Jawapan Mt Kertas 1 Pahang 2008

JPNP PAHANG

SULIT 3472/1 Additional Mathematics Kertas 1 Peraturan Pemarkahan September 2008

PEPERIKSAAN PERCUBAAN SPM TAHUN 2008

ADDITIONAL MATHEMATICS KERTAS 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

1

JPNP PAHANG

Question 1 (a)

Working / Solution q=5

Marks 1

Total

2

1 (b)

one to one or 1 to 1

1

2 (a)

3+ x or equivalent 4

2

3−

2(b)

3− x 2 2

B1

2

3 - 2x

B1

x = 3 - 2y or equivalent 3

p<

4

q +1 4(1 − q)

4p <

3

(q + 1) 2 or equivalent (gathering of q terms) 1− q2

(q + 1) 2 − 4 p(1 − q 2 ) 4

B1

−

1 1 ≤ x ≤ or − 0.3333 ≤ x ≤ 0.5 3 2

−

1 1 <x< 3 2

OR

B2

−

1 3

3

4 1 2

x

B3

Must indicate the range correctly by shading or other ways 1 1 and x = 3 2 (both values must be seen) x=−

Accept ‘=’ or any inequality signs ‘>’, ‘<’, ‘ ≤ ’, ‘ ≥ ’

(3x + 1)(2x – 1)

2

B2

B1

4

JPNP PAHANG

5

1

3

log10

x4 y6 x3 y 6 = 4 or log + log10 x = 4 or 10 y2 y2

B2

x =4 y2 or equivalent (combining at least two terms of log correctly in a correct equation) log10 x 4 y 6 − log10 y 2 = 4 or log10 x 3 y 6 + log10

log10 x 3 y 6 or log10 y 2 or log10 10 4 6

3

4

17.65 or 17.63 x=

B1

log10 25 2 log10 5 or x = or equivalent log10 1.2 log10 6 − log10 5

B3

x

7

8 (a)

8(b)

⎛6⎞ log10 ⎜ ⎟ = log10 25 or log10 6 x = log10 5 x + 2 or ⎝5⎠ log10 (3 x × 2 x ) = ( x + 2) log10 5 or equivalent

B2

6 x or 5 x (5 2 ) or log10 (3 x × 2 x ) = log10 5 x + 2

B1 3

5(13p + 11q) or equivalent 10 [2(2 p + q) + (10 − 1)( p + q)] 2

B2

3p + 2q – (2p + q) or 4p + 3q – (3p + 2q)

B1

3

2

13122 a = 2 or r = 3 or seen

B1

2 or equivalent 99

2

1

4

B1

a = 0.02 and r = 0.01 or seen

3

4

JPNP PAHANG

9

10

11

(10, 7)

3

4(0) + 1(h) 4(2) + 1(k ) = 2 or =3 1+ 4 1+ 4

B2

P(0 , 2)

B1

-6

3

mPQ = 2

B2

P(4, 2)

B1

⎛ 4⎞ ⎜⎜ ⎟⎟ ⎝5⎠

B1

B2

4i + 3j + 11i + 5j −

B1

3

4

1 x + 4y 2

2 1 (6y – 3x) + (3x) or equivalent 3 2

B3

2 AS = (6y – 3x) 3

B2

(6y – 3x) or

2

3

1 (15i + 8j) or equivalent 17 15 2 + 8 2

13

3

2

⎛ 3⎞ ⎛ 2 ⎞ 2⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ ⎝ 1⎠ ⎝ − 3⎠ 12

3

1 2 (3x) or AS = QS 2 3

4

B1

4

JPNP PAHANG

14

108

3

3(−12)(5 − 3 x) 2 (−3) or equivalent

B2

4(5 − 3x) 3 (−3) or equivalent 15

16

B1

10

3

3

2 ⎞ dp ⎛ = ⎜3 − 2 ⎟ × 4 dt ⎝ 2 ⎠

B2

dp 2 = 3 − 2 or equivalent dq q

B1

y = 10x2

3

4

log10 y = log10 x 2 + log10 10 or log10

y =1 x2

B3

or equivalent B2

log10 y = 2 log10 x + 1 mPQ = 2 17

B1

3

4

3 3

⎡ ⎤ k 3 + [x ]2 = 7 or equivalent ⎢ 3⎥ ⎣ (2 x − 5) ⎦ 2

∫

3

2

18

3

g ( x)dx + ∫ dx = 7

B1

3

2

1 1 y = − (3 − 2 x) 4 + 2 or equivalent 8 8 2=−

B2

(3 − 2 × 1) 4 + c or equivalent 8

(3 − 2 x) 4 or equivalent − 2( 4)

3

B2

B1

5

3

JPNP PAHANG 19

20

3

⎛3⎞ ⎛3⎞ 4 + 4⎜ ⎟ + 4 + 4⎜ ⎟ or equivalent ⎝2⎠ ⎝2⎠ 12 = 20

21

1 2 (4 )θ 2

B1

B2

Any one of a , b or c correct

B1

90o, 123.69o (123o 41’), 270o, 303.69o (303o 41’)

4

3 2

B3

cos x(3 cos x + 2 sin x) = 0

3

B2

3 cos 2 x + 2 sin x cos x = 0

23

3

3

3 ,c=0 2 Any two of a , b or c correct

a=2,b=

cos x = 0 and tan x = −

22

B2

B1

4

3

277200 12

C 4 ×8 C 3 × 5 C 2 or 495 × 56 × 10 or equivalent

B2

12

C 4 or 8 C 3 or 5 C 2 or 495 or 56 or 10 or equivalent

B1

1 or an equivalent single fraction 15

3

3 2 2 × × 6 5 6

B2

3 2 2 or or 6 5 6

B1

6

3

3

JPNP PAHANG

14 or 2.8 5

1

24(a) 24(b)

1.296

2

2 ⎛ 2⎞ 7 × × ⎜1 − ⎟ or equivalent 5 ⎝ 5⎠

B1

25 (a)

1.1

1

25(b)

61.2

2

70 − µ = *1.1 (his k) 8

B1

7

3

3