Skema Add Math P1 Trial Spm Zon A 2009

SULIT 3472/1 Additional Mathematics Paper 1 Sept 2009

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5 2009

ADDITIONAL MATHEMATICS Paper 1 MARKING SCHEME

This marking scheme consists of 7 printed pages

2 PAPER 1 MARKING SCHEME Number

Solution and marking scheme

3472/1 Sub Marks

Full Marks 2

1 (a)

4

1

(b)

One to many relation

1

2 (a)

4 x 2 + 20 x + 21

2

(2 x + 5)2 − 4

(b) 3 (a)

(b)

4

5

B1

−3

1

5x − 2 3

2

y=

3x + 2 5

3

B1

1

11 5

49 8

3

49 − 8p > 0

B2

(−5)2 − 4(2)(p − 3) > 0

B1

x2 + 4x − 18 = 0

3

3α + 3β = −4 or 3α(3β) = −18

B2

p<

3

α+β=−

4 and αβ = −2 3

3

3 B1

3

Number

Solution and marking scheme

Sub Marks

6 (a)

5

1

(b)

13

1

(c)

x=3

1

x ≤ −1, x ≥ 2

7

Full Marks

3

3 3

8

9

( x − 2)( x + 1) ≥ 0

B2

x2 − x − 2 ≥ 0

B1

5 3

3

2x + 5 = −x

B2

33 × 32( x +1) = 3− x or 33+2 x+ 2 = 3− x

B1

x=−

x =1

3

x+3 =2 or IE 3x − 1 x +3 log 2 =1 or IE 3x − 1 10 (a)

21

B2

3

B1

2

3 1 d =  7(2) 2 − 3(2)  − 2 × 7(1)2 − 3(1)  or IE 2 2

(b)

3

B1

3

1

216

11 3

7 33

3 B2

0.21 1 − 0.01

a = 0.21 and

r = 0.01

B1

4

Number

Solution and marking scheme

Sub Marks

Full Marks

12 r=2 q = −6

B1 B1

y = x + 3* x

B1 B1

p=3

M ( 1, 5 )

13

14

3

 3(−3) + 2(7) 3(3) + 2(8)  M ,  3+ 2 3+ 2  

B2

AM : BM = 2 : 3

B1

a=−

3

3

4 b

B2

 b − a ×  −  = −1  4 m1 = − a or

15 (a)

B1

2 or

5 − (6v) − 3u + 6v % % 3 %

Use TL

15   6v + ( −3u% − 4v% ) 23 %

b)

or

uuur PQ = 12 i − 5 j % %

24 i − 10 j % % 13 12 i − 5 j % % or 13

4

B1

2

−3u + v % %

16 a)

3

b m2 = − 4

−3u − 4v % %

(b)

4

Use TL

B1

1

2 2×

12 i − 5 j % % 13

B1

3

5

Number

Solution and marking scheme

Sub Marks

Full Marks

17

210o , 270o330o

4

1 sin θ = − and sin θ = −1 2

B3

4

B2 (2sin x + 1)(sin x + 1) = 0 B1

2sin 2 x + 3sin x + 1 = 0 18 (a)

4.8 cm

2

8 × 0.6

B1

81.33

2

4

(b)

1 (8)2(π − 0.6) 2

or

π − 0.6

B1

19 3

 3 151   ,   4 16 

3 2x −

3 = 0 or 2

dy 3 = 2x − dx 2

x=

3 4

B2

B1

20 dx = −10 dt 10 = (3 −

3 3 16 dx ). 2 2 dt

dy 16 = 3− 2 dx x

B2

B1

6

Number 21 (a) (b)

Solution and marking scheme

Sub Marks

−6

1

16 27

2

Full Marks

3

6

 kx 2  2+  = 10  2 3

B1

22 Mean = 13 AND

Standard deviation = 8

Mean = 13 AND

Variance (new) = 64

Mean = 13 OR

Variance (new) = 64

3 B2

3

B1

23 (a)

7 20

1 3

(b)

29 60

2

2 7  3 5   × + ×   5 12   5 12  24 (a)

(b)

B1

P3

30 240 8

25 (a)

2

336 8

B1

C 3 ×10 C 2 ×12 C1

0.2266

4

2 B1 2 4

 P Z 

128 − 125  ≥  OR P[Z ≥ 0.75] 4

B1

7

Number (b)

Solution and marking scheme 127.1 m − 125   P Z ≥ = 0.3 OR  4 

m − 125 = 0.524 4

Sub Marks 2

B1

Full Marks