;954/1,950/1 .\ Mathematics TIS , Paper 1 2009 3 hours

JABATAN PELAJARAN NEGERI NEGERI SEMBILAN DARUL KHUSUS

PEPERIKSAANPERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA

2009

MATHEMATICST MATHEMATICSS

Paper 1

Three hours

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO

Instructions to candidates:

Answer all questions. Answers may be written in either English or Bahasa Malaysia.

All necessary working should be shown clearly.

Non-exact numerical 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 the question.

This question paper has 3 printed pages

954/1,950/ 1 © 2009 Copyright PPD Seremban

[Turn Over CONFIDENTIAL *

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1. Find the solution set of the inequality I 3x - 2 I < x + 5 .

2. Find the value of a if one of the roots of the equation 6x3 + ax2 - 9 x - 2 = 0 is - ~ .

2

With this value of a, find the remaining roots of the equation.

3. Show that the matrix A ~ G ~ ) is a root of the equation A' - 7 A + 21 ~ 0,

where I and 0 are the 2 x 2 identity matrix and null matrix respectively.

Hence, find the matrix A-I.

4. Express 3x2

- 6x - 5 in partial fractions. (x + 1)( x-I) 2

5. Solve the equation ~ + .JIi+-; = .J- 4x - 3 .

6. The equation eX = 2 - x2 has one root between 0 and 1.

[4]

[4]

[5]

[5]

[5]

By using the Newton Raphson method, with the first approximation of Xo = 0.5 , find

the root correct to three decimal places.

7. The function lis periodic with period 2 andl is defined by I (x) = X 2 for 0 ~ x < I

I(x) = 3 - 2x for 1 ~ x < 2

(i) Sketch the graph I for - 2 ~ x < 4

(ii) Find lim I(x) and lim I(x) X-41+ X-41-

Determine whether I is continuous at x = 1

8. A curve is defined by the parametric equations 2

x = (3 - 6t + 4, y = t - 3 + -t

[6]

[3]

[5]

Find the equations of the normals to the curve at the points where the curve meets the x -axis. [7]

Hence, find the coordinates of the point of intersection of the normals. [2]

2

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9. The variable point P(x, y) moves in such a way that AP 2 = 4BP 2 where A is the

point (1,3) and B is the point (4, -3).

(a) Find the equation ofthe locus of P which is a circle. State the radius and the centre of

thecirc1e.

(b) Find the equation of the tangent to the circle at the point (1,-3).

( c) The line OT is a tangent to the circle, where 0 is the origin.

Calculate the length of OT.

[5]

[4]

[2]

10. Expand (1-xr2 (1+ ax2 )'I. in ascending powers ofx up to and including the term in x3. [5]

(a) Find the value of a if the first four terms in the above expansion are

1 + 2x + 4x2 + 6x3•

State the set of values of x for which the expansion is valid. [4]

(b) By taking x = VB , find the value of Vfl correct to three significant figures. [3]

1 L Find the coordinates of the stationary points on the curve y = x2(x-3) and determine their nature. [5]

Sketch the curve. . [2]

Find the area of the region bounded by the curve and the x-axis. Calculate the volume of the solid formed when the region is rotated through 211: radian about the x-axis. [8]

12. (a) The third term of an arithmetic progression is 10, the nth term is 8 and the

(2n)th term is ! . Find the sum of the first 2n terms. [8] 2

(b) In a geometric progression, the first term is 2 and the sum of the first three terms is

equal to thirteen times of the third term. Find two possible values of the common

ratio r.

The sum of the first n terms and the sum to infinity of the geometric series with r > 0

are Sn and Soorespectively.Determine the value of Soo and the smallest value ofn such

that Soo - Sn < 0.002. [8]

3

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MARKING SCHEME MATHS TliS 1, 2009 L 3x-2 > -(x+5)

2.

3.

x > -% and

3x-2 < x+5 x < 3Yz

- % < x < 3Yz {x:-% <x<3Yz}

:: a=--7

6X 3 -7X2 -9x-2=O

(2x + 1)(3x2 '-5x::':".2)=0

(2x+ 1)(3x + l)(x - 2)= 0

1 ... the other roots are -- and 2.

3

2 . . (19 14) (3 2). (1 °1.J = (·.0°.· · OO. ~ .. A -7 A +21 = 35 26- 7 5 4 + 20 )

. . ·isa Toot of the equation. (32J' 5 4

A2 -7A +21=0 A2 A-I _ 7 A A-I + 21 A"l = 0 A-I

A - 7 I + 2 A"I = 0

( 2 -IJ ·-Ii %

Ml

Ml

1'v11 Al

Ml

Al

MI

Al

MI

Al

Ml

Ml

Al

[4]

[4]

[5J

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4. 3x 2 - .6x-5 ABC . . ' = '. +-. -' -+--.,,--

(x+l)(x-1)2 ( x +l) ,' (x~l) (x-l) 2 , , 2 • .... . 2

3x - 6x - 5.= A(x -1) + B(x + l)(x -1) + C(x + 1) Substitute x =1 , C = -4

x=-l, A=1 Comparing coefficients, B '~ 2

3x 2 -6x-5 1 2 4 --~-- = + -- ~ ---:-(x+l)(x-l) 2 (x+l) (x-I) (x-I)2

5. 4iy:' 1 '- ·~-t, '~,/~l ~+~' .\~ ;tit}, ~j,~~~<t' " "o;, 3

1 .... ,. }~~ +" 2'"\;(f'-~~ ~~~\

x ;:;;

6.

Xo =0.5 _ '. e0

5+ 0.52 - 2 _ 'l? '

, Xl - 0.5 - 0 5 - 0.5.)8-. e ' + 2(0.5)

X2 = 0.5382~ 0.0025%.7893 =0.5373

X3 = 05373 - 0.00007/ =0.5373 . / 2,7860

x = 0.537 (3 d.p.)

2

M1

Al MI

Al

Al

Ml

M1

Al

Al

B1

MIAI

Al

MI

Al

[5]

[5]

[6]

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7, (i)

Oi)

8.

lirn lex )==lim(3 - 2x) == 1 .17 .... 1+X0 I+

Iim!tx}==·limx2 =='1 x .... 1- . ·.17 .... 1-

/(l};::;; .~' . t . , - .

lim! ex), d leI) x01 ..• .

is continuous at ~ L

dx = 3t2 -6, dt

dy t 2 -2 ii .

= == dx 3t 2

- 6 3t 2

Gradient of normal :

.t ; ~;' §Fadiept ,of normal= --]; 'EggE(ti,qt1of nofm~i ' :'

c'-' 3:iK', .. :;±~ :" ~p >1~/3·;,~:,~ , f)- -'

'i~ra.qient ofindrri181: ~= -Equation of nor mar :

Solving:

Point of intersection :

. Curves Lines

All correct

Dl Dl Dl

Bl

BI

Ml

Ml

AI

MI

Al

Ml

Al

Bl

Al

[5]

AI [7]

Ml

Al [2J

[3]

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. X2 + y 2 -lDx + lOy +30 = 0

Radius =..JiO Centre ~(5,-5)

(bf + ..... '.' dy ....... dy . 2x+2y--lO+1O-.. =0

dx dx at

2~6 dy -10+ 10dy =0

dx dx

. dy =2 dx

. Equati9n.of tailg'enf :

y+3=2 , x-J

(c) C(5,-5)

y

-F 36

4

Ml

Al MI

Al Al

MI

Al

MI

Al

MI

[5]

[4]

Al [2]

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11.

10 Et_~X2~! ."t ('2)('1'J m~,ji2~~ 3)c-d + ~ 2( ~ :')(~4\X)J + Ml

, =1+ 2x +3x2',+ 4x3 + .. , ...... ... ... : .. ;". Al

( , 2 ) I/. _ , ' ,/ > 2 , " , 1+ ax - L+Y4 ax + ...•. , .. " ... i . ' BI

(1-x)-2(1-t-ax2 ) Y< ,,7" 1,+2x-f-::3xL + 4x3 + %ax2 + y; ;x3 + '" : .... .. M1 , " = 1+2x+(3+ ~ a) x2 + (4+ y; a) x3 + ......... Al [5J

. (a) ' Equating 3+Y4 a = 4 , ' a =4

V~lid, l~xli <l " ,'14~2r< 1

{x ::~ <x< Yz}

C' b',)' :" ('·" 1~1 / )c2("1":"'" 4/ "' )'1.= ,1 ' 2(1/ ) 4' ' If' 2 6(11.)3 ' ' " /8" ' + / 82 : . ' ,j- /8 + l iS) + I S + ..... ... ..

cUrl ( 1 /(6)'I.~' 1.3242 . - . , . . - . .

Vfl = ,32x (1.3242) = 2.0277 49 " '

= 2.03(3 s:f.)

fx(x-2) = 0

:x=2 y= O y= -4

Thest?-tionarypointsar~(O, 0) and (2, -4) ":: ','. , " " """ 'iz,BY ~;()f': 6, 0; -' ..

dx 2

d2

y 0 -< dx 2

Max. pt, (O,O) .

y

d2 y , - > 0 'dx 2

Min. pt.C2,-4)

5

Ml Al

1'-11

Al [4]

M1

Ml

A1 [3]

Ml

Al

M l

AlAl [5]

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12. (a)

(1)3(2)

(2) -(3)

(4) . .;..

Area = J: (x 3 - 3x 2 )dx

~ [:' -3~'I

~ ,,[< _ 6d' + 9n: (

37

6 9(3)5 J = 1t ·~-3 +-5- -0

729TC 35

2d= a+ en -l)d == :8

~=~ -1'1 7

n=7 1

d =--2

a 11

= 108:':5

(5)

6

any 2

DIDI [2]

MI

Ml

Ml

Al

Ml

M1

Ml

Al

Bl

M1

Ml

Al

Al

Al

M1

Al [8]

[8]

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(b) 2 + 2[+ 21'2 = 13(2r2) (3r 't: )(4r+ l)= 0

r '== Y3 ) r = ..: y".

S. 2 .0 =1-'-l{

=3 s;; ~ Sri < 0.002

211-c!3'rJ 3· - < 0.002

1-){ (Y3t < Y3(0.002)

nJogO.3J3 · < log 0.000667 n > 6.65 n=7

7

Ml Ml Al

Ml

Al

Ml

Ml Al [8]

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