Docstoc

Additional Mathematics - SMK Subang Jaya - PowerPoint

Document Sample
Additional Mathematics - SMK Subang Jaya - PowerPoint Powered By Docstoc
					 MENJAWAB SOALAN
MATEMATIK TAMBAHAN

  SMK SUBANG JAYA
  Jalan SS14/6, Subang Jaya, Selangor


         19 JULAI 2012
               oleh:

             Ng K.L.
       SMK Tinggi Kajang
      PAPER 1
( Jawab Semua Soalan )
Question 1:
(a) { (1,p), (2,r), (3,s), (4,p) } √ 2

     Any one pair correct √ A1

    Common mistakes:
    * No bracket and wrong type of bracket
    * No comma
    * Wrong pairing
    * Wrong position of object-image
Question 1:

(b) many – to – one √ 1

  Common mistakes:
  * one – to – many
  * many – to – one, one – to – many
  * fancy ways of writing the type of relation
  * mistaken as notation of relation
Question 2:
(a)   7 √1

(b) f(x) = x + 2
    or
    f:x      x+2 √1
      Common mistakes:
      • Mistaken as the type of relation.
      • Weakness in deducing the f(x) from
        the diagram of the relation.
Question 3:
(a) 2, 4 √ 1

(b) { 1, 2, 3, 4 } √ 1

   Common mistakes:
   * No bracket and wrong type of bracket
   * No comma
   * Listed objects as the answer or
     mistaken objects as the range
Question 4:
      2
(a)   3   √1   from: 3m – 2 = 0, substitute x as m




(b) 0 ≤ f(x) ≤ 13 √ 2    Common mistakes:

    13 (seen) √ A1       * 2 ≤ f(x) ≤ 13
                         * x = 0, f(x) = 2
                         *0 ≤ x ≤5
                         * Do not understand the
                           meaning of range
Question 5:
(a) p = 8, q = −3       √ 3 (both correct)

   4x + 8x – 3 √ A2
      2                      gf(x) = 4x 2 + px + q

   (2x + 1)2 + 2(2x + 1) – 6 √ A1       gf(x) = g(2x + 1)

   Weaknesses:
   * Not knowing that the question is about
     finding the composite function or do not
     understand in finding the composite function.
   * Not realise to compare the composite
      functions to get the value of p and q.
Question 6:
                                 • Majority candidates did
                                   not include this term or
                                   did not know how to find.
(a)   3/2 √ A1

          –1           3x - 5       1
(b)   f        (x) =          , x     √2
                       1  2x       2


          y5
      x        √ A1              • Show the process of
         3  2y
                                    finding the inverse
      Weakness:                     function
      * Do not know how to find the inverse
        function.
Question 7:
5x 2 – 3x – 4 = 0 √ A1
(convert into general form of the quadratic equation)


    ( 3)  ( 3)2  4(5)( 4)
x
               2(5)             √ A2

x = 1.243, x = − 0.6434 √ 3 (both correct –
                             in 4 significnt
                             figures)
Question 8:
                           Common mistakes:
k<2    √3                  • 36 - 4k -4(3) > 0

                           • 36 – 12k + 12 > 0
6 – 4(k + 1)(3) > 0 √ A2
 2

                           • −12k > − 48
                                 k > - 48/-12
6 – 4(k + 1)(3)
 2
                            therefore: k > 2
or b2 – 4ac > 0
or a = (k+1), b = 6, c = 3 √ A1
Question 9:                 Common mistakes:
                            •α+β =−5
 2
x – 5x + 6 = 0 √ 3          •α+β =5
                            • α + β = − 5/2
                            • αβ = 3
4αβ = 6                     • x 2 + 5x + 6 = 0

2(α + β) = 5 √ A2 (both correct)

α + β = 5/2 √ A1 (either one)
αβ = 3/2
Question 10:
 2                  2
b – 4ac > 0 or 4 – 4(2)(m) √ A1
42 – 4(2)(m) > 0 √ A2
16 – 8m > 0
16 > 8m
8m < 16               Common mistakes:
                      * − 8m > −16/−8
m<2 √3                     m>2

                        • 4 – 4(2)(m) > 0
                          (careless)
Question 11:
            2
y = a(x + p) + q            minimum or maximum value


         axis of symmetry

q=−1√1
p = − 2 √ 1 from: x + p = 0  2 + p = 0
                      2
At (0,3): 3 = a(0 – 2) + (− 2)
           5 = 4a
           a = 5/4 √ 1
Question 12:
      2
5x – x – 4 ≤ 0 √ A1
x2 – 5x + 4 ≥ 0
       2
Let x – 5x + 4 = 0
                                    −     +

(x – 1)(x – 4) = 0
                                                 +
                                    −     −      +
                                x                    x
                        1   4       + 1
                                    1     −   4 4+


x = 1, x = 4 √ A2              √ A2
Then: x ≤ 1          or (x – 1)(x – 4) ≥ 0
         x≥4√3           x ≥ 1, x ≥ 4 ?
(based on the graph)    Then: x ≤ 1, x ≥ 4 √ 3
Question 13:
                  2
f (x) = 5 + 4x – x
       5 + 4x – x 2 ≤ 8
          2
      − x + 4x – 3 ≤ 0 √ A1        √ A2
       x2 – 4x + 3 ≥ 0         1          3
                                              x


         2
Let x – 4x + 3 = 0
       (x – 1) (x – 3) = 0
        x = 1 , x = 3 ? √ A2
Then: x ≤ 1, x ≥ 3 √ 3
Question 14:                          Common mistakes:
                                      * = lg2 x lg2 x lg5 ÷ lg2
            lg20                        = 0.3 x 0.3 x 0.7 ÷ 0.3
log2 20 =    lg2
                      √ A1            * = 0.3 + 0.3 + 0.7 – 0.3

            lg (2 x 2 x 5)               lg 2 2  lg 5
       =                         or    =     lg 2
                 lg 2
            lg 2  lg 2  lg 5              2 lg 2  lg 5
       =                                =                   √ A2
                   lg 2                         lg 2
                                 √ A2
            0.3  0.3  0.7             2(0.3)  0.7
       =                         √ A3 =              √ A3
                   0.3                          0.3
            13
       =         √4
            3
Question 15:
                                          Wrong Concept:
              log5 125x 2
logx125x2   =               √ A1
                 log5 x                    log5 5 3 x 2
                                             log5 x
              log5 5 3  log5 x 2
            =                     √ A2
                    log5 x                 3log5 x 2
                                         =
              3log5 5  2log5 x             log5 x
            =                   √ A3
                   log5 x
              3  2k
            =        √4
                 k
Question 16:
                    2n   
2 n .2 3  2 n  5       
                    2     √ A1
                         

    n     5
2  8 - 1   √ A2
          2
    19 
    n
2     
   2 
    n -1
2          (19) √ 3
Question 17:
(a) / AOC = 2π – 2.675 = 3.614
    r=   7.2 = 1.992 ==> Use s = rθ
        3.614
                   √2
            √ A1

(b) Area Sector COB = ½ (1.992)2 x 2.675 √ A1
                    = 5.307 √ 2
                                Use A = ½ r2θ
Question 18: Coordinates Geometry
                (Area of polygon)


          3 1 5 3       √ A1
 12 = ½
          h 0 −h h

 24 =     (0 – h + 5h) – (−3h + 0 + h) √ A2

 6h = 24 , 6h = − 24
  h=4       h = − 4 √ 3 (Both correct)
Question 19:
Line 1: 6x + 3y – 5 = 0  y = −2x + 5/3
        m1 = −2 √ A1
Line 2: m2 = ½
        x-intercept = a; y-intercept = −3

       Then: −(−3) = ½ √ A1
               a

                    a=6 √3
Question 20:
      dy                du
(a)         = 10u or         =3     √ A1 (either one)
      du                dx
      dy
            = 10(3x + 2) x 3   or    90x + 60 √ 2
      dx

(b) δx = 0.01 √ A1
                                               δy dy
      δy = 30(3x2 + 2) x 3 x 0.01        Use     =
                                               δx dx
           = 7.2   √2                           x=2
Question 21: Differentiation
           2
  y = 3x – 2x
 dy                                       dy
      = 6x – 2 √ A1      At turning point    =0
 dx                                       dx
 At turning point; 6x – 2 = 0

                         x = ⅓ √ A2
      y = 3(⅓)2 – 2(⅓)

       =−⅓
  Coordinate of turning point = (   ⅓, −⅓ )       √3
Question 22:
     N = 9, x = 12   Use the formula to find
         Σx          mean of ungrouped data:
(a) 12 =                       Σx
          9                 x=
                                N
   Σx = 108 √ 1
(b) N = 8, x = 11
   Σx = 108 – p
        108 − p
   11 =          √ B1 p = −30 √ 2
           8
     PAPER 2
SECTION A, B and C
SECTION A & B
Question 1:
y = 2x – 1 √ K1
  2
x + 2(2x – 1) + x(2x – 1) = 5 √ M1
    2
3x + 3x – 7 = 0
    3  3 2  4(3)( 7)
x                       √ M1
           2(3)

                                Follow question’s directive –
x = 1.107, − 2.107 √ A1         whether in 3 decimal places,
y = 1.214, − 5.214              or did not mention, then leave
                                the answer in 4 significant figures
                       √ A1
Question 2
x + 3y + 10 = 0    ...(1)     (y + 4) (y + 10) = 0
                                y = − 4,     √ M1
4    12
   +     + 1 = 0 ...(2)         y = −10 √ A1
x     y
                              Sub. y (4):
4y + 12x + xy = 0 ...(3)        y=−4
                                x=2
x = − 3y – 10     ...(4) √ K1   y = −10
Sub. (4) into (3)               x = 20 √ A1

4y + 12(− 3y – 10) + (−3y – 10)y = 0 √ M1
−3y 2 – 42y – 120 = 0
3y 2 + 42y + 120 = 0
Question 3:
     dy
 (a)    = px2 – 4x ;
     dx

     Line: y + x – 5 = 0
                   y=−x+5
   Gradient tangent = −1 √ K1

    dy
    dx   = −1, ==> px − 4x = −1 √ M1
                       2




    At (1, 3);   p(1)2 – 4(1) = −1
                            p = 3 √ A1
    dy
(b)    = 3x2 – 4x
    dx

    y=  ( 3 x2  4 x )dx

    y = x3 – 2x2 + C        √ K1


    At (1, 3); 3 = (1)3 – 2(1)2 + C
               C=4       √ M1

    y=x        3
                    – 2x + 4 √ A1
                        2
Question 4:
                         Axis of symmetry, from x value
                         of the coordinate given, (2, k)
(a) x = 2 √ A1
(b) f(x) = x2 + 2hx + h2 – h2 – 5 √ M1
         = (x + h) – h – 5
                  2    2
                                    √ K1
      h = −2 √ A1           From: x + h = 0
                            x = 2; 2 + h = 0
      k = −h – 5
              2

         = −4 – 5           minimum value = k
                            From the y value of
         = −9 √ A1          coordinate given, (2, k)
Question 4(c):
                                From: f(x) = x2 + 2hx – 5

• a = 1 > 0; a minimum graph
                                  f(x)
• y-intercept = 5
• axis of symmetry, x = 2                x=2
• minimum value = −9                                    √ A1
• minimum point = (2, −9)
                                                         x
                                  0      2


                         √ A1    −5
                                 −9
                                             (2, −9)   √ A1
Question 5: Statistics
     f              x              fx               fx2
                ( midpoint )

     7            133             931            123823
     8            138             1104         152352
    13            143             1859         265837
    10            148             1480         219040
     8            153             1224         187272
     4            158              632          99856
 Σ f = 50          -           Σfx = 5 890 Σfx2= 1 048 180
                                    √ M1            √ M1
Based on the formula of mean and variance, prepare the above
table for a much easier procedure of solving the problem
Question 5: (continue)
                     √ M1
             5 890                                  Σfx
(a) mean =           = 117.8 √ A1              x=
                                                    Σf
              50
               1 048 180
(b) variance =                     (117.8)2 √ M1
                  50
              = 7 086.76 √ A1                  2  Σfx2
                                              σ =      − x2
                                                   Σf

(c) Std. dev. = 84.18       √ A1
                                   σ = √ σ2
Question 6: Coordinate Geometry
       y–3 6–3
(a)         =       √ K1
       0 – 6 15 – 6

            y = 1 √ A1
      then; A = (0, 1)
                         7–1
      Gradient AE, mAE =     =2      √ M1
                           3–0

      Eqn. AE: y – 7 = 2( x – 3 )
                   y = 2x + 1 √ A1
Question 6: (continue)
          2(x) + 3(3)          E (3, 7)
(b) 5 =
            2+3                2      B (5, 3)

           √ M1 (either one)           3
                                                 D (x, y)
          2(y) + 3(7)
    3=
             2+3

    x=8,       y = −3 √ A1         (either one)



    D ( 8, – 3 )   √ A1
Question 6: (continue)
                    5 15    8    5   √ K1
(C) Area ∆BCD = ½
                    3   6   −3   3


             = ½ (30 – 45 + 24) – (−15 + 48 + 45) √ M1
             = 34.5 √ A1
Question 7: Linear Law
(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

                                          √ A1



•The value of log y must be ≥ 2 decimal places

• If table not shown, the A1 mark can be given on
  all the points plotted correctly on the graph
       Plot graph log y against x log y
       log y
                            Not using the given scales – minus 1
                            Scales not uniform – minus 1
                            Not using graph paper – minus 1

              √ M1
       1.0                                                                 +
                                                                   +
       0.8                                                 +
                                     √ A1
                                            +
       0.6           √ A1        +
                       +
       0.4
                                                                   √ M1
√ M1                                            √ A1
       0.2 √ A1
                                     √ A1
       0.0                                                                     x
          0      1     2        3      4         5     6       7       8   9
                                                  √ M1
(b) Plot log10 y against x
     (correct axes and uniform scales)
     6 points plotted correctly            √ A1

     Lines of best fit √ A1
 (c) (i) y = 4.8; log 4.8 = 0.68,
                                      √ A1
                       x = 5.6 ~ 5.7
     log10 y = log10 h + 2x log10 k √ K1
      (ii) c = log10 h √ M1
                            √ A1
            h = 1.78
      (iii) m = 2 log10 k √ M1
            k = 1.09 to 1.12     √ A1
Question 8. Circular Measures
             o       o            o
  (a) θ = 60 + 60 = 120               √ K1

             B

                 60 o            π
                          = 120     = 2.095               √ A1
         C       A              180

             D

  (b) Area ABCD = 2 ( ½ ) (4)2 sin60o                    √ M1

                = 13.86 √ A1
                        Use Solution of Triangles Principles
(c) Area shaded region       √ M1

    = Area Sector ABC – Area ∆ ABC
           2
    = ½ (4) (2.095) – ½ (Area ABCD)
           2
    = ½ (4) (2.095) – ½ (13.86)
               √ M1 √ M1
    = 9.832               √ M1


          √ A1
Question 9: Index Number
          Q2010
(a) Use               x 100 = 140
          Q2008
                           √ P1
                        140
   (i)    Q2010 =           x 8 = 112 √ A1
                        100

   (ii)           I        W           IW                Σ IW
                                                   Use        = 132
          P   140          30         4200                ΣW
          Q       m        20         20m           9890 + 20m
          R   110          15         1650                     = 132
                                                        100
          S   104          10         1040         √ M1
                                                             m = 165
          T   120          25         3000
                                                               √ A1
                        ΣW =100 ΣIW = 9890 + 20m
Question 9 (continue…)
(b)   Q2012
              x 100 = 110 √ P1
      Q2010
                     √ P1
                  110
      Q2012 =         x 60 = 66       √ A1
                  100


      Q2008       125         Q2010       132
(c)           =         and           =
      Q2006       100         Q2008       100

       Q2010         132 125
             x 100 =    x    x 100 = 165 √ A1
       Q2006         100 100
                        √M1   √M1
Question 10: Solution of Triangle
               A
(a)            30o                Use Cos Rule.                √ M1
          9               7
                                  BC2 = 9 2 + 72 – 2(9)(7) cos 30o
      B                       C
                                       = 20.881 √ A1
                                  BC = 4.570 √ A1
(b)
      A
                                  Use Sin Rule.
      7            25 o       D    Sin / CAD       Sin 25
                                               =            √ M1
              11                     11              7
      C
                                  sin / CAD = 0.6641 √ A1

                                  / CAD = 41.61o √ A1
Question 10: (continue)

(c) Area ABCD √ M1
    = ½(9)(7) sin30o + ½(7)(11) sin113.39o
                    √ M1     √ M1
   = 39.84   √ A1
    TAMAT
SELAMAT BERJAYA
TERIMA KASIH

				
DOCUMENT INFO
Shared By:
Stats:
views:1102
posted:7/19/2012
language:
pages:47