Question

The number of integral solutions of the inequality $|1-\frac{|x|}{1+|x|}|\ge \frac{1}{2},$ is

Answer 1

$|1-\frac{|x|}{1+|x|}|\ge \frac{1}{2}\Rightarrow \left|\frac{1}{1+|x|}\right|\ge \frac{1}{2}\Rightarrow \frac{1}{1+|x|}\ge \frac{1}{2}(\because 1+|x|\ge 1)\Rightarrow 1+|x|\le 2\Rightarrow |x|\le 1\therefore x\in [-1,1]$
Hence, number of integral solutions is $3.$