If | x + 3 | + x x + 2 1 , then x

Question

If $\frac{| x + 3 | + x}{x + 2} > 1$ , then $x \in$

Smart Achievers · Accepted Answer

(3)

$We have,$ $\frac{| x + 3 | + x}{x + 2} > 1$

$\Rightarrow$ $\frac{| x + 3 | + x}{x + 2} - 1 > 0 \Rightarrow \frac{| x + 3 | - 2}{x + 2} > 0$

$Now, two cases arise:$

$Case I: When x + 3 \geq 0, i.e., x \geq - 3 .$

$Then,$ $\frac{| x + 3 | - 2}{x + 2} > 0 \Rightarrow \frac{x + 3 - 2}{x + 2} > 0 \Rightarrow \frac{x + 1}{x + 2} > 0$

$\Rightarrow$ ${x + 1 > 0 and x + 2 > 0}$ $or$ ${x + 1 < 0 and x + 2 < 0}$

$\Rightarrow$ ${x > - 1 and x > - 2}$ $or$ ${x < - 1 and x < - 2}$

$\Rightarrow$ $x > - 1 or x < - 2$

$\Rightarrow$ $x \in (- 1, \infty) or x \in (- \infty, - 2)$

$\Rightarrow$ $x \in [- 3, - 2) \cup (- 1, \infty)$ $[Since x \geq - 3]$ ...(i)

$Case II: When x + 3 < 0, i.e., x < - 3 .$

$Then, \frac{| x + 3 | - 2}{x + 2} > 0 \Rightarrow \frac{- x - 3 - 2}{x + 2} > 0$

$\Rightarrow \frac{- (x + 5)}{x + 2} > 0 \Rightarrow \frac{x + 5}{x + 2} < 0$

$\Rightarrow$ ${x + 5 < 0 and x + 2 > 0}$ $or$ ${x + 5 > 0 and x + 2 < 0}$

$\Rightarrow$ ${x < - 5 and x > - 2}$ $which is not possible,$

$or$ ${x > - 5 and x < - 2}$

$\Rightarrow$ $x \in (- 5, - 2)$

$\Rightarrow$ $x \in (- 5, - 3)$ $[Since x < - 3]$ ...(ii)

$Combining (i) and (ii), the required solution is$

$x \in (- 5, - 2) \cup (- 1, \infty)$

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