stpm trial 2009 matht&s1 q&a (n9)

8/14/2019 STPM Trial 2009 MathT&S1 Q&A (N9)

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;954/1,950/1 .\Mathematics TIS ,Paper 1 20093 hours

JABATAN PELAJARAN NEGERINEGERI SEMBILAN DARUL KHUSUSPEPERIKSAANPERCUBAANSIJIL TINGGI PERSEKOLAHAN MALAYSIA

2009

MATHEMATICSTMATHEMATICSS

Paper 1Three hours

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SOInstructions 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 ofangles in degrees, unless a different level o faccuracy is specified in the question.

This question paper has 3 printed pages

954/1,950/ 1 2009 Copyright PPD Seremban

[Turn OverCONFIDENTIAL *

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1. Find the solution set of the inequality I3x - 2 I< x + 5 .2. Find the value of a if one of the roots of the equation 6x3 +ax2 - 9x - 2 =0 is - .2

With this value of a, find the remaining roots of the equation.3. Show that the matrix A G) s a root of the equation A' - 7A + 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 3x 2 - 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 , findthe root correct to three decimal places.

7. The function l i s periodic with period 2 andl is defined byI (x) = X 2 for 0 x < II (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 whetherI is continuous at x = 18. A curve is defined by the parametric equations

2x = (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 thex -axis. [7]Hence, find the coordinates of the point of intersection of the normals. [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 thepoint (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 ofthecirc1e.

(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 ofOT.

[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 + 6x3State the set ofvalues 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 determinetheir nature. [5]Sketch the curve. . [2]Find the area of the region bounded by the curve and the x-axis. Calculate the volumeof 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 he first 2n terms. [8]2

(b) In a geometric progression, the first term is 2 and the sum of the first three terms isequal to thirteen times of the third term. Find two possible values of the commonratio r.The sum of the first n terms and the sum to infinity of the geometric series with r > 0are Sn and Soorespectively.Determine the value of Soo and the smallest value ofn suchthat Soo - Sn < 0.002. [8]

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

2.

3.

x > -%and

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

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


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4. 3x2 - .6x-5 ABC. . ' = '. + -- ' +--. , , - -(x+l)(x-1)2 ( x +l) ,' ( x ~ l ) (x - l ) 2 ,, 2 . ... . 23x - 6x - 5.= A(x -1) + B(x + l)(x -1) + C(x + 1)

Substitute x =1 , C = -4x=-l , A=1

Comparing coefficients, 23x2 -6 x -5 1 2 4- - ~ - = +-- ---:-(x+l)(x-l) 2 (x+l) (x-I) (x-I) 25. 4iy:' 1 '- ~ - t / ~ l + ~ ' ;ti t} , ~ 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 . 5 3 8 2 ~ 0.0025%.7893 =0.5373

X3 = 05373 - 0.00007/ =0.5373. / 2,7860x = 0.537 (3 d.p.)

2

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[5]

[5]

[6]

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7, (i)

Oi)

8.

lirn lex ==lim(3 - 2x) == 1.17 ....1+X 0 I+Iim !tx}==limx 2 == '1x ...1- . 17 . . . .1-/(l};::;; t

. , - .l im! ex),d leI)x0 1 .. .

is continuous at L

dx =3t2 -6 ,dtdy

t 2 -2ii .= ==dx 3t 2 - 6 3t 2Gradient of normal :

a d i e p t ,of normal= --];'EggE(ti,qt1of n ~ c'-'3iK', . : ; ~ > 1 ~ : , ~ , f)- -

' i ~ r a . q i e n t ofindrri181: -Equation of normar :

Solving:

Point of intersection :

. CurvesLines

All correctDlDlDlBlBIMlMlAI

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[5]

AI [7]

Ml

Al [2J

[3]

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. X2 + y 2 -lDx + lOy +30 = 0Radius =..JiOCentre ~ ( 5 , - 5 )

(bf +..... '.' dy ... . . . dy .2x+2y - - lO+1O- . =0dx dxat2 ~ 6 dy -10+ 10 dy =0dx dx

. dy =2dx

.Equati9n.of tailg'enf :y+3=2, x-J

(c) C(5,-5)

y

-F 36

4

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11.

10 2 ~ ( ' 2 ) ( ' 1 ' j i 2 ~ ~ 3)c-d + 2( : ' ) ( ~ 4 \ X ) J + Ml, = 1 + 2x +3x2' + 4x3 +.. , ...... ... ... : .. ;". Al

( , 2 ) I/ . _ , ' ,/ > 2 , " ,1+ ax - L+Y4 ax + ... . " ... . ' 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 ~ l i d , l ~ x l , ~ 2 r < 1{x :


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12. (a)

(1)3(2)(2) -(3)(4) .;..

Area = J: (x 3 - 3x 2 )dx[:'-3 ~ ' I ,,[


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(b) 2 + 2[+ 21'2 = 13(2r2)(3r 't:) ( 4 r+ l )= 0r '== Y3 ) r = ..:y".S. 2.0=1-'-l{

=3s;; Sri < 0.002

211-c!3'rJ3- < 0.0021-){(Y3t < Y3 (0 .002)nJogO.3J3 < log 0.000667n > 6.65n=7

7

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stpm trial 2009 matht&s1 q&a (n9)

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