Solve the following

Solution:

$A s, \frac{1}{|x|-3} \leq \frac{1}{2}$

$\Rightarrow \frac{1}{|x|-3}-\frac{1}{2} \leq 0$

$\Rightarrow \frac{2-(|x|-3)}{2(|x|-3)} \leq 0$

$\Rightarrow \frac{2-|x|+3}{2(|x|-3)} \leq 0$

$\Rightarrow \frac{5-|x|}{|x|-3} \leq 0$

Case I : When $x \geq 0,|x|=x$,

$\frac{5-x}{x-3} \leq 0$

$\Rightarrow(5-x \leq 0$ and $x-3>0)$ or $(5-x \geq 0$ and $x-3<0)$

$\Rightarrow(x \geq 5$ and $x>3)$ or $(x \leq 5$ and $x<3)$

$\Rightarrow x \geq 5$ or $x<3$

$\Rightarrow x \in(0,3) \cup[5, \infty)$

Case II : When $x<0,|x|=-x$,

$\frac{5+x}{-x-3} \leq 0$

$\Rightarrow \frac{x+5}{x+3} \geq 0$

$\Rightarrow(x+5>0$ and $x+3>0)$ or $(x+5<0$ and $x+3<0)$

$\Rightarrow(x>-5$ and $x>-3)$ or $(x<-5$ and $x<-3)$

$\Rightarrow x>-3$ or $x<-5$

$\Rightarrow x \in(-\infty,-5) \cup(-3, \infty)$

So, from both the cases, we get

$x \in(-\infty,-5) \cup(-3, \infty) \cup(0,3) \cup[5, \infty)$

$\therefore x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty)$