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9.   y                                           12.  (a)  When x = 1
                                                                          3(1)  – 6(1)  – 3(1) + 6 = 0
                                                                              3
                                                                                   2
                                                                                          x  – x – 2
                                                                                           2
                                                                                       3
                                                                                           2
                                                                                 x – 1  x  – 2x  – x + 2
                                                                                       3
                                                                                          2
                                                                                      x  – x
                                                                                          2
                       15                                                               –x  – x
                                                                                          2
                                                                                        –x  + x
                                                                                           –2x + 2
                                                     x
                            0.25        3         5                                        –2x + 2
                                                                          y = |3x  – 6x  – 3x + 6|
                                                                                    2
                                                                                3
                                                                            = |3(x  − 2x  − x + 2)|
                                                                                     2
                                                                                3
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                                                                            = |3(x − 1)(x  − x − 2)|
                                                                                      2
                                                                            = |3(x − 1)(x + 1)(x − 2)|
                                                            [3]
                                                                          y = x  – 7x + 10
                                                                              2
                    10.  (a)   y                                          y = (x – 5)(x – 2)

                                                                                y
                                                                              10
                                                              x
                                   1    2           4
                          –24                                                 6
                                                            [3]
                                                                                                       x
                      (b)   y                                               –1    1  2       5

                                                                                                             [8]
                                                                       (b)  3(x – 1)(x + 1)(x – 2) = (x – 5)(x – 2)
                                                                                      3x  – 3 = x – 5
                                                                                        2
                          24                                                       3x  – x + 2 = 0 (rejected)
                                                                                     2
                                                            x              3(x – 1)(x + 1)(x – 2) = –(x – 5)(x – 2)
                                 1     2           4
                                                                                      3x  – 3 = –x + 5
                                                                                        2
                                                                                   3x  + x – 8 = 0
                                                                                     2
                                                                                           x = 1.47
                                                                                           y = (1.475)  – 7(1.475) + 10
                                                                                                   2
                                                            [2]                            y = 1.85
                                                                          or
                    11.                y                                                   x = –1.81
                                                                                           y = (–1.808)  – 7(–1.808) + 10
                                                                                                   2
                                                                                           y = 25.92
                                                                          Intersection points
                                                                          (1.47, 1.85), (–1.81, 25.92) and (2, 0)
                                                                                                             [4]
                                                                     13.  y = –(x + 1)(x – 1)(x – 2)
                                                                       y = –(x  – 1)(x – 2)
                                                                            2
                                                         x
                        –5            0              5                 a = –1, b = –1, c = –2
                                                                                                             [3]
                                                            [3]
                                                                     14.  2 = a(2)(–2)(–3)
                                                                       a =  1
                                                                          6
                                                                       b = 2
                                                                       c = –2
                                                                       d = –3


                         Cambridge IGCSE
                                          TM
                  172     Ace Your Additional Mathematics






         Answers Add Math.indd   172                                                                             14/03/2022   12:29 PM
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