add mth 12 dan skema f4 final mlk 2009

8/4/2019 Add Mth 12 Dan Skema f4 Final Mlk 2009

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NO.KAD PENGENALAN/I.C NUMBER- -

PEPERIKSAAN SELARAS AKHIR TAHUNSEKOLAH-SEKOLAH MENENGAH NEGERI MELAKA

KelolaanPEJABAT PELAJARAN DAERAH

JASIN * ALOR GAJAH * MELAKA TENGAHDengan kerjasama :

JABATAN PELAJARAN NEGERI MELAKATINGKATAN 4 2009

ADDITIONAL MATHEMATICSPaper 12 hours

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Tuliskan nombor kad pengenalan, nama dan tingkatan anda pada ruang yang disediakan.

Write your I.C. number, name andclass in the space provided.

2. Calon dikehendaki membaca arahandi halaman 2 dan halaman 3

Candidates are required to read theinstructions 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

9 310 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

3472/1Form FourAdditional MathematicsPaper 12009

2 hours Nama Calon :

Tingkatan : .


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SULIT 3472/1

3472/1 SULIT

2MAKLUMAT UNTUK CALON

1. Kertas soalan ini mengandungi 25 soalan.

2. Jawab semua soalan.

3. Bagi setiap soalan berikan SATU jawapan 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 anda

untuk 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. Answer ALL questions.

3. Give only ONE answer 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.


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SULIT 3472/1

[See overleaf]3472/1 SULIT

3

ALGEBRA

1 a

acbb x

242

= 8 ab

bc

ca log

loglog =

2 a m x a n = a m + n 9 Tn = a + (n -1)d

3 a m a n = a m n 10 S n = [ ]2 ( 1)2n

a n d +

4 ( a m )n = a m n 11 T n = 1nar

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

, 11 1

n n

n

a r a r S r

r r

= =

6 nmnm

aaa logloglog = 13 , 11a

S r r

=


37/43

7 (a) Midpoint of QR = (5,4) or m QR = 1 y = x + c passes through (5,4) y = x + 9 or equivalent

or

22 )5()6( + y x or 22 )3()4( + y x x2 12 x + 36 + y2 10 y + 25 = x2 8 x + 16 + y2 6 y + 9 y = x + 9 or equivalent

(b) 22 )3()4( + y x = 2x2 + y 2 8x 6y + 21 = 0

(c) x 2 + (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

111

1

11

11

111

11

3

2

3

2

8 (a) h =2

24

xor curved surface area of cone = 2 x2 or curved

surface area of cylinder = 2 x(2

24 x

)

Total surface area = 2 x2 + 2 x( x24

) + x2

= 3 (x2 + x16 )

(b) 6 x -2

48

x

= 0

x = 2A = 36

(c)dt dA

dAdx

dt dx

=

=

2

486

1

x x

(42 )

= 2

1

1

1

1

11

1

1

3

3

2


38/43

(d)2

486(

x x A

= )(0.003)

= 0.063

11 2

9 (a) Identify Q 1 class = 6 10 or L = 5.5

5.5 + )5(5

3)17(41 + k

= 7.5

k = 3

(b) x =20

255 or 2 x = 4175

2)20

255(

204175

=

= 6.796

(c) Complete histogramUsing X to determine the modeMode = 11.75

1

1

1

1

1

1

211

3

3

4

10 (a) 60

(b) s = 8(20 x180

)

= 2.973 cm

(c) AB 2 = 8 2 + 8 2 2(8)(8)Cos 100AB = 12.26 cm

Perimeter = 26.21 cm

1

11

111

1

2

3

Mode=11.75


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(d) Segment 1 = (8 2)(1.745) (8)(8) sin 100Segment 2 = (8 2)(1.047) (8)(8) sin 60Shaded region = 24.33 5.791

= 18. 54 cm 2

1111 4

11(a) AC m = 2 or ABm =

2

1

1** = AB AC mm 11 2

(b) midpoint AC = (0 , 0)

D ( x , y) =

+

21

,2

7 y x

02

7=

+ xor 0

21

= y

x = 7 or y = 1 D = ( 7, 1)

Area of parallelogram =21

12

71

12

71

=21 *

1629

=2

13unit 2

(c) Centre = midpoint BC

= E (3 ,23

)

Distance EC or Distance EB or Distance ED

( )2

2

23

3

++ y x = ( )2

2 223

13

+++ or

( )2

2

23

3

++ y x = ( )2

2

23

137

++

053622 =++ y x y x

1

1

1

1

1

1

1

1

2

2

4


40/43

12 (a) PR 2 = 12 2 + 20 2 2(12)(20) Cos 72 0

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

(b) Use16

48Sin*89.19

SinS 0=

S = 67.49 0 , 112.51 0 (must 2 d.p, correct round off

(c) (i) P = 180 48 0 67.49 0 *

= 64.51 0

48Sin16

*49.67Sin*89.19

*51.64SinRS

== or use Cosine

Rule in PRS

RS = 19.43

( ii) 00 72sin)12)(20(x21

or48sinx*)43.19*)(89.19(x21

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

P32RM

06or equivalent

P06 = RM25.60

ii) Use I 08/04 =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


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i) += x IW 209550

Use100

IWI 06 / 08 = *

100209550

5.124 x+

=

x = 145

ii) Use 100xCostCost

I06

0806 / 08 = or equivalent

or Cost 08 =100

14x5.124RM

= RM 17.43

1

1

1

1

15

14 (a) i) PR 2 = 6 2 + 14.5 2 2(6)(14.5) Cos 48 0

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

ii)*39.11

110Sin8.4

SinPRQ0

= or equivalent

S = 23.33 0

(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 = 180 110 0 23.33 *

= 46.67 0

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 =

= 23.34 0

Area PQR = *034.23sinx*)82.8()8.4(x21

= 8.386

1

1

1

1 6

15(a) Use I 04/00 = 100x

PP

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*50400

I 00 / 04 =

= 140

ii) Use 100xCostCost

I00

0400 / 04 = or

Cost 00 =140

100x50.128RM

= RM20.36

1

1, 0

1

1

1

1

1

2

5


43/43

(c) 130I 04 / 08 =

Use I 08/04 =100

xII 00 / 0404 / 08 or

=100

140x130

= 182

1

1

13

add mth 12 dan skema f4 final mlk 2009

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