4e5n Amath P2 Prelim 2009 With Ans

Name: _______________________________ (

)

Class: _______

MONTFORT SECONDARY SCHOOL PRELIMINARY EXAMINATION 2009

Secondary 4 Express/ 5 Normal Academic

ADDITIONAL MATHEMATICS PAPER 2 Tuesday

15 September 2009

4038/02 0830 – 1100

2 hours 30 minutes

INSTRUCTIONS TO CANDIDATES Write your name, register number and class on the question paper and writing paper provided. Attempt all questions. Write your answers and working on the separate writing paper provided. 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 INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question of part question. The total number of 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 6 printed pages including the cover page. Setter: R Tang

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Mathematical Formulae 1.

ALGEBRA

Quadratic Equation For the equation

ax 2 + bx + c = 0 ,

x=

− b ± b 2 − 4ac 2a

Binomial Theorem

(a + b )n

n n n = a n +  a n−1b +  a n − 2 b 2 + ... +  a n − r b r + ... + b n , 1  2 r

where n is a positive integer and

n n!   = .  r  (n − r )!r! 2.

TRIGONOMETRY

Identities

sin 2 A. + cos 2 A = 1 sec 2 A = 1 + tan 2 A

Formulae for

∆ABC

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 m sin A sin B tan A ± tan B tan ( A ± B ) = 1 m tan A tan B sin 2 A = 2 sin A cos A 2 cos 2 A = cos A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A 2 tan A tan 2 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

a b c = = . sin A sin B sin C a 2 = b 2 + c 2 − 2bc cos A. ∆=

1 bc sin A. 2

2

1

2

The value, V dollars of an investment-linked fund after t years, is given by V = V0 ln(t + 5) , where V0 dollars is the original value of the money invested in the fund. Calculate (i) the amount of money an investor would get after 10 years of investment, if he invests $20000, (ii) the number of years needed for an investment value to be tripled.

[1] [3]

The diagram shows part of the graph y = 5 − 2 x − 1 . The graph cuts the x-axis at A and B and the vertex is at C. (i) Find the coordinates of the points A, B and C. [3] (ii) Hence, write down the range of values of x for which y is negative. [1] y

C

A

3

Solve the equation 25 x − 60(5 x−1 ) + 27 = 0 .

4

It is given that sin x =

B

x

[5]

5 4 , cos y = − , where x , y both lie between 0° and 360° . Given also 13 5 that x lies in the same quadrant as y, evaluate without using calculators the exact value of (i) sin(2 y − x) 1 (ii) cos x [5] 2

3

5

6

7

cos x π . Find the x-coordinate, where 0 < x < , of the point 3 − sin x 2 at which the tangent to the curve is parallel to the line y = π . [5] The equation of a curve is y =

Express

(ii)

Hence find the exact value of

(i) (ii)

9

10

∫

−2

−3

[4]

x 2 + 3x dx . (1 − x)(1 + x) 2

[3]

The roots of the quadratic equation 2 x 2 + 4 x − 5 = 0 are α and β . (i) State the value of α + β and αβ . (ii)

8

x 2 + 3x in partial fractions. (1 − x)(1 + x)2

(i)

[2]

Find the quadratic equation in x whose roots are α β and αβ . 3

3

1 − cos 2 x + cos 4 x − cos 6 x = tan 3 x sin 2 x − sin 4 x + sin 6 x 1 − cos 2 x + cos 4 x − cos 6 x Hence, solve the equation = −4 , for 0 ≤ x ≤ 3 . sin 2 x − sin 4 x + sin 6 x Prove the identity

The function f is defined, for 0 ≤ x ≤ π , by f ( x) = 3sin 2 x − 1.5 . (i) State the amplitude of f. (ii) State the period of f. (iii) State the coordinates of the minimum point of f(x). (iv) Find the coordinates of the points where f(x) = 0. (v) Sketch the graph of f(x). (vi) Sketch the graph of f ( x) .

[5]

[4] [3]

[1] [1] [1] [3] [2] [2]

A particle moves in a straight line so that, t seconds after leaving a fixed point O, its velocity, v ms −1 , is given by v = 5(1 + e−0.5t ) . (i) Find the acceleration of the particle when v = 7. [3] (ii) Calculate, to the nearest metre, the displacement of the particle from O when t = 2. [3] (iii) State the value which v approaches as t becomes very large. [1] (iv) Sketch the velocity-time graph for the motion of the particle. [2]

4

11

The diagram shows part of the curve y = 1 + 5 x , intersecting the y-axis at A. The tangent to the curve at the point P(3, 4) intersects the y-axis at B. Find the area of the shaded region ABP. [8] y

y = 1 + 5x P(3,4) B

A x

12

The diagram shows the sideview of a crane which is used to transport bricks from one point to another. The arms AB and BC of the crane are 5 metres and 15 metres respectively. The angle between the two arms is 90 ° and the angle made by the arm BC with the ground is θ . Show that S, the distance between the bricks and the crane, is given by 5sin θ + 15cos θ metres. Express S in the form of R sin(θ + α ) and hence find the value of of θ (i) if S is a maximum, (ii) if S is 12 metres. [10] B 5m

15 m

A

S

θ

C

ground

Bricks Side view of crane

5

13

The diagram, which is not drawn to scale, shows a quadrilateral ABCD in which A is (0, 8). The equation of line AB is 3 y = 2 x + 24 and B lies on the line y = 2x. D lies on the perpendicular bisector of AB. The line AC is parallel to the x-axis and the line CD is parallel to the y-axis. Given that the area of triangle ABC is 14 square units, find (i) the coordinates of B, and of C, [4] (ii) the equation of perpendicular bisector of AB and hence the coordinates of D, [4] (iii) the area of the quadrilateral ABCD. [2] y=2x y B

3y=2x + 24

C

A (0, 8)

D

x

14

In the diagram, AB is the diameter of the circle and BP is tangent to the circle at B. Prove that (i) ∠PQR + ∠PSR = 180° , (ii) AS × AP = AR × AQ , (ii)

AS × AP = AB 2 .

P

[4] [2] [3]

S R

A

Q

B

End-of-Paper 6

Answer Key 1i)

$54,161.00

ii)

2i)

A(-2, 0)

3)

1.37 , 0.683

5)

0.340

7i)

α +β =−

8ii)

0.606, 1.65, 2.70

9i)

3

B(3, 0) 4i)

C(0.5, 5)

ii)

253 325

ii)

1 26

ii)

4 1 ln − 3 2

1 1 − 1 − x (1 + x)2

6i) 5 2

15.1 years x < -2 , x > 3

αβ = −

5 2

ii) π

iii)

(

10i)

a = −1 ms −2

ii)

16 m

11)

63 sq. units 80

12)

s = 5 10 sin(θ + 71.6°)

i)

18.4°

ii)

59.1°

13i)

B (6, 12)

ii)

D (7, 4)

iii)

28 sq. units

14i)

Let ∠QBR = x , then ∠BAR = x (alt. seg. theorem). Since ∠QBA = 90° (tan ⊥ radius), then ∠PQR = 90° + x (sum of int. ∠RBA = 90° − x = ∠PSR (ext. of a cyclic quad.) ∴∠PQR + ∠PSR = 90° + x + 90° − x = 180°

C (7, 8)

ii)

3π , −4.5) 4

16 x 2 + 360 x + 625 = 0

iv)

π 5π ( , 0) , ( , 0) 12 12

iii)

5 ms −1

ii)

∆ASR is similar to ∆AQP since ∠A is common and ∠ASR = ∠AQP . AS AR Then = , therefore AS × AP = AR × AQ . AQ AP

iii)

AB 2 = AQ 2 − BQ 2

= = = =

AQ 2 − QR × QA AQ ( AQ − QR ) AQ × AR AS × AP

s of a ∆ = ext. ∆ )

(Pythagoras Theorem) (Tangent – Secant Theorem)

(from part ii)

7