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```									                                  JABATAN PELAJARAN PERAK
DARUL RIDZUAN
SOLAF 2011 / 1
PERATURAN PEMARKAHAN

PAPER      1
No.                             Important steps / Answers   Marks

1     (a)   p= 4                                              1
q=–1                                              1
(b)   many to one                                       1

2     (a)   1 – h(4) = – 1                                    1
1
h=
2                                     1
(b) – 3                                                 1

3     (a)   5                                                 1
(b)   f(x) – 7 = 2x – 1                                 1
2                                                 1

4            2  h                                           1
(a)            1
2
4                                                1
(b)    –2≤ x ≤ 4                                        1

5     82 – 4(p)(6) > 0 or 62 – 4(3)( – p + 1 ) > 0            1
8
p <
3                                                   1
p > –2                                                  1
8
–2< p <
3                                         1

6     (a)   h=–1                                              1
k=7                                               1
(b)   2                                                 1
No.                                              Important steps / Answers          Marks

7                   1
x            2
2 = N                                                                           1
3
N    2
1

8     4x = 3                                                                          1
lg 3
x=
lg 4                                                                         1
0.7925                                                                          1

9     log8    2 log b    log 3
         
log b    log 4    log 27                                                        1
3
2             log 3
log 2 3                                                                       1
2        log 3
log 2
9                                                                               1

10             4x  1
log5             1
7  x                                                                  1
4x  1
 5
7  x                                                                           1
x = 4                                                                           1

11    (a)      a=–3                                                                   1
b = 18                                                                1
(b)      – 3 + ( n – 1 ) ( 7 ) = 46                                             1
8                                                                      1

12              ar 4     1
(a)                                                                           1
ar      27
1
                                                                     1
3

(b)      9                                                                      1

13             11                                     4
(a)           [ 2( – 7 ) + 10d ] –                  [ 2( – 7 ) + 3d ] = 441
2                                      2                               1
10                                                                     1
(b)      – 7 + 5( 10 )                                                          1
43                                                                     1
No.                                    Important steps / Answers   Marks

14    m2 = 9                                                         1
m=–3                                                           1
3
mn = 
2                                              1
1
n=
2                                                        1

15    lg y = 1 + 3 lg x                                              1
k = 1 + 3( 2 )                                                 1
7                                                              1

16                   1 3  q
(a)        (     ,     )
2   2                                           1

1        3  q
(b)        6(     )  4(       )  19
2          2                                   1

5                                                   1

17    (a)       h=6                                                  1

(b)       P( 0 , 3 )                                           1
6 6  15
( ,       )
3   3                                               1
(2,7)                                                1

18    10( 2.5 ) or 5 (π )                                            1
10( 2.5 ) + 5 (π ) + 10                                        1
5π + 35                                                        1

19    1 2                     1 2 
(8 )( )           or     ( 6 )( )
2      2                 2       3                             1

1 2                      1 2 
(8 )( )            –      ( 6 )( )
2      2                  2       3                            1

10π                                                            1

20     0 f ( x) dx  4
2                                                            1
1
2( 7 ) – 4                                                     1
10
No.                          Important steps / Answers   Marks

21      3                                 5
[ x 2  2 x ] 21   1 f ( x) dx 

2
1
2                                 2
3                          5
6  4  (  2 )   1 f ( x) dx 
2
1
2                          2
8
1

22    dA
 2r                                            1
dr
1
 3                                             1
2r
1
1
3

23    3x2 – 12 = 0                                         1
x=–2                                                 1
( – 2 , 24 )                                         1

24    (a)   2 + 7 + 4 + 11 + 5 + n = 6(6)                  1
7                                              1

(b)   6                                              1

25    ∑ x = 28 – 8                                         1
∑x 2 = 170 – 82                                      1
106     20
 ( )2                                        1
4       4
1.225
1

END OF MARKING SCHEME

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