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Skema Jawapan & Pemarkahan Add Maths Kertas 1&2 SPM

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									MENJAWAB SOALAN SPM
ADDITIONAL MATHEMATICS
Kertas 1 dan 2
Skema Jawapan & Pemarkahan
Siri Ceramah Di:
 (i) SMK Dato’ Klana Putra, Lenggeng, Negeri Sembilan,
 (ii) SMK Engku Husain, Semenyih, Selangor,
(iii) SMK Dengkil, Kuala Langat, Selangor,
(iv) SMK Taman Jasmin 2, Kajang, Selangor,
(v) SMK Jalan 4, Bandar Baru Bangi, Selangor,
(vi) SMK Dato’ Ahmad Razali, Amapang, Selangor.
Q.1:

                       {-
(a) -1, 1 / -1 and 1 / {-1, 1}
                            ….√(1)
       reject: 1 or -1 / (-1, 1) / [-1, 1]


    many-to-
(b) many-to-one /                     ….√(1)
    many with one /
    many → one
Q.2(a):
(a)    f(x) = 4x + 2
      f -1(x) = y
       f(y) = x
              = 4y + 2   √ -----------(1)
           x = 4y + 2

              x-2
         y = ———         √------------(2)
               4
                      Q.2(b):
                      Q.2(b):
    gf(2)
(b) gf(2) = g[f(2)]
          = g[4(2) + 2]     √………..………..(1)
          = g(10)
          = 102 – 3(10) – 1
          = 69              √…………………..(2)

    f(2) = 4(2) + 2
         = 10            √ ………………….(1)
Q.3:
Q.3:
   gh(x) = g[h(x)]
          = mx2 + n – 3 ……..(P1)
          = 3x2 + 7
    Bandingkan:
      m=3             ……...(1)
    n–3=7             ………(P1)
      n = 10          ………(2) (Kedua-duanya)
Q.4
Q.4:
   2x2 + p + 2 = 2px + x2
     x2 – 2px + p + 2 = 0             √ ………..(P1)
    a = 1, b = - 2p, c = p +2
    b2 - 4ac > 0
   (-2p)2 – 4(1)(p + 2) > 0
     4p2 – 4p – 8 > 0                √ ………..(P2)
       p2 – p – 2 > 0                               x
                                    -1       2
          Let: p  2 – p – 2 = 0

                 (p – 2)(p+ 1) = 0
                 p = 2 atau p = - 1
   Therefore: p > 2 atau p < -1 √ ..…….(3)
Q.5:
Q.5:
   (x – ⅔)(x + 4) = 0      ….…….(P1)
    x2 – ⅔ x – 4x – 8/3 = 0
    3x2 + 10x – 8 = 0       …………(2)

    OR:
    x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(P1)
    3x2 + 10x – 8 = 0           …………(2)
Q.6:
Q.6:

 (a) p = 3 (axis of symmetry) √...(1)
 (b) x = 3                    √...(1)
 (c) (3, -1)                   √...(1)

   Note: Minimum value = -1
    Q.7:
   32x + 1 = 4x
   (2x + 1)log10 3 = xlog10 4          √ …..…….(P1)
   2xlog10 3 + log10 3 = xlog10 4

   2x(0.4771) – x(0.6021) = - 0.4771   √.………..(P2)
   0.9542x – 0.6021x = - 0.4771
   0.3521x = - 0.4771
                   - 0.4771
          x = ————
                0.3521

             = - 1.355                 √ ………...(3)
Q.8:
   2log2M = 2 + 4log4N
                  4log2N
    2log2M = 2 + ———          √……….(P1)
                   log24

    log2M2 = log24 + 2log2N   √..……..(P2)

    log2M2 = log24N2          √..….…(P3)
       M2 = 4N2
       M = √4N2
       M = 2N                 √..…....(4)
Q.9:
 (y – 1) – x = (5 + 2x) – (y – 1) √....(P1)
 y – 1 – x = 5 + 2x – y + 1
         2y = 3x + 7             √….(P2)
               3x + 7
          y = ———                √….(3)
                 2
Q.10(a):
   a = 108

    ar2 = 12

    ar2   =   12    √ ……….(P1)
     a        108


     r2 = ¹/9
     r = ¹/3        √..……….(2)
Q.10(b):
        108
S∞    = ———       √…….….(P1)
        1 – ¹/3

       108
  = ———
       ⅔

  =   162         √…..…….(2)
Q.11:

               x
   y =     ————
             a + bx
                 a + bx
    ¹/y = ————
                     x

    ¹/y =   a/       +b   √ ……….(P1)
                 x

     a = gradient
       = - 6/4 = - 3/2    √ ……….(2)

     b = y-intercept
       = 6                √……….(1)
Q.12:

 x/y - y/4 = 1
  4x - 2y =        8        √..……(P1)
         2y =      4x - 8
            y =    4/ x – 4
                     2
  gradient, m =    2        √..……(P2)
        y–7=       2(x – 2)
             y =   2x + 3 √.……(3)
Q.13:
              4m + 9s                        2m + 6t        √…. (P1)
    S   =                      or      t=
                       5                          5
            P(2m, m)                Q(s, t)           R(3s, 2t)

                           3                  2


        5s – 9s = 4m,           or 5t – 6t = 2m
            - 4s = 4m,                 - t = 2m            √……(P2)
              - s = m,
            - 2s = - t
                s = t/2             √…...…………………...(3)
                               Q.14:
                               Q.14:
   PA : PB = 2 : 1                       √..…(P1)
    reject:   PA/      = 2/1
                    PB
   PA = 2 √(x – 5)2 + (y – 0) 2 √…..(P1) or
    PB = √(x + 2)2 + (y + 3)2               √….(P1)
   PA2 = 4[(x – 5)2 + y 2 ] =
    PB2 = (x + 2)2 + (y + 3)2                √....(P2) or
   4[(x – 5)2 + y 2 ] = (x + 2)2 + (y + 3)2 √…..(P2)
   3x2 + 3y2 - 44x + 6y + 87 = 0 √....(3)
Q.15
          D                  C
                                 AC = AB + BC
                                    = AB - CB                 .........(P1)
A                 B                 = (i + 2j) – (-5i – 6j)

                                    = 6i + 8j                 ….…(2)

                  6i + 8j
Unit vector AC = ————                       ……..….(P1)
                 √ 62 + 82

                 6i + 8j 3i + 4j
              = ———— or ————                …..……(2)
                   10       5
Q.16
                        PQ = PO + OQ
        P(1, 2)
                            = - OP + OQ
 O                          = - (i + j) + (8i + 4j)
                            = 7i + 3j               ….(1)
            Q(8, -4)

7i + 3j = ma + nb
        = m(i + j) + n(2i – 3j)
        = (m + 2n)i + (m – 3n)j   ……………….…..(P1)

Compare:   m + 2n = 7              5n = 4        .….(P1)
           m – 3n = 3               n = 4/5 )..(2 either)
                                    m = 27/5 )..(3)(both)
                           Q.17:
                           Q.17:
   sec θ = 1/cos θ
                                         1
                                             1 – t2

         = 1/t   ………(1)
                                         t




    cos (90 – θ) = sin θ       ……….(1)

                  = √ 1 – t ……….(2)
                           2
                                                     Q.18:
                                                     Q.18:
   2 sec2θ – 3tanθ = 4
               3tanθ
   2(1 + tan2θ) – 3tanθ = 4
                   3tanθ                     …..…..(1)
     2 + 2 tan2θ – 3tanθ = 4
                   3tanθ
     2tan2θ - 3tanθ – 2 = 0
              3tanθ
   (2tanθ 1)(tanθ
    (2tanθ + 1)(tanθ – 2) = 0                ………..(2)
   tanθ = -½ atau tanθ = 2
    tanθ              tanθ
    θ = (360o – 26o 34’), (180o - 26o 34’)
   θ = 333o 26’, 153o 26’                   ………...(3)
    θ = 63o 26’, (180o + 63o 26’)
   θ = 630 26’, 2430 26’                 ….....….(3)
   θ = 630 26’, 153o 26’, 2430 26’, 333o 26’. …….….(4)
Q.19:
   ̸̱ BOC = π – 1
            = 3.142 – 1
            = 2.142       ………..(P1)

        s = jθ
       BC = 15 x 2.142    ………..(P2)
          = 32.13         ………..(3)
Q.20:
Q.20:
                           2
 dy     (x – 4)(4x) – (2x + 3)(1)
 — = ———————————                     ……(P2)
 dx             (x - 4)2
                               4x   …….(P1)

          2
        2x – 16x - 3
      = ———————                     ..……(3)
              (x - 4)2
Q.21:
Q.21:
   dy/dx = 4x – 3                 ………..(1)
      δx = 0.01                   ………..(1)
       x = 2


     δy∕      ≈ dy∕dx
           δx

          δy ≈ dy∕dx (δx)

             = [4(2) – 3](0.01)   …………(2)
             = 5(0.01)
             = 0.05               …………(3)
Q.22:
     3            3
   ∫ f(x) + ∫ (kx)dx = 20
    1            1
                            3
        8 + [ kx2 ∕ 2 ]        = 20    …..(P1)
                            1

                            3
             [ kx2 ∕ 2 ]       = 12 …..(P2)
                            1

              9k/2 – k/2 = 12,         8k/2 = 12   …..(P3)
                                         k = 3     …..(4)
Q.23:
Q.23:

    6                      4
        P           or         P          ….(1)
            4                      3


6               4
    P x P = (6 x 5 x 4 x 3) x (4 x 3 x 2) ..(P2)
                              (4
        4           3


                         = 360 x 24
                         = 86400       ……...(3
                                       ……...(3)
Q.24:
Q.24:
           9               5
   (a)        C       x       C                                           ……….(P1)
                   6               4

          atau 9 x 8 x 7
                 3x2
     = 84                                                                 …………(2)

           5           9               5       9           5       9
   (b)        C x C               +   C x C           +   C x C           …...….(P1)
                5              5           4       6           3       7

          = 948                                                            …………(2)
Q.25:

    0.5 – 0.225   ……….(P1)

=   0.275         ……….(2)
                        PAPER 2
Q1:
                            x+4
      x = 3y – 4   or   y = ———        ………………. √(P1)
                             3

      Eliminate x or y:
      (3y – 4)2 + y(3y – 4) – 40 = 0   …………………. √(M1)
        3y2 -7y – 6 = 0
      (3y + 2) (y – 3)= 0 or using quadratic equation …. √(M1)
                           or completing the square method.
       y = - 2/3 , y = 3.     ………………………………√(A1)

      x = - 6, x = 5      ………………………………√(A1)
 Q3:
(a)   r = 1.05                     √….(P1)
      Use T6 = ar 5
      T6 = 18000 x (1.05)5
                    (1 05)        √….(M1)
         =   RM 22 973.00
                                   √….(A1)

(b) Use ar n-1 ≥ 36 000            √….(M1)
    18000 x (1.05)n-1 ≥ 36 000
            (1 05)
                   n ≥ 16          √….(A1)

      No working minus 1 mark if answer is
      correct
                   n
                a(r – 1)
(c)   Use   S6 = ———
                 r-1

                                  6
      Use   S6=18000{(1.05) – 1}..√(M1)
                       1.05 – 1

              = RM 122 434            √..(A1)
Q8:
(a)
         x    1.5 3.0 4.5 6.0 7.5 9.0
      log10 y 0.40 0.51 0.64 0.76 0.89 1.00

                                   √.... (P1)
        Plot graph log y against x2
  log y
       √ P1
1.0                                                   +
                                              +       √ G1
0.8                                   +
                      √ P1
                          +
0.6
                  +
              +
0.4

0.2

0.0                                                          x
   0      1   2   3    4      5   6       7       8   9
(b) Plot log10 y against x
   (correct axes and uniform scales)…..√(P1)
    6 points plotted correctly …………..√(P1)
    Lines of best fit               …………..√(G1)

(c) log10 y = log10 h + 2x log10 k …………….√(P1)
          c = log10 h            .................... √(M1)
          h = 1.78                   ………….√(A1)
         m = 2 log10 k               ………….√(M1)
          k = 1.09 – 1.12            ………….√(A1)
Q13:
                  P1
(a)   Use   I = —— x 100
                   P0
                  1.80
            h   = —— x 100 ………………..√(M1)
                  1.50

                = 120           ………………….√(A1)

                  0.90
      112.5 =
      112.        —— x 100 ……………… √(M1)*
                    k

            k = 0.8 or 80 sen   ……...……………√(A1)
    I      W         IW
    150     30       4500
    120     45       5400
    112.5
    112.    15       1687.5
                     1687.
    105     10       1050
            100     12637.5
                    12637.    ……………√(P1)
                              ……………

                  12637.5
                  12637.

            I = —— x 100      …………√(M1)
                   100


              =      126.38   …………√(A1)
(c) (i)             150
          126.38 = —— x 100 ……√(M1)
                   100

                 189.
               = 189.6     ……√(A1)

          P1
(c) (ii) —— x 100 = 189.6 ……√(M1)
         25

                      47.
                    = 47.39 ……√(A1)
   MENJAWAB SOALAN
ADDITIONAL MATHEMATICS

         SPM
      PAPER 1 & 2


   THE END

								
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