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SULIT 1

3472/1 ZON A KUCHING 2011 SULIT

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM 2011

Kertas soalan ini mengandungi 16 halaman bercetak

For examiner’s use only

Question Total Marks Marks

Obtained 1 2

2 3

3 4

4 3

5 3

6 3

7 3

8 4

9 3

10 3

11 4

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 2011 2 Jam

http://www.chngtuition.blogspot.comhttp://tutormansor.wordpress.com/

SULIT 3472/1

ZON A KUCHING 2011 [ Lihat sebelah 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 ,

2 v

uy = ,

2

du dvv udy dx dx

dx v

−= ,

dx

duv

dx

dvu

dx

dy +=

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 ZON A KUCHING 2011 [ 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 2011 [ Lihat sebelah SULIT

4

THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION 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

dzzfzQ )()( P(X > k) = Q(k)

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

Q(z)

z

f

O k


SULIT 3472/1


5

Answer all questions.

1. Diagram 1 shows the graph of the function f(x) = (x − 1)2.

Diagram 1 State (a) the type of relation, (b) the value of k.

[2 marks] Answer : (a)

(b)

2. The function 1f − is defined by 1 3( )

2f x

x− =

−, x k≠ .

(a) State the value of k. (b) Find the function f . [3 marks] Answer : (a)

(b)

3

2

2

1

For examiner’s

use only

x

f(x) = (x − 1)2

0

f(x)

k


SULIT 3472/1


6

3. Given the function : 2 3f x x→ − and composite function 2: 6 4 1fg x x x→ − + . Find (a) g(x), (b) the value of gf(−1).

[4 marks] Answer : (a)

(b)

4. Given the equation 2 2x x k+ = − has two distinct roots, find the range of values of k.

[3 marks] Answer :

For examiner’s

use only

4

3

3

4


SULIT 3472/1


7

5. Diagram 5 shows the graph of function y = (x − 2)2 + q, where q is a constant. Given that the line y = 3 is the tangent to the curve.

Diagram 5

(a) State the equation of axis of symmetry.

(b) State the value of q.

(c) Find the value of k. [3 marks] Answer : (a) (b) (c) ___________________________________________________________________________

6. Find the range of values of x which satisfies 4x − 5x2 ≤ −1 [3 marks] Answer :

3

5

3

6

For examiner’s

use only

x

y

O

k y = (x − 2)2 + q

y = 3


SULIT 3472/1


8

7. Solve the equation 3 164 8 4x x x+ += . [3 marks] Answer :

8. Given that 2log m r= and 2log n t= , express 38log16

m

n

in terms of r and / or t.

[4 marks] Answer :

9. If 3, x, y and 15 are consecutive terms of an arithmetic progression, find the value of x

and y. [3 marks]

Answer :

3

7

3

9

4

8

For examiner’s

use only


SULIT 3472/1


9

10. The third and sixth terms of a geometric progression are 1 and 8 respectively. Find the fi rst term and common ratio of the progression. [3 marks]

Answer :

11. Express 0.363636... in the form of p

qwhere p and q are positive integers. Hence express

2.363636... as a single fraction. [4 marks] Answer :

3

10

4

11

For examiner’s

use only


SULIT 3472/1


10

12. The variables x and y are related by the equation 227 xxy −= . A straight line graph

is obtained by plotting x

yagainst x, as shown in Diagram 12.

Diagram 12 Find the value of h and of k. [3 marks] Answer :

13. Diagram 13 shows a quadrilateral PQRS.

Diagram 13

Given the area of the quadrilateral is 80 unit2, find the value of a. [3 marks] Answer :

3

13

3

12

For examiner’s

use only

(7, h)

x

y

(k, 1)

O x

x

y

O

Q(4a, 3a)

R(6, −1)

S(−2, −4)

P(−5, 4)


SULIT 3472/1


11

14. Given A(−5, k), B(−1, 6), C(1, −5). Find the possible values of k if AB = 2BC. [3 marks] Answer :

15. Diagram 15 shows vector OA��

drawn on a Cartesian plane. Diagram 15

(a) Express OA��

in the form x

y

.

(b) Find the unit vector in the direction of OA��

. [3 marks] Answer : (a)

(b)

3

15

3

14

For examiner’s

use only

2

4

6

0

2

4 6 8 10 12 x

A

y


SULIT 3472/1


12

16 . Given that (2 1) 3k= − +a i j and 4 5= +b i j .

Find the value of k if 2 3+a b is parallel to y-axis. [3 marks] Answer :

. ___________________________________________________________________________

17. It is given that 12

5tan =θ and θ is an acute angle.

Find the value of each of the following (a) ( )tan θ− ,

(b) θθ sinsec + . [4 marks] Answer : (a) (b)

4

17

For examiner’s

use only

3

16


SULIT 3472/1


13

18.

Diagram 18

Diagram 18 above shows a sector POQ with centre O. The perimeter of sector POQ is 40 cm. Given that the radius of the sector is 15 cm, find the value ofθ , 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

O

P

Q

θ


SULIT 14 3472/1

3472/1 ZON A KUCHING 2011 Lihat sebelah SULIT

20. Given 5

2( ) 6f x dx=∫ and

2

0( )f x dx−∫ = 2. Find ( )0

5f x dx∫ . [3 marks]

Answer :

___________________________________________________________________________ 21. Diagram 21 shows the graph of y2 = (x − 3) and x = 5. Diagram 21

Find the volume generated when the shaded region is rotated through 360° about x-axis. [3 marks]

Answer :

22. Given that the mean and the standard deviation of a set of numbers are 7 and 2. If each of

the numbers is multiplied by 3, find

(a) the mean, (b) the variance of the new set of numbers. [3 marks] Answer :

3

20

3

21

3

22

For examiner’s

use only

3 5

x

y y2 = x − 3

O


SULIT 15 3472/1


23. The number of ways in which a group of 4 men and 3 women can be seated in a row of (a) 8 chairs, (b) 8 chairs if the first two chairs in the row are occupied by the men. [4 marks] Answer :

___________________________________________________________________________ 24. A box contains 40 marbles. The colours of the marbles are yellow and blue. If a marble

is drawn from the box, the probability that a yellow marble drawn is 2

5.

Find the number of blue marbles that have to be added to the box such that the

probability of obtaining a blue marble becomes 5

7. [3 marks]

Answer :

For examiner’s

use only

3

24

4

23


SULIT 16 3472/1


25. The continuous random variable X is distributed normally with mean µ and variance 25. Given that ( 20) 0.7881P X < = , find the value of µ . [4 marks]

Answer :

END OF QUESTION PAPER

4

25

For examiner’s

use only


3472/1 Matematik Tambahan Kertas 1 2 jam Sept 2011

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011

MATEMATIK TAMBAHAN

Kertas 1

Dua jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

Skema Pemarkahan ini mengandungi 7 halaman bercetak

MARKING SCHEME


2

MARKING SCHEME FOR PAPER 1 -2011 ZON A

No Solution and marking scheme Sub Marks Total Marks

1.

(a) many to one relation (b) 1

1 1

2

2. (a) k = 2

(b) 3 2

( ) , 0.x

f x xx

+= ≠

3

2y

x=

−

1

2

B1

3

3. (a) 2( ) 3 2 2g x x x= − + 22 ( ) 3 6 4 1g x x x− = − + @ f −1(fg(x)) = f −1(6x2 −4x + 1) (b) 87 ( 1) 5f − = −

2

B1 2

B1

4

4. k > 1

4 4k− > − or 4 < 4k

2(2) 4(1)( ) 0k− − >

3

B2

B1

3


3

No Solution and marking scheme Sub Marks Total Marks 5.

(a) x = 2 (b) q = 3 (c) k = 7

1 1 1

3

6.

1

, 15

x x≤ − ≥

(5 1)( 1) 0x x+ − ≥

3

B2

B1

3

7. x = −16 6 18 5 2x x+ = + 26(x+3) = 23x 22x+2

3

B2

B1

3

8.

4 3

3

r t− −

2 24log 2 3log

3

r n− −

3

2 2 2log log 16 log3

m n− −

2 3

32

log16

log 2

m

n

4

B3

B2 B1

4

9. x = 7, y = 11 Solving equation or x = 7 @ y = 11 x − 3 = y − x or x – 3 = 15 – y or d = 4

3

B2

B1

3

− 1

5

1

+ +

−


4

No Solution and marking scheme Sub Marks Total Marks

10.

r = 2 , a = 1

4

r = 2 @ a = 1

4

ar2 = 1 ------------ (1) or ar5= 8 -------------(2)

3

B2

B1

3

11.

26

11

2.363636… = 2+ 0.363636… = 2 + 4/11

S∞ = 0.36

1 0.01− @

4

11

a = 0.36 and r = 0.0036/0.36 = 0.01

4

B3

B2

B1

4

12. 7−=h , k = 3 h = −7 @ k = 3

xx

y27−= or 72 +−= x

x

y

3

B2

B1

3

13. a = 2

( ) 8054532

1 =+a

[ ]15( 4) ( 2)( 1) 6 3 4 4 ( 2)4 6( 4) 4 ( 1) ( 5)(3 ) 80

2a a a a− − + − − + × + × − − − − − − − − =

3

B2

B1

3

14. k = −16, 28

484)6( 2 =−k @ 226 ±=−k @ 0)28)(16( =−+ kk

2 2 2 2( 5 ( 1)) ( 6) 2 ( 1 1) (6 ( 5))k− − − + − = − − + − −

3

B2

B1

3


5No Solution and marking scheme Sub Marks Total

Marks 15. .

(a) OA��

= 8

4

or 8 4+i j

(b) 81480

@

8 4

80

+i j

2 28 4OA = +��

= 54 @ 80

1

2

B1

3

16.

2

5−=k

0104 =+k ba 32 + = ( ) jik 21104 ++

3

B2

B1

3

17. (a) ( )tan tanθ θ− = −

5

12= −

(b) 73

1156

or 156

229

12

13sec =θ or

13

5

12

13sinsec +=+ θθ

13

5sin =θ @

13

12cos =θ

1

3

B2

B1

4

18.

23

radθ =

10 15θ= sPQ = 10 cm

3

B2

B1

3


6No Solution and marking scheme Sub Marks Total Marks 19.

0.4y∂ ≈ (12(1) 4)(0.05)y∂ ≈ −

12 4 or 1.05 1 0.05dy

x xdx

= − ∂ = − =

1

B2

B1

3

20. −4

( ) ( )2 2

0 5

f x dx f x dx−∫ ∫

( ) ( )2 5

0 2

f x dx f x dx+∫ ∫

3

B2

B1

3

21. 2

2 25 3

3 5 3 35 5

− × − − ×

2 5

3

32

xx

−

3

B2

B1

3

22. variance 36, mean 21= = Variance = 36 @ Mean = 21 Variance = 2 23 2× @ Mean =3 7× @ SD = 3×2 @ σ = 6

3

B2

B1

3

23. (a) 40320

8

7P

(b) 8640 4 6

52P P×

2

B1

2

B1

4


7No Solution and marking scheme Sub Marks Total Marks 24.

16

168 7 200 5 x x+ = + @ 24 5

40 7

x

x

+ =+

n(B) = 24 + x and n(S) = 40 + x

3

B2

B1

3

25. 16µ =

20

0.85

µ− = (from table)

20

P(Z ) 0.21195

µ−≥ =

P( 20) 0.2119X ≥ =

4

B3

B2

B1

4


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