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My Possible Answer - Add Maths Paper 1 - SPM 2011

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My Possible Answer - Add Maths Paper 1 - SPM 2011 Powered By Docstoc
					  My Possible Answer & Marking
          (Q13 to Q25)
(It has nothing to do with any official answer and marking scheme by the concern authority)




                Additional Mathematics
                        Paper 1
                       SPM 2011
13. (a) Gradient = − 3 √ 1
                             √M1
    (b) m┴ = ⅓
        Eqn: y – 3 = ⅓ (x – 1)
              3y = x + 8 √ 2
14. sin 2θ = cos θ
    2 sinθ cosθ = cosθ    √M1     √M2 Each basic angle
    cosθ (2 sinθ – 1) = 0         √M3 Both basic angles
    cosθ = 0 √M1  θ = 90o, 270o
    sinθ = ½       θ = 30o, 150o
              √M1
                   θ = 30o, 90o, 150o, 270o √ 4
                                                         - 24
15. (a) cot θ = 4              5
                                                     ┌
                                                            B
                3    √1                     3   -7
                                A       ┌                  25
                                    4

    (b) sin (A + B) = sinA cosB + cos AsinB
                √M1 = (3/5)(- 24/25) + (4/5)(-7/25)
                    = - 4/5 √ 2

                                                 E
16. DF = − OD + DE √M1                      F                   D
       = −(3i + 2j) + (−5i + 3j) √M2                      O
       = −8i + j √ 3                                 DE = OF
                     8       h
17. (a) r + s =   ( -2 ) + ( 7 )
              =        +
                  ( 8 5 h) √ 1

    (b) | r + s | = 13
        √ (8 + h)2 + 52 = 13 √M1
           (8 + h)2 + 25 = 169
                   8 + h = √144
                       h = 4 √ 2 (h = positive value)
18. (a) PR = 6 √M1
        tan θ = 6/8 = 0.75
            θ = 0.6436 rad. √ 2
   (b) PQ = 10(0.6436) = 6.436 √M1
       Perimeter shaded region
       = 6 + 2 + 6.436
       = 14.44 √ 2
          5x ,     dy
19.   y= 2              = g(x)  ∫ g(x)dx = y
         x +1      dx
      3                     5x            3
      ∫ 2g(x)dx = 2y = 2[             ]       √M1
      0                     x2   +1       0


                          5(3)
                     = 2[      − 0 ] √M2
                          9+1

                     = 3 √3
                  12
20. y = 10 –
                   x
    dy       12                 10x – xy = 12
         =        √M1           x(10 – y) = 12
    dx       x2                 y = 4 then x = 2
             12
         =        =   3   √M1
             22
    ∂y = (4 + p ) – 4 = p √M1
    ∂x   dx            dx ( ∂y )
       =       ∂x =
    ∂y dy              dy
                                  p
                  √M2 = 1 ( p ) =      √3
                        3         3
          a
21.       ∫ (x + 1) dx
          4


            x2      a
      =   [    + x]        √M1
            2       4


            a2        42
      =   [    + a]–[    + 4 ] √M2
            2         2

           a2
      =    2
                + a − 12    √3
22. ∑ x = 240 , ∑x2 = 9654

            240
     (a) x = 6 = 40 √ 1

                 ∑x2
     (b) σ = √         −   x2
                  N

              9654
           = √ 6 − 402 √M1
           =3     √2
23. (a)   7P    = 2520     √2
            5
          √M1

            √M1
    (b) 2 x 5 x 5 P3 = 600           √2



          OR       2   5    4    3        5   √M1
                  = 600
                            √2
24. (a) n(S) = 20, n(M) = 6

              6    3
       P(M) =    =    √1
              20   10

   (b) P(M and N) = P(M ∩ N)
                      n(M ∩ N) 2
                    =         = √M1
                       n(S)    20
                      1
                    = 10 √ 2
25. (a) P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3)
                 = (12/27) + (6/27) + (1/27) √M1
                 = 19/27 √ 2

    (b) m = P(X = 0) = 1 – P(X ≥ 1)
            = 1 – 19/27
            = 8/27 √ 1
the end

				
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