Add Maths - Paper 1 - Answers & Marking

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					MENJAWAB SOALAN
      SPM
   ADDITIONAL
  MATHEMATICS
  SMK TINGGI KAJANG
      5 OGOS 2011
  SMK CHERAS JAYA
      11 OGOS 2011
FORMAT KERTAS SOALAN
PAPER 1
   25 soalan – jawab semua soalan.
   80 markah.
   Markah setiap soalan antara 2 hingga 4.
   Masa 2 jam (120 minit).
   Purata masa menjawab 1 soalan = 4.8 minit.
   Condong ke arah format objektif.
   Menguji tahap pengetahuan dan kefahaman.
   Soalan-soalan tumpu kepada konsep-konsep
    asas tajuk-tajuk yang diuji.
      Pemarkahan Kertas 1
1. Diagram 1 shows the relation between set P
   and Q.
   State
   (a) object of 5,
   (b) type of relation.            [2 marks]



                    Answer: (a) …………….(1)
                            (b) …………….(1)
6. Diagram 2 shows the graph of function
   y = (x – p) – 1, where p is constant.
   Find,
   (a) the value of p,
   (b) the equation of axis of symmetry,
   (c) the minimum point.                [3 marks]


                      Answer: (a) …………….(1)
                              (b) …………….(1)
                              (c) …………….(1)
17. Given cos θ = t, 00 < θ < 900,
    express in terms of t.
    (a) sec θ,
   (b) cos (900 – θ)           [3 marks]



                 Answer: (a) …………….(1)
                         (b) …………….(2)
11. Diagram 3 shows a straight line graph
    of 1/y against 1/x .
    Variables x and y ere related by equation
    y = …………..
    Find the values of a and b.       [3 marks]




                 Answer: (a) a = …………….(2)
                         (b) b = …………….(1)
16. Given P(1, 2), Q(8, - 4), a = i + j and
    b = 2i – 3j. If PQ = ma + nb, with m and
    n as constant.
    Find the values of m and n. [3 marks]




             Answer: m = ………..} (3
                     n = ………..} (both)
9. Given the first three terms of an
   arithmetic progresssion are x, y – 1
   and 5 + 2x. Express y in terms of x.
                                  [3 marks]




                      Answer: ............. (3)
10. The first term and the third term of a
    geometric progression are 108 and 12
    respectively. Find
    (a) the common ration,
    (b) the sum of infinity of the
        progression.               [4 marks]


                Answer: (a) …………….(2)
                        (b) …………….(2)
2. Given f(x) = 4x + 2 and g(x) = x2 – 3x – 1,
   find
   (a) f -1(x),
   (b) gf(2)                            [4 marks]




                      Answer: (a) …………….(1)
                              (b) …………….(3)
8. Given 2 log2M = 2 + 4 log4N, express
   M in terms of N.           [4 marks]




                    Answer: ……….. (4)
PAPER 2
   100 markah.
   3 Sections, A, B dan C.
   Masa 2 jam 30 minit (150 minit).
   Menguji tahap pengetahuan, kefahaman
     dan aplikasi.
   Langkah kerja matematik yang sistematik
    perlu ditunjukkan.
   Markah diberi kepada jawapan dengan
    langkah kerja matematik yang betul.**
     Markah diberi untuk jawapan yang betul sahaja



     Markah diberi untuk kerja yang betul mengikut
     kepada kesilapan sebelumnya


K1   Markah diberi untuk tahap pengetahuan



M1   Markah diberi untuk kaedah kerja



A1   Markah diberi untuk nilai
1.   2.   3.   4.   5.
K1   K1        K1   M1




A1   M1
          M1   A1   A1




A1   A1        A1   A1
SECTION A:
 40 markah.
 6 soalan – jawab SEMUA soalan.
 Setiap soalan antara 4 hingga 7
  markah.
 Purata masa setiap soalan
  = 0.4 x 150 / 6 = 10 minit.
SECTION B:
 40 markah.
 5 soalan tetapi jawab mana-mana 4
  soalan.
 Boleh jawab semua 5 soalan, markah
  akan dikira dari 4 soalan dengan
  markah tertinggi.
 Setiap soalan 10 markah.
 Purata masa setiap soalan
  = 0.4 x 150 / 4 = 15 minit.
SECTION C:
 20 markah.
 4 soalan. Jawab mana-mana 2 soalan.
 Boleh jawab semua 4 soalan tetapi markah
  dikira dari 2 soalan dengan markah tertinggi.
 Setiap soalan 10 markah.
 Purata masa setiap soalan
  = 0.2 x 150 / 2 = 15 minit.
 Soalan digubal hanya dari 4 tajuk yang telah
  ditetapkan, 2 tajuk dari tajuk Form 4 dan 2
  lagi dari tajuk Form 5.
PAPER 1
Tajuk-Tajuk TIADA Dalam Paper 1

    Form 4:
   Simultaneous Equation
   Solution of Triangles
   Index Number

 Form 5:
 Motion in Straight Line
 Linear Programming
PAPER 2
Section A: (40 marks)       Section B: (40 Marks)
Form 4:                     Form 4:
* Functions                  * Coordinate Geometry
* Quadratic Functions        * Circular Measures
* Simultanous Equations      * Differentiation
* Coordinate Geometry
* Statistics
* Circular Measures
* Differentiation

Form 5:                     Form 5:
* Progression                * Linear Law
* Integration               * Integration
* Vectors                   * Vectors
* Trigonometric Functions   * Probability Distribution
PAPER 2
Section C: (20 Marks)

Form 4:
• Solution of Triangles
• Index Numbers


Form 5:
• Motion in Straight Line
• Linear Programming


Tajuk-Tajuk TIADA dalam PAPER 2:
• Quadratic Equations
• Indices & Logarithms
• Permutations & Combinations
• Probabilty
        CONTOH
JAWAPAN DAN PEMARKAHAN

     Additional Mathematics
              SPM


      KERTAS 1
Q.1:

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


(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   √ -----------(Z1)
           x = 4y + 2

              x-2
         y = ———         √------------(2)
               4
                      Q.2(b):
(b) gf(2) = g[f(2)]
         = g[4(2) + 2]     √………..………..(Z1)
         = g(10)
         = 102 – 3(10) – 1
         = 69              √…………………..(2)

    f(2) = 4(2) + 2
         = 10             √ ………………….(Z1)
Q.3:
   gh(x) = g[h(x)]
          = mx2 + n – 3 ……..(Z1)
          = 3x2 + 7
    Bandingkan:
       m=3            ……...(Z2) - either one
    n–3=7             ………(Z2)
        n = 10        ………(3) (both m and n correct)
Q.4:
   2x2 + p + 2 = 2px + x2
    x2 – 2px + p + 2 = 0           √ ………..(Z1)
    a = 1, b = - 2p, c = p +2
    b2 - 4ac > 0
   (-2p)2 – 4(1)(p + 2) > 0
     4p2 – 4p – 8 > 0              √ ………..(Z2)
      p2 – p – 2 > 0
          Let: p2 – p – 2 = 0                       x
                (p – 2)(p+ 1) = 0     -1         2


                p = 2 atau p = - 1

   Therefore: p > 2 atau p < -1 √ ..…….(3)
Q.5:
   (x – ⅔)(x + 4) = 0      ….…….(Z1)
    x2 – ⅔ x – 4x – 8/3 = 0
    3x2 + 10x – 8 = 0       …………(2)

    OR:
    x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(Z1)
    3x2 + 10x – 8 = 0           …………(2)
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       √ …..…….(Z1)
   2xlog10 3 + log10 3 = xlog10 4

   2x(0.4771) – x(0.6021) = - 0.4771 √.………..(Z2)
   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 + ———          √……….(Z1)
                   log24

    log2M2 = log24 + 2log2N   √..……..(Z2)

    log2M2 = log24N2          √..….…(Z3)
       M2 = 4N2
       M = √4N2
       M = 2N                 √..…....(4)
Q.9:
 (y – 1) – x = (5 + 2x) – (y – 1) √....(Z1)
 y – 1 – x = 5 + 2x – y + 1
         2y = 3x + 7             √….(Z2)
               3x + 7
          y = ———                √….(3)
                 2
Q.10(a):

    a = 108

    ar2 = 12

    ar2 = 12    √ ……….(Z1)
    a     108


    r2 = ¹/9

     r = ¹/3    √..……….(2)
Q.10(b):
     108
S∞ = ——— √…….….(Z1)
     1 – ¹/3

      108
  = ———
      ⅔

  =   162   √…..…….(2)
Q.11:

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

    ¹/y =   a/       +b   √ ……….(Z1)
                 x

     a = gradient
       = - 6/4 = - 3/2    √ ……...(Z2) – either a or b correct

     b = y-intercept
       = 6                √……….(3) – both a and b are correct
Q.12:

 y/2 - y/4 = 1
  4x - 2y =        8      √..……(Z1)
         2y =      4x - 8
            y =    4/ x – 4
                     2
  gradient, m =    2      √..……(Z2)
        y–7=       2(x – 2)
             y =   2x + 3 √.……(3)
Q.13:
              4m + 9s                        2m + 6t √…. (Z1)
    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 √……(Z2)
              - s = m,
            - 2s = - t
                s = t/2             √……………………...(3)
                                Q.14:
   PA : PB = 2 : 1                       √..…(Z1)
    reject:   PA/      = 2 /1
                    PB
   PA = 2 √(x – 5)2 + (y – 0) 2 √…..(Z1) or
    PB = √(x + 2)2 + (y + 3)2               √….(Z1)
   PA2 = 4[(x – 5)2 + y 2 ] =
    PB2 = (x + 2)2 + (y + 3)2                √....(Z2) or
   4[(x – 5)2 + y 2 ] = (x + 2)2 + (y + 3)2 √…..(Z2)
   3x2 + 3y2 - 44x + 6y + 87 = 0 √....(3)
Q.15
          D                   C
                                  AC = AB + BC
                                     = AB - CB                 .........(Z1)
A                  B                 = (i + 2j) – (-5i – 6j)

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

                   6i + 8j
Unit vector AC = ————                        ……..….(Z1)
                 √ 6 2 + 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   ……………….…..(Z1)

Compare:   m + 2n = 7              5n = 4        .….(Z1)
           m – 3n = 3              n = 4/5 )..(Z2 either)
                                   m = 27/5 )..(3 - both)
                           Q.17:
   sec θ = 1/cos θ
                                          1
                                              1 – t2

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




    cos (90 – θ) = sin θ       ……….(Z1)

                  = √ 1 – t ……….(2)
                           2
                                                    Q.18:
   2 sec2θ – 3tanθ = 4
   2(1 + tan2θ) – 3tanθ = 4                 …..…..(Z1)
     2 + 2 tan2θ – 3tanθ = 4
     2tan2θ - 3tanθ – 2 = 0
   (2tanθ + 1)(tanθ – 2) = 0                ………..(Z2)
   tanθ = -½ atau tanθ = 2
    θ = (360o – 26o 34’), (180o - 26o 34’)
   θ = 333o 26’, 153o 26’                   ………...(Z3)
    θ = 63o 26’, (180o + 63o 26’)
   θ = 630 26’, 2430 26’                 ….....….(Z3)
   θ = 630 26’, 153o 26’, 2430 26’, 333o 26’. …….….(4)
Q.19:
   ̸̱ BOC = π – 1
            = 3.142 – 1
            = 2.142       ………..(Z1)

        s = jθ
       BC = 15 x 2.142    ………..(Z2)
          = 32.13         ………..(3)
Q.20:
                           2
 dy     (x – 4)(4x) – (2x + 3)(1)
 — = ———————————                         ….…(Z2)
 dx             (x - 4)2
                               4x or 1   …….(Z1)

          2
        2x – 16x - 3
      = ———————                          ..……(3)
              (x - 4)2
Q.21:
   dy/dx = 4x – 3                 ………..(Z1)
      δx = 0.01                   ………..(Z1)
       x = 2


     δy∕      ≈ dy∕dx
           δx

          δy ≈ dy∕dx (δx)

             = [4(2) – 3](0.01)   …………(Z2)
             = 5(0.01)
             = 0.05               …………(3)
Q.22:
    3           3
   ∫ f(x) + ∫ (kx)dx = 20 .......(Z1)
    1           1
                           3
       8 + [ kx2 ∕ 2 ]        = 20    …..(Z2)
                           1

                           3
            [ kx2 ∕ 2 ]       = 12 …..(Z2)
                           1

             9k/2 – k/2 = 12,         8k/2 = 12   …..(Z3)
                                        k = 3     …..(4)
Q.23:

    6                      4
        P           or    P             ….(Z1)
            4                  3


6               4
    P x P = (6 x 5 x 4 x 3) x (4 x 3 x 2) ..(Z2)
        4           3


                         = 360 x 24
                         = 86400      ……...(3)
Q.24:
           9               5
   (a)        C       x       C                                           ……….(Z1)
                   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           …...….(Z1)
                5              5           4       6           3       7

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

   0.5 – 0.225   ……….(Z1)
 = 0.275         ……….(2)
TEKNIK MENJAWAB SOALAN
ADDITIONAL MATHEMATICS
          SPM
        PAPER 1


  THE END
TEKNIK MENJAWAB SOALAN
ADDITIONAL MATHEMATICS
          SPM
        PAPER 1



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