First slide

Introduction to sets

Question

$The set of all real numbers x for which x^{2} - | x + 2 | + x > 0 is$

Moderate

A

$(- \infty, - 2)$
B

$(- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$
C

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

$(\sqrt{2}, \infty)$

Solution

$\begin{aligned} Case I: x + 2 \geq 0 or x \geq - 2 \\ ∴ x^{2} - 2 > 0 \\ \Rightarrow (x - \sqrt{2}) (x + \sqrt{2}) > 0 \\ \Rightarrow x < - \sqrt{2} or x > \sqrt{2} \end{aligned}$
$∴ x \in [- 2, - \sqrt{2}) \cup (\sqrt{2}, \infty) - - - - - (1)$
$Case II: x + 2 < 0 or x < - 2$
$∴ x^{2} + 2 x + 2 > 0, which is true for real x .$
$∴ x \in (- \infty, - 2) - - - - (2)$
From (1) and (2),
$x \in (- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$

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