matematik tambahan kertas 1 dua jam · 3472/1 matematik tambahan kertas 1 sept 2010 2 jam . sulit...

SULIT 1 3472/1

3472/1 ZON A KUCHING 2010 [Lihat sebelah SULIT

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM 2010

Kertas soalan ini mengandungi 16 halaman bercetak

For examiner’s use only

Question Total Marks Marks

Obtained 1 3

2 4

3 4

4 3

5 2

6 3

7 3

8 3

9 4

10 3

11 3

12 3

13 3

14 3

15 3

16 3

17 4

18 3

19 3

20 3

21 3

22 3

23 4

24 3

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 Sept 2010 2 Jam

SULIT 3472/1

3472/1 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

3472/ 1 ZON A KUCHING 2010 [Lihat sebelah SULIT

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 ∓ sinA sinB

11 tan (A± B) = BA

BA

tantan1

tantan

∓

±

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

3472/1 ZON A KUCHING 2010 SULIT

4

THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DIS TRIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)

z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

Minus / Tolak

0.0

0.1

0.2

0.3

0.4

0.5000

0.4602

0.4207

0.3821

0.3446

0.4960

0.4562

0.4168

0.3783

0.3409

0.4920

0.4522

0.4129

0.3745

0.3372

0.4880

0.4483

0.4090

0.3707

0.3336

0.4840

0.4443

0.4052

0.3669

0.3300

0.4801

0.4404

0.4013

0.3632

0.3264

0.4761

0.4364

0.3974

0.3594

0.3228

0.4721

0.4325

0.3936

0.3557

0.3192

0.4681

0.4286

0.3897

0.3520

0.3156

0.4641

0.4247

0.3859

0.3483

0.3121

4

4

4

4

4

8

8

8

7

7

12

12

12

11

11

16

16

15

15

15

20

20

19

19

18

24

24

23

22

22

28

28

27

26

25

32

32

31

30

29

36

36

35

34

32

0.5

0.6

0.7

0.8

0.9

0.3085

0.2743

0.2420

0.2119

0.1841

0.3050

0.2709

0.2389

0.2090

0.1814

0.3015

0.2676

0.2358

0.2061

0.1788

0.2981

0.2643

0.2327

0.2033

0.1762

0.2946

0.2611

0.2296

0.2005

0.1736

0.2912

0.2578

0.2266

0.1977

0.1711

0.2877

0.2546

0.2236

0.1949

0.1685

0.2843

0.2514

0.2206

0.1922

0.1660

0.2810

0.2483

0.2177

0.1894

0.1635

0.2776

0.2451

0.2148

0.1867

0.1611

3

3

3

3

3

7

7

6

5

5

10

10

9

8

8

14

13

12

11

10

17

16

15

14

13

20

19

18

16

15

24

23

21

19

18

27

26

24

22

20

31

29

27

25

23

1.0

1.1

1.2

1.3

1.4

0.1587

0.1357

0.1151

0.0968

0.0808

0.1562

0.1335

0.1131

0.0951

0.0793

0.1539

0.1314

0.1112

0.0934

0.0778

0.1515

0.1292

0.1093

0.0918

0.0764

0.1492

0.1271

0.1075

0.0901

0.0749

0.1469

0.1251

0.1056

0.0885

0.0735

0.1446

0.1230

0.1038

0.0869

0.0721

0.1423

0.1210

0.1020

0.0853

0.0708

0.1401

0.1190

0.1003

0.0838

0.0694

0.1379

0.1170

0.0985

0.0823

0.0681

2

2

2

2

1

5

4

4

3

3

7

6

6

5

4

9

8

7

6

6

12

10

9

8

7

14

12

11

10

8

16

14

13

11

10

19

16

15

13

11

21

18

17

14

13

1.5

1.6

1.7

1.8

1.9

0.0668

0.0548

0.0446

0.0359

0.0287

0.0655

0.0537

0.0436

0.0351

0.0281

0.0643

0.0526

0.0427

0.0344

0.0274

0.0630

0.0516

0.0418

0.0336

0.0268

0.0618

0.0505

0.0409

0.0329

0.0262

0.0606

0.0495

0.0401

0.0322

0.0256

0.0594

0.0485

0.0392

0.0314

0.0250

0.0582

0..0475

0.0384

0.0307

0.0244

0.0571

0.0465

0.0375

0.0301

0.0239

0.0559

0.0455

0.0367

0.0294

0.0233

1

1

1

1

1

2

2

2

1

1

4

3

3

2

2

5

4

4

3

2

6

5

4

4

3

7

6

5

4

4

8

7

6

5

4

10

8

7

6

5

11

9

8

6

5

2.0

2.1

2.2

2.3

0.0228

0.0179

0.0139

0.0107

0.0222

0.0174

0.0136

0.0104

0.0217

0.0170

0.0132

0.0102

0.0212

0.0166

0.0129

0.00990

0.0207

0.0162

0.0125

0.00964

0.0202

0.0158

0.0122

0.00939

0.0197

0.0154

0.0119

0.00914

0.0192

0.0150

0.0116

0.00889

0.0188

0.0146

0.0113

0.00866

0.0183

0.0143

0.0110

0.00842

0

0

0

0

3

2

1

1

1

1

5

5

1

1

1

1

8

7

2

2

1

1

10

9

2

2

2

1

13

12

3

2

2

2

15

14

3

3

2

2

18

16

4

3

3

2

20

16

4

4

3

2

23

21

2.4 0.00820 0.00798 0.00776 0.00755 0.00734

0.00714

0.00695

0.00676

0.00657

0.00639

2

2

4

4

6

6

8

7

11

9

13

11

15

13

17

15

19

17

2.5

2.6

2.7

2.8

2.9

0.00621

0.00466

0.00347

0.00256

0.00187

0.00604

0.00453

0.00336

0.00248

0.00181

0.00587

0.00440

0.00326

0.00240

0.00175

0.00570

0.00427

0.00317

0.00233

0.00169

0.00554

0.00415

0.00307

0.00226

0.00164

0.00539

0.00402

0.00298

0.00219

0.00159

0.00523

0.00391

0.00289

0.00212

0.00154

0.00508

0.00379

0.00280

0.00205

0.00149

0.00494

0.00368

0.00272

0.00199

0.00144

0.00480

0.00357

0.00264

0.00193

0.00139

2

1

1

1

0

3

2

2

1

1

5

3

3

2

1

6

5

4

3

2

8

6

5

4

2

9

7

6

4

3

11

9

7

5

3

12

9

8

6

4

14

10

9

6

4

3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4

Example / Contoh:

−= 2

2

1exp

2

1)( zzf

π If X ~ N(0, 1), then

Jika X ~ N(0, 1), maka

( ) ( ) k

yQ z f z dz

x

∞ ∆=∆∫ P(X > k) = Q(k)

P(X > 2.1) = Q(2.1) = 0.0179

Q(z)

z

f

O k

SULIT 3472/1

3472/1 ZON A KUCHING 2010 [ Lihat sebelah SULIT

5

Answer all questions.

1. Diagram 1 shows the quadratic function h.

DIAGRAM 1 (a) State (i) the objects of 64, (ii) the value of k. (b) Using the function notation, express h in terms of x. [3 marks] Answer : (a) (i)………………

(ii)........................

(b) ………………... 2. Given that f : x→2x − 1 and g : x→ x2 + 3x + 5 , find (a) f −1(x), (b) gf (x). [4 marks]

Answer : (a) ………………..

(b) …………………

8 ● 4 ● k ● −8 ●

● 16 ● 64

x h(x)

4

2

3

1

For examiner’s

use only

x

SULIT 3472/1


6

3. Given that h : x→px + q , g : x→( x – 3)2 +7 and hg : x→−2( x – 3)2 + 10, find

(a) the value of p and of q,

(b) the value of gh(−3). [4 marks]

Answer : (a) p =......... q =..............

(b) ……………………... 4. Given that α andβ are the roots of the equation x2 + x – 2 =0. Form a quadratic equation which has the roots 2α and 2β .

[3 marks]

Answer : .........…………………

For examiner’s

use only

4

3

3

4

SULIT 3472/1


7

5. If the graph of the quadratic function 2( ) 2f x x x p= + − touches the x-axis at only one point, find the value of p. [2 marks]

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

___________________________________________________________________________

6. Find the range of the values of p for which ( 3) 2 6p p p− ≤ − [3 marks]

Answer : ……........................

2

5

3

6

For examiner’s

use only

SULIT 3472/1


8

7. Solve the equation 32(8)2x = 1. [3 marks]

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

8. Solve the equation log 2 x – log 4 25 = 0 [3 marks]

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

9. If 1 + log a 8

a = 2 log a x + log a 2, express x in terms of a.

[4 marks]

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

3

7

4

9

3

8

For examiner’s

use only

SULIT 3472/1


9

10. The first three terms of an arithmetic progression are ,,5 x− and 9. Find

(a) value of x

(b) the sum from the 7th to the 11th terms

of the arithmetic progression. [3 marks]

Answer : (a) ……………………..

(b) ……………………….

11. The first and the second term of an geometric progression is m and 4

3m.

Find the sum of the first 4 terms of the progression if m = 6 [3 marks]

Answer : …...…………..….......

3

10

For examiner’s

use only

3

11

SULIT 3472/1


10

12. The variables x and y are related by the equation 15

1.yx

= −

DIAGRAM 12

A straight line graph is obtained by plotting xy against x, as shown in Diagram 12. Find the value of a and of b.

[3 marks]

Answer: a =...….………. b = ....................

13. Given the coordinates of points S and T are (−3, 2) and (7, −3) respectively. Point P

divides the line segment ST in the ratio 3 : 2, find the coordinates of P. [3 marks]

Answer : ………………..…….

3

13

3

12

For examiner’s

use only

O B(20, b)

A(0, a)

x

xy

SULIT 3472/1


11

14. Given that the vertices of a quadrilateral ABCD are A(0, −1), B(3, 3), C(6, k) and D(5, −1) and its area is 15 unit2, find the value of k.

[3 marks] Answer : .…………………

15. Given ( 1,3) , (2, 1).P Q− − Find, in terms of unit vectors � and � ,

(a) PQ→

(b) the unit vector in the direction of PQ→

. [3 marks]

Answer : (a)…...…………..….......

(b) ....................................

3

15

3

14

For examiner’s

use only

SULIT 3472/1


12

16. Diagram 16 shows a triangle OPQ.

Given OP p=→

, OQ q=→

and point M lies on PQ such that PM : PQ = 1 : 4. Express

OM→

in terms of p and q .

[3 marks]

Answer : OM→

= …….…………...

. ___________________________________________________________________________

17. Solve the equation cosec2 2 0x+ = for 0˚ ≤ x ≤ 360˚. [4 marks]

Answer : …...…………..….......

4

17

For examiner’s

use only

3

16

�

� �

�

DIAGRAM 16

P

Q O

M

SULIT 3472/1


13

18. Diagram 18 shows the sector OAD and the sector OBC with a common centre O. C

DIAGRAM 18 Given OD = 4 cm, OC = 3 cm and the ratio of the length of arc AD to the length of arc BC is 2 : 3. Find the value of θ in radians. [3 marks] Answer : ……………………..

19. Given that 26 4y x x= − , find the small approximate change in y when x increases

from 1 to 1.05.

[3 marks]

Answer : ………………………

3

19

3

18

For examiner’s

use only

D

θ B O A

SULIT 14 3472/1

3472/1 2010 Hak Cipta Zon A Kuching [Lihat sebelah SULIT

20. Given that the gradient of the tangent to the curve 1 2

hy

x=

− at the point where x = 2

is −2, calculate the value of h. [3 marks]

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

21. Given that 3

1

( ) 5f x dx=∫ . Find the value of the constant k if

2 3

1 2

( ) [ ( ) ] 15.f x dx f x kx dx+ + =∫ ∫ [3 marks]

Answer : ……………………..

22. A set of integers 4, 5, 8, 9, 10, 15, 18, m and nhas a mean 12 and mode 18. Find the value of mand of n if n > m. [3 marks] Answer : …………………. .

3

20

3

21

3

22

For examiner’s

use only

SULIT 15 3472/1

3472/1 ZON A KUCHING 2010 Lihat sebelah SULIT

23. A debate team consists of 2 Form Four and 4 Form Five students. If they are chosen from 10 Form Four students and 12 Form Five students, find

(a) the number of ways the debate team can be formed.

(b) the number of ways the debate team can be arranged in a row to take photograph with the 2 Form Four students sitting next to each other.

[4 marks]

Answer : (a) ……………………..

(b) .……………..……… ___________________________________________________________________________ 24. Robert and David’s favourite drink is Coca-cola. If they were asked to choose a drink

from a box that contains 4 cans of Coca-cola and 5 cans of Orange juice. Find the probability that

(a) both of them got their favourite drink,

(b) only one of them got their favourite drink. [3 marks]

Answer : (a) ……………………..

(b) .……………..………

For examiner’s

use only

3

24

4

23

SULIT 16 3472/1

3472/1 2010 Hak Cipta Zon A Kuching [Lihat sebelah SULIT

25. Diagram 25 shows a standard normal distribution graph. DIAGRAM 25 The probability is represented by the area of the shaded region is 0.7019 (a) Find the value of k. (b) X is a continuous random variable which is normally distributed with a mean of 45

and a standard deviation of 5. Find the value of X when the z-score is k.

[4 marks]

Answer : (a) ……………………..

(b) .……………..………

END OF QUESTION PAPER

4

25

For examiner’s

use only

( )f z

z k 0

3472/1 Matematik Tambahan Kertas 1 2 jam Sept 2010

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010

MATEMATIK TAMBAHAN

Kertas 1

Dua jam


Skema Pemarkahan ini mengandungi 7 halaman bercetak

MARKING SCHEME

2 MARKING SCHEME FOR PAPER 1 -2010 ZON A

No Solution and marking scheme Sub Marks Total Marks 1. (a) (i) 8, −8

(ii) −4 (b) 2 2( ) or :h x x h x x= →

1 1 1

3

2. (a) 1 1

( )2

xf x− +=

1

2

yx

+=

(b) 2( ) 4 2 3gf x x x= + +

( ) ( )2(2 1) 2 1 3 2 1 5g x x x− = − + − +

2

B1 2

B1

4

3. (a) p = −2 q = 24 (b) ( 3) 736gh − = ( 3) 30h − =

1 1 2

B1

4

4. 2 2 8 0x x+ − =

SONR = 2 2 2( ) 2( 1)α β α β+ = + = − PONR = 2 2 4 4( 2)α β αβ× = = − SOR = 1α β+ = − POR = 2αβ = −

3

B2

B1

3

3No Solution and marking scheme Sub Marks Total Marks 5.

1p = −

2(2) 4(1)( ) 0

or

4 4 0

p

p

− − =

+ =

2

B1

2

6. 2 3p≤ ≤ 2 3 ( 2)( 3) 0p p− − ≤

3

B2 B1

3

7.

5

6x = −

3(2x) = −5 or 5 + 3(2x) = 0 or IE

3(2 ) 52 2x −= or 25×(23)2x = 20

3

B2

B1

3

8. 5x =

2 22log log 25x = or x2 = 25

22

2

log 25log 0

log 4x − =

3

B2

B1

3

9.

4

ax =

2

228

ax=

2

2log log 28a aa

x=

22

2log @ log 2 @ log8a a aa

x x

4

B3

B2

B1

4


(a) 2x =

(b) 255 33011 =S or 756 =S or

[ ])7)(10()5(22

11 +− or [ ])7)(5()5(22

6 +−

1 2

B1

3

11. 4920

( )4

4

6 9 1

9 1S

−=

−

6 or 9a r= =

3

B2

B1

3

12. 15, 5a b= = − a = 15 or b = −5 15xy x= − +

3

B2

B1

3

13. P ( 3, −1)

( ) ( ) ( ) ( )

+−+

++−=

32

3322,

32

7332P

SP : PT = 3 : 2 or

3

B2

B1

3

14. k = 2 2 26 30k− − = −

1

[0(3) 3 6( 1) 5( 1) 3( 1) 6(3) 5 0( 1)] 152

k k+ + − + − − − − − − − = −

3

B2

B1

3

• T S P

2 3

5No Solution and marking scheme Sub Marks Total Marks

15. (a) 3 4 i j−

(b) 4 3 43

@5 5 5

j i ji −−

2 2

3 4

(3) ( 4)

i j−

+ −

1 2

B1

3

16.

qp4

1

4

3 +

( )qpp +−+

4

1

1

4OM OP PQ= +→ → →

3

B2

B1

3

17. 105 , 165 , 285 , 345x = � � � � 2x =210˚, 330˚, 570˚, 690˚ 30˚

22sin

1 −=x

or 2

12sin −=x

4

B3

B2

B1

4

18.

.3

radπθ =

12 6( )θ π θ= −

4 2

3( ) 3

θπ θ

=−

3

B2

B1

3


0.4y∂ = (12 4)(0.05)y x∂ = −

12 4 or 1.05 1 0.05dy

x xdx

= − ∂ = − =

3

B2

B1

3

20. 9h = −

2

22

[1 2(2)]

h = −−

2

2

(1 2 )

dy h

dx x=

− or equivalent

3

B2

B1

3

21. 4

32

2

5 152

kx + =

∫ ∫ =+2

1

3

2

5)()( dxxfdxxf

3

B2

B1

3

22. 18, 21m n= = 39m n+ =

4 5 8 9 10 15 18

129

x m nx

N

+ + + + + + + += = =∑

3

B2

B1

3


(a) 22275 10 12

2 4C C×

(b) 240

5!2!

2

B1 2

B1

4

24.

(a) 4 3 1

9 8 6× =

(b) 5

9

4 5 5 4

( ) ( )9 8 9 8

× + ×

1 2

B1

3

25. (a) k = 0.53 1 – 0.7019 (b) X = 47.65

45

0.535

X − =

2

B1 2

B1

4

SULIT 1 3472/2


3472/2 Matematik Tambahan Kertas 2 2 ½ jam 2010



MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit


1. This question paper consists of three sections : Section A, Section B and Section C. 2. Answer all question in Section A , four questions from Section B and two questions from

Section C.

3. Give only one answer / solution to each question..

4. Show your working. It may help you to get marks.

5. The diagram in the questions provided are not drawn to scale unless stated. 6. The marks allocated for each question and sub-part of a question are shown in brackets..

7. A list of formulae is provided on pages 2 to 3.

8. A booklet of four-figure mathematical tables is provided.

9. You may use a non-programmable scientific calculator.

Kertas soalan ini mengandungi 11 halaman bercetak

SULIT 3472/2


2

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

ALGEBRA

1 x = a

acbb

2

42 −±−

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

xi yjr

x y

∧ +=+

GEOM ETRY

SULIT 3472/2


3

STATISTICS

TRIGONOMETRY

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) = rnr

rn 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

= ×

9 sin (A± B) = sinA cosB ± cosA sinB

10 cos (A± B) = cosA cosB ∓ sinA sinB

11 tan (A± B) = BA

BA

tantan1

tantan

∓

±

12 C

c

B

b

A

a

sinsinsin==

13 a2 = b2 + c2 − 2bc cos A

14 Area of triangle = Cabsin2

1

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

−

SULIT 4 3472/2


THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DIST RIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)

z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

Minus / Tolak

0.0

0.1

0.2

0.3

0.4

0.5000

0.4602

0.4207

0.3821

0.3446

0.4960

0.4562

0.4168

0.3783

0.3409

0.4920

0.4522

0.4129

0.3745

0.3372

0.4880

0.4483

0.4090

0.3707

0.3336

0.4840

0.4443

0.4052

0.3669

0.3300

0.4801

0.4404

0.4013

0.3632

0.3264

0.4761

0.4364

0.3974

0.3594

0.3228

0.4721

0.4325

0.3936

0.3557

0.3192

0.4681

0.4286

0.3897

0.3520

0.3156

0.4641

0.4247

0.3859

0.3483

0.3121

4

4

4

4

4

8

8

8

7

7

12

12

12

11

11

16

16

15

15

15

20

20

19

19

18

24

24

23

22

22

28

28

27

26

25

32

32

31

30

29

36

36

35

34

32

0.5

0.6

0.7

0.8

0.9

0.3085

0.2743

0.2420

0.2119

0.1841

0.3050

0.2709

0.2389

0.2090

0.1814

0.3015

0.2676

0.2358

0.2061

0.1788

0.2981

0.2643

0.2327

0.2033

0.1762

0.2946

0.2611

0.2296

0.2005

0.1736

0.2912

0.2578

0.2266

0.1977

0.1711

0.2877

0.2546

0.2236

0.1949

0.1685

0.2843

0.2514

0.2206

0.1922

0.1660

0.2810

0.2483

0.2177

0.1894

0.1635

0.2776

0.2451

0.2148

0.1867

0.1611

3

3

3

3

3

7

7

6

5

5

10

10

9

8

8

14

13

12

11

10

17

16

15

14

13

20

19

18

16

15

24

23

21

19

18

27

26

24

22

20

31

29

27

25

23

1.0

1.1

1.2

1.3

1.4

0.1587

0.1357

0.1151

0.0968

0.0808

0.1562

0.1335

0.1131

0.0951

0.0793

0.1539

0.1314

0.1112

0.0934

0.0778

0.1515

0.1292

0.1093

0.0918

0.0764

0.1492

0.1271

0.1075

0.0901

0.0749

0.1469

0.1251

0.1056

0.0885

0.0735

0.1446

0.1230

0.1038

0.0869

0.0721

0.1423

0.1210

0.1020

0.0853

0.0708

0.1401

0.1190

0.1003

0.0838

0.0694

0.1379

0.1170

0.0985

0.0823

0.0681

2

2

2

2

1

5

4

4

3

3

7

6

6

5

4

9

8

7

6

6

12

10

9

8

7

14

12

11

10

8

16

14

13

11

10

19

16

15

13

11

21

18

17

14

13

1.5

1.6

1.7

1.8

1.9

0.0668

0.0548

0.0446

0.0359

0.0287

0.0655

0.0537

0.0436

0.0351

0.0281

0.0643

0.0526

0.0427

0.0344

0.0274

0.0630

0.0516

0.0418

0.0336

0.0268

0.0618

0.0505

0.0409

0.0329

0.0262

0.0606

0.0495

0.0401

0.0322

0.0256

0.0594

0.0485

0.0392

0.0314

0.0250

0.0582

0..0475

0.0384

0.0307

0.0244

0.0571

0.0465

0.0375

0.0301

0.0239

0.0559

0.0455

0.0367

0.0294

0.0233

1

1

1

1

1

2

2

2

1

1

4

3

3

2

2

5

4

4

3

2

6

5

4

4

3

7

6

5

4

4

8

7

6

5

4

10

8

7

6

5

11

9

8

6

5

2.0

2.1

2.2

2.3

0.0228

0.0179

0.0139

0.0107

0.0222

0.0174

0.0136

0.0104

0.0217

0.0170

0.0132

0.0102

0.0212

0.0166

0.0129

0.00990

0.0207

0.0162

0.0125

0.00964

0.0202

0.0158

0.0122

0.00939

0.0197

0.0154

0.0119

0.00914

0.0192

0.0150

0.0116

0.00889

0.0188

0.0146

0.0113

0.00866

0.0183

0.0143

0.0110

0.00842

0

0

0

0

3

2

1

1

1

1

5

5

1

1

1

1

8

7

2

2

1

1

10

9

2

2

2

1

13

12

3

2

2

2

15

14

3

3

2

2

18

16

4

3

3

2

20

16

4

4

3

2

23

21

2.4 0.00820 0.00798 0.00776 0.00755 0.00734

0.00714

0.00695

0.00676

0.00657

0.00639

2

2

4

4

6

6

8

7

11

9

13

11

15

13

17

15

19

17

2.5

2.6

2.7

2.8

2.9

0.00621

0.00466

0.00347

0.00256

0.00187

0.00604

0.00453

0.00336

0.00248

0.00181

0.00587

0.00440

0.00326

0.00240

0.00175

0.00570

0.00427

0.00317

0.00233

0.00169

0.00554

0.00415

0.00307

0.00226

0.00164

0.00539

0.00402

0.00298

0.00219

0.00159

0.00523

0.00391

0.00289

0.00212

0.00154

0.00508

0.00379

0.00280

0.00205

0.00149

0.00494

0.00368

0.00272

0.00199

0.00144

0.00480

0.00357

0.00264

0.00193

0.00139

2

1

1

1

0

3

2

2

1

1

5

3

3

2

1

6

5

4

3

2

8

6

5

4

2

9

7

6

4

3

11

9

7

5

3

12

9

8

6

4

14

10

9

6

4

3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4

Example / Contoh:

−= 2

2

1exp

2

1)( zzf

π If X ~ N(0, 1), then

Jika X ~ N(0, 1), maka

( ) ( )

k

yQ z f z dz

x

∞ ∆=∆∫ P(X > k) = Q(k)

P(X > 2.1) = Q(2.1) = 0.0179

4

Q(z)

z

f

O k

SULIT 3472/2


5

SECTION A

[40 marks]

Answer all questions in this section.

1 Solve the simultaneous equations 2x y− = and 2 25 2 8x xy y− + = . [5 marks]

2 ( ) 0f x = is a quadratic equation which has the roots 2 and p− that are distinct from

each other. (a) Write ( ) 0f x = in the form 2ax bx c+ + . [2 marks] (b) The curve ( )y kf x= cuts the y-axis at the point (0, 12). Given that the value of p = 3, calculate (i) the value of k. [2 marks] (ii) the coordinates of the maximum point of the curve. [3 marks] 3 A number of wires are arranged as shown in the Diagram 3.

The longest wire is 40 cm long. The length of each subsequent wire is 20% shorter than the previous wire.

Find (a) the length of the 6th wire [3 marks] (b) the total length of the first 5 wires. [2 marks] (c) the sum to infinity of the length of the wires. [2 marks]

DIAGRAM 3

SULIT 3472/2


6 4 (a) Prove that ).2cos1(tan2sin xxx += [2 marks]

(b) (i) Sketch the graph of 12sin −= xy for 0 ≤ x ≤ π.

(ii) Hence, determine the value of p such that px 22sin = has only two real

solutions for 0 ≤ x ≤ π. [6 marks]

5 Table 1 shows the distribution of marks obtained by 100 students in a test.

Marks 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69

Number of students

8 16 30 25 14 7

(a) Using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 5 students on the

vertical axis, construct a histogram for the given data and from the histogram, estimate the value of mode of the marks of the students.

[3 marks]

(b) Without drawing an ogive, calculate the median mark. [3 marks] 6 Diagram 6 shows a pentagon PQTRS.

(a) Express in terms of x and y .

(i) PR→

,

(ii) QS→

. [2 marks]

(b) Given PT→

= m PR→

and QT→

= nQS→

, express PT→

in terms of (i) m, x and y ,

(ii) n, x and y .

Hence, find the values of m and of n. [5 marks]

TABLE 1

DIAGRAM 6

4x

3y

x

T

Q

R S

P

SULIT 3472/2


7

SECTION B

[40 marks]

Answer four questions from this section. 7 (a) A closed cylindrical water tank, with a radius of r cm is to be constructed using

aluminium sheets to hold 250π m3 of water. If the total surface area of the water

tank, A m2, is given by 2 5002A r

r

ππ= + . Calculate the minimum total surface area.

[3 marks] (b) Diagram 7 shows part of the curve )5( xxy −= intersecting the straight line y = 5 – x

at point P(1, 4). (i) Find the area of the shaded region. [4 marks] (ii) The region enclosed by the curve and the x-axis is rotated through 360o about

the x-axis. Find the volume generated, in terms of π . [3 marks]

x O

y

P(1, 4)

)5( xxy −=

y = 5 − x DIAGRAM 7

SULIT 3472/2


8

8 Use graph paper to answer this question.

Table 2 shows the values of two variables, x and y , obtained from an experiment.

The variables x and y are related by the equation 2−= xaby where a and b are constants.

x 5 6 7 9 10 12 y 7.24 8.71 10.9 17.4 21.4 32.3

(a) Plot logy against (x − 2), by using a scale of 2 cm to 1 unit on the (2−x )-axis and 2 cm to 0.2 unit on the log y -axis.

Hence, draw the line of best fit. [5 marks] (b) Use your graph from 7(a) to find the value of

(i) ,a

(ii) .b [5 marks]

9 Diagram 9 shows a sector of a circle with centre O. ABC is a segment with the height, MC

of 3 m. AB is a chord of the segment with the length of 8 m and M is the mid point of AB. Given

that r is the radius of the sector, find

(a) (i) the length of OM in terms of r.

(ii) the value of r, [3 marks]

(b) ∠AOB in radian, [3 marks]

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

TABLE 2

A B

O

C

M

3 m

r DIAGRAM 9

SULIT 3472/2


9

10 Solution by scale drawing is not accepted. Diagram 10 shows a trapezium ABCD. Given that the equation of the line AB is 3 4 0x y− + = , find (a) the value of m, [3 marks] (b) the equation of AD and hence, the coordinates of point A, [5 marks] (c) the equation of the locus of P if P moves in such a way that BPD∠ is always a right

angle. [2 marks] 11 (a) The school hockey team held a training session on penalty shooting. Each player was

given 5 trials. After the session, it was found that on the average, the mean for number of goals scored by a player is 3. If a player is chosen at random, find the probability that a player

(i) fails to score any goal, (ii) scores at least 4 goals. [5 marks] (b) The body mass of 483 pupils in a certain school follows a normal distribution with

mean of 35 kg and a standard deviation of 10 kg.

(i) If a pupil is chosen at random, find the probability that his body mass is between 30 kg and 60 kg.

(ii) A pupil will be placed under obesity list if his body mass exceeds 60 kg.

Estimate the number of pupils whose names will appear in the list. [5 marks]

O C(m, 0) x

y

D(3 , −1)

A

B(8, 4)

.

.

.

. DIAGRAM 10

SULIT 3472/2


10

SECTION C

[20 marks]

Answer two questions from this section. 12 A particle moves in a straight line and passes through a fixed point O.

Its velocity, v ms 1− , is given by 1582 +−= ttV , where t is the time, in seconds, after

leaving O . [Assume motion to the right is positive.]

Find

(a) the initial velocity, in ms−1, [1 mark]

(b) the minimum velocity, in ms−1, [3 marks]

(c) the range of values of t at which the particles moves to the right, [3 marks]

(d) the distance, in m, travelled by the particle in the third second. [3 marks]

13 The table 13 shows the monthly expenditure of a family on different items in the years 2008 and 2009 and their respective weightages, with total weightage of 15.

Item Expenditure (RM)

Weightage Year 2008 Year 2009

Electricity 180 210 4 Water 30 40 2

Telephone and internet 120 No change x Others 450 540 6

(a) Using 2008 as the base year, calculate the price index for the

(i) expenditure of electricity,

(ii) expenditure of telephone and internet

in the year 2009. [2 marks]

(b) Calculate the composite index for the expenditure on these items in the year 2009 based on the year 2008. [3 marks]

(c) The rate of increase for the expenditure on electricity and others from the year 2009

to 2011 is expected to be 10% while that of water and telephone and internet remain unchanged.

Calculate

(i) the expected expenditure on electricity in the year 2011,

(ii) the composite index in the year 2011 based on the year 2008. [5 marks]

TABLE 13

SULIT 3472/2


11

14 The diagram 14 shows a quadrilateral PQRS.

Given that the area of ∆QRS is 28 cm2 and ∠QRS is acute. Calculate

(a) ∠QRS, [2 marks]

(b) the length of QS in cm, [2 marks]

(c) ∠PQS [3 marks]

(d) the area of quadrilateral PQRS. [3 marks]

DIAGRAM 14

P Q

S R

7.5 cm

14 cm

30o

10 cm

SULIT 3472/2


12 15 Use the graph paper provided to answer this question.

A factory produces two types of commodities, A and B. The commodities produced by the factory satisfy the following constraints.

I : The ratio of the number of commodities A and B produced in a day must not be

less than 1 : 2. II : The total number of commodities A and B produced in a day must be at least 7

units. III : The number of commodity A must exceed the number of commodity B

produced in a day by not more than 3.

(a) Taking x as the number of units of commodity A and y as the number of units of commodity B produced in a day, write down three inequalities, other than 0x ≥ and

0y ≥ , which satisfy the above conditions. [3 marks]

(b) By using a scale of 2cm to 1 unit of commodity on both axes, construct and shade the region R that satisfies all the above conditions. [3 marks]

(c) By using your graph in (b), find

(i) the minimum number of units of commodity A produced if the number of units of commodity B produced is 5 units,

(ii) the minimum cost of production of these commodities each day, if the cost of production for one units of commodity A and one unit of commodity B in a day are RM100 and RM80 respectively.

[4 marks]

END OF QUESTION PAPER

3472/2 Matematik Tambahan Kertas 2 2 ½ jam Sept 2010



MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit


Skema Pemarkahan ini mengandungi 15 halaman bercetak

MARKING SCHEME

2

ADDITIONAL MATHEMATICS MARKING SCHEME

Zon A Kuching 2010 – PAPER 2

QUESTION

NO. SOLUTION MARKS

1

2 2

2

(2 ) 5 (2 ) 2 8 0

( 1)( 2) 0

x y

y y y y

y y

= +

+ − + + − =

+ + =

1, 2

@

1, 0

y y

x x

= − = −

= =

5

2

(a)

(b) (i)

(ii)

2

( ) ( )( 2)

( 2) 2

f x x p x

x p x p

= + −

= + − −

12 [ 2(3)]k= −

2k = −

2 2

2

2

1 12[ 6]

2 2

1 252( )

2 2

1 25 1 1max point , or ,12

2 2 2 2

y x x

x

= − + + − −

= − + +

∴ = − −

2

5

Solve the quadratic equation by using the factorization @ quadratic formula @ completing the square must be shown

Eliminate orx y

Note : OW−−−− 1 if the working of solving quadratic equation is not shown. 5

7

P1

K1

K1

N1

N1

K1

K1

N1

N1

K1

K1

N1

3

QUESTION NO.

SOLUTION MARKS

3

(a)

(b)

(c)

a = 40 and r = 0.8 |T6 = 40(0.8)5 = 13.1072

3

( )8.01

8.0140 5

5 −−=S

= 134.464

2

8.01

40

−=∞S

= 200

2

4

(a)

2 sincos2 2cos 1 or tan

cos

LHS to RHS or RHS to LHS

xx x x

x= − =

2

N1

K1

N1

K1

N1

P1

K1

7

N1

K1

4

QUESTION NO.

SOLUTION MARKS

(b)

(i)

(ii)

12 −= py

2

1=p

4

2

5 (a)

Refer to the graph Mode = 37

3

(b)

m = 29.5 +

10024

2 (10)30

−

=38.17

3

y ¼ π ½ π ¾ π π x O -1 Shape of sine curve Modulus Amplitude or period Translation

P1

P1

P1

P1

K1

6 N1

N1

8

K1

N1

P1

Lower boundary OR

10024

2 (10)30

−

5

QUESTION NO.

SOLUTION MARKS

6

(a)

(i) 3PR y x= +��

(ii) 4 3QS x y= − +��

2

(b)

3PT my mx= +��

STPSPT += = )34)(1(3 yxny −−+

= ynxn 3)44( +−

Compare coefficient of x and y

4

5n m= =

5

N1

N1

7

N1

K1

K1

N1

N1

6

19.5 29.5 39.5 49.5

5

20

30

0

10

15

marks

Number of students

Correct both axes (Uniform scale) K1 All points are plotted correctly N1

9.5 59.5 69.5

25

7

* y-intercept= 5 must be correct.

QUESTION NO.

SOLUTION MARKS

7

(a)

2500

4dA

rdr r

ππ= −

2500

4 0rr

ππ − =

r = 5 A = 678.58 m2

3

(b) (i)

(ii)

Finding area of trapezium

A1 = )1)(45(*2

1 + or A1 = 1

0

2

25

− x

x

Integrate dxxx )5( 2∫ −

A2 =

−

32

5 32 xx

Using limits ∫1

0 into A2

Area = A1 – A2 = 3

7@

3

12 @ 2.333 unit2

4

Integrate

dxxx 22 )5(∫ −π

+−

54

10

3

25 543 xxxπ

Use the limits ∫5

0into

+−

54

10

3

25 543 xxx

Volume generated = π6

625@ π

6

1104 @ 104.17π unit3

Note: OW − 1 once only for correct answer without showing the process of intergration.

3

N1

K1

K1

K1

K1

K1

N1

K1

K1

N1

10

8

Q8

2 3 4 5

0.2

×

0

×

x − 2

log10 y

×

2−x 3 4 5 7 8 10 log y

0.86 0.94 1.04 1.24 1.33 1.51

(a) Each set of values correct (log10 y must be at least 2 decimal places) N1, N1

Y = mX + c log 10 y = ( 2−x )log 10 b + log 10 a K1 where Y = log 10 y, X = (x − 2), m = log 10 b and c = log 10 a (c) log 10 b = gradient

log 10 b = 1.46 0.57

9.4 0

−−

= 0.09468 K1

b = 1.244 N1 log 10 a = Y-intercept log 10 a = 0.57 K1 a = 3⋅715 N1

Correct both axes (Uniform scale) K1 All points are plotted correctly N1 Line of best fit N1

1 6 7 8

×

×

10

N1

N1

0.4

0.6

0.8

1.0

1.2

1.4

1.6

×

9

QUESTION NO.

SOLUTION MARKS

9 (a) (i)

(ii)

3−r (r − 3)2 + 42 = r2 Using Pythagoras theorem on ∆BMO

r =6

25 (or 4.1667)

3

(b)

4.167 3 1.167

,

1 4tan

2 1.1671

1.287 rad.2

2.574 rad.

OM cm

let AOBθ

θ

θ

θ

= − == ∠

=

=

=

3

Area of sector OAB =21

(6

25)2 (2.574) = 22.34 cm2

Area OAB∆ =21

(8)( )36

25 − = 4.667 cm2

Area of shaded region =22.34 − 4.667 = 17.673 cm2

4

K1

P1

N1

K1

P1

N1

10

K1

K1

N1

K1

10

QUESTION NO.

SOLUTION MARKS

10 (a)

(b)

(c)

1

3

12

3

CD ABm m

y x

= =

= −

m = 6

3

3

1 3(3)

3 8

ADm

c

y x

= −

− = − +

= − +

Solving the equations y = −3x + 8 and 3 4 0x y− + = A(2, 2)

5

2 2

( 1) 41

3 8

11 3 20 0

y y

x x

x y x y

− − −× = −− −

+ − − + =

2

10

N1

K1

K1

K1

N1

N1

N1

K1

K1

N1

11

QUESTION

NO. SOLUTION MARKS

11

(a) (i)

(ii)

3

, 15

p p q= + =

52

( 0)5

P X = =

or IE

0.01024= ( 4) ( 4) ( 5)P X P X P X≥ = = + =

4 1 5

54

3 2 3

5 5 5C

= +

= 0.337

3

(b) (i)

(ii)

30 35 60 35

or 10 10

− −

( )0.5 2.5 1 ( 0.5) ( 2.5)P Z P Z P Z− ≤ ≤ = − ≥ − ≥

or 1 − 0.3085 − 0.00621 or R(−0.5) − R(2.5) = 0.6853 or 0.68525 Number of pupils = P( 60) 483X ≥ × = 3

7

N1

K1

K1

10

K1

N1

K1

K1

K1

N1

N1

12

QUESTION NO.

SOLUTION MARKS

12 (a)

(b)

(c)

(d)

v = 15 ms−1

1

a = 2t − 8 and a = 0 or dv

dt = 0

t = 4 vmin = −1 ms−1

3

(t − 3)(t − 5) > 0 0 ≤ t < 3, t > 5

2

324 15

3

ts t t= − +

| s3 − s2 | OR 3

2v dt∫

2

18 163

−

OR 3 3

2 23 24(3) 15(3) 4(2) 15(2)

3 3

− + − − +

11

3 m

4

10

K1

K1

K1

K1

N1

N1

P1

K1

N1

K1

13

QUESTION NO.

SOLUTION MARKS

13

(a) (i)

(ii)

(b)

(c) (i)

(ii)

Electricity : I = 2.1

1001.8

× = 116.7

Telephone and internet : I = 100

2

116.7(4) 133.3(2) 100(3) 120(6)

15

116.9

I+ + +=

=

3

110

210100

231

RM

RM

×

=

110 110(116.7 )(4) 133.3(2) 100(3) (120 )(6)

100 10015

124.8

× + + + ×

=

5

10

P2, 1

N1

K1

K1

K1

K1 K1

N1

N1

Any one of the price indices correct.

Any one of the price indices formula correct.

14

QUESTION

NO. SOLUTION MARKS

14 (a)

(b)

(c)

(d)

1

28 (14) (10) sin2

QRS= ∠

23 35'oQRS∴ ∠ =

2

2 2 2( ) 10 14 2(10)(14)cos23 35'

6.276

oQS

QS

= + −

=

2

6.276 7.5

sin sin30QPS=

∠ �

∠QPS = 24°44′ ∠PQS = 125°16′

3

Area of quadrilateral

128 (7.5)(6.276)sin125 15'

2

= 47.22

o

PQRS

= +

3

10

K1

N1

K1

N1

N1

K1

N1

K1 K1

N1

15

y

Answer for question 15

1 2 3 4 5 6 7 0 8

1

2

7

6

5

8

4

3

(3, 4) •

(a) I. 2y x≤

II. 7x y+ ≥ III. 3y x≥ −

(b) Refer to the graph, 1 or 2 graph(s) correct 3 graphs correct Correct area (c) i) xmin = 3 ii) max point (3, 4) k = 100x + 80y Maximum Profit = RM 100(3) + RM 80(4) = RM 620

10

N1

N1

N1

N1

N1

N1

K1

N1

K1

N1

R

matematik tambahan kertas 1 dua jam · 3472/1 matematik tambahan kertas 1 sept 2010 2 jam . sulit...

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