Q50 / 75 · VITEEE 2021 · Mathematics
If $\frac{|x+3|+x}{x+2}>1$, then $x \in$
- A$(-5,-2)$
- B$(-1, \infty)$
- Correct answerC$(-5,-2) \cup(-1, \infty)$
- DNone of these
Solution
We have $\frac{|x+3|+x}{x+2}>1$
$$
\Rightarrow \frac{|x+3|+x}{x+2}-1>0 \Rightarrow \frac{|x+3|-2}{x+2}>0
$$
Now, two cases arise :
Case I : When $x+3 \geq 0$, i.e. $x \geq-3$. Then,
$$
\begin{aligned}
& \frac{|x+3|-2}{x+2}>0 \Rightarrow \frac{x+3-2}{x+2}>0 \\
& \Rightarrow \frac{x+1}{x+2}>0 \\
& \Rightarrow\{(x+1)>0 \text { and } x+2>0\} \\
& \text { or }\{x+1 < 0 \text { and } x+2 < 0\} \\
& \Rightarrow\{x>-1 \text { and } x>-2\} \\
& \text { or }\{x < -1 \text { and } x < -2\} \\
& \Rightarrow x>-1 \text { or } x < -2 \\
& \Rightarrow x \in(-1, \infty) \text { or } x \in(-\infty,-2) \\
& \Rightarrow x \in(-3,-2) \cup(-1, \infty)[\text { Since } x \geq-3]
\end{aligned}
$$
Case II : When $\mathrm{x}+3 < 0$, i.e. $\mathrm{x} < -3$
$$
\begin{aligned}
& \frac{|x+3|-2}{x+2}>0 \Rightarrow \frac{-x-3-2}{x+2}>0 \\
& \Rightarrow \frac{-(x+5)}{x+2}>0 \Rightarrow \frac{x+5}{x+2} < 0 \\
& \Rightarrow(x+5 < 0 \text { and } x+2>0) \text { or }(x+5>0) \\
& \text { and } x+2 < 0) \\
& \Rightarrow(x < -5 \text { and } x>-2) \text { or }(x>-5 \text { and } x < -2)
\end{aligned}
$$
it is not possible.
$$
\Rightarrow \mathrm{x} \in(-5,-2)
$$
Combining (i) and (ii), the required solution is $x \in(-5,-2) \cup(-1, \infty)$.
Hence, the correct answer is option: 'C'.