Modul 5-jawapan Kertas1 .pdf

ADDITIONAL MATHEMATICS Paper 1

JABATAN PENDIDIKAN NEGERI PULAU PINANG PROGRAM PENINGKATAN AKADEMIK MODUL 5 SPM 2015 MARK SCHEME

ADDITIONAL MATHEMATICS Paper 1

2

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ADDITIONAL MATHEMATICS MODULE 5 PAPER 1 2015 Question 1

Solution and Mark Scheme

Sub Marks 1

Total Mark

2

(a)

3

(b)

One to one

1

11

2

2

8 = 3

B1 2

5 3

p(3)2 + 3(3)  6 = 0

B1

m = - 5, n = 1 and p = 3

4

B2

n = 1 or p = 3

B1

5

or and 

2

2x2 - x – 11 = 0

2

4

B3 3

and  = 5

2

+ = 

3

or  = 5

2

-2

243

7

log3 q = 5

B2

B1 2

4

B1

2

3 B2 B1

8

(a)

- 40

(b)

x = - 8 or y = - 32 4096

3

2 B1 2 B1

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3

11

+ = 

6

2

3

- 2 = m(0  1)2 + 3

1

2

4

3

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522

9

T12 = 720 + 11(-18)

B2

720 , 702, 684,... or d = -18

B1

192

10

11

3

(a)

3

4

a = 3 and r =2

B3

a = 3 or r = 2

B2

3

B1 2

4

B1 (b)

–1

2 B1

or q = 3 – 4 (1) t = –3

12

2 or

|

B1

|

h = 26 and - 22

13

B2

|

|

x = 114.30° and 155.71° x = 114.30° or 155.71° sin 2x =

15

(a)

16

B1

B3

B2 B1

k = - 16

2

)

(

)

1:3

1

(a)

11i + (26 -k)j

1

(b)

26

2

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4

B1

(b)

√

3

4

sec x =

(

2

3

h = 26 or h = - 22

14

4

= 11

B1

3

3

4 5

17

3 (5)

B2

(5)

B1

p=2

18

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3

3

-6p(-13) + 11 = 167

B2

-6px + 11

B1

3

3

19

B2 B1 20

(a)

2

3

2 B1

(b)

21

20

2

(a)

0.3948 / 0.395

(b)

1.680 / 1.679 – 1.682

B1 1 3

( )

B2

0.7896

B1

1.68

22

4

4

4 B3

B2 or

23

(a)

280 8C 5 C 3 4

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2.8

or

equivalent

B1

2 B1

4

5 (b)

708

2

8C 5 C + 8C 5 C + 8 C 5 2 6 1 7

24

B1

(a)

3 16

1

(b)

31 48

2

 1 4  1 5       2 6  2 8

25

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(a)

- 0.94

174.83 = * – 0.94

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3

2

0.5 – 0.3264 (b)

B1

4

B1 2 B1

4