Question

Let a function $f:\left(2,\in fty\right)\to \left[0,\in fty\right)$ defined as $f\left(x\right)=\left|\frac{x-3}{x-2}\right|$ , then $f$ is

• A. injective & surjective
• B. not injective but surjective
• C. injective but not surjective
• D. neither injective nor surjective

Answer 1

Answer: (B)

$f(x)=\left|1-\frac{1}{\left(x-2\right)}\right|$
$\because 2<x<\infty$
$0<x-2<\infty \Rightarrow \infty >\frac{1}{x-2}>0$
$-\infty <-\frac{1}{\left(x-2\right)}<0\Rightarrow -\in fty<1-\frac{1}{\left(x-2\right)}<1$
$\because 0\le \left|1-\frac{1}{\left(x-2\right)}\right|<\infty$
$\therefore$ Range of $f\left(x\right)\in \left[0,\in fty\right)=$ co-domain
Hence, $f(x)$ is surjective
Let, $f\left(x\right)=\frac{1}{2}\Rightarrow \left|1-\frac{1}{x-2}\right|=\frac{1}{2}$
$1-\frac{1}{\left(x-2\right)}=\frac{1}{2},1-\frac{1}{\left(x-2\right)}=-\frac{1}{2}$
$\frac{1}{x-2}=\frac{1}{2},\frac{1}{x-2}=\frac{3}{2}$
$x=4,x=\frac{8}{3}$
$\therefore f\left(4\right)=f\left(\frac{8}{3}\right)=\frac{1}{2}$
$\therefore f\left(x\right)$ is many-one (not injective)