Functions Ques 12

  1. 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



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