Let f, g : R R be defined as : f(x) = |x - 1| and g(x) = cases ex, & x 0 \ x + 1, & x 0. cases Then the function f(g(x)) is
- neither one-one nor onto.
- one-one but not onto.
- both one-one and onto.
- onto but not one-one.
Answer
(1) neither one-one nor onto.
Solution
f(g(x)) = |g(x) - 1| = cases ex - 1, & x 0 \ -x, & x 0 cases
both pieces are 0 range = [0, ) R not onto
f(g(1)) = e - 1 and f(g(1 - e)) = e - 1 two inputs, one output not one-one
(Graph: -x falls to 0 on (- , 0], ex - 1 rises from 0 on [0, ) — a V shape.)
Answer: (1)