Ujian Setara 2 - Add Maths - Answer & Marking Scheme by nklye

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									472/1                                                                                                                      SULIT
1 Ogos 2012
Nama: …………………………………………………..                                                        Kelas: ………………………….

                SEKOLAH MENENGAH KEBANGSAAN TINGGI KAJANG
                   SEKOLAH MENENGAH KEBANGSAAN TINGGI KAJANG
                                      JALAN SEMENYIH, 43000 KAJANG, SELANGOR
                                    JALAN SEMENYIH, 43000 KAJANG, SELANGOR
                                        JALAN SEMENYIH, 43000 KAJANG, SELANGOR
                                                UJIAN SETARA 2 / 2012
                                              ADDITIONAL MATHEMATICS
                                                     MATEMATIK TAMBAHAN

                                        ANSWER AND MARKING SCHEME

 1. Kertas soalan ini adalah dalam dwibahasa. Soalan dalam bahasa Inggeris mendahului soalan yang
    sepadan dalam bahasa Melayu.
 2. Calon dibenarkan menjawab keseluruhan atau sebahagian soalan sama ada dalam bahasa Inggeris
    atau bahasa Melayu.
 3. Jawab semua soalan pada ruang jawapan yang telah disediakan pada kertas soalan ini.
 4. Tunjukkan langkah penyelesaian anda. Ia mungkin membantu anda mendapat markah.
 5. Rajah-rajah bagi soalan-soalan yang berkaitan tidak dilukis mengikut skala kecuali dinyatakan.
 6. Anda boleh menggunakan kalkulator saintifk yang tidak diprogramkan.

        Prepared by :                                                                              Checked and verified by:

      ...........................                                                                  ……………………….
      ( NG KOK LYE )                                                                               ( SOH BOON CHUAN )



The following formulae may be helpful in answering the questions. The symbols given are the ones commonly
used.
Berikut adalah formula-formula yang mungkin boleh membantu dalam soalan-soalan Simbol-simbol yang diberi adalah
yang sudah biasa digubakan.
                                                               ALGEBRA

            b  b 2  4ac                                                 5.    log a mn = log a m + log a n
 1.     x=
                 2a
                                                                                          m
 2.     am  an = a m + n                                                  6.     log a     = log a m  log a n
                                                                                          n
 3.     am  an = a m  n
 4.     (am) n = a nm                                                      7.    log a mn = n log a m

                                                              GEOMETRY              log c b
                                                                      8. log a b =
                                                                                    log c a
                                                                       3. A point dividing a segment of a line
 1.     Distance =           ( x2  x1 ) 2  ( y 2  y1 ) 2                                 nx1  mx2 ny1  my2 
                                                                                ( x, y) =            ,          
 2.     Midpoint :                                                                          mn         mn 
                       x1  x 2   y  y2 
        (x , y) =               , 1                                      4. Area of triangle =
                       2            2                                         1
                                                                                  ( x1 y 2  x2 y3  x3 y11 )  ( x2 y1  x3 y 2  x1 y3 )
                                                                                2
                                                              STATISTICS

 1.   x   =
              x                                                           2.  =
                                                                               2          x
                                                                                               2
                                                                                                   x
                                                                                                        2
                N                                                                          N
     4721/1/2                                                                                                          SULIT

                                                        Answer all questions.
                                                        Jawab semua soalan.

1.      The following information refer to the set A and set B.
        Maklumat berikut adalah berkaitan dengan Set A dan Set B.
                     Set A = { -3, -2, 2, 3 }
                     Set B = { 4, 9 }
        The relations between set A and set B is defined by the set of ordered pairs {(-3, 9) (-2, 4) (2, 4) (3, 9)}.
        Hubungan antara set A dan set B ditakrifkan oleh set pasangan bertertib {(-3, 9) (-2, 4) (2, 4) (3, 9)}.

        (a)     State the type of relations.
                Nyatakan jenis hubungan.
        (b)     Using the functions notation, write a relation between set A and set B.
                Dengan menggunakan tatatanda fungsi, tulis satu hubungan antara set A dan set B.

                                                                                                            [2 marks/markah]
        Answer / Jawapan:

        (a)     Many–to–one √ 1


        (b)     f(x) = x2   or        f:x          x2 √ 1                                                      2


2.     Given that the functions f : x → 2x + 6 and f -1 : x → kx + p , where k and p are constant, find the
       value of k and of p.
        Diberi fungsi f : x → 2x + 6 dan f -1 : x → kx + p , dengan keadaan k dan p ialah pemalar, cari nilai k dan
        nilai p.                                                                                 [3 marks/markah]

        Answer / Jawapan:

                      x-6               x
        f -1(x) =                or         − 3 √ A1
                       2               2
       Compare with f -1 : x            kx + p

        k=½

         p=–3        √ 3 (both correct , A2 if one correct)                                                        3



3.      Form the equation which has roots −4 and ⅓. Give your answer in the form ax2 + bx + c = 0, where
        a, b, c are constants.
        Bentukkan persamaan kuadratik yang mempunyai punca-punca −4 dan ⅓. Berikan jawapan dalam bentuk
        ax2 + bx + c = 0 dengan keadaan a, b dan adalah pemalar.
                                                                                                          [2 marks/markah]
        Answer / Jawapan:

        SOR: - 4 + 1/3 = - 11/3

        POR: (- 4)(1/3) = - 4/3             √ A1            or        using factorisation method:

        Equation: x2 − ( - 11/3) x + (- 4/3) = 0                      ( x + 4)(3x + 1) = 0 √ A1

                       3x2 + 11x – 4 = 0 √ 2
                                                                                                               2
4.   Diagam 1 shows the graph of a quadratic functions f(x) = 3(x + p) 2 + 5, where p is a constant.
     Rajah 1 menunjukkan graf kuadratik f(x) = 3(x + p) 2 + 5 , dengan keadaan p ialah pemalar.
                                                               y



                                                                   y = f(x)


                                                       ●
                                                  (-2, q)
                                                                          x             Diagram 1
                                                           0                             Rajah 1

     The curve y = f(x) has the minimum point (-2 , q) where q is a constant.
     Lengkung y = f(x) mempunyai titik minimum (-2 , q) dengan keadaan q ialah pemalar.
     State / Nyatakan

     (a) the value of p,
           nilai p,
     (b)    the value of q,
            nilai q,
     (c)   the equation of the axis of symmetry.
            persamaan paksi simetri.                                                                [3 marks/markah]

     Answer / Jawapan:

     (a) 2      √1

     (b) 5      √1

     (c) x = −2 √ 1                                                                                    3


5.   Given that the graph of f (x) = x2 – 5x + p touches the x-axis at one point, find the value of p.
     Diberi bahawa graf f(x) = x2 – 5x + p menyentuh paksi-x pada satu titik, cari nilai p.
                                                                                                    [2 marks/markah]
     Answer / Jawapan :

       b2 – 4ac = 0
       (-5)2 – 4(1)(p) = 0      √ A1
                                                                                                       2
       p = 25/4 √ 2




6. Solve the equation 52x – 1 = 12 x                                                                [3 marks/markah]
     Selesaikan persamaan 5 2x – 1 = 12 x
     Answer / Jawapan:
      log10 5 2x – 1 = log10 12 x √ A1
     (2x – 1)log10 5 = x log10 12
     (2x – 1)(0.6990) = x(1.0792) √ A2
                                                                                                       3
     x = 2.193 √ 3
7.   Given that log2 3 = p and log2 5 = q, express log2 7.5 in terms of p and q.                   [3 marks/markah]
     Diberi bahawa log2 3 = p dan log2 5 = q, ungkapkan log2 7.5 dalam sebutan p dan q.

     Answer / Jawapan:

     log2 15 = log2 3 x 5 √ A1
          2            2
             = log2 3 + log2 − log2 2 √ A2

                = p+q–1 √3
                                                                                                       3




8. Find the equation of the straight line that passes through point (2, 5) and is perpendicular to
   2y = –3x + 6.
     Cari persaman garis lurus yang melalui titik (2, 5) dan berserenjang dengan 2y= − 3x + 6.
                                                                                                   [3 marks/markah]
     Answer / Jawapan:
            3
     y= -     x+3
            2
     m1 = −3/2
     m2 = 2/3 √ A1
     Equation: y – 5 = 2/3 ( x – 2) √ A2
                   y = 2/3 x + 11/3         or     3y = 2x + 11 √ 3                                    3




9 . Given that point A(6, –4) and B(–2, 4). Point P moves such that PA : PB = 2 : 3. Find the equation of
    the locus of P.
     Diberi titik A (6, −4) dan B (−2, 4). Titik P bergerak dalam keadaan PA : PB = 2: 3. Cari persamaan lokus P.
                                                                                                      [3 marks/markah]
     Answer / Jawapan:

     3PA = 2PB

      3 √(x – 6)2 + (y + 4)2     = 2 √(x + 2)2 + (y − 4)2 √ A1

     9[ x 2 – 12x + 36 + y 2 + 8y + 16 ] = 4[ x2 + 4x + 4 + y2 − 8y + 16 ] √ A2

      5x2 + 5y2 − 124x +104y + 388 = 0 √ 3

                                                                                                       3
10. Find the mode, mean and the variance for the set of data 3, 5, 6, 9, 6, 7.
    Cari mode, min dan varian bagi set data 3, 5, 6, 9, 6, 7.                                    [4 marks/markah]


    mode = 6 √ 1

    mean = 3 + 5 + 6 + 9 + 6 + 7
                   6
         = 6 √1
                 2   2   2   2   2   2
     variance = 3 + 5 + 6 + 9 + 6 + 7 − 6 2 √ A1
                           6                                                                          4
                = 3.333 √ 2




11. Solve the simultaneous equations x – y = 3 and x2 + 3y = 7. Give your answer correct to three
    decimal places.
    Selesaikan persamaan serentak x – y = 3 dan x2 + 3y = 7. Berikan jawapan anda betul kepada tiga tempat perpuluhan.
                                                                                                   [5 marks/markah]


    x = 3+y          √ P1                   or                  y = x – 3 √ P1

    (3 + y)2 + 3y = 7 √ M1                                      x2 + 3(x – 3) = 7      √ M1

    y2 + 9y + 2 = 0                                             x2 + 3x – 16 = 0

           - 9  √ 92 – 4(1)(2)                                      - 3  √ 32 – 4(1)( -16 )
     y =                              √ M1                      x=                                 √ M1
                  2(1)                                                          2(1)


    y = − 0.228, − 8.772          √ A1                          x=     2.772,     1.424 √ A1
    x=      2.772,     1.424      √ A1                          y = − 0.228, − 8.772 √ A1



                                                                                                      5
12. Solutions to this question by scale drawing will not be accepted.
     Penyelesaian untuk soalan ini dengan menggunakan lukisan berskala adalah tidak diterima.
                                                                                y

     The diagram shows a trapezium ABCD such that A lies                            B
     on the y-axis and C lies on the x-axis. The equations of AD
     and CD are 2y + 3x = 6 and 2y = 3x – 18 respectively.                  A
     Rajah menunjukkan sebuah trapezium ABCD dengan A terletak                              C
     pada paksi-y dan C terletak pada paksi-x. Persmaan bagi AD dan                             x
                                                                            O
     CD adalah 2y + 3x = 6 dan 2y = 3x – 18 masing-masing.
                                                                                        D
     Find / Cari
     (a) the coordinates of point D,
         koordinat titik D,
     (b) the equation of line AB.
          persamaan garis AB.                                                                        [7 mark/markah]


     Answer / Jawapan:

     (a) 2y = 3x – 18 ........(1)                 (b) mBC = mAD = − 3/2 √ P1
           2y + 3x = 6 .........(2)                            mAB = 3/2        √ P1

           (1)     (2)                                  Coordinate A: x = 0 , 2y + 3(0) = 6
           3x – 18 + 3x = 6                                                           y=3
                     6x = 24                                          A = (0, 3) √ A1
                         x = 4 √ M1                     Equation AB:       y – 3 = 3/2 ( x – 0) √ M1
           x=4         (1)                                                      y = 3/2 x + 3 √ A1
                       y=−3
           Therefore D = (4, − 3) √ A1

                                                                                                     7




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                    KERTAS SOALAN dan SKEMA JAWAPAN & PEMERKAHAN TAMAT

								
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