Pahang 2009 Spm Trial - Add Maths

PERCUBAAN PAHANG SPM 2009

SULIT

3472/1

NAMA ANGKA GILIRAN

PEPERIKSAAN PERCUBAAN SPM TAHUN 2009

3472/1

ADDITIONAL MATHEMATICS Kertas 1 September 2009 2 jam

Dua jam

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 dibenarkan menjawab keseluruhan atau sebahagian soalan dalam bahasa Inggeris atau bahasa Malaysia.

4.

Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.

Untuk Kegunaan Pemeriksa Kod Pemeriksa: Markah Markah Soalan Penuh Diperoleh 1 2 2 4 3 3 4 3 5 3 6 4 7 3 8 4 9 3 10 3 11 2 12 4 13 4 14 2 15 4 16 4 17 3 18 3 19 3 20 3 21 4 22 3 23 3 24 3 25 3 Jumlah 80

Kertas soalan ini mengandungi 20 halaman bercetak. [Lihat sebelah SULIT

3472/1

1

SULIT

3472/1

INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON 1.

This question paper consists of 25 questions. Kertas soalan ini mengandungi 25 soalan.

2.

Answer all questions. Jawab semua soalan.

3.

Give only one answer for each question. Bagi setiap soalan beri satu jawapan sahaja.

4

Write your answers in the spaces provided in this question paper. Jawapan anda hendaklah ditulis pada ruang yang disediakan dalam kertas soalan ini.

5.

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.

6.

If you wish to change your answer, cross out the answer that you have done. Then write down the new answer. Jika anda hendak menukar jawapan, batalkan dengan kemas jawapan yang telah dibuat. Kemudian tulis jawapan yang baru.

7.

The diagrams in the questions provided are not drawn to scale unless stated. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.

8.

The marks allocated for each question are shown in brackets. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan.

9.

A list of formulae is provided on pages 3 to 5. Satu senarai rumus disediakan di halaman 3 hingga 5.

10.

A four-figure table for the Normal Distribution N(0, 1) is provided on page 2. Satu jadual empat angka bagi Taburan Normal N(0, 1) disediakan di halaman 2.

11.

You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram.

12.

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

2

SULIT

3472/1

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

2

3

4

7

8

9

0.4641

4

8

12

16

20

24

28

32

36

0.4286

0.4247

4

8

12

16

0.3897

0.3859

4

8

12

15

20

24

28

32

36

19

23

27

31

35

0.3557

0.3520

0.3483

4

7

11

0.3192

0.3156

0.3121

4

7

11

15

19

22

26

30

34

15

18

22

25

29

0.2877

0.2843

0.2810

0.2776

3

7

32

10

14

17

20

24

27

31

0.2578

0.2546

0.2514

0.2483

0.2451

3

0.2266

0.2236

0.2206

0.2177

0.2148

3

7

10

13

16

19

23

26

29

6

9

12

15

18

21

24

27

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

3

5

8

11

14

16

19

22

25

3

5

8

10

13

15

18

20

23

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1379

2

5

7

9

12

14

16

19

21

0.1170

2

4

6

8

10

12

14

16

18

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

0.1003

0.0985

2

4

6

7

9

11

13

15

17

0.0838

0.0823

2

3

5

6

8

10

11

13

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

14

0.0708

0.0694

0.0681

1

3

4

6

7

8

10

11

13

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0594

0.0582

0.0571

0.0559

1

2

4

5

6

7

8

10

11

0.0485

0..0475

0.0465

0.0455

1

2

3

4

5

6

7

8

9

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0401

0.0392

0.0384

0.0375

0.0367

1

2

3

4

4

5

6

7

8

0.0322

0.0314

0.0307

0.0301

0.0294

1

1

2

3

4

4

5

6

6

1.9

0.0287

0.0281

0.0274

0.0268

2.0

0.0228

0.0222

0.0217

0.0212

0.0262

0.0256

0.0250

0.0244

0.0239

0.0233

1

1

2

2

3

4

4

5

5

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

0

1

1

2

2

3

3

4

4

2.1

0.0179

0.0174

0.0170

2.2

0.0139

0.0136

0.0132

0.0166

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

0

1

1

2

2

2

3

3

4

0.0129

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

0

1

1

1

2

2

2

3

3

2.3

0.0107

0.0104

0.0102

0

1

1

1

1

2

2

2

2

0.00990

0.00964

0.00939

0.00914

3

5

8

10

13

15

18

20

23

2

5

7

9

12

14

16

16

21

z

0

1

2

3

4

5

6

7

8

9

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3228

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.8

0.2119

0.2090

0.2061

0.2033

0.9

0.1841

0.1814

0.1788

0.1762

1.0

0.1587

0.1562

0.1539

1.1

0.1357

0.1335

0.1314

1.2

0.1151

0.1131

1.3

0.0968

0.0951

1.4

0.0808

1.5 1.6

2.4

0.00820

0.00798

0.00776

0.00755

5

6

Minus / Tolak

0.00889

0.00866

0.00842

0.00734

2

4

6

8

11

13

15

17

19

0.00714

0.00695

0.00676

0.00657

0.00639

2

4

6

7

9

11

13

15

17

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2

3

5

6

8

9

11

12

14

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

1

2

3

5

6

7

9

9

10

2.7

0.00347

0.00336

0.00326

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00264

1

2

3

4

5

6

7

8

9

2.8

0.00256

0.00248

0.00240

0.00233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

1

1

2

3

4

4

5

6

6

2.9

0.00187

0.00181

0.00175

0.00169

0.00164

0.00159

0.00154

0.00149

0.00144

0.00139

0

1

1

2

2

3

3

4

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

 1  exp − z 2  2π  2  1

f ( z) =

Example / Contoh:

f (z)

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

∞

Q( z ) = ∫ f ( z ) dz

P(X > k) = Q(k)

k

Q(z)

3 O

k

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

z

SULIT

3472/1

SULIT

3

3472/1

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

y=

u dy , v dx

3

dy dy d = × dx du d 5 Volume generate d

ALGEBRA 1

x=

2

− b ± b 2 − 4ac 2a

log c b log c c

8

log a b =

a m × a n = a m+n

9

Tn = a + (n − 1)d

3

a m ÷ a n = a m− n

10

n S n = [2a + (n − 1)d ] 2

4

(a m ) n = a mn

11

Tn = ar n −1

5

log a mn = log a m + log a n

12

Sn =

a (r n − 1) a (1 − r n ) = ,r ≠ 1 r −1 1− r

6

log a

13

S∞ =

a , r <1 1− r

7

log a m n = n log a m

m = log a m − log a n n

CALCULUS / KALKULUS 1

y = uv ,

dy dv du =u +v dx dx dx

4

Area under a curve Luas di bawah lengkung

4

SULIT

3472/1

STATISTICS / STATISTIK 1

x=

∑x N

3

∑ fx ∑f ∑ ( x − x) σ=

4

σ=

2

5

6 7

x=

∑x

2

N

∑ f ( x − x) ∑f

=

N

2

=

−x

∑ fx ∑f

1   N −F C m = L+ 2  fm      Q1 I= × 100 Q0

I=

2

∑W I ∑W

i i

2

2

−x

2

8

n

Pr =

n! (n − r )!

9

n

Cr =

n! (n − r )!r!

10

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

11

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

12

Mean / Min , µ = np

13

σ = npq

14

Z=

i

X −µ

σ

GEOMETRY / GEOMETRI 1

Distance / Jarak

5

r = x2 + y2

6

^ xi + y j r= x2 + y2

= ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 ^

2

Midpoint / Titik tengah

 x + x 2 y1 + y 2  ( x, y ) =  1 ,  2   2 3

A point dividing a segment of a line Titik yang membahagi suatu tembereng garis  nx + mx 2 ny1 + my 2  ( x, y ) =  1 ,  m+n   m+n

4

Area of triangle / Luas segitiga 1 = ( x1 y 2 + x 2 y + x3 y1 ) − ( x 2 y1 + x3 y 2 + x1 y 3 ) 2

5

SULIT

3472/1 5 TRIGONOMETRY / TRIGONOMETRI 1

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

2

Area of sector, A =

Luas sector, L =

3

1 2 r θ 2

8

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

9

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

1 2 j θ 2

sin 2 A + cos 2 A = 1

kos ( A ± B ) = kosA kosB ∓ sin A sin B tan A ± tan B 1 ∓ tan A tan B

10

tan( A ± B ) =

11

tan 2 A =

12

a b c = = sin A sin B sin C

sin 2 A + kos 2 A = 1 4

sec 2 A = 1 + tan 2 A sek 2 A = 1 + tan 2 A

5

cosec 2 A = 1 + cot 2 A

2 tan A 1 − tan 2 A

kosek 2 A = 1 + kot 2 A 6

sin 2 A = 2 sin A cos A sin 2 A = 2 sin A kosA

13

a 2 = b 2 + c 2 − 2bc cos A a 2 = b 2 + c 2 − 2bc kosA

7

cos 2 A = cos 2 A − sin 2 A

14

Area of triangle / Luas segitiga 1 = ab sin c 2

= 2 cos 2 A − 1 = 1 − 2 sin 2 A kos 2 A = kos 2 A − sin 2 A = 2 kos 2 A − 1 = 1 − 2 sin 2 A

6

SULIT

3472/1

For Examiner’s Use

Answer all questions. Jawab semua soalan.

1

Diagram 1 shows the relation between set P and set Q. Rajah 1 menunjukkan hubungan antara set P dan set Q. f x

y

. 6. 10 . 12 .

.2 .4 .6 .w

Set P

Set Q

2

Diagram 1 Rajah 1

State Nyatakan (a)

(b)

the type of relation between set P and set Q. jenis hubungan antara set P adan set Q. x +1. 2 x nilai bagi w jika f : x → + 1 . 2

the value of w if f : x →

[ 2 marks] [2 markah]

Answer / Jawapan: 1

(a) ……………………...………… (b) w =………………………….…

2

7

SULIT

3472/1

2

For Examiner’s Use

Given the function g : x → 3( x − 1) , find Diberi fungsi g : x → 3( x − 1) , cari (a)

the value of g 2 (4), nilai bagi g 2 (4),

[2 marks] [2 markah] [2 markah]

(b)

the function of f if gf(x) = 6x. fungsi f jika gf(x) = 6x.

[ 2 marks] [2 markah] [2 markah]

Answer / Jawapan:

(a)

……………………..……...

2

(b) ……………………………. ______________________________________________________________________

3

4

The quadratic equation x2 - (3 – p)x + p - 3 = 0, where p is constant, has two equal roots. Find the possible values of p. [3 marks]

Persamaan kuadratik x2 - (3 – p)x + p - 3 = 0, dengan keadaan p ialah pemalar, mempunyai dua punca sama. Cari nilai-nilai p yang mungkin. [3 marks] [3 markah]

3 Answer / Jawapan: p = …………………………...…...

[Lihat sebelah 8

3

For Examiner’s Use

4

Find the range of values of x for which ( x − 2) 2 ≤ 8 − x. Cari julat nilai x bagi ( x − 2) ≤ 8 − x. 2

[3 marks] [3 markah]

4 4

Answer / Jawapan:

5

……………………………...…...…..

Given that log 4 3 y − log 4 6 x = 2 , express y in terms of x. Diberi log 4 3 y − log 4 6 x = 2 , ungkapkan y dalam sebutan x.

[3 marks] [3 markah]

5 3

Answer / Jawapan:

……………………………...…...…..

SULIT

3472/1 For Examiner’s Use

6

Given that log 4 x = p, log32 y = q and

Diberi log 4 x = p, log32 y = q dan

x = 2mp + nq , find the value of m and n. y [4 marks]

x = 2mp + nq , cari nilai bagi m dan n. [4 markah] y

6 4 Answer / Jawapan: m = ……….. n = ………… ______________________________________________________________________ 7

A piece of string of length 12 m is cut into 20 pieces in such a way that the lengths of the pieces are in arithmetic progressions. If the length of the longest piece is five times of the length of the shortest piece, find the length of the longest piece. [3 marks] Seutas dawai yang panjangnya 12 m dipotong kepada 20 keratan dengan keadaan ukuran keratan membentuk satu janjang aritmetik. Jika ukuran keratan terpanjang ialah lima kali keratan terpendek, cari ukuran keratan terpanjang. [3 markah]

7 Answer / Jawapan:

……....……………………………. 3 [Lihat sebelah

10

SULIT

3472/1

For Examiner’s Use

8

The fourth and seventh terms of a geometric progression are 18 and 486 respectively. Find the third term. [4 marks] Sebutan keempat dan ketujuh bagi satu janjang geometri masing-masing ialah 18 dan 486. Cari sebutan ketiga. [4 markah]

8 Answer / Jawapan:

4

..……………………...…...…………

________________________________________________________________________ 9

Point P(h, 7) divides line the segment joining the points E(3, 10) and F(8, k) internally such that EP : PF = 1: 4. Find the values of h and k. [3 marks] Titik P(h, 7) membahagi dalam tembereng garis yang menyambungkan titik E(3,10) dan F(8, k) dengan keadaan EP : PF =1 : 4. Cari nilai bagi h dan k. [3 markah]

9 3

11

SULIT

3472/1 Answer / Jawapan:

10

h = ………… k = ………….

Solution to this question by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak diterima. In Diagram 2, OABC is a quadrilateral. The equation of the straight line AB is x y + = 1. 6 4 Dalam Rajah 2, OABC adalah sebuah sisiempat. Persamaan bagi garis lurus AB x y ialah + = 1. 6 4 y A

. x

.B(9, -2)

O

.

C (2, -3)

Diagram 2 Rajah 2

Find the area of the quadrilateral OABC. Cari luas bagi sisempat OABC.

Answer / Jawapan:

[3 marks] [3 markah]

.......……………………………….

10 10 3 3

Lihat sebelah 12

SULIT

3472/1 11

For Examiner’s Use

.

1 a − b , express c in the form x i + y j . ~ ~ ~ ~ ~ ~ ~ 2~ ~ ~ ~ ~ [2 marks] 1 Diberi a = 4 i − 6 j , b = i − j dan c = a − b , ungkapkan c dalam bentuk x i + y j ~ ~ ~ ~ ~ ~ ~ 2~ ~ ~ ~ ~ [2 markah]

Given a = 4 i − 6 j , b = i − j and c =

Answer / Jawapan:

11

c = ~

…………………..…..

________________________________________________________________________ 2

12

The vector OF has a magnitude of 10 unit and has the same direction as OE .  3   x Given that OE =   and OF =   , find the value of x and y. [3 marks]  − 4  y Vector OF mempunyai magnitude 10 unit dan mempunyai arah yang sama dengan  3   x OE . Diberi OE =   dan OF =   , cari nilai x dan nilai y. [3 markah]  − 4  y

12 Answer / Jawapan: 4

13

x =………… y = …………..

SULIT

3472/1 For Examiner’s Use

13

Diagram 3 shows a triangle ABC. Rajah 3 menunjukkan sebuah segitiga ABC. A

.E Diagram 3 Rajah 3

.

B

C

D

The point E is the midpoint of AC and D lies on the line BC such that BC = 5DC. Given AB = x and BC = 5 y , express in term of x and y , ~

~

~

~

Titik E ialah titik tengah bagi AC dan D terletak pada garis BC dengan keadaan BC = 5DC. Diberi AB = x dan BC = 5 y , ungkapkan dalam sebutan ~

~

x dan y , ~

~

(a) AD , (b) DE . [4 marks] [4 markah]

Answer / Jawapan:

(a) .……….………………..…..

13

(b) .……….………………..…..

14

x 2 − 3x . x→2 x2 − 4 x + 3

Find the value of lim

2 Cari nilai bagi had 2x − 3x x →2

x − 4x + 3

4

[2 marks]

.

[2 markah]

14 Answer / Jawapan:

…….………………………..……..

[Lihat sebelah 14

3

S

SULIT

3472/1 15

2 Volume, V cm3, of a solid is given by V = 8πr 2 + πr 3 , r is the radius. Find the 3 approximate change in V if r increases from 3 cm to 3.005 cm. (Give your answers in terms of π ). 2 Isipadu, V cm3, bagi sebuah pepejal diberi oleh V = 8πr 2 + πr 3 , r ialah jejari. 3 Cari perubahan hampir bagi V jika r bertambah daripada 3 cm kepada 3.005 cm. (Beri jawapan anda dalam sebutan π ). [4 marks] [4 markah]

15 4

Answer / Jawapan : ……………………………… ________________________________________________________________________

16

Diagram 4 shows part of a straight line graph drawn to represent linear form of the 625 equation y = . x Rajah 4 menunjukkan sebahagian daripada graph garis lurus yang dilukis untuk 625 mewakili bentuk linear bagi persamaan y = . x log5 y P (0, h)

Q (k, 1) log5 x

O Diagram 4 Rajah 4 Find the values of h and k. Cari nilai bagi h dan k.

[4 marks] [4 markah]

16 4

Answer / Jawapan:

15

h = …………… k = ……………..

SULIT

3472/1 For Examiner’s Use

17

(6 + x)(6 − x) dx . x4 (6 + x)(6 − x) dx . x4

Find the value of Cari nilai

∫

1 −1

∫

1

[3 marks]

−1

[3 markah]

17 3 Answer / Jawapan:

18

…..……………………………….

The gradient function of a curve passing through (1, 2) is given by Find the equation of the curve.

Fungsi kecerunan suatu lengkung yang melalui (1, 2) diberi oleh persamaan lengkung itu.

1 . (3 x − 4) 2 [3 marks]

1 . Cari (3 x − 4) 2 [3 markah]

18 Answer / Jawapan:

……………………………...…...….. 3

[Lihat sebelah 16

SULIT

3472/1

For Examiner’s Use

19

In Diagram 5, OAC is a right-angled triangle and OAB is a sector of a circle with centre A. Dalam Rajah 5, OAC ialah sebuah segitiga tegak dan OAB ialah sebuah sektor bulatan berpusat A. A

6 cm

0.927 rad.

O

B

8 cm

C

Diagram 5 Rajah 5

Given that OA = 6 cm , OC = 8 cm and ∠OAB = 0.927 rad , find the area of the shaded region. [3 marks] Diberi OA = 6 cm, OC = 8 cm dan ∠OAB = 0.927 rad , cari luas rantau berlorek. [3 markah]

19 3

Answer / Jawapan: …………………………..….. ________________________________________________________________________

20

Given that cos 700 = h and sin 350 = k , express in terms of h and/or k (a)

cos1400 ,

(b)

sin 1050 [3 marks]

Given that cos 700 = h and sin 350 = k , express in terms of h and/or k, (a) cos1400, (b) sin 1050.

[3 markah]

20 3

Answer / Jawapan:

(a) ………………………… (b) …………………………

17

SULIT

3472/1 For Examiner’s Use

21

Solve the equation 3 sec 2 x − 4 tan x − 2 = 0 for 00 ≤ x ≤ 3600 . Selesaikan persamaan 3sek 2 x − 4 tan x − 2 = 0 bagi 00 ≤ x ≤ 3600

[4 marks] [4 markah]

21 Answer / Jawapan:

22

……….………………………..…..

4

Five boys and four girls are to stand in a line. Calculate the number of possible arrangements if (a) there is no restriction, (b) no two boys are to stand beside each other. [3 marks]

Lima orang lelaki dan empat orang perempuan berdiri pada satu baris. Kira bilangan susunan yang mungkin jika (a) tiada syarat yang dikenakan, (b) tiada dua orang lelaki yang berdiri sebelah menyebelah. [3 markah]

Answer / Jawapan: (a) …………………………………. (b) …………………………………

[Lihat sebelah

18

22 3

SULIT

3472/1

23 Lee will play against players E, F and G in a badminton competition. The 5 3 2 probabilities that Lee will beat E, F and G are , and respectively. Calculate 6 4 3 the probability that Lee will beat at least two of the three players. [3 marks] Lee akan berlawan dengan pemain E, F dan G dalam satu pertandingan badminton. Kebarangkalian bahawa Lee akan mengalahkan E, F dan G masing5 3 2 masing ialah , dan . Hitungkan kebarangkalian bahawa Lee akan 6 4 3 mengalahkan sekurang-kurang dua daripada tiga orang pemain. [3 markah]

23 Answer / Jawapan : ……………………………… 3 ________________________________________________________________________

24

In an examination, 40 % of the students passed. If a sample of 10 students is randomly selected, find the probability that less than 2 students passed. [3 marks] Dalam satu peperiksaan,didapati 40 % daripada pelajar lulus. Jika satu sampel 10 orang pelajar dipilih secara rawak, cari kebarangkalian bahawa kurang daripada [3 makah] 2 orang pelajar lulus.

24 3

Answer / Jawapan:

…….……………………...…...…..

19

SULIT

3472/1 For Examiner’s Use

25

Diagram 7 shows a standardised normal distribution graph. Rajah 7 menunjukkan satu graf taburan normal piawai. f (z)

z

k O Diagram 7 Rajah 7

Given that the area of the shaded region is 30.5 % of the total area under the curve, find Diberi bahawa luas rantau berlorek ialah 30.5 % daripada keseluruhan luas rantau dibawah lengkung, cari (a) P ( z < k ) , (b) the value of k, cari nilai k. [3 marks] [3 markah]

25 Answer / Jawapan:

(a)

……………………..……..

(b)

…………………………….

END OF QUESTION PAPER KERTAS SOALAN TAMAT

20

3

SULIT

3472/1

SULIT 3472/1 Additional Mathematics Kertas 1 Peraturan Pemarkahan August/September 2009

PEPERIKSAAN PERCUBAAN SPM TAHUN 2009

ADDITIONAL MATHEMATICS KERTAS 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

21

SULIT Question 1 (a) 1 (b)

3472/1 Working / Solution One- to- one or 1- to- 1 or 1-1

Marks 1

w=7

Total

1 2

2 (a)

24

2

g (4) = 9 2(b)

B1

2 x +1

2

3( f ( x) − 1) = 6 x

B1 4

3

3 and 7

3

( p − 3)( p − 7) = 0 or equivalent

B2

[−(3 − p )]2 − 4(1)( p − 3) = 0

4

B1 3

−1 ≤ x ≤ 4

−1

4

x

4

or

−1

4

x

B3

Must indicate the range correctly by shading or other ways

( x + 1)( x − 4) ≤ 0 or x 2 − 3x − 4 ≤ 0

or

(− x + 4)( x + 1) ≥ 0

B2

− x 2 + 3x + 4 ≥ 0

B1 3

22

SULIT 5

3472/1 y = 32 x

3

3y = 42 6x

B2

log 4

3y =2 6x

B1 3

6

m = 2 and n = -5

4

m = 2 or n = -5

B3

22 p −5q

B2

4 p or 32 q

B1 4

7

1 m or 100 cm

3

20 4a [2a + 19( )] or 2 19 20 4a 12 = [2a + 19( )] or 2 − 95 12 =

d=

4a 19

or

a + 5a

20 [a + 5a ] or 2 20 1 12 = [a + a ] 2 5 12 =

or d = −

4a 95

or

1 a+ a 5

B2

B1 3

8

6

4 B3

18 2 (3) 27 r =3 and a =

B2

18 27

B1

ar 3 = 18 and ar 6 = 486

4

23

SULIT 9

10

3472/1 h = 4 and k = -5

3

h = 4 or k = -5

B2

4(3) + 1(8) 4(10) + 1(k ) or 1+ 4 1+ 4

B1 3

29.5

3 B2

1 [0 × (−3)... or equivalent 2 10 2 9 0 0 2 0 −3 −2 4 0

or other correct arrangement

B1 3

11

i− 2 j ~

2

~

1 (4 i − 6 j ) − ( i − j ) ~ 2 ~ ~ ~

B1 2

12

3

x = 6 and y = -8

 x 1 3    = 10 ×   5  − 4  y 1 3  =   OE 5  − 4  OE

13 (a)

(b)

or

B2

OF = 10

OE

B1

OE

4

x +4y

1

~

~

−

1 3 x− y ~ 2 2~

3

B2

1 CE = y + (−5 y − x) ~ 2 ~ ~

CA = −5 y − x ~

14

~

B1

1 or DE = DC + CA 2

4

2

2 x x→2 x − 1

lim

B1 2

24

SULIT 15

3472/1

0.33π

4

(16π (3) + 2π (3) 2 ) × 0.005

B3

dv = 16πr + 2πr 2 dr

B2

16πr or 2πr 2 or

δr = 0.005

B1 4

16

h = 4 and k = 3 h=4

or

4

4 −1 = −1 0−k

h −1 = −1 or 0−k

B3

h = log 5 625

B2

log5 y = log 5 625 − log5 x

B1 4

17

3

-22 1   12 1   12  +  − 3 +  −  −  1 1   (−1) (−1) 

B2

1

 36 x −3 x −1   − 3 − −1   −1 18

y=−

1 5 + or equivalent 3(3 x − 4) 3

y=−

1 + c or equivalent 3(3 x − 4)

B1 3 3

B2

(3 x − 4) −1 or equivalent 3(−1) B1 3

25

SULIT

3472/1 7.314 cm2

19

20(a)

(b)

3

1 1 × 6 × 8 − × 6 2 × 0.927 or equivalent 2 2

B2

1 2 × 6 × 0.927 or equivalent 2

B1 3

2h − 1

1

2

2

k 1 − h 2 + h 1 − k 2 or equivalent

B1

sin 70 0 cos 35 0 + cos 70 0 sin 35 0

3 o

21

0

’

o

o

o

18.43 (18 26 ) , 45 , 198.43 (198 26’), 225

o

18.43o and 45o

4 B3

(3 tan x − 1)(tan x − 1) = 0

B2

3(tan 2 x + 1) − 4 tan x − 2 = 0

B1 4

22 (a) (b)

362880

1

2880

2

5! × 4!

or

4

P4 × 5P5

or 544332211 B1 3

23

3

61 72 5 3 2 5 3 1 5 1 2 1 3 2 × × + × × + × × + × × 6 4 3 6 4 3 6 4 3 6 4 3

B2

5 3 2 5 3 1 5 1 2 1 3 2 × × or × × or × × or × × 6 4 3 6 4 3 6 4 3 6 4 3

B1

26

3

SULIT 24

3472/1 0.04636

3

10

C0 (0.4)0(0.6)10 – 10C1(0.4)1(0.6)9

10

C0 (0.4)0(0.6)10

or

10

C1 (0.4)1(0.6)9

B2

or

B1

P(x=0)-P(x=1) 3 25 (a)

0.805

1

25(b)

0.86

2

0.195

B1 3

27

SULIT

3472/1

28

SULIT

3472/1

29

SULIT

3472/1

30

SULIT

3472/1

31

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

NAMA DAN LOGO 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.

2

CONFIDENTIAL

3472/2

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

ALGEBRA

1

− b ± b 2 − 4ac x= 2a

8

log a b =

2

am x an = a m + n

9

Tn = a + (n − 1)d

3

am ÷ an = a m – n

10. S n =

4

( am )n = a m n

11 Tn = a r n −1

5

log a mn = log a m + log a n

6

log a

7

log a mn = n log a m

m = log a m − log a n n

12 S n =

log c b log c a

n [ 2a + (n − 1)d ] 2

(

) = a (1 − r ) , r ≠ 1

a r n −1

n

r −1 1− r a 13 S∞ = , r <1 1− r

CALCULUS KALKULUS

4 1

2

y = uv ,

dy dv du =u +v dx dx dx

u dy y= , = v dx

v

Area under a curve Luas di bawah lengkung b

=

du dv −u dx dx v2

5

∫ x dy

a

a

Volume generated Isipadu janaan

b

3

dy dy du = × dx du dx

= π y 2 dx

∫ a

b

∫ y dx or (atau )

b

or (atau )

∫π x

2

dy

a

CONFIDENTIAL

3

CONFIDENTIAL

3472/2

STATISTICS STATISTIK 1

x=

Σx N

7

2

x=

Σ fx Σf

8

n

Pr =

n! (n − r )!

9

n

Cr =

n! (n − r )!r !

Σ(x − x ) 3 σ= = N 2

Σx 2 −x2 N

Σ f (x − x ) = Σf

Σ fx 2 −x2 Σf

2

4

σ=

5

1   N −F  C m = L+  2  fm     

I=

Σ Wi I i ΣWi

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

11 P ( X = r ) = nCr p r q n − r , p + q = 1 12 Mean / min, µ = np 13 σ =

6

Q I = 1 × 100 Q0 14 Z =

npq x−µ

σ

GEOMETRY GEOMETRI 1 =

Distance/jarak

(x1 − x 2 )

2

+ ( y1 − y 2 )

4 Area of a triangle/ Luas segitiga =

1 (x1 y 2 + x 2 y 3 + x3 y1 ) − (x 2 y1 + x 3 y 2 + x1 y 3 ) 2

2

2 Mid point / Titik tengah

(x, y ) =  x1 + x2 , y1 + y 2  

3

2

2



A point dividing a segment of a line Titik yang membahagi suatu tembereng garis

(x , y ) =  nx1 + mx2 , ny1 + my2  m+n   m+n

5

r = ~

^

6 r = ~

x2 + y2

x i+ y j ~

~

x + y2 2

CONFIDENTIAL

4

CONFIDENTIAL

3472/2

TRIGONOMETRY TRIGONOMETRI

1

2

8

Arc length, s = rθ Panjang lengkok, s= jθ Area of a sector, A =

sin ( A ± B ) = sin A kos B ± ko s A sin B 1 2 rθ 2

9

tan ( A ± B ) =

11

tan 2 A =

2

12

a b c = = sin A sin B sin C

sin 2A = 2 sin A cos A

13

a 2 = b 2 + c 2 − 2bc cos A

sec2 A =1 + tan 2 A se k 2 A = 1 + tan 2 A

5

2 tan A 1 − tan 2 A

co sec 2 A = 1 + cot 2 A ko se k A = 1 + k ot A 2

6

tan A ± tan B 1 ∓ tan A tanB

10

sin 2 A + k os 2 A = 1 4

cos ( A ± B ) = cos A cos B ∓ sin A sin B ko s ( A ± B ) = k os A k os B ∓ sin A sin B

1 2 jθ 2 sin 2 A + cos 2 A = 1

Luas sektor, L = 3

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

a 2 = b 2 + c 2 − 2bc kos A

sin 2A = 2 sin A 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 ab sin C 2

CONFIDENTIAL

5

CONFIDENTIAL

3472/2

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

2

3

4

5

7

8

9

24

28

32

36

24

28

32

36

19

23

27

31

35

19

22

26

30

34

15

18

22

25

29

32

14

17

20

24

27

31

10

13

16

19

23

26

29

9

12

15

18

21

24

27

5

8

11

14

16

19

22

25

3

5

8

10

13

15

18

20

23

2

5

7

9

12

14

16

19

21

0.1170

2

4

6

8

10

12

14

16

18

0.0985

2

4

6

7

9

11

13

15

17

0.0838

0.0823

2

3

5

6

8

10

11

13

14

0.0694

0.0681

1

3

4

6

7

8

10

11

13

0.0582

0.0571

0.0559

1

2

4

5

6

7

8

10

11

0..0475

0.0465

0.0455

1

2

3

4

5

6

7

8

9

0.0392

0.0384

0.0375

0.0367

1

2

3

4

4

5

6

7

8

0.0314

0.0307

0.0301

0.0294

1

1

2

3

4

4

5

6

6

0.0256

0.0250

0.0244

0.0239

0.0233

1

1

2

2

3

4

4

5

5

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

0

1

1

2

2

3

3

4

4

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

0

1

1

2

2

2

3

3

4

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

0

1

1

1

2

2

2

3

3

0

1

1

1

1

2

2

2

2

3

5

8

10

13

15

18

20

23

2

5

7

9

12

14

16

16

21 19

z

0

1

2

3

4

5

6

7

8

9

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681

0.4641

4

8

12

16

20

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.4286

0.4247

4

8

12

16

20

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3897

0.3859

4

8

12

15

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.3557

0.3520

0.3483

4

7

11

15

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3228

0.3192

0.3156

0.3121

4

7

11

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.2877

0.2843

0.2810

0.2776

3

7

10

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

0.2546

0.2514

0.2483

0.2451

3

7

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

3

6

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

3

0.9

0.1841

0.1814

0.1788

0.1762

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

1.0

0.1587

0.1562

0.1539

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1379

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

1.2

0.1151

0.1131

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

1.4

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0.0708

1.5

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0594

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0322

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

2.0

0.0228

0.0222

0.0217

0.0212

2.1

0.0179

0.0174

0.0170

0.0166

2.2

0.0139

0.0136

0.0132

0.0129

2.3

0.0107

0.0104

0.0102 0.00990

2.4

0.00820

0.00798

0.00776

0.00755

0.00964

0.00939

0.00914 0.00889

0.00866

0.00842

0.00734

6

Minus / Tolak

2

4

6

8

11

13

15

17

0.00714

0.00695

0.00676

0.00657

0.00639

2

4

6

7

9

11

13

15

17

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2

3

5

6

8

9

11

12

14

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

1

2

3

5

6

7

9

9

10

2.7

0.00347

0.00336

0.00326

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00264

1

2

3

4

5

6

7

8

9

2.8

0.00256

0.00248

0.00240

0.00233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

1

1

2

3

4

4

5

6

6

2.9

0.00187

0.00181

0.00175

0.00169

0.00164

0.00159

0.00154

0.00149

0.00144

0.00139

0

1

1

2

2

3

3

4

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

f (z)

Example / Contoh:

 1  exp − z 2  2π  2  1

f ( z) =

If X ~ N(0, 1),

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

Q(z)

∞

Q( z ) = ∫ f ( z ) dz

P(X > k) = Q(k)

k

Q(2.1) = 0.0179

O

k

P(X z >

2.1)

CONFIDENTIAL

=

6

CONFIDENTIAL

3472/2

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 : 2 1 + = x + 3y = 5 x y

[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. D A

•E

Diagram 1 Rajah 1 B

F

C

1 2 BC and DE = DF , 4 5 1 2 Diberi BA = 24 x , BF = 10 y , CD = 30 x − 30 y , BF = BC dan DE = DF , 4 5

Given that BA = 24 x , BF = 10 y , CD = 30 x − 30 y , BF =

(a) express in terms of x and/or y , ungkapkan dalam sebutan x dan/atau y ,

(i)

AC

(ii)

DF

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

[3 marks] [3 markah] [3 marks] [3 markah]

CONFIDENTIAL

7

CONFIDENTIAL

3.(a) Prove that

1 + cos x + cos 2 x = cot x . sin 2 x + sin x

Buktikan bahawa

1 + cos x + cos 2 x = cot x . sin 2 x + sin x

3472/2

[3 marks]

[3 markah]

(b)(i) Sketch the graph of the trigonometric function y = 3 cos x + 1 for the domain 0 ≤ x ≤ 2π . Lakar graf bagi fungsi trigonometri y = 3 cos x + 1 untuk domain 0 ≤ x ≤ 2π . (ii) On the same axes, sketch the graph of a suitable straight line that can be used to solve the equation 3π cos x = 3 x − π . State the number of solutions to the equation 3π cos x = 3 x − π for 0 ≤ x ≤ 2π . Pada paksi yang sama, lakar graf bagi satu garis lurus yang sesuai digunakan untuk menyelesaikan persamaan 3π cos x = 3 x − π . Nyatakan bilangan penyelesaian bagi persamaan 3π cos x = 3 x − π untuk 0 ≤ x ≤ 2π . [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 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60

Number of students 2 3 5 10 K 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]

CONFIDENTIAL

8

CONFIDENTIAL

3472/2

(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.

5. (a)

[3 markah]

Cari persamaan normal kepada lengkung y = 3 x +

(b)

1 at (1 , 4). x

[3 marks]

1 pada (1 , 4). x

[3 markah]

Find the equation of the normal to the curve y = 3 x +

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.

r cm

Diagram 2 Rajah 2 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] –1

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

π, kadar perubahan isipadu air dalam bekas itu pada ketika jejari permukaan air adalah 20 cm.

[3 markah]

CONFIDENTIAL

CONFIDENTIAL 6.

9

3472/2

Diagram 3 shows a few circles. The first circle is the largest circle with a radius of 2 R cm. The second circle has a radius of R cm . The third circle has a radius which is 3 2 of the radius of second circle and this process is continued indefinitely. 3 Rajah 3 menunjukkan beberapa bulatan. Bulatan pertama adalah bulatan terbesar dan 2 mempunyai jejari R cm. Bulatan kedua mempunyai jejari R cm. Bulatan ketiga 3 2 daripada jejari bulatan kedua dan proses ini mempunyai jejari yang merupakan 3 diteruskan sehingga ketakhinggaan.

R cm

Diagram 3 Rajah 3 (a) Show that the perimeters of the circles form a geometric progression with common 2 ratio by using first three circles. [2 marks] 3 Tunjukkan bahawa perimeter bulatan-bulatan itu membentuk satu janjang geometri 2 dengan nisbah sepunya dengan menggunakan tiga bulatan pertama. [2 marks] 3 (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]

CONFIDENTIAL

10

CONFIDENTIAL

3472/2

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 y =

p x+2 , where p and q are constants. q

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 p x+2 , dengan keadaan p dan q adalah pemalar. Satu daripada nilai y telah salah q direkodkan. x –1 0 1 2 3 4 y 8.4 10.1 12.1 13.2 17.4 20.9 y=

Table 2 Jadual 2 (a) Plot log10 y against ( x + 2) , using a scale of 2 cm to 1 unit on the ( x + 2) - axis and 2 cm to 0.05 unit on the log10 y -axis. [Start the log10 y -axis with the value 0.8]. Hence, draw the line of best fit. [4 marks] Plot log10 y melawan ( x + 2) , dengan menggunakan skala 2 cm kepada 1 unit pada paksi- ( x + 2) dan 2 cm kepada 0.05 unit pada paksi- log10 y . [Mulakan paksi- log10 y 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 . nilai p dan nilai q.

[6 marks] [6 markah]

CONFIDENTIAL

11

CONFIDENTIAL 8.

3472/2

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. y A(-4 , 2) x

O

•

C Diagram 4 Rajah 4

B(6 , -8)

(a)

Calculate the area, in unit2, of triangle OAB. Hitungkan luas, dalam unit2, bagi segitiga OAB.

(b)

Find the equation of the perpendicular bisector of line segment AB. Cari persamaan pembahagi dua sama serenjang bagi tembereng garis AB.

[2 marks] [2 markah] [3 marks] [3 markah]

4 of the distance of point B from the perpendicular 5 bisector of the line segment AB, find the coordinates of point C. [2 marks] 4 Diberi panjang BC ialah daripada jarak titik B dari pembahagi dua sama 5 serenjang bagi tembereng garis AB, cari koordinat bagi titik C. [2 markah]

(c)

Given that the length BC is

(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]

CONFIDENTIAL

12

CONFIDENTIAL 9.

3472/2

Diagram 5 shows the straight line y = 2x which passes through the maximum point of a quadratic curve y = −( x − α )( x − β ) , where α and β are constants. Rajah 5 menunjukkan garis lurus y = 2x yang melalui titik maksimum suatu garis lengkung kuadratik, y = −( x − α )( x − β ) , dengan keadaan α and β ialah pemalar.

y

y = 2x

Q P

x

O

4

Diagram 5 Rajah 5 (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.

(b)

[2 markah]

Calculate the area of the shaded region P.

[4 marks]

Hitungkan luas rantau berlorek P.

(c)

[4 markah]

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]

CONFIDENTIAL

CONFIDENTIAL

13

3472/2

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]

CONFIDENTIAL

14

CONFIDENTIAL

3472/2

11. Use π = 3.142 in this question. Gunakan π = 3.142 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. B

A

5 cm

Diagram 6 Rajah 6

1.2 rad

O (a)

(b)

C

Find Cari (i)

the length, in cm, of the arc AC. panjang, dalam cm, lengkok AC.

[2 marks] [2 markah]

(ii)

the length, in cm, of radius AB. panjang, dalam cm, jejari AB.

[2 marks] [2 markah]

(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]

CONFIDENTIAL

15

CONFIDENTIAL

3472/2

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 A B C D E

Price index for the year 2007 based on the year 2005 Percentage of ingredient Peratus bahan (%) Indeks harga pada tahun 2007 berasaskan tahun 2005 140 30 x 20 110 15 104 10 120 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. [2 marks] [2 markah]

(ii) the value of x. nilai x.

[2 marks] [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. [3 marks] [3 markah] (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]

CONFIDENTIAL

CONFIDENTIAL

16

3472/2

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 P adalah selari. 110o 10 cm S 50o R

Q

Diagram 7 Rajah 7

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. panjang, dalam cm, PR dan QR.

[3 marks] [3 markah]

(b) the length, in cm, of diagonal QS. panjang, dalam cm, perpenjuru QS.

[3 marks] [3 markah]

(c) the area, in cm2, of PQRS. luas, dalam cm2, PQRS.

[4 marks] [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, halaju maksimum zarah itu,

[ 3 marks ] [3 markah]

(b) the value of k, nilai k,

[2 marks] [2 markah]

(c) the distance travelled in the third second, jarak yang dilalui dalam saat ketiga.

[2 marks] [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] CONFIDENTIAL

CONFIDENTIAL

17

3472/2

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 x ≥ 0 and y ≥ 0 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

CONFIDENTIAL

18

CONFIDENTIAL

3472/2

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

A

B

C

Soalan

Soalan Dijawab

1

Markah Penuh 6

2

6

3

8

4

7

5

6

6

7

7

10

8

10

9

10

10

10

11

10

12

10

13

10

14

10

15

10

Markah Diperoleh (Untuk Kegunaan Pemeriksa)

SULIT

CONFIDENTIAL

CONFIDENTIAL

19

3472/2

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

CONFIDENTIAL

20

CONFIDENTIAL Question 1.

3472/2

Working/Solution 2 1 + = 5 and x + 3 y = 5 x y 2 1 or any correct pairs from + = x + 3 y = 5 x y 5− x x = 5 − 3y or y= 3

Marks P1

P1

5− x into the non-linear 3 equation to obtain a quadratic equation in terms of x or y 15 y 2 − 26 y + 5 = 0 or 5 x 2 − 24 x + 10 = 0 Substitute x = 5 − 3 y or y =

K1

y = 1.513 ; y = 0.220 or x = 4.339 ; x = 0.461

N1 N1 K1

Use triangle law correctly to find AC or DF (i) AC = AB + BC

6

(ii) DF = DC + CF

DF = −30 x

AC = −24 x + 40 y 2(b)

K1

Solve quadratic equation using formula or completing the squares

x = 0.461 ; x = 4.340 or y = 0.220 ; y = 1.513 2(a)

Total

N1 N1

Find vector CE or EC or AE or EA by correct triangle law.

K1

CE = CF + FE 3 = CF + FD 5 3 = CF − DF 5 = 6(3 x − 5 y ) 1 CE 6 Find AC in term of CE or vice versa or any equation that can conclude A, C, E are collinear. AC = −24 x + 40 y 3x − 5 y =

K1

AC = −8(3 x − 5 y ) 4 AC = − CE 3 4 ∴A, C, E are collinear because AC = − CE 3

N1

CONFIDENTIAL

6

21

CONFIDENTIAL

Question 3(a)

3(b)(i)

3472/2

Working/Solution Use cos 2x or sin 2x and factorize 1 + cos x + cos 2 x 1 + cos x + (2 cos 2 x − 1) = sin 2 x + sin x 2 sin x cos x + sin x cos x + 2 cos 2 x = sin x(2 cos x + 1) cos x(2 cos x + 1) = sin x(2 cos x + 1) cos x = sin x = cot x 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), ( (π , –2), (

3π , 1) and (2π , 4). 2 y

y=

4

π

2

, 1),

Marks K1

Total

N1

N1 P1 OR P2 OR P3

3x

π

y = 3 cos x + 1

1 O

π

π

3π 2

2

2π

x

-2 3(b)(ii)

Get the correct linear equation and draw a straight line. 3π cos x = 3 x − π

K1

3π cos x + π = 3 x

π (3 cos x + 1) = 3x 3 cos x + 1 =

3x

π

The equation of straight line : y =

3x

π

Straight line drawn correctly and number of solutions = 1

N1

CONFIDENTIAL

8

CONFIDENTIAL

Question 4.(a)

4(b)

5(a)

5(b)

22

Working/Solution (i) Use median formula with at least two of the L, N, F, f and c correctly substituted.  22 + k  − 10   (10) 34.5 = 30.5 +  2 10       k =6 (ii) median mark = 42.5 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

3472/2

Marks K1

N1 N1 P1 K1

Correct way of finding the value of mode.

K1

Modal mark = 35.0 1 y = 3x + x dy 1 = 3− 2 dx x

N1

Find gradient of normal and use y − y1 = m( x − x1 ) dy at (1 , 4), =2 dx 1 Gradient of normal = − 2 Equation of normal : 1 y − 4 = − ( x − 1) 2 x + 2 y − 9 = 0 or equivalent dV Get the expression for V and find to determine the dr dV value of at r = 20 cm. dr 2 V = π r3 3 dV = 2π r 2 dr dV dV dr Use = × dt dr dt dV = 2π (20) 2 (−0.1) dt = −80π cm 3 s −1

Total

7

P1

K1

N1

K1

K1

N1

CONFIDENTIAL

6

23

CONFIDENTIAL

Question 6.

3472/2

Working/Solution (a) Perimeters of the first three circles : 4 8 2πR , πR , πR 3 9 T T Check the ratios 2 and 3 T1 T2

Marks

Total

K1

Make a correct conclusion : T T 2 Since 2 = 3 = , the perimeters of the first three T1 T2 3 circles form a geometric progression with common 2 ratio . 3

N1

(b)

πR 2 = 900π R = 30

P1

(i) Use the formula Tn = ar n−1 :

K1

9

2 T10 = (60π )  3 = 1.561π cm

N1

a : 1− r 60π S∞ = 2 1− 3 = 180π cm y correct.

(ii) Use the formula S ∞ =

7(a)

All values of log10

x+2 log10 y

1 0.92

2 1.00

3 1.08

4 1.12

K1

N1

5 1.24

6 1.32

N1

Plot log10 y against ( x + 2 ) with correct axes, uniform scales and at least one point plotted correctly.

K1 N1

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 .)

N1

CONFIDENTIAL

7

24

CONFIDENTIAL 7(b)

(i) Recognize the wrong recorded value of y and use graph to find the should be value of y. log 10 y = 1 .16

y = 14 .. 45 (ii) log10 y = ( x + 2) log10 p − log10 q Use − log10 q = c or log10 q = −0.84 p = 0.1445 ; 8(a)

Use log10 p = m log10 p = 0.08 p = 1.202

1 x1 x 2 x3 x1 correctly to find the area of triangle 2 y1 y 2 y 3 y1 Area = 10 unit2 Either midpoint of AB or gradient of AB correct. Mid point of AB = (1 , –3) Gradient of AB = –1 Use

8(b)

8(c)

8(d)

3472/2 K1

N1 P1

K1 N1 N1 10 K1 N1 P1

Use y − y1 = m( x − x1 ) with his midpoint of AB and his gradient of normal. y + 3 = 1(x -1) y=x–4 Use Ratio Theorem follow his midpoint  4(1) + 1(6) 4(−3) + 1(−8)  C= ,  5 5   C = ( 2 , − 4) Use of distance formula correctly for PB or PC

K1

PB = ( x − 6) 2 + ( y + 8) 2 ; PC = ( x − 2) 2 + ( y + 4) 2

K1

N1 K1 N1

Use BP = 2 PC ( x − 6) 2 + ( y + 8) 2 = 2 ( x − 2) 2 + ( y + 4) 2 3 x + 3 y − 4 x + 16 y − 20 = 0 (i) (2 , 4) y = − x( x − 4) or eqivalent (ii) Correct method of finding area under curve or area of ∆ 2

9(a) 9(b)

4

2

K1 N1

10

P1 P1 K1

2

 2x 2   4x 2 x3  1 − or   or × 2 × 4   3 2 2  2  2 0 Find integration value using correct limits  2(2) 2 2(0) 3   4(4) 2 4 3   4(2) 2 2 3  − − − or −       3  2 3 2   2  2 Find the sum of two area 28 unit2 or equivalent 3

K1

K1 N1

CONFIDENTIAL

25

CONFIDENTIAL

9(c)

Correct method of finding volume of revolution or volume of cone. 2 2 1 π ∫ 16 x 2 − 8 x 3 + x 4 dx or π 4 x 2 dx or π (4) 2 (2) 0 0 3 Find integration value using correct limits  16(2) 3 8(2) 4 2 5    4( 2) 3  π  − +  − 0 or π  − 0 4 5    3   3 Find the difference of two volume 32 π cm3 or equivalent 5 2 3 (i) Both p = and q = or equivalent 5 5

(

10(a)

3472/2

)

3

8

K1

∫

K1

K1 N1

10

P1

5

 2  3 C 3     or equivalent  5 5

K1

0.2787 or other more accurate answers. 0

8

1

7

N1 2

 2 3  2 3  2 3 (ii) 1 - C 0     + 8C1     + 8C 2      5 5  5 5  5 5

6

8

3

5

4

4

8

 2  3  2 3  2 3 or C 3     + 8C 4     + ...+ 8C8      5 5 5 5  5 5 All terms must be correct and completed in full

K1 0

8

10(b)

11(a)

0.6846 or other more accurate answers. 60 − 50   (i) P Z ≥  16   0.2660 m − 50  m − 50    (ii) P Z <  = 0.28 or P Z ≥ −  = 0.28 16  16    or equivalent m − 50 − = 0.583 16 40.672 kg (i) SAC = 5(1.2) 6 cm (ii) Correct method of finding AB 1 AC 1 .2 2 = sin rad or equivalent OR cosine rule 5 2 AB = AC = 5.646 cm (5.64578271...)

N1 K1 N1

K1

K1 N1 K1 N1

10

K1

N1

CONFIDENTIAL

26

CONFIDENTIAL 11(b)

(i)

∠BAC = 1.2 +

π + 1 .2

or equivalent

2 2.171 rad or equivalent

1 (5.646)2(2.171) 2 1 Area of segment AC = (5) 2 (1.2 − sin 1.2rad ) 2 or equivalent

3472/2 K1 N1

(ii) Area of sector BAC =

Find difference of the two area 1 1 (5.646) 2 (2.171) − (5) 2 (1.2 − sin 1.2rad ) 2 2

K1 K1

K1

Area of shaded region = 31.25 cm2 (31.25261028...) N1 12(a)

7 (i) × 100 = 140 or equivalent A05 RM5 (ii)

12(b)

12(c)

140(30) + x(20) + 110(15) + 104(10) + 120(25) = 132 30 + 20 + 15 + 10 + 25 165.5

Q07 × 100 = 132 40 Q07 = RM 52.80 Q09 = RM 58.08 Q05 Q 104 × 100 = 125 and 07 = Q03 Q05 100 Q Q 125 Or 07 × 100 = 104 and 05 = Q05 Q03 100 Q I 07,03 = 07 × 100 Q03 Q Q 125 104 I 07,03 = 07 × 05 × 100 or × 104 or 125 × or Q05 Q03 100 100

10

K1 N1

K1 N1 K1 N1 N1

P1

K1

125 × 104 100 = 130

N1

10

CONFIDENTIAL

27

CONFIDENTIAL 13(a)

13(b)

13(c)

Using correct rules to find PR and QR. PR 10 QR 10 = or = o o o sin 110 sin 20 o sin 50 sin 20

K1

PR = 22.40 cm (22.39764114...) QR = 27.47 cm (27.47477419...) Use cosine rule to find QS. QS2 = 42 + 27.472 – 2(4)(27.47)cos (110 + 20)o QS = 30.20 cm Use formula correctly to find area of triangle PQR or PRS. 1 Area ∆PQR = (27.47)(10) sin 50 o 2 1 or Area ∆PRS = (22.40)(4) sin 110 o 2

N1 N1

Use Area PQRS = sum of two area Area PQRS = 147.32 cm2 14(a)

14(b)

14(c)

3472/2

Find expression for v. 2t 2 v = 2t − +c 2 Find t for maximum v and substitute in his v. a = 2 - 2t = 0 2(1) 2 and v max = 2 − +3 2 vmax = 4 ms-1 Try solving v = 0 2k - k2 + 3 = 0 (3 - k)(1 + k) = 0 or equivalent k=3s Correct method in finding distance traveled.

K1 N1 N1 K1

N1

K1 N1 K1

10

K1

N1 K1

N1 K1

3

14(d)

 2t 2 t 3  1 s= − + 3t  or s = t 2 − t 3 + 3t and s3 − s 2 3 3  2 2 5 s= m 3 Find expression for s. t3 s = t 2 − + 3t 3 Solve his s = 0 t3 t 2 − + 3t = 0 3 3t2 - t3 + 9t = 0 t2 - 3t -9 = 0 t = 4.854 s

N1 K1

K1

N1

10

CONFIDENTIAL

CONFIDENTIAL

15(a)

15(b)

15(c)

28

3472/2

x + y ≤ 160 note: ss-1 for all answers in terms of P and Q 40 x ≤ 100 y or equivalent y ≤ x + 80 or equivalent 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. (i) Drawing of the straight line, x = 50, completely across the shaded region R. 20 ≤ y ≤ 110 (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

N1 N1 N1 K1 K1 N1 K1 N1 K1

N1

10

CONFIDENTIAL

29

CONFIDENTIAL

3472/2

Graph For Question 4(b) Number of students

10

9

8

7

6

5

4

3

2

1

0.5

10.5

20.5

30.5 40.5 Modal mark = 35.0

50.5

60.5

marks

CONFIDENTIAL

30

CONFIDENTIAL

3472/2

No.7(a)

log10 y

x

1.30 (6 , 1.32) 1.25 x

1.20 1.16 1.15 x

1.32 − 0.92 6 −1 m = 0.08 c = 0.84

1.10

m=

x

1.05

x

1.00

0.95 x (1

, 0.92)

0.90

0.85

0.80

0

1

2

3

4

5

6

7

CONFIDENTIAL

x+2

CONFIDENTIAL

31

3472/2

Qn. 15

CONFIDENTIAL