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2
                              1
                          k =  –  3 2 2 2  – 2                       20.  (a)  2x  + 3x − 7
                                                                                 3
                                                                                    2
                                                                             1
                                                                              2
                              1                                           = 2 x  +  x  – 7
                            =                                                    2
                              4                                                     2
                                                                             31
                                                            [4]           = 2 x +   3 2   –   9 4  – 7
                      (b)  x =  0 + 3  = 1.5                [1]                   4  2  16
                                                                             1
                              2                                           = 2 x +  3 2   –  65
                                         ,
                      (c)  Maximum point  1 3 1 2           [1]           k = –  3  4  8
                                        2 4
                                                                               4
                    14.  a = 1, b = 3                                                     3   65             [3]
                                                                                        1
                                                                                          4
 =  1                                                       [2]        (b)  Minimum point  –  , –   8 2
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 2                  15.   3x  − 5x + p = 0                                                                   [1]
                          2
 = √3                       b  – 4ac , 0                               (c)
                             2
                      (−5)  − 4(3)(p) , 0                                                         y
                         2
 = √(7)  –                 25 − 12p , 0
 2
                               12p . 25                                                 –    3 , 65
                                                                                         4 8
 2
 2
 =  1  8 2  + 3                  p .  25                    [3]
                                     12
 =                        –(–7) ±  (−7)  − 4(2√6 )(√6 )
                                     2
 =                  16.  x =       2(2√6)                                                        7
                      x =   2    or  x =   3
                         √6        2√6                                                                      x

                                 3

                      a = 2  or  a =   (rejected)                            – 3 –  65             – 3 +  65
                                                                               4
                                                                                                      4
                      b = 6      2
                                                            [3]                                              [3]
                    17.  x  – kx + 2k – 3 = 0                        21.  (a)   y
                       2
                                b  − 4ac . 0
                                 2
                      (−k)  − 4(1)(2k − 3) . 0                                                   x
                         2
                             k  − 8k + 12 . 0                               –1              7
                              2
                           (k − 6)(k − 2) . 0
                                     k , 2, k . 6
                                                            [5]
                    18.  mx − 8 = x  − 3x − 4                                 –7
                               2
                          x  − (3 + m)x + 4 = 0
                           2
                                 b  − 4ac . 0
                                  2
                       [−(3 + m)]  − 4(1)(4) . 0                           2                                 [3]
                               2
                          9 + 6m + m  − 16 . 0                         (b)  x  − 6x − 7
                                   2
                                                                                 2
                              m  + 6m − 7 . 0                             = (x − 3)  − 16
                               2
                           (m + 7)(m − 1) . 0                             m > 16
                                      m , −7, m . 1                                                          [2]
                                                            [5]      22.  (a)  y = –2(x  – 4x) – 2
                                                                                 2
                                                                                     2
                    19.  (a)  9 + 4x − 2x                                   = −2[(x − 2)  − 4] − 2
                                  2
                                                                                    2
                         = −2x  + 4x + 9                                    = −2(x − 2)  + 6
                              2
                         = −2[(x − 1)  − 1] + 9                                 y
                                   2
                         = −2(x − 1)  + 11                                           (2, 6)
                                  2
                         = 11 − 2(x − 1)
                                     2
                                                            [2]
                         a = 11, b = 2, c = −1
                      (b)  Maximum point (1, 11)            [1]                                   x


                                                                              –2  2 –   3   2 +   3
                      (c)  −21 > f(x) > 11                  [1]
                      (d)  No, it does not have an inverse because it is not
                         a one-to-one function.             [1]
                                                                                                             [5]
                         Cambridge IGCSE
                                          TM
                  170     Ace Your Additional Mathematics
         Answers Add Math.indd   170                                                                             14/03/2022   12:29 PM
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