Spm Trial 2014 Addmaths Qa Kedah

8/20/2019 Spm Trial 2014 Addmaths Qa Kedah

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SULIT

Kertas soalan ini mengandungi 22 halaman bercetak dan 2 halaman kosong.

1. Tulis nama dan tingkatan anda pada

ruangan yang disediakan.

2. Kertas soalan ini adalah dalamdwibahasa.

3. Soalan dalam bahasa Inggeris

mendahului soalan yang sepadan

dalam bahasa Melayu.

4.

Calon dibenarkan menjawab

keseluruhan atau sebahagian soalan

sama ada dalam bahasa Inggeris atau

bahasa Melayu.

5. Calon dikehendaki membacamaklumat di halaman belakang kertas

soalan ini.

MODUL PENINGKATAN PRESTASI TINGKATAN 5TAHUN 2014

MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)

ADDITIONAL MATHEMATICS ( MODULE 2 )

Kertas 1

Ogos 2014

2 jam Dua jam

3472 / 1

Untuk Kegunaan Pemeriksa

Soalan MarkahPenuh

MarkahDiperolehi

1 3

2 3

3 4

4 3

5 3

6 3

7 3

8 39 4

10 4

11 2

12 4

13 3

14 2

15 3

16 4

17 4

18 4

19 3

20 3

21 3

22 3

23 3

24 3

25 3

TOTAL 80

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

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

SULIT

http://www.chngtuition.blogspot.com


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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 ac

x a

− ± −

=

2 am × an = a m + n

3 am ÷ a

n = 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 logab = a

b

c

c

log

log

9 T n = a + (n−1)d

10 S n = ])1(2[2

d nan

−+

11 T n = arn – 1

12 S n =r

r a

r

r ann

−

−=

−

−

1

)1(

1

)1( , (r ≠ 1)

13 r

aS −=∞ 1 , r


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SULIT

3

STATISTICS

1 Arc length, s = r θ

2 Area of sector , A = 21

2r θ

3 sin 2 A + cos2 A = 1

4 sec2 A = 1 + tan2 A

5 cosec2 A = 1 + cot2 A

6 sin 2 A = 2 sin A cos A

7 cos 2 A = cos2 A – sin

2 A

= 2 cos2 A − 1

= 1 − 2 sin2 A

8 tan 2 A = A

A2tan1

tan2

−

TRIGONOMETRY

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

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

11 tan ( A ± B) = B A

B A

tantan1

tantan

±

12C

c

B

b

A

a

sinsinsin==

13 a2 = b

2 + c2 − 2bc cos A

14 Area of triangle = C absin2

1

1 x = N

x∑

2 x =∑∑

f

fx

3 σ =( )

2 x x

N

−∑ = 2

2

x N

x−

∑

4 σ =∑

∑ − f

x x f 2)(

= 22

x f

x f −

∑∑

5 m = C f

F N

Lm

−

+ 2

1

6 1

0

100Q

I Q

= ×

7 i i

i

W I I

W

∑=

∑

8)!(

!

r n

nPr n

−=

9 !)!(

!

r r n

nC r

n

−=

10 P( A ∪ B) = P( A) + P( B) − P( A ∩ B)

11 P ( X = r ) = r nr r

n q pC − , p + q = 1

12 Mean µ = np

13 σ npq=

14 Z =σ

X − µ



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Answer all questions.

Jawab semua soalan.

1 Diagram 1 shows the relation between set H and set K .

Rajah 1 menunjukkan hubungan antara set H dan set K.

The relation is defined by the set of ordered pairs .

Hubungan itu ditakrifkan oleh set pasangan tertib .

State

Nyatakan

(a) the value of m and of n .

nilai m dan nilai n .

(b) the type of the relation.

jenis hubungan itu .

[3 marks][3 markah]

Answer/ Jawapan:

(a)

(b)

( ) ( ) ( ){ }2,9 , 1, , , 9m n−

( ) ( ) ( ){ }2,9 , 1, , , 9m n−

For

examiner’s

use only

3

1

1

̶ 2

2

5

3

Diagram 1

Rajah 1

9

Set H Set K

7



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2 The following information refers to the functions f and g and the composite function

.

Maklumat berikut adalah berkaitan dengan fungsi f dan g dan fungsi gubahan .

Find the value of p .

Cari nilai p .

[3 marks]

[3 markah]

Answer/ Jawapan:

1 f g

−

1 f g−

( )1

: 5 3

6:

3

f x x

g x x

f g p−

→ −

→

=

For

examiner’s

use only

3

2



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3 Given that function and . Find the value of

Diberi fungsi dan . Cari nilai bagi

(a) ,

(b) k if

k jika .

[4 marks]

[4 markah]

Answer/ Jawapan:

(a)

(b)

4 The quadratic equation , where p is a constant, has two equal

roots. Find the possible values of p.

Persamaan kuadratik , dengan keadaan p ialah pemalar ,

mempunyai dua punca yang sama. Cari nilai-nilai yang mungkin bagi p .

[3 marks]

[3 markah]Answer/ Jawapan:

( ) 3 f x x k = − ( )1 2 3g x x− = −

( ) 3 f x x k = − ( )1 2 3g x x− = −

( )3g

( )1 1 8g f − =

( )1 1 8g f − =

2 2 3 2 0 x px p+ + − =

2 2 3 2 0 x px p+ + − =

3

4

For

examiner’s

use only

4

3



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5 Diagram 5 shows the graph of a quadratic function , where

p is a constant, has a minimum point at .

Rajah 5 menunjukkan graf fungsi kuadratik , dengan keadaan p

ialah pemalar , mempunyai titik minimum di .

(a) Find the value of p and of q .Cari nilai p dan nilai q .

(b) State the equation of the axis of symmetry of the curve. Nyatakan persamaan paksi simetri bagi lengkung itu.

[3marks]

[3markah]

Answer/ Jawapan:

(a)

(b)

( ) ( )2 5 f x x p= − −

( )3 , q−

( ) ( )2 5 f x x p= − −

( )3 , q−

3

5

f ( x)

x

( − 3 , q )·O

Diagram 5

Rajah 5

For

examiner’s

use only



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6Find the range of values of x for 2

93

2

x x

−− ≤ . [3 marks]

Cari julat nilai x bagi 29

32

x x

−− ≤ . [3 markah]

Answer/ Jawapan:

7 Given that and , express in terms of a and b .

[3 marks]

Diberi dan , ungkapkan dalam sebutan a dan b .

[3 markah]

Answer/ Jawapan:

2log h a= 2log k b=3

8log 16 k h

2log h a= 2log k b=3

8log 16 k h

3

6

3

7

For

examiner’s

use only



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8 Solve the equation :

Selesaikan persamaan :

( )11

7 49343

x− =

[3 marks]

[3 markah]

Answer/ Jawapan:

9 An arithmetic progression consists of 26 terms. Given the first term is 2 and the sum of

the last 8 terms is 532. Find the15th term of the progression.

[4marks]

Suatu janjang aritmetik mengandungi 26 sebutan. Diberi sebutan pertama ialah 2 dan

hasil tambah 8 sebutan terakhir ialah 532 .Cari sebutan ke-15 bagi janjang itu.[4markah]

Answer/ Jawapan:

4

9

3

8

For

examiner’s

use only



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10The first term and second term of a geometric progression is and m respectively.

Sebutan pertama dan kedua suatu janjang geometri masing-masing ialah dan m.

Find

Cari

(a) the values, other than zero, that are not possible for m.nilai-nilai yang tidak mungkin bagi m selain daripada sifar .

(b) sum of the first 5 terms of the geometric progression if m = 9

hasil tambah 5 sebutan yang pertama bagi janjang geometri itu jika m = 9

[4 marks]


(a)

(b)

11Given that

19 3 1 ...

3+ + + + is an infinite series of a geometric progression. Find the

sum to infinity of the series. [2 marks]

Diberi

1

9 3 1 ...3+ + + + ialah satu siri takterhingga bagi suatu janjang geometri. Carihasil tambah hingga sebutan ketakterhinggaan bagi siri itu. [2 markah]

Answer/ Jawapan:

9

3m

9

3m

2

11

4

10

For

examiner’s

use only



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12 Diagram 12 shows a straight line graph obtained by plotting against .

Rajah 12 menunjukkan graf garis lurus yang diperoleh dengan memplot

melawan .

Diagram 12 Rajah 12

The variables x and y are related by the equation . Find the value of

Pembolehubah x dan y dihubungkan oleh persamaan . Cari nilai bagi

(a) a

(b) b

[4marks][4markah]

Answer/ Jawapan:

(a)

(b)

xy2 x

xy

2 x

b y ax

x− =

b y ax

x− =

Q ( 7 , 13 )

xy

x2 O

P( 2 , 3 )

For

examiner’s

use only

4

12



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13 Given that and . A point moves such that .

Find the equation of the locus of point P.

Diberi titik dan titik . Titik bergerak dengan keadaan

. Cari persamaan lokus bagi titik P.[3 marks]

[3 markah]

Answer/ Jawapan:

14 Given that the points ( ),2S k , ( )3, 4T and ( )11,8U are collinear. Find the value of k .

Diberi titik-titik ( ),2S k , ( )3, 4T dan ( )11,8U adalah segaris. Cari nilai bagi k .[2 marks]

[2 markah]

Answer/ Jawapan:

( )3, 2 M ( )0,4 N ( ),P x y : 1 : 2PM PN =

( )3, 2 M ( )0,4 N ( ),P x y

: 1 : 2PM PN =

2

14

3

13

For

examiner’s

use only



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15 Diagram 15 shows the vector drawn on a Cartesian plane.

Rajah 15 menunjukkan vektor AB

yang dilukis pada suatu satah Cartesan.

Diagram 15

Rajah 15

(a) Express in the form

Ungkapkan dalam bentuk

(b) Find the unit vector in the direction of .

Cari vektor unit dalam arah .

[3 marks]


(a)

(b)

AB

AB j yi x +

AB j yi x +

AB

AB

0 2

3

6

6

5

4 53

4

2

1

1 A

B

y

x

For

examiner’s

use only

3

15



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16 Given that , and 3OC OB=

, where

. Find the value of m .

Diberi , dan 3OC OB=

, di mana

. Cari nilai m

[4 marks][4 markah]

Answer/ Jawapan:

17 Solve the equation 3cos 2 8sin 5 x x= − for .

Selesaikan persamaan 3cos 2 8sin 5 x x= − bagi .

[4 marks]

[4 markah]

Answer/ Jawapan:

3 8OA i j= +

3 4 AB i j= +

0OA AB mOC + + =

3 8OA i j= +

3 4 AB i j= +

0OA AB mOC + + =

0 360 x≤ ≤

0 360 x≤ ≤

4

17

For

examiner’s

use only

4

16



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18

Diagram 18 shows a sector AOB of a circle with centre O and radius of 12 cm.

Given that point C is the midpoint of OA. Find

(a) AOB∠ , in radians,(b) the area, in cm², of the shaded region.

[4 marks]

Rajah 18 menunjukkan sebuah sektor AOB bagi sebuah bulatan berpusat O dan

berjejari 12 cm. Diberi bahawa titik C ialah titik tengah bagi garis OA. Cari

(a) AOB∠ , dalam radian , (b) luas , dalam cm² , kawasan berlorek .

[4 markah]

Answer/ Jawapan:

(a)

(b)

4

18

For

examiner’s

use only

O A

B

C

Diagram 18

Rajah 18

12 cm



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19 Given that the mean of a set of seven numbers is 10.(a) Find the sum of the set of the numbers.

(b) A number k is added into the set of the numbers, the new mean is 11.Find the value of k .

[3 marks]

Diberi bahawa min bagi satu kumpulan tujuh nombor ialah 10.(a) Cari hasil tambah bagi kumpulan nombor itu.

(b) Satu nombor k ditambah ke dalam kumpulan nombor itu , min baru ialah11. Cari nilai bagi k .

[3 markah]

Answer/ Jawapan:

(a)

(b)

20 Given that , evaluate .

Diberi bahawa , nilaikan

[3 marks]

[3 markah]

Answer/ Jawapan:

2)12()( −= x x xh )1(h ′′2)12()( −= x x xh )1(h ′′

3

20

For

examiner’s

use only

3

19



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21 Given the gradient to the curve 2 8 11 y x x= − + is −2 at point ( ), 4k − . Find the

value of k .

Diberi kecerunan bagi lengkok 2 8 11 y x x= − + ialah −2 pada titik ( ), 4k − . Cari

nilai bagi k .

[3 marks]

[3 markah]

Answer/ Jawapan:

22Given that and , find the value of .

[3 marks]

Diberi bahawa dan , cari nilai bagi .

[3 markah]

Answer/ Jawapan:

( ) 23 1

1

x y

x

+=

−

( )dy

h x

dx

= ( )2

0

1

5

h x d x

∫

( ) 23 11

x y

x

+=

−( )

dyh x

dx= ( )

2

0

1

5h x d x∫

3

22

3

21

For

examiner’s

use only



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23 In FIFA World Cup 2014, the probability of Team Y success to enter the final is

3

5, while the probability that Team Y will win in the final is

5

7. Find the probability

that

(a) Team Y will fail to enter the final.

(b) Team Y will fail to become the winner in FIFA World Cup 2014.[3 marks]

Dalam Piala Dunia FIFA 2014 , kebarangkalian bagi Pasukan Y berjaya memasuki

pusingan akhir ialah3

5

, manakala kebarangkalian bagi Pasukan Y akan menang

dalam pusingan akhir ialah5

7. Cari kebarangkalian bahawa

(a) Pasukan Y akan gagal memasuki pusingan akhir.

(b) Pasukan Y akan gagal menjadi juara dalam Piala Dunia FIFA 2014.

[3 markah]

Answer/ Jawapan:(a)

(b)

For

examiner’s

use only

3

23



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24 A teacher wants to form a team of 8 students to collect the donation from each class.These 8 students are chosen from 4 monitors, 3 assistant monitors and 5 prefects.

Calculate the number of different ways the team can be form if

(a) there is no restriction,

(b) the team contains only 3 monitors and 2 assistant monitors.[3 marks]

Seorang guru ingin membentuk satu kumpulan 8 orang murid untuk mengutip derma

dari setiap kelas. Kumpulan 8 orang murid itu mesti dipilih daripada 4 orang ketuakelas, 3 orang penolong ketua kelas dan 5 orang pengawas. Hitungkan bilangan cara

yang berlainan kumpulan itu boleh dibentuk jika

(a) tiada syarat dikenakan

(b) kumpulan itu hanya terdiri daripada 3 orang ketua kelas dan 2 orang penolong

ketua kelas.

[3 markah]

Answer/ Jawapan:

(a)

(b)

3

24

For

examiner’s

use only



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25 Diagram 25 shows a standard normal distribution graph.

Rajah 25 menunjukkan satu graf taburan normal piawai.

The probability represented by the area of the shaded region is 0 3835⋅ . Find the valueof k .

Kebarangkalian yang diwakili oleh luas kawasan berlorek ialah 0 3835⋅ . Cari nilai k .

[3 marks][3 markah]

Answer/ Jawapan:

END OF QUESTION PAPER

KERTAS SOALAN TAMAT

For

examiner’s

use only

f ( z)

z k O

0∙3835

0∙14

Diagram 25

Rajah 25

3

25



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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 91 2 3 4 5 6 7 8 9

Minus / Tolak

0.00.1

0.2

0.3

0.4

0.50000.4602

0.4207

0.3821

0.3446

0.49600.4562

0.4168

0.3783

0.3409

0.49200.4522

0.4129

0.3745

0.3372

0.48800.4483

0.4090

0.3707

0.3336

0.48400.4443

0.4052

0.3669

0.3300

0.48010.4404

0.4013

0.3632

0.3264

0.47610.4364

0.3974

0.3594

0.3228

0.47210.4325

0.3936

0.3557

0.3192

0.46810.4286

0.3897

0.3520

0.3156

0.46410.4247

0.3859

0.3483

0.3121

44

4

4

4

88

8

7

7

1212

12

11

11

1616

15

15

15

2020

19

19

18

2424

23

22

22

2828

27

26

25

3232

31

30

29

3636

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

Q(z)

z

f ( z)

O

Example / Contoh:

If X ~ N(0, 1), then P( X > k ) = Q(k )

Jika X ~ N(0, 1), maka P( X > k ) = Q(k )

−= 2

2

1exp

2

1)( z z f

π

∫∞

=k

dz z f zQ )()(



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HALAMAN KOSONG



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INFORMATION FOR CANDIDATES

MAKLUMAT UNTUK CALON

1. This question paper consists of 25 questions.

Kertas soalan ini mengandungi 25 soalan.

2.


Jawab semua soalan.

3. Write your answers in the spaces provided in the question paper.

Tulis jawapan anda dalam ruang yang disediakan dalam kertas soalan.

4.

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

Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu

anda untuk mendapatkan markah.

5. If you wish to change your answer, cross out the answer that you have done.

Then write down the new answer.

Sekiranya anda hendak menukar jawapan, batalkan jawapan yang telah dibuat.

Kemudian tulis jawapan yang baru.

6. The diagrams in the questions provided are not drawn to scale unless stated.

Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.

7. The marks allocated for each question are shown in brackets.

Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan.

8. A list of formulae is provided on pages 2 and 3.

Satu senarai rumus disediakan di halaman 2 dan 3.

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

Sebuah buku sifir matematik empat angka disediakan.

10.

You may use a non-programmable scientific calculator.

Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram.

11.

Hand in this question paper to the invigilator at the end of the examination.Serahkan kertas soalan ini kepada pengawas peperiksaan di akhir peperiksaan.



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3472/2 Additi onal M athematics Paper 2 [Lihat halaman sebelah

SULIT

1

MAJLIS PENGETUA SEKOLAH MALAYSIA

NEGERI KEDAH DARUL AMAN

MODUL PENINGKATAN PRESTASI TINGKATAN LIMA 2014

MATEMATIK TAMBAHAN

KERTAS 2

MODUL 2

2

12 jam

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. This question paper consists of three sections : Section A, Section B and Section C.

2. Answer all questions 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 your marks.

5. The diagrams provided are not drawn according to scale unless stated.

6. The marks allocated for each question and sub - part of a question are shown in brackets.

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

8. A list of formulae is provided in page 2 and 3.

Kertas soalan ini mengandungi 18 halaman bercetak dan 2 halaman kosong.



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3472/2 Additi onal M athematics Paper 2 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 ac x

a

8.

a

bb

c

ca

log

loglog

2. m n m na a a 9. d naT n )1(

3. m n m na a a 10. ])1(2[

2d na

nS n

4. ( )m n mna a 11. 1 nn ar T

5. nmmn aaa logloglog 12.

r

r a

r

r aS

nn

n

1

)1(

1

)1(, r ≠ 1

6. log log loga a a

m m nn

13. r

aS

1 , r < 1

7. mnm an

a loglog

CALCULUS

1. y = uv,dx

duv

dx

dvu

dx

dy

4 Area under a curve

= b

a dx y or

= b

a dy x

2. y =v

u ,

2v

dx

dvu

dx

duv

dx

dy

5. Volume of revolution

= b

a dx y2 or

= b

a dy x2

3. dx

du

du

dy

dx

dy

GEOMETRY

1. Distance = 2122

12 )()( y y x x 4. Area of triangle

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

1( ) ( )

2 x y x y x y x y x y x y

2. Mid point

( x , y ) =

2,

2

2121 y y x x 5. 22 y xr

3. Division of line segment by a point

( x , y ) =

nm

myny

nm

mxnx 2121 ,

6. 2 2

ˆ xi yj

r x y



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SULIT

3

STATISTICS

1. N

x x

7

i

ii

W

I W I

2. f

fx x 8 )!(

!r n

n P r n

3. N

x x 2)(

= 22

x N

x 9 !)!(

!

r r n

nC r

n

4.

f

x x f 2)( = 2

2

x f

fx

10 P(AB) = P(A) + P(B) – P(AB)

11 P ( X = r ) = r nr r n q pC , p + q = 1

5. m = L + C f

F N

m

21

12 Mean , = np

13 npq

6. 1000

1 Q

Q I 14 Z =

X

TRIGONOMETRY

1. Arc length, s = r 8. sin ( A B ) = sin A cos B cos A sin B

2. Area of sector, A = 2

2

1r

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

3. sin ² A + cos² A = 110 tan ( A B ) =

B A

B A

tantan1

tantan

4. sec ² A = 1 + tan ² A 11 tan 2 A =

A

A

2tan1

tan2

5. cosec ² A = 1 + cot ² A 12

C

c

B

b

A

a

sinsinsin

6. sin 2 A = 2sin A cos A 13 a² = b² + c² – 2bc cos A

7. cos 2 A = cos ² A – sin ² A = 2 cos ² A – 1

= 1 – 2 sin ² A

14 Area of triangle =1

sin2

ab C



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4

Section A

Bahagian A

[ 40 marks ]

[ 40 markah ]


Jawab semua soalan.

1 Solve the simultaneous equations 2 1 x y and 2 2(2 1) 50 x y . [5 marks]

Selesaikan persamaan serentak 2 1 x y dan 2 2(2 1) 50 x y . [5 markah]

2

Diagram 2 shows part of a structure made up of rectangular blocks. The first column has one

block. For each of the other columns, the number of blocks is doubled the previous column.

(a) Find the number of blocks in the 8th

columns, [2 marks]

(b) Calculate

(i) the total volume of the blocks if there are 10 columns of blocks.

(ii) the total cost of the 10 columns of blocks if each block cost RM 0 80 . [5 marks]

Rajah 2 menunjukkan susunan suatu struktur yang terdiri daripada blok yang berbentuk segi

empat tepat. Lajur pertama mempunyai satu blok. Bagi setiap lajur berikutnya, bilangan blok

adalah dua kali ganda daripada lajur sebelumnya.

(a) Carikan bilangan blok bagi lajur ke 8. [2 markah]

(b) Hitungkan

(i) jumlah isipadu blok jika terdapat 10 lajur bagi struktur itu.

(ii) jumlah kos bagi 10 lajur blok jika setiap blok RM 0 80 . [5 markah]

Diagram 2 / Rajah 2

5 cm3 cm

7 cm



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5

3 (a) Sketch the graph of cos3 y x for 0 x . [3 marks]

(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions

for the equation cos3 0 x x for 0 x .

State the number of solutions. [3 marks]

(a) Lakar graf bagi cos3 y x untuk 0 x . [3 markah]

(b) Seterusnya, dengan menggunakan paksi yang sama, lakar satu garis lurus yang sesuai

untuk mencari bilangan penyelesaian bagi persamaan cos3 0 x x untuk 0 x .

Nyatakan bilangan penyelesaian itu.

[3 markah]

4 (a) Given that a set X has score 1 2 3 10, , ............... x x x x . The mean and standard deviation of

set X are 10 and 4 respectively. Find x and 2 x for set X . [4 marks]

(b) Another set Y has score 3 101 2 3 33 3

, , ................

2 2 2 2

x x x x .

Find the mean and variance for set Y . [3 marks]

(a) Diberi bahawa set X mempunyai skor 1 2 3 10, , ............... x x x x . Min dan sisihan piawai

ialah 10 dan 4 masing-masing. Carikan x and 2 x bagi set X . [4 markah]

(b) Satu lagi set Y mempunyai skor 3 101 2 3 33 3

, , ................2 2 2 2

x x x x .

Carikan min dan varians bagi set Y . [3 markah]



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6

5 (a) Given that 5 125log log 1 K V , express V in terms of . K [3 marks]

(b) Given that the function : 2 f x k mx , and 1 1

: 38

f x x , calculate

(i) the value of k and of m ,

(ii) the value of p if 1 1

( )2

f p . [5 marks]

(a) Diberi bahawa 5 125log log 1 K V , ungkapkan V dalam sebutan . K [3 markah]

(b) Diberi bahawa fungsi : 2 f x k mx , dan 1 1

: 38

f x x , hitungkan

(i) nilai-nilai bagi k dan m ,

(ii) nilai bagi p jika 1 1

( )2

f p . [5 markah]

6 (a) Given that29 1

( )3 1

x f x

x

, find '( ). f x [2 marks]

(b) A curve has a gradient function of 2 3kx x , the tangent to the curve at the point

(2,12) is parallel to the straight line 2 4 9 y x , find

(i) the value of k ,

(ii) the equation of the normal to the curve at point (2,12) . [5 marks]

(a) Diberi bahawa 29 1

( )3 1

x f x

x

, cari '( ). f x [2 markah]

(b) Fungsi kecerunan suatu lengkung ialah 2 3kx x , tangen pada lengkung di titik (2,12)

adalah selari kepada garis lurus 2 4 9 y x , cari

(i) nilai bagi k ,

(ii) persamaan normal pada lengkung di titik (2,12) . [5 markah]



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SULIT

7

Section B

Bahagian B

[ 40 marks ]

[ 40 markah ]

Answer four questions from this section. Jawab empat soalan daripada bahagian ini.

7 Use graph paper to answer this question.

Gunakan kertas graf untuk menjawab soalan ini.

x 1 2 3 4 5 6

y 3 5 22 0 67 5 152 0 287 5 486 0

Table 7/ Jadual 7

Table 7 shows the values of two variables, x and y, obtained from an experiment. Variables x

and y are related by the equation 23 xk

pkx y , where k and p are constants.

(a) Plot2 x

y against x , using a scale of 2 cm to 1 unit on the x axis and 1 cm to 1 unit

on the2 x

y - axis. Hence, draw the line of best fit. [4 marks]

(b) Use the graph in 7 (a) to find the value of

(i) k

(ii) p

(iii) y when x = 2 5 . [6 marks]

Jadual 7 menunjukkan nilai-nilai bagi dua pembolehubah, x dan y, yang diperoleh daripada

satu eksperimen. Pembolehubah x dan y dihubungkan oleh persamaan 23 xk

pkx y ,

dengan keadaan k dan p ialah pemalar.

(a) Plot2

x y melawan x , dengan menggunakan skala 2 cm kepada 1 unit pada paksi x dan

1 cm kepada 1 unit pada paksi-2 x

y. Seterusnya, lukis garis lurus penyuaian terbaik.

[4 markah]

(b) Gunakan graf di 7(a) untuk mencari nilai

(i) k,

(ii) p,

(iii) y apabila x = 2 5 . [6 markah]



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8

8

Diagram 8 shows a quadrilateral PQRT and a triangle RST . M is a midpoint of TR. PQ =9 x , PT = 8 y , 2 PQ =3TS and PT = 2 QR .

(a) Express the following vectors in terms of x and y .

(i) TR

(ii) PS

(iii) MS

Hence, Show that P , M and S are collinear.[6 marks]

(b) It is given that1

2 x and

3

4 y , find PS . [4 marks]

Rajah 8 menunjukkan sebuah sisiempat PQRT dan segitiga RST. M ialah titik tengah TR.

PQ = 9 x , PT = 8 y , 2 PQ =3TS dan PT = 2 QR .

(a) Ungkapkan vektor yang berikut dalam sebutan x dan y

(i) TR

(ii) SR

(iii) MS

Seterusnya, tunjukkan bahawa P , M dan S adalah segaris. [6 markah]

(b) Diberi bahawa 1

2 x dan

3

4 y , cari PS . [4 markah]

S

P

Q

R

T

M

Diagram 8/ Rajah 8



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9

9

Diagram 9 shows the straight line y = 3 x + 4 intersecting the curve y = x2 at the points K .

Find

(a) the coordinates of K , [3 marks]

(b) the area of the shaded region B, [3 marks]

(c) the volume generated, in terms of π , when the shaded region A is revolved through 360o

about the y-axis.

[4 marks]

Rajah 9 menunjukkan garis lurus y = 3 x +4 yang menyilang lengkung y = x2 pada titik K .

Cari

(a) koordinat K , [3 markah] (b) luas rantau berlorek B, [3 markah]

(c) isipadu janaan, dalam sebutan π , apabila rantau berlorek A dikisarkan melalui 360o

pada paksi-y.

[4 markah]

Diagram 9/ Rajah 9

O

y

x

y = x 2

y = 3 x + 4

K

B

A



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10

10

Diagram 10 shows a semicircle ABC with centre O and a sector ABO with centre A. The radius

of semicircle ABC and sector ABO is 8 cm.

[ Use π = 3 142 ]

Calculate

(a) the value of , in radian, [2 marks]

(b) the perimeter, in cm, of shaded region , [4 marks]

(c) the area, in cm2 , of the shaded region . [4 marks]

Rajah 10 menunjukkan sebuah semi bulatan ABC dengan pusat O dan sector ABO dengan

pusat A. Jejari bagi semi bulatan ABC dan sektor bulatan ABO ialah 8 cm.

[ Guna π = 3 142 ]

Hitung

(a) nilai , dalam radian, [2 markah]

(b) perimeter, dalam cm, kawasan berlorek , [4 markah]

(c) luas, dalam cm2, kawasan berlorek . [4 markah]

B

C A

O

Diagram 10 / Rajah 10



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SULIT

11

11 (a) In an examination, 85% of the candidates passed Mathematics. If a sample of 6 candidates

is chosen at random, find the probability that

(i) all the candidates passed Mathematics,

(ii) at least 2 candidates failed Mathematics.[5 marks]

(b) The body mass of 500 students in a school follows a normal distribution with a mean of

52 kg and a standard deviation of 10 kg.

(i) If a student is chosen at random, find the probability that his body mass is between

40 kg and 60 kg.

(ii) Calculate the number of students whose body mass are between 40 kg and 60 kg.

[5 marks]

(a) Dalam suatu peperiksaan, 85 % calon lulus Matematik . Jika 6 calon dipilih secara rawak ,

cari kebarangkalian bahawa

(i) kesemua calon itu lulus Matematik,

(ii) sekurang-kurangnya 2 calon gagal Matematik .

[5 markah]

(b) Jisim badan 500 pelajar sebuah sekolah adalah mengikut taburan normal dengan min

52 kg dan sisihan piawai 10 kg.

(i) Jika seorang pelajar dipilih secara rawak, carikan kebarangkalian bahawa

jisim badannya berada di antara 40 kg dan 60 kg.

(ii) Hitung bilangan pelajar yang mempunyai jisim badan di antara 40 kg dan 60 kg.

[5 markah]



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12

Section C

Bahagian C

[ 20 marks ]

[ 20 markah ]

Answer any two questions from this section.

Jawab mana-mana dua soalan daripada bahagian ini.

12 A particle moves along a straight line from a fixed point O. Its velocity, v ms-1

, is given by

v = kt – 3t 2 , where k is a constant and t is the time, in seconds, after leaving the point O. The

velocity of the particle is maximum when t = 2 5 s .

[Assume motion to the right is positive.]

Find

(a) the value of k , [2 marks]

(b) the value of t when the particle passes O again, [3 marks]

(c) the time, in seconds, when the particle stops instantaneously, [2 marks]

(d ) the distance travelled, in m, by the particle in the first 7 seconds. [3 marks]

Suatu zarah bergerak di sepanjang suatu garis lurus dari satu titik tetap O. Halajunya, v ms-1

,

diberi oleh v = kt – 3t 2 , dengan keadaan k ialah pemalar dan t ialah masa, dalam saat,

selepas meninggalkan titik O. Halaju zarah itu adalah maksimum pada t = 2 5 s

[ Anggapkan gerakan ke arah kanan sebagai positif.]

Cari

(a) nilai k , [2 markah]

(b) nilai bagi t apabila zarah itu melalui titik O semula , [3 markah]

(c) masa, dalam saat, apabila zarah berhenti seketika, [2 markah]

(d ) jarak yang dilalui, dalam m , oleh zarah itu dalam tujuh saat pertama. [3 markah]



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13

13 Diagram 13 shows a quadrilateral KLMN .

Diagram 13/ Rajah 13

Calculate

(a) the length, in cm, of KM ,. [2 marks]

(b) KMN , [2 marks]

(c) LKM , [3 marks]

(d ) the area, in cm2, of quadrilateral KLMN . [3 marks]

Rajah 13 menunjukkan sisiempat KLMN.

Hitungkan

(a) panjang, dalam cm, KM , [2 markah]

(b) KMN , [2 markah]

(c) LKM , [3 markah]

(d ) luas, dalam cm2 , bagi sisiempat KLMN . [3 markah]

N

53 cm

M

115

K

66 cm

30 cm

L

85



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14

14 Table 14 shows the prices and the price indices of five components P , Q, R, S and T needed to

produce a certain type of digital camera. Pie chart 14 shows the relative quantity of

components needed in producing the camera.

Component

Komponen

Price (RM)

per unit Harga (RM)

per unit

Price index for the year

2012 based on the year

2010

Indeks harga pada tahun2012 berasaskan tahun

20102010 2012

P 1·10 1·21 110

Q 1·80 x 120

R 3·20 4·00 150

S 2·50 3.05 122

T 2·00 2·80 y

Table 14 / Jadual 14

Pie chart 14 / Carta pai 14

80º35º

160º

40º

PQ

R

S

T



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SULIT

15

(a) Find the value of x and y. [4 marks]

(b) Calculate the composite index for the production cost of the camera in the year 2012

based on the year 2010. [3 marks]

(c) The price of each component increases by 10% from the year 2012 to the year 2014.

Given that the production cost of the camera in year 2010 is RM500, calculate the

corresponding cost in year 2014. [3 marks]

Jadual 14 menunjukkan harga dan indeks harga bagi lima komponen P, Q, R, S dan T yang

diperlukan untuk menghasilkan sejenis kamera digital. Carta pai 14 menunjukkan kuantiti

relatif bagi komponen yang diperlukan dalam penghasilan kamera digital itu.

(a) Cari nilai x dan y. [4 markah]

(b) Hitungkan indeks gubahan bagi kos penghasilan kamera digital itu pada tahun 2012

berasaskan tahun 2010. [3 markah]

(c) Harga setiap komponen meningkat 10% dari tahun 2012 ke tahun 2014. Diberi kos

penghasilan kamera digital itu dalam tahun 2010 ialah RM500 , hitungkan kosnya yang

sepadan pada tahun 2014. [3 markah]



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16

15 Use a graph paper to answer this question.

A factory produces two types of pillow, type A and type B. In a day, it can produce x pillows

of type A and y pillows of type B. The time taken to produce a pillow of type A is 20 minutes

and a pillow of type B is 30 minutes.

The production of the pillow per day is based on the following constraints.

I. : The time taken to make pillows of type A is not more than the time taken to make

pillows of type B.

II. : The total number of pillows produced is not more than 500.

III. : The number of pillows of type B must exceed the number of pillows of type A by

at most 200.

(a) Write three inequalities, other than which satisfy all the above

constraints. [3 marks]

(b) By using the scale of 2 cm to 100 pillows on both axes, construct and shade the region

R which satisfies all the above constraints. [3 marks]

(c) Use graph from 15(b), to find

(i) the maximum number of pillows of type A if 280 of pillows of type B produced.

(ii) maximum profit that can be obtained, if the profit of selling pillow A is RM20 and

pillow B is RM12 00. [4 marks]



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SULIT

17

END OF QUESTION PAPER

KERTAS SOALAN TAMAT

Gunakan kertas graf untuk menjawab soalan ini.

Sebuah kilang menghasilkan 2 jenis bantal, jenis A dan jenis B. Dalam satu hari, kilang itu

boleh menghasilkan x biji bantal jenis A dan y biji bantal jenis B. Masa yang diambil untuk

menghasilkan sebiji bantal jenis A ialah 20 minit dan sebiji bantal jenis B ialah 30 minit.

Pengeluaran bantal dalam satu hari adalah berdasarkan kekangan yang berikut:

I: Masa yang diambil untuk membuat bantal A tidak melebihi masa yang diambiluntuk membuat bantal jenis B.

II: Jumlah bantal yang dihasilkan tidak melebihi 500 bji.

III: Bilangan bantal jenis B mesti melebihi bilangan bantal jenis A selebih-lebihnya

200 biji.

(a) Tulis tiga ketaksamaan, selain , yang memenuhi semua kekangan di

atas. [3 markah]

(b) Menggunakan skala 2 cm kepada 100 biji bantal pada kedua-dua paksi, bina dan lorek

rantau R yang memenuhi semua kekangan di atas. [3 markah]

(c) Gunakan graf anda daripada 15(b) untuk mencari

(i) bilangan maksimum bantal jenis A jika 280 biji bantal jenis B dihasilkan.

(ii) keuntungan maksimum yang boleh diperolehi, jika keuntungan jualan bagi bantal A

ialah RM20 dan bantal B ialah RM12 00.

[4markah]



42/63

SULIT 3472/2

August 2014


18

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 91 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

Q(z)

z

f (z )

O

Example / Contoh:

If X ~ N(0, 1), then P ( X > k ) = Q(k )Jika X ~ N(0, 1), maka P ( X > k ) = Q(k )

2

2

1exp

2

1)( z z f

k

dz z f z Q )()(



43/63

SULIT 3472/2

August 2014


SULIT

19

HALAMAN KOSONG



44/63

SULIT 3472/2

August 2014


20

HALAMAN KOSONG



45/63

SULIT 3472/1

3472/1 Additional Mathematics Paper 1

SULIT

1

Additional

Mathematics

Paper 1

Ogos, 2014

MODUL PENINGKATAN PRESTASI TINGKATAN 5

TAHUN 2014

ANJURAN

MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)

ADDITIONAL MATHEMATICS

MARKING SCHEME

Paper 1

MODUL 2

.



46/63

SULIT 3472/1


SULIT

2

PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2014

Marking Scheme Module 2

Additional Mathematics Paper 1

Question Solution/ Marking Scheme Answer Marks

1

(a) B1 : 3 2m or n= = (a) 3m = AND 2n =

(b) many – to – one

2

1

2

B2 : ( )6 63 12 f or p p

= =

B1 : 1 6

3 f p

− =

1

2 p = 3

3

(a) B1 : 2 3 3 y− =

(b) B1 : ( )2 3 3 8k − − =

(a) ( ) 1

3

3

g = −

(b) 5k =

2

2

4

B2 :( )( )1 2 0

1 2

p p or

p or p

− − =

= =

B1 : ( ) ( )( )22 4 1 3 2 p p− −

1 2 p AND p= = 3

5

(a) B1 : 3 5 p or q= − = − (a) 3 p = − AND 5q = −

(b) 3 x = −

2

1



47/63

SULIT 3472/1


SULIT

3


6B2 :

or5

2 x = , 3 x = −

B1: (2 5)( 3) x x− +

53

2 x− ≤ ≤ 3

7

B2 :

3

2 2 2

2

log 16 log loglog 8

h k + +

B1 :

3

2 2

2 2

log 16 log

log log 8

or h or

k or

4 3

3

a b+ + 3

8

B2: 6221 −=−+ x

B1 : ( )2

2( 1) 37(7) 7 x or − −

25− 3

9

B3 : ( )2 14 3+

B2 :

( ) ( )26 18

2 2 25 2 2 17 5322 2

3

d d

or d

+ − + =

=

B1 :

( ) ( )26 18

2 2 25 2 2 172 2

d or d + +

44 4

52 3−



48/63

SULIT 3472/1


SULIT

4


10

(a) B1: 1

9

3 =

m

m

(b) B1:

9

11

9

1181

5

5

−

−

=S

(a) 3 3m AND m= = −

(b) 91.123

2

2

11

B1:1

3r =

272

2

12

(a) B1: 2 xy ax b= +

(b) B1 :

( )

( )

13 7* 2

3 2* 2

b

or

b

= +

= +

(a) 2

(b) 1b = −

2

2

13

B2 :

( ) ( ) ( )2 2 224 3 2 4 x y x y − + − = + −

B1 :

( ) ( ) ( ) ( )2 2 2 2

3 2 0 4 x y or x y− + − − + −

2 23 3 24 8 36 0 x y x y+ − − + = 3

14

B1:

( ) ( ) ( ) ( ) ( ) ( )1

11 4 3 2 8 8 3 4 11 2 02

k k × + × + × − × + × + × =

OR

8 4 4 2

11 3 3 k

− −=

− −

1k = − 2



49/63

SULIT 3472/1


SULIT

5


15

(b) B1 : 5 or22

43 +

(a) ji 43 +

(b)5

43 ji +

1

2

16

B3 : 6 2 0 12 4 0m OR m+ = + =

B2 :3 3 6 0

8 4 12 03

m + + =

B1 :6

12OB

=

3m = − 4

17

B3 : 41 81 138 19 x or x= ⋅ = ⋅

B2 :

26sin 8sin 8 0

2sin 2 sin

3

x x or

x or x

− − + =

= − =

B1 : ( )23 1 2sin 8sin 5 x x− = −

41 81

138 19

x

AND

x

= ⋅

= ⋅

4

18

(b) B2: ( ) ( ) ( )( )21 1

12 * 1.047 6 10.3922 2

−

B1: ( ) ( ) ( )( )21 1

12 * 1.047 6 10.3922 2

or

(a) 1.047rad

(b) 44.21

1

3



50/63

SULIT 3472/1


SULIT

6


19

(b) B1: 70 118

k ∗

+ =

(a) 70

(b) 18k =

1

2

20

B2 : (1) 24 8h′′ = −

B1 : 1812)( 2 +−=′ x x xh

16 3

21

B2 : 2 8 2 3k or x− = − =

B1 : 2 8 x −

3k = 3

22

B2 :( )

( )2

2

0

3 11 149 1

5 1 5

xor

x

+− − −

B1 : ( ) ( ) 23 1

1

xh x d x

x

+=

−∫

10 3

23

(b) B1:2 3 2 3 5

15 5 7 5 7

or

+ × − ×

(a)2

5

(b)4

7

1

2

24

(b) B1 : 4 3 53 2 3C C C × ×

(a) 495

(b) 120

1

2

25

B2 : p( z>k ) = 0.0608

B1 : P( z > 0. 14) = 0.4443

k = 1.548 3

END OF MARKING SCHEME



51/63

1

Nama Pelajar : ………………………………… Tingkatan 5 : …………………….

3472/2

Additional

Mathematics

August 2014

MODUL PENINGKATAN PRESTASI TINGKATAN 5

TAHUN 2014

ADDITIONAL MATHEMATICS

Paper 2

( MODUL 2 )

MARKING SCHEME



52/63

2

SULIT 3472/2

MARKING SCHEME

ADDITIONAL MATHEMATICS PAPER 2 2014

N0. SOLUTION MARKS

1 2 1 x y or 12 x y

2 2 50 x x 2 25 x

5 5 x x

5 x and 5 x (both)

2 y and 3 y (both)

P1

K1 Eliminate x/y

K1 Solve quadratic equation

N1

N1

5

2(a)

(b)

(i)

(ii)

7

8 (1)(2)

128

T

10

10

(1)(2 1)

2 1

1023

1023 (3)(7)(5)

107415

S

V

1023 0.8

818.4

K1

N1

K1

K1

N1

K1

N1

7

3

(a)

(b)

y = x

draw the straight line y = x

Number of solutions = 3

P1 cos shape correct.

P1 Amplitude = 2 [ Maximum = 1

and Minimum = -1 ]

P11

12

cycle in 0 x or

N1 For equation

K1 Sketch the straight line

N1

6

-4



53/63

3

4

(a)

(b)

1010

100

x

x

2

2 2

2

4 1010

1160

x

x

10 3

2

6.5

mean

or2

42

2

4

K1

N1

K1

N1

K1

N1 N1

7

5

(a)

(b)

5 125

55

3

5 5

3

5

3

log log 1

loglog 1

3

log log 3

log 3

125

K V

V K

K V

K

V

K V

i)

1( )

2

1 13

2 8 2

4 24

x k f x

m

k

m m

m and k

ii)

1 1( ) 3

8 2

20

p

p

K1

K1

N1

K1

K1

N1

K1

N1

8



54/63

4

6

(a)

(b)

(3 1)(3 1)( ) 3 1

3 1

' ( ) 3

x x f x x

x

f x

i)

2

2

3

2 (2) 3(2)

2

dykx x

dx

k

k

ii)

1

2

112 (2)

2

111

2

normal m

c

y x

K1

N1

K1

N1

P1

K1

N1

7



55/63

5

7

(a)

(b)

(c)

(i)

(ii)

(iii)

x 1 2 3 4 5 6

2

y

x 3.5 5.5 7.5 9.5 11.5 13.5

2

y

x

2

y

x= kx+

p

k

k = *gradient

k = 2.0

p

k = *y-intercept

p = 3.0

y = 40

N1 6 correct

values of 2

y

x

K1 Plot2

y

x vs x.

Correct axes &

uniform scale

N1 6 points plotted

correctly

N1 Line of best-fit

P1

K1

N1

K1

N1

N1

10

1.5

0 x



56/63

6

N0. SOLUTION MARKS

8(a)

i)

ii)

iii)

b)

2(9 )

3TS x

2

PT QR

TR TP PQ QR

PS PT TS

MS MR RS

PS k MS

PS PT TS

6 8 x y =

k =4

4 PS MS and S is a common point or equivalent

2 21 3

6( ) 8( )2 4

PS

= 45

K1 (TS or QR )

N1

K1

N1

K1

N1

K1

N1

K1

N1

8 9 4

9 4

y x y

x y

3 42

9 43 4

2

3 22

TR x y

x y x y

x y

6 8 x y

3( 2 )

2k x y

= 6 x

= 4 y

6 8 x y



57/63

7

9

a)

b)

c)

2

2

3 4

3 4 0

( 1)( 4) 0

4, 16

(4,16)

x x

x x

x x

x y

K

4

2

y

x

22

0

23

0

3

2

(4)(2)

83

28

3

16

3

x dx

x

cm

16

2

4

162

4

2 2

1(4) (12)

3

642

16 4642 2

256 1664

2 2

56

ydy

y

K1 for solvingquad.eqn.

N1

N1

K1 use area of

rectangle - ( ) y dx

K1 integrate

correctly

and Sub.

the limit

correctly

N1

K1

K1 correct limit

K1 integrate

correctly

N1

10

Area B

Volume A



58/63

8

N0. SOLUTION MARKS

10

(a)

(b)

(c)

60o

1.047 rad

8(1.047)OB

S or 8(2.095) BC S OR 8(3.142) AC S

= 8.38 = 16.76 = 25.14

Perimeter = 8.38+16.76+8 or Perimeter = 25.14 + 8

= 33.14 = 33.14

Area of OAB = 21

(8) (1.047)2

= 33.50 cm

2

Area of triangle OAB = 21

(8) sin 602

= 27.71

Area of the shaded region = 33.50 – 27.71

= 5.79 cm2

P1

N1

K1 Use s r

N1

K1

N1

K1 Use formula

21

2

A r

K1

K1

N1

10



59/63

9

N0. SOLUTION MARKS

11

(a)

(i)

(ii)

(b)

(i)

(ii)

X= Students passed Mathematics

p = 0.85 , q = 1- 0.85 = 0.15 , n = 6

P(X =6) = 6 6 06 (0.85) (0.15)c

=0.3772

P (Y≥2) = 1 – P(Y=0) – P(Y = 1)

Or = ( 2) ( 3) ......... ( 6) P Y P Y P Y

= 1 -

6 1 5

1(0.15) (0.85)c -

6 0 6

0 (0.15) (0.85)c

=0.2235

µ= 52 , σ =10

P( 40 < X < 60 ) = P (40 52

10

< Z <

60 52

10

)

= P( -1.2 < Z < 0.8)

= 0.6731

n = 0.6731 x 500

n = 337

P1

K1 Use P (X

=r ) =r nr

r

n q pC

N1

K1

N1

K1 Use Z =

X

K1

N1

K1

N1

10



60/63

10

N0. SOLUTION MARKS

12

(a)

(b)

(c)

(d)

6a k t

6(2.5) 0k

k = 15

2 315

2 s t t

2 315 02

t t

t = 7.5 s

215 3 0t t

t = 5

Total distance

=

5 72 2

3 3

0 5

15 15

2 2

t t t t

d = 62.5 + 38

= 100.5

K1

N1

K1

K1

N1

K1

N1

K1 (for

Integration;

either one

and

substitute

the limit

5 7

0 5

or )

K1

(for use and

summation)

N1

10



61/63

11

N0. SOLUTION MARKS

13

(a)

(b)

(i)

(ii)

(iii)

(i)

KI (PetuaKosinus)

N1

K1 (Petua Sinus)

N1

K1 (GunaPetua Sinus)

K1

N1

K1

K1

N1

10



62/63

12

N0. SOLUTION MARKS

14(a)

(b)

(c)

(i)

Lihat 45º

I 2012/2010 =

= 124.64

I 2014/2010 = 124.64 = 137.10

Q 2014 = = RM685.50

K1

N1

K1

N1

K1

K1

N1

K1

K1

N1

10

i)



63/63

N0. SOLUTION MARKS

15

(a)

(b)

(c)

i) ii) iii)

Sekurang-kurangnya 1 garislurusdilukis dengan betul yang

melibatan x dan y.

Semuagarislurusdilukisbetul.

Kawasandilorek dengan betul

i) Bilanganmaksimumbantal A = 220

ii) Titikmaksimum (300, 200)

Keuntunganmaksimum;

k = RM8 400

N1

N1

N1

K1

K1

N1

N1

P1

K1

N1

650

600

550

500

450

400

350

300

250

200

150

100

50

-50

-100

-400 -300 -200 -100 100 200 300 400 500 600 700 800 900 1000 1100 1200

220

(300, 200)

Type B

Type A

Spm Trial 2014 Addmaths Qa Kedah

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