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SULIT 1 3472/1

3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM 2009

Kertas soalan ini mengandungi 15 halaman bercetak

For examiner’s use only

Question Total Marks Marks Obtained

1 2

2 3

3 3

4 3

5 3

6 3

7 3

8 3

9 3

10 3

11 3

12 4

13 3

14 3

15 4

16 3

17 4

18 4

19 3

20 3

21 3

22 3

23 3

24 4

25 4

TOTAL 80

MATEMATIK TAMBAHAN Kertas 1 Dua jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1 This question paper consists of 25 questions. 2. Answer all questions. 3. Give only one answer for each question. 4. Write your answers clearly in the spaces provided in

the question paper. 5. Show your working. It may help you to get marks. 6. If you wish to change your answer, cross out the work

that you have done. Then write down the new answer.

7. The diagrams in the questions provided are not

drawn to scale unless stated. 8. The marks allocated for each question and sub-part

of a question are shown in brackets. 9. A list of formulae is provided on pages 2 to 3. 10. A booklet of four-figure mathematical tables is provided. . 11 You may use a non-programmable scientific calculator. 12 This question paper must be handed in at the end of

the examination .

Name : ………………..…………… Form : ………………………..……

3472/1 Matematik Tambahan Kertas 1 2009 2 Jam

http://mathsmozac.blogspot.com

http://tutormansor.wordpress.com/

SULIT 3472/2

3472/1 2009 Hak Cipta Zon A Kuching SULIT

2

The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.

ALGEBRA

1 2 4

2

b b acx

a

− ± −=

2 am × an = a m + n 3 am ÷ an = a m − n

4 (am)n = a mn 5 log a mn = log a m + log a n

6 log a n

m = log a m − log a n

7 log a mn = n log a m

8 log a b = a

b

c

c

log

log

9 Tn = a + (n − 1)d

10 Sn = ])1(2[2

dnan −+

11 Tn = ar n − 1

12 Sn = r

ra

r

ra nn

−−=

−−

1

)1(

1

)1( , (r ≠ 1)

13 r

aS

−=∞ 1

, r <1

CALCULUS

1 y = uv , dx

duv

dx

dvu

dx

dy +=

2 v

uy = ,

2

du dvv udy dx dx

dx v

−= ,

3 dx

du

du

dy

dx

dy ×=

4 Area under a curve

= ∫b

a

y dx or

= ∫b

a

x dy

5 Volume generated

= ∫b

a

y2π dx or

= ∫b

a

x2π dy

5 A point dividing a segment of a line

(x, y) = ,21

++

nm

mxnx

++

nm

myny 21

6 Area of triangle

= 1 2 2 3 3 1 2 1 3 2 1 3

1( ) ( )

2x y x y x y x y x y x y+ + − + +

1 Distance = 221

221 )()( yyxx −+−

2 Midpoint

(x , y) =

+2

21 xx ,

+2

21 yy

3 22 yxr +=

4 2 2

ˆxi yj

rx y

+=+

GEOMETRY



SULIT 3472/1


3

STATISTIC

1 Arc length, s = rθ

2 Area of sector , A = 21

2r θ

3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A

6 sin 2A = 2 sinA cosA 7 cos 2A = cos2A – sin2 A = 2 cos2A − 1 = 1 − 2 sin2A

8 tan 2A = A

A2tan1

tan2

−

TRIGONOMETRY

9 sin (A± B) = sinA cosB ± cosA sinB

10 cos (A± B) = cosA cosB m sinA sinB

11 tan (A± B) = BA

BA

tantan1

tantan

m

±

12 C

c

B

b

A

a

sinsinsin==

13 a2 = b2 + c2 − 2bc cosA

14 Area of triangle = Cabsin2

1

7 1

11

w

IwI

∑

∑=

8 )!(

!

rn

nPr

n

−=

9 !)!(

!

rrn

nCr

n

−=

10 P(A∪ B) = P(A) + P(B) − P(A∩ B)

11 P(X = r) = rnrr

n qpC − , p + q = 1 12 Mean µ = np

13 npq=σ

14 z = σ

µ−x

1 x = N

x∑

2 x = ∑∑

f

fx

3 σ = 2( )x x

N

−∑ = 2

2xx

N−∑

4 σ = 2( )f x x

f

−∑∑

= 2

2fxx

f−∑

∑

5 m = Cf

FNL

m

−+ 2

1

6 1

0

100Q

IQ

= ×



SULIT 3472/1


4

Answer all questions.

1 Diagram 1 shows a graph of the relation between two variables x and y.

State (a) the object of 8,

(b) the type of relation between x and y.

[ 2 marks ]

Answer : (a) ……………………..

(b) ……………………...

2 Given that the function ( ) 2 5f x x= + , 2( ) 4g x x= − , find

(a) ( )gf x

(b) ( 2)gf − [ 3 marks ]

Answer : (a) ……………………..

(b) ……………………...

3

2

2

1

For examiner’s

use only

x

• •

•

•

•

0 2 4 6 8 10

4 8 12 16

24

y

DIAGRAM 1

20



SULIT 3472/1


5

3 Given that the function 3 2

:5

xf x

+→ , find

(a) 1( )f x−

(b) the value of x such that 1( ) 3f x− = [ 3 marks ]

Answer : (a) ……………………..

(b) ……………………... 4 The quadratic equation 2x2− 5x+ p − 3 = 0 has two different roots, find the range of

values of p. [ 3 marks ]

Answer : .........…………………

For examiner’s

use only

3

3

3

4



SULIT 3472/1


6

5 Given α and β are the roots of the quadratic equation 3x2 + 4x − 6 = 0, form the quadratic equation whose roots are 3α and 3β .

[ 3 marks ]

Answer : .................................

___________________________________________________________________________

6 Diagram 2 shows the graph of the quadratic function y = 2(x – 3)2− p which has a minimum value of −5.

Find (a) the value of p, (b) the value of q, (c) the equation of the axis of symmetry.

[ 3 marks ]

Answer : (a) ……........................

(b) ……........................

(c)..................................

3

5

3

6

For examiner’s

use only

DIAGRAM 2

q

x

y

O



SULIT 3472/1


7

7 Find the range of values of x for which (2x − 3)(x + 1) ≥ x +1. [ 3 marks ]

Answer : ..................................

8 Solve the equation 27×9x + 1 = 1

3x .

[ 3 marks ]

Answer : ...................................

9 Solve the equation log 2 (x + 3) = 1 + log 2 (3x − 1).

[ 3 marks ]

Answer : ......................................

3

7

3

9

3

8

For examiner’s

use only



SULIT 3472/1


8

10 The sum of the first n terms of an arithmetic progression is given by [ ]nnSn 372

3 2 −= .

Find (a) the common difference, (b) the eleventh term of the progression [ 3 marks ]

Answer : (a) ……………………..

(b).……………..………

11 Express the recurring decimal 0.21212121... as a fraction in its simplest form.

[ 3 marks ]

Answer : …...…………..….......

3

10

For examiner’s

use only

3

11



SULIT 3472/1


9

12 Diagram 3 shows the graph pxxy += 2 .

Based on the graph above a table of x

y against x is obtained as show in table 1

x

y −3 r

x q −1

Calculate the values of p, q and r.

[ 4 marks ]

Answer : p =…...….………..….......

q= ....................................

r= ....................................

___________________________________________________________________________

13 Find the coordinates of point M which divides line segment joining the points ( 3, 3)A − and (7, 8)B such that AM : AB = 2 : 5. [ 3 marks ]

Answer : ………………..…….

3

13

4

12

DIAGRAM 3

For examiner’s

use only

−3

x

y

0

pxxy += 2

TABLE 1



SULIT 3472/1


10

14 Given the straight lines 3y ax+ = and 4 4y bx+ = are perpendicular to each other. Express a in terms of b.

[ 3 marks ]

Answer : .…………………

15 In Diagram 4, QR is parallel to PS and T is the midpoint of QR. [ 4 marks ]

Given that PS : QR = 3 : 5, PQ→

= 3u and PS→

= 6v, express, in terms of u and v, of

(a) RS→

,

(b) TS→

. [ 4 marks ]

Answer : (a)…...…………..…....... (b) ....................................

4

15

3

14

For examiner’s

use only

Q

S

R

T

DIAGRAM 4

P



SULIT 3472/1


11

16 Given 2 3OP i j= − +→

% %

and 10 2OQ i j= −→

% %

. Find, in terms of the unit vectors, i% and j

%

,

(a) PQ→

(b) the vector whose magnitude is 2 units and in the direction of PQ→

.

[ 3 marks ]

Answer : (a) PQ→

= …….…………...

(b) ……………………….. ___________________________________________________________________________

17 Solve the equation 2 sin x + xsin

1 = –3 for 0° ≤ x ≤ 360°. [ 4 marks ]

Answer : …...…………..….......

4

17

For examiner’s

use only

3

16



SULIT 3472/1


12

18 Diagram 5 shows a semicircle with centre O.

The diameter of the circle is 16 cm and ∠POQ = 0.6 radian. Calculate

(a) the length of arc QP,

(b) area of the shaded region. [ 4 marks ]

Answer : (a) ……………………..

(b) .……………..……… ___________________________________________________________________________

19 Find the coordinates of the minimum point of the curve 2 310

2y x x= − + .

[ 3 marks ]

Answer : ………………………

3

19

4

18

For examiner’s

use only

DIAGRAM 5 P O

Q

0.6 rad

R



SULIT 13 3472/1


20 Two variables, x and y, are related by the equation .16

3x

xy += Given that y increases

at a constant rate of 10 unit per second, find the rate of change of x when x = 2. [3 marks]

Answer : …...…………..…....... ___________________________________________________________________________

21 Given that 6

3( ) 2,g x dx=∫ find

(a) 363 ( )g x dx∫ ,

(b) the value of k if 6

3[ ( ) ] 10g x kx dx+ =∫ .

[ 3 marks ]

Answer : (a) ……………………..

(b) .……………..………

22 Given that the mean and variance of a set of n numbers x1, x2, . . . , xn are 3 and 2.56

respectively. Find the mean and standard deviation of the new set of n numbers 5x1 − 2, 5x2 − 2, . . . , 5xn − 2.

[ 3 marks ]

Answer : Mean = ……………………..

Standard deviation = .……………..………

3

20

3

21

3

22

For examiner’s

use only



SULIT 14 3472/1


23 The probability that Kamal is chosen as a school librarian is 5

2 whereas the

probability that Alisa is chosen as a school librarian is 12

5.

Find the probability that

(a) neither of them is chosen as a school librarian,

(b) only one of them is chosen as a school librarian. [ 3 marks ]

Answer : (a) ……………………..

(b) .……………..……… ___________________________________________________________________________ 24 Mathematics Club of a school has 8 Form 5 students, 10 Form 4 students and 12 Form 3

students.

(a) A teacher wants to choose Form 5 students to form a committee consisting a president, a vice president and a secretary, find the number of ways the committee can be formed.

(b) A team is to be formed to take part in a Mathematics competition. How many different teams, each comprising 3 Form 5 students, 2 Form 4 students and 1 Form 3 student can be formed? [ 4 marks ]

Answer : (a) ……………………..

(b) .……………..………

For examiner’s

use only

4

24

3

23



SULIT 15 3472/1


25 The mass of a packet of biscuit is normally distributed with a mean of 125 g and a variance of 16 g2.

(a) Find the probability that a packet of biscuit chosen at random from a sample will

have mass not less than 128 g. (b) If 30% of the packets chosen at random have mass more than m g, find the value

of m. [ 4 marks ]

Answer : (a) ……………………..

(b) .……………..………

END OF QUESTION PAPER

4

25

For examiner’s

use only



SULIT 3472/1 Additional Mathematics Paper 1 Sept 2009

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5

2009

ADDITIONAL MATHEMATICS

Paper 1

MARKING SCHEME

This marking scheme consists of 7 printed pages



2

PAPER 1 MARKING SCHEME 3472/1

Number Solution and marking scheme Sub

Marks Full

Marks 1

(a)

(b)

4 One to many relation

1

1

2

2 (a)

(b)

24 20 21x x+ +

2(2 5) 4x + −

3−

2

B1

1

3

3 (a)

(b)

5 2

3

x −

3 2

5

xy

+=

11

5

2

B1

1

3

4

p < 498

49 − 8p > 0 (−5)2 − 4(2)(p − 3) > 0

3

B2

B1

3

5

x2 + 4x − 18 = 0 3α + 3β = −4 or 3α(3β) = −18

α + β = − 43

and αβ = −2

3 B2

B1

3



3


Marks Full

Marks 6

(a)

(b)

(c)

5

13

x = 3

1

1

1

3

7

x ≤ −1, x ≥ 2

2

( 2)( 1) 0

2 0

x x

x x

− + ≥− − ≥

3

B2

B1

3

8

53

x = −

2x + 5 = −x

2( 1)3 3 2 23 3 3 or 3 3x x x x+ − + + −× = =

3

B2

B1

3

9

2

1

3=2 or IE

3 1+3

log =1 or IE 3 1

x

x

xx

x

=+−

−

3

B2

B1

3

10 (a)

(b)

21

2 23 17(2) 3(2) 2 7(1) 3(1)

2 2d = − − × − or IE

216

2

B1

1

3

11

7

33

0.21

1 0.01−

a = 0.21 and r = 0.01

3

B2

B1

3



4


Marks Full

Marks 12

r = 2 q = −6 y

x = x + 3*

p = 3

B1

B1

B1

B1

4

13 ( )1, 5M

3( 3) 2(7) 3(3) 2(8)

,3 2 3 2

M− + +

+ +

AM : BM = 2 : 3

3

B2

B1

3

14 a = − 4

b

14

ba − × − = −

1 2or 4

bm a m= − = −

3

B2

B1

3

15 (a)

(b)

3 4u v− −% %

5

(6 ) 3 63

v u v− − +% % %

3u v− +% %

( )1 56 3 4

2 3v u v + − −

% % %

2

B1

2 B1

4

16 a)

b)

12 5PQ i j= −uuur

% %

24 10

13

i j−% %

12 5

13

i j−% % or

12 52

13

i j−× % %

1 2 B1

3

or Use TL

or Use TL



5


Marks Full

Marks 17

2

210 ,270 330

1sin and sin 1

2

(2sin 1)(sin 1) 0

2sin 3sin 1 0

x x

x x

θ θ= − = −

+ + =

+ + =

o o o

4

B3

B2

B1

4

18 (a)

(b)

4.8 cm 8 × 0.6 81.33 1

2(8)2(π − 0.6) or π − 0.6

2

B1

2

B1

4

19 3 151

,4 16

2x − 3

2 = 0 or x =

3

4

3

22

dyx

dx= −

3

B2

B1

3

20

10−=dt

dx

10 = dt

dx).

2

163(

2−

2

163

xdx

dy −=

3

B2 B1

3



6


Marks Full

Marks 21 (a)

(b)

−6 16

27

102

26

3

2

=

+ kx

1

2

B1

3

22 Mean = 13 AND Standard deviation = 8 Mean = 13 AND Variance (new) = 64 Mean = 13 OR Variance (new) = 64

3

B2

B1

3

23

(a)

(b)

7

20

29

60

×+

×12

5

5

3

12

7

5

2

1

2 B1

3

24 (a)

(b)

336

38 P 30 240

112

210

38 CCC ××

2

B1

2

B1

4

25 (a)

0.2266

128 125

4P Z

− ≥ OR P[Z ≥ 0.75]

2

B1

4



7


Marks Full

Marks (b) 127.1

125

0.34

mP Z

− ≥ = OR

1250.524

4

m − =

2

B1



add math p1 trial spm zon a 2009 - tutor mansor ... 3472/1 3472/1 2009 hak cipta zon a kuching [...

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