Page 40 - ACE YR IGCSE A TOP APPR' TO ADD MATH
P. 40

3
                         2
                    15.   6x  + 8x – 4   = 10                          (b)  √x y 2
                           x 2                                                1
                                                                              3
                                                                                 2
                        6x  + 8x − 4 = 10x 2                              = (x )(y )
                         2
                                                                               1
                         x  − 2x + 1 = 0                                  = (2p)  (8 )
                         2
                                                                               3
                                                                                  q 2
                           (x − 1)  = 0                                      p
                                2
                                x = 1                                     = 2  · 2
                                                                             3
                                                                                6q
 =  1                                                       [4]           = 2 p 3  + 6q
 2                  16.     8 · 5 x − 1  + 2 = 5 x + 1                                                       [3]
                                   x
                            −1
                         x
                                      1
 = √3                 8 · 5  · 5  + 2 = 5  · 5                             1
                                                                            x
                      Let 5  = y                                     21.  Let e  = y
                                                                           2
                          x
 = √(7)  –            8y                                                        13y   = 14 + 19y
 2
                                                                                   2
                         log x Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
                                                                          2
                       5   + 2  = 5y                                   13y  − 19y − 14 = 0
                                                                                        7
 2
 2
 =  1 8  + 3             17y  = 10                                                y  = –  13  or y = 2
                              10
 2
 1                         y  =  17                                               1 2 x  7           1 2 x
 = 2  2                       10                                                 e   = –  13  (rejected) or e  = 2
                           x
 =                        5   =  17                                                                  1 x = ln2

                           x  = −0.3297                                                              2 x = 2 ln 2
                                                            [4]                                              [4]
                    17.  y = 2 , log y = b                           22.  2x log  e = log √2 −  3x log(2e  + 1)
                          b
                              2
                                                                                               3
                          x
                           2
                                                                            4
                                                                                  3
                      log 2 3 4                                        2x log  e + 3x log(2e  + 1) = log √2
                                                                                       3
                          8y
                                                                                               3
                                                                            4
                                                                                      3
                      = log x  − log 8y                                x[2 log e + 3 log(2e  + 1)] = log √2
                            2
                                                                                                3
                                                                            4
                          2
                                 2
                                    3
                      = 2 log x − [log 2  + log y]                     x =       log √2
                                                                                   3
                                  2
                                         2
                           2
                      = 2a − 3 − b                                         2 log e + 3 log(2e  + 1)
                                                                                        3
                                                                               4
                                                            [3]          = 0.0502
                                       2
                    18.  9e  − e  x  = 27 − 3x                                                               [3]
                            3x
                              2
                        3x
                        e (9 − x ) = 3(9 − x )
                              2
                                      2
                        3x
                              3x
                                             2
                             e  = 3     or  9 − x  = 0               23.  (a)
                              x =  ln 3   or   x = ±3                     f(x)
                                  3                         [3]           8
                    19.  Let y = log x                                    7
                              5
                                    3 log 5
                         2 log x = 1 +   log x                            6     f(x) = 2In (4x – 3)
                                        5
                             5
                             2y = 1 +  3  5                               5
                                    y
                      2y  – y – 3 = 0                                     4                         e    + 3
                        2
                                                                                                     x
                                                                                                     2
                                 3
                                                                                               –1
                             y =   or y = –1                              3                    f (x) =  4
                                 2
                             x = √5  or x =  1                            2
                                   3
                                         5                  [4]           1
                    20.  (a)  log 16y                                                                       x
                            x
                        = log 16 + log y                                  0    1  2   3   4   5   6   7  8
                                   x
                            x
                        4 log 2   log y                                                                      [2]
                        =   2    +   2
                                                                               −1
                                    2
                            2     log x                                (b)  Let f (x) = y
                             =  4  +  3q                                  2 ln (4y − 3)  = x
                               p   p                                                     x
                             =  4 + 3q                                      ln (4y − 3)  = ln e  2
                                 p                                                      x 2
                                                            [4]                 f (x)  =    e  + 3
                                                                                –1
                                                                                         4
                                                                                                             [3]
                         Cambridge IGCSE
                                          TM
                  180     Ace Your Additional Mathematics
         Answers Add Math.indd   180                                                                             14/03/2022   12:29 PM
   35   36   37   38   39   40   41   42   43   44   45