Question

If $\frac{|x-2|}{x-2} \geq 0$, 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, $\frac{|x-2|}{x-2} \geq 0$
Case I If $x<2$
$\frac{|x-2|}{x-2} \geq 0$
$\Rightarrow \frac{-(x-2)}{x-2} \geq 0 [\because|x-2|=-(x-2) \text {, if } x<2]$
$\Rightarrow -1 \geq 0$
$\text { which is not true. }$
$\text { Case II If } x >2$
$\frac{|x-2|}{x-2} \geq 0$
$\Rightarrow \frac{x-2}{x-2} \geq 0 (\because|x-2|=x-2, \text { if } x>2)$
$\Rightarrow 1 \geq 0$
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., $x \in(2, \infty)$