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NO.KAD PENGENALAN/I.C NUMBER

- -

PEPERIKSAAN SELARAS AKHIR TAHUN

SEKOLAH-SEKOLAH MENENGAH NEGERI MELAKA

Kelolaan

PEJABAT PELAJARAN DAERAH

JASIN * ALOR GAJAH * MELAKA TENGAH

Dengan kerjasama :

JABATAN PELAJARAN NEGERI MELAKA

TINGKATAN 4 2009

ADDITIONAL MATHEMATICS

Paper 1

2 hours

JANGAN BUKA KERTAS SOALAN INI

SEHINGGA DIBERITAHU

1. Tuliskannombor kad pengenalan,nama dantingkatan anda pada ruang

yang disediakan.

Write your I.C. number,name and

class in the space provided.

2. Calon dikehendaki membaca arahandi halaman 2 dan halaman 3

Candidates are required to read the

instructions on page 2 and 3 .

Kertas soalan ini mengandungi 15 halaman bercetak

Examiners Code

Full MarksQuestion

Marks Acquired

1 2

2 3

3 3

4 4

5 3

6 3

7 3

8 3

9 3

10 3

11 3

12 3

13 4

14 4

15 3

16 4

17 3

18 3

19 3

20 4

21 322 3

23 4

24 3

25 3

Total 80

3472/1

Form Four

Additional Mathematics

Paper 1

2009

2 hoursNama Calon :

Tingkatan : .


2/43

SULIT 3472/1

3472/1 SULIT

2

MAKLUMAT UNTUK CALON

1. Kertas soalan ini mengandungi 25 soalan.2. Jawab semua soalan.3. Bagi setiap soalan berikan SATUjawapan sahaja.4.

Jawapan hendaklah ditulis dengan jelas dalam ruang yang disediakan dalam kertas soalan.

5. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu andauntuk mendapatkan markah.

6. Sekiranya anda hendak menukarkan jawapan, batalkan kerja mengira yang telah dibuat.Kemudian tuliskan jawapan yang baru.

7. Rajah yang mengiringi soalan tidak dilukiskan mengikut skala kecuali dinyatakan.8. Markah yang diperuntukkan bagi setiap soalan dan ceraian soalan ditunjukkan dalam

kurungan.

9. Satu senarai rumus disediakan di halaman 4 hingga 6.10.Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram.11.Kertas soalan ini hendaklah diserahkan di akhir peperiksaan.

INFORMATIONS FOR CANDIDATES

1. This question paper consists of 25 questions.2. AnswerALLquestions.3. Give only ONEanswer for each question.4. Write your answer clearly in the spaces provided in the question paper.5. Show your working. It may help you to get marks.6. If you wish to change your answer, cross out the work that you have done. Then write down the

new answer.

7. The diagram in the questions provided are not drawn to scale unless stated.8. The marks allocated for each question and sub-part of a question are shown in brackets.9. A list of formulae is provided on pages 4 to 6.10.You may use a non-programmable scientific calculator.11.This question paper must be handed in at the end of the examination.


3/43

SULIT 3472/1

[See overleaf]

3472/1 SULIT

3

ALGEBRA

1a

acbbx

2

42 = 8

a

bb

c

ca

log

loglog =

2 am x a

n= a

m + n9 Tn = a + (n -1)d

3 am an = a m n 10 Sn = [ ]2 ( 1)2

na n d+

4 ( am )n = a m n 11 Tn =1n

ar

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

, 11 1

n n

n

a r a r S r

r r

= =

6 nmn

maaa logloglog = 13 , 1

1

aS r

r =


37/43

7 (a) Midpoint of QR = (5,4) or mQR = 1

y = x + c passes through (5,4)

y = x + 9 or equivalent

or

22)5()6( + yx or 22 )3()4( + yx

x2 12x + 36 +y2 10y + 25 =x2 8x + 16 +y2 6y + 9

y = x + 9 or equivalent

(b) 22 )3()4( + yx = 2

x2

+ y2

8x 6y + 21 = 0

(c) x2

+ (9 x)2

8x 6(9 x) + 21 = 0

(x 4)(x 6) = 0

(4,5), (6,3)

(d) Replacing point (5,4) to equation of locus P.

No

1

1

1

1

11

1

1

1

1

1

1

1

3

2

3

2

8(a) h =

2

24

xor curved surface area of cone = 2x

2 or curved

surface area of cylinder = 2x(2

24

x)

Total surface area = 2x2 + 2x(

x

24) + x

2

= 3(x2 +

x

16)

(b) 6x -2

48

x

= 0

x = 2A = 36

(c)dt

dA

dA

dx

dt

dx=

=

2

486

1

xx

(42)

= 2

1

1

1

1

11

1

1

3

3

2


38/43


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(d) Segment 1 = (82)(1.745) (8)(8) sin 100

Segment 2 = (82)(1.047) (8)(8) sin 60

Shaded region = 24.33 5.791

= 18. 54 cm2

1

1

1

1 4

11(a) ACm = 2 or ABm =

2

1

1**

= ABAC mm 11 2

(b) midpointAC= (0 , 0)

D (x ,y) =

+

2

1,

2

7 yx

02

7=

+xor 0

2

1=

y

x = 7 ory = 1

D = ( 7, 1)

Area of parallelogram =2

1

12

71

12

71

=2

1 *1629

=2

13unit 2

(c) Centre = midpointBC

=E(3 ,2

3 )

DistanceECor DistanceEB or DistanceED

( )2

2

2

33

++ yx = ( )

2

22

2

313

+++ or

( )2

2

2

33

++ yx = ( )

2

2

2

3137

++

053622 =++ yxyx

1

1

1

1

1

1

1

1

2

2

4


40/43

12 (a) PR2 = 122 + 202 2(12)(20) Cos 720

PR = 19.89 (at least 3 s.f, correct round off)

(b) Use16

48Sin

*89.19

SinS 0=

S = 67.490

, 112.510 (must 2 d.p, correct round off

(c) (i) P = 180 480 67.490 *

= 64.510

48Sin

16

*49.67Sin

*89.19

*51.64Sin

RS== or use Cosine

Rule in PRS

RS = 19.43

( ii) 00 72sin)12)(20(x2

1or48sinx*)43.19*)(89.19(x

2

1

Area = 143.60 + 114.13

Area = 257.73

1

1

1

1,1

1

1

1

1

1

2

3

5

13(a) i) 125 = 100x

P

32RM

06

or equivalent

P06 = RM25.60

ii) Use I08/04 =100

xII 04/0606/08 or equivalent

=100

140x115

= 161

1

1

1

1

1 5

(b)

Item I W IW

P 125 30 3750

Q 120 10 1200

R x 20 20x

S 115 40 4600

= 100W += x209550IW


41/43

i) += xIW 209550

Use100

IWI 06/08

= *

100

209550

5.124

x+

=

x = 145

ii) Use 100xCost

CostI

06

0806/08 =

or equivalent

or Cost08 =100

14x5.124RM

= RM 17.43

1

1

1

1

15

14 (a) i) PR2 = 62 + 14.52 2(6)(14.5) Cos 480

PR = 11.39 (at least 3 s.f, correct round off)

ii)*39.11

110Sin8.4

SinPRQ0

= or equivalent

S = 23.330

(must 2 d.p, correct round off, decimal form better than

minutes form)

1

1

1

1 4

(b) i)

Must correct labeled, angle P is obtuse

ii) P = 180 1100 23.33*

= 46.670

1

1

Q

P

R


42/43

*33.23Sin

8.4

*33.133Sin

QR

*67.46Sin

QR==

QR = 8.817 / 8.82

000 33.2333.133180QR'P =

= 23.340

Area PQR = *034.23sinx*)82.8()8.4(x2

1

= 8.386

1

1

1

1 6

15(a) Use I04/00 = 100x

P

P

00

04

x = 125, y = 2.08, z = 0.72

(b) i)

Item I W IW

A 125* 80 10000*B 130 100 13000

C 160 140 22400

D 125 40 5000

= 360W = 50400IW

= 50400IW

Use360

IWI 00/04

= or

360

*50400I 00/04 =

= 140

ii) Use 100xCost

CostI

00

0400/04 =

or

Cost00 =140

100x50.128RM

= RM20.36

1

1, 0

1

1

1

1

1

2

5


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