Question

The number of integral solutions of the inequality ${\left(\frac{1}{3}\right)}^{\frac{|x+2|}{2-|x|}}>9$ is

Answer 1

${\left(\frac{1}{3}\right)}^{\frac{|x+2|}{2-|x|}}>9$
$\Rightarrow 3^{\frac{|x+2|}{2-|x|}}>3^2$
$\Rightarrow \frac{|x+2|}{2-|x|}>2$
$\Rightarrow \frac{|x+2|}{2-|x|}<-2$

$\text{Case I:}x<-2$
$\frac{-(x+2)}{2+x}<-2$
$\Rightarrow -1<-2$, which is not true.

$\text{Case II:}-2<x<0$
$\frac{(x+2)}{2+x}<-2$
$\Rightarrow 1<-2$, which is not true.

$\text{Case III:}x\ge 0$
$\frac{x+2}{2-x}<-2$
$\Rightarrow \frac{x+2}{2-x}+2<0$
$\Rightarrow \frac{6-x}{2-x}<0$
$\Rightarrow 2<x<6$
Hence, the solution is $(2,6)$.