Question

Solve $\frac{|x|-1}{|x|-2}\ge 0,x\ne \pm 2$

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

Answer 1

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

$\frac{|x|-1}{|x|-2}\ge 0$
Case $1:$
$|x|-1\ge 0$ and $|x|-2>0$
$\Rightarrow x\in (-\infty ,-2)\cup (2,\infty )$

Case $2:$
$|x|-1\le 0$ and $|x|-2<0$
$\Rightarrow x\in [-1,1]$

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