The solution set of inequality $\frac{| x + 2 | - x}{x} < 2 is$

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$(- \infty, 0) \cup (1, \infty)$

[0, 2)

(0, 1)

$(- \infty, 0) \cup (2, \infty)$

answer is C.

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Detailed Solution

$\frac{| x + 2 | - x - 2 x}{x} < 0$

$(| x + 2 | - 3 x) x < 0 ........... (1)$
Case (i) : $x + 2 \geq 0 \Rightarrow x \geq - 2$
Then (1) becomes
$(- 2 x + 2) x < 0 \Rightarrow (x - 1) x > 0$
$\Rightarrow (- \infty, 0) \cup (1, \infty) \Rightarrow [- 2, 0) \cup (1, \infty) (∴ x \geq - 2)$
Case (ii): $x + 2 < 0 \Rightarrow x < - 2$
Then (1) becomes
$(- x - 2 - 3 x) x < 0 \Rightarrow (- 4 x - 2) x < 0$
$\Rightarrow (2 x + 1) x > 0 \Rightarrow x \in (- \infty, - \frac{1}{2}) \cup (0, \infty)$
$x \in (- \infty, - 2) (∵ x < - 2)$
Hence $x \in (- \infty, 0) \cup (1, \infty)$