8/7/2019 Perc STPM 2010 MTS1

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SMJK Yu HuaSTPM Trial Examination 2010

Mathematics T/SPaper 1

Form : 6AS1, 6AS2, 6AS3, 6AK1, 6AK2 Prepared by : Pn RosmayaDate : 18th

October 2010 Checked by : Pn. Chew Aun SimTime : 10.10 am 1.10 am Verified by : En. Pradeep

Instructions to candidates :

Answer all questions. All necessary working should be shown clearly. Non-exactnumerical answers may be given correct to three significant figures, or one decimal place

in the case of angles in degrees, unless a different level of accuracy is specified in thequestion.

1. Use the laws of algebra of sets to prove that[A (A B )] - (A B ) = f [4 marks]

2. Given that z1 = -3 + i and z2 = 2 + 4i , find the modulus and argument of z1z2.[4 marks]

3. The sum of the first n terms of a series is given as

Show that the series is a geometric progression.[5 marks]

4. Given that the line y = mx cuts the circle x2 + y2 + 4x + 2y 20 = 0 at the point Aand B. If P is the midpoint of the chord AB, find the coordinates of P, in terms ofm. Hence, find the equation of the locus P, as m varies.

[6 marks]

5. Given that y = ln (sin3 2x) , find [3 marks]Hence , show that 3 + + 36 = 0 [3 marks]

6. The function g is defined by(x 3)

2

, x 3b , x > 3x

Given that g(x) is continuous at x = 3, find the value of b . [3 marks]

Sketch the graph of y = g(x). [4 marks]

7. Express in partial fractions. [5 marks]

g(x) =1 - _

dy

dxd

2y dy

2

dx2

dx

6x2

+ 3x - 1(x + 1)(2x - 1)

10 -2

n + 1

3n - 1

____

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Hence, find the value of dx , give your answer correct

to three decimal places. [4 marks]

8. Expand 1 + in ascending powers of x up to and including the term in x3.State the range of x for which this expansion is valid. [6 marks]

Without using calculators or mathematical tables, evaluate (1.004)3

correct tofive decimal places. [4 marks]

9. a) Given that P = 2 1 3 , find the matrix Q such that Q = P2 5P 3I,- 1 2 13 1 1

where I is a 3 3 identity matrix. [3 marks]

b) Find PQ and hence P-1

[4 marks]c) Use your answer in b) to solve the system of equations :

2x y + 3z = 23

-x + 2y + z = -463x + y + z = 69 [4 marks]

10.a) Find the area of the region bounded by the curve y = + 1and the graph y = x + 1 [5 marks]

b) Find the volume of the solid generated when the region bounded by the curve

y = + 1 and the graph y = x + 1 for x > 0 is rotated 360o about

the y-axis. [6 marks]

11.Using the substitution y = x + , express f(x) = 6x4 + 5x3 38x2 + 5x + 6 = 0as a polynomial in y. [3 marks]

Hence, solve the equation f(x) = 0. [10 marks]

12.Sketch on the same coordinate axes, the graphs of y = ex and y = 10x.Show that the equation e

x

10x = 0 has a root in the interval [0,1][4 marks]

Using the Newton-Raphson method and 0.2 as the first approximation, find the

root correct to three decimal places.[5 marks]

The other root of the equation lies in the interval [n, n + 1], where n is an integer.State the value of n.

[5 marks]

6x2

+ 3x - 1(x + 1)(2x - 1)

2

1

2x

3

3

2

x2

23

2

x2

2

3

2

1

x

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Mark Scheme for STPM Trial examination

1. [A (A B) ] ( A B )= [(A A) (A B )] ( A B ) Distributive Law= [f (A B ) ] (A B) De Morgan s Law

= (A B) (A B) Identity Law= f proven

2. z1z2 = (-3 + i)(2 + 4i)= -10 - 10i

z1z2 = (-10 )2 + (-10)2 = 200 = 10 2

3. Sn = 10 2n+13

n 1

Using Tn = Sn Sn 1

= 10 2n+1

- 10 2n3

n 13

n 2

= 2n

2n+1

3n 2 3n 1

= 2n

1 23

n 23

= 2n

13

n 23

= 2n

3n 1

= 3 2n

3

Tn = 3 2n

Tn 1 33 2

n 1

3= 2

3Therefore, it is a geometric progression with r = 2

3

-10

-10

arg (z1z2) = tan-1 -10

-10

= _ 3p rad4

________

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4.

P = x1 + x2 y1 + y22 2

= ( X , Y )

Substitute y = mx into circle equation :x

2+ (mx)

2+ 4x + 2mx 20 = 0

(1 + m2) x

2+ 2(2 + m)x 20 = 0

SOR = x1 + x2 = _ 2(2 + m)

(1 + m2)

SOR 2 = x1

+ x2

= _ (2 + m) = X

2 (1 + m2)Substitute point P (X,Y) into y = mx,

Y = mX = m _ (2 + m) = _ m(2 + m)(1 + m

2) (1 + m

2)

Therefore point P = _ (2 + m) , _ m(2 + m)(1 + m

2) (1 + m

2)

From point P, x = _ (2 + m)

(1 + m2)

Substitute m = y into above equation : x = _ 2 + y

x x1 + y

2

x2

x + y2

= -2 yx

2x

x2

+ y2

+2x + y = 0

5. y = ln (sin3 2x)Using chain rule :dy = 1 . d (sin

32x)

dx sin3

2x dx

= 6 sin2

2x. cos 2xsin

32x

= 6 cos 2x = 6 cot 2xsin 2x

y = mxP

A

B

(x1,y1)

(x2,y2)

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d2y = -12 cosec2 2x

dx2

= -12 (1 + cot2 2x)

d2y + 12 cot

22x + 12 = 0

dx2

3 d

2y + 36 cot

22x + 36 = 0

dx2

3 d2y + ( 6 cot 2x )

2+ 36 = 0

dx2

3 d2y + dy

2+ 36 = 0

dx2

dx

6. (x 3)2 , x 3g(x) = 1 a , x > 3

x

lim g(x) = lim (x - 3)2

= 0

x3- x3-

lim g(x) = lim 1 - a = 1 a

x3+ x3+ x 3

Given that g(x) is continuous at x = 3

Therefore : lim g(x) = lim

x3- x3+0 = 1 a

3a = 13

a = 3

9

1

0 3 x

y

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7. 6x2 + 3x 1 = 6x2 + 3x 1 = 3 + 2 .

(x + 1)(2x 1) 2x2

+ x 1 2x2

+ x 1= 3 + 2 .

(x + 1)(2x 1)

2 . = A . + B .(x + 1)(2x 1) (x + 1) (2x 1)2 = A(2x 1) + B(x + 1)

When x = -1 , A = -23

When x = , B = 43

6x2

+ 3x 1 = 3 - 2 . + 4 .(x + 1)(2x 1) 3(x + 1) 3(2x 1)

26x

2+ 3x 1 =

23 - 2 . + 4 . dx

1 (x + 1)(2x 1) 1 3(x + 1) 3(2x 1)= 3x 2 lnx + 1 + 2 ln2x 1 2

3 3 1= 3x + 2 ln 2x 1

2

3 x + 1 1= 6 + 2 ln 1 3 + 2 ln 1

3 3 2= 3 + 2 ln 2

3= 3.462 (3 decimal places)

8. 1 + 2x = 1 + C1 2x + C2 2x 2 + C3 2x 3 + .3 3 3 3

= 1 + x + 1 x2

1 x3

(up to x3)

6 54For expansion to be valid :

2x < 13

\ 3 < x < 32 2

(1.004)3

= (1 + 0.004)3

= (1 + 0.004)

= 1 + 0.006 + 1 (0.006)2

1 (0.006)3

2x = 0.004

6 54 3= 1 + 0.006 + 0.000006 0.000000004 x = 0.006

= 1.00601 (5 decimal places)

3

2

3

2

3

2

3

2

32

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9. a)Q = P2 5P 3I= 2 1 3 2 1 3 2 1 3 1 0 0

1 2 1 1 2 1 5 1 2 1 3 0 1 03 1 1 3 1 1 3 1 1 0 0 1

= 14 1 8 10 5 15 3 0 01 6 0 5 10 5 0 3 08 0 11 15 5 5 0 0 3

= 1 4 7

4 7 57 5 3

b) PQ = 2 1 3 1 4 7

1 2 1 4 7 53 1 1 7 5 3

= 23 0 0

0 23 00 0 23

PQ = 23I

P1

PQ = 23P1

IIQ = 23P

1

_ 1 Q = P1

23

P1

= _ 1 Q23

= _ 1 1 4 723 4 7 5

7 5 3

c) 2 1 3 x 231 2 1 y = 46

3 1 1 z 69

x 2 1 31

23y = 1 2 1 46

z 3 1 1 69

_ 1 1 4 7 23= 23 4 7 5 46

7 5 3 691 4 7 1

= - 4 7 5 27 5 3 3

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28 28

= - 3 = 3 x = 28, y = 3, z = 1212 12

10.a)

The area is symmetrical about the y-axis :

HenceA = 2 [ ( x + 1) ( + 1) ] dx

= 2 [ x ] dx

= 2 [ x2

]

= 4 unit2

b) Volume = p x2 dy Volume of cone

= p 2(y 1) dy 1 p (32) (11 1)3 2

= p [(y 1)2] 27 p2

= 27 p unit3= 6 p unit3

11.

6x

4+ 5x

338x

2+ 5x + 6 = 0

6x2

+ 5x 38 + 5 + 6 = 0x x

2

6 (x2

+ 1 ) + 5 ( x + 1 ) 38 = 0x

2x

6 ( y2

2 ) + 5y 38 = 06y

2+ 5y 50 = 0

(3y +10)(2y 5 ) = 0

y

x0

1

A

y = x2

+ 1

y =3/2x + 1

Coordinates of A?x

2+ 1 = 3x + 1

2 2

\ x = 3, y = 112

A 3 , 11

2

3

0

3 x2

2 2

30

3 x2

2 23 x

2

4 6

3

0

11

/2

1

11

/2

1 11/2

1

y = x + 1x

y2

= x2

+ 2 + 1

x2

y2

2 = x2

+ 1

x2

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y = _ 10 or y = 53 2

x + 1 = _ 10 x + 1 = 5x 3 x 2

3x2

+ 10x + 3 = 0 2x2

5x + 2 = 0

(3x + 1)(x + 3) = 0 (2x 1)(x 2) = 0x = _ 1 , x = -3 x = , x = 23

12.

ex

10x = 0let f(x) = e

x10x

when x =1, f(1) = e1

10(1) = 7.28 negative

when x = 0, f(0) = e0

10(0) = 1 positive

f(x) = ex

10xf (x) = e

x10

Using Newton-Raphson method to find the root in the interval [0,1] :

First approximation xo = 0.2,x1 = 0.2 e

0.210(0.2) = 0.111307

e0.2

10x2 = 0.111307 e

0.11130710(0.111307) = 0.1118

e0.111307

10x3 = 0.1118 e

0.111810(0.1118) = 0.112

e0.1118 10\ the root is 0.112 ( 3 decimal places)

To find n if the other root is in the interval [n, n+1] :

f(2) = e2

10(2) = 12.61f(3) = e

310(3) = 9.91 negative

f(4) = e4

10(4) = 14.60 positive

Therefore , n = 3

0

1

y

x

y = 10x

y = ex

Therefore , there is a root

in the interval [0,1]