Question

Solve $\left|\frac{x-1}{x+2}\right|<2,x\ne -2$

• A. $(-\infty ,-5)\cup (-1,\infty )$
• B. $(-\infty ,-5)\cup (1,\infty )$
• C. $(-1,\infty )$
• D. $(-\infty ,-5)\cup (-1,1)$

Answer 1

The correct option is A $(-\infty ,-5)\cup (-1,\infty )$

$\frac{|x-1|}{|x+2|}<2,x\ne -2\Rightarrow \frac{|x-1|-2|x+2|}{|x+2|}<2,x\ne -2$

Here, $|x+2|>0$ always
So $|x-1|-2|x+2|<0$
For $x<-2,-x+1+2x+4<0$
$\Rightarrow x\in (-\infty ,-5)$
For $-2\le x<1,-x+1-2x-4<0$
$\Rightarrow x\in (-1,1)$
For $x\ge 1,x-1-2x-4<0$
$\Rightarrow x\in [1,\infty )$

$\therefore x\in (-\infty ,-5)\cup (-1,\infty )$