Add Math (4047/02)

Prepared by Mr Ang, Nov 2016 1

Requirement : Answer all questions

Total marks : 100

Duration : 2 hour 30 minutes

1. The price, $P, of a company share on 1st January has been increasing each year from 1995

to 2015. The company claims that this increase is exponential and so can be modelled by

an equation of form

0ektP P

where P0 and k are constants and t is the time in years since 1st January 1995.

The table below gives values of P and t for some of the years 1995 to 2010.

(i) Plot a suitable straight line graph to show that the model is valid for the years 1995 to

2010. [2]

(ii) Estimate the value of P0 and k. [3]

(iii) Assuming that the model is still appropriate, estimate the price of a share on

1st January 2015. [2]

Solution:

(i) Given that 0ektP P ,

0ln lnP P kt

To plot ln P against t.

Year 1995 2000 2005 2010

t years 0 5 10 15

$P 2.00 2.44 3.00 3.65

Year 1995 2000 2005 2010

t years 0 5 10 15

$P 2.00 2.44 3.00 3.65

ln P 0.69 0.89 1.10 1.29

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(ii) When t = 0, ln 0.7P , 00.7 ln P , 0.7

0 e 2.01P .

the gradient 1.5 0.7

0.039920.05

k

(3 s.f.)

(iii) When t = 20, ln 1.475P ,

1.475e 4.37P (3 s.f.)

the price of a share on 1st January 2015 is $4.37

O 5 10 15 20

0.5

1

1.5

•

•

•

•

t

ln P

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2. (i) Given that the coefficient of x2 in the expansion of 2 6

1 2 1x px is 16, find the

two possible values of the constant p. [5]

(ii) For each value of p, find the coefficient of x3 in the expansion of 6

1 px . [3]

Solution:

From 2 21 2 1 4 4x x x

6 6 0 5 1 4 2 0 6

6 0 6 1 6 2 6 61 1 1 1 1px C px C px C px C px

the coefficient of x2 , 24 24 15 16p p

215 24 12 0p p

25 8 4 0p p

5 2 2 0p p

2

5p or 2p

(ii)

When 2p , the coefficient of x3, 3 3 3

6 3 1 20 2 160C p

When 2

5p , the coefficient of x3,

33 3

6 3

21 20 1.28

5C p

1 ‒4x 4x2

1 4x2

‒6px 24px2 2 215 p x 2 215 p x

Add Math (4047/02)

Prepared by Mr Ang, Nov 2016 4

3. (i) Using cos3 cos 2x x x , show that cos3x may be expressed as 2cos 1 4sinx x .

[3]

(ii) Find all the values of x between 0° and 360° for which 2cos3 15sin cosx x x . [5]

Solution:

(i) Let cos3 cos 2x x x ,

cos 2 cos sin 2 sinx x x x

2 2cos sin cos 2sin cos sinx x x x x x

2 2cos cos 3sinx x x

2cos 1 4sinx x

(ii) 2cos3 15sin cosx x x

22cos 1 4sin 15sin cosx x x x

2cos 2 15sin 8sin 0x x x

cos 2 sin 1 8sin 0x x x

cos 0x or sin 2x or 1

sin8

x

When cos 0x , 90x or 270x

When sin 2x , no solution as sin 1x

When 1

sin8

x ,

1 1sin

8x

Principal angle, 7.2x (1 d.p.)

7.2x or 180 7.2 172.8x (1 d.p.)

Therefore, 7.2x or 90x or 172.8x or 270x

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4. The roots of the quadratic equation 2 2 5 0x x are and . Find

(i) the value of 3 3 , [5]

(ii) a quadratic equation with roots 2

and

2

. [3]

Solution:

(i) 2

5

3 2 2 2 3 3 2 22 3 3

33 3 2 23 3

33 3 3

33 3 2 3 5 2 22

(ii)

3 3

22 2

22

25

2 2

1 1

5

2 22 10

25 5x x

or 225 22 5 0x x

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5.

The diagram shows a circle passing through the vertices of a triangle ABC.

The tangents to the circles at A and B intersect at the point P.

The point D lies on AC such that DC = DB.

(i) Prove that angle APB + angle ADB = 180° [5]

(ii) Given that A, P, B and D lie on a circle, prove that PD and BC are parallel. [2]

Solution:

(i) Let O be the centre of the circle.

180APB AOB (tangent from external point P)

1

2ACB AOB ( s at circumference = ½ of at the centre of the circle)

180 2CDB ACB ( CDB is an isosceles)

180CDB AOB

180 180CDB APB

CDB APB

Since 180CDB ADB (straight angle)

180APB ADB

\ /

C

P

A

B

D

\ /

C

P

A

B

D

O

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(ii)

PDB PAB ( s in the same segment)

1

2ACB DCB AOB ( s at circumference = ½ of at the centre of the circle)

1

2DCB DBC AOB ( CDB is an isosceles)

Since 2 180PAB APB ,

1 1

180 1802 2

PAB AOB AOB (tangent from external point P)

Therefore, 1

2PDB AOB DBC , (alternate s)

Hence , PD and BC are parallel

\ /

C

P

A

B

D

O

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6. It is given that 3

2 2 5y x x .

(i) Obtain an expression for d

d

y

x in the form

22 5ax b x , where a and b are

integers, [3]

(ii) Determine the values of x for which y is a decreasing function. [1]

The variables x and y are such that, when x = 3, y is increasing at a rate of 0.35 units per

second.

(iii) Find the rate of change of x when x = 3. [2]

It is given further that the variable z is such that z = y2.

(iv) Show that, when x =3, z is increasing at twice the rate of y. [2]

Solution:

(i) Given that 3

2 2 5y x x ,

3 2d

2 5 3 2 2 5 2d

yx x x

x

2d

2 5 2 5 6 2d

yx x x

x

2d

8 17 2 5d

yx x

x

(ii) for which y is a decreasing function, d

0d

y

x

2

8 17 2 5 0x x

1

28

x

(iii) d d d

d d d

y y x

t x t

. d

0.35d

y

t , when 3x .

when 3x , 2d

24 17 6 5 7d

y

x

d

0.35 7d

x

t

d0.05

d

x

t

(‒)

x (‒) 12

8

12

2

(+) (+)

Add Math (4047/02)

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7. (i) Given that 2xu , express 2 1 22 2 6x x as an equation in u. [3]

(ii) Hence, find the values of x for which 2 1 22 2 6x x , giving your answer, where

appropriate, to 1 decimal place. [4]

(iii) Explain why the equation 2 1 22 2x x k has no solution if 8k . [3]

Solution:

(i) 2 1 22 2 6x x

21

2 4 2 6 02

x x

2 8 12 0u u

(ii) 2 8 12 0u u

2 6 0u u

2u or 6u

When 2u , 2 2x , 1x

When 6u , 2 6x , lg 6

2.6lg 2

x (1 d.p.)

(iii) Given that 2 1 22 2x x k , 2 8 2 0u u k

For the equation to have real solutions, the discriminant 0 .

2

8 4 1 2 0k

64 8 0k

8k

For the equation 2 1 22 2x x k to have no solution, then 8k .

Add Math (4047/02)

Prepared by Mr Ang, Nov 2016 10

8. It is given that 3 2f 3 4 12x x x x .

(i) By showing clearly your working factorise f x . [3]

(ii) Explain why the equation f 0x has only one real root and state its value. [2]

(iii) Find the value of the constant k for which the graph of fy x kx has a stationary

point at which 2

2

d0

d

y

x . [5]

Solution:

(i) 23f 3 3 3 3 4 3 12 0 , 3x is a factor of f x .

By long division,

2f 3 4x x x

(ii) f 0x ,

3 20 3 4 12x x x

20 3 4x x

Since 2 4 4 0x ,

3 0x

3x

(iii) fy x kx

3 23 4 12y x x x kx

2d3 6 4

d

yx x k

x

2

2

d6 6

d

yx

x

For 2

2

d0

d

y

x , 1x

When 1x ,

d

3 6 4d

yk

x

d

1d

yk

x

To be a stationary point at 1x , d

0d

y

x

0 1 k

1k

Add Math (4047/02)

Prepared by Mr Ang, Nov 2016 11

9.

The diagram above shows part of the graph of 3 22 3y x x x .

The x-coordinate of the point A is 2

3.

(i) Find the gradient of the curve at A. [4]

The tangents to the curve at the points A and B are parallel.

(ii) Find the x-coordinate of point B. [3]

(iii) Showing all your working, find the total area of the shaded region bounded by the

curve, the x-axis and the lines from A and B perpendicular to the x-axis. [4]

Solution:

(i) 2d3 4 3

d

yx x

x

When 2

3x ,

2d 2 2

3 4 3 1d 3 3

y

x

the gradient of the curve at A is 1.

(ii) 21 3 4 3x x

23 4 4 0x x

3 2 2 0x x

2

3x or 2x

the x-coordinate of point B is ‒2.

y

x O

A

B

3 22 3y x x x

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(iii) total area of the shaded region

2

0 33 2 3 2

2 0

2 3 d 2 3 dx x x x x x x x

4 4

3 2 3 2

202 3 2 3

324 3 2 4 3 2

0

x xx x x x

4

4 3 23 2

2

2 2 3 2 2 3 232 2

4 3 2 4 3 3 2 3

16 4 16 2

4 63 81 81 3

22 34

3 81

22 34

3 81

628

81

61

781

unit2

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10. A car, driven along a straight road, passes a signpost, A, with a speed of p km/h.

A little later the car passes a second signpost, B, with a speed of 80 km/h.

Between signposts A and B, the speed, v km/h, of the car is given by 2530e 20tv ,

where t, the time after passing A, is measured in hours.

(i) State the value of p. [1]

(ii) Calculate, to the nearest second, the time taken to travel from A to B. [3]

(iii) Calculate the distance between A and B. [5]

(iv) Obtain an expression, in terms of t, for the acceleration of the car between A and B.

[2]

Solution:

(i) When 0t , 030e 20 50p

(ii) At B, let t T ,

2580 30e 20T

252 e T

25 ln 2T

ln 2

0.027725

T hours (3 s.f.)

the time taken to travel from A to B, 100 seconds

(iii) the distance between A and B

ln 2

2525

0

30e 20 dt t

ln 2

2525

0

3025e 20 d

25

t t

25

ln 26

e 20 255

0

t t

ln 2 06 ln 2 6e 20 e

5 25 5

6 4ln 2 6

25 5 5

6 4ln 2

5 5

1.75 (3 s.f.)

the distance between A and B is 1.75 km

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(iv) the acceleration of the car between A and B,

25d30 25 e

d

tv

t

25d750e

d

tv

t

Add Math (4047/02)

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11. The equation of a circle, C1, with centre A, is 2 2 4 2 95x y x y .

(i) Find the coordinates of A and the radius of C1. [4]

(ii) Show that the point P(10, 7) lies on C1. [1]

(iii) Find the equation of the tangent to C1 at P. [3]

A second circle, C2, has diameter AP.

(iv) Find the equation of C2. [3]

(v) Find the equation of the tangent to C2 at P. [1]

Solution:

(i) 2 2 4 2 95x y x y

2 24 4 2 1 95 4 1x x y y

2 2 22 1 10x y

2,1A , radius = 10

(ii) P(10, 7)

2 2 210 2 7 1 10

2 2 28 6 10

2 210 10

point P(10, 7) lies on C1

(iii) 2,1A , P(10, 7)

Gradient of AP, 1 7 3

2 10 4APm

Gradient of tangent to C1 at P, 4

3APm

the equation of the tangent to C1 at P,

7 10

4 3

y x

4 61

3 3y x

(iv) Midpoint of AP 2 10 1 7

, 6,42 2

radius10

52

the equation of C2 , 2 2 26 4 5x y

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(v) the equation of the tangent to C2 at P = the equation of the tangent to C1 at P

4 61

3 3y x