Skema jawapan Kertas 1 Matematik Tambahan SPM Number Solution by act50979

VIEWS: 2,591 PAGES: 4

									                               Skema jawapan Kertas 1 Matematik Tambahan SPM

                                                                                Sub   Fu
Number                                  Solution and Marking Scheme
                                                                               Marks Mar
  1      7                                                                       2
          2(1)  5                                                             B1     2

  2      g ( x)  2 x  1                                                        3
         g ( y )  2( x  3)  5                                                B2
                                                                                      3
         y  x3                                                                B1
  3             3               1
         m  and k                                                            3
                4               4
                3             1                                                 B2
         m  or k  
                4             4
                     3 x
         f 1 ( x )                                                           B1    3
                     4 4
 4 (a)   2x  x  1  0
            2                                                                    2
                                                                                B1
         2( x 2  2 x  1)  5 x  3

  (b)    h9                                                                     2
         2(4)  h(4)  4  0
              2                                                                 B1    4
  5      p 1                                                                    3
          4 p  4                                                             B2
                                                                                      3
         (2) 2  4(1)( p )  0                                                 B1
 6 (a)   k=2                                                                    1
   (b)   x=2                                                                    1
   (c)   (2, 3 )                                                               1     3

          1                                                                     3
  7      pq
                 1
                                                                                B2
         log a 5  log a 7
            1
         log a 35                                                               B1    3

  9      x2 + y2 – 4x – 14y + 37 = 0.                                           3
         (x – 2)2 + ( y – 7)2 = 42                                              B2
         or equivalent x2 – 4x + 4 + y2 – 14y + 49 = 16

         AP = 4     or       ( x  2) 2  ( y  7) 2  4                        B1
                                                                                      3

                                    www.cikgurohaiza.com
                                                                           Sub   Fu
Number                                      Solution and Marking Scheme
                                                                          Marks Mar
  10           3                                                            3
         y=      x4
               4
                                                           3
         Gradient of line perpendicular to AB, m =                         B2
                                                           4
                                        4
         Gradient of AB:                                                  B1    3
                                        3
  11     244                                                               3

         2 3
                  x         2

                                  72
                                                                           B2
                         4
                                                                           B1
         x= 7                                                                    3
  12
            15
         r    cm                                                          3
            
             10
         r
            2                                                              B2
              
            3
                    4    2
         AOB  2   OR                                                B1
                                                                                 3
                    3    3
  13      3x  1                                                           3
           2x 1
                          x
           2x 1                                                          B2
                         2x 1
                                       1
                       1             
           2 x  1  x( )(2)(2 x  1) 2                                    B1    3
                       2
  14            1                                                          3
         P(2, )
                2
         2(2 x  5)  2 or x  2                                          B2
         dy
              2(2 x  5)                                                  B1    3
         dx
  15          27                                                           2
         y 2
              x
                     3
               3    93                  3
         y  x   or    or a  x and r                                 B1    2
               x    x x                 x




                                        www.cikgurohaiza.com
                                                                      Sub   Fu
Number                                 Solution and Marking Scheme
                                                                     Marks Mar
  16     17                                                            3
         x6                                                          B2
                                                                             3
         (4 x  1)  (3 x  1)  (6 x  3)  (4 x  1)                B1
   17    47                                                           3
         99
           0.47
                                                                      B2
         1  0.01
         0.47  0.0047  0.000047  ...                               B1    3
  18     k  8, h  1                                                  3
         k  1 6   2 or 3  1 h   2                             B2
         y                                                                  3
             2 x                                                    B1
         x2
19 (a)   5                                                           1

   (b)   12.5                                                        3
                      4
              3 
         10   x 2                                                  B2
               2 1
              4           4
         2  g ( x)dx   3xdx                                        B1    4
           1              1
  20       10i   13 j                                                 3
               
           269    269

         3a  b  10 2  13 2  269                                   B2

         10 
          
         13                                                         B1    3
          
  21       
                 3                                                 3
         PY = b  a
                    2
           
                    1
         PY   a  b  ( a )                                        B2
                       2
           
                         1
         PQ   a  b or QY   a                                     B1    3
                                2

 22      210 , 330                                                  4
                     1
         sin x =  , sin x =  2 ( both)                              B3
                     2
         (2sin x  1)(sin x  2)  0
                                                                      B2
                                 www.cikgurohaiza.com
                                                                     Sub   Fu
Number                                Solution and Marking Scheme
                                                                    Marks Mar

         cos 2 x  1  2sin 2 x                                      B1    4


23 (a)   924                                                         1

   (b)   112                                                         3
         5
           C4 7 C2  5C5 7 C1
                                                                     B2
         5
             C4 7 C2 or 5C5 7 C1
                                                                     B1    4
24 (a)                                                                2
         13
         25                                                          B1
         3 3 2 2
            
         5 5 5 5
                                                                     2
          24
  (b)                                                                B1    4
         125
                    2
            2 3
         2   
           5 5
25 (a)   1.667                                                       2
            54  50
         Z                                                          B1
               2.4


  (b)    0.9505                                                      2
         1  0.00177  0.04776
             43  50      54  50
         P(          Z          )
               2.4          2.4                                      B1    4




                                  www.cikgurohaiza.com

								
To top