Page 6 - Pra U STPM 2022 Penggal 1 - Mathematics (T)
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Mathematics Term 1  STPM  Chapter 1 Functions

            Set P is the domain and set Q is the codomain. The set of images {1, 4, 9} is the range of this function.

                                    P             'is the square of'    Q
                                                                                                          1
                                   –2
                                   –1                                   1
                                    1                                   4
                                    2                                   9
                                    3                                   16    Range
                                  Domain                             Codomain
                                                      Figure 1.3


                 Example 1


                                            X                           Y
                                           –2
                                                                        0
                                           –1
                                                                        1
                                            0
                                                                        2
                                            1
                                                                        5
                                            2
                                                          f
              (a)  Show that the diagram above defines a function f from set X to set Y.
              (b)  Find f(x).
              (c)  State the domain of f and its range.
              Solution:           (a)  For each element x in set X, there exists a unique image y in set Y. Hence, f
                                      can be defined as a function from set X to set Y.
                                  (b)  f(0) = 1     = 0  + 1
                                                       2
                                      f(–1) = f(1) = 2 = 1  + 1
                                                       2
                                      f(–2) = f(2) = 5 = 2  + 1
                                                       2
                                      Hence, f(x) = x  + 1.
                                                   2
                                  (c)  Domain of f is {–2, –1, 0, 1, 2}
                                      Range of f is {1, 2, 5}


                 Example 2

              Determine  whether  each  equation defines  y  as a  function of  x.  If so, determine  whether  the  function is
              one-to-one.
              (a)  y = x + 2,
                      2
              (b)  y = x  + 3x + 2,
              (c)  y  = x – 2.
                   2
              Solution:           (a)  y = x + 2
                                      For each value of x  R, the value of y is unique.
                                      Thus y is a function of x.
                                      No two values of x  R have the same image.
                                      Thus the function is one-to-one.



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     01a STPM Math T T1.indd   3                                                                    3/28/18   4:20 PM
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