Pierce Prelim 2009 Am2 + Answer

Class Candidate Name

Index Number

_______________________________________

PEIRCE SECONDARY SCHOOL Department of MATHEMATICS GCE ‘O’ Level Preliminary Examination II for Secondary Four Express / Five Normal Academic Additional Mathematics Paper 2 Monday

4038/2

14 Sep 2009

0800 – 1030

Additional materials: A4 writing paper (8 sheets)

TIME

2 hours 30 minutes

INSTRUCTIONS TO CANDIDATES Write your name, class and register number on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Show your working on the same page as the rest the answer. Do notall use staples, paper clips, highlighters, glue or of correction fluid. Omission of essential working will result in loss of marks. Answer all questions. Write your answers on the separate Answer Paper provided. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of Angles angle in degrees, unless a different level of accuracy is specified in the question. The use of a scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. 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 number of marks for this paper is 100.

This question paper consists of 5 printed pages, including this cover page.

[Turn over Final Copy by Ms V Tan

2 Mathematical Formulae

1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0 x=

 b  b 2  4ac 2a

Binomial expansion n  n n (a + b)n = an +   an  1b +   an  2b2 + ......... +   an  r br + … + b n , 1  2 r  n n! n(n  1)...(n  r  1) where n is a positive integer and   = = r!  r  r !(n  r )! 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 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 tan(A  B) = 1  tan A tan B sin 2A = 2 sinA cosA 2 cos 2A = cos A  sin2 A = 2 cos2 A  1 = 1  2 sin2 A 2 tan A tan 2A = 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 Formulae for ABC a b c = = sin A sin B sin C 2 2 2 a = b + c  2bc cos A 1  = ab sin C 2

3 Answer all questions

1

A certain micro-organism grows from an initial population of 500 to size P at the end of t months. It is given that P  400  100e kt where k is a constant and that the population doubled at the end of 5 months. Find

2

3

(a)

the value of k,

[2]

(b)

how long it will take for the population to reach 15 000, leaving your answer to the nearest month,

[3]

(c)

the rate of change of the population at the end of 1 year.

[2]

(a)

State the amplitude and period of 2 sin 3x .

[2]

(b)

Given that y  2sin 3 x for 0  x   , (i)

sketch the graph of y,

[3]

(ii)

state the range of y,

[1]

(iii)

state the coordinates of the maximum point(s) of y.

[2]

2 x. 

(c)

On the same diagram, draw the line y 

(d)

Hence state the number of solutions to 2sin 3x 

[1] 2 x. 

[1]

A particle is moving such that its velocity, v ms-1 is given by v  k cos 2t  1 . It starts moving from point O with an initial velocity of 10 ms-1. Find (a)

the value of k,

[2]

(b)

the time when the particle is first at rest and its acceleration at this instant,

[4]

(c)

the distance travelled by the particle in the first two seconds.

[4]

4 4

OABC is a trapezium with OA parallel to CB, where O is the origin. It is given that A(5,10) , B(7, 5) , C ( x, y ) and the area of OABC is 36 unit2. y Find (a)

the equation of CB,

(b)

the coordinates of C.

[2]

A (5,10)

[3]

AB and OC produced meet at D. Find (c)

the coordinates of D.

B (7,5) O

x

[6]

C (x , y )

5

(a)

Solve sin x  sin 3 x  sin 2 x  0 for 0  x   , leaving your answers in terms of π. [6]

(b)

(i)

Solve 2sin x  4 cos x  3 for 0  x  2 .

(ii)

Hence state the least possible integer value of k for which 2 sin x  4 cos x  k has solutions.

6

[5]

[2]

The diagram shows a tangent drawn to the curve y  x3  2 x 2 at A(1,1) . Find (a)

the equation of the tangent to the curve at A,

[3]

(b)

the coordinates of the point where the tangent at A cuts the curve again,

[3]

(c)

the area bounded by the tangent at A and the curve,

[3]

(d)

the x-coordinate of the point(s) on the curve where the normal is parallel to this tangent.

[3] A (– 1,1)

5 The term with the highest power in a polynomial f ( x) is x3 . The roots to f ( x)  0 are

7

1, 2 and 3 .

8

(a)

Express f ( x) in descending powers of x.

(b)

Find, correct to 2 decimal places, the coordinates of the turning points of y  f ( x) and determine the nature of each point.

[6]

(c)

Sketch the graph of y  f ( x) , showing clearly the x and y intercepts.

[2]

(d)

State the range of values of x for which f ( x) is decreasing.

[1]

(a)

(i)

Show that

d 3x  c x 2x  1  , where c is an integer. dx 2x  1

(ii)

Hence find



(i)

Express

(ii)

Hence find

(b)

9

[3]





6x  2 dx . 2x 1

x2 in partial fractions. ( x  2)( x  1) 2



x2 dx . ( x  2)( x  1) 2

[2] [2]

[4]

[4]

The circle C1 has equation x 2  y 2  8 x  14 y  52  0 . Its centre, X, lies on the line 2 y  3 x  k . (a)

Find the centre, X and show that the value of k is 2.

[3]

(b)

The line cuts the circle at A( x1 , y1 ) and B( x2 , y2 ) where x1  x2 . Find the coordinates of B.

[4]

(c)

Find the equation of the tangent to the circle at B.

[3]

(d)

Find the equation of C2, the reflection of C1 about the tangent to the circle at B.

[3]

End of paper

6 PEIRCE SECONDARY SCHOOL MATHEMATICS DEPARTMENT GCE O LVL PRELIMIINARY EXAMINATION 2009 ADDITIONAL MATHEMATICS 4038/02 MARKING SCHEME 1(a) 400  100e5k  1000 e5k  6 ln 6 k  0.3583 5 k = 0.358 (b) 400  100e0.3583t  15000 e0.3583t  146 t  14 dP (c)  35.83e 0.3583t dt dP  35.83e(120.3583)  2639.6 dt = 2640 2(a) amplitude = 2 2 period = 120 or 3 (bi) sine graph mod sine graph values (ii) 0  y  2        5  (iii)  , 2  ,  , 2  ,  , 2  6  2   6 

9 s  sin 2t  t 2 t = 2, s = – 1.405 t = 0.8412, s = 5.313 diatance travelled = 2 x 5.313 + 1.405 = 12.03 = 12.0

A1 M1 M1 A1 M1

A1 B1 B1 B1 B1 B1 B1 one correct B1 all correct B2

straight line B1

A1

4(a) gradOA= 2 y = 2x + c Sub (7,5), c = – 9 y = 2x – 9 10 x 7 5 0 (b)  36 2 0 y 5 10 0 5x – 7y = 27 Sub y = 2x – 9 in 5x – 7y = 27 5x – 7(2x – 9) = 27 x=4 y=–1 C (4, – 1) (c) eqn AB y  10 5  x 5 2 2 y  5 x  45 eqn OC y 1  x 4 4 y  x 2(– 5x + 45) = – x x = 10 1 y = 2 2 1 D (10, 2 ) 2 5(a) sin 3x  sin x  sin 2 x  0 2 sin 2 x cos x  sin 2 x  0 sin 2 x(2 cos x  1)  0

sin 2 x  0 (d) 6 solutions

M1

3(a) 10 = k + 1 k=9 (b) 9 cos 2t  1  0 1 cos 2t   9 basic angle = 1.459 2t    1.459 t = 0.8412 = 0.841 a  18sin 2t a = –18sin 1.682 = – 17.88 = – 17.9 (c) s   9cos 2t  1 dt

M1 A1 M1

9 sin 2t  t  c 2 t = 0, s = 0, c = 0

M1 M1

M1

(c)



B1

A1 B1 A1

 x  0, ,  2 (bi) 20 sin( x  1.107)  3

M1

M1

A1 M2

M1 M1 M1 A1 M1 M1

1 2  x  3 2 x 3

3

20 x  1.107  0.7353, 2.406, 7.018 x = 1.299, 5.911 = 1.30, 5.91 (ii) For no soltutions k  1 20 k  4.47 least k = – 4

6(a) y’ = 3x2 + 4x At A, y’ = – 1

B1

cos x  

2 x  0,  , 2

sin( x  1.107) 

B1

M1 M2 A1 B2

basic angle B1

A1,A1 M1 A1 M1

7 Sub (– 1, 1) in y=–x+c c=0 y=–x (b) x3 + 2x2 = – x x (x2 + 2x + 1) = 0 x = 0, x = – 1 When x = 0, y = 0 Ans: (0, 0) (c)

Area =



0

M1

A1 M1 A1 A1 3

1

C=–1 Sub x = 0, A + 2B – 2C = 0 B=–3 4 3 1 Ans:   x  2 x  1 ( x  1) 2

2

 x  ( x  2 x ) dx

M1

(ii)

4

3

 x  2  x  1  ( x  2)

M1

1 1 2 = 0  (   ) 2 4 3 1 = 12 2 (d) 3x + 4x = 1 3x2 + 4x – 1 = 0

A1 M1

4  16  12 6 x  0.215, 1.55 x

A1,A1

7(a) f ( x)  ( x  1)( x  2)( x  3)

M2

= x3  2 x2  5 x  6 (b) f '  3 x 2  4 x  5  0 x = 2.119, –0.786= 2.12, –0.79 y = – 4.06, 8.21 f ''  6 x  4 (2.12, – 4.06), f’’>0, min pt (– 0.79, 8.21), f’’ < 0, max pt (c)

B1 M1 A1 A1 M1 A1 A1

y

6

1

–2

3

Shape x and y intercept values (d) 0.79  x  2.12 8(ai) y '  2 x  1  y' 

(ii)

(bi)

x 2x  1

3x  1

2x  1 2(3x  1)  2 x  1 dx  2 x 2 x  1  c x2  ( x  2)( x  1) 2 A( x  1)2  B( x  2)( x  1)  C ( x  2) ( x  2)( x  1) 2 Sub x = 2, 4 = A Sub x = 1, 1 = – C

x

B1 B1 B1 M1 B1 B1+B1

M1

B A1 B1

dx

= 4 ln( x  2)  3ln( x  1)  ( x  2) 1  c = 4 ln( x  2)  3ln( x  1) 

0

 1 x 4 2 x3  =  x2    4 3  1  2

2

A&C A1

B1

B1

1 c x2 B1 B1

9(a) ( x  4) 2  ( y  7) 2  117 Centre X (4,7) Sub in eqn: 2(7) – 3(4) = 2 k=2 2  3x (b) y  2 2 2  3x    2  3x  x2     8 x  14    52  0  2   2  x 2  8 x  20  0 (x – 10)(x + 2) = 0 x = 10 B (10, 16) 3 (c) grad XB = 2 2 grad tangent =  3 2 y   xc 3 2 Sub (10,16), c  22 3 2 2 y   x  22 3 3 (d) B is the mid pt of 2 centres x4 y7  10  16 2 2 x = 16 y = 25 C2 : ( x  16) 2  ( y  25)2  117

M1 B1 B1

M1 M1 A1 A1

M1

M1 A1

M1 A1 A1