# sijil pelajaran malaysia 2010 matematik tambahan kertas 1 skema jawapan peperiksaan percubaan sijil...

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• SULIT

3472/1 © 2010 Hak Cipta Jabatan Pelajaran Selangor SULIT

PROGRAM PENINGKATAN PRESTASI AKADEMIK 3472/1 SIJIL PELAJARAN MALAYSIA 2010 Matematik Tambahan Kertas 1

SKEMA JAWAPAN PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2010

MATEMATIK TAMBAHAN

Kertas 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan Pemarkahan ini mengandungi 4 halaman bercetak

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• SULIT 2 3472/1

3472/1 © 2010 Hak Cipta Jabatan Pelajaran Selangor SULIT

1 (a)

2 3

t

(b) 0 ≤ f(x)≤13

1

2

6

p = – 2 and q = – 5

p = – 2 or q = – 5

(x + 2)2 – 5

3

B2 B1

3

2

2

(a)

2 11)4( f

2 324  x

(b) 1k

2 1

2 3)2(2 

k

B1

2

B1

7

2x

9 13 x

23 3333  xx

or equivalent

B2

B1

3

(a) 16x –5

4(4x – 1) – 1 (b) a =8 and b = 5 a =8 or b = 5

1

2 B1

4

2x2–11x– 6=0 (x– 6)(x +

2 1 ) = 0 or equivalent

2

B1

8 9

x = 243

5log 3 x

5log 22 3 x

2 3

y = 9 or G.P : 8, 12, 18 y2 – 8y – 9 = 0 or

3 2

1 3

 

 

y y

y y

3

B2

B1

3

B2

B1

5

2 < k < 6 (k – 2)(k – 6) < 0 k2–4(1)(2k – 3)< 0

3

B2

B1

10

n = 5

729)3(9 1 n

2

B1

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• SULIT 3 3472/1

3472/1 © 2010 Hak Cipta Jabatan Pelajaran Selangor SULIT

11

–1

[ 8] – [9] S4 = [24-16] or equivalent OR S3 = [18 – 9] or equivalent

3

B2

B1

15

(a) 1 : 2

37   

nm nm

(b) 4

2

B1

1

12

(a) 1

1cos 2 

 p

(b) 21 2

p p

  

   

  

   

 1 1

1 2

22 pp p or

equivalent

1 2 B1

16

8

bamba 9)2()64(  ,

2

9 4 6

 

m

or equivalent

2

B1

13

h = 3 1 and k = –2

h = 3 1 or k = –2

231  kor h

gradient = 2 or

h

kx y

11 

4

B3

B2

B1

17

(a) ba 68  (b) ba 34 

 abbKH 86 2 16 

or equivalent

1

2

B1

14

(a) 3 (b) 05422  yyx 3)( 22  yx

1

2

B1

18

(a) 2. 5 radian

45.92)6.8( 2 1 2 

(b) 21.5cm 5.26.8 

2 B1 2 B1

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• SULIT 4 3472/1

3472/1 © 2010 Hak Cipta Jabatan Pelajaran Selangor SULIT

19

8 3

a and b = 3

8a + 2b = 3 or 8a + b = 0 4ax + b = 0

3 B2 B1

23

(a) 84 (b) 30 2

4 4

5 CC  2

4 4

5 CorC

1 3 B2 B1

20 (a) 2

3 p

2

12 

 p

(b) 4

2)( 1

xpdx dy

 

2 B1 2 B1

24

(a) 18 5

(b) 18 11

6 5

3 2  +

6 1

3 1 

6 5

3 2  or

6 1

3 1 

1 3 B2 B1

21

(a) – 4 (b) k = 2 0)2)(4(  kk or

kk  



  

2 )2(

2 4

2 1 22

k xor

2

2

2 4

2 1

 

 

 

1 3 B2 B1

25

(a) k = 1.03 8485.01)(  kzP (b) 198

1515.0 30

 n

2 B1 2 B1

22

(a) m = 14 (b) std deviation = 4.598

212 7

1156 

1

2

B1

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