Additional Mathematics Paper 1

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					ADDITIONAL MATHEMATICS
              SPM 2012
               PAPER 1
              ( PART ONE )


LAST KOPEK REVISION
      TRY TO ANSWER THESE QUESTIONS
AND CHECK YOUR ANSWERS FROM THE FOLLOWING
       SCHEME OF ANSWER AND MARKING
                  PAPER 1

1. Diagram shows part of the relation f(x)
   State                          f(x) f(x) = (x – 2)2 − 3
   (a) the type of
       the relation,
                                                     x
   (b) the range.                       .
                                           (2, -3)


   Answer:
   (a) Many-to-one     √1

   (b) f(x) ≥ -3 √ 1
2.   Given the function f(x) = 3 – x, g(x) = px2 – q and
     gf(x) = 3x2 – 18x + 5. Find
     (a) gf(- 2),
     (b) the value of p and q.

Answer:
(a) f (-2) = 3 – (−2) = 5 √ M1
    gf(- 2) = g(5) = 3(5)2 – 18(5) + 5
                   = −10 √ 2
(b) gf(x) = p(3 – x)2 – q
           = px2 – 6px + 9p - q √ M1
     Compare to 3x2 – 18x + 5
     p = 3, 9p – q = 5  9(3) – q = 5
     q = 22 √ 2 Both correct
3. A function g is defined as g(x) = 7x – 4. Find
   (a) g -1,
   (b) the value of h if g -1(h) = 2

Answer:
                                  Using the principle:
                x+4      √1                    cx – b
(a) g -1 =                        ax + b
                                          =
                 7                  c            a


                    h+4
(b) g   -1(h)   =         =2      √ M1
                     7

                         h = 10    √2
4.   A quadratic equation with roots p and q is
     x2 + mx + m = 0 where p, q and m are constants.
     Express p in terms of q.

Answer:

(x – p) (x – q) = 0                    OR:
                                       Use SOR/POR
x2 – (p + q) x + pq = 0 √ M1
                                       method:
Compare to x2 + mx + m = 0
                                       SOR = p + q
− (p + q ) = m and pq = m              POR = pq


pq = − (p + q)   √ M2    pq + p = −q
                                     −q
                               p=            √3
                                    1+q
 5.   Find the range of values of x for 4x2 ≥ 3 – 4x

Answer:
                                    a = 4 > 0  minimum graph

4x2 + 4x – 3 ≥ 0 √ M1
Let 4x2 + 4x – 3 = 0
                                                            x
     (2x – 1) (2x + 3) = 0             -³   2       ½
                                                         √ M2
     x=½; x=−³2 ?            √ M2


      x≤−³2; x≥½       √3
6.   Diagram shows the graph of a quadratic function y =
     f(x). The straight line y = −15 is a tangent of the curve
     y = f(x).
     (a) State the equation of the axis of symmetry of the
         curve.
     (b) Express f(x) in the form of (x + p) 2 + q, where p
         and q are constants.
 Answer:                                                     y
        −2 + 6    √ M1
                         the mid-point of the two                y = f(x)
(a) x =                  points on the curve that
          2              intersect at x-axis
                                                                       x
                                                        −2   0     6
                                                                       y = −15
     x=2     √2

                         (x + p)² + q
                         x + p = 0  x = −p (axis of symmetry)
(b) (x – 2)² − 15 √ 2    q = optimum value (either minimum or
    q = −15 √ M1         maximum based on the graph)
                                                               2
                    7.    Solve the equation: 16x =
                                                              8 2−x

 Answer:
√ M1
                2
 24(x)   =                √ M1
             23(2 − x)
 24x = 2 [2 −3(2 − x)]

 24x = 21 – 6 + 3x
                                 Use the concept of the rules of
 4x = – 5 + 3x           √ M2    equation of indices
  x = − 5 √3
     8. Solve the equation: 2 logx4 + logx8 = −7

Answer:
    √ M1
logx 42 + logx 8 = −7           Converting log to indices


logx 16(8) = −7     √ M1   OR   logx 16(8) = logx x −7
                                Use the concept of the rules of
                                equation of logarithms.
     128 = x −7     √M2
        27 = x −7          OR    2 7 = ( 1 /x ) 7
    (½ )−7 = x −7

           x=½      √3
9.   The nth term of an arithmetic progression is given by
     Tn = 12 – 3n. Find the common difference of the
     progression.

     Answer:

     a = T1 = 12 – 3(1)
            = 9 √ M1
         T2 = 12 – 3(2)
            = 6
          d = T2 − a
            =6–9       √ M2
            = −3       √3
10.   Given that 12, 6, 3 … is a geometric progression, find
      the sum of the first 7 terms after the 3 th. term of the
      progression

      Answer:                      OR:

                                   T4 = 12 ( ½ )3 = 12/8
      a = 12
                                   S7* = (12/8) [1 – ( ½ )7]
      r = ½ √ M1                      = 2.977
      S7* = S10 – S3
              12 [1 – ( ½ )10] – 12 [1 – ( ½ )3]
          =                                          √ M2
                  1–½                1–½
          = 2.977    √3                S7* = Sum from T4 to T10 or
                                       first 7 terms after the 3rd. term
                                                   k
11.   Given 0.471 + 0.000471 + 0.000000472 + … = -------
      Find the value of k.                        333


      Answer:

      a = 0.471
       r = 0.001    √ M1

              0.471            k            Use the concept of
      S∞ =                 =         √ M2
                                            the sum of infinity.
             1 – 0.001         333


         =    157
              333
       k = 157     √3
12. Diagram shows a circle with center O. The length of
    minor arc AB is 4.8 cm and the angle of minor sector
    BOC is 1 rad.
    Using π = 3.142, find
    (a) the value of θ,
    (b) the length, in cm, of the major arc AC.
    Answer:
              4.8
   (a) θ =        = 0.75 rad.   √1                    A
              6.4                             O   θ

                                                      B
   (b) θAC (major) = 2π – (1 + 0.75)
                                       √ M1
                   = 4.534 rad.                   C

       SAC (major) = 6.4 ( 4.534 )
                    = 29.02 cm. √ 2
                                                                      dy
     13.       Given that y =   x3   (3x +   1)4,   find the value of
                                                                      dx
               when x = −1.

     Answer:
          dy
             = (3x + 1)4 3x2 + x3 (4)(3x + 1)3 (3)          √ M1
          dx

          dy
x = −1;        = [3(−1) + 1]4 3(−1)2 + (−1)3 (4)[3(−1) + 1]3(3)
          dx
                       √3                                          √ M2
               = 144

				
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