Functions Ques 12
- Let a function $f:(0, \infty) \longrightarrow(0, \infty)$ 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
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Answer:
Correct Answer: 12.(d)
Solution: (d) We have, $f(x)=\frac{|x-1|}{x}= \begin{cases}-\frac{(x-1)}{x}, & \text { if } 0<x \leq 1 \\ \frac{x-1}{x}, & \text { if } x>1\end{cases}$
$ = \begin{cases}\frac{1}{x}-1, & \text { if } 0<x \leq 1 \\ 1-\frac{1}{x}, & \text { if } x>1\end{cases} $
Now, let us draw the graph of $y=f(x)$
Note that when $x \rightarrow 0$, then $f(x) \rightarrow \infty$,
when $x=1$, then $f(x)=0$, and when $x \rightarrow \infty$, then $f(x) \rightarrow 1$
Clearly, $f(x)$ is not injective because if $f(x)<1$, then $f$ is many one, as shown in figure.
Also, $f(x)$ is not surjective because range of $f(x)$ is $[0, \infty[$ and but in problem co-domain is $(0, \infty)$, which is wrong.
$\therefore f(x)$ is neither injective nor surjective
BETA
