Jurongville Prelim 2009 Am P1

  • June 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Jurongville Prelim 2009 Am P1 as PDF for free.

More details

  • Words: 944
  • Pages: 5
JVSS AM Paper 1 Prelim 2009 1.

Find the range of values of x for which x 2  12  4 x  0 .

2.

Given that (225 p w ) 2  (243 p 3 )

3.

A cuboid has a square base of sides (3  2 2) cm and a volume of

3



2 3



[4]

3x5 y , evaluate w, x and y. p

[4]

(18  11 2) cm3 . Find, without using a calculator and giving your answer in

the form a  b 2 , where a and b are integers.

4.

5.

6.

(i) (ii)

the area of the square base of the cuboid, the height of the cuboid.

[2] [3]

(i) (ii)

Solve the equation 2 2 x  2  2 x  3 . Hence or otherwise, solve the equation 2 2 x  2  2  x  3 .

[4] [1]

 3 2  , find M-1 and hence, solve the simultaneous Given that M =   4 1   equations 3 y x3 2 4 x  y  14

x4 in partial fractions. (3x  5)( x  3)

(i)

Express

(ii)

Hence or otherwise, find the gradient of the curve y  at the point where x = 2.

3

[6]

[3]

x4 (3x  5)( x  3) [3]

JVSS AM Paper 1 Prelim 2009 7.

The figure below shows a square of sides 1 cm each. A point P on AB and a point Q on BC are such that QC = x cm and PB = kx cm, where k is a positive constant. (i)

Express the area of triangle OPQ in terms of k and x and show that it is equal to

(ii)

1 (kx2  x + 1) cm2 2

[2]

Find the minimum area of triangle OPQ in terms of k. A

P

kx cm

[4]

B

Q 1 cm

x cm

O

C 1 cm

8.

(i)

Find the term which is independent of x in the expansion of 16

1   2 x  6  . 2x   (ii)

[2]

Write down the first three terms of the expansion in ascending powers of x of (a)

 3x  1   2  

(b)

2  x 5

5

Hence, obtain the coefficient of x2 in the expansion of 5

3 2   2  2x  x  . 2  

[4]

4

JVSS AM Paper 1 Prelim 2009

9.

x2  1 with respect to x. 2x  3

(i)

Differentiate

(ii)

Hence evaluate 

(i)

Differentiate 3 cos[( x 2  1)(3x  1) 2 ] with respect to x.

(ii)

Find the gradient of the curve y  ln(1  cos 3 x) where x 

(i)

Prove the identity 1 – cos2 + cos4 – cos6 = 4sin cos2 sin3. [4]

(ii)

Hence find all the angles  between 0 and  inclusive which satisfy the equation cos2 – cos4 + cos6 = 1. [4]

2

10.

11.

12.

x 2  3x

0 3  2 x  3

2

[2]

dx .

[5]

[4]

 . 2

[3]

The figure shows a trapezium ABCD in which AD is parallel to BC and AB is parallel to the x-axis. The point A, C and D are (4, 2), (7, 16) and (0, 10) respectively. The point X lies on BC such that AXB = 90. (i) (ii) (iii)

Find the equation of BC and of AX. Find the coordinates of B and X. Calculate the area of the trapezium ABCD.

y

[3] [3] [2]

C(7,16)

D(0,10)

X

B

A(4,2)

x

13.

(i)

The table below shows experimental values of two variables x and y. 5

JVSS AM Paper 1 Prelim 2009

x y

1 0

2 0.245

3 0.349

4 0.423

5 0.482

6 0.521

The two variables x and y are known to be related by the equation y  lg ax  b , where a and b are constants. Express the equation y  lg ax  b in a form suitable for drawing a straight line graph and explain how the graph can be drawn. Hence, by using suitable scales, draw the straight line graph. [4] (ii) (iii)

Use the graph to estimate the value of a and b. [2] If the x-coordinate of a point on the straight line is 4.5, evaluate the corresponding value of y. [2]

END OF PAPER

6

JVSS AM Paper 1 Prelim 2009 Answer Key 1. x  6 or 2  x  3 1 2 2. x  6 , y  3 , w  3 3 3(i) 17  12 2 (ii) 42  29 2 4. -0.42, 0.42 5. x  3.09 ,  1.64 7 17 6(i)  (ii) 11 4( x  3) 4(3x  5) 4k  1 7(ii) 8k 15 45 2 8(i) 113.75 (ii)(a) 1  x  x  ... (b) 32  80 x  80 x 2  ... , 200 2 2 2 2( x  3 x  1) 2 9(i) (ii) 2 3 (2 x  3) 10(i) 6(1  3x)(6 x 2  x  3) sin[( x 2  1)(3x  1) 2 ] (ii) 3   2 3 11(ii) 0, , , , , 4 3 3 4 1 12(i) BC: y  2 x  30 AX: y  x (ii) B(14,2), X(12,6) (iii) 110 2 13(ii) a= 2.0, b = -1.0 (iii) y = 0.452

7

Related Documents