, A1]
10]
[M1 ,A1]
(ii) (10-1)! 10 = 3628800
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9 (i) Let r.v. X r be the number of respondents who favour greater consumer protection out of r respondents. [M1] - distn X 30 ~ B(30, 0.95) P( X 30 = 25) = 0.0124 [A1] (ii) X n ~ B(n, 0.95) P( X n 40) > 0.98 1 – P( X n 39) > 0.98 Using GC, least n = 46
[M1] [M1] [A1]
(iii)Let r.v. Y be the number of respondents who do not favour greater consumer protection out of 60 respondents. Y ~ B(60, 0.05) Y ~ Po(3) approximately. P( X 60 58) = P(Y 2) = 1 – P(Y 1) = 0.801
[M1] [B1] [M1] [A1]
10 (i) Shape Label axes and 2 end points (6.6,119) & (13.5,141)
[B1] [B1]
Based on the scatter diagram, the linear model is not suitable even though r (= 0.906) is quite close to 1.
[B1]
(ii) Choose Model (b): y = axb Reason: The graph of y = axb fits the scatter diagram better. : From the scatter diagram, we see that as x increases, y increases at a decreasing rate.
[B1] [B1 – any 1 reason]
(iii) r = 0.932
[B1]
y = axb ln y = ln(axb) ln y = ln a + b ln x ln y = 4.1912 + 0.3056 ln x ln a = 4.1912 a = 66.1 b = 0.306
[M1] [A1] [A1] [B1 + B1]
(iv) when y = 110, x = 5.29. Extrapolation hence not reliable.
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11 (a) Let A be the random variable that denotes the weight, in grams, of a bar of Brand A chocolate A ~ N 180,100 Let B be the random variable that denotes the weight, in grams, of a bar of Brand B chocolate B ~ N 240,400 A1 A2 A3 2 B ~ N 60,1900
[M1]
P A1 A2 A3 2 B P A1 A2 A3 2 B 0
[M1]
P A1 A2 A3 2 B 0 0.916 ( to 3 s.f.)
[A1]
(b)(i)
A ~ N 180,
100
[M1]
60
P A 179 0.781( to 3 s.f.)
[A1]
(b)(ii) No need to use Central Limit Theorem because the population follows a Normal Distribution.
[B1]
(c)(i) 0.02 A ~ N 0.02180, 0.02 100 2
0.02 A ~ N 3.6,0.04
[M1] [A1]
P 0.02 A 3.80 0.1586552596 0.159 ( to 3 s.f.)
(c)(ii) Let X be the random variable that denotes the number of bars of Brand A chocolate, in a box, that cost more than $3.80. [M1] X ~ B 100,0.1586552596 n 100, n 30 np 15.86552596, np 5 [M1] nq 84.13447404, nq 5 [M1] X ~ N 15.86552596,13.34837682 approx P X 5 cc P X 5.5
[M1] [A1]
P X 5.5 0.9977237843 0.998 ( to 3 s.f.)
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