tutor mansor burhan – mathematics & additional mathematics...3472/1 2012 hak cipta zon a kuching...

SULIT 1 3472/1

3472/1 2012 HAK CIPTA ZON A KUCHING SULIT

M,

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM 2012

Kertas soalan ini mengandungi 16 halaman bercetak

For examiner’s use only

Question Total Marks Marks

Obtained 1 3

2 4

3 3

4 3

5 2

6 3

7 3

8 3

9 3

10 3

11 4

12 3

13 3

14 3

15 3

16 3

17 4

18 3

19 4

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 2012 2 Jam

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

SULIT 3472/1

[ Lihat sebelah 3472/1 SULIT

2 The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah biasa digunakan.

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

SULIT 3472/1

3472/1 2012 HAK CIPTA ZON A KUCHING [Lihat sebelah SULIT

3

STATISTICS STATISTIK

TRIGONOMETRY TRIGONOMETRI

7 i i

i

w II

w

∑=∑

8 !

( )!n

rn

Pn r

=−

9 !

( )! !n

rn

Cn r r

=−

10 P(A∪ B) = P(A) + P(B) − P(A∩ B)

11 P(X = r) = n r n rrC p q

− , p + q = 1

12 Mean / Min µ = np

13 npqσ =

14 z = X − µ

σ

1 x = N

x∑

2 x = ∑∑

f

fx

3 σ = 2( )x x

N

−∑ = 2

2x xN

−∑

4 σ = 2( )f x x

f

−∑∑

= 2

2fx xf

−∑∑

5 m = 12

m

N FL C

f

− +

6 1

0

Q100

QI = ×

8 tan 2A = 22 tan

1 tan

A

A−

9 sin (A± B) = sin A cos B ± cos A sin B

sin (A± B) = sin A kos B ± kos A sin B

10 cos (A± B) = cosA cosB ∓ sinA sinB kos (A± B) = kos A kos B ∓ sin A sin B

11 tan (A± B) = BA

BA

tantan1

tantan

∓

±

12 C

c

B

b

A

a

sinsinsin==

13 a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc kos A 14 Area of triangle / Luas segi tiga

= Cabsin2

1

1 Arc length, s = rθ Panjang lengkok, s =jθ

2 Area of sector , A = 21

2r θ

Luas sector, L = 21

2j θ

3 sin2 A + cos2 A = 1 sin2 A + kos2 A = 1

4 sec2 A = 1 + tan2 A sec2 A = 1 + tan2 A

5 cosec2 A = 1 + cot2 A kosek2 A = 1 + kot2 A

6 sin 2A = 2 sin A cos A sin 2A = 2 sin A kos A

7 cos 2A = cos2 A – sin2 A = 2 cos2 A − 1 = 1 − 2 sin2 A kos 2A = kos2 A – sin2 A = 2 kos2 A − 1 = 1 − 2 sin2 A


SULIT 3472/1

3472/1 ZON A KUCHING 2012 [ 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:

−= 22

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 an arrow diagram representing the mapping of x onto f(x). State Rajah 1 menunjukkan satu gambar rajah anak panah yang mewakili pemetaan x kepada f(x). Nyatakan

(a) the range of the relation, julat hubungan hubungan tersebut,

(b) the object of 3, objek kepada 3,

(c) Using function notation, write a relation for the above mapping. Dengan menggunakan tatatanda fungsi, tulis satu hubungan untuk pemetaan di atas. [3 marks] Answer / Jawapan: (a)

(b)

(c)

2 Given the functions 23: −→ xxf and 32: 2 −→ xxg Diberi fungsi 23: −→ xxf dan fungsi 32: 2 −→ xxg

Find Carikan

(a) )4(1−f , (b) )(xgf . [4 marks]

[4 markah] Answer / Jawapan: (a)

(b)

x f(x)

4

9

16

25

2

3

4

5

Diagram 1 Rajah 1

3

1

For examiner’s

use only

[3 markah]

4

2


SULIT 3472/1


6

3 Given the function 2( ) , (4) 16f x ax bx f= + = − , and (2) 6f = − , find the value of a and of b. Diberi fungsi 2( ) , (4) 16f x ax bx f= + = − , dan (2) 6f = − , cari nilai a dan nilai b

[3 marks] [3 markah]

Answer / Jawapan: 4 The quadratic equation 23 0x kx h− + = has roots −4 and 3. Find the value of k and of h. Persamaan kuadratik 23 0x kx h− + = mempunyai punca-punca −4 dan 3. Cari nilai k dan nilai h.


Answer / Jawapan:

For examiner’s

use only

3

3

3

4


SULIT 3472/1


7

5 Diagram 5 shows the graph of a quadratic function ( ) ( )23 5 27f x x= − − . Rajah 5 menunjukkan graf bagi fungsi kuadratik ( ) ( )23 5 27f x x= − − .

Find Cari

(a) the equation of the axis of symmetry of the curve, persamaan paksi simetri lengkung itu,

(b) the value of h. nilai h.

[2 marks] [2 markah] Answer / Jawapan:

(a)

(b) ___________________________________________________________________________

6 Find the range of the values of x for ( ) 156 −≤−xx . [3 marks] Cari julat nilai x bagi ( ) 156 −≤−xx . [3 markah] Answer / Jawapan:

x 0

.

Diagram 5 Rajah 5

f(x)

2

( ) ( )23 5 27f x x= − −

. h

3

6

For examiner’s

use only

2

5


SULIT 3472/1


8

7 Solve the equation 3 5 11

24

xx

−+= . [3 marks]

Selesaikan persamaan 3 5 11

24

xx

−+= . [3 markah]

Answer / Jawapan:

8 Solve the equation 4 43 log ( 2) logx x+ − = . [3 marks] Selesaikan persamaan 4 43 log ( 2) logx x+ − = . [3 markah]

Answer / Jawapan:

9 The first three terms of an arithmetic progression are 3,2,7− , find Tiga sebutan pertama suatu janjang aritmetik ialah 3,2,7− , cari

(a) the common difference of the progression, beza sepunya janjang itu,

(b) the sum of the first 15 terms. hasil tambah 15 sebutan pertama. [3 marks] [3 markah] Answer / Jawapan: (a) (b)

3

7

3

9

3

8

For examiner’s

use only


SULIT 3472/1


9

10 The fourth term of a geometric progression is 216 and the sum of the fourth and the fifth term is 864. Find

Sebutan keempat suatu janjang geometri ialah 216 dan hasil tambah sebutan keempat dan sebutan kelima adalah 864. Cari

(a) the first term, sebutan pertama,

(b) the common ratio, nisbah sepunya,

of the progression. janjang itu.


Answer / Jawapan:

(a) (b)

11 Three consecutive terms of a geometric progression are x, −12, 6, ... . Find Tiga sebutan berturut-turut suatu janjang geometri ialah x, −12, 6, ... . Cari

(a) the value of x, nilai x,

(b) the sum to infinity of the progression. hasil tambah ketakterhinggaan janjang itu. [4 marks] [4 markah]

Answer / Jawapan:

(a)

(b)

3

10

4

11

For examiner’s

use only


SULIT 3472/1


10

12 The variables xand y are related by the equation xbyax =+ 22 , where a, b are constants.

Diagram 12 shows the straight line obtained by plotting x

y2against x .

Pembolehubah-pembolehubah x dan y dihubungkan oleh persamaan xbyax =+ 22 , dengan keadaan a, b adalah pemalar. Rajah 12 menunjukkan garis lurus yang diperoleh dengan

memplot x

y2melawan x.

Diagram 12 Rajah 12

Find the value of a and of b. [3 marks]

Cari nilai a dan nilai b. [3 markah]

Answer / Jawapan :

13 The coordinates of points P , Q and R are ( 2,1)− , (3, 4)− and (8,12) respectively. Find the

equation of the straight line that passes through point R and is perpendicular to the linePQ . [3 marks] Koordinat bagi titik P, Q dan R adalah ( 2,1)− , (3, 4)− and (8,12) masing-masing. Cari

persamaan garis lurus yang melalui titik R dan berserenjang dengan garis lurus PQ. [3 markah]

Answer / Jawapan:

3

12

For examiner’s

use only

3

13

x

0

4

( )6, 1− •

x

y2


SULIT 3472/1


11

14

Diagram 14 Rajah 14 Diagram 14 shows a straight lineBCwhich intersects the x-axis at pointC and point D is the midpoint of .BC Given the equation of BC is2 7 12 0,x y+ − = find the distance ofAD . [3 marks]

Rajah 14 menunjukkan satu garis lurus BC yang bersilang dengan paksi-x di titik C dan titik D ialah titik tengah BC. Diberi persamaan BC ialah 2 7 12 0,x y+ − = cari jarak AD.

[3 markah] Answer / Jawapan:

15 OPQR is a parallelogram such that 3 5OP i j= +��

ɶ ɶand 4 3OR i j= − +��

ɶ ɶ. Find

OPQR ialah sebuah segi empat selari dengan 3 5OP i j= +��

ɶ ɶ dan 4 3OR i j= − +

��

ɶ ɶ. Cari

(a) OQ��

,

(b) the unit vector in the direction of OQ��

.

vektor unit dalam arah OQ��

. [3 marks]

[3 markah] Answer / Jawapan:

(a) (b)

3

15

3

14

For examiner’s

use only

(3,8)A• ( 8,4)B −

D

0 x

y

C


SULIT 3472/1


12

16 Diagram 16 shows OA a=��

ɶ, OB b=��

ɶ, OP��

and PT��

drawn on a square grid. .

Rajah 16 menunjukkan OA a=��

ɶ, OB b=��

ɶ, OP��

dan PT��

dilukis pada satah grid segi empat sama.

Express in terms of a

ɶ and b

ɶ:

Ungkapkan dalam sebutan aɶ

dan bɶ

:

(a) OP��

,

(b) PT��

. [3 marks] [3 markah] Answer / Jawapan: (a) (b) .

17 Solve the equation 3)cos(sin4 2 =+ xx for .3600 °≤≤° x [4 marks]

Selesaikan persamaan 4(sin x + kos x)2 = 3 untuk .3600 °≤≤° x [4 markah]

Answer / Jawapan:

4

17

3

16

Diagram 16 Rajah 16

aɶ

bɶ

P

T

O

B

A


SULIT 3472/1


13

18 Diagram 18 shows a sector OPQ with centre O. Rajah 18 menunjukkan sebuah sektor OPQ berpusat O.

Diagram 18

Rajah 18

Given that the area of the sector POQ is 358.43 cm2 and the angle of the sector POQ is 0.354 radian. Using 3.142,π = find Diberi luas sektor POQ ialah 358.43 cm2 dan sudut sektor POQ ialah 0.354 radian. Guna π = 3.142, cari

(a) ∠ POQ in degrees, ∠ POQ dalam darjah,

(b) the radius of the sector POQ. jejari sektor POQ. [3 marks]

[3 markah]

Answer / Jawapan: (a)

(b)

19. Given that 22 4 3y x x k= + + has a minimum value of 13, find the value of k Diberi 22 4 3y x x k= + + mempunyai 13 sebagai nilai minimum, cari nilai k.

[ 4 marks] [4 markah] Answer / Jawapan:

4

19

3

18

For examiner’s

use only

P

O

Q

0.354 rad


SULIT 14 3472/1

3472/1 ZON A KUCHING 2012 Lihat sebelah SULIT

20 Given that 216 18 5y x x= − + , Diberi 216 18 5y x x= − + ,

(a) find dy

dx in terms of x

cari dy

dx dalam sebutan x.

(b) If x changes from 2.01 to 2, find the corresponding small change in y. Jika x berubah dari 2.01 kepada 2, cari perubahan kecil dalam y yang sepadan. [3 marks] [3 markah] Answer / Jawapan:

(a) (b)

21 Given that

4

0( ) 3f x dx=∫ and

4

0( ) 5g x dx=∫ . Find

Diberi

4

0( ) 3f x dx=∫ dan

4

0( ) 5g x dx=∫ . Cari

(a)

4 0

0 4( ) ( ) f x dx g x dx×∫ ∫

(b)

4

0[3 ( ) 1]f x dx−∫ .

[3 marks] [3 markah] Answer / Jawapan:

(a) (b)

22 The first quartile and standard deviation of a set of number 1 2 3 4, , , ,..., nx x x x x are 5 and 9 respectively. Find the first quartile and variance of the new set of number 1 2 3 43, 3, 3, 3,..., 3.nx x x x x− − − − − [3 marks] Kuartil pertama dan sisihan piawai bagi satu set nombor 1 2 3 4, , , ,..., nx x x x x ialah 5

dan 9. Cari kuartil pertama dan varians bagi satu set nombor yang baru 1 2 3 43, 3, 3, 3,..., 3.nx x x x x− − − − − [3 markah]

Answer / Jawapan:

3

20

3

21

For examiner’s

use only

3

22


SULIT 15 3472/1


23 How many 4-digit number that can be formed from the digits 1, 3, 5, 7, 8 and 9 without repetition.

Berapa 4-digit nombor yang boleh dibentuk daripada digit 1, 3, 5, 7, 8 dan 9 tanpa ulangan.

(a) if there is no restriction, jika tanpa apa-apa syarat,

(b) 4 digit even number greater than 7000. nombor genap 4-digit yang lebih daripada 7000. [4 marks] [4 markah]

Answer / Jawapan:

(a) (b)

_________________________________________

24 The probability of Amin and Atan wins a game of tennis are 1 3

and 7 5

respectively.

Find the probability that

Kebarangkalian Amin and Atan menenangi suatu pertandingan tennis masing-masing

ialah 1 3

dan 7 5

, cari kebarangkalain bahawa

(a) both of them win, kedua-duanya menang,

(b) only one of them wins. hanya seorang yang menang. [3 marks] [3 markah]

Answer / Jawapan:

(a) (b)

For examiner’s

use only

3

24

4

23


SULIT 16 3472/1


25 X is a continuous random variable of a normal distribution with a mean of 12.45 and a variance of 9.

X ialah pemboleh ubah rawak selanjar bagi suatu taburan normal dengan min 12.45 dan varians 9.

Find Cari

(a) the value of X when the z-score is 1.35, nilai X apabila skor-z ialah 1.35,

(b) the value of k when P(−k < z < k) = 0.7881. nilai k apabila P(−k < z < k) = 0.7881.


Answer / Jawapan:

(a) (b)

END OF QUESTION PAPER

4

25

For examiner’s

use only


3472/1 Matematik Tambahan Kertas 1 2 jam Sept 2012

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2012

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 -2012 ZON A

No Solution and marking scheme Sub Marks Total Marks 1. (a) {2, 3, 4, 5}

(b) 9

(c) ( )f x x=

1 1 1

3

2. (a) 2)4(1 =−f

3

2+= yx

(b) 218 24 5x x− + 3)23(2)( 2 −−= xxgf

2

B1 2

B1

4

3.

b = − 12

and a = −2

(−3 − 2b) + 4b = −4 OR Other method or b = − 12

or a =

−2 4a + 16b = −16 or 2a + 4b = −6 or equivalent

3

B2

B1

3

4. k = −3 and h = −36 k = −3 or h = −36

( 4)(3)3

h− = or 4 33

k− + =

3

B2

B1

3


3

No Solution and marking scheme Sub Marks Total Marks 5.

(a) 5x = (b) 8h =

1 1

2

6.

1 1

3 2x≤ ≤

1

3

1

2

(3 1)( 2 1) 0x x− − ≤

3

B2

B1

3

7.

3

5x =

3 5 2( 1)

3 5 2 2

2 2x x

x x

− − +

− = − −

=

3

B2

B1

3

8.

128

63x =

3

4 4 4

64( 2)

log 4 log ( 2) log

x x

x x

− =

+ − =

3

B2

B1

3

9. (a) 5 (b) 480

[ ]15 15 2( 3) (15 1)(5)2S = − + −

1 2

B1

3


4

No Solution and marking scheme Sub Marks Total Marks 10. (a) 8a =

(b) 3r =

3

3

(1 ) 648

216

ar r

ar

+ =

1 2

B1

3

11. (a) 24 (b) 16

24

11

2+

1

2

B1

3

12.

1 1

and 8 4

a b= − =

1 1

or 8 4

a b= − =

2 1y a

xx b b

= − + or 12

a

b− = or 1 4

b=

3

B2

B1

3

13.

4y x= + 12 1( 8)y x− = − or any correct substitution to find value of c 1PQm = − or m × (−1) = −1 or m = 1

3

B2

B1

3

14. 7.211 or 52 ( )1, 2D − C(6, 0)

3

B2

B1

3


5

No Solution and marking scheme Sub Marks Total Marks

15. (a) 8OQ i j= − +

��

ɶ ɶ

(b) 8

65

i j− +ɶ ɶ

65 or 2 28

( 1) 8

i j− +

− +ɶ ɶ

1 2

B1

3

16. ( ) +3a OP b a=

��

ɶ ɶ

( ) b a b− −

ɶ ɶ

= 3 2

PT PO OT

a b a

= +− − +

��

ɶ ɶ ɶ

or equivalent

1

2

B1

3

17. 97.24 ,172.76 , 277.24 , 352.76x = ° ° ° ° x = 97.24°, 172.76°

4

12sin −=x

4

3coscossin2sin 22 =++ xxxx or

4

1cossin2 −=xx

4

B3

B2

B1

4

18. (a) 20.28° (b) 45

358.43 = 21

(0.354)2

r or r = 2 358.43

0.354

×

1 2

B1

3


6

No Solution and marking scheme Sub Marks Total Marks 19.

5k = 213 2( 1) 4( 1) 3k= − + − = 1x = −

4 4 0dy

xdx

= − =

4

B3

B2

B1

4

20.

(a) 32 18dy

xdx

= −

(b) −0.4632 46.32 ( 0.01)yδ ≈ × − or δx = −0.01

1

2

B1

3

21. (a) −15 (b) 5

[ ]

4 40

03 ( ) or or 3(3)f x dx x∫

1 2

B1

3

22. First quartile = 2 Variance = 81

2(9)

1

2

B1

3

23. (a) 360 6 4 or 6 5 4 3P × × ×

(b) 24

422 1 or 2 4 3 1× × × × ×P

2

B1 2

B1

4


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

(a) 3 1 3

5 7 35× =

(b) 4

7

2 1 3 6

5 7 5 7 × + ×

1 2

B1

3

25. (a) 16.5

1.35 = 12.45

3

X −

(b) k = 1.248 P(z > k) = 0.1060

2

B1 2

B1

4


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