Question

Let S denote the set of real values of x for which
$|x^3-x|\le x\left(1\right)and\frac{2}{|x-2|}>\frac{3}{|1-2x|}\left(2\right)$
then S equals

• A. $(2,\infty )$
  
• B. $(\frac{8}{7},\sqrt{2})$
  
• C. $(\sqrt{2},\infty )$
  
• D. $(\frac{8}{7},\infty )$

Answer 1

The correct option is B

$(\frac{8}{7},\sqrt{2})$

For x<0, (1) is not satisfied.
For x=0, $\frac{1}{2}$, 2, (2) is not satisfied.
For x>0, (1) becomes
$|x^2-1|\le 1or0<x^2\le 2\iff 0<x\le \sqrt{2}.$
Now, for 0<x< $\frac{1}{2}$, (2) 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) $\iff x>\frac{8}{7}.$
thus, $xϵ(\frac{8}{7},\sqrt{2})$