Question

Solve $\frac{-1}{|x|-2}\ge 1,x\ne \pm 2$

• A. $[1,2)$
• B. $(-2,-1]\cup [1,2)$
• C. $[-1,1]$
• D. $(-2,-1]$

Answer 1

The correct option is B $(-2,-1]\cup [1,2)$

$\frac{-1}{|x|-2}\ge 1\Rightarrow 1+\frac{1}{|x|-2}\le 0\Rightarrow \frac{|x|-1}{|x|-2}\le 0$
$\Rightarrow |x|-1\le 0$ and $|x|-2>0$
$\Rightarrow |x|\le 1$ and $|x|>2$ Not possible

Or, $|x|-1\ge 0$ and $|x|-2<0$
$\Rightarrow |x|\ge 1$ and $|x|<2$
$\Rightarrow x\in (-\infty ,-1]\cup [1,\infty )\text{and}x\in (-2,2)$

$\therefore x\in (-2,-1]\cup [1,2)$