question_answer 21)

Show that the signum function given by is neither one-one nor onto.

Answer:

      Let x1 = 1 and x2 = 3 be two real numbers       f(x1) = f(1) = 1; f(x2) = f(3) = 1              is not one-one.       Also, let y = 2             But range of f is {?1, 0, 1}       Clearly, there is no value of x for which f(x) = 2        is not onto.       Hence function is neither one-one nor onto.