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SEKOLAH

PEPERIKSAAN PERCUBAAN SPM TAHUN 2009

ADDITIONAL MATHEMATICS

Paper 2 Two hours and thirty minutes

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Kertas soalan ini adalah dalam dwibahasa.

2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam Bahasa

Malaysia.

3. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.

4. Calon dikehendaki menceraikan halaman 18 dan ikat sebagai muka hadapan

bersama-sama dengan buku jawapan.

Kertas soalan ini mengandungi 19 halaman bercetak.

3472/2 Form Five Additional Mathematics Paper 2 September 2009 2 ½ hours

http://mathsmozac.blogspot.com

http://tutormansor.wordpress.com/

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

ALGEBRA

1 a

acbbx

2

42 −±−= 8 a

bb

c

ca log

loglog =

2 am x an = a m + n 9 ( 1)nT a n d= + −

3 am ÷ an = a m – n 10. [ ]2 ( 1)2n

nS a n d= + −

4 ( am )n = a m n 11 1nnT a r −=

5 nmmn aaa logloglog += 12 ( ) ( )1 1

, 11 1

n n

n

a r a rS r

r r

− −= = ≠

− −

6 nmn

maaa logloglog −= 13 , 1

1

aS r

r∞ = <−

7 log a mn = n log a m

CALCULUS KALKULUS

1 y = uv , dx

duv

dx

dvu

dx

dy +=

4 Area under a curve Luas di bawah lengkung

= ( )b b

a a

y dx or atau x dy∫ ∫

2 2

,v

dx

dvu

dx

duv

dx

dy

v

uy

−==

5 Volume generated Isipadu janaan

= 2 2( )b b

a a

y dx or atau x dyπ π∫ ∫

3 dx

du

du

dy

dx

dy ×=



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STATISTICS STATISTIK

1 N

xx

Σ=

7 i

ii

W

IWI

ΣΣ

=

2 f

fxx

ΣΣ=

8 !

( )!n

r

nP

n r=

−

3 ( ) 2

22

xN

x

N

xx −Σ=−Σ=σ

9 !

( )! !n

r

nC

n r r=

−

4 ( ) 2

22

xf

fx

f

xxf −Σ

Σ=Σ

−Σ=σ

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

5 Cf

FNLm

m

−+= 2

1

11 ( ) , 1n r n rrP X r C p q p q−= = + =

12 / min,Mean npµ =

6 1

0100

QI

Q= ×

13 npqσ =

14 x

Zµ

σ−=

GEOMETRY GEOMETRI

1 Distance/jarak

= ( ) ( )221

221 yyxx −+−

4 Area of a triangle/ Luas segitiga =

( ) ( )3123121332212

1yxyxyxyxyxyx ++−++

2 Mid point / Titik tengah

( )

++=2

,2

, 2121 yyxxyx

5 2 2

~r x y= +

3 A point dividing a segment of a line

Titik yang membahagi suatu tembereng garis

( )

++

++=

nm

myny

nm

mxnxyx 2121 ,,

6 ^

~ ~

2 2~

x i y jr

x y

+=

+



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TRIGONOMETRY TRIGONOMETRI

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

8 ( )sin sin cos cos sinA B A B A B± = ±

( )sin sin s sinA B A kos B ko A B± = ±

2 Area of a sector, 21

2A r θ=

Luas sektor, L = 21

2j θ

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

( )s os os sin sinko A B k A k B A B± = ∓

3 2 2cos 1sin A A+ =

2 2os 1sin A k A+ =

10 ( ) tan tantan

1 tan

A BA B

A tanB

±± =∓

4 2 2sec 1 tanA A= +

2 2se 1 tank A A= +

11 2

2 tantan 2

1 tan

AA

A=

−

5 2 2sec 1 cotco A A= +

2 2se 1 otko k A k A= +

12 sin sin sin

a b c

A B C= =

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

13 2 2 2 2 cosa b c bc A= + −

2 2 2 2a b c bc kos A= + −

7 cos 2A = cos2 A – sin2 A = 2 cos2A – 1 = 1 – 2 sin2 A kos 2A = kos2 A – sin2 A = 2 kos2A – 1 = 1 – 2 sin2 A

14 Area of triangle/ Luas segitiga

= 1

sin2

ab C



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

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

Minus / Tolak

0.0

0.1

0.2

0.3

0.4

0.5000

0.4602

0.4207

0.3821

0.3446

0.4960

0.4562

0.4168

0.3783

0.3409

0.4920

0.4522

0.4129

0.3745

0.3372

0.4880

0.4483

0.4090

0.3707

0.3336

0.4840

0.4443

0.4052

0.3669

0.3300

0.4801

0.4404

0.4013

0.3632

0.3264

0.4761

0.4364

0.3974

0.3594

0.3228

0.4721

0.4325

0.3936

0.3557

0.3192

0.4681

0.4286

0.3897

0.3520

0.3156

0.4641

0.4247

0.3859

0.3483

0.3121

4

4

4

4

4

8

8

8

7

7

12

12

12

11

11

16

16

15

15

15

20

20

19

19

18

24

24

23

22

22

28

28

27

26

25

32

32

31

30

29

36

36

35

34

32

0.5

0.6

0.7

0.8

0.9

0.3085

0.2743

0.2420

0.2119

0.1841

0.3050

0.2709

0.2389

0.2090

0.1814

0.3015

0.2676

0.2358

0.2061

0.1788

0.2981

0.2643

0.2327

0.2033

0.1762

0.2946

0.2611

0.2296

0.2005

0.1736

0.2912

0.2578

0.2266

0.1977

0.1711

0.2877

0.2546

0.2236

0.1949

0.1685

0.2843

0.2514

0.2206

0.1922

0.1660

0.2810

0.2483

0.2177

0.1894

0.1635

0.2776

0.2451

0.2148

0.1867

0.1611

3

3

3

3

3

7

7

6

5

5

10

10

9

8

8

14

13

12

11

10

17

16

15

14

13

20

19

18

16

15

24

23

21

19

18

27

26

24

22

20

31

29

27

25

23

1.0

1.1

1.2

1.3

1.4

0.1587

0.1357

0.1151

0.0968

0.0808

0.1562

0.1335

0.1131

0.0951

0.0793

0.1539

0.1314

0.1112

0.0934

0.0778

0.1515

0.1292

0.1093

0.0918

0.0764

0.1492

0.1271

0.1075

0.0901

0.0749

0.1469

0.1251

0.1056

0.0885

0.0735

0.1446

0.1230

0.1038

0.0869

0.0721

0.1423

0.1210

0.1020

0.0853

0.0708

0.1401

0.1190

0.1003

0.0838

0.0694

0.1379

0.1170

0.0985

0.0823

0.0681

2

2

2

2

1

5

4

4

3

3

7

6

6

5

4

9

8

7

6

6

12

10

9

8

7

14

12

11

10

8

16

14

13

11

10

19

16

15

13

11

21

18

17

14

13

1.5

1.6

1.7

1.8

1.9

0.0668

0.0548

0.0446

0.0359

0.0287

0.0655

0.0537

0.0436

0.0351

0.0281

0.0643

0.0526

0.0427

0.0344

0.0274

0.0630

0.0516

0.0418

0.0336

0.0268

0.0618

0.0505

0.0409

0.0329

0.0262

0.0606

0.0495

0.0401

0.0322

0.0256

0.0594

0.0485

0.0392

0.0314

0.0250

0.0582

0..0475

0.0384

0.0307

0.0244

0.0571

0.0465

0.0375

0.0301

0.0239

0.0559

0.0455

0.0367

0.0294

0.0233

1

1

1

1

1

2

2

2

1

1

4

3

3

2

2

5

4

4

3

2

6

5

4

4

3

7

6

5

4

4

8

7

6

5

4

10

8

7

6

5

11

9

8

6

5

2.0

2.1

2.2

2.3

0.0228

0.0179

0.0139

0.0107

0.0222

0.0174

0.0136

0.0104

0.0217

0.0170

0.0132

0.0102

0.0212

0.0166

0.0129

0.00990

0.0207

0.0162

0.0125

0.00964

0.0202

0.0158

0.0122

0.00939

0.0197

0.0154

0.0119

0.00914

0.0192

0.0150

0.0116

0.00889

0.0188

0.0146

0.0113

0.00866

0.0183

0.0143

0.0110

0.00842

0

0

0

0

3

2

1

1

1

1

5

5

1

1

1

1

8

7

2

2

1

1

10

9

2

2

2

1

13

12

3

2

2

2

15

14

3

3

2

2

18

16

4

3

3

2

20

16

4

4

3

2

23

21

2.4 0.00820 0.00798 0.00776 0.00755 0.00734

0.00714

0.00695

0.00676

0.00657

0.00639

2

2

4

4

6

6

8

7

11

9

13

11

15

13

17

15

19

17

2.5

2.6

2.7

2.8

2.9

0.00621

0.00466

0.00347

0.00256

0.00187

0.00604

0.00453

0.00336

0.00248

0.00181

0.00587

0.00440

0.00326

0.00240

0.00175

0.00570

0.00427

0.00317

0.00233

0.00169

0.00554

0.00415

0.00307

0.00226

0.00164

0.00539

0.00402

0.00298

0.00219

0.00159

0.00523

0.00391

0.00289

0.00212

0.00154

0.00508

0.00379

0.00280

0.00205

0.00149

0.00494

0.00368

0.00272

0.00199

0.00144

0.00480

0.00357

0.00264

0.00193

0.00139

2

1

1

1

0

3

2

2

1

1

5

3

3

2

1

6

5

4

3

2

8

6

5

4

2

9

7

6

4

3

11

9

7

5

3

12

9

8

6

4

14

10

9

6

4

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

Example / Contoh:

−= 2

2

1exp

2

1)( zzf

π If X ~ N(0, 1),

then Jika X ~ N(0, 1), maka

∫∞

=k

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

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

Q(z)

z

f (z)

O k



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Section A Bahagian A

[40 marks]

[40 markah]

Answer all questions. Jawab semua soalan

1. Solve the following simultaneous equations , give your answers correct to three decimal

places. Selesaikan persamaan serentak berikut dengan memberi jawapan anda tepat kepada

tiga tempat perpuluhan :

5312 =+=+ yxyx

[6 marks]

[6 markah] 2. In Diagram 1, ABCD is a quadrilateral. BFC and DEF are straight lines. Dalam Rajah 1, ABCD ialah sebuah sisi empat. BFC dan DEF adalah garis lurus.

Given that xBA 24= , yBF 10= , yxCD 3030 −= , BCBF4

1= and DFDE5

2= ,

Diberi xBA 24= , yBF 10= , yxCD 3030 −= , BCBF4

1= dan DFDE5

2= ,

(a) express in terms of x and/or y ,

ungkapkan dalam sebutan x dan/atau y ,

(i) AC (ii) DF [3 marks] [3 markah]

(b) show that the points A, E and C are collinear. [3 marks] tunjukkan bahawa titik A, E dan C adalah segaris . [3 markah]

Diagram 1 Rajah 1

A

B C

D

E ••••

F



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3.(a) Prove that xxx

xxcot

sin2sin

2coscos1 =+++

. [3 marks]

Buktikan bahawa xxx

xxcot

sin2sin

2coscos1 =+++

. [3 markah]

(b)(i) Sketch the graph of the trigonometric function 1cos3 += xy for the domain π20 ≤≤ x . Lakar graf bagi fungsi trigonometri 1cos3 += xy untuk domain π20 ≤≤ x .

(ii) On the same axes, sketch the graph of a suitable straight line that can be used to solve the equation ππ −= xx 3cos3 . State the number of solutions to the equation ππ −= xx 3cos3 for π20 ≤≤ x .

Pada paksi yang sama, lakar graf bagi satu garis lurus yang sesuai digunakan untuk menyelesaikan persamaan ππ −= xx 3cos3 . Nyatakan bilangan penyelesaian bagi persamaan ππ −= xx 3cos3 untuk

π20 ≤≤ x . [5 marks] [5 markah]

4. Table 1 shows the frequency distribution of the Additional Mathematics marks of a

group of students. Jadual 1 menunjukkan taburan kekerapan markah Matematik Tambahan bagi

sekumpulan pelajar.

Marks Number of students 1 – 10 2 11 – 20 3 21 – 30 5 31 – 40 10 41 – 50 K 51 – 60 2

Table 1 Jadual 1

(a) Given that the median mark is 34.5,

Diberi markah median adalah 34.5,

(i) calculate the value of k, hitungkan nilai k,

(ii) find the median mark if the mark of each student is increased by 8.

cari markah median jika markah setiap pelajar ditambahkan sebanyak 8. [4 marks] [4 markah]



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(b) Given that k = 4, draw a histogram to represent the frequency distribution of the

mark by using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1

student on the vertical axis.

Diberi k = 4, lukis sebuah histogram untuk mewakili taburan kekerapan markah

dengan menggunakan skala 2 cm kepada 10 markah pada paksi ufuk dan 2 cm

kepada 1 pelajar pada paksi tegak.

Hence, find the modal mark. [3 marks]

Seterusnya, cari markah mod. [3 markah]

5. (a) Find the equation of the normal to the curve x

xy1

3 += at (1 , 4). [3 marks]

Cari persamaan normal kepada lengkung x

xy1

3 += pada (1 , 4). [3 markah]

(b) Diagram 2 shows a leaking hemispherical container with a radius of r cm.

Rajah 2 menunjukkan sebuah bekas bocor yang berbentuk hemisfera dengan jejari

r cm.

Given that the radius of the water surface is decreasing at the rate of 0.1 cm s–1, find

in terms of π, the rate of change of the volume of water in the container at the instant

the radius of the water surface is 20 cm. [3 marks]

Diberi jejari permukaan air menyusut dengan kadar 0.1 cm s–1, cari dalam sebutan

π, kadar perubahan isipadu air dalam bekas itu pada ketika jejari permukaan air

adalah 20 cm. [3 markah]

Diagram 2 Rajah 2

r cm



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6. Diagram 3 shows a few circles. The first circle is the largest circle with a radius of

R cm. The second circle has a radius of R3

2cm . The third circle has a radius which is

3

2 of the radius of second circle and this process is continued indefinitely.

Rajah 3 menunjukkan beberapa bulatan. Bulatan pertama adalah bulatan terbesar dan

mempunyai jejari R cm. Bulatan kedua mempunyai jejari R3

2cm. Bulatan ketiga

mempunyai jejari yang merupakan 3

2 daripada jejari bulatan kedua dan proses ini

diteruskan sehingga ketakhinggaan.

(a) Show that the perimeters of the circles form a geometric progression with common

ratio 3

2by using first three circles. [2 marks]

Tunjukkan bahawa perimeter bulatan-bulatan itu membentuk satu janjang geometri

dengan nisbah sepunya 3

2dengan menggunakan tiga bulatan pertama. [2 marks]

(b) Given that the area of the largest circle is 900π cm2, find in terms of π,

Diberi bahawa luas bulatan yang terbesar ialah 900π cm2, cari dalam sebutan π, (i) the circumference of the tenth circle, Ukurlilit bagi bulatan yang kesepuluh, (ii) the sum of the circumference of infinite number of circles formed. [5 marks] jumlah ukurlilit bagi semua bulatan yang dapat dibentuk sehingga

ketakterhinggaan [5 markah]

R cm

Diagram 3 Rajah 3



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Section B Bahagian B

[40 marks]

[ 40 markah]

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

7. Use graph paper to answer this question. Gunakan kertas graf untuk menjawab soalan ini.

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

Variables x and y are related by the equation q

py

x 2+= , where p and qare constants.

One of the values of y is incorrectly recorded. Jadual 2 menunjukkan nilai-nilai bagi dua pembolehubah, x dan y , yang diperoleh daripada satu eksperimen. Pembolehubah x dan y dihubungkan oleh persamaan

q

py

x 2+= , dengan keadaan p dan qadalah pemalar. Satu daripada nilai y telah salah

direkodkan.

Table 2 Jadual 2

(a) Plot y10log against )2( +x , using a scale of 2 cm to 1 unit on the )2(+x - axis

and 2 cm to 0.05 unit on the y10log -axis.

[Start the y10log -axis with the value 0.8].

Hence, draw the line of best fit. [4 marks] Plot y10log melawan )2( +x , dengan menggunakan skala 2 cm kepada 1 unit

pada paksi- )2( +x dan 2 cm kepada 0.05 unit pada paksi- y10log .

[Mulakan paksi- y10log dengan nilai 0.8]

Seterusnya, lukis garis lurus penyuaian terbaik. [4 markah]

(b) Use your graph from 7(a), find Gunakan graf anda di 7(a), cari

(i) the correct value of y that is wrongly recoded. nilai yang betul bagi nilai y yang salah direkodkan. (ii) the values of p and q . [6 marks] nilai p dan nilai q. [6 markah]

x –1 0 1 2 3 4 y 8.4 10.1 12.1 13.2 17.4 20.9



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8. Solutions to this question by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak diterima.

Diagram 4 shows the triangle OAB where O is the origin. Point C lies on the straight line AB. Rajah 4 menunjukkan segitiga OAB dengan O ialah titik asalan. Titik C terletak pada garis lurus AB.

(a) Calculate the area, in unit2, of triangle OAB. [2 marks] Hitungkan luas, dalam unit2, bagi segitiga OAB. [2 markah] (b) Find the equation of the perpendicular bisector of line segment AB. [3 marks] Cari persamaan pembahagi dua sama serenjang bagi tembereng garis AB. [3 markah]

(c) Given that the length BC is 5

4 of the distance of point B from the perpendicular

bisector of the line segment AB, find the coordinates of point C. [2 marks]

Diberi panjang BC ialah 5

4daripada jarak titik B dari pembahagi dua sama

serenjang bagi tembereng garis AB, cari koordinat bagi titik C. [2 markah] (d) A point P moves such that its distance from point B is always twice its distance

from point C. Find the equation of the locus of P. [3 markah] Satu titik P bergerak dengan keadaan jaraknya dari titik B adalah sentiasa dua kali jaraknya dari titik C. Cari persamaan lokus bagi P. [3 markah]

x

y

B(6 , -8)

A(-4 , 2)

O

Diagram 4 Rajah 4

C •



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9. Diagram 5 shows the straight line y = 2x which passes through the maximum point of a

quadratic curve ))(( βα −−−= xxy , where βα and are constants.

Rajah 5 menunjukkan garis lurus y = 2x yang melalui titik maksimum suatu garis

lengkung kuadratik, ))(( βα −−−= xxy , dengan keadaan βα and ialah pemalar.

(a) State

Nyatakan

(i) the coordinates of the maximum point,

koordinat titik maksimum itu,

(ii) the equation of the quadratic curve. [2 marks]

persamaan garis lengkung kuadratik itu. [2 markah]

(b) Calculate the area of the shaded region P. [4 marks]

Hitungkan luas rantau berlorek P. [4 markah]

(c) Find the volume of the solid generated, in terms of π , when the

region Q is revolved through 360o about the x-axis. [4 marks]

Cari isipadu pepejal yang dijanakan, dalam sebutan π , apabila rantau

Q dikisarkan melalui 360o pada paksi-x. [4 markah]

x

y y = 2x

O 4

P

Q

Diagram 5 Rajah 5



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10. (a) In a house to house check carried out in Taman Maju, aedes mosquitoes were

found in 2 out of every 5 houses. If 8 houses in Taman Maju are chosen at random,

calculate the probability that

Dalam suatu pemeriksaan dari rumah ke rumah di Taman Maju, nyamuk aedes telah

dijumpai dalam 2 daripada setiap 5 buah rumah. Jika 8 buah rumah di Taman Maju

dipilih secara rawak, hitung kebarangkalian bahawa

(i) exactly 3 houses are infested with aedes mosquitoes,

tepat 3 buah rumah akan dijumpai dengan nyamuk aedes,

(ii) more than 2 houses are infested with aedes mosquitoes.

lebih daripada 2 buah rumah akan dijumpai dengan nyamuk aedes.

[5 marks]

[5 markah]

(b) A study on the body mass of a group of students is conducted and it is found that the

mass of a student is normally distributed with a mean of 50 kg and a variance of

256 kg2.

Satu kajian jisim badan dijalankan ke atas sekumpulan pelajar dan didapati jisim

seorang pelajar adalah mengikut taburan normal dengan min 50 kg dan varians

256 kg2.

(i) If a student is selected randomly, calculate the probability that his mass is more

than 60 kg.

Jika seorang pelajar dipilih secara rawak, hitungkan kebarangkalian bahawa

jisimnya adalah lebih daripada 60 kg.

(ii) Given that 28% of the students weigh less than m kg, calculate the value of m.

Diberi bahawa 28% daripada pelajar itu mempunyai jisim kurang daripada

m kg, cari nilai m.

[5 marks]

[5 markah]



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14

11. Use 142.3=π in this question. Gunakan 142.3=π dalam soalan ini.

Diagram 6 shows a circular sector OAC with a radius of 5 cm and ∠AOC is 1.2 radian. BC is an arc of the circle with centre A. Rajah 6 menunjukkan satu sektor bulatan OAC dengan jejari 5 cm dan ∠AOC adalah 1.2 radian. BC adalah lengkok bulatan dengan pusat A.

(a) Find Cari

(i) the length, in cm, of the arc AC. [2 marks] panjang, dalam cm, lengkok AC. [2 markah]

(ii) the length, in cm, of radius AB. [2 marks]

panjang, dalam cm, jejari AB. [2 markah]

(b) (i) Show that ∠BAC = 2.171 radian. Tunjukkan ∠BAC = 2.171 radian.

(ii) Hence, calculate the area, in cm2, of the shaded region.

Seterusnya, hitungkan luas, dalam cm2, rantau yang berlorek. [6 marks]

[6 markah]

O

A

B

C

5 cm

1.2 rad Diagram 6

Rajah 6



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15

Section C

Bahagian C

[20 marks] [20 markah]

Answer two questions from this section.

Jawab dua soalan daripada bahagian ini.

12. Table 3 shows the price indices and the percentage of usage of 5 different ingredients A, B, C, D and E needed to make a cake. The composite index number for the cost of making the cake in the year 2007 based on the year 2005 is 132. Jadual 3 menunjukkan indeks harga dan peratus penggunaan lima jenis bahan A, B, C, D dan E yang diperlukan untuk membuat sejenis kek. Nombor indeks gubahan kos membuat kek itu pada tahun 2007 berasaskan tahun 2005 ialah 132.

Ingredients Jenis bahan

Price index for the year 2007 based on the year 2005 Indeks harga pada tahun 2007 berasaskan tahun 2005

Percentage of ingredient Peratus bahan (%)

A 140 30 B x 20 C 110 15 D 104 10 E 120 25

Table 3 Jadual 3

(a) Calculate

Hitungkan

(i) the price of A in the year 2005 if its price in the year 2007 is RM7. harga A pada tahun 2005 jika harganya pada tahun 2007 ialah RM7.


(ii) the value of x. [2 marks]

nilai x. [2 markah]

(b) The cost of the cake increased 10% from the year 2007 to the year 2009. Find the price of the cake in the year 2009 if its price in the year 2005 is RM40. Kos penghasilan kek itu bertambah 10 % dari tahun 2007 ke tahun 2009. Cari harga kek itu pada tahun 2009 jika harganya pada tahun 2005 ialah RM40.


(c) Find the price index of D in the year 2007 based on the year 2003 if its price index

in the year 2005 based on the year 2003 is 125. [3 marks] Carikan indeks harga bagi D pada tahun 2007 berasaskan tahun 2003 jika indeks harganya pada tahun 2005 berasaskan tahun 2003 ialah 125. [3 markah]



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16

13. Diagram 7 shows a quadrilateral PQRS where the sides PQ and RS are parallel.

Rajah 7 menunjukkan sebuah sisiempat PQRS dengan keadaan sisi PQ dan sisi RS adalah selari.

Given that PQ = 10 cm, ∠RPQ = 1100 , ∠PQR = 500 and SR : PQ = 2 : 5, calculate Diberi PQ = 10 cm, ∠RPQ = 1100 , ∠PQR = 500 dan SR : PQ = 2 : 5, hitungkan (a) the length, in cm, of PR and QR. [3 marks]

panjang, dalam cm, PR dan QR. [3 markah]

(b) the length, in cm, of diagonal QS. [3 marks] panjang, dalam cm, perpenjuru QS. [3 markah]

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

luas, dalam cm2, PQRS. [4 markah]

14. A particle moves along a straight line and passes through a fixed point O with a velocity of 3 ms-1. Its acceleration, a ms-2, is given by a = 2 – 2t, where t is the time, in seconds, after passing through O. The particle stops momentarily at time, t = k s.

Suatu zarah bergerak di sepanjang suatu garis lurus dan melalui satu titik tetap O dengan halaju 3 ms-1. Pecutannya, a ms-2, diberi oleh a = 2 - 2t, dengan keadaan t ialah masa, dalam saat, selepas melalui O. Zarah itu berhenti seketika pada masa, t = k s.

Find Cari (a) the maximum velocity of the particle, [ 3 marks ]

halaju maksimum zarah itu, [3 markah]

(b) the value of k, [2 marks] nilai k, [2 markah]

(c) the distance travelled in the third second, [2 marks] jarak yang dilalui dalam saat ketiga. [2 markah]

(d) the value of t , correct to two decimal places, when the particle passes O again. [3 marks] nilai t, betul kepada dua tempat perpuluhan, apabila zarah itu melalui titik O semula. [3 markah]

Diagram 7 Rajah 7

10 cm

P

Q R 50o

110o

S



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17

15. Use the graph paper provided to answer this question Gunakan kertas graf yang disediakan untuk menjawab soalan ini.

A factory produces two types of school bags, type P and type Q. In a day, it can produce x bag of type P and y bag of type Q. The time taken to produce a bag of type P is 40 minutes and a bag of type Q is 50 minutes.

Sebuah kilang menghasilkan dua jenis beg sekolah, jenis P dan jenis Q. Dalam satu hari, kilang itu boleh menghasilkan x beg jenis P dan y beg jenis Q. Masa yang diambil untuk menghasilkan satu beg jenis P ialah 40 minit dan satu beg jenis Q ialah 50 minit.

The production of the bags per day is based on the following constraints : Pengeluaran beg dalam satu hari adalah berdasarkan kepada kekangan berikut :

I : The total number of bags produced is not more than 160.

Jumlah bilangan beg yang dihasilkan tidak melebihi 160.

II : The time taken to make bag P is not more than twice the time taken to make bag Q.

Masa yang diambil untuk membuat beg P tidak melebihi dua kali ganda masa yang diambil untuk membuat beg Q.

III : The number of bag Q exceed the number of bag P by at most 80. Bilangan beg Q melebihi bilangan beg P selebih-lebihnya 80.

(a) Write down three inequalities, other than 0≥x and 0≥y which satisfy all the

above constraints. [3 marks ] Tulis tiga ketaksamaan, selain x ≥ 0 dan y µ 0, yang memenuhi semua kekangan di atas. [3 markah]

(b) By using a scale of 2 cm to 20 bags on both axes, construct and shade the region R

that satisfies all the above constraints. [3 marks] Menggunakan skala 2 cm kepada 20 beg pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan di atas. [3 markah]

(c) Use your graph in 15 (b) to answer the following :

Gunakan graf anda di 15 (b) untuk menjawab yang berikut :

(i) Find the range of the number of bag Q that can be produced if the number of bag P is 50.

Cari julat bilangan beg Q yang boleh dihasilkan jika bilangan beg P ialah 50.

(ii) If the profit of selling bag P is RM20 and bag Q is RM30, find the maximum profit that can be obtained.

Jika untung jualan bagi beg P ialah RM20 dan beg Q ialah RM30, cari keuntungan maksimum yang boleh diperolehi. [4 marks] [4 markah]

END OF QUESTION PAPER KERTAS SOALAN TAMAT



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18

NAMA:

KELAS :

NO. KAD PENGENALAN: ANGKA GILIRAN Arahan Kepada calon Tulis nama, kelas, nombor kad pengenalan dan angka giliran anda pada ruang yang

disediakan.

Tandakan ( √ ) untuk soalan yang dijawab.

Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan kertas jawapan.

Kod Pemeriksa

Bahagian Soalan Soalan Dijawab

Markah Penuh

Markah Diperoleh (Untuk Kegunaan Pemeriksa)

A

1 6

2 6

3 8

4 7

5 6

6 7

B

7 10

8 10

9 10

10 10

11 10

C

12 10

13 10

14 10

15 10

SULIT



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19

3472/2[PP] Additional Mathematics Paper 2 September 2009

SEKTOR PENGURUSAN AKADEMIK JABATAN PELAJARAN PAHANG

PEPERIKSAAN PERCUBAAN SPM

TAHUN 2009

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 12 printed pages



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20

Question Working/Solution Marks Total 1.

512 =+yx

and 53 =+ yx

or any correct pairs from 5312 =+=+ yxyx

yx 35−= or 3

5 xy

−=

Substitute yx 35−= or 3

5 xy

−= into the non-linear

equation to obtain a quadratic equation in terms of x or y

052615 2 =+− yy or 010245 2 =+− xx Solve quadratic equation using formula or completing the squares

220.0;513.1 == yy or 461.0;339.4 == xx

340.4;461.0 == xx or 513.1;220.0 == yy

P1

P1

K1

K1

N1

N1

6 2(a) Use triangle law correctly to find AC or DF

(i) BCABAC += (ii) CFDCDF +=

yxAC 4024 +−= xDF 30−=

K1

N1 N1

2(b) Find vector CE or EC or AE or EA by correct triangle law.

)53(65

35

3

yx

DFCF

FDCF

FECFCE

−=

−=

+=

+=

CEyx6

153 =−

Find AC in term of CE or vice versa or any equation that can conclude A, C, E are collinear.

CEAC

yxAC

yxAC

3

4

)53(8

4024

−=

−−=

+−=

∴A, C, E are collinear because CEAC3

4−=

K1

K1

N1

6



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21

Question Working/Solution Marks Total

3(a) Use cos 2x or sin 2x and factorize

xx

x

xx

xx

xx

xx

xxx

xx

xx

xx

cotsin

cos

)1cos2(sin

)1cos2(cos

)1cos2(sin

cos2cos

sincossin2

)1cos2(cos1

sin2sin

2coscos1

2

2

=

=

++=

++=

+−++=

+++

K1

N1

N1

3(b)(i) Correct shape of cosine function with amplitude of 3 units or Correct shape of cosine function and translated 1 unit

OR Correct shape of cosine function with amplitude of 3 units and translated 1 unit.

OR Correct shape of cosine function with amplitude of 3 units

and translated 1 unit and passing through (0 , 4), (2

π , 1),

(π , –2), (2

3π , 1) and (2π , 4).

P1

OR

P2

OR

P3

3(b)(ii) Get the correct linear equation and draw a straight line.

π

πππ

ππ

xx

xx

xx

xx

31cos3

3)1cos3(

3cos3

3cos3

=+

=+=+−=

The equation of straight line : πx

y3=

Straight line drawn correctly and number of solutions = 1

K1

N1

8

4

1

-2 2

π 2

3π

π 2πx

y

O

πx

y3=

1cos3 += xy



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22


4.(a) (i) Use median formula with at least two of the L, N, F, f and c correctly substituted.

)10(10

102

22

5.305.34

−+

+=

k

6=k (ii) median mark = 42.5

K1

N1

N1

P1

4(b) Correct axes and uniform scales with all the lower and upper boundaries correctly labeled and the Height of at least three bars are proportional to the frequency Correct way of finding the value of mode. Modal mark = 35.0

K1

K1

N1

7 5(a)

xxy

13 +=

2

13

xdx

dy −=

Find gradient of normal and use )( 11 xxmyy −=−

at (1 , 4), 2=dx

dy

Gradient of normal = 2

1−

Equation of normal :

)1(2

14 −−=− xy

092 =−+ yx or equivalent

P1

K1

N1

5(b) Get the expression for V and find

dr

dV to determine the

value of dr

dV at r = 20 cm.

3

3

2rV π=

22 rdr

dV π=

Use dt

dr

dr

dV

dt

dV ×=

13

2

80

)1.0()20(2

−−=

−=

scm

dt

dV

π

π

K1

K1

N1

6



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23


6. (a) Perimeters of the first three circles :

Rπ2 , Rπ3

4, Rπ

9

8

Check the ratios 1

2

T

T and

2

3

T

T

Make a correct conclusion :

Since 3

2

2

3

1

2 ==T

T

T

T , the perimeters of the first three

circles form a geometric progression with common

ratio 3

2.

(b)

30

9002

==

R

R ππ

(i) Use the formula 1−= nn arT :

cm561.1

3

2)60(

9

10

π

π

=

=T

(ii) Use the formula r

aS

−=∞ 1

:

cm1803

21

60

π

π

=

−=∞S

K1

N1

P1

K1

N1

K1

N1

7

7(a) All values of y10log correct.

Plot y10log against ( 2+x ) with correct axes, uniform

scales and at least one point plotted correctly. 6 points plotted correctly. Line of best fit, ( passes through as many points as possible and balance in terms of numbers point appear above and below the line, if any .)

2+x 1 2 3 4 5 6 y10log 0.92 1.00 1.08 1.12 1.24 1.32

N1

K1

N1

N1



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24

7(b) (i) Recognize the wrong recorded value of y and use graph to find the should be value of y.

45..14

16.1log10

==

y

y

(ii) qpxy 101010 loglog)2(log −+=

Use cq =− 10log or Use mp=10log

84.0log10 −=q 08.0log10 =p

p = 0.1445 ; p = 1.202

K1

N1

P1

K1

N1 N1

10 8(a)

Use 1

1

3

3

2

2

1

1

2

1y

x

y

x

y

x

y

x correctly to find the area of triangle

Area = 10 unit2

K1

N1

8(b) Either midpoint of AB or gradient of AB correct. Mid point of AB = (1 , –3)

Gradient of AB = –1 Use )( 11 xxmyy −=− with his midpoint of AB and his gradient of normal.

y + 3 = 1(x -1) y = x – 4

P1

K1

N1

8(c) Use Ratio Theorem follow his midpoint

−+−+=5

)8(1)3(4,

5

)6(1)1(4C

)4,2( −=C

K1

N1

8(d) Use of distance formula correctly for PB or PC 22 )8()6( ++−= yxPB ; 22 )4()2( ++−= yxPC

Use BP = 2 PC

2222 )4()2(2)8()6( ++−=++− yxyx

02016433 22 =−+−+ yxyx

K1

K1 N1

10

9(a) (i) (2 , 4) (ii) )4( −−= xxy or eqivalent

P1 P1

9(b) Correct method of finding area under curve or area of ∆ 4

2

32

32

4

− xx

or

2

0

2

2

2

x or 42

2

1 ××

Find integration value using correct limits

−−

−

3

2

2

)2(4

3

4

2

)4(4 3232

or

−

2

)0(2

2

)2(2 32

Find the sum of two area

3

28 unit2 or equivalent

K1

K1

K1

N1



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25

9(c) Correct method of finding volume of revolution or volume

of cone.

( )dxxxx∫ +−2

0

432 816π or dxx∫2

0

24π or )2()4(3

1 2π

Find integration value using correct limits

−

+− 0

5

2

4

)2(8

3

)2(16 543

π or

− 0

3

)2(4 3

π

Find the difference of two volume

π5

32 cm3 or equivalent

K1

K1

K1

N1

10

10(a) (i) Both

5

2=p and 5

3=q or equivalent

53

38

5

3

5

2

C or equivalent

0.2787 or other more accurate answers.

(ii) 1 - 62

28

71

18

80

08

5

3

5

2

5

3

5

2

5

3

5

2

+

+

CCC

or 08

88

44

48

53

38

5

3

5

2...

5

3

5

2

5

3

5

2

++

+

CCC

All terms must be correct and completed in full 0.6846 or other more accurate answers.

P1

K1

N1

K1

N1

10(b) (i)

−≥16

5060ZP

0.2660

(ii) 28.016

50 =

−< mZP or 28.0

16

50 =

−−≥ mZP

or equivalent

583.016

50 =−− m

40.672 kg

K1 N1

K1

K1 N1

10 11(a) (i) SAC = 5(1.2)

6 cm (ii) Correct method of finding AB

radAC

2

2.1sin

52

1

= or equivalent OR cosine rule

AB = AC = 5.646 cm (5.64578271...)

K1 N1

K1

N1



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26

11(b) (i)

2

2.12.1

++=∠ πBAC or equivalent

2.171 rad or equivalent

(ii) Area of sector BAC = 2

1(5.646)2(2.171)

Area of segment AC = ( )rad2.1sin2.1)5(2

1 2 −

or equivalent Find difference of the two area

)2.1sin2.1()5(2

1)171.2()646.5(

2

1 22 rad−−

Area of shaded region = 31.25 cm2 (31.25261028...)

K1

N1

K1

K1

K1

N1

10 12(a)

(i) 1401007

05

=×A

or equivalent

RM5

(ii) 1322510152030

)25(120)10(104)15(110)20()30(140 =++++

++++ x

165.5

K1 N1

K1 N1

12(b) 132100

4007 =×

Q

Q07 = RM 52.80 Q09 = RM 58.08

K1 N1 N1

12(c) 125100

03

05 =×Q

Q and

100

104

05

07 =Q

Q

Or 10410005

07 =×Q

Q and

100

125

03

05 =Q

Q

10003

0703,07 ×=

Q

QI

10003

05

05

0703,07 ××=

Q

Q

Q

QI or 104

100

125× or 100

104125× or

100

104125×

= 130

P1

K1

N1

10



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27

13(a) Using correct rules to find PR and QR.

oo

PR

20sin

10

50sin= or

oo

QR

20sin

10

110sin=

PR = 22.40 cm (22.39764114...) QR = 27.47 cm (27.47477419...)

K1

N1 N1

13(b) Use cosine rule to find QS. QS2 = 42 + 27.472 – 2(4)(27.47)cos (110 + 20)o QS = 30.20 cm

K1 N1 N1

13(c) Use formula correctly to find area of triangle PQR or PRS.

oPQR 50sin)10)(47.27(2

1Area =∆

or oPRS 110sin)4)(40.22(2

1Area =∆

Use Area PQRS = sum of two area Area PQRS = 147.32 cm2

K1

N1

K1

N1

10 14(a) Find expression for v.

ct

tv +−=2

22

2

Find t for maximum v and substitute in his v. a = 2 - 2t = 0

and 32

)1(22

2

max +−=v

vmax = 4 ms-1

K1

K1

N1

14(b) Try solving v = 0 2k - k2 + 3 = 0 (3 - k)(1 + k) = 0 or equivalent k = 3 s

K1

N1

14(c) Correct method in finding distance traveled. 3

2

32

332

2

+−= t

tts or ttts 3

3

1 32 +−= and 23 ss −

3

5=s m

K1

N1

14(d) Find expression for s.

tt

ts 33

32 +−=

Solve his s = 0

033

32 =+− t

tt

3t2 - t3 + 9t = 0 t2 - 3t -9 = 0 t = 4.854 s

K1

K1

N1

10



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28

15(a) 160≤+ yx note: ss-1 for all answers in terms of P and Q

yx 10040 ≤ or equivalent 80+≤ xy or equivalent

N1 N1 N1

15(b) Axes correct and one correct straight line. All three straight lines are correct. The shaded region of R is correct. Please refer to attached graph.

K1 K1 N1

15(c) (i) Drawing of the straight line, x = 50, completely across the shaded region R. 11020 ≤≤ y (ii) Construction of the line 20x + 30y = k at (40, 120) or any clear indication of point (40, 120) only or 30(120) + 20(40) = k maximum profit = RM4400

K1

N1

K1

N1

10



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29

10

8

1

3

4

5

6

2

7

9

Graph For Question 4(b) Number of students

0.5 10.5 20.5 30.5 40.5 50.5 Modal mark = 35.0

marks 60.5



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30

1 2 3 4 5 6 x + 2

0.85

0.90

0.95

1.00

1.05

1.10

1.15

y10log

x

x

x

x

x

x

No.7(a)

0 7 0.80

1.20

1.25

1.30

(1 , 0.92)

(6 , 1.32)

84.0

08.016

92.032.1

==

−−=

c

m

m

1.16



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31

Qn. 15



tutor mansor burhan...peperiksaan percubaan spm tahun 2009 additional mathematics paper 2 two hours...

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