Question

If $f\left(x\right)={\log }_{0.1}\left(\frac{1-|x|}{1+|x|}\right),$ then the number of integral values of $x,$ satisfying $f\left(x\right)\ge f\left(x^2\right)$ is

Answer 1

$f\left(x\right)\ge f\left(x^2\right)$
$\Rightarrow {\log }_{0.1}\left(\frac{1-|x|}{1+|x|}\right)\ge {\log }_{0.1}\left(\frac{1-x^2}{1+x^2}\right)$
$\Rightarrow \frac{1-|x|}{1+|x|}\le \frac{1-x^2}{1+x^2}$
$\Rightarrow \frac{1-|x|}{1+|x|}-\frac{(1-|x|)(1+|x|)}{1+x^2}\le 0$
$\Rightarrow (1-|x|)[\frac{1}{1+|x|}-\frac{1+|x|}{1+x^2}]\le 0$
$\Rightarrow (1-|x|)\left[\frac{-2|x|}{(1+|x|)(1+x^2)}\right]\le 0$
$\Rightarrow (1-|x|)\ge 0$
$\Rightarrow (|x|-1)\le 0$
$\Rightarrow x\in [-1,1]$
But for $x=\pm 1,f\left(x\right)$ is not defined.
$\therefore x=0$ is the only solution.