Question

The solution set of $\frac{x-2}{x+5}>2$ is

• A. $(-\infty ,-12)$
• B. $[-12,-5]$
• C. $(-12,-5)$
• D. $(-5,\infty )$

Answer 1

Answer: (C)

We have, $\frac{x-2}{x+5}>2$
Subtracting 2 from both sides,
$\frac{x-2}{x+5}-2>0$
$\Rightarrow \frac{x-2-2(x+5)}{x+5}>0$
$\Rightarrow \frac{x-2-2x-10}{x+5}>0$
$\Rightarrow \frac{-(x+12)}{x+5}>0$
Multiply by $(-1)$ on both sides,
$\frac{x+12}{x+5}<0$
Given inequality will be less than 0 in two cases.
Case I
$(x+12)>0\text{ and }x+5<0$
$\Rightarrow x>-12\text{ and }x<-5$
$(\because \frac{a}{b}>0\Rightarrow a\text{ and }b\text{ are of opposite signs })$
Case II
$(x+12)<0\text{ and }x+5>0$
$\text{ or }x<-12\text{ and }x>-5\text{ (not possible) }$
$\therefore -12<x<-5$
$\text{ i.e., }x\in (-12,-5)$