Chij Prelim Am 1 2009

Class

Register Number

Name

CHIJ SECONDARY (TOA PAYOH) PRELIMINARY EXAMINATION 2009

SECONDARY FOUR (SPECIAL /EXPRESS) & FIVE(NORMAL)

ADDITIONAL MATHEMATICS PAPER 1

4038/01 22 September 2009 2 hours

Additional Materials: Writing Paper

READ THESE INSTRUCTIONS FIRST Write your name, register number and class 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 on the separate answer papers provided. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the 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.

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This document consists of 6 printed pages including the cover page. [Turn over

chijsectp.4S/E&5N.prelim.amath1.2009

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Mathematical Formulae 1.

ALGEBRA

Quadratic Equation For the equation ax2 + bx + c = 0, x=

− b ± b 2 − 4ac . 2a

Binomial expansion  n  n −1  n  n −2 2  n  n −r r n ( a + b) n = a n +  1  a b +  2  a b + ....... +  r  a b + ....... + b ,       n  n! n(n −1).......( n − r +1) where n is a positive integer and  r   = r!(n − r )! = 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 sin A cos A. cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A 2 tan A tan 2 A = . 1 − tan 2 A 1 1 ( 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

sin A + sin B = 2 sin

Formulae for ΔABC

a b c = = sin A sin B sin C

.

a 2 = b 2 + c 2 − 2bc cos A .

Δ= 1

(a)

1 bc sin A . 2

The diagram shows a triangle ABC in which BC = ( 2 3 +5 ) cm and ACˆ B =60°. Find the exact length of AB.

A [2]

chijsectp.4S/E&5N.prelim.amath1.2009

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60° B (2 (b)

2

Find the value of AC 2, expressing your answer in the form where a and b are integers.

a +b 3

[2]

Solve the simultaneous equations y − 3 x + 2 = 0 2 x 2 + 2 y 2 +11 x + y − 66 = 0 .

3

C 3 +5 ) cm

7 x 2 − 4x + 5 ( x − 2) x 2 +1

(i)

Express

(ii)

Hence, evaluate

(

)

in partial fractions.

[4]

[3]

7 x 2 − 4x +5 ∫3 ( x − 2)( x 2 +1)dx 4

[3] 4

Solve the following equations,

5

(a)

cos 3 x − cos x = sin x cos x , for 0 0 ≤ x ≤ 360 0 .

(b)

tan 2 x =

Given that y =

3 π tan 4 4

for 0 < x < 6 .

[3]

sin θ , find the value of θ for 0 ≤ θ ≤ 2π for which y is cos θ + 4

stationary. 6

[3]

[4]

(i)

Differentiate 2 xe 5−2 x with respect to x.

(ii)

Hence find the value of

3

∫

1

xe

5 −2 x

dx

[2] .

[3]

7

(a)

Given that (2 + x) 6 (1 + ax − 2 x 2 ) = 64 + 512 x + bx 2 + ...... , find the value of a and of b. [4]

chijsectp.4S/E&5N.prelim.amath1.2009

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8

(b)

8

(a)

k  In the binomial expansion of  4 x 2 +  , where k is a positive constant , the x  1 − coefficient of x 5 is . Find the value of k. [4] 4

Find the range of values of k for which the graph of y = x 2 − 2(2k −1) x + 4 lies above the x-axis for all values of x.

(b)

Given that y = 4 x 2 ln 2 x and x > 0, find the range of values of x for which y is increasing.

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(i)

[3]

[3] 5 1

Find the inverse of 

−2 . 2  

[1] (ii)

Hence solve the simultaneous equations 5 x =11 + 2 y 2 y + x −1 = 0

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The equation of a curve is y = (i)

3x − 7 . 4x + 5

Differentiate y with respect to x and write your answer in the form dy = dx

(ii)

[3]

A + Bx

( 4 x + 5) 3

, where A and B are integers.

Find the x-coordinate of the point on the curve at which the tangent to the curve is parallel to the x-axis

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[2]

[2]

A particle P travels in a straight line so that its distance, s m, from a fixed point

chijsectp.4S/E&5N.prelim.amath1.2009

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O is given by s = 2t +

18 , where t is the time in seconds measured from the t +1

start of the motion. Calculate

12

(i)

the initial acceleration of P,

[2]

(ii)

the velocity of P when it is next at starting point ,

[3]

(iii)

the value of t when the particle is instantaneously at rest,

[2]

(iv)

the total distance traveled in the first 5 seconds.

[3]

The table below shows experimental values of two variables x and y. x y

2 8.48

4 5.99

6 4.90

8 4.24

It is known that x and y are connected by the equation x n y = k , where k and n are constants. (a)

By taking a scale of 2 cm to represent 0.5 unit on each axis , draw the graph of ln y against ln x for the given data.

(b)

[3]

Use your graph to estimate (i)

the value of n and of k,

[4]

(ii)

the value of x when y = e 2.

[2]

13 y cm

chijsectp.4S/E&5N.prelim.amath1.2009

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10x cm 1 5x cm

5x cm

4x cm

The cross section of a thin metallic container with an open rectangular top is an isosceles trapezium whose parallel sides are 10x cm and 4x cm and the equal sides are 5x cm. The total amount of metal sheet used is 1050 cm2. (i)

Given that the length of the container is y cm, express y in terms of x.

(ii)

Show that the volume of the container, V cm3, is given by V = 2100x – 112 x3.

(iii)

[3] [2]

Find the value of x for which V has a stationary value. Find this value of V and determine whether this is a maximum or minimum. [5]

END OF PAPER

Numerical solutions

1

(a)

AB =

6 +5 3

cm

(b)

AC2 = 148 + 80

3

chijsectp.4S/E&5N.prelim.amath1.2009

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2

x = 2 , y = 4 or x = − 1.5 , y = − 6.5

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(i)

5 2x + 2 x − 2 x +1

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(a) (b)

x = 0° , 90° , 180°, 194.5°, 270°, 345.5°, 360° x = 0.322, 1.89 , 3.46 , 5.03 radian

5

θ = 1.82 , 4.46 radian

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(i) (ii)

(ii)

4.00

2e 5−2 x − 4 xe 5−2 x 3 3 5 e − or 14.4 4 4e

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(a)

a = 5 , b = 1072

8

(a)

9

(i)

10

(i)

11

(i)

36 m/s2

(iii)

t=2

1 3
6 x + 29

( 4 x + 5) 3

(ii)

1 2

12

(b)(i) k ≈ e 2..5 or 12.2 , n ≈

13

(i)

y=

(ii)

V = 2100x – 112x3 ( To be shown )

(iii)

x = 2.5 Stationary value of V = 3500 cm3 V is maximum when x = 2.5

1 2

(b)

k=

(b)

x>

(ii)

x=2,y= −

(ii)

x = −4

1

1 2 e

or 0.303 1 2

5 6

7 m/s 9

(iv)

11 m

(ii)

x ≈ 2.72

75 − 4 x 2 x

chijsectp.4S/E&5N.prelim.amath1.2009