Question

Let a function $f:\left(2 , \in fty\right) \rightarrow \left[0 , \in fty\right)$ 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)

$f\left(\right.x\left.\right)=\left|1 - \frac{1}{\left(x - 2\right)}\right|$
$\because 2 < x < \infty$
$0 < x-2 < \infty\Rightarrow \infty>\frac{1}{x - 2}>0$
$-\infty < -\frac{1}{\left(x - 2\right)} < 0\Rightarrow -\in fty < 1-\frac{1}{\left(x - 2\right)} < 1$
$\because 0\leq \left|1 - \frac{1}{\left(x - 2\right)}\right| < \infty$
$\therefore $ Range of $f\left(x\right)\in \left[0 , \in fty\right)=$ co-domain
Hence, $f\left(\right.x\left.\right)$ 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)