Let f, g : R → R be defined as : f(x) = |x - 1| and g(x) = {{e^x, x ≥ 0}, {x+1, x ≤ 0}.

Question

Let \(f, \mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}\) be defined as : \(f(\mathrm{x})=|\mathrm{x}-1|\) and \(\mathrm{g}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{e}^{\mathrm{x}}, & \mathrm{x} \geq 0 \\ \mathrm{x}+1, & \mathrm{x} \leq 0\end{array}\right.\). Then the function \(f(\mathrm{g}(\mathrm{x}))\) is

(1) neither one-one nor onto.

(2) one-one but not onto.

(3) both one-one and onto.

(4) onto but not one-one.

Answer 1

Correct option is (1) neither one-one nor onto.

\(f(x) = \begin{cases} 1-x,& x<1\\x-1,& x\ge 1\end{cases}\)

\(g(x) = \begin{cases}e^x,& x\ge 0\\x+1,& x\le 0\end{cases}\)

\(f(g(x)) =\begin{cases}1-(x+1), & x< 0\\e^x -1,&x\ge 0 \end{cases}\)

neither one-one nor onto.