question_answer

Let a function\[f:(0,\infty )\to (0,\infty )\]be defined by\[f(x)=\left| 1-\frac{1}{x} \right|.\]Then f is

A) injective only

B) both injective as well as surjective

C) not injective but it is surjective

D) neither injective nor surjective

E) None of these

Correct Answer: E

Solution :

Here, \[f(x)=\left| 1-\frac{1}{x} \right|=\left| \frac{x-1}{x} \right|\]\[=\left\{ \begin{align}   & \frac{1-x}{x},\,\,\,\,\,0<x<1 \\  & \frac{x-1}{x}\,\,\,\,\,\,\,x\ge 1 \\ \end{align} \right.\] \[\therefore \]f(x) is not injective but range of function is\[[0,\infty )\] Note: If co-domain is \[[0,\infty )\], then f(x) will be surjective.