Question

If $\frac{|x-3|}{x-3}$ > 0 , then

• A. $x \in (-3 , \infty)$
• B. $x \in (3 , \infty)$
• C. $x \in (2 , \infty)$
• D. $x \in (1 , \infty)$

Answer 1

Answer: (B)

$x \in (3 , \infty)$

Answer 2

We have,
$\frac{|x-3|}{x-3}>0$
Now, $\frac{|x-3|}{x-3}=\begin{cases} \frac{x-3}{x-3}=1, & x \geq 3 \\\ \frac{-(x-3)}{x-3}=-1, & x<3 \end{cases}$
$\therefore \frac{|x-3|}{x-3}>0$ only holds when $x \in(3, \infty)$