Question

If $\left|\frac{|x|+2}{|x^2|+2|x|}\right|>\frac{1}{2}$, then $x\in$

• A. $(-2,2)-\left\{0\right\}$
• B. $(-2,0)$
• C. $(0,2)$
• D. $\left\{0\right\}$

Answer 1

The correct option is A $(-2,2)-\left\{0\right\}$

$\left|\frac{|x|+2}{|x^2|+2|x|}\right|>\frac{1}{2},x\ne 0$
$\Rightarrow \frac{|x|+2}{|x^2|+2|x|}>\frac{1}{2}$
$\Rightarrow \frac{|x^2|+2x}{|x+2|}<2$
$\Rightarrow |x^2|+2|x|<2|x|+4$
$\Rightarrow |x^2|<4\Rightarrow 0\le x^2<4\Rightarrow -2<x<2$
$\therefore x\in (-2,2)-\left\{0\right\}$