Skema Add Math P1 Trial Spm Zon A Kuching 2008

SULIT 3472/1 Additional Mathematics Paper 1 Sept 2008

SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5 2008

ADDITIONAL MATHEMATICS Paper 1 MARKING SCHEME This marking scheme consists of 6 printed pages

2 PAPER 1 MARKING SCHEME Number 1 (a) (b) 2

Solution and marking scheme 0

(b)

x − 5 or

f : x → x − 5 or f(x) = x − 5

a = 1 and b = −1

5

1

Full Marks

2

3 3

gf(x) = x + x − 1

B2

(x − 1)2 + 3(x − 1) + 1

B1

1

1 2

3

x3 1 ,x   2x 1 2

2

x 3 1 2x

B1

y= 4

Sub Marks 1

2

3 (a)

3472/1

t>1

3

–12t < 12 or equivalent

B2

(6)2 – 12 (t +2)(3) < 0

B1

x2 + 3x + 14 = 0

3

2 (2 ) = 14

 

and

2 + 2 = 3

3 7 and  = 2 2

3

B2 3 B1

3

Number 6

7

Solution and marking scheme

Sub Marks

Full Marks

3

3

1 2 a(−2)² + 3 = 5

B2

p = −2 and q = 3

B1

a=

−3 ≤ x ≤ 5

3 3

8

(x − 5)(x + 3) ≤ 0

B2

x² − 2x − 15 ≤ 0

B1

x  4

3

x 1 2x  3

or

10 (a)

B2 3

3x 12 x  33 or

9

equivalent 3

x 1 2x

3

 33

5 1

B1

3

lg 2  lg17 lg 2

B2

lg 34 lg 2

B1

d 5

2

a @ T1 = 4

or

T2 = 9

585

1

11

18, 54, 162

3

a=

or

54

or 18 or equivalent (Solving)

2 and r = 3 19683

3

B1

(b)

n = 12

3

B2 B1

3

4

Number 12 (a)

Solution and marking scheme k = 1000000 log10 y = 4 log10 x + log10 k

(b)

h = 22

14

h = ±2

3

1  h  2h  0  h  5h  0   9 2

B2

1  h  2h  0  h  5h  0  2

B1

p6

m1 

16 (a)

(b)

4

B1

3

3

p 5     1 5  6

15

B1

Full Marks

2

h−6 =4 4−0 13

Sub Marks 2

p 5

or

or

equivalent

m2  

5 6

B2 B1

m = 1

3

1 m 2  5 5

B2

a  b  5 i   1  m  j % % % u uur % BD  i  5 j % % uuur uuu r uuur uuu r uuur BD  BA  AD or BA  BC 50 uuur AC  7 i  j %%

3

3

B1 2 B1 2 B1

4

5

Number 17

Solution and marking scheme 66.42o, 293.58o cos 

2 5

1  5cos  3 18

or equivalent

Sub Marks 3

Full Marks

B2

3

B1 3

25 7

3 1

1 2

2

B2

3     5

19 (a)

1 cos 2x

B1

r = 16

2

 AOC = 0.45 (b)

20

or

344576 or 33458 1 (16) 2 (2 692) 2

7.2 = r (0.45)

B1 4

or

(2, 2) and (0, 2)

3346

2 B1 4

x = 0, 2

B3

dy = 0 or 3x(x + 2) = 0 dx

B2

dy = 3x2 + 6x dx

B1

4

6

Number 21 (a)

Solution and marking scheme dy  24 x  36 dx dy  6m dm

(b)

22

or equivalent dm 2 dx

and

Sub Marks 2

4 B1

108

2

 y  [24(3)  36]  0 01

B1

1

3

2(1  3x ) 2  c 1 3(1  3 x) 2

1  3  2

Full Marks

3 B2

c

1

3(1  3 x) 2

or

1  3  2

3

11 15

23

3

4 1 1  5 3 4 5

B1

or

or

equivalent

1 3

B2

B1

24 (a)

6

1

(b)

120

2

6

C3  4C2

25

  1.789

3

B1 3 3

51  50  0.559 

B2

0.559

B1