St Gabriels Prelim 2009 Am P1 W Solutions

Name: …………………………………………(

)

Class: Sec ………………..

St. Gabriel’s Secondary School 2009 ‘O’ Preliminary Examination Subject Paper Level/Stream Duration Date Setters

: : : : : :

Additional Mathematics 4038/01 4 Express / 5 Normal 2 hours 4 September 2009 Teo Chen Nee

Additional materials: Writing paper Graph paper String



Section A: You must answer all parts of Question 1. READ THESE INSTRUCTIONS FIRST

 Write Section B: Answer oneregister question. your name, class and number on all the work you hand in. 

Write in dark blue or black pen. You are advised to spend no longer than 50 minutes on Section A. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid.

INFORMATION FOR CANDIDATES

Answer all the questions. Write your answers on the separate Answer Paper provided. The number of marks of each part-question is shown in brackets. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in case of angles 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 80. Target Set:

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

a  b

n

n n n  a n    a n  1 b    a n  2 b 2  ....    a n  r b r  ....  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  B) 

tan A  tan B 1  tan A tan B

sin 2A = 2 sinA cos A 2

cos 2A = cos A  sin2 A = 2 cos2 A  1 = 1  2 sin2 A

tan 2 A 

2 tan A 1  tan 2 A

sin A  sin B  2sin 12 ( A  B ) cos 21 ( A  B) sin A  sin B  2cos 12 ( A  B)sin 12 ( A  B) cos A  cos B  2cos 12 ( A  B) cos 12 ( A  B) cos A  cos B   2sin 12 ( A  B ) sin 12 ( A  B )

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

3 1

Calculate the values of x for which

5  2x 

Hence state the range of values of x for which

2

2 x  1. 3 5  2x 

2 x  1. 3

Find the range of values of k for which the curve y   2k  1 x 2  3kx  2k  4 lies entirely below the x-axis and has a maximum point.

3

4

Find all the angles between 0 and  which satisfy the equation 1 sin 3x + sin x = cosec x. 4

Prove that tan   45  

6

7

 cos x  sin x  dy . y  ln   , find dx  sin x  cos x 

Given that

(ii)

Hence evaluate

(a)

Given that

(b)

Solve the equation 2 lg e x  1  lg  3  e x  .



sec 2 x dx .

 log 2   log p   3 , 5

[4]

[4]

(i)

π 6 0

[4]

tan   1 . 1  tan 

Hence show that cot15  2  3 .

5

[4]

2

[3]

[3]

find the value of p.

[3] [4]

The degree of a polynomial f(x) is 3. The coefficient of the term containing the highest power of x is 2. Given that the equation f(x) = 0 has a repeated root 2 and another root q, (i)

show that f ( x)  2 x 3  2(4  q ) x 2  8(1  q ) x  8q ,

[3]

(ii)

find the value of q if f(x) has a remainder of 5 when divided by x – 1,

[2]

(iii) find the remainder when f(x) is divided by x 2  2 x  1 .

[3]

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4 8

l 12x 5x

The diagram shows a prism of length l cm. The vertical cross-section is in the shape of a right-angled triangle of sides 5x cm and 12x cm. Given that the prism has a volume of 450 cm3, (i) (ii)

express l in terms of x,

[2] 2

show that the total surface area, A cm , of the prism is given by 450 A  60 x 2  . x

[2]

Given that x can vary,

9

(iii) find the value of x for which A has a stationary value.

[3]

(iv) Determine whether this stationary value is a maximum or a minimum.

[1]

A curve has the equation y 

(a)

3x  4 1 , x . 2x  1 2 dy . dx

(i)

Obtain an expression for

(ii)

Find the acute angle made by the tangent to the curve at x = 2 with the x-axis.

[2]

[3] (b)

Given that x and y vary with time t, dx dy find the values of x for which  11 . dt dt

[3]

5 n

10

(a)

(b)

4  In the expansion of  x 3   in descending powers of x, x  the thirteenth term is independent of x. Find the value of n. (i)

(ii)

11

x  Express the first three terms of the expansion of 1   3  in ascending powers of x, in terms of n.

n

[1]

x  Given that the first two non-zero terms of the expansion of 1  mx  1   3  28 2 are 1 and  x , where n is a positive integer, 9 find the value of m and of n.

n

[4]

Given that A(10, 5) and B(2, 7) are two points lying on a circle, centre O(c, 2), find (i)

the equation of the perpendicular bisector of the chord AB,

[4]

(ii)

the value of c,

[2]

(iii) the equation of the circle.

12

[3]

[3]

The table shows the population P, in millions, of a country in the month of January every five years since 1989. t

5

10

15

20

P

32.4

73.2

126.2

188.9

It is known that P and t are related by the equation P  10  at n , where a and n are constants. (i)

Using graph paper, draw the graph of lg(P – 10) against lg t.

[3]

Use the graph to estimate (ii)

the value of a and of n,

[4]

(iii) the year in which the population reached 100 millions

[3]

END OF PAPER

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6

St Gabriel’s Secondary School 2009 Preliminary Examination Sec 4E5N Additional Mathematics Paper 1 Answer Key

1

1 1 or 4 2 2 1 1 1  x  4 2 2

x=1

2

k < 4

3

π 5π 7π 11π , , , 12 12 12 12

9

10

5

6

7

(i)

(a)

(i)

11 (2 x  1) 2

(ii)

23.7

(b)

x = 5 or 6

(a)

n = 16

(b)

(i)

2  n  x   n  x  1             1  3   2  3 

(ii)

m=

2 sec 2x

(ii)

0.658

(a)

p = 125

(b)

x = 0.693

(ii)

q= 

11

3 2

(iii) 11 – 4x

12

(i)

y + 4x + 18 = 0

(ii)

c = 5

(iii)

x 2  y 2  10 x  4 y  5  0

(ii)

a  2.00, n  1.5

(iii) Year 2001 8

(i)

l=

15 x2

(iii) x = 1.55 (iv) Minimum

7 , n=7 3

2009 A Maths 4E5N Prelim Paper 1 Marking Scheme