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					                                  JABATAN PELAJARAN PERAK
                                       DARUL RIDZUAN
                                       SOLAF 2011 / 1
                                 ADDITIONAL MATHEMATICS
                                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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posted:10/4/2011
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