Question

Let $S$ denote the set of real values of $x$ for which $|x^3-x|\le x\&\frac{2}{|x-2|}>\frac{3}{|1-2x|},$ then $S$ is equal to

Answer 1

We have given
$|x^3-x|\le x$
For $x<0,$ It is not satisfied.
and
For $x>0$
$\Rightarrow |x^2-1|\le 1\Rightarrow 0<x^2\le 2\Rightarrow 0<x\le \sqrt{2}\cdots \left(1\right)$
Also we have given
$\frac{2}{|x-2|}>\frac{3}{|1-2x|}$

For $x=0,\text{and}(\frac{1}{2},2)$ is not satisfied.

Now, for $0<x<\frac{1}{2},\left(2\right)$ becomes
$2(1-2x)>3(2-x)$ $\iff$ x<-4
and for $\frac{1}{2}<x\le \sqrt{2},$ (2) reads as
$2(2x-1)>3(2-x)$
$\Rightarrow x>\frac{8}{7}$
thus, $x\in (\frac{8}{7},\sqrt{2})$