Question

If $\frac{|x-2|}{x-2}\ge 0$, then

• A. $x\in [2,\infty )$
• B. $x\in (2,\infty )$
• C. $x\in (-\infty ,2)$
• D. $x\in (-\infty ,2]$

Answer 1

Answer: (B)

We have, $\frac{|x-2|}{x-2}\ge 0$
Case I If $x<2$
$\frac{|x-2|}{x-2}\ge 0$
$\Rightarrow \frac{-(x-2)}{x-2}\ge 0[\because |x-2|=-(x-2)\text{, if }x<2]$
$\Rightarrow -1\ge 0$
$\text{ which is not true. }$
$\text{ Case II If }x>2$
$\frac{|x-2|}{x-2}\ge 0$
$\Rightarrow \frac{x-2}{x-2}\ge 0(\because |x-2|=x-2,\text{ if }x>2)$
$\Rightarrow 1\ge 0$
which is true.
Also, given expression is not defined at $x=2$, since denominator at $x=2$ is 0 . There is no solution at $x=2$.
Hence, $x>2$ is the required solution. i.e., $x\in (2,\infty )$