Question

The solution set of the inequation $\frac{|x+3|+x}{x+2}>1$ is

• A. $(-5,-2)\cup (-1,\infty )$
• B. $(-5,-2)$
• C. $(-1,\infty )$
• D. None of these

Answer 1

The correct option is A $(-5,-2)\cup (-1,\infty )$

We have,
$\frac{|x+3|+x}{x+2}$
$\Rightarrow \frac{|x+3|+x}{x+2}-1>0$
$\Rightarrow \frac{|x+3|+x-x-2}{x+2}>0$
$\Rightarrow \frac{|x+3|-2}{x+2}>0$
Now, two cases arise:
Case I: When $x+3\ge 0$, i.e., $x\ge -3$
In this case, we have
$|x+3|=x+3$
Now, $\frac{|x+3|-2}{x+2}>0$
$\Rightarrow \frac{x+3-2}{x+2}>0$
$\Rightarrow \frac{x+1}{x+2}>0$
$\Rightarrow xϵ(-\infty ,-2)\cup (-1,\infty )$
But, $x\ge -3$
$\therefore xϵ[-3,-2)\cup (-1,\infty )$
Case II When $x+3<0$, i.e., $x<-3$
In this case, we have
$|x+3|=-(x+3)$
Now, $\frac{|x+3|-2}{x+2}>0$
$\Rightarrow \frac{-(x+3)-2}{x+2}>0$
$\Rightarrow \frac{-x-5}{x+2}>0$
$\Rightarrow \frac{x+5}{x+2}<0$
$\Rightarrow -5<x<-2$
But $x<-3$.
$\therefore xϵ(-5,-3)$
Hence, the solution set of the given inequation is $(-5,-2)\cup (-1\infty )$.