Question

If $\left|\frac{|x|-1}{|x|-2}\right|=4$, then $x\in$

• A. $\{-\frac{9}{5},-\frac{7}{3},\frac{7}{3},\frac{9}{5}\}$
• B. $\{\frac{7}{3},\frac{9}{5}\}$
• C. $\{-\frac{7}{3},\frac{7}{3}\}$
• D. $\{-\frac{9}{5},\frac{9}{5}\}$

Answer 1

The correct option is A $\{-\frac{9}{5},-\frac{7}{3},\frac{7}{3},\frac{9}{5}\}$

$\left|\frac{|x|-1}{|x|-2}\right|=4$
$\Rightarrow |x|-1=\pm 4(|x|-2),x\ne \pm 2$
$\Rightarrow 5|x|=9$ or $3|x|=7,x\ne \pm 2$
$\Rightarrow x=\pm \frac{9}{5}$ or $x=\pm \frac{7}{3},x\ne \pm 2$
$\therefore x=\pm \frac{7}{3},\pm \frac{9}{5}$