Amk Prelim 2009 Am2

ANG MO KIO SECONDARY SCHOOL PRELIMINARY EXAMINATION 2009 SECONDARY 4 EXPRESS/5 NORMAL ACADEMIC ADDITIONAL MATHEMATICS FRIDAY

04 SEPTEMBER 2009

Name of Setter:

Miss Deepa Sivasothy

Additional Materials:

Answer paper

4038/02 2 hours 30 minutes

READ THESE INSTRUCTIONS FIRST Write your class, index number and name in the spaces on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Write your answers and working on the separate Answer Paper provided. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal places in the case of angles in degrees, unless a different level of accuracy is specified in the question. At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 100. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers.

This document consists of 7 printed pages. [Turn over

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax 2  bx  c  0,

x

 b  b 2  4ac 2a

Binomial expansion n  n  n (a  b) n  a n   a n1b   a n 2 b 2  ...   a n r b r  ...  b n , 1  2 r n n(n  1) ... (n  r  1) n! where n is a positive integer and     r!  r  r!(n  r )!

2. TRIGONOMETRY Identities sin 2 A  cos 2 A  1 sec 2 A  1  tan 2 A cos ec 2 A  1  cot 2 A sin( A  B)  sin A cos B  cos A sin B cos( A  B)  cos A cos B  sin A sin B

tan A  tan B 1  tan A tan B sin 2 A  2 sin A cos A

tan( A  B) 

cos 2 A  cos 2 A  sin 2 A  2 cos 2 A  1  1  2 sin 2 A 2 tan A 1  tan 2 A 1 1 sin A  sin B  2 sin ( A  B) cos ( A  B) 2 2 1 1 sin A  sin B  2 cos ( A  B) sin ( A  B) 2 2 1 1 cos A  cos B  2 cos ( A  B) cos ( A  B) 2 2 1 1 cos A  cos B  2 sin ( A  B) sin ( A  B) 2 2 tan 2 A 

Formulae for ABC

a b c   sin A sin B sin C a 2  b 2  c 2  2bc cos A 1   ab sin C 2

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

3 1

In an experiment, a liquid was heated to a high temperature. It was then allowed to cool. Its temperature, 

Degree Celsius, when it had been cooling for time t

minutes, is given by the equation   75e



t 3

 20 .

(i)

State the initial temperature of the liquid.

[1]

(ii)

Calculate the value of t when  = 65.

[2]

(iii) Find the rate at which  is decreasing when t = 3.

2

[2]

(iv)

Find the value which  approaches as t becomes very large.

(a)

The function f is defined, for 0  x  360 , by f ( x)  3 cos Ax  B , where

[1]

A and B are constants. (i)

Given that the minimum value of f is  2 , state the value of B.

[1]

(ii)

State the amplitude of f.

[1]

(iii) Given that the period of f is 180  , state the value of A. (iv)

(b)

3

Sketch the graph of y = f(x) for 0  x  360 .

[2]

2 x  1. 3

(i)

Solve 3 

(ii)

Hence sketch the graph of y  3 

[2] 2 x  1 for  1  x  9 . 3

[3]

Given that f(x) = 2 x 3  ax 2  18 x  b , (i)

find the value of a and of b for which 2x2 – 5x – 3 is a factor of f(x),

[4]

(ii)

state the third factor,

[1]

(iii) hence, solve the equation 6( y + 2)3 + 3(y + 2)2 – 54y – 135 = 0.

4

[1]

(i)

If one root of the equation 4 x 2  8 x  p  0 is three times the other root,

[2]

[4]

find the value of the constant p and the two roots. (ii)

If  and  are the roots of the equation 5x 2  7 x  3  0 , find the value of

1 1  .  1  1

[Turn over

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

[3]

4 5

(a)

Given that log p 2  x and log p 3  y , find in terms of x and y

(i)

log p (ii) (b)

If

[3]

27 p 5 . 32

Solve the equation log 9 ( x 2  3 x  2)  log 3 (3 x  2)  3 log 8 2 .

[4]

 a  b 7 , where a and b are rational numbers, find the

[4]

9 4 7

values of a and of b.

6

(i) (ii)

sin x  cosec x  cot x . 1  cos x Find all the angles between 0 and 360 which satisfy the equation

Prove the identity

10  5 tan x 

7

(a)

[4]

2 . cos 2 x

The total surface area of a spherical ice-ball is decreasing at a rate of 2 cm 2 /s . Find the rate of change of the volume when its radius is 0.5 cm.

(b)

[3]

The gradient of a curve, at the point ( x, y ) on the curve, is 2 

[4]

k . If it has x2

a turning point at (1, 2), find

8

(i)

the value of k ,

[2]

(ii)

the the equation of the curve.

[3]

A circle, C passes through the points A(5, 6) and B(4, −1) and has its centre lying on the line 4y = x – 11. Find (i)

the equation of the perpendicular bisector of AB,

[3]

(ii)

the equation of circle C.

[4]

[Turn over

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

5 9 P

Q O

S

R

T

In the circle, O is the centre and PS is the diameter. SR is a tangent to the circle and QRS is a right angle. The lines PQ and SR are produced to meet at T. (i)

Prove that triangles PQS and SRQ are similar.

[3]

(ii)

Prove that SQ 2  SP  RQ .

[2]

(iii) Prove that TQ 2  TP  TQ  SP  RQ .

[4]

[Turn over

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

6

10 y

y  2x3  2

C B(b, 18)

0

A

x

The diagram shows part of the curve y  2 x 3  2 and the point B(b, 18) lies on the curve. The line passing through B intersects the x and y axes at A and C respectively. Given that the area of the shaded region is 66 units2, find (i) the value of b.

[1]

(ii) the coordinates A and C.

[5]

11 A particle starts from rest at 5 m from a fixed point O and moves in a straight line with a velocity, v = 12t – 3t2 m/s where t is the time in seconds after leaving from the initial rest position. (i) Calculate the acceleration when the particle is next at rest.

[3]

(ii) Calculate the maximum velocity.

[2]

(iii) Express the displacement, s, from point O in terms of t.

[2]

(iv) Hence, find the total distance travelled in the first five seconds.

[2] [Turn over

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

7

12 Ceiling

Y θ

40

30 Z

X

Floor P

Q

The diagram shows a L-shaped rod XYZ, where XYZ  90 , XY = 40 cm and YZ = 30 cm. It is hung from the ceiling at Y such that YZ makes an angle of  with the ceiling. P and Q are two points on the floor which are vertically below X and Z respectively. Given that  varies, (i) show that PQ = 40 sin   30 cos  ,

[2]

(ii) express PQ in the form of R sin     ,

[3]

(iii) state the range of PQ,

[1]

(iv) calculate the values of  when PQ = 45 cm and

[2]

(v) find the perpendicular distance of X from the ceiling when PQ has a

[4]

maximum length.

END OF PAPER

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

8

Answer Scheme

Question Answer 1(i)

When t = 0,

  75e 0  20   95C  ( B1)

1(ii)

75e e

t  3



t 3

 20  65

 0.6

t  ln 0.6  ( M1) 3 t  1.53 min  ( A1) 

t

1(iii)

 d 1    75e 3  ( M1) dt 3 d 75 When t  3 ;   e 1 dt 3   9.20C / min  ( A1)

1(iv)

  75e    20   0  20   20C  ( B1)

2(a)(i)

B = 1  ( B1)

2(a)(ii)

Amplitude = 3  ( B1)

2(a)(iii)

A = 2  ( B1)

2(a)(iv)

f ( x )  3 cos 2 x  1 y, x intercepts  ( A1) turning points  (A1)

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

9

2(b)(i)

2 x 1 3 2 2 3  x  1 or 3  x  1  ( M1) 3 3 x  3 or x  6  (A1)

2(b)(ii)

Correct shape  ( A1)

3

Correct vertex, end points  ( A1) Correct x, y intercepts  ( A1)

3(i)

2x2 – 5x – 3 = (2x + 1)(x – 3)

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

10 f (3)  2(3) 3  a (3) 2  18(3)  b  9a  b 99a  b  0    (1)  ( M1) 2

 1  1  1  1 f     2    a     18    b  2  2  2  2 3 a 8   b 4 4 a 3  b  8  ( M1) 4 4 a  4b  35    (2 ) (1)  4 : 36a  4b  0    (3) (3)  ( 2) : 35a  35 a  1  ( A1), b  9  (A1)

3(ii)

f ( 3)  0  x  3 is the third factor.  ( B1)

3(iii)

6( y  2) 3  3( y  2) 2  54 y  135  0 6( y  2) 3  3( y  2) 2  54( y  2)  27  0 2( y  2) 3  ( y  2 ) 2  18( y  2 )  9  0  ( M1)  y  2   y  2

4(i)

1 or y  2  3 or y  2  3 2

1 or y  1 or y  5  ( A1) 2

   (3 )  3 2 p 3 2   ( M1) 4 8   3   ( M1) 4 1  2 2

1 p  4  3  2 p  3  ( A1) 1 3 Two roots  ,  ( A1) 2 2

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

11 4(ii)

 

7 5

3  ( M1) 5 1 1   1  1  1 1        1 7 2 5  ( M1)  3 7   1 5 5 3    ( A1) 5

  

5(a)(i)

log p

27 p 5 32 5

 log p 27  log p p 2  log p 32  (M1)  log p 33 

5  log p 2 5  (M1) 2

1  3 y  2  5 x  (A1) 2

5(a)(ii)

log 9 ( x 2  3 x  2)  log 3 (3x  2)  3 log 8 2 log 3 ( x 2  3 x  2)  log 3 (3 x  2)  log 8 2 3  (M1) log 3 9 log 3 ( x 2  3 x  2)  log 3 (3 x  2)  1 2 log 3 ( x 2  3 x  2)  2 log 3 (3 x  2)  2  x 2  3x  2    2  (M1) log 3  2   (3 x  2)  x 2  3x  2  3  2  (M1) 9 x 2  12 x  4 9 x 2  27 x  18  9 x 2  12 x  4 15 x  14 x

14  (A1) 15

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

12 5(b)

9 4 7  4 7 4 7 36  9 7  (M1) 16  7 36  9 7  9  4  7  (M1) 

(4  7 ) 2  16  8 7  7  (M1)  23  8 7 a  23, b  8  (A1)

6(i)

RHS  cosec x  cot x 1 cos x  sin x sin x 1  cos x   (M1) sin x sin x(1  cos x)  sin 2 x sin x(1  cos x)   (M1) 1  cos 2 x sin x(1  cos x)  (1  cos x)(1  cos x) sin x   (A1) 1  cos x  LHS(Proven) 

6(ii)

2 cos 2 x 10  5 tan x  2 sec 2 x 10  5 tan x 

2 sec 2 x  5 tan x  10  0 2(1  tan 2 x)  5 tan x  10  0  (M1) 2tan 2 x  5 tan x  8  0  (M1)  5  25  4(2)(8) 2(2) tan x  1.108 or tan x  3.608  (M1) x  47.9, 105.5, 227.9, 285.5  (A 1) tan x 

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

13 7(a)

A  4r dA  8r dr dr 1 dA   dt dA dt dr 1   2 8 (0.5) 1   (M1) 2 4 V  r 3 3 dV  4r 2  (M1) dr dV dV dr   dt dr dt 1  4 (0.5) 2    (M1) 2 1   cm3 / s  (A1) 2

7(b)(i)

dy k 2 2 dx x dy At (1,2) ;  2k dx 2  k  0  (M1) k  2  (A1)

7(b)(ii)

y   2  2 x 2 dx 2 x 1 c 1 2 y  2 x   c  (M1) x 2  22c y  2x 

c  2  (M1) 2 y  2 x   2  (A1) x

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

14 8(i)

Midpoint = (4.5, 2.5) Gradient of AB = 7 Gradient of perpendicular bisector = 

1 7

1 y   x  c  (M1) 7 At (4.5, 2.5); 1 2.5   (4.5)  c  (M1) 7 1 c3 7 1 1 y   x  3  (A1) 7 7

8(ii)

1 1 1 11 x  3  x   (M1) 7 7 4 4 x  15, y  1



centre (15,1)  (M1) radius  (15  5) 2  (6  1) 2  125  (M1) Equation of C : ( x  15) 2  ( y  1) 2  125  (A1)

9(i)

PQS  90( in semi circle)  (M1) PQS  SRQ  90 SPQ  QSR( in alt segment)  (M1)  PQS and SRQ are similar  (A1)

9(ii)

From (a) SP QS   (M1) QS RQ SQ 2  SP  RQ  ( A1)

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

15 9(iii)

ST 2  TP  TQ (tan  sec theorem)  (M1) ST 2  TQ 2  SQ 2 (pythagoras theorem)  (M1) TQ 2  SQ 2  TP  TQ  (M1) TQ 2  (SP  RQ)  TP  TQ [from(ii)] TQ 2  TP  TQ  SP  RQ  (A1)

10(i)

When y = 18; 2 x 3  2  18 x2 b  2  ( B1)

10(ii)

2 3 0 (2 x  2)dx 2

 2x4    2 x  (M1)  4 0  12unints 2 Area of triangle  66  12  54 Let A(a, 0) 1  (a  2)  18  54  (M1) 2 a 8 A(8, 0)  (A1) Equation of AC : y 0 18  x 8 2 8 y  3 x  24  (M1) At y  axis, x  0 y  24 C (0, 24)  (A1)

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

16 11(i)

v0 12t  3t 2  0 3t (4  t )  0 t  4  (M1) a  12  6t  (M1) When t  4; a  12  6( 4 ) a  12m / s 2  (A1)

11(ii)

dv  12  6t dt 12  6t  0 t  2  (M1) v  12(2)  3(2) 2  12m / s  (A1)

11(iii)

12t 2 3t 3  c 2 3 s  6t 2  t 3  c  ( M1) At t  0; s  5, c5 s

 s  6t 2  t 3  5  (A1)

11(iv)

When t  0, s  5m t  4, s  37m t  5, s  30m  (M1) Total Distance  32  7  39m  (A1)

12(i)

PY 40 PY  40 sin 

sin  

YQ 30 YQ  30 cos  both PYand YQ  (M1) cos  

PQ  PY  YQ  40 sin   30 cos  (shown)  (A1)

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02

17 12(ii)

R  40 2  30 2 R  50  (M1) 30 40   36.9  (M1) PQ  50 sin(  36.9)  (A1) tan  

12(iii)

30  PQ  50  (B1)

12(iv)

50 sin(  36.9)  45 sin(  36.9)  0.9   36.9  64.15...  (M1)

  27.3, 78.9  (A1)

12(v)

Maximum length of PQ = 50 Maximum length of PQ  50  (M1) sin(  36.9)  1 Maximum   53.1  (M1) AX cos 53.1   (M1) 40 AX  24.0cm  (A1)

AMKSS 4E_5NA_Additional Math Prelim Exam 2009, 4038/02