P1+P2+Skema 2015 kedah add math

8/17/2019 P1+P2+Skema 2015 kedah add math

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Modul ini mengandungi 23 halaman bercetak dan 1 halaman kosong.

1.

Tulis nama dan tingkatan anda pada

ruangan yang disediakan.

2. Kertas soalan ini adalah dalam

dwibahasa.

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 membaca

maklumat di halaman belakang kertas

soalan ini.

PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2015MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)

ADDITIONAL MATHEMATICS

Kertas 1

Ogos 2015

2 jam Dua jam

3472 / 1

Untuk Kegunaan Pemeriksa

Soalan Markah

Penuh

Markah

Diperolehi

1 2

2 3

3 3

4 3

5 3

6 3

7 3

8 39 4

10 3

11 2

12 4

13 3

14 3

15 3

16 3

17 4

18 4

19 3

20 4

21 2

22 4

23 3

24 4

25 4

TOTAL 80

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

JANGAN BUKA MODUL INI SEHINGGA DIBERITAHU

SULIT


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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 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 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 a nn

1

)1(

1

)1( , (r 1)

13 r

a

S 1 , r


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3

STATISTICS

1 Arc length, s = r

2 Area of sector , A = 21

2r

3 sin 2 A + cos 2 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 – sin2 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 = b2 + c2 − 2bc cos A

14 Area of triangle = C ab sin2

1

1 x = N

x

2 x =

f

fx

3 =

2 x x

N

=

2

2

x N

x

4 =

f

x x f 2)( = 2

2

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

n P r 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

nq pC , p + q = 1

12 Mean µ = np

13 σ npq

14 Z =σ

X


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Answer all questions.

Jawab semua soalan.

1 Function f maps elements in set A to elements in set B . Given that the function

: 3 4 f x x and A{ -1, 0, 1, 2},

Fungsi f memetakan unsur-unsur dalam set A kepada unsur-unsur dalam set B. Diberi

fungsi sebagai : 3 4 f x x dan A{ -1, 0, 1,2}.

State/nyatakan

(a) the range of the function f ( x) .

julat bagi fungsi f ( x) .

(b) the object has unchanged under function f ( x) .

objek yang tidak berubah dibawah fungsi f ( x) . [2 marks]

[2 markah]

Answer/ Jawapan:

(a)

(b)

Forexaminer’s

use only

2

1


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2 Given function :h x k x m , 0k and

2 : 4 15h x x . Find the values of k and m.

Diberi fungsi :h x k x m , 0k

dan

2 : 4 15h x x . Cari nilai bagi k dan m.[3 marks]

[3 markah]

Answer/ Jawapan:

3Given that function

3( ) , f x x p

x p

and x x g 52)( .

Diberi fungsi3

( ) , f x x p x p

dan x x g 52)( .

FindCari

(a) )(1 x g ,

(b) the value of p if 1( 3) ( 4) g g p f p .

nilai p jika 1( 3) ( 4) g g p f p .[3 marks]

[3 markah]

Answer/ Jawapan:

(a)

(b)

Forexaminer’s

use only

3

2

3

3


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4

Diagram 4 shows the graph of the quadratic function 2( ) ( ) g x x p q .

Rajah 4 menunjukkan graf fungsi kuadratik 2( ) ( ) g x x p q

.

(a) State the value of q.

Nyatakan nilai bagi q.

(b) Find the value of p.

Cari nilai bagi p .

[3marks]

[3markah]

Answer/ Jawapan:

(a)

(b)

3

4

Forexaminer’s

use only

g ( x)

x

4

5

Diagram 4

Rajah 4


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5 Find the range of values of x for 2 5 12 9 1 x x x . [3 marks]

Cari julat nilai x bagi 2 5 12 9 1 x x x . [3 markah]

Answer/ Jawapan:

6

Given1

2 is a root of the quadratic equation 2 2 5 0 x k x k , where k is a

constant. Find

Diberi1

2 ialah satu punca bagi persamaan kuadratik 2 2 5 0 x k x k , dengan

keadaan k ialah pemalar . Cari

(a) the value of k ,nilai bagi k ,

(b) the product of roots of the equation.

hasil darab punca-punca bagi persamaan tersebut .

[3 marks]

[3 markah]

Answer/ Jawapan:

(a)

(b)

3

5

Forexaminer’s

use only

3

6


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7 Solve the equation :

Selesaikan persamaan : 1

4 80 4 x x [3 marks]

[3 markah]Answer/ Jawapan:

8Given 3log 17a and 3log 8b , find the value of

2

9log a

b .

Diberi 3log 17a dan 3log 8b , cari nilai bagi2

9log a

b.

[3 marks]

[3 markah]

Answer/ Jawapan:

3

7

3

8

Forexaminer’s

use only


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9 John was looking for a shop lot to start his business. There are two shop lots offered for

rent. The yearly rental for shop lot A is RM 9 000 with 20% yearly increment. While the

rental for shop lot B is RM 20 000 per year without increment.

State the different of the total rental for 10 years between shop lot A and shop lot B.

( Round off your answer to the nearest ringgit )

John decided to choose the shop lot which offered lower total rental for 10 years. Which

shop lot should John choose ?

[4 marks]

John ingin menyewa sebuah kedai untuk memulakan perniagaannya. Terdapat dua buah

kedai menawarkan sewa. Sewa tahunan bagi kedai A ialah RM 9 000 dengan peningkatan

20% setiap tahun. Manakala sewa bagi kedai B ialah RM 20 000 per tahun tanpa

peningkatan.

Nyatakan perbezaan jumlah sewa untuk 10 tahun antara kedai A dengan kedai B.

( Bundarkan jawapan anda kepada ringgit terdekat )

John merancang untuk memilih kedai yang menawarkan jumlah sewa yang lebih rendah

untuk 10 tahun. Kedai manakah yang harus John pilih ?

[4 markah]

Answer/ Jawapan:

4

9

Forexaminer’s

use only


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10 It is given that 29, 22, 15, k , … 76 is an arithmetic progression. Diberi 29, 22, 15, k , … 76 adalah suatu janjang aritmetik .

Find

Cari

(a) the value of k ,

nilai k ,

(b) the number of term of the progression .

bilangan sebutan bagi janjang tersebut .

[3 marks]

[3 markah]

Answer/ Jawapan:

(a)

(b)

11Given the value of the sum to infinity of a geometric progression is five times of the first

term of the progression. Find the common ratio of the progression.

[2 marks]

Diberi hasil tambah ketakterhinggaan bagi suatu janjang geometri adalah sama dengan

lima kali sebutan pertama janjang tersebut . Cari nilai nisbah sepunya bagi janjang

tersebut.

[2 markah]

Answer/ Jawapan:

2

11

3

10

Forexaminer’s

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12 In Diagram 12, OST and OUV are sector of two circles with centre O. The ratio of the area

of sector OUV to the area of shaded region is 1 : 8 .

Dalam Rajah 12, OST dan OUV ialah sektor bagi dua bulatan berpusat O. Nisbah bagi

luas sektor OUV kepada luas kawasan berlorek ialah 1 : 8 .

Given that 4 cmOU and the area of sector OUV is 26 cm , find

Diberi 4 cmOU dan luas sektor OUV ialah26 cm , cari

(a) SOT in radians,SOT dalam radian,

(b) the length of SU . panjang SU .

[4 marks]

[4 markah]

Answer/ Jawapan:

(a)

(b)

4

12

O

T

S

VDiagram 12

Rajah 12

4 cm

U

Forexaminer’s

use only


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13 Given that 4, 6 , 1, 0 , 10, 2 A B C and D are the vertices of a parallelogram.

Find the coordinates of point D .

[3 marks]

Diberi bahawa 4, 6 , 1, 0 , 10, 2 A B C dan D ialah bucu-bucu bagi sebuah segi

empat selari. Cari koordinat bagi titik D .

[3 markah]

Answer/ Jawapan:

14 Given that the equation of a locus P with centre , A h k and diameter 13 cm is

012522 y x y x . Find the coordinate of point A.

Diberi persamaan lokus P berpusat , A h k dan berdiameter 13 cm ialah

012522 y x y x . Cari koordinat titik A.

[3 marks]

[3 markah]

Answer/ Jawapan:

3

14

3

13

Forexaminer’s

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15

Diagram 15 shows a line of best fit obtained by plotting 2 x y against2

x . Express y in

terms of x.

Rajah 15 menunjukkan graf garis lurus penyesuaian terbaik yang diperoleh dengan

memplot 2 x y melawan2

x . Ungkapkan y dalam sebutan bagi x.

[3 marks]

[3 markah]

Answer/ Jawapan:

3

15

Forexaminer’s

use only

( 4 , 1 )

2 x y

xO

( 2 , 3 )

Diagram 15

Rajah 15


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16

Number of smart phone

Bilangan telefon pintar1 2 3 4

Number of families Bilangan keluarga

30 k 48 16

Table 16

Jadual 16

Table 16 shows the frequency distribution of the number of smart phone in the families of

a survey.

Jadual 16 menunjukkan taburan kekerapan bilangan telefon pintar dalam keluarga bagi

satu kajian.

(a) Find the range of the data.

Cari julat bagi data tersebut .

(b) Find the range of k if

Cari julat bagi k jika

i) the mode of the number of smart phone in the families is 2 ,

mod bagi bilangan telefon pintar dalam keluarga ialah 2 ,

ii)

the median of the number of smart phone in the families is 2.median bagi bilangan telefon pintar dalam keluarga ialah 2.

[3 marks]

[3 markah]

Answer/ Jawapan:

(a)

(b) (i)

(ii)

Forexaminer’s

use only

3

16


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17The daily profit , P , of a candle factory is given by

21454

P x x , where x is the

number of boxes of candle. How many boxes of candle produced will give maximum

profit and what is the maximum profit of the factory ?

Keuntungan harian , P , bagi kilang lilin diberi oleh2145

4 P x x , dimana x ialah

bilangan kotak lilin. Berapa kotak lilin dihasilkan akan memberi keuntungan maksimum

dan berapakah keuntungan maksimum kilang tersebut ?

[4 marks]

[4 markah]

Answer/ Jawapan:

18 Given that 2

0

2

3 f x dx , find the values

Diberi 2

0

2

3 f x dx , cari nilai bagi.

(a) 0

23 f x dx ,

(b) 2

0

15

2 f x dx

.

[4 marks]

[4 markah]

Answer/ Jawapan:

(a)

(b)

4

18

Forexaminer’s

use only

4

17


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19

Diagram 19 shows the curve y f x and straight line y g x intersect at point

3, 3 A . Given 3

03 f x dx , find the area of the shaded region.

[3 marks]

Rajah 19 menunjukkan lengkung y f x dan garis lurus y g x bersilang pada

titik 3, 3 A . Diberi 3

03 f x dx , cari luas rantau berlorek .

[3 markah]

Answer/ Jawapan:

Forexaminer’s

use only

y f x

xO

Diagram 19

Rajah 19

A ( 3, 3)

y

y g x

6

3

19


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20 Diagram 20 shows a triangle PQR. T is a midpoint of PQ . Given 9 PR y , 3 PT x and

2 RS SQ . Express in terms of x and y

Rajah 20 menunjukkan sebuah segitiga PQR . T ialah titik tengah bagi PQ . Diberi

9 PR y , 3 PT x dan 2 RS SQ . Ungkapkan dalam sebutan x dan y

(a) RQ

(b) ST

[4 marks][4 markah]

Answer/ Jawapan:

(a)

(b)

4

20

Forexaminer’s

use only

R

S

T Q PDiagram 20

Rajah 20


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21

Given that2

5OA

,

2

k OC

and2

2 AC

. Find the value of k .

Diberi bahawa 25

OA

,2

k OC dan 2

2 AC

. Cari nilai k .

[2 marks]

[2 markah]

Answer/ Jawapan:

22 Given the equation 3 cos 2 y for 180 .

Diberi suatu persamaan 3 cos 2 y bagi 180 .

(a) State the amplitude.

Nyatakan amplitude .

(b) Solve the equation 3cos2 y when 2 y . Selesaikan persamaan 3cos2 y apabila 2 y .

[4 marks]

[4 markah]

Answer/ Jawapan:

(a)

(b)

4

22

2

21

Forexaminer’s

use only


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23 Ali took part in shooting competition at the state level. Each participant will be given

the opportunity to shoot of 20 times. It was found that estimates for Ali shoot the target

is 4.

Ali mengambil bahagian dalam suatu pertandingan menembak peringkat negeri. Setiap

peserta akan diberi peluang menembak sebanyak 20 kali . Didapati bahawa anggarantembakan Ali kena pada sasaran adalah 4 .

(a)

Find the probability Ali hits the target.

Cari kebarangkalian tembakan Ali kena pada sasaran.

(b) If Ali was given another 3 chances to shoot, find the probability that 2 shots hit

the target.

Jika Ali diberi 3 kali lagi peluang menembak, cari kebarangkalian 2 daripada

tembakan tersebut kena pada sasaran.

[3 marks][3 markah]

Answer/ Jawapan:

(a)

(b)

3

23

Forexaminer’s

use only


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24 A committee of 5 students is to be form from 5 boys and 7 girls. Find

Satu jawatankuasa terdiri daripada 5 pelajar dapat dibentuk daripada 5 pelajar lelaki

dan 7 pelajar perempuan. Cari

(a) the number of ways, the committee can be form if the committee must consists of 3

boys and 2 girls,

bilangan cara jawatankuasa itu dibentuk jika jawatankuasa tersebut mestilah

terdiri daripada 3 lelaki dan 2 perempuan,

(b) the number of ways the committee members can be arranged in a row for a

group photograph, if the girls need to be sit separately .

bilangan cara menyusun jawatankuasa itu dalam satu sesi bergambar , jikakedua-dua pelajar perempuan itu duduk berasingan.

[4 marks]

[4 markah]

Answer/ Jawapan:

(a)

(b)

4

24

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25

X 0 1 2 3 4 5

P(X= x)

243

1

243

10

243

40 k

243

80

243

32

Table 25 shows the probability distribution for number of son in a family.

Jadual 25 menunjukkan taburan kebarangkalian bagi bilangan anak lelaki dalam sebuah

keluarga.

FindCari

(a) the value of k.,

nilai k.,

(b) the probability of getting son.

kebarangkalian mendapat anak lelaki.

[4 marks]

[4 markah]

Answer/ Jawapan:

(a)

(b)

END OF QUESTION PAPERKERTAS SOALAN TAMAT

4

25

Table 25

Jadual 25

Forexaminer’s

use only


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THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0,1)

KEBARANGKALI AN 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

k

dz z f z Q )()(


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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.

The Upper Tail Probability Q( z ) For The Normal Distribution N(0, 1) Table is

provided on page 22.

Jadual Kebarangkalian Hujung Atas Q( z ) Bagi Taburan Normal N(0, 1) disediakan

di halaman 22 .

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

PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2015

MATEMATIK TAMBAHAN

KERTAS 2

MODUL 2

2

12 jam

Dua jam tiga puluh minit

JANGAN BUKA MODUL INI SEHINGGA DIBERITAHU

1. This module 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 onlyone

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. The Upper Tail Probability Q(z) For The Normal Distribution N(0,1) Table is provided on

Page 20 .

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

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

Modul ini mengandungi 20 halaman bercetak.


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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

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 am

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 x

2

3. dx

du

du

dy

dx

dy

GEOMETRY

1. Distance =2

122

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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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 5 x y and 23 2 3 x y .

Give your answer correct to three decimal places [5 marks]

Selesaikan persamaan serentak 2 5 x y dan 23 2 3 x y .

Beri jawapan anda betul kepada tiga tempat perpuluhan. [5 markah]

2 Given that the function 22 f x x nx p has a minimum point at 1, 7 .

(a) Find the value of n and of p. [3 marks]

(b) Sketch the graph of the function f x . [2 marks]

(c) Hence, find the range of value of h if the function f x h has two distinct roots.

[2 marks]

Diberi bahawa fungsi 22 f x x nx p mempunyai titik minimum pada 1, 7 .

(a) Cari nilai n dan p. [3 markah]

(b) Lakar graf fungsi f x . [2 markah]

(c) Seterusnya, cari julat bagi nilai h jika fungsi f x h mempunyai dua punca yang berbeza.

[2 markah]


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6

5

T

S R

P

Q

U

Diagram 5 / Rajah 5

In Diagram 5, PQU is right angled triangle and PQRS is a quadrilateral. The straight lines PS

and QU intersect at point T . It is given 10 PU x , 6 PQ y ,1

2 RS PU , 15 6QR x y ,

: 1 : 2QT QU , : : PT TS m n , 4 x units and 5 y units.

(a) Find QU .

(b) Express in terms of x and / or y

(i) UT

(ii) PS

(c) Find :m n .

[8 marks]


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7

Dalam Rajah 5, PQU ialah sebuah segitiga tegak dan PQRS ialah sebuah sisi empat. Garis

lurus PS dan QU bersilang di titik T. Diberi bahawa 10 PU x , 6 PQ y ,1

2 RS PU ,

15 6QR x y , : 1 : 2QT QU , : : PT TS m n , 4 x unit dan 5 y unit .

(a) Cari QU .

(b) Ungkapkan dalam sebutan x dan / atau y

(i) UT

(ii) PS

(c) Cari :m n .

[8 markah]

6

Diagram 6 / Rajah 6

A farmer needs to build security fence along the remaining 3 sides of front compound of the farm

house as shown in Diagram 6. Find the maximum area of compound that can be enclosed if the

farmer has only 220 m of fencing.

[6 marks]

Seorang penternak perlu membina pagar keselamatan sepanjang 3 sempadan bagi halaman

hadapan rumah ternakan seperti ditunjuk dalam Rajah 6. Cari luas maksimum halaman yang

boleh dikelilingi jika penternak itu hanya mempunyai 220 m pagar.

[6 markah]

x

y

fence

farm house


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8

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.

Guna kertas graf untuk menjawab soalan ini.

x 1 0 1 5 2 0 2 5 3 0 3 5

y 5 01

3 55

2 50

1 77

1 26

0 88

Table 7/ Jadual 7

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

(a) Based on Table 7, construct a table for the values of10log y . [1 mark ]

(b) Plot10log y against x , using a scale of 2 cm to 0.5 unit on the x axis and 2 cm to 0.1

unit on the10log y -axis. Hence, draw the line of best fit. [3 marks]

(c) Use the graph in 7 (b),

(i) express y in terms of x ,

(ii) find the value of x when y = 2 . [6 marks]

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

satu eksperimen.

(a) Berdasarkan Jadual 7 , bina satu jadual untuk nilai-nilai10log y .

[1 markah]

(b) Plot 10log y melawan x , dengan menggunakan skala 2 cm kepada 0.5 unit pada paksi x

dan 2 cm kepada 0.1 unit pada paksi-10log y . Seterusnya, lukis garis lurus penyuaian

terbaik.

[3 markah]

(c) Gunakan graf di 7(b),

(i) ungkapkan y dalam sebutan x,

(ii) cari nilai x apabila y=2. [6 markah]


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9

8

Diagram 8 / Rajah 8

Retapy Sdn Bhd wants to build a tunnel with a curved top. The curve is an arc of circle from the

bottom of the tunnel. The width of the tunnel is 6 m and the height of the vertical wall is 8 24 m.

[ Use 3 142 ]

(a) What is the length of the curved top of the tunnel?

[6 marks]

(b) Find the area of the cross section of the tunnel.

[4 marks]

Retapy Sdn Bhd ingin membina sebuah terowong yang melengkung di atas. Lengkungan itu

ialah suatu lengkok bulatan daripada dasar terowong. Lebar terowong itu ialah 6 m dan tinggi

dinding mencancang ialah 8 24 m. [ Guna 3 142 ]

(a) Apakah panjang lengkungan atas terowong itu?

[6 markah]

(b) Cari luas keratan rentas terowong itu.

[4 markah]


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10

9 Diagram 9 shows a quadrilateral OPQR . Point S lies on the line PQ.

Rajah 9 menunjukkan sebuah sisi empat OPQR. Titik S terletak pada garis PQ.

Diagram 9 / Rajah 9

(a) Find the distance of RS .[2 marks]

(b) Point ,T x y moves such that its distance from point S is always 5 units. Find theequation of the locus of point T .

[2 marks]

(c) Given that point P and point Q lies on the locus T , calculate

(i) the value of h,(ii) the coordinates of Q.

[4 marks]

(d ) Find the area, in unit2, of the quadrilateral OPQR.

[2 marks]

(a) Cari jarak RS.[2 markah]

(b) Titik ,T x y bergerak dengan keadaan jaraknya dari titik S sentiasa 5 unit. Cari persamaan locus bagi titik T.[2 markah]

(c) Diberi titik P dan titik Q terletak pada lokus T , hitungkan

(i) nilai h, (ii) koordinat bagi Q.

[4 markah]

(d ) Cari luas, dalam unit2, sisiempat OPQR.

[2 markah]


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11

10 Diagram 10 shows part of the curve2

16 y

x . The straight line 2 y x intersects the curve at

point A.

Rajah 10 menunjukkan sebahagian daripada lengkung 216

y x . Garis lurus 2 y x meyilang

lengkung itu pada titik A.

Diagram 10/ Rajah 10

(a) Find the coordinates of point A. [2 marks]

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

(c) Calculate the volume generated, in terms of π, when the shaded area G rotated through 360⁰

about the x-axis.

[4 marks]

(a) Cari koordinat titik A. [2 markah]

(b) Hitung luas rantau berlorek H. [4 markah]

(c) Hitung isipadu yang dijanakan, dalam sebutan π, apabila rantau G dikisarkan melalui 360⁰

pada paksi-x.

[4 markah]

y =

H

G

0

x = 4

x

y = 2 x

A

y


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12

11 (a) In a survey, it is found that 65% of households in Malaysia have internet at home.

A sample of 20 households is chosen at random.

(i) What is the standard deviation of the household?

(ii) Find the probability that exactly 12 households have internet at home.

[5 marks]

(b) The mass of durians from a farm have a normal distribution with a mean of 2 kg and a

standard deviation of 0.8 kg. Calculate

(i) the probability that a durian chosen at random from this farm has a mass of more than

1 kg.

(ii) the value of m if 68% of the durian have masses less than m kg.

[5 marks]

(a) Dalam satu kajian, didapati bahawa 65% penghuni rumah di Malaysia mempunyai

internet di rumah. Satu sample 20 penghuni rumah dipilih secara rawak.

(i) Apakah sisihan piawai penghuni rumah ?

(ii) Cari kebarangkalian tepat 12 penghuni rumah mempunyai internet di rumah.

[5 markah]

(b) Jisim bagi buah durian dari sebuah ladang mempunyai taburan normal dengan min 2 kg

dan sisihan piawai 0.8 kg. Hitung(i) kebarangkalian bahawa sebiji durian yang dipilih secara rawak dari ladang ini

mempunyai jisim lebih daripada 1 kg.

(ii) nilai m jika 68% daripada durian mempunyai jisim kurang daripada m kg.

[5 markah]


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13

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 = 22 +7t – 2t 2 , where t is the time, in seconds, after leaving the point O.

[Assume motion to the right is positive.]

Find

(a) the velocity of the particle when the acceleration is zero, [3 marks]

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

(c) the distance from O when the particle is stop instantaneously, [2 marks]

(d ) the total 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 = 22 +7t – 2t 2 , dengan t ialah masa, dalam saat , selepas meninggalkan titik O.

[ Anggapkan gerakan ke arah kanan sebagai positif.]

Cari

(a) halaju zarah apabila pecutannya sifar , [3 markah]

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

(c) jarak dari O apabila zarah itu berhenti seketika, [2 markah]

(d ) jumlah jarak yang dilalui, dalam m , oleh zarah itu dalam 7 saat pertama. [3 markah]


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14

13 Diagram 13 is a bar chart indicating the weekly cost of the items P , Q, R, S and T for

the year 2005. Table 13 shows the prices and the price indices for the items.

Rajah 13 ialah carta bar yang memaparkan kos mingguan bagi bahan-bahan P , Q, R, S dan T

bagi tahun 2005. Jadual 13 menunjukkan harga-harga dan harga indeks untuk bahan-bahan.

Weekly cost / Kos mingguan (RM)

34

32

25

16

13

0 Items /

Bahan-bahan

Diagram 13 / Rajah 13

Items Bahan-

bahan

Price in / Harga pada

2005(RM)

Price in / Harga pada

2010(RM)

Price Index in 2010 based on 2005

Indeks harga pada

2010 berasaskan

2005

P x 700 175Q 002 502 125

R 004 505 yS 006 009 150T 502 3 00 120

Table 13 / Jadual 13

(a) Find the value of

(i) x, (ii) y.

[3 marks]

P Q R S T


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15

(b) Calculate the composite index for the items in the year 2010 based on the year 2005.

[3 marks]

(c) The total monthly cost of the items in the year 2005 is RM456.

Calculate the corresponding total monthly cost for the year 2010.

[2 marks]

(d ) The cost of the items increases by 20% from the year 2010 to the year 2014.

Find the composite index for the year 2014 based on the year 2005.

[2 marks]

(a) Cari nilai bagi

(i) x,

(ii) y.

[3 markah]

(b) Hitung indeks gubahan bagi bahan-bahan pada tahun 2010 berasaskan tahun 2005.

[3 markah]

(c) Jumlah kos bulanan bahan-bahan pada tahun 2005 ialah RM456.

Hitung jumlah kos bulanan yang sepadan pada tahun 2010.

[2 markah]

(d ) Kos bahan-bahan meningkat 20% dari tahun 2010 ke tahun 2014.

Cari indeks gubahan bagi tahun 2014 berasaskan tahun 2005.

[2 markah]


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16

14 Use graph paper to answer this question.

A factory produces two types of electronic devices P and Q by using machines A and B.

Table 14 shows the time taken to produce devices P and Q respectively.

Device

Peranti

Time taken (minutes)

Masa diambil (minit)

Machine A Mesin A

Machine B Mesin B

P 50 20

Q 25 40

Table 14/ Jadual 14

In any given week, the factory produces x units of device P and y units of device Q.

The production of the electronic devices per week is based on the following constraints:

I : Machine A operates not more than 2500 minutes.

II : Machine B operates at least 1600 minutes.

III : The number of device Q produced is not more than three times the number of device P

produced.

(a) Write three inequalities, other than x 0 and y 0, which satisfy all the above constraints.

[3 marks]

(b) Using a scale of 2 cm to 10 units on both axes, construct and shade the region R which

satisfies all of the above constraints.

[3 marks]

(c) Use your graph in 14(b) to find

(i) the maximum number of device P that could be produced, if the factory plans to

produce only 30 units of device Q,

(ii) the maximum profit per week if the profit from a unit of device P is RM20 and

from a unit of device Q is RM30.

[4 marks]


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17

Guna kertas graf untuk menjawab soalan ini.

Sebuah kilang menghasilkan dua peranti elektronik P dan Q dengan menggunakan mesin A

dan B. Jadual 14 menunjukkan masa yang diambil untuk menghasilkan peranti P dan Q.

Dalam mana-mana satu minggu, kilang tersebut menghasilkan x unit bagi peranti P dan y unit bagi

peranti Q. Penghasilan peranti-peranti tersebut adalah berdasarkan kekangan berikut:

I : Mesin A beroperasi tidak melebihi 2500 minit.

II : Mesin B beroperasi sekurang-kurangnya 1600 minit.

III : Bilangan peranti Q yang dihasilkan tidak melebihi tiga kali ganda bilangan peranti P

yang dihasilkan.

(a) Tuliskan tiga ketaksamaan, selain x 0 dan y 0 , yang memenuhi semua kekangan di

atas.

[3 markah]

(b) Menggunakan skala 2 cm kepada 10 unit pada kedua-dua paksi, bina dan lorekrantau R yang memenuhi semua kekangan di atas.

[3 markah]

(c) Gunakan graf anda di 14(b) untuk mencari

(i) bilangan maksimum bagi peranti P yang boleh dihasilkan jika kilang tersebut

bercadang untuk menghasilkan 30 unit peranti Q sahaja,

(ii) keuntungan maksimum seminggu jika keuntungan yang diperoleh dari satu unit

peranti P ialah RM20 dan dari satu unit peranti Q ialah RM30.

[4 markah]


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18

15 Diagram 15 shows a quadrilateral PQRS such that ∠ PQR is acute.

Diagram 15 / Rajah 15

(a) Calculate

(i) ∠ PQR [2 marks]

(ii) ∠ RSP . [2 mark s]

(iii) the area, in cm2, of quadrilateral PQRS . [4 marks]

(b) A triangle PQ’R has the same measurement as triangle PQR, that is PR = 15 cm,

RQ’ = 9 cm and ∠Q’PR = 30o, but is different in shape to triangle PQR.

(i) Sketch the triangle PQ’R,

(ii) State the size of ∠ P Q’ R. [2 marks]

Q

R

S

P

8 cm

15 cm

10 cm

30o

9 cm


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19

END OF QUESTION PAPER

KERTAS SOALAN TAMAT

Rajah 15 menunjukkan sebuah sisiempat PQRS dengan ∠ P Q R ialah sudut tirus.

(a) Hitungkan

(i)∠

PQR, [2 markah]

(ii)∠ RSP [2 markah]

(iii) luas, dalam cm2, bagi sisiempat PQRS . [4 markah]

(b) Satu segi tiga PQ’ R mempunyai sukatan yang sama dengan segitiga PQR, dengan

PR = 15 cm, RQ’ = 9 cm dan ∠Q’PR = 30o, tetapi mempunyai bentuk yang berbeza dengan

segitiga PQR.

(i) Lakarkan segitiga PQ’R,

(ii) Nyatakan saiz ∠ P Q’ R. [2 markah]


44/64

SULIT 3472/2

August 2015


20

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

k

dz z f z Q )()(


45/64

SULIT 3472/1

3472/1 Additi onal M athematics Paper 1

SULIT

1

Additional

Mathematics

Paper 1

August, 2015

PROGRAM PENINGKATAN PRESTASI AKADEMIK

SPM 2015

ANJURAN

MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)


MARKING SCHEME

Paper 1

MODUL 2

.


46/64

SULIT 3472/1


SULIT

2

PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2015

Marking Scheme

Additional Mathematics Paper 1

Question Solution/ Marking Scheme Answer Marks

1

(a) }2,1,4,7{

(b)

2

1

1

2 B2: 4 15k k x m m x or2 5k or m

B1:

k k x m m

2k and 5m 3

3

(b) B1: 3

34

p p p

(a) 5

2)(1

x x g

(b)

4

9 p

1

2

4

(b) B1 : 25 (0 ) 4 p

(a) 4q

(b) 1 p

1

2

5

B2 : 2 7 3 0 x x or

or7

2 x , 3 x

B1: 22 21 x x

3 x

2

7 x

33 7

2


47/64

SULIT 3472/1


SULIT

3

6

(a) B1:

21 1

2 52 2

k k

2

7k

(b) 2 POR

2

1

7

B2:1

4 1 804

x

B1 : 14

4

x

3 x 3

8B2 :

23 3

3 3

log log

log 9 log 9

a b

B1 : 29 9log loga b or2

233 3

3 3

loglog log

2 log 9 log 9

a

a bb or or

13 3

9

B3:

109000 1.2 1

20 000 101.2 1

B2: 109000 1.2 1

1.2 1

B1: 1.2 20 000 10r or RM

33 628 RM

and

Shop lot B4

10

(b) B1: 76)7)(1(29 n

(a) 8

(b) 16

1

2

11

B1: aa

551

5

4r

2


48/64

SULIT 3472/1


SULIT

4

12

(b) B2 : 544

3

2

1 2

r or 12r

B1 :*

21 36

2 4r

(a) 34

rad

(b) 8

1

3

13

B2: 7 8 x or y

B1:1 4 10 0 6 2

2 2 2 2

x yor

7, 8 3

14

B2: 52 h or 122 k

B1:

2

22

2

13)()(

k yh x A

6,

2

5 3

15

B2: 2 22 7 x y x

B1: 2 7m or c 2

72 y

x 3

16

(a) 3

(b) 48k

(c) 34k

1

1

1

17 B3 : 90 x

B2:1

45 02

x

B1:1

452

x

max 2025 P 4


49/64

SULIT 3472/1


SULIT

5

18(a) B1 :

2

03 f x dx

(b) B1 :

2 2

0 0

15

2dx f x dx

or

x5

(a) 2

(b)2

9 3

2

2

19

B2: 3

0

16 3 3

2 f x dx

B1: 1

6 3 3 32

or any number 10·5 3

20(a) B1 : RQ = PQ RP

(b) B1: )3()(3

2 x RQ or

)3()69(3

2 x x y

x y 69

x y 6

2

2

21

OC AO AC

k B

25

2

7

2:1 4 2

22

)(b B2: 81.311,12.482

B1: cos3

22 or 3cos 22

)(a 3

( ) 24 06,

155 91

b

1

3


50/64

SULIT 3472/1


SULIT

6

23

)(b B1:

4

4

5

1 2

2

3

c or 3

5

4

5

1

5

1

)(a5

1 p

12512

1

2

24

(a) B1:

(b)

5 7

3 2C C

(b) B1:5! (2 4!)

(a) 210

(b) 72

2

2

25

(a) B1 :243

32

243

80

243

40

243

10

243

11

(b) B1:24332)()( 05

5

5 q pC or243325 p

243

80k

(b)32

2

2

END OF MARKING SCHEME


51/64

1

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

3472/2

Additional

Mathematics

August 2015

PROGRAM PENINGKATAN PRESTASI AKADEMIK

SPM 2015


Paper 2

( MODULE 2 )

.

MARKING SCHEME


52/64

2

MARKING SCHEME

ADDITIONAL MATHEMATICS TRIAL EXAMINATION AUGUST 2015

MODULE 2 ( PAPER 2 )

N0. SOLUTION MARKS

1 5 2 y x or 5 2 y x

23 2(5 2 ) 3 x x 23 4 13 0 x x

24 4 4(3)( 13)

2(3) x

1 519 x and 2 852 x (both)

1 962 y and 10 704 y

P1

K1 Eliminate x/y

K1 Solve quadratic equation

N1

N1

5

2

(a)

(b)

(c)

2 2

24 8

n x x p

2

74 8

n n x or p

n = 4,

p = -5

2

4 4 2 5 0h

7h

K1

N1

N1

P1 (Shape)

P1 ( Min point)

K1

N1

7


53/64

3

N0. SOLUTION MARKS

3

(a)

(b)

3

4log3log

4log3log3

4log3log

4log33log3

4log3log

4log3log 33

nn

nn

nn

nn

nn

nn

nn

y y x

y y x

aaa

aaa

2

13

log

2

1loglog3

logloglog 3

K1

K1

N1

K1

K1

N1

6

4

(a)

(b)

(i)

2

2 2

2

2 2

sin

cos cos cos

cos sin

cos 2

x

x x x

x x

x

K1 for

2

2

sin

cos

x

x

N1

P1 for - cosine curve

P1 for amplitude 3 and -3

P1 for cycle 0 to

K1 for1

2

x y


54/64

4

(ii)

2 2 1

3 cos 1 tan2

13cos2

2

1

2

x x x

x x

x y

Number of solution = 2

N1

N1

8

5

(a)

(b)(i)

(ii)

(c)

6 10

QU QP PU

y x

2 230 40

50

QU

units

1

2

16 10

2

3 5

UT UQ

y x

y x

6 15 6 520 12

PS PQ QR RS

y x y x x y

10 3 5

5 3

PT PU UT

x y x

x y

K1 find (a) triangle law

OR b(ii) quadrilateral law

K1

N1

N1

N1

K1 find 10 3 5 PT x y x

OR

3 5 15 6 5TS y x x y x


55/64

5

3 5 15 6 5

15 9

TS TQ QR RS

y x x y x

x y

:5 3 : 15 9

1 : 3

PT TS x y x y

K1

N1

8

6

2 220

220 2

x y

y x

2

220 2

220 2

A xy

A x x

x x

220 4 0

55

dA x

dx x

2

2 4 0

d A

dx

Maximum

max

2

55 110

55 110

6050

x y

A

m

P1

K1

K1

K1

N1

N1

6


56/64

6

N0. SOLUTION MARKS

7

(a)

(b)

(c)

(i)

(ii)

x 1 2 3 4 5 6

10log y 0.70 0.55 0.40 0.25 0.10 -0.06

10log y

*gradient

= - 0.30

y-intercept = 1.0

10

0.3 1

log 0.3 1

10 x

y x

y

10

log 0.30 y

x = 2.35

N1 6 correct

values of 10log y

K1 Plot 10log y vs

x.

Correct axes &

uniform scale

N1 6 points plotted

correctly

N1 Line of best-fit

K1 finding gradient

K1 for y-intercept

K1

N1

K1 finding x

N1

10

x

0

1.0


57/64

7

N0. SOLUTION MARKS

8

(a)

(b)

3 m

8.24 m

r

s

1 8 24tan

3

180 2 70 40

2 23 8 24

8 769

r

m

408 769

180

6 12

s

m

1 3 8 24 24 72 A

2

2

1 408 769 26 8415

2 180 A

Area of the cross section of the tunnel

2

24 72 26 8415

51 56

A

m

K1

K1

K1

K1 Use s r

K1 in rad

N1

K1

K1 Use formula

21

2 A r

K1

N1

10


58/64

8

N0. SOLUTION MARKS

9

(a)

(b)

(c)

(i)

(ii)

(d)

Distance

2 27 1 1 7

10 units

Locus T

2 2

2 2

2 2

1 7 5

2 1 14 49 25 0

2 14 25 0

x y

x x y y

x y x y

2

2

5

25 10 14 25 0

14 40 0

4 10

x

h h

h h

h h

4h

[ Use distance PS ]

2

2

10

100 2 140 25 0

2 15 0

5 3

y

x x

x x

x x

OR

int

5 41 7

2 2

3 10

mid po

x yand

x y

3,10Q

[ Use mid-point / distance QS ]

Area OPQR

0 7 3 5 01

0 1 10 4 02

170 12 3 50

2

129 1@ 64

2 2

K1

N1

K1

N1

K1

N1

K1

N1

K1

N1

10


59/64

9

N0. SOLUTION MARKS

10

(a)

(b)

(c)

2

162 x

x

Coordinate A = (2,4)

1

4 8 22

4

2

2

16 x dx

4

2

1612

x

8

24

2

2

16dx

x

21

4 23

28 32

3 3

20

K1

N1

K1 Area of

trapezium

K1 integrate and sub.

the limit correctly

K1

N1

K1 integrate and sub.

the limit correctly

K1 volume of cone

K1

N1

10


60/64

10

N0. SOLUTION MARKS

11

(a)

(i)

(ii)

(b)

(i)

(ii)

Standard deviation,

20 0 65 0 35

2 133

20 12 81212 0 65 0 35

0 1614

P X C

µ= 2 , σ = 0.8

P( X > 1 ) = P (Z >1 2

0.8

)

= P( Z > -1.25)

= 1.25 P Z

= 1 – 0.1056

= 0.8944

P( X < m) = 0.68

P( X > m) = 1 − 0.68 = 0.32

m − 2

0.8 = 0.468

m = 2.374

P10 65

0 35

p and

q

K1

N1

K1

N1

K1 Use Z =

X

N1

K1

K1

N1

10


61/64

11

N0. SOLUTION MARKS

12

(a)

(b)

(c)

(d)

dva

dt 0

7 – 4t = 0

t 7

4

128

8v

(2t – 11 ) (t + 2) = 0

11

2t

, t = – 2 (not accepted)

3

27 2222 3

11

2

t S t t

t

s 23

11524

2 37

7 222(7) (7) (7)

2 3S

Total distance =23 23 5

115 (115 96 )24 24 6

=1

13512

m

K1

K1 sub t into v

N1

K1

N1

(for11

2t only)

K1(for integration)

N1

K1

K1

(for summation)

N1

10


62/64

12

N0. SOLUTION MARKS

13

(a)

(b)

(c)

(d)

.

x

0 7 175 100 (or formula finding y /z )

x = 0.40

y = 137.5

W = 16 , 32 , 25 , 34 , 13

( ) ( ) ( . ) ( ) ( ) x x x x x I

175 16 125 32 137 5 25 150 34 120 13

120

= 140.81

. x456 140 81

100

= RM 642.09

. x140 81 120

100 (or 140.81 + 140.81x0.2)

= 168.97

K1

N1

N1

P1

K1

N1

K1

N1

K1

N1

10


63/64

13

N0. SOLUTION MARKS

14

(a)

(b)

(c)

(d)

I : 50 x + 25 y 2500 or

2 x + y 100

II : 20 x +40 y 1600 or

x + 2 y 80

III : y 3 x

(20, 60)

(35, 30)

2x + y = 100

y = 3x

x + 2y = 80

y = 30

100

90

80

70

60

50

40

30

20

10

10080604020 9070503010

x

y

At least one straight line is drawn correctly from inequalities involving

x and y.

All the three straight lines are drawn correctly

Region is correctly shaded

35

Maximum point (20, 60)

Maximum profit = 20(20) + 30(60)

= RM 2200

N1

N1

N1

K1

N1

N1

N1

N1

K1

N1

10


64/64

N0. SOLUTION MARKS

15

(a)(i)

(ii)

(iii)

(b)(i)

(ii)

sin sin o

PQR

30

15 9

PQR = 56.44o

cos( )( )

RSP

2 2 2 8 10 15

2 8 10

RSP = 112.41o

PRQ = 93.56o

Area = ( )( )sin . ( )( )sin .o o 1 1

9 15 93 56 8 10 112 41 2 2

= 67.37 + 36.98

= 104.35

123.56o

K1

N1

K1

K1

K1

K1, K1

(for using

area= ½absinc

andsummation)

N1

N1

N1

R P

Q’

P1+P2+Skema 2015 kedah add math

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