3U HSC Questions by Topic

3 UNIT MATHEMATICS (HSC) – METHODS OF INTEGRATION – HSC

Methods of Integration 3U97-1e)! 1

Using the substitution u = 2x + 1, or otherwise, find

4x

 2x + 1 dx .¤ 0

« 2 - Ln3 »

3U97-4b)! 1 2

By using the substitution x = sin t, or otherwise, evaluate



1  x 2 dx .¤

0

«

3 3  2 » 24

3U96-1f)! Using the substitution u = ex, find

ex  1 e2x dx .¤ « tan-1 ex + c »

3U96-2b)! Use the table of standard integrals to show that

6

15

dx x2  64

 Ln 2 .¤ « Proof »

3U95-1e)! 1

2

Use the subsitution u = 9 - x to find



6 x 9  x2 dx . ¤

0

« 2(27 - 16 2 ) » 3U94-1b)! 10

Evaluate

 2

x dx using the substitution x = t² + 1. ¤ x 1 « 21 13 »

3U94-3b)!

 4

Evaluate



3 sin x cos2 x dx . ¤

0

« 7 »

8

3U93-1c)! 1

Evaluate



4t (2t  1) 5 d t by using the substitution u = 2t - 1. ¤

1 2

« 13 »

42

3U92-1c)! 3

Find the exact value of

 1

1 4  x2

dx . ¤ «  »

6

3U92-4a)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

3 UNIT MATHEMATICS (HSC) – METHODS OF INTEGRATION – HSC 1

Evaluate

2x dx by using the substitution u = 2x + 1. ¤ (2x  1) 2

 0

« 1 ln 3  1 »

2

3

3U91-1a)! Evaluate: 1

x dx x 1



i.

2

0 1

 (1  5x )

4

ii.

dx , using the substitution u = 1 + 5x. ¤

-1

« i)

1 2

ln 2 ii) 352 »

3U90-1a)! Evaluate: 1

1 dx 1 x2



i.

0 1

x dx , using the substitution u = 1 + x. ¤ 1 x



ii.

0

« i)  ii) 2 (2  2) »

4

3U90-4a)! Find: i.



ii.



3

ln 2 x dx , using the substitution u = ln 2x x

cos2 2 x dx . ¤ « i)

1 2

  (ln 2x)2 + c ii) 1  x sin 4 x   c »

2

4



3U89-1b)! Evaluate: 2

1



i.

4  x2

1 0

x

ii.

dx ,

1+ x dx , using the substitution u = 1 + x. ¤

-1

« i) 3U88-2b)! Find

x

 3

ii) - 4 »

15

2  x2 dx using the substitution u = 2 + x². ¤ «

(2  x2 )3 c » 3

3U87-4i)! Find the volume of the solid formed when the region bounded by the x-axis and the curve y = x(8 - x3)4 between x = 0 and x = 2 is rotated about the x-axis. (You may need to use the substitution u = 8 - x3 to evaluate the integral involved.) ¤ ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

3 UNIT MATHEMATICS (HSC) – METHODS OF INTEGRATION – HSC

«

  227 units3 » 27

3U86-1iii)! 1

Use the substitution u = x² + 2 to evaluate

 0

x dx . ¤ (x  2 ) 2 2

« 1 »

12

3U85-1ii)! 1

Find

 x(1  x

2 2

) dx . ¤

0

«

7 » 6

3U85-4i)! 1

4

Using the substitution u = x , or otherwise, show that

 0

x3  dx  . ¤ 8 1 x 16 « Proof »

¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

3 UNIT MATHEMATICS (HSC) – PRIMITIVE OF SIN2X AND COS2X – HSC

2

Primitive of sin x and cos2x 3U96-3bc)! i.

 2

Show that  cos2 x dx   4

ii.

 1  .¤ 8 4

The function g(x) is given by g(x) = 2 + cos x. The graph y = g(x) for

   x  is rotated 4 2

about the x-axis. Find the volume of the solid generated. (You may use the result of part i.)¤ « i) Proof ii) 3U95-5a)! i.

(9  30  16 2 ) units3 » 8

Solve the equation sin 2x = 2sin2 x for 0 < x < .

 , then sin 2x > 2sin2 x. 4

ii.

Show that if 0 < x <

iii.

Find the area enclosed between the curves y = sin 2x and y = 2sin2 x for 0  x 

3U91-3b)!

 .¤ 4    « i) x = ii) Proof iii) 1-  units2 »  4 4

 2

Evaluate  sin2 3x dx . ¤ 0

« 0.25 » 3U90-4a)! Find: i.



ii.



ln 2 x dx , using the substitution u = ln 2x x

cos2 2x dx . ¤ « i)

1 2

  (ln 2x)2 + c ii) 1  x+ sin 4 x  + c »

2

4



3U88-1c)! Evaluate: 1

i.

x 0

2x dx , 1

2



ii.

 sin2 x dx . ¤ 0

« i) loge2 ii) 0.5 » 3U84-4ii)! a. b.

Noting that 2cos²x  1 + cos 2x, prove that 8cos4x  3 + 4cos 2x + cos 4x. Sketch, on the same diagram, the curves y = cos x, y = cos²x, for 0  x 

 . Find the 2

area enclosed between these curves and the volume generated when this area is rotated about the x-axis. ¤ ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

3 UNIT MATHEMATICS (HSC) – PRIMITIVE OF SIN2X AND COS2X – HSC

y 1 y = cos x y = cos²x 0

« a) Proof b)


 2

x

Area = 1 -

 2 units3 » units², Volume = 16 4

3 UNIT MATHEMATICS (HSC) – APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD – HSC

Equation dN  k (N - P) dt

Velocity and Acceleration as a Function of x Projectile Motion Simple Harmonic Motion 3U97-2b)! A particle is moving in simple harmonic motion. Its displacement x at time t is given by x = 3sin(2t + 5). i. Find the period of the motion. ii. Find the maximum acceleration of the particle. iii. Find the speed of the particle when x = 2.¤ « i)  ii) 12 iii) 2 5 » 3U97-5a)! A particle moves along the x axis, starting at x = 0.1 at time t = 0. The velocity of the particle is 2

described by v = 2xe -x , x  0.1, where x is the displacement of the particle from the origin. 2





i.

Show that the particle has acceleration given by a = e -2x 1  4x 2 , x  0.1.

ii. iii.

Hence find the fastest speed attained by the particle. Show that T, the time taken to travel from x = 1 to x = 2, can be expressed as 2

T= 1

iv.

1 x2 e dx . 2x

Use the trapezoidal rule with three function values to obtain an approximate value for T.¤ « i) Proof ii) e-0.25 iii) Proof iv) 10 »

3U97-7a)!

y P

V

h

0

x

A particle is projected horizontally from a point P, h metres above O, with a velocity of V metres per x = 0 and y = - g . second. The equations of motion of the particle are  i. Using calculus, show that the position of the particle at time t is given by x = Vt,

1 y = h - gt 2 . 2 A canistor containing a life raft is dropped from a plane to a stranded sailor. The plane is travelling at a constant velocity of 216 km/h, at a height of 120 metres above sea level, along a path that passes above the sailor. ii. How long will the canister take to hit the water? (Take g = 10 m/s2.) iii. A current is causing the sailor to drift at a speed of 3.6 km/h in the same direction as the plane is travelling. The canister is dropped from the plane when the horizontal distance from the plane to the sailor is D metres. What values can D take if the canister lands at most 50 metres from the stranded sailor?¤ ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


« i) Proof ii) 4.9 sec. iii) 239.04 m or 339.04 m » 3U96-5a)! A cup of hot coffee at temperature TC loses heat when placed in a cooler environment. It cools according to the law

dT  k(T  T0 ) , dt where t is time elapsed in minutes, at T0 is the temperature of the environment in degrees Celsius. i. A cup of coffee at 100C is placed in an environment at -20C for 3 minutes, and cools to 70C. Find k. ii. The same cup of coffee, at 70C, is then placed in an environment at 20C. Assuming k remains the same, find the temperature of the coffee after a further 15 minutes.¤ « i) -0.09589 (to 5 d.p.) ii) 31.86 » 3U96-5b)! A particle is moving along the x-axis. Its velocity v at position x is given by

v  10x  x2 . Find the acceleration of the particle when x = 4.¤ « 1 » 3U96-6)! x

O A trolley is moving in simple harmonic motion about the origin O. The displacement, x metres, of the centre of the trolley form O at time t seconds is given by

  x  6sin 2t   .  4

a.

State the period and amplitute of the motion.

b.

Sketch the graph of x  6sin 2t   for 0  t  2.

c. d. e.

Find the velocity of the trolley when t = 0. Find the first time after t = 0 when the centre of the trolley is at x = 3.

 

 4

w P

A particle P, on top of the trolley, is moving in simple harmonic motion about the centre of the trolley. Its displacement, w metres, from the centre of the trolley at time t seconds, is given by w = sin(2t). The displacement, y metres, of P from the origin is the sum of the two displacements x and w, so that

  y  6 sin 2t    sin(2t) .  4

i.

Show that P is moving in simple harmonic motion about O.



ii.

Find the amplitute of this motion.¤

x 6 3 2

 2

3 2



-6

« a) Period = , Amplitude = 6 b)

2 t

c) 6 2 m/s

d)

7 sec e) i) Proof ii) 6.74 m » 24

3U95-7)! y V



P h

h R

O

F

r

x C

A cap C is lying outside a softball field, r metres from the fence F, which is h metres high. The fence is R metres from the point O, and the point P is h metres above O. Axes are based at O, as shown. At time t = 0, a ball is hit from P at a speed V metres per second and at an angle  to the horizontal, towards the cap. a. The equations of motion of the ball are x 1 = 0, y = -g. Using calculus, show that the position of the ball at time t is given by x = Vt cos  y = Vt sin  - 21 gt2 + h. b.

Hence show that the trajectory of the ball is given by y = h + x tan  - x2

c.

The ball clears the fence. Show that V2 

d.

gR . 2sin  cos 

After clearing the fence, the ball hits the cap C. Show that tan  

e.

g . 2 V cos2  2

Rh . (R r) r

Suppose that the ball clears the fence, and that V  50, g = 10, R = 80, and h = 1. What is the closest point to the fence where the ball can land? ¤ « a) b) c) d) Proof e) 0.16m (to 2 d.p.) »

3U94-4c)!



The acceleration of a particle P moving in a straight line is given by

d2 x  3x(x 4) , where x metres dt 2

is the displacement from the origin O and t is the time in seconds. Initially the particle is at O and its velocity is 4 2 m/s. i.

Using

d2 x  d  1 2 , show that 2  2 3  6 2 16 , where v is the velocity of P. x x  v  v  dt 2 dx  2 

ii. Calculate the velocity and acceleration of P at x = 2. iii. In which direction does P move from x = 2? Give a reason for your answer. iv. Briefly describe the motion of P after it moves from x = 2. ¤ « i) Proof ii) v = 0, x = -12ms-2 iii) Towards O. When x = 2, the velocity is O and the acceleration is towards O. iv) The particle accelerates to the origin and then slows to a stop. It then moves off in the opposite direction. » 3U93-3c)! The velocity v ms-1 of a particle moving in simple harmonic motion along the x axis is given by v² = 8 + 2x - x². i. Between which two points is the particle oscillating? ii. What is the amplitude of the motion? iii. Find the acceleration of the particle in terms of x. iv. Find the period of the oscillation. ¤ « i) x = -2 and 4 ii) 3 metres iii) x = (1 - x)ms-2 iv) 2 seconds » 3U93-6a)! Let T be the temperature inside a room at time t and let A be the constant outside air temperature. Newton's law of cooling states that the rate of change of the temperature T is proportional to (T - A). i. Show that T = A + Cekt (where C and k are constants) satisfies Newton's law of cooling. ii. The outside air temperature is 5C and a heating system breakdown causes the inside room temperature to drop from 20C to 17C in half an hour. After how many hours is the inside room temperature equal to 10C? ¤ « i) Proof ii) 2 hours and 28 minutes » 3U93-7b)! A projectile is fired from the origin O with velocity V and with angle of elevation , where  

 . 2

You may assume that x = Vt cos and y  - 1 gt2  Vt sin  , where x and y are the horizontal and

2

vertical displacements of the projectile in metres from O at time t seconds after firing. i. Show that the equation of flight of the projectile can be written as

y  x tan   ii. iii.

1 2 V2 = h . x 1  tan2  , where 2g 4h

Show that the point (X, Y), where X  0, can be hit by firing at two different angles 1 and 2 provided X² < 4h(h - Y). Show that no point above the x axis can be hit by firing at two different angles 1 and 2, satisfying 1 <

  and 2 < .¤ 4 4

« Proof » 3U92-2b)! The displacement x metres of a particle moving in simple harmonic motion is given by x = 3cos t, where the time t is in seconds. i. What is the period of oscillation? ii. What is the speed v of the particle as it moves through the equilibrium position? ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


iii.

Show that the acceleration of the particle is proportional to the displacement from the equilibrium position. ¤ « i) 2 seconds ii) 3 ms-1 iii) Proof »

3U92-5b)! In a flock of 1000 chickens, the number P infected with a disease at time t years is given by

P

1000 , where c is a constant. 1  ce-1000 t

i. ii.

Show that, eventually, all the chickens will be infected. Suppose that when time t = 0, exactly one chicken was infected. After how many days will 500 chickens be infected?

iii.

Show that

dP  P(1000  P) . ¤ dt « i) Proof ii) Approximately 2 21 days iii) Proof »

3U92-7a)! Consider the function y = f(), where f()  cos  

 6

i.

Verify that f ( )  0 .

ii.

Sketch the curve y = f() for 0 <  

1 . 4 3sin 

 given that f () < 0. On your sketch, write the 2

coordinates of the turning point in exact form and label the asymptote.¤ f()

asymptote,  = 0 1   ,  6 3

 1   ,  2 3

« i) Proof ii)

»

3U92-7b)! y V  O

d

initial position of target

x

T

A projectile, of initial speed V m/s, is fired at an angle of elevation  from the origin O towards a target T, which is moving away from O along the x axis. You may assume that the projectile's 2 trajectory is defined by the equations x = Vt cos and y  - 1 gt  Vt sin  , where x and y are the

2

horizontal and vertical displacements of the projectile in metres at time t seconds after firing, and where g is the acceleration due to gravity. i. ii.

2V sin  seconds. g 2 V2 sin  cos  Show that the horizontal range of the projectile is metres. g Show that the projectile is above the x axis for a total of



iii.

At the instant the projectile is fired, the target T is d metres from O and it is moving away at a constant speed of u m/s. Suppose the projectile hits the target when fired at an angle of elevation . Show that u  V cos 

In parts (iv) and (v), assume that gd  iv.

gd . 2Vsin 

V2 . 2 3

By using (iii) and the graph of part (a), show that if u >

V the target cannot be hit by the 3

projectile, no matter at what angle of elevation  the projectile is fired. v.

Suppose that u <

V . Show that the target can be hit when it is at precisely two distances 3

from O. ¤ « Proof » 3U91-4b)! The acceleration of a particle moving in a straight line is given by

d2 x  2 x 3 where x is the dt 2

displacement, in metres, from the origin O and t is the time in seconds. Initially the particle is at rest at x = 4. i. If the velocity of the particle is v m/s, show that V2  2 x2  3x 4 . ii. Show that the particle does not pass through the origin. iii. Determine the position of the particle when v = 10. Justify your answer. ¤ « i) Proof ii) Proof iii) x = 9 is the only solution since the particle does not pass through the origin » 3U91-6b)! y



building 50m

O

200m

x

FIGURE NOT TO SCALE. The diagram shows the path of a projectile launched at an angle of elevation , with an initial speed of 40 m/s, from the top of a 50 metre high building. The acceleration due to gravity is assumed to be 10 m/s².

d2 x  0 and d2 y  -10 , show that x = 40t cos  and y = -5t² + 40t sin  + 50, dt2 dt2

i.

Given that

ii.

where x and y are the horizontal and vertical displacements of the projectile in metres from O at time t seconds after launching. The projectile lands on the ground 200 metres from the base of the building. Find the two possible values for . Give your answers to the nearest degree. ¤ « i) Proof ii)  = 31, 45 »

3U90-2c)!



V

y

 O

x

The path of a projectile fired from the origin O is given by x = Vt cos , y = Vt sin  - 5t² where V is the initial speed in metres per second and  is the angle of projection as in the diagram and t is the time in seconds. i. Find the maximum height reached by the projectile in terms of V and . ii. Find the range in terms of V and . iii. Prove that the range is maximum when  = 45. ¤

2 2  m ii) V2 sin 2 m iii) Proof » « i) V sin

20

10

3U90-3b)! The velocity v ms-1 of a particle moving in simple harmonic motion along the x axis is given by v² = -5 + 6x - x², where x is in metres. i. Between which two points is the particle oscillating? ii. Find the centre of motion of the particle. iii. Find the maximum speed of the particle. iv. Find the acceleration of the particle in terms of x. ¤ « i) x = 1 and 5 ii) x = 3 iii) 2ms-1 iv) a = (3 - x)ms-2 » 3U90-6b)! Assume that the rate at which a body warms in air is proportional to the difference between its temperature T and the constant temperature A of the surrounding air. This rate can be expressed by the differential equation i. ii.

iii.

dT  k(T A) where t is the time in minutes and k is a constant. dt

Show that T = A + Cekt, where C is a constant, is a solution of the differential equation. A cooled body warms from 5C to 10C in 20 minutes. The air temperature around the body is 25C. Find the temperature of the body after a further 40 minutes have elapsed. Give your answer to the nearest degree. By referring to the equation for T, explain the behaviour of T as t becomes large. ¤ « i) Proof ii) 17 iii) T approaches 25 »

3U89-2c)!

d  1 2 dv v  . dx  2  dt

i.

Show that

ii.

The acceleration of a particle moving in a straight line is given by  x = - 2 e-x where x metres is the displacement from the origin. Initially, the particle is at the origin with velocity

iii.

2 ms-1. Prove that v = 2e 2 . What happens to v as x increases without bound? ¤ « i) Proof ii) Proof iii) The velocity of the particle approaches zero »

-

x

3U89-7a)! A "six" is scored in a cricket game when the ball is hit over the boundary fence on the full as in the diagram. A ball is hit from O with velocity V = 32ms-1 at an angle  to the horizontal and towards the 1 metre high boundary fence 100 metres away.



y V  O

1m fence x

100m

i.

Derive the equations of motion for the ball in flight using axes as in the diagram. (Air resistance is to be neglected and the acceleration due to gravity is taken as 10 ms-2.) ii. Show that the ball just clears the boundary fence when 50000 tan² - 102400 tan  + 51024 = 0. iii. In what range must  lie for a "six" to be scored? iv. If, during the flight of the ball, its velocity is reduced by piercing an extremely thin "board", show by a sketch how the path is altered. Without further calculation, discuss qualitatively the effect of air resistance on your answer in (iii). ¤ y = -10, y = -10t + 32 sin, y = -5t2 + 32t sin. Horizontal Motion: x = 0, « i) Vertical Motion:  y board

original path

x

altered path x = 32cos, x = 32t cos. ii) Proof iii) 4035'    50 iv) The altered path is parabolic in shape, however the horizontal velocity of the ball is reduced. Under air resistance, both the horizontal and vertical velocities are reduced. As a result, the time of flight, horizontal distance and vertical height are less than those of the same hit and no air resistance. » 3U88-3a)! A particle undergoes simple harmonic motion about the origin O. Its displacement x centimetres   from O at time t seconds, is given by x  3cos 2t    .  3 i. Express the acceleration as a function of displacement. ii. Write down the amplitude of the motion. iii. Find the value of x for which the speed is a maximum and determine this speed. ¤ « i) x = -4x ii) 3cm iii) x = 0, max. speed = 6cm/s » 3U87-4ii)! A pebble is projected from the top of a vertical cliff with velocity 20ms-1 at an angle of elevation of 30. The cliff is 40 metres high and overlooks a lake. y 30

CLIFF

O

40m

LAKE

FIGURE NOT TO SCALE ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

x


a.

Take the origin O to be the point at the base of the cliff immediately below the point of projection. Derive expressions for the horizontal component x(t) and vertical component y(t) of the pebble's displacement from O after t seconds. (Air resistance is to be neglected.) b. Calculate the time which elapses before the pebble hits the lake and the distance of the point of impact from the foot of the cliff. [Assume the acceleration due to gravity is 10ms-2.] ¤ « a) x(t) = 10 3 t metres, y(t) = - 21 gt2 + 10t + 40 metres b) 4 seconds, 40 3 metres from the origin. » 3U87-6i)! The rate at which a body cools in air is assumed to be proportional to the difference between its temperature T and the constant temperature S of the surrounding air. This can be expressed by the differential equation a. b.

dT  k(T S) where t is the time in hours and k is a constant. dt

Show that T = S + Bekt, where B is a constant, is a solution of the differential equation. A heated body cools from 80C to 40C in 2 hours. The air temperature S around the body is 20C. Find the temperature of the body after one further hour has elapsed. Give your answer correct to the nearest degree. ¤ « a) Proof b) 32C »

3U86-5ii)! A particle is oscillating in simple harmonic motion such that its displacement x metres from a given origin O satisfies the equation a. b.

c.

d2 x  4 x , where t is the time in seconds. dt 2

Show that x = a cos(2t + ß) is a possible equation of motion for this particle, where a and ß are constants. The particle is observed at time t = 0 to have a velocity of 2 metres per second and a displacement from the origin of 4 metres. Show that the amplitude of oscillation is 17 metres. Determine the maximum velocity of the particle. ¤ « a) Proof b) Proof c) 2 17 ms-1 »

3U85-5i)! Firefighters are forced to stay 60m away from a dangerous fire burning in a low open tank on horizontal ground. They have two pumps. One, which can eject water in any direction at 30ms-1, is on the ground, while the other, which can eject water at 40ms-1 but only horizontally, is on a vertical stand 5m high. Can both pumps reach the fire? Justify your answer. (Assume that g = 10ms-2, and that all frictional forces, including air resistance, can be neglected.) ¤ « Only the first pump can reach the fire. The first pump has a maximum range of 90m whilst the range of the second pump is fixed at 40m. » 3U85-5ii)! A scientist found that the amount, Q(t), of a substance present in a mineral at time t  0 satisfies the equation 4 a. b. c.

d2 Q  4 dQ  Q  0 . dt dt2

Verify that Q(t) = A(1 + t)e-0.5t satisfies this equation for any constant A > 0. If Q(0) = 10mg, find the maximum value of Q(t) and the time at which this occurs. Describe what happens to Q(t) is t increases indefinitely. ¤ « a) Proof b) The maximum value of Q(t) = 20e-0.5, at t = 1 c) Q(t) approaches zero. »

3U84-2iii)! A particle executes simple harmonic motion with period T seconds and amplitude A cm. What is its maximum velocity? ¤ « ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

2 A cm/s » T

3 UNIT MATHEMATICS (HSC) – INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS – HSC

Inverse Functions and Inverse Trigonometric Functions 3U97-1d)! 2

Evaluate

dx

 4+ x

2

.¤

0

« 3U97-6a)! The function f(x) = sec x for 0  x 

 » 8

 , and is not defined for other values of x. 2

i.

State the domain of the inverse function f -1(x).

ii.

Show that f -1(x) = cos-1   .

iii.

Hence find

 1  x

d -1 f ( x) .¤ dx « i) x  1 ii) Proof iii)

1 x x2  1

»

3U96-3a)! y

y = f(x)

1 4 -4

Let f(x)  i. ii.

-2

O

2

4

x

1 . The graph of y = f(x) is sketched above. 16  x2

Show that f(x) is an even function. Find the area enclosed by y = f(x), the x-axis, x = -2 and x = 2.¤ « i) Proof ii)

3U96-3d)! The function h(x) is given by h(x) = sin-1x + cos-1x, 0  x  1. i. Find h(x). ii. Sketch the graph of y = h(x).¤


 units2 » 3


h

 2

« i) 0 ii)

1 x

0

»

3U96-7b)! Consider the function f(x) = i. ii. iii. iv. v. vi.

1 [(x – 1)2 + 7]. 4

Sketch the parabola y = f (x), showing clearly any intercepts with the axes, and the coordinates of its vertex. Use the same scale on both axes. What is the largest domain containing the value x = 3, for which the function has an inverse function f -1 (x)? Sketch the graph of y = f -1 (x) on the same set of axes as your graph in part (i). Label the two graphs clearly. What is the domain of the inverse function? Let a be a real number not in the domain found in part (ii). Find f -1 (f(a)). Find the coordinates of any points of intersection of the two curves y = f(x) and y = f -1 (x).¤ y = f (x)

y

y = f -1 (x)

4 3 2 1 -3

-2 -1 0

(1, 1.75) (1.75, 1) 1

2

3

4

5

x

ii) x  1 iv) x  1.75 v) f -1 (f (a)) = 2 - a vi) (2, 2), (4, 4) »

« i) iii) 3U95-4)!

ex Consider the function f(x) = 3 e x

Note that ex is always positive, and that f(x) is defined for all real x. a. Show that f(x) has no stationary points.

3e x (3  e x ) (3  e x ) 3

b.

Find the coordinates of the point of inflexion, given that f (x) =

c. d.

Show that 0 < f(x) < 1 for all x. Describe the behaviour of f(x) for very large positive and very large negative values of x, i.e. as x tends to  and x tends to -. Sketch the curve y = f(x). Explain why f(x) has an inverse function. Find the inverse function y = f -1(x). ¤

e. f. g.



« a) Proof b) (ln 3, 0.5 1) c) Proof d) As x tends to , f(x) tends to 1 and as x tends to -, f(x) tends to. y 1 2 1 4

x

ln 3

e)

f) f(x) is an increasing function and so any horizontal

 3x   »  1- x 

line will cut it in one point only. g) y = ln  3U95-6a)! P

T

R Q

PT is a tangent to the circle PRQ, and QR is a secant intersecting the circle in Q and R. The line QR intersects PT at T. Copy or trace the diagram into your Writing Booklet. i. Prove that the triangles PRT and QPT are similar. ii. Hence prove that PT2 = QT  RT. ¤ « Proof » 3U95-6b)! P

L

cable clip

3 Q



R

O

T

x

wheel

A long cable is wrapped over a wheel of radius 3 metres and one end is attached to a clip at T. The centre of the wheel is at O, and QR is a diameter. The point T lies on the line OR at a distance x metres from O. The cable is tangential to the wheel at P and Q as shown. Let POR =  (in radians). The length of cable in contact with the wheel is L metres; that is, the length of the arc between P and Q is L metres.

3 . x

i.

Explain why cos  =

ii.

Show that L = 3  - cos-1    .  x

iii.

Show that

 

 3  

dL -9 = . dx x x2 - 9

What is the significance of the fact that ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

dL is negative? dx


iv. v.

Let s = L + PT. Using part (a), or otherwise, express s in terms of x. The clip at T is moved away from O along the line OR at a constant speed of 2 metres per second. Find the rate at which s changes when x = 10. ¤

OP ii) Proof iii) Proof. As x increases, the length of the OT 3 91  cable decreases. iv) S = 3  - cos-1  + x2 - 9 v) m/sec »  x 5

« i) PT is a tangent  OPT = 90 and cos  =

3U94-6a)! Consider the function f(x) = 3x - x3. i. Sketch y = f(x), showing the x and y intercepts and the coordinates of the stationary points. ii. Find the largest domain containing the origin for which f(x) has an inverse function, f -1 (x). iii. State the domain of f -1(x). iv. Find the gradient of the inverse function at x = 0. ¤ y

(1, 2)

 3

 3

x (-1, -2)

« i)

ii) -1  x  1 iii) -2  x  2 iv) 1/3 »

3U93-3a)!

x 3

Consider the function f(x) = 2 cos-1 . i. ii. iii.

Evaluate f(0). Draw the graph of y = f(x). State the domain and range of y = f(x). ¤ y

2



« i)  ii)

3 x

-3

iii) Domain: -3  x  3, Range: 0  f(x)  2 »

3U92-1c)! 3

Find the exact value of

 1

1 4  x2

dx . ¤ « /6 »

3U92-3b)! Consider the function f(x) = 2tan-1 x. i. Evaluate f( 3) . ii. Draw the graph of y = f(x), labelling any key features. iii. Find the slope of the curve at the point where it cuts the y axis. ¤



y  y = 2tan -1x (0, 0) x (point of inflexion)

« i)

2 ii) 3

-

iii) 2 »

3U91-5a)!

x 2

Consider the function f(x)  3 sin-1 . i. ii. iii.

Evaluate f(2). Draw the graph of y = f(x). State the domain and range of y = f(x). ¤ f(x) 3 2

-2

2 

« i) 3 ii) 2

x

3 2

iii) Domain: -2  x  2 Range: - 3  f(x)  3 »

2

2

3U91-7b)! B

3

A



1

P

O

x

FIGURE NOT TO SCALE. In the diagram, a vertical pole AB, 3 metres high, is placed on top of a support 1 metre high. The pole subtends an angle of  radians at the point P, which is x metres from the base O of the support.

4 x

1 x

i.

Show that  = tan-1  tan-1 .

ii.

Show that  is a maximum when x = 2.

iii.

Deduce that the maximum angle subtended at P is  = tan-1 3 . ¤

4

« Proof » 3U90-1a)!

Evaluate: 1

i.

1

 1 x 0


2

dx

3 UNIT MATHEMATICS (HSC) – INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS – HSC 1

ii.

 0

x dx , using the substitution u = 1 + x. ¤ 1 x « i) /4 ii) 2 (2  2) »

3

3U90-4c)! i. ii. iii.

State the domain and range of the function given by y = cos-12x. Sketch the graph of the function given by y = cos-12x. Find the slope of the curve at the point where it cuts the y axis. ¤ y

  2

3U89-1b)!

« i) Domain: -0.5  x  0.5, Range: 0  y   ii) Evaluate: 2

i.

 1

1 4  x2

1

-2

1 2

x

iii) -2 »

dx ,

0

ii.

x

1+ x dx , using the substitution u = 1 + x. ¤

-1

« i) /3 ii) -4/15 » 3U89-6a)! y

O

x

FIGURE NOT TO SCALE The figure shows a sketch of the curve y = (x - 1)² - 3. i. Find the largest positive domain such that the graph defines a function f(x) which has an inverse. ii. Find this inverse function and state its domain. iii. State a domain for which the function does not have an inverse. Give a brief reason for your answer. ¤ « i) x  1 ii) y = 1 + x 3 , Domain: x  -3 ii) Any domain which includes values of x on both sides of the vertex (1, -3). The corresponding inverse relation would be the parabola (y - 1)2 = x + 3, which is not a function. » 3U89-7b)! i.

1 x

Differentiate y  tan-1 , x  0 , and hence show that


d  -1 1  tan x tan-1   0 . dx  x


1 x

Sketch the curve y  tan-1 x tan-1 . ¤

ii.

y  2

x

« i) -

1 , Proof ii) x +1

  2

»

2

3U88-1a)! Differentiate: i. x cos x ii. tan-13x. ¤ « i) cos x - x sin x ii) 3U88-2a)! i. ii.

Draw a sketch of y = sin-1x. State the domain and range. A region R is bounded by the curve y = sin-1x, the x-axis and the line x = 1. Use Simpson's rule with three function values to find an approximation for the area of R. Give your answer correct to 2 decimal places. ¤

y

y = sin-1 x

 2

-1

1 

x

 2

« i) 3U87-1ii)! Write down primitive functions of: a. (3x + 2)10 b.

3 » 1+ 9 x 2

Domain: -1  x  1, Range: -0.5 y  0.5 ii) 0.61 units2 »

5 .¤ 2 + x2 « a)

(3x+ 2 )11 5  x + c b) tan-1   + c » 33  2 2

3U87-2iii)! a.

Differentiate x sin -1 x + 1- x 2 . 1

b.

Hence evaluate

 sin

1

x dx . ¤

0

« a) sin-1x b) 0.5 - 1 » 3U87-2iv)! State the domain and range of y = 2sin-1(3x). ¤ « Domain: -1/3  x  1/3, Range -  y   » 3U86-6i)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


a. b.

Sketch the graph of the function y = tan-1x stating clearly the range and domain. Given that x² + 4x + 5  (x + a)² + b², find a, b.

c.

Using the result in (b), find



1 dx . ¤ x + 4 x+5 2

y  2  4

x

  4   2

« a)

Range: -0.5 < y < 0.5, Domain: All real x b) a = 2, b = ±1 c) tan-1(x + 2) + c »

3U85-4i)!

x3  0 1  x8 dx  16 . ¤ 1

Using the substitution u = x4, or otherwise, show that

« Proof » 3U84-1i)! Find the derivative (with respect to x) of a.

sin-1 2x, for x < 1 ,

2

2x

b.

e .¤ x2 + 3 « a)

2 1- 4 x2

b)

2 e2 x (x2  x 3) » (x2  3)2

3U84-6i)!

d F(u) when u is a function of x. dx 3 1 x 2 1 tan 3  -1 x  Differentiate (with respect to x)  tan  and hence find the exact value of  dx . ¤   0 x2  9 3 Write down a formula for calculating

 x 6 tan-1    3  d d du « , » F(u) = F(u)  , 2 216 dx du dx x +9


3 UNIT MATHEMATICS (HSC) – INDUCTION – HSC

Induction 3U97-5b)! i.

 n  n   n + 1    , where  r   r + 1  r + 1

For positive integers n and r, with r < n, show that    

 n n! . Do not use induction.    r  r!(n - r)! ii.

Use mathematical induction to prove that, for n  3,

 j -1  n     .¤ 2   3 j= 3 n



« Proof » 3U94-3c)! Prove by mathematical induction that n3 + 2n is divisible by 3, for all positive integers n. ¤ « Proof » 3U93-5a)! For n = 1, 2, 3, . . ., let Sn = 1² + 2² + ... + n². i. Use mathematical induction to prove that, for n = 1, 2, 3, . . ., Sn  16 n(n 1)(2 n 1) ii. By using the result of (i) estimate the least n such that Sn  109. ¤ « i) Proof ii) 1442 » 3U92-4b)! Let Sn = 1  2 + 2  3 + . . . + (n - 1)  n. Use mathematical induction to prove that, for all integers n with n  2, Sn  13 (n 1) n(n 1) . ¤ « Proof » 3U91-4a)! Use mathematical induction to prove that, for all positive integers n, 1 + 2 + 4 + ... + 2n - 1 = 2n - 1. ¤ « Proof » 3U90-7a)! Use mathematical induction to prove that, for every positive integer n, 13  6n + 2 is divisible by 5. ¤ « Proof » 3U89-5b)! i. By considering the sum of the terms of an arithmetic series, show that

(1  2... n )2  1 n2 (n 1)2 . 4 ii.

By using the Principle of Mathematical Induction prove that 13 + 23 + ... + n3 = (1 + 2 + ... + n)², for all n  1. ¤ « Proof »

3U88-3b)! Prove by mathematical induction that for n  1, 12  32 ...(2 n 1)2  13 n(2 n 1)(2 n 1) . ¤ « Proof » 3U86-7i)! Prove by mathematical induction that 1  20 + 2  21 + 3  2² + . . . + n  2n - 1 = 1 + (n - 1)2n for all integers n  1. ¤ « Proof » 3U85-4iii)! Use the Principle of Mathematical Induction to prove that 5n + 2(11n) is a multiple of 3 for all positive integers n. ¤ « Proof » ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

3 UNIT MATHEMATICS (HSC) – INDUCTION – HSC

3U84-7ii)! It is given that A > 0, B > 0 and n is a positive integer. a. Divide An + 1 - AnB + Bn + 1 - BnA by A - B, and deduce that An + 1 + Bn + 1  AnB + BnA.

 A B An  Bn Using (a), show by mathematical induction that  .¤    2  2 n

b.

« a) An - Bn b) Proof »


3 UNIT MATHEMATICS (HSC) – BINOMIAL THEOREM – HSC

Binomial Theorem 3U97-5b)! i.

 n  n   n + 1    , where  r   r + 1  r + 1

For positive integers n and r, with r < n, show that    

 n n! . Do not use induction.    r  r!(n - r)! ii.

Use mathematical induction to prove that, for n  3,

 j -1  n     .¤ 2   3 j= 3 n



« Proof » 3U97-7b)!

 n -1  n -1  n -1   n   ...  n .  1   2   n - 2

i.

Simplify n

ii.

Find the smallest positive integer n such that n

 n -1  n -1  n -1   n   ...  n   20 000 .¤  1   2   n - 2 « i) 2n(2n-2 - 1) ii) 12 »

3U96-7a)! Using the fact that (1 + x)4(1 + x)9 = (1 + x)13, show that 4 9 C0 C4 + 4C19C3 + 4C29C2 + 4C39C1 + 4C49C0 = 13C4.¤ « Proof » 3U95-3b)! Find the value of the term that does not depend on x in the expansion of 6

 2 3 x   . ¤  x « 1215 » 3U92-6c)!

 n  1

 n  n  2  n  n  n  n Show that 1       . . .(1) n    0 .  1  2  n  n  n 1  n 1 Show that 1  1    1    . . .(-1)n .¤   2  1 3  2 n 1  n n 1

Consider the binomial expansion 1    x   x2 . . .  xn  (1  x)n . i. ii.

« Proof » 3U90-6c)!

 

1 x

n

i.

Show that xn (1  x)n 1    (1  x)2 n .

ii.

 n  n  n  2n Hence prove that 1       . . .     . ¤  1  2  n  n

2

2

2

« Proof » 3U89-3b)!

 

Find the constant term in the expansion of  x

20

1   .¤ 2 x3 « -484 21 »


3 UNIT MATHEMATICS (HSC) – BINOMIAL THEOREM – HSC

3U88-6b)! Suppose (7  3x)25 

25

t

k

xk .

k= 0

i.

Use the Binomial Theorem to write an expression for tk, 0  k  25.

ii.

Show that

iii.

Hence or otherwise find the largest coefficient tk. You may leave your answer in the form

t k+1  3(25  k) . 7(k 1) tk

 25 c d  7 3 .¤  k

 25 25 - k k 3 ii) Proof iii) t7 = 7  k

« i) tk = 

 25 18 7  7 3 »  7

3U87-3ii)! 11

1  Find the coefficient of x in the expansion of  2 x  . ¤  x 3

« 42240 » 3U86-4ii)! Factorise a² + 3a + 2 and hence or otherwise find the coefficient of a4 in (a² + 3a + 2)6. ¤ « a² + 3a + 2 = (a + 1)(a + 2), 9420 » 3U85-6i)! When (3 + 2x)n is written out as a polynomial in x, the coefficients of x5 and x6 have the same value. Find n. ¤ « 14 » 3U85-6ii)!

10  2

10  4

10  6

10  8

Prove that 1    32    34    36    38  310  29 (210  1) . ¤ « Proof » 3U84-7i)! Assume that, for all real numbers x and all positive integers n, (1  x )n 

n

 n

  r  x . r

r= 0

Show that n

a.

 n

 (-1)  r  , and find simple expressions for r

0 =

r= 0

 n  ,  r r= 0 n  n r .¤  r = 0  r n

b. c.

2

r

« a) Proof b) 3n c) n  2n - 1 »


3 UNIT MATHEMATICS (HSC) – FURTHER PROBABILITY – HSC

Further Probability 3U97-3c)! In each game of Sic Bo, three regular, six-sided dice are thrown once. i. In a single game, what is the probability that all three dice show 2? ii. What is the probability that exactly two of the dice show 2? iii. What is the probability that exactly two of the dice show the same number? iv. A player claims that you expect to see three different numbers on the dice in at least half of the games. Is the player correct? Justify your answer.¤ « i)

5 5 1 5 ii) iii) iv)Yes, since the probability of three different numbers on the dice is which is 72 9 216 12 1 greater than . » 2

3U96-5c)! Mice are placed in the centre of a maze which has five exits. Each mouse is equally likely to leave the maze through any one of the five exits. Thus, the probability of any given mouse leaving by a particular exit is

1 . 5

Four mice, A, B, C and D, are put into the maze and behave independently. i. What is the probability that A, B, C and D all come out the same exit? ii. What is the probability that A, B and C come out the same exit, and D comes out a different exit? iii. What is the probability that any three of the four mice come out the same exit, and the other comes out a different exit? iv. What is the probability that no more than two mice come out the same exit?¤ « i)

1 4 16 108 ii) iii) iv) » 125 125 125 125

3U95-5b)! In a Jackpot Lottery, 1500 numbers are drawn from a barrel containing the 100 000 ticket numbers available. After all the 1500 prize-winning numbers are drawn, they are returned to the barrel and a jackpot number is drawn. If the jackpot number is the same as one of the 1500 numbers that have already been selected, then the additional jackpot prize is won. The probability that the jackpot prize is won in a given game is thus p=

1500 = 0.015. 100 000

i.

Calculate the probability that the jackpot prize will be won exactly once in 10 independent lottery games. ii. Calculate the probability that the jackpot prize will be won at least once in 10 independent lottery games. iii. The jackpot prize is initially $8000, and it increases by $8000 each time the prize in NOT won. Calculate the probability that the jackpot prize will exceed $200 000 when it is finally won. ¤ « i) 0.131 (to 3 d.p.) ii) 0.140 (to 3 d.p.) iii) 0.685 (to 3 d.p.) » 3U94-3a)! New cars are subjected to a quality check, which 75% pass. Calculate the probability that of the next ten cars checked, more than seven will pass. Leave your answer in unsimplified form. ¤

10  8

10  9

10 10

«   (0.25)2 (0.75)8    (0.25)(0.75)9    (0.75)10 » ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


3U92-1d)! The probability that any one of the thirty-one days in December is rainy is 0.2. What is the probability that December has exactly ten rainy days? Leave your answer in index form. ¤

 31 10

«   (0.2)10(0.8)21 » 3U92-6b)! A total of five players is selected at random from four sporting teams. Each of the teams consists of ten players numbered from 1 to 10. i. What is the probability that of the five selected players, three are numbered '6' and two are numbered '8'? ii. What is the probability that the five selected players contain at least four players from the same team? ¤

1 ii) 28 » 703 27417

« i) 3U91-2b)! When Mendel crossed a tall strain of pea with a dwarf strain of pea, he found that were tall and

3 of the offspring 4

1 were dwarf. Suppose five such offspring were selected at random. Find the 4

probability that: i. all of these offspring were tall ii. at least three of these offspring were tall. Leave your answers in index form. ¤ 5

3   « i)  3  ii) 34  35 » 

4

4

3U90-6a)! Sam sits for a multiple choice examination which has 10 questions, each with four possible answers only one of which is correct. What is the probability that Sam answers exactly six questions correctly by chance alone? ¤ « 8505 »

524288

3U89-3a)! A committee of 3 is to be elected from a club of 8 members. i. How many different committees can be formed? ii. If there are 4 Queenslanders in the club, what is the probability that a randomly selected committee of 3 contains only Queenslanders? ¤ « i) 56 ii) 1 »

14

3U88-5b)! A meeting room contains a round table surrounded by ten chairs. These chairs are indistinguishable and equally spaced around the table. i. A committee of ten people includes three teenagers. How many seating arrangements are there in which all three sit together? Give brief reasons for your answer. ii. Elections are held for the position of Chairperson and Secretary in a second committee of ten people seated around this table. What is the probability that the two people elected are sitting directly opposite each other? Give brief reasons for your answer. ¤ « i) 30240 ii) 1 »

9

3U87-3iii)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


One fifth of all jellybeans are black. A random sample of ten jellybeans is chosen. a. What is the probability that this sample contains exactly two black jellybeans? Give your answer correct to 3 decimal places. b. What is the probability that the sample contains fewer than two black jellybeans? Give your answer correct to 3 decimal places. c. Which is more likely: the sample contains fewer than 2 black jellybeans, or the sample contains more than 2 black jellybeans? Give reasons for your answer. ¤ « a) 0.302 b) 0.376 c) The probability that the sample contains more than 2 black jellybeans is 0.322.  It is more likely the sample contains less than 2 black jellybeans. » 3U86-6ii)! A given school in a certain State has 3 mathematics teachers. The probability in that State that a mathematics teacher is female is 0.4. a. What is the probability that in the given school there is at least one female mathematics teacher? b. In the same State the probability that a mathematics teacher (male or female) is a graduate is 0.7. What is the probability that in the given school none of the three mathematics teachers is a female graduate? (Give your answer correct to 3 decimal places.) ¤ « a) 0.784 b) 0.373 » 3U85-6iii)! David plays a game in which the probability that his score is s is

1  8 , for s = -4, -3, -2, -1;   1  4    (0.3) s (0.7) 4  s , for s = 0,1, 2, 3, 4;  2  s  0 , for all other values of s. What is a. his most likely score? b. the probability (expressed as a decimal correct to three places) that his score is positive? c. the probability (expressed as a decimal correct to three places) that, after playing the game twice, his total score is -3? ¤ « a) 1 b) 0.380 c) 0.113 » 3U84-3ii)! A box contains ten tennis balls of which four have never been used. For the first game two balls are selected at random and, after play, are returned to the box. For the second game two balls are also selected at random from the box. Find the probability of each of the following events: a. precisely one of the balls selected for the first game has been used before; b. neither ball selected for the first game has been used before, but both balls selected for the second game have been used before the second game. ¤ « a) 8 b) 56 »

15


675

3 UNIT MATHEMATICS (HSC) – ITERATIVE METHODS FOR NUMERICAL ESTIMATION OF THE ROOTS OF A POLYNOMIAL EQUATION – HSC

Iterative Methods for Numerical Estimation of the Roots of a Polynomial Equation 3U97-3b)!

P  B

C

O

A

r

NOT TO SCALE The point P lies on the circumference of a semicircle of radius r and diameter AB, as shown. The point C lies on AB and PC is perpendicular to AB. The arc AP subtends an angle  at the centre O, and the length of the arc AP is twice the length of PC. i. Show that 2sin = . ii.

Taking  = 1.8 as an approximation for the solution to the equation 2sin =  between

 2

and , use one application of Newton’s method to give a better approximation.¤ « i) Proof ii) 1.9 » 3U96-2a)! The function f(x) = x3 - Ln(x + 1) has one root between 0.5 and 1. i. Show that the root lies between 0.8 and 0.9. ii. Hence use the halving-the-interval method to find the value of the root, correct to one decimal place.¤ « i) Proof ii) 0.9 » 3U95-2a)! Let f(x) = x3 + 5x2 + 17x - 10. The equation f(x) = 0 has only one real root. i. Show that the root lies between 0 and 2. ii. Use one application of the `halving the interval' method to find a smaller interval containing the root. iii. Which end of the smaller interval found in part (ii) is closer to the root? Briefly justify your answer. ¤ « i) Proof ii) The root lies between 0 and 1 iii) f(0.5) = -0.125,  the root lies between 0.5 and 1. The root is closer to 1 than 0. » 3U94-4b)! Taking x = 1.0 as the first approximation, use Newton's method to find a second approximation to the root of x - 3 + e2x = 0. ¤ « 0.7 (to 1 d.p.) » 3U93-5c)! y

O


y = f(x)

a

b x

3 UNIT MATHEMATICS (HSC) – ITERATIVE METHODS FOR NUMERICAL ESTIMATION OF THE ROOTS OF A POLYNOMIAL EQUATION – HSC

Consider the above graph of y = f(x). The value a shown on the axis is taken as the first approximation to the solution b of f(x) = 0. Is the second approximation obtained by Newton's method a better approximation to b than a is? Give a reason for your answer. ¤ « No. There is a stationary point between the first approximation and the solution to f(x) = 0. » 3U92-2c)! Use Newton's method to find a second approximation to the positive root of x - 2sin x = 0. Take x = 1.7 as the first approximation. ¤ « 1.93 (to 2 d.p.) » 3U91-3a)! Taking x = 0.5 as a first approximation to the root of x + ln x = 0, use Newton's method to find a second approximation. ¤ « 0.56 (to 2 d.p.) » 3U90-3c)! You are given that 3.5 is an approximate root of the equation x3 - 50 = 0. Using one application of Newton's method, find a better approximation. ¤ « 3.69 (to 2 d.p.) » 3U87-5)! The polynomial equation f(x) = 8x3 + 12x2 - 18x - 20 = 0 has a root at x = -2. a. Find all roots of f(x) = 0. b. Draw a sketch of the graph of y = f(x) showing the coordinates of its points of intersection with the axes and all stationary points. c. Apply Newton's method once to approximate a root of f(x) = 0 beginning with an initial approximation x1 = 1. d. Willy chose an initial approximation of x1 = 0.49 and used Newton's method a number of times in order to approximate a root of f(x) = 0. State, giving reasons, the root of f(x) = 0 to which Willy's approximations are getting closer. (It is not necessary to do additional calculations.) ¤ y (- 32 , 7) -2

1

1

21

21

x

4

4

-20

« a) x = -2,


1  21 b) 4

( 12 , -25)

c) 1.6 d) x = -2 »

3 UNIT MATHEMATICS (HSC) – HARDER APPLICATIONS OF HSC 2 UNIT TOPICS – HSC

Harder Applications of HSC 2 Unit Topics 3U97-1a)! Differentiate e3x cos x.¤ « e3x(3cos x - sin x) » 3U97-1c)!

 b  c

Given that logab = 2.8 and logac = 4.1, find loga   .¤ « -1.3 » 3U97-3a)! i. ii.

On the same set of axes, sketch the graphs of y = 2sin and y =  for      . Use your sketch to find the number of solutions of the equation 2sin =  for      .¤

y

y=

2 y = 2 sin

1 -

 2

0

 2

-1





-2 « i)

ii) 3 solutions »

3U97-4a)!

B  

 A

D

C

The triangle ABC is isosceles, with AB = BC, and BD is perpendicular to AC. Let ABD = CBD =  and BAD =  , as shown in the diagram. i. Show that sin   cos  . ii. By applying the sine rule in ABC, show that sin2 = 2sin cos . iii.

 , show that the limiting sum of the geometric series 2 sin2 + sin2 cos2   sin2 cos4   sin2 cos6   ... is equal to 2cot .¤

Given that 0   

« Proof » 3U97-4c)!



P

0.9km  S

Q

x km

A searchlight on the ground at S detects and tracks a plane P that is due east of the searchlight. The plane is flying due west at a constant velocity of 240 kilometres per hour and maintains a constant height of 900 metres above ground level. Let (t) radians be the angle of elevation of the plane at time t seconds and let x(t) kilometres be the distance from S to the point Q on the ground directly below P.

dx 09 .  2 . d sin 

i.

Show that

ii.

Show that the rate of change of the angle of elevation of the plane when  

 is equal to 4

1 radians per second.¤ 27 « Proof » 3U97-6b)! An amount $A is borrowed at r% per annum reducible interest, calculated monthly. The loan is to be

 

repaid in equal monthly instalments of $M. Let R = 1+

r   and let $Bn be the amount owing 1200

after n monthly repayments have been made. i.

 R n  1 .  R -1 

Show that Bn  AR n  M

Pat borrows $30 000 at 9% per annum reducible interest, calculated monthly. The loan is to be repaid in 60 equal monthly instalments. ii. Show that the monthly repayments should be $622.75. iii. With the twelfth repayment, Pat pays an additional $5000, so this payment is $5622.75. After this, repayments continue at $622.75 per month. How many more repayments will be needed?¤ « i) ii) Proof iii) 37 instalments » 3U96-4b)!



h

r Grain is poured at a constant rate of 0.5 cubic metres per second. It forms a conical pile, with the angle at the apex of the cone equal to 60. The height of the pile is h metres, and the radius of the base is r metres.

h . 3

i.

Show that r 

ii.

Show that V, the volume of the pile, is given by V 

iii.

Hence find the rate at which the height of the pile is increasing when the height of the pile is 3 metres.¤

h3 . 9

« i) Proof ii) Proof iii)

1 m/s » 6

3U95-1b)! 4

Evaluate

 y dx if xy = 1. ¤ 1

« 2ln 2 » 3U95-1c)! Find

lim x0

sin x .¤ 5x « 1 »

5

3U95-2b)! y

6

3

xy = 3 O

1

x

The shaded area is bounded by the curve xy = 3, the lines x = 1 and y = 6, and the two axes. A solid is formed by rotating the shaded area about the y axis. Find the volume of this solid by considering separately the regions above and below y = 3. ¤ ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


«

9 units3 » 2

3U95-4)! Consider the function f(x) =

ex . 3  ex

Note that ex is always positive, and that f(x) is defined for all real x. a. Show that f(x) has no stationary points.

3 ex  3  ex 

b.

Find the coordinates of the point of inflexion, given that f (x) =

c. d.

Show that 0 < f(x) < 1 for all x. Describe the behaviour of f(x) for very large positive and very large negative values of x, i.e. as x approaches  and x approaches -. Sketch the curve y = f(x). Explain why f(x) has an inverse function. Find the inverse function y = f -1(x). ¤

e. f. g.

 3  ex 

3

.

« a) Proof b) (ln 3, 1 ) c) Proof d) As x tends to , f(x) tends to 1 and as x tends to -, f(x) tends to 0.

2

y 1 2 1 4

e)

x

ln 3

f) f(x) is an increasing function and so any horizontal

 3x   »  1- x 

line will cut it in one point only. g) y = ln  3U94-1a)!

 8

Using the table of standard integrals, find the exact value of

 sec 2 x tan 2 x dx . ¤ 0

«

1 2(

2 -1) »

3U94-5b)! y S h x² = 32y O

x

The part of the curve x² = 32y between y = 0 and y = h is rotated about the y axis. Show that the volume enclosed is given by V = 16h². ¤ « Proof » 3U94-5c)!



y S h x² = 32y O

x

The diagram represents the water in a dam on a farm. The depth of the water is h metres, the volume of water in the dam is V m3, and the area of the surface of the water is S m². The water in the dam evaporates according to the rule

dV  - kS , where k is a positive constant, and t is the time in hours. dt

i.

Describe in words what the rule says about the rate of evaporation.

ii.

Show that

dh  -k dt

Initially the dam contains 64 m3 of water. Calculate how long it will take for the dam to empty by evaporation when k = 0.001. ¤ « i) The rate of evaporation is directly proportional to the surface area of the water. ii) Proof iii) 2000 hours » 3U94-6b)! iii.

bottles bin

1st row 2nd row 3rd row

The figure shows a bottle-storage rack. It consists of n rows of 'bins' stacked in such a way that the number of bins in the r th row is r, counting from the top.

n(n 1) . 2

i.

Show that the total number of bins in the storage rack is

ii.

Each bin in the r th row contains c + r bottles, where c is a constant. (For example, each bin in the third row contains c + 3 bottles). Find an expression for the total number of bottles in the storage rack. [You may assume that 1² + 2² + ... + n² = 16 n(n 1)(2 n 1) ).] Enzo notices that c = 5 and that the average number of bottles per bin in the storage rack is 10. Calculate the number of rows in the storage rack. ¤ « i) Proof ii) 16 n(n + 1)(3c + 2n + 1) iii) 7 »

iii.

3U94-7a)! i. ii. iii.

Sketch carefully on the same set of axes the graphs of y = x4 and y = cos(x) for 0  x  1.5 (Your diagram should be at least half a page in size.) On the same diagram, sketch the graph of y = x4 + cos(x). Label clearly the three curves on your diagram. Using the graph, determine the number of positive real roots of the equation x4 + cos(x) = 0. ¤



y 5 y = x4

y = x4 + cos(x)

1 x

1 0.5 1.5 y = cos( x)

« i) ii)

iii) 2 »

3U93-1b)!  2



Find the exact value of sec2 x dx . ¤  6

«

2 3 2 or » 3 3

3U93-1d)! If y = cos(Ln x) find: i. ii.

dy dx d2 y ¤ dx2 « i)

- sin(ln x) sin(ln x)  cos(ln x) ii) » x x2

3U93-2c)! Suppose the cubic f(x) = x3 + ax2 + bx + c has a relative maximum at x =  and a relative minimum at x = ß. i. Prove that  ß  - 23 a . ii.

Deduce that the point of inflexion occurs at x 

 ß .¤ 2

« Proof » 3U93-4c)! i. ii.

Sketch the graph of y = cos x for 0  x  2. By using (i), or otherwise, find those values of x satisfying 0  x  2 for which the geometrical series 1 + 2cos x + 4cos²x + 8cos3x + ... has a limiting sum. ¤ y 1

y = cos x 

2 x

-1

« i) 3U93-6b)!


ii)  < x < 2 or 4 < x < 5 »

3

3

3

3


A(L, d)  d  B

P

O

x

L

In the diagram, the x axis represents a major blood vessel, whilst the line PA represents a minor blood vessel that joins the major blood vessel at P. The point A has coordinates (L, d) and PA makes an angle  <

 with the normal to the x axis at P, as shown in the diagram. 2

It is known that the resistance to flow in a blood vessel is proportional to its length, where the constant of proportionality depends upon the particular blood vessel. Let R be the sum of the resistances to flow in OP and PA. i. Show that R = c1(L - d tan) + c2d sec, where c1 and c2 are constants. ii. The blood vessel PA is joined to the blood vessel Ox in such a way that R is minimized. If

c2 = 2 find the angle  that minimizes R. (You may assume that L is large compared to c1 d.) ¤

« i) Proof ii)  »

6

3U92-4c)!

L 2

B

r d

A

L

The diagram shows a cylindrical barrel of length L and radius r. The point A is at one end of the barrel, at the very bottom of the rim. The point B is at the very top of the barrel, half-way along its length. The length of AB is d.

 L  2 L2  d   . 4  4

i.

Show that the volume of the barrel is V 

ii.

Find L in terms of d if the barrel has maximum volume for the given d. ¤ « i) Proof ii) L =

3U92-7a)! Consider the function y = f(), where f()  cos   i.

 6

Verify that f ( ) = 0 .


1 . 4 3sin 

2d » 3


ii.

Sketch the curve y = f() for 0 <  

 given that f () < 0. On your sketch, write the 2

coordinates of the turning point in exact form and label the asymptote.¤ asymptote,  = 0 f()  1   ,   6 3

 1    ,   2 4 3

« i) Proof ii)

»

3U91-2a)! Consider y = ekx where k is a constant.

dy d2 y . and dx dx2

i.

Find

ii.

Determine the values of k for which y = ekx satisfies the equation

dy d2 y  7  12 y = 0 . ¤ 2 dx dx dy d 2 y = k2ekx ii) k = -3, -4 » « i) = kekx, dx dx 2

3U91-3c)! If y = 10x, find

dy when x = 1. ¤ dx « 10 Ln10 »

3U91-3d)! The volume, V, of a sphere of radius r mm is increasing at a constant rate of 200 mm3 per second.

dr in terms of r. dt

i.

Find

ii.

Determine the rate of increase of the surface area, S, of the sphere when the radius is 50mm. ( V = 43  r3 , S = 4 r2 )¤ « i) dr  502 ii) 8 mm2s-1 »

dt

3U91-6a)! I. ii. iii.

r

On the same axes, sketch the curves y = sin x, y = cos x, and y = sin x + cos x, for 0  x  2. From your graph, determine the number of values of x in the interval 0  x  2 for which sin x + cos x = 1. For what values of the constant k does sin x + cos x = k have exactly two solutions in the interval 0  x  2? ¤ y

y = sin x + cos x

 2

y = cos x

« i) 3U90-7b)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995



3 2

2 x y = sin x

ii) 3 iii) - 2  k < 2 , k  1 »


C 

A

k



B

Points A, B and C lie on a circle. The length of the chord AB is a constant k. The radian measures of ABC and BCA are  and  respectively. i. Let L equal the sum of the lengths of chords CA and CB. Show that L is given by

L

k [sin   sin(  )] . sin 

ii.

Why is  a constant?

iii.

Evaluate

iv.

Hence show that L is a maximum when  

dL   when    . d 2 2

   . 2 2

¤ « i) Proof ii)  is subtended by a chord of fixed length, k. This is independent of the position on the circumference of C. iii) 0 iv) Proof » 3U89-4a)! A circular plate of radius r is heated so that the area of the plate expands at a constant rate of 3.2 cm² min-1. At what rate does r increase when r = 10cm? ¤ « 0.051cm min-1 (to 2 significant figures) » 3U88-1a)! Differentiate: i. x cos x ii. tan-13x. ¤ « i) cos x - x sin x ii)

3 » 1  9 x2

3U88-1c)! Evaluate: 1

i.

x 0 

ii.

2x dx , 1

2

 sin x dx . ¤ 2

0

« i) loge2 ii)  »

2

3U88-5a)! i. ii. iii.

Find the stationary points for the curve y = x - 2sin x for 0  x  2. Determine whether they are relative maxima or minima. Find the co-ordinates of those points on the curve corresponding to x = 0,  and 2. Hence draw a careful sketch of the curve. ¤



5 5 « i) (  ,  - 3 ) is a relative minimum turning point, ( ,  3 ) is a relative maximum turning point 3 3

3

3

 5 5   ,  3  3 3 

y

ii) (0, 0), (, ), (2, 2) iii)

    ,  3 3 3 



2

x

»

3U88-7a)! 30cm

30cm h cm

S

Water is poured into a conical vessel at a constant rate of 24cm3 per second. The depth of water is h cm at any time t seconds. What is the rate of increase of the area of the surface S of the liquid when the depth is 16cm? ¤ « 3cm2/s » 3U88-7b)! A parcel, in the shape of a rectangular prism, has sides x cm, x cm and y cm. The girth is the smallest distance around the parcel. A Courier Company will only deliver parcels for which the longest side L cm and the girth g cm satisfy L + g  100. Find the dimensions of the parcel of largest volume, for which L + g = 100, that the Courier Company will deliver. ¤ « 16 23 cm  16 23 cm  33 13 cm » 3U87-1i)! Differentiate: a. b.

1 3  x2

exloge(2x). ¤ « a)

-2 x x 1 b) e {  ln(2x)} » 2 x (3  x2 )

3U86-3i)! Find the volume of the solid obtained when the region between the curves y = x3 and y = x², from x = 0 to x = 1, is rotated about the x axis. ¤

« 2 units3 »

35

3U86-7ii)!



An irrigation channel is to have a cross-section in the shape of a trapezium as in the accompanying figure. The bottom and sides are each L metres long. Suppose that the sides of the channel make an angle    with the horizontal.

2

L metres

L metres 

 L metres

Find the area of the cross-section of the channel as a function of . For what angle  is the area of the cross-section a maximum? ¤

a. b.

« a) L²sin(1 + cos) m² b)  »

3

3U85-1i)! Find the value of the derivative of tan²x at x =

 .¤ 4 « 4 »

3U85-1ii)! 1

Find  x(1+ x 2 ) 2 dx . ¤ 0

« 7 »

6

3U85-3ii)!

 t  . Find  15

The function x(t) is given by x(t)  4  60sin a. b. b.

M, the maximum value of x(t); the least positive value of t for which x(t) = M; the values of x(t) for which x (t)  2 . ¤ « a) 64 b)

45 c) 4 ± 30 3 » 2

3U85-7)! B

P

C

A

The diagram shows a straight road BC running due East. A four-wheel drive ambulance is in open country at A, 3km due South of B. It must reach C, 9km due East of B, as quickly as possible. The driver knows that she can travel at 80km per hour in open country and at 100km per hour along the road. She intends to proceed in a straight line to some point P on the road and then to continue along the road to C. She wishes to choose P so that the total time for the journey APC is a minimum. a. If the distance BP is x km, derive an expression for t(x), the total journey time from A to C via P, in terms of x. ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


b.

Show that the minimum time for the total journey APC is 6 43 minutes. ¤ « a) t(x) 

9  x2 9  x b) Proof »  80 100

3U84-2ii)! A

C

B F D

In the figure, BC a. b. c.

E

DE and AB:BD = 3:5. Show that

ABC is similar to ADE, BFC is similar to EFD, DF : FC = 8 : 3. ¤ « Proof »

3U84-3i)! The carbon isotope C14 decays at a rate proportional to its mass. Tree ring experiments suggest that 50% decay takes 5580 years. A fossil contains 30% of the amount of C14 in a similarly sized living organism. Estimate the age of the fossil. ¤ « 9690 years (to the nearest 10 years) » 3U84-4i)! A spherical bubble is expanding so that its volume is increasing at the constant rate of 10 mm3 per second. What is the rate of increase of the radius when the surface area is 500 mm²? ( V = 43  R3 , S = 4 R2 .)¤ « 1 mm/s »

50


3 UNIT MATHEMATICS (PRELIMINARY) – OTHER INEQUALITIES – HSC

Other Inequalities 3U97-2c)! The polynomial P(x) = x3 + bx2 + cx + d has roots 0, 3 and -3. i. Find b, c and d. ii. Without using calculus, sketch the graph of y = P(x). iii.

Hence, or otherwise, solve the inequality

-3

x2  9  0.¤ x y

0

3

x

« i) b = 0, c = -9, d = 0 ii)

iii) -3 < x < 0 or x > 3 »

3U96-1e)! Solve the inequality

2  1.¤ x 1

« x < 1 or x  3 »

3U94-2c)!

1 2 and x = 2 satisfy the equation 7  3x  . 3 x

i.

Verify that x 

ii.

On the same set of axes, sketch the graphs of y = 7 - 3x and y 

iii.

Using part (ii), or otherwise, write down all values of x for which 7  3x  y 8

2 x

2 .¤ x

1   , 6 3 

(2, 1)

y

2 . x

2

y

2 x

x

1 3 y = 7 - 3x

iii) 0 < x < 1 or x > 2 »

« i) Proof ii)

3

x 4 > 0. ¤ x 2

3U90-1b)!

Solve the inequality

3U89-1d)!

4 Solve for x: 1. ¤ 5 x

« -2 < x < 0 or x > 2 »

3U85-1iii)!

Find all positive values of x for which

« 1  x < 5 »

6 > x 1 . ¤ x « 0 < x < 3 or x < -2 »


3 UNIT MATHEMATICS (PRELIMINARY) – CIRCLE GEOMETRY – HSC

Circle Geometry 3U97-2a)!

26 O

C

D y A

B

The points A, B and C lie on a circle with centre O. The lines AO and BC are parallel, and OB and AC intersect at D. Also, ACB = 26 and BDC = y, as shown in the diagram. Copy or trace the diagram into your Writing Booklet. i. State why AOB = 52. ii. Find y. Justify your answer.¤ « i) The angle subtended by an arc at the centre of a circle is twice the angle subtended by the arc at the circumference ii) y = 102 » 3U96-2c)! The points O and P in the plane are d cm apart. A circle centre O is drawn to pass through P, and another circle centre P is drawn to pass through O. The two circles meet at S and T, as in the diagram. S

O

P

i.

T Show that triangle SOP is equilateral.

ii.

Show that the size of the angle SOT is

iii.

Hence find the area of the shaded region in terms of d.¤

2 . 3 « i) Proof ii) Proof iii)

4  3 3 2 d » 6

3U95-6a)! P

T

Q

R

PT is a tangent to the circle PRQ, and QR is a secant intersecting the circle in Q and R. The line QR intersects PT at T. Copy or trace the diagram into your Writing Booklet. i. Prove that the triangles PRT and QPT are similar. ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


ii.

Hence prove that PT = QT  RT. ¤ 2

« Proof » 3U94-2b)! B  

A

X P

Y C

Q

Let ABPQC be a circle such that AB = AC, AP meets BC at X, and AQ meets BC at Y, as in the diagram. Let BAP =  and ABC = ß. i. Copy the diagram into your Writing Booklet and state why AXC =  + ß. ii. Prove that BQP = . iii. Prove that BQA = ß. iv. Prove that PQYX is a cyclic quadrilateral. ¤ « i) AXC is the exterior angle of AXB and as such is equal to the sum of the two interior opposite angles. ii) iii) iv) Proof » 3U93-4a)! G

A

E O  B

F

D

C

In the diagram, AB is a diameter of the circle, centre O, and BC is tangential to the circle at B. The line AED intersects the circle at E and BC at D. The tangent to the circle at E intersects BC at F. Let EBF = . i. Copy the diagram into your Writing Booklet. ii.

Prove that FED =

iii.

Prove that BF = FD. ¤

 - . 2

« Proof » 3U92-3c)! M

S P



O

N

R

Q

In the diagram P, Q, R, and S are points on a circle centre O, and POQ = . The lines PS and QR intersect at M and the lines QS and PR intersect at N. i. Explain why  PRM =   21  . ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


ii.

Show that PNQ + PMQ = . ¤ « Proof »

3U91-2c)!

A

C B

The diagram shows two circles intersecting at A and B. The diameter of one circle is AC. Copy this diagram into your examination booklet. i. On your diagram draw a straight line through A, parallel to CB, to meet the second circle in D. ii. Prove that BD is a diameter of the second circle. iii. Suppose that BD is parallel to CA. Prove that the circles have equal radii. ¤ A D C B

« i)

ii) Proof iii) Proof »

3U90-3a)! A

B T

C

P

(FIGURE NOT TO SCALE) AB is a diameter of a circle ABC. The tangents at A and C meet at T. The lines TC and AB are produced to meet at P. Copy the diagram into your examination booklet. Join AC and CB. i. Prove that CAT = 90 - BCP. ii. Hence, or otherwise, prove that ATC = 2BCP. ¤ « Proof » 3U89-2a)! Q 6 P x R 5 S

FIGURE NOT TO SCALE ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


PQ is a tangent to a circle QRS, while PRS is a secant intersecting the circle in R and S, as in the diagram. Given that PQ = 6, RS = 5, PR = x, find x. ¤ « x = 4 » 3U89-5a)! AB and CD are two intersecting chords of a circle and CD is parallel to the tangent to the circle at B. i. Draw a neat sketch of the above information in your writing booklet. ii. Prove that AB bisects CAD. ¤ A

C

« i)

P

D

B

Q

ii) Proof »

3U88-4a)! i. M

O

S

N

SM and SN are tangents drawn from an external point S to a circle with centre O. The points of contact of these tangents with the circle are M and N. Copy this diagram into your writing booklet. By proving triangle OMS and ONS are congruent show that SM = SN. ii.

P

A

C

T

Q B

Two circles touch externally at C. The circles, which have centres P and Q, are touched by a common tangent at A and B respectively. The common tangent at C meets AB in T. . Copy this diagram in your writing booklet. Using the result from (i) prove that AT = TB. ß. Show that ACB is a right angle. ¤ « Proof » 3U87-2i)!



B

C

FIGURE NOT TO SCALE Two circles cut at points B and C as shown in the diagram. A diameter of one circle is AB while BD is a diameter of the other. a. Draw a neat sketch in your answer book showing the given information. b. Prove that A, C and D are collinear, giving reasons. ¤ B

A

D

C

« a)

b) Proof »

3U86-2ii)! Two circles with centres X and Y intersect at two points A and B. a. Draw a neat sketch joining XA, XB, YA, YB, XY, AB. Let P be the point where XY meets AB. b. Prove that the triangles AXY and BXY are congruent. c. Prove that AP = BP. d. Given that XA is also a tangent to the circle with centre Y, prove that XAYB is a cyclic quadrilateral. ¤ A X

Y B

Proof » 3U85-2i)! P, Q are points on a circle and the tangents to the circle at P, Q meet as S. R is a point on the circle so that the chord PR is parallel to QS. a. Draw a neat sketch in your answer book, showing the given information. b. Giving reasons, prove carefully that QP = QR. ¤ «

P

R

« a) 3U84-2i)!


S

Q

b) Proof »


X

A

B

Y

C

Given that AB = AC and XY is tangent to circle ABC at A, prove that XY is parallel to BC. ¤ « Proof »


3 UNIT MATHEMATICS (PRELIMINARY) – FURTHER TRIGONOMETRY (SUMS AND DIFFERENCES, T FORMULAE, IDENTITIES AND EQUATIONS) – HSC

Further Trigonometry (sums and differences, t formulae, identities and equations) 3U96-4a)! Prove that

sin 3 cos 3   2 (for sin   0, cos   0).¤ sin  cos  « Proof »

3U96-4c)!

Q

A

B X

P

Z

D

Y

R

C

S NOT TO SCALE The figure shows the net of an oblique pyramid with a rectangular base. In this figure, PXZYR is a straight line, PX = 15cm, RY = 20cm, AB = 25cm, and BC = 10cm. Further, AP = PD and BR = RC. When the net is folded, points P, Q, R, and S all meet at the apex T, which lies vertically above the point Z in the horizontal base, as shown below.

T

B Y A

C Z

X D i. ii. iii.

Show that triangle TXY is right-angled. Hence show that T is 12cm above the base. Hence find the angle that the face DCT makes with the base.¤ « i) Proof ii) Proof iii) 67 23(to nearest minute) »

3U94-2a)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


Prove the following identity:

2 tan A  sin 2 A . ¤ 1 + tan2 A « Proof »

3U94-5a)! Find all angles , where 0    2, for which

3sin   cos  1 . ¤ «  =

 3

or  »

3U94-7c)! Uluru is a large rock on flat ground in Central Australia. Three tourists A, B, and C are observing Uluru from the ground. A is due north of Uluru, C is due east of Uluru, and B is on the line-of-sight from A to C and between them. The angles of elevation to the summit of Uluru from A, B, and C are 26, 28, and 30, respectively. Determine the bearing of B from Uluru. ¤ « 00432' (to the nearest minute) » 3U93-2a)! Prove the following identity:

sin A sin A   tan 2 A . ¤ cos A sin A cos A sin A « Proof »

3U92-2a)! Solve the equation 2sin² = sin 2 for 0    2. ¤

«  = 0,  , , 5 , 2 »

4

4

3U92-3a)! P

C 20 A

100m

60 600m

B

FIGURE NOT TO SCALE Two yachts A and B subtend an angle of 60 at the base C of a cliff. From yacht A the angle of elevation of the point P, 100 metres vertically above C, is 20. Yacht B is 600 metres from C. i. Calculate the length AC. ii. Calculate the distance between the two yachts. ¤ « i) 274.7m (to 1 d.p.) ii) 520m (to the nearest metre) » 3U90-2a)! i. Factorize a3 + b3. ii. 3U90-5a)!

Hence, or otherwise, show that

2 sin3 A 2 cos3 A  2  sin 2 A , if sin A + cos A  0. ¤ sin A cos A

« i) (a + b)(a2 - ab + b2) ii) Proof » Find all the angles  with 0    2 for which sin  + cos  = 1. ¤ «  = 0,  , 2 »

2

3U89-2b)! Find all angles  with 0    2 for which sin 2 = sin . ¤

«  = 0,  , , 5 or 2 »

3


3


3U89-3c)! P

h 12 11

Q

A

1000m B

FIGURE NOT TO SCALE The angle of elevation of a tower PQ of height h metres at a point A due east of it is 12. From another point B, the bearing of the tower is 051T and the angle of elevation is 11. The points A and B are 1000 metres apart and on the same level as the base Q of the tower. i. Show that AQB = 141. ii. Consider the triangle APQ and show that AQ = h tan 78. iii. Find a similar expression for BQ. iv. Use the cosine rule in the triangle AQB to calculate h to the nearest metre. ¤ « i) Proof ii) Proof iii) BQ = h tan 79 iv) 108m » 3U88-3c)! Given that 0  x 

    cos x sin x , prove that tan   x  .¤  4  cos x sin x 4

« Proof » 3U87-3i)! Find, for 0  x  2, all solutions of the equation sin 2x = cos x. ¤

« x =  ,  , 5 , 3 »

6 2

6

2

3U87-7)! The rectangular piece of paper PQRS shown on the next page is folded along a line AB, where A and B lie on edges PQ and PS respectively. This line is so positioned that, after folding, P coincides with a point P' which lies on the edge QR. This fold line AB makes an acute angle  with the edge PQ. The length of AB is L and that of PQ is w. a. Show that P'AQ = ( - 2). b.

Prove that L =

c.

w . cos (1- cos 2)

More than one fold line exists such that P coincides with a point on QR after folding. Find the value of  which corresponds to the fold line of minimum length. d. Let CD be the fold line of minimum length, where C lies on PQ and D lies on PS. Calculate the length of CP. (Remove the next page and attach it to your Writing Booklet for this question.)



R

S

B

P' L



Q

3U86-1ii)! a. b.

P « a)AProof b) Proof c) 5444' (to the nearest minute) d) 3w »

4

Write down, in surd form, the values of sin 45, cos 45, sin 30, cos 30. Hence show that sin 75

1 3 .¤ 2 2 « a)

1 1 1 3 , , , b) Proof » 2 2 2 2

3U86-5i)! Find all values of  with 0     such that 2sin  + cos  = 1. ¤ «  = 0, 2.214 (to 4 sig. figs) » 3U84-6ii)! a.

b.

  + A =  sin A . 2 

Consider the statement cos

For which sign is this statement true for all A? For which A is the statement true for both signs? Taking A = 5 in (a) write down a value of B such that -sin 5 = cos B. Hence find the least value of  between 0 and 2 such that sin 5 + cos 8 = 0. ¤



« a) Negative, A = 0, n, where n is an integer b)


3 » 26

3 UNIT MATHEMATICS (PRELIMINARY) – ANGLES BETWEEN TWO LINES & INTERNAL AND EXTERNAL DIVISION OF LINES INTO GIVEN RATIOS – HSC

Angles between 2 Lines. Internal and External Division of Lines into Given Ratios 3U96-1b)! Find the acute angle between the lines 2x + y = 4 and x - y = 2, to the nearest degree.¤ « 72 » 3U96-1c)! Let A(-1, 2) and B(3, 5) be points in the plane. Find the coordinates of the point C which divides the interval AB externally in the ratio 3 : 1.¤ « (5, 6.5) » 3U95-3c)! y 

x2 y= 4

P(2a, a2 ) S(0, 1)  x

O

Let P(2a, a2) be a point on the parabola y =

x2 , and let S be the point (0, 1). The tangent to the 4

parabola at P makes an angle of ß with the x axis. The angle between SP and the tangent is . Assume that a > 0, as indicated. i. Show that tan ß = a. ii.

 1 Show that the gradient of SP is 1  a-  .

iii.

Show that tan  =

iv. v.

Hence find the value of  + ß. Find the coordinates of P if  = ß. ¤

2

a

1 . a

« i) ii) iii) Proof iv) 90 v) P(2, 1) » 3U94-1c)! The interval AB has end-points A(-2, 3) and B(10, 11). Find the coordinates of the point P which divides the interval AB in the ratio 3 : 1. ¤ « P(7, 9) internally or P(11, 9.5) externally » 3U94-1d)! The graphs of y = x and y = x3 intersect at x = 1. Find the size of the acute angle between these curves at x = 1. ¤ « 2634' (to the nearest minute) » 3U93-5b)! i. Prove that the graph of y = ln x is concave down for all x > 0. ii. Sketch the graph of y = ln x. iii. Suppose 0 < a < b and consider the points A(a, ln a) and B(b, ln b) on the graph of y = ln x. Find the coordinates of the point P that divides the line segment AB in the ratio 2 : 1. iv.

By using (ii) and (iii) deduce that 1 ln a + 2 ln b < ln 1 a + 2 b . ¤


3

3

3

3


y

y = ln x 1 1

e

x

« i) Proof ii)

iii) P 1 a + 2 b, 1 ln a + 2 ln b iv) Proof » 3 3 3 3

3U92-1e)! y y = mx

y

45 O

1 x 2

x

FIGURE NOT TO SCALE The angle between the lines y = mx and y  21 x is 45 as shown in the diagram. Find the exact value of m. ¤ « 3 » 3U91-1c)! The point P(-3, 8) divides the interval AB externally in the ratio k : 1. If A is the point (6, -4) and B is the point (0, 4), find the value of k. ¤ « -3 » 3U90-1c)! The parabolas y = x² and y = (x - 2)² intersect at a point P. i. Find the coordinates of P. ii. Find the angle between the tangents to the curves at P. Give your answer to the nearest degree. ¤ « i) P(1, 1) ii) 53 » 3U89-1c)! Find the coordinates of the point which divides the interval AB with A(1, 4) and B(5, 2) externally in the ratio 1 : 3. ¤ « (-1, 5) » 3U88-1b)! Find the coordinates of the point P which divides the interval AB with end points A(2, 3) and B(5, -7) internally in the ratio 4 : 9. ¤ « P( 2 12 , - 1 ) »

13

3U87-1iii)! a. b.

3

13

Find the equation of the normal n to the curve y = x4 + 4 x2 at the point A(1, 5). Find, to the nearest degree, the size of the acute angle between the line n and the line L with equation 2x + 3y - 7 = 0. ¤ « a) x + 10y - 51 = 0 b) 28 »

3U86-2i)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


Find the coordinates of the point P which divides the interval AB internally in the ratio 2 : 3 where A and B have coordinates (1, -3) and (6, 7) respectively. ¤ « P(3, 1) » 3U85-1iv)! Find the acute angle between the lines y = -x and 3 y  x . ¤ « 75 »


3 UNIT MATHEMATICS (PRELIMINARY) – TRIGONOMETRIC FUNCTIONS – HSC

Parametric Representation 3U95-3c)!

y x2 y= 4

 P(2a, a2 )

S(0, 1)  O

Let P(2a, a2) be a point on the parabola y =

x

x2 , and let S be the point (0, 1). The tangent to the 4

parabola at P makes an angle of ß with the x axis. The angle between SP and the tangent is . Assume that a > 0, as indicated. i. Show that tan ß = a. 1 2

 a  1 .    a

ii.

Show that the gradient of SP is

iii.

Show that tan  =

iv. v.

Hence find the value of  + ß. Find the coordinates of P if  = ß. ¤

1 . a

« i) ii) iii) Proof iv) 90 v) P(2, 1) » 3U94-3d)! Two points P(2ap, ap²) and Q(2aq, aq²) lie on the parabola x² = 4ay. i. Show that the equation of the tangent to the parabola at P is y = px - ap². ii. The tangent at P and the line through Q parallel to the y axis intersect at T. Find the coordinates of T. iii. Write down the coordinates of M, the midpoint of PT. iv. Determine the locus of M when pq = -1. ¤ « i) Proof ii) T(2ap, 2apq - ap2) iii) M(a(p + q), apq) iv) The directrix of x2 = 4ay, ie y = -a » 3U93-7a)! Consider the parabola 4ay = x² where a > 0, and suppose the tangents at P(2ap, ap²) and Q(2aq, aq²) intersect at the point T. Let S(0, a) be the focus of the parabola. i. Find the coordinates of T. (You may assume that the equation of the tangent at P is y = px - ap².) ii. Show that SP = a(p² + 1). iii. Suppose P and Q move on the parabola in such a way that SP + SQ = 4a. Show that T is constrained to move on a parabola. ¤ « i) T(a(p + q), apq) ii) iii) Proof: the parabola x2 = 2a2 + 2ay » 3U92-5a)!



R

y

T

P(2at, at²) x² = 4ay

S(0, a) O

Q D

x y = -a

The diagram shows the parabola x² = 4ay with focus S(0, a) and directrix y = -a. The point P(2at, at²) is an arbitrary point on the parabola and the line RP is drawn parallel to the y axis, meeting the directrix at D. The tangent QPT to the parabola at P intersects SD at Q. i. Explain why SP = PD. ii. Find the gradient m1 of the tangent at P. iii. Find the gradient m2 of the line SD. iv. Prove that PQ is perpendicular to SD. v. Prove that RPT = SPQ. ¤ « i) A parabola is the locus of a point P which moves such that its distance from a fixed point S (the focus) is equal to its distance from a fixed line (the directrix, y = -a). PD is perpendicular to the directrix.  SP = PD ii) t iii) - 1 iv) Proof v) Proof »

t

3U91-5b)! i.

Sketch the parabola whose parametric equations are x = t and y = t². On your diagram mark the points P and Q which correspond to t = -1 and t = 2 respectively.

ii.

Show that the tangents to the parabola at P and Q intersect at R( 1 , - 2) .

iii.

Let T(t, t²) be the point on the parabola between P and Q such that the tangent at T meets QR at the mid-point of QR. Show that the tangent at T is parallel to PQ. ¤ y

2

Q(2, 4) P(-1, 1)

« i)

x

ii) Proof iii) Proof »

3U90-5c)! The point P(2ap, ap²) lies on the parabola x2 = 4ay. The focus S is the point (0, a). The tangent at P meets the y axis at Q. i. Find the coordinates of Q. ii. Prove that SP = SQ. iii. Hence show that PSQ + 2SQP = 180. ¤ « i) Q(0, -ap2) ii) Proof iii) Proof » 3U89-6b)! Two points P(2ap, ap²) and Q(2aq, aq²) lie on the parabola x² = 4ay. i. Derive the equation of the tangent to the parabola at P. ii. Find the coordinates of the point of intersection T of the tangents to the parabola at P and Q. iii. You are given that the tangents at P and Q in (ii) intersect at an angle of 45. Show that p - q = 1 + pq. ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


iv.

3U88-6a)! i. ii. iii.

By evaluating the expression x² - 4ay at T, or otherwise, find the locus of T when the tangents at P and Q intersect as given in (iii). ¤ « i) y = px - ap2 ii) T(a(p + q), apq) iii) Proof iv) x2 - 4ay = (a + y)2, an hyperbola » Show that the equation of the tangent to the parabola x² = 4Ay at any point P(2Ap, Ap²) is given by px - y - Ap² = 0. S is the focus of the parabola and T the point of intersection of the tangent and the y-axis. Prove that SP = ST. Hence show that SPT is equal to the acute angle between the tangent and the line through P parallel to the axis of the parabola. ¤ « Proof »

3U87-6ii)! Two points P(2Ap, Ap²) and Q(2Aq, Aq²) lie on the parabola x² = 4Ay, where A > 0. The chord PQ passes through the focus. a. Show that pq = -1. b. Show that the point of intersection T of the tangents to the parabola at P and Q lies on the line y = -A. 2

c.

 1 Show that the chord PQ has length A  p  . ¤  p « Proof »

3U85-3i)! a. b.

Derive the equation of the normal to the parabola x = 2At, y = At² at the point where t = T. For the parabola x = 2t, y = t², find the values (if any) of T for which the normal at the point where t = T passes through (0, 6). ¤ « a) x + Ty = AT3 + 2AT b) T = -2, 0, 2 »

3U84-5)! The straight line y = mx + b meets the parabola x = 2At, y = At² in real distinct points P, Q which correspond respectively to the values t = p, t = q.

b . A

a.

Prove that pq  -

b.

Prove that p2  q2  4 m2 

c. d.

Show that the equation of the normal to the parabola at P is x + py = 2Ap + Ap3. The point N is the point of intersection of the normals to the parabola at P and Q. Prove that the coordinates of N are (-Apq{p + q}, A{2 + p² + pq + q²}), and express these coordinates in terms of A, m, and b. Suppose that the chord PQ is free to move while maintaining a fixed gradient. Show that the locus of N is a straight line and verify that this straight line is a normal to the parabola. ¤ « a) Proof b) Proof c) Proof d) Proof, N(2mb, 2A + 4Am² + b) e) Proof »

e.


2b . A

3 UNIT MATHEMATICS (PRELIMINARY) –PERMUTATIONS AND COMBINATIONS – HSC

Permutations and Combinations 3U96-1d)! A committee of 3 men and 4 women is to be formed from a group of 8 men and 6 women. Write an expression for the number of ways this can be done.¤ « 8C3  6C4 » 3U95-3a)! A

B

C

D

E

F

G

H

A security lock has 8 buttons labelled as shown. Each person using the lock is given a 3-letter code. I. How many different codes are possible if letters can be repeated and their order is important? ii. How many different codes are possible if letters cannot be repeated and their order is important? iii. Now suppose that the lock operates by holding 3 buttons down together, so that order is NOT important. How many different codes are possible? ¤ « i) 512 ii) 336 iii) 56 » 3U94-7b)!

L

The figure shows 9 points lying in the plane, 5 of which lie on the line L. No other set of 3 of these points is collinear. i. How many sets of 3 points can be chosen from the 5 points lying on L? ii. How many different triangles can be formed using the 9 points as vertices?¤ « i) 10 ii) 74 » 3U93-1e)! A class consists of 10 girls and 12 boys. How many ways are there of selecting a committee of 3 girls and 2 boys from this class? ¤ « 7920 » 3U93-4b)! Five travellers arrive in a town where there are five hotels. i. How many different accommodation arrangements are there if there are no restrictions on where the travellers stay? ii. How many different accommodation arrangements are there if each traveller stays at a different hotel? iii. Suppose two of the travellers are husband and wife and must go to the same hotel. How many different accommodation arrangements are there if the other three can go to any of the other hotels? ¤ « i) 3125 ii) 120 iii) 320 » 3U91-4c)! Containers are coded by different arrangements of coloured dots in a row. The colours used are red, white, and blue. In an arrangement, at most three of the dots are red, at most two of the dots are white, and at most one is blue. i. Find the number of different codes possible if six dots are used. ii. On some containers only five dots are used. Find the number of different codes possible in this case. Justify your answer. ¤ « i) 60 ii) 60 » 3U90-4b)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995

3 UNIT MATHEMATICS (PRELIMINARY) –PERMUTATIONS AND COMBINATIONS – HSC

There are three identical blue marbles and four identical yellow marbles arranged in a row. i. How many different arrangements are possible ? ii. How many different arrangements of just five of these marbles are possible ? ¤ « i) 35 ii) 25 » 3U89-4c)! Let each different arrangement of all the letters of DELETED be called a word. i. How many words are possible? ii. In how many of these words will the D's be separated? ¤ « i) 420 ii) 300 » 3U88-1d)! Find the number of six-letter arrangements that can be made from the letters in the word SYDNEY. ¤ « 360 »


3 UNIT MATHEMATICS (PRELIMINARY) – POLYNOMIALS – HSC

Polynomials 3U97-2c)! The polynomial P(x) = x3 + bx2 + cx + d has roots 0, 3 and -3. i. Find b, c and d. ii. Without using calculus, sketch the graph of y = P(x). iii.

Hence, or otherwise, solve the inequality

-3

x2  9  0.¤ x y

0

3

x

« i) b = 0, c = -9, d = 0 ii) 3U96-1a)! (x - 2) is a factor of the polynomial P(x) = 2x3 + x + a. Find the value of a.¤

iii) -3 < x < 0 or x > 3 »

« a = -18 » 3U95-2c)! Consider the equation x3 + 6x2 - x - 30 = 0. One of the roots of this equation is equal to the sum of the other two roots. Find the values of the three roots. ¤ « -5, -3 and 2 » 3U94-4a)! When the polynomial P(x) is divided by (x + 1)(x - 4), the quotient is Q(x) and the remainder is R(x). i. Why is the most general form of R(x) given by R(x) = ax + b? ii. Given that P(4) = -5, show that R(4) = -5. iii. Further, when P(x) is divided by (x + 1), the remainder is 5. Find R(x). ¤ « i) The degree of R(x) must be less than the degree of (x + 1)(x - 4) ii) Proof iii) R(x) = -2x + 3 » 3U93-3b)! When the polynomial P(x) is divided by x² - 1 the remainder is 3x - 1. What is the remainder when P(x) is divided by x - 1? ¤ « 2 » 3U92-6a)! Show that (x - 1)(x - 2) is a factor of P(x) = xn(2m - 1) + xm(1 - 2n) + (2n - 2m) where m and n are positive integers. ¤ « Proof » 3U91-1b)! The polynomial P(x) = x3 + ax + 12 has a factor (x + 3). Find the value of a. ¤ « -5 » 3U90-2b)! A polynomial is given by p(x) = x3 + ax² + bx - 18. Find values for a and b if (x + 2) is a factor of p(x) and if -24 is the remainder when p(x) is divided by (x - 1). ¤ « a = 2, b = -9 » 3U89-4b)! i. The polynomial equation P(x) = 0 has a double root at x = a. By writing P(x) = (x - a)²Q(x), where Q(x) is a polynomial, show that P'(a) = 0. ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


ii.

Hence or otherwise find the values of a and b if x = 1 is a double root of x4 + ax3 + bx² - 5x + 1 = 0. ¤ « i) Proof ii) a = -5, b = 8 »

3U88-2c)! If , ß and  are the roots of x3 - 3x + 1 = 0 find: i. +ß+ ii. ß iii.

1 1 1   .¤    « i) 0 ii) -1 iii) 3 »

3U88-4b)! i. ii. iii.

Divide the polynomial f(x) = 2x4 - 10x3 + 12x² + 2x - 3 by g(x) = x² - 3x + 1. Hence write f(x) = g(x)q(x) + r(x) where q(x) and r(x) are polynomials and r(x) has degree less than 2. Hence show that f(x) and g(x) have no zeros in common. ¤ « ii) 2x4 - 10x3 + 12x2 + 2x - 3 = (x2 - 3x + 1)(2x2 - 4x - 2) - 1 iii) Proof »

3U87-5)! The polynomial equation f(x) = 8x3 + 12x2 - 18x - 20 = 0 has a root at x = -2. a. Find all roots of f(x) = 0. b. Draw a sketch of the graph of y = f(x) showing the coordinates of its points of intersection with the axes and all stationary points. c. Apply Newton's method once to approximate a root of f(x) = 0 beginning with an initial approximation x1 = 1. d. Willy chose an initial approximation of x1 = 0.49 and used Newton's method a number of times in order to approximate a root of f(x) = 0. State, giving reasons, the root of f(x) = 0 to which Willy's approximations are getting closer. (It is not necessary to do additional calculations.) ¤ y  3    , 7  2 

-2

1  21 4

1  21 4

x

-20 1   ,  25 2 

1  21 b) « a) x = -2, 4

c) 1.6 d) x = -2 »

3U86-3ii)! Consider the curve y = x4 + 4x3 - 16x + 1. a. Verify that the curve has a minimum at x = 1. b.

Factorise

dy completely, and hence determine the location and nature of any other dx

stationary points of the curve. ¤ « a) Proof b)

dy = 4(x - 1)(x + 2)², (-2, 17) is a point of inflexion » dx

3U85-2iii)! Given that there is a constant c such that (x4 + y4) = (x² + cxy + y²)(x² - cxy + y²) identically in x and y, find c. ¤ «  2 » 3U85-4ii)! ¤©BOARD OF STUDIES NSW 1984 - 1997 ©EDUDATA: DATAVER1.0 1995


a. b.

3

Factorise completely the polynomial p(x) = x - x² - 8x + 12, given that the equation p(x) = 0 has a repeated root. The polynomial q(x) has the form q(x) = p(x)(x + a), with p(x) as in (a) and where the constant a is chosen so that q(x)  0 for all real values of x. Find all possible values of a. ¤ « a) p(x) = (x - 2)²(x + 3) b) a = 3 »

3U84-1ii)! Define the function f by f(x) = x3 + 3x² - 9x - 27. a. Show that (x - 3) is a factor of x3 + 3x² - 9x - 27, and factorise this expression completely. b. Find where the graph of y = f(x) meets the axes. c. Find the stationary points of f and determine their nature. d. Find the point(s) of inflexion (if any) of f. ¤ « a) Proof, (x - 3)(x + 3)² b) (-3, 0), (3, 0) and (0, -27) c) (-3, 0) is a relative maximum turning point and (1, -32) is a relative minimum turning point. d) (-1, -16) »


3 UNIT MATHEMATICS (PRELIMINARY) – HARDER APPLICATIONS OF THE PRELIMINARY 2 UNIT COURSE – HSC

Harder Applications of the Preliminary 2 Unit Course 3U97-1b)! Find the perpendicular distance from the point (1, 2) to the line y = 3x + 4.¤ «

10 » 2

3U95-1a)! On a number plane, indicate the region specified by y  x - 1 and y  1. ¤ y = |x - 1|

y

y=1

1 0

1

x

« 3U95-1d)! Factorize 2n + 1 + 2n, and hence write

»

21001  21000 as a power of 2. ¤ 3 « 3(2n), 21000 »

3U93-1a)! Solve the inequality x2 - 3x < 4. ¤ « -1 < x < 4 » 3U93-2b)! i. ii.

On the same diagram, sketch the graphs of y = x and y = 2x - 1. By using (i) or otherwise, determine for what values of c the equation 2x - 1 = x + c has exactly two solutions. ¤ y y = |2x - 1| 1

 1 1  ,   3 3

(1, 1)

x y=x

ii) c > - 1 »

« i)

2

3U92-1a)! Solve x2 - x - 2 > 0.¤ « x > 2, x < -1 » 3U92-1b)! Differentiate

1 .¤ 1  x2 «

3U91-1d)! i. ii.

Sketch the graph of y = x - 2. For what values of x is x - 2 < x?¤


-x (1+ x2 )3

»


y 2 1 4 x

2

« i)

ii) x > 1 »

3U91-7a)! Let f(x) = i. ii. iii. iv.

x . x -1 2

For what values of x is f(x) undefined? Show that y = f(x) is an odd function. Show that f (x) < 0 at all values of x for which the function is defined. Hence sketch y = f(x). ¤ y

-1

1

x

« i) ±1 ii) Proof iii) Proof iv) 3U90-5b)! Consider the circle x² + y² - 2x - 14y + 25 = 0. i. Show that if the line y = mx intersects the circle in two distinct points, then (1 + 7m)² - 25(1 + m²) > 0. ii. For what values of m is the line y = mx a tangent to the circle? ¤

»

« i) Proof ii) m = - 4 , 3 »

3

4

3U89-1a)! Factorize 2m3 - 128. ¤ « 2(m - 4)(m2 + 4m + 16) » 3U87-1i)! Differentiate: a. b.

1 3  x2

exloge(2x). ¤ « a)

3U87-1iii)! a. b.

1 -2 x b) e x {  ln(2x)} » x (3  x 2 ) 2

Find the equation of the normal n to the curve y  x4  4 x2 at the point A(1, 5). Find, to the nearest degree, the size of the acute angle between the line n and the line L with equation 2x + 3y - 7 = 0. ¤ « a) x + 10y - 51 = 0 b) 28 » 3

3U87-2ii)! Find the values of x for which (x - 2)²  4. ¤ « x  4 or x  0 » 3U86-1i)! Write

1 7 in the form a b 7 where a and b are rational. ¤ 3 7 « 5 + 2 7 »



3U86-1iv)! Find all real numbers x such that x² + 4x > 5. ¤ « x < -5 or x > 1 » 3U86-4i)! a. b.

Find the coordinates of the vertex and the focus and the equation of the directrix of the parabola y = x² - 4x. Draw a sketch of the curve. A line whose equation is y = mx - 4 passes through the point (0, -4) and is a tangent to the parabola y = x² - 4x. Find the two possible values of m. ¤ y

y = x2 - 4x

4x

0

« a) V(2, -4), F(2, -3 43 ) and y = -4 41 ,

(2, -4)

b) 0, -8 »

3U85-2ii)! A circle has equation x² + y² - 4x + 2y = 0. a. Find the centre and radius of the circle. b. The line x + 2y = 0 meets this circle in two points, A, B. . Find the co-ordinates of A and B. ß. Calculate the distance AB. ¤ « a) centre (2, -1), radius = 5 units b) ) A(0, 0), B(4, -2) ß) 2 5 units » 3U85-2iii)! Given that there is a constant c such that (x4 + y4) = (x² + cxy + y²)(x² - cxy + y²) identically in x and y, find c. ¤ «  2 » 3U84-2ii)! A

C

B F D

In the figure, BC a. b. c.

E

DE and AB:BD = 3:5. Show that

ABC is similar to ADE, BFC is similar to EFD, DF : FC = 8 : 3. ¤ « Proof »


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