Spm Addmath1 Ans (kedah)

2

SULIT

3472/1

SPM Trial Examination 2008 Kedah Darul Aman Marking Scheme Additional Mathematics Paper 1 Question 1 (a)

2

Solution/ Marking Scheme B1: p(2) − 3 = 1

x2 (a) B1: 3y −2 = 2 or g 1 ( x)  3

Answer (a) 2

Marks 2

(b) −5 4 (a) 3

1 2

(b) x 2  2

2

(a) 7

1

(b) 3

1

(c) 5

1

(b) B1: 3 f ( x)  2  3x 2  4 3

4

B2:  1(3  1)  

2k  1 2

B1 :   3   4 or  (3 )  



(a) 2

1

(b) 14

1

(c) 4

1

B2:

 52

B1: 7

3

2k  1 2

5

6

7 2

5 5 
3

x=2

3

5 2

(5 – 2p)(5 + 2p)

B2 : 3 x  3 2 or 3 x  9 B1 : 27  3 x  3 x  234 or 3 x  33  3 x  234

3472/1

Additional Mathematics Paper 1

SULIT

3

SULIT Question 8

9

Solution/ Marking Scheme

1 B2: 2 log 8 2  log 8 5 2

3472/1 Answer

Marks

(a) x  64

1

(b) x  73

1

1 2p q 2

3

(a) 2

3

(b) −1

1

(a) 8

1

(b) 4

3

6

3

p = 2, q = 5

4

(i) 2:3

1

(ii) 4

2

B1: log 8 4 5  log 8 4  log 8 5 10

(a) B2:

y2 y  y6 y2

B1: log 2 ( y  2)  log 2 ( y  6)  log 2 y  log 2 ( y  2) 11 (a) T1  S1  8 (b) B2: T2  20  8 or 12 B1 : S 2  2  22  3 or 20 12

13

n

1 1 B2:    64 2 64(1  0.5 n ) B1: 126  1  0.5

B3 : p  2  1 B2 : 2    1  p B1 : m1  2, or m2  

1 or 1  p (2)  q p

14 t 6 2 2 1  3  6 (ii) B1 :  or t  1 t 3 5

m=2

15 B1: m  2 or n  1

3472/1

2

n = −1

Additional Mathematics Paper 1

SULIT

4

SULIT Question 16

3472/1

Solution/ Marking Scheme

Answer

Marks

6 2 x+ y 5 5

3

ED  EB  BD =2y +

2 BC 5

= 2y +

2 (3x − 4y) 5

= 2y +

6 8 x− y 5 5

=

B1 : BD 

2 BC 5

B2

6 2 x+ y 5 5

OR

ED  EA  AD = −2y + CD 3 = −2y+ AC  CB 5

B1 :

3 =−2y + 3x+ (4y-3x) 5

= −2y + 3x + =

CD 

3 CB 5

B2

12 9 y− x 5 5

6 2 x+ y 5 5

17 B3: 45 0 , 75 0 58' (75.96 o ) (both)

45 0 , 225 0 , 0

B2 : tan x  1 , tan x  4 (both)

4 o

75 58' (75.96 ), 255 0 58' (255.96 o )

B1 : 3 tan 2 x  5 tan x  4(tan 2 x  1) 18 1 2 1 (7) (0.5236)  (5) 2 (0.5236) or 2 2 1 2  1 2  (7) ( )  (5) ( ) 2 6 2 6 1 2 1 2 B1: (7) (0.5236) or (5) (0.5236) or 2 2 1 2  1  (7) ( ) or (5) 2 ( ) 2 6 2 6

B2 :

3472/1

(a) 0.5236 rad 1 ( ) 6

1

(b) 6.283 cm2 (2  )

3

Additional Mathematics Paper 1

SULIT

5

SULIT Question 19

Solution/ Marking Scheme dr 1 dr B2 :   50 or 50  4 (5) 2  dt 100 dt

3472/1 Answer

Mark

0.5 cms-1

3

3

3

(a) −4.5

1

(b) 35

3

18 x (2  3 x 2 ) 4

3

2

2

(a) 924

1

(b) 917

2

48  45 6

(a) 0.3085

2

k  45  0.5 6

(b) 42

2

B1 :

dV dV  4r 2 or  100 dr dr

B2:

k k 6    3(3  4  5)  3(3  2  5) 7

20

k (3x  5) 1 B1:  1 3 21

(b) B2 : 10  p[4  2]  80

B1 :

 g ( x) dx   pdx

B2:

dy dy  3(2  3 x 2 ) 4 (6 x ) or  3u  4 (6 x ) dx dx

or 10 or px

22

B1:  3(2  3 x 2 ) 4 or  6 x or  3u 4

23 B1:

24

(b) B1: 12 C 6 5

25

1  10 5

or

C1  7 C 5  5C 2  7 C 4  5C 3  7 C 3  5C 4  7 C 2  5C 5  7 C1

(a) B1 : Z 

(b) B1 :

END OF MARKING SCHEME 3472/1

Additional Mathematics Paper 1

SULIT