Docstoc

Add Maths - Questions, Answers & Marking Scheme - UJian Setara 2/2010

Document Sample
Add Maths - Questions, Answers & Marking Scheme - UJian Setara 2/2010 Powered By Docstoc
					                                                                                           1

3472/1/2                                                      Name : ……………………………...
US2
17 Aug 2010                                                   Class: Form 4 ……………………..


                                    UJIAN SETARA 2
                              (ANSWER & MARKING SCHEME)

                                  ADDITIONAL MATHEMATICS

                                           Paper 1 & 2
                                            One hour


                 SEKOLAH MENENGAH KEBANGSAAN TINGGI KAJANG


INFORMATION FOR CANDIDATES:

1.   This paper consists of two parts, namely Paper 1 and Paper
2.   Answer all questions in both parts in this booklet.
3.   The diagrams shown in the questions are not drawn to scale unless stated.
4.   You may use a non-programmable scientific calculator.
5.   The following formulae may be helpful in answering the questions. The symbols given are
     the ones commonly used.

                 b  b 2  4ac                                         log c b
     5.1. x                                           5.8. log a b 
                      2a                                                log c a

     5.2. a m  a n  a mn                                     a     b       c
                                                       5.9.              
                                                              sin A sin B   sin C
     5.3. a m ÷ a n = a m-n
                                                       5.10. a 2 = b 2 + c 2 – 2bc cos A
     5.4. ( a m ) n = a m n
                                                       5.11. Area of triangle
     5.5. loga mn = loga m + loga n
                                                                1
                                                             = ab sin C
                 m                                              2
     5.6. loga     = loga m – loga n
                 n
     5.7. loga mn = n loga m


                          This question paper consist of 7 printed pages


Prepared by: …………………………..                                     Verified by: ………………………...
                                                                                                      2

                                                   Paper 1

                                          Answer all questions.
                                              Jawab semua soalan.

1.   The following information refers to set K and set L.
     Maklumat berikut adalah berkaitan dengan set K dan set L.


                                 K = { -1, 0, 1, 2 }
                                 L = { 2. 3. 6 }


     The relation between set K and set L is defined by the following set of ordered pairs
     { (-1, 3), (0, 2), (1, 3), (2, 6) }.
     Hubungan di antara set K dan set L ditakrifkan oleh set pasangan tertib yang berikut
     { (-1, 3), (0, 2), (1, 3), (2, 6) }.

     State
     Nyatakan

     (a)     the image of 1,
             imej bagi 1,
     (b)     the type of the relation.                                                        [2 marks]
             jenist bagi hubungan tersebut.



                                                                                   3          √1
                                                                    Answer : (a) ………………………...
                                                                                  many-to-one √1
                                                                             (b) ……...……………........


            1
2.   Given     and –2 are the roots of a quadratic equation. Write the quadratic equation in
            2
     general form.                                                                [2 marks]
     Diberi ½ dan –2 adalah punca-punca bagi satu persamaan kuadratik. Tuliskan persamaan kuadratik itu
     dalam bentuk am.

     (x – ½)(x + 2) = 0              OR          S.O.R = ½ + (-2) = −3/2
                       √ P1                      P.O.R = ½ x (-2) = −1 √ P1




                                                                                2x2 − 3x – 2 = 0    √2
                                                                    Answer : …..………………………...
                                                                                                         3

3.   The quadratic equation x2 – 5x+ p = 0 has two distinctive different roots. Find the range
     of value p.
     Persamaan kuadratik x2 – 5x + p = 0 mempunyai dua punca berlainan yang nyata. Cari julat nilai p.

                                                                                               [2 marks]
             a = 1, b = −5, c = p

             b2 – 4 ac > 0

           (-5)2 – 4(1)(p) > 0 √ P1

             25 > 4p

                                                                          p < 25/4 √2
                                                                Answer : ….………………………...


4.   Diagram 1 shows the graph of function y = (x – p)2 + 5.
     Rajah 1 menunjukkan graf fungsi y = (x – p)2 + 5.                                         [3 marks]
     State                                                                 y
     Nyatakan
     (a)     the value of p,
             nilai p.
     (b)     the equation of axis of symmetry.
             persamaan bagi paksi simetri,
                                                                                                  x
     (c)     the maximum value.                                        0                4
             nilai maksimum.


                                                                                 2     √1
                                                                Answer : (a) p = …………………...
                                                                              x=2      √1
                                                                         (b) ……...…………….......
                                                                                    5     √1
                                                                               (c) ……………………….

5.   Solve the equation.
     Selesaikan persamaan

     32x + 1 = 92x                                                                             [3 marks]


             32x + 1 = (32)2x   √ P1
             2x + 1 = 4x        √ P2



                                                                             ½      √3
                                                                Answer: x = ……………………….
                                                                                                         4

6.    Given log4 x – log2 (y + 2) = 1. Express y in terms of x.
      Diberi log4 x – log2 (y + 2) = 1. Nyatakan y dalam sebutan x .                             [4 marks]

             log 2 x
      √ P1            log 2 (y  2)  1
             log 2 4
             log 2 x
                       log 2 (y  2)  1
               2
             log 2 x 1/2  log 2 (y  2)  1
     √ P2              x 1/2
             log 2           1
                     (y  2)
     √ P3    x 1/2
                    21
             y2
             x 1/2  2y  4
                     x 1/2  4                                         x 4
     √4      y                             OR                y
                          2                                             2

                                                                                     x 4 √4
                                                                                  y
                                                                                      2
                                                                       Answer : ……………………………


7.    Find the range of values of x for which x(2x – 7)  7 – 2x.                                [4 marks]
      Cari julat nilai x bagi x(2x – 7)  7 – 2x.



             2x2 – 5x – 7  0         √ P1                                               √ P3

      Let 2x2 – 5x – 7 = 0
                                                                                         x
             (2x – 7)(x + 1) = 0 √ P2                            −1           7/2

             x = 7/2, x = − 1        √ P3




                                                                              x ≤ − 1, x  7/2        √4

                                                                       Answer : ……………………………
                                                                                                     5

                                               Paper 2

                                      Answer all questions.
                                        Jawab semua soalan


1.   Solve the simultaneous equations x + 2y = 3 and x2 + y2 + xy = 7. Give the answers
     correct to three decimal places.
     Selesakan persamaan serentak x + 2y = 3 dan x2 + y2 + xy = 7. Beri jawapan betul kepada tiga angka
     perpuluhan.
                                                                                            [5 marks]


             x = 3 – 2y                          √ M1

             (3 – 2y)2 + y2 + (3 – 2y)y = 7      √ M1

             3y2 – 9y + 2 = 0

                 ( 9 )  9 2  4( 3 )( 2 )
             y=                                  √ M1
                           2( 3 )


             y1 = 2.758,    y2 = 0.2417         √ A1

             x1 = −2.516, x2 =2.517             √ A1
                                                                6

2.   Given that f(x) = 2x + 3 and g(x) = x2 + 2x + 1.
     Diberi f(x) = 2x + 3 dan g(x) = x2 + 2x + 1.
     Find,
     Cari,
     (a)     f -1(x)                                    [1 mark]
               -1
     (b)     f g(x)                                     [2 marks]
     (c)     h(x) such that fh(x) = 2x – 5.             [2 marks]

                         x3
     (a)     f −1(x) =                        √ P1
                          2

     (b)     f −1g(x) = f −1(x2 + 2x + 1) √ M1
                           x2  2x  2
                      =                   √ A1
                                2

     (c)     Let h(x) = y
                 f(y) = 2x – 5
                 f(y) = 2y + 3
              2y + 3 = 2x – 5                √ M1
                            2x  8
                       y=
                              2
                    h(x) = x – 4             √ A1
                                                                                               7

3.   Diagram 2 shows a quadrilateral ABCD.
                                                        A      10 cm    B

     (a)   Calculate,                                         52o

           (i)    Length of BC.

           (ii)  BCD                   [5 marks]            29 cm

     (b)   Point C’ lies on BC such that                                       98o
           DC = DC’.                                                                D
                                                                                7.6cm
           (i) Sketch the triangle BC’D.                                C

           (ii) Calculate the area, in cm2, triangle BC’D.                     DIAGRAM 2
                                                  [5 marks]



     (a)   (i)    BC2 = 102 + 292 – 2(10)(29) cos52o √ M1
                  BC = 24.16                         √ A1
                  sin CBD sin 98 o
           (ii)                                            √ M1
                      7.6   24.16

                   CBD = 18.15o                            √ P1

                   BCD = 63.85o                            √ A1


     (b)   (i)      B

                          18.15o
                                            (ii)  BC’D = 116.15o,  BDC’= 45.70o √ P1
                                                    sin 116.15 o sin 18.15 o
                                                                                       √ M1
                              √ P1                       BD          7.6

                    C’             45.70o           BD = 21.9
                     7.6 cm
                                   D                Area BC’D = ½(7.6)(21.9) sin 45.70o √ M1
                                                             = 59.56                    √ A1

				
DOCUMENT INFO