# skema jawapan peperiksaan percubaan kertas 1 skema jawapan peperiksaan percubaan kertas 1 2017...

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• SKEMA JAWAPAN

PEPERIKSAAN PERCUBAAN KERTAS 1

2017

SOALAN 1

BIL JAWAPAN SUB MARK MARKAH

PENUH

1

1)32(log:B1

5)32(:B2

5

1





xx

xx

2

5 x

3 3

2 1

3

1

2 

x

x

 

  

 2 log

x

x

3

B2

B1

3

3

2527:1

25133:2

213

logloglog

loglog

aaa

aa

aB

B

nm







3

4 3 x – 3y

B2 : y

x

3

2

1

2

3

)3(

B1 : m = 9 x atau n = 33

y

3 3

5. 1

B2 :    

 

3

1 3

log log

5

5

B1 :

3 3

• SOALAN 2

1

1

7

7

7

7

1

7

777

7log

)753(log

7

1 log

105log :B1

7log

7log5log3log :B2

1

1 :B3









atau

kh

1kh

4

4

2

kh

hk



2

12

3log2log2

5log2log3log2

55

555



5log2log2log 55 2

5  or 3log2log 5 2

5  or

5log2log3log 1212 2

12 

12log

90log

5

5 or 2 log 5 3 or 2 log52

4

B3

B2

B1

4

3 (a)

p

1

(b) 12

25

p

p

13log

23log 2

5

m

m

m

m

m

m

9log

243log 2

1

3

B2

B1

4

• 4

2

23 xy 

B3 : 2

log8loglog2 222 bc 

B2: 4log

8loglog

2

2

2

2 bc 

B1: bc 8loglog 4 2

4  @ 4log

8 log

2

2

2 b

c

4 4

5

q

pq

1

B3 : log2 3(1 – p) = pq

B2 : log2 3 = p(log2 4 + log2 3)

B1 : 12log

3log

2

2 = p

4 4

SOALAN 3

1 15

B2 : )2)(1()5)(1()42(  nxnx

B1 : )5)(1()42(  nx atau

)2)(1(  nx

3 3

2 n = 42

87)2)(1(5  n

d = 2

3

B2

B1

3

3 1256

)168(2))8(4)200(2( 2

5 5 S

)8(42005 T

3

B2

B1

3

• 4 a) k = 2h + 1

k -3h = (h + 2) – k

b) 9 – 6h

2

B1

1

3

5 a) d = 6

b) 5

B1 :  

  

 

   )5(283)6(283 )5()6(

22

1

2

3

SOALAN 4

1 7 1 3

4

B1 : 282 T atau 7

28

2

2

6

1

3

1 1

9

1

3

1 r

3

B2

B1

3

3 3

2

6 r dan p = 6

3 2

6 r atau p = 6

p

p

p

ppp

18

2

36

6,0,6

2



3

B2

B1

3

• 4 x1 = 3 dan x2 = –1 B2 : (x – 3)(x + 1) = 0

B1 : 3x 2 – 6x – 9 = 0

3

5 -24

B2 :

 

  

 

4

3 1

6

B1 : a = - 6 , 4

3 r

3 3

SOALAN 5

1 (a) 5

B1 : )2(

)4(

)4(

)2(

 

x

x

x

x

2 4

(b)

2

243

B1 : a = T1 = 81 ...seen atau

3

1 1

81

2

2 (a)

3

2

5

4

27

8  or equivalent 

  

 

2

3

1

2

T

T

T

T

(b) 3

4 or equivalent

• 3

2 1

9

4

3 a) x = 6

B1 : 12

2412 

x

b) 6096

B1 :    

12

16

12

16 22 310

 

1

1

4

4 atau 5.5

B3 :  

  

 

2

1 11

2T

B2 : a = 11

B1 : 22

2

1 1

a

4 4

5 a) a = 8 , - 2

B1 : 4

224

 

a

a

a

a

b) 16

2187 atau 136.6875

B1 :  

  

 

2

3 8

9 3

16 T

1

1

4

SOALAN 6

1 y = x3

3

3

• B2 : 4 = 4 + c atau 12 = 12 + c atau c = 0

B1 : m = 1

2 p = 2 and 1q

p = 2 or 1q

04

)5(3

 p or 55 q

qpxxy 52 2 

3

B2

B1

3

3 s = 3 and t = 5

7 = 2s + 1 or t = 2(2) + 1

3

B1

3

4 r = 1, p = 0.001 (BOTH)

r = 1 or p = 0.001 (either one)

=-3 or r = or

3

5 m =

2

3 dan n = 3

B1 : –2 = m

3 

y = –2x

2 + 3x

2

1

3

SOALAN 7

BIL JAWAPAN SUB MARK MARKAH PENUH

B3 : 0)2)(2(  

4

4

• B2 : 100 2

40

2

1 2

 

  

 

B1 : 40 rrr

2 (a) 842.1

5.65 

(b) 23.025

)842.1()5(

2

1 2 * (candidate’s  from a)

2

B1

2

B1

4

3 (a) 10.8 cm

(b) 19.44 cm2

) 142.3

180 9.0tan(612

2

1 9.012

2

1 2 

6

9.0tan h

1

3

B2

B1

4

4 (a) θ = 0.842 rad

kos θ = 15

10

(b) 126.708 cm 2

Area = 88)18.11)(10(

2

1 )841.0()15(

2

1 2 

2

B1

2

B1

4

SOALAN 8

• BIL JAWAPAN SUB MARK MARKAH PENUH

1 (a)

15

17

(b) 221

21

 

  

  

  

 

  

 

  

 

13

5

17

15

13

12

17

8 B1

1

2

3

2 (a) tan A =

12

5 

(b) sin (A – B) = 65

56

 

  

  

  

 

  

  

  

5

3

13

12

5

4

13

5

1

2

B1

3

3 θ = 45o, 63.43o, 206.57o, 225o

θ = 63.43o, 206.57o OR θ = 45o, 225o

(tan θ - 2)(tan θ - 1) = 0

3

B2

B1

3

4

7

17 atau 2.4286

B2 :

)1)( 5

12 (1

1 5

12





atau tan (112.62 – 45)

B1 : tan  = 5

12  atau tan 45 = 1

3 3

5 (a)

p

1 

1

2

3

• (b) p

p21 

21 p seen

B1

SOALAN 9

BIL JAWAPAN SUB MARK MARKAH PENUH

1 x = 0 o , 30

o , 150

o , 180

o , 360

o (all correct)

B3 : 0 o and 30

o (both)

B2 : sin x (-2sinx +1 )= 0

B1 : (1-2sin 2 x) + sin x - 1 = 0

4 4

2 0, 120, 180, 240, 360

B3 : 0, 180, 360 atau 120, 240

B2 : sin x(2 kos x + 1) = 0

B1 : 2 sin x kos x + sin x = 0

4 4

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