Question

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

• A. $( - \infty, - 2) \cup (2,\infty)$
• B. $( - \infty, - \sqrt{2}) \cup (\sqrt{2},\infty)$
• C. $( - \infty, - 1) \cup (1,\infty)$
• D. $(\sqrt{2},\infty)$

Answer 1

Answer: (B)

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

Answer 2

Sol. Case I: If $x + 2 \geq 0$ i.e. $x \geq - 2$, we get

$x^{2} - x - 2 + x > 0$ ⇒ $x^{2} - 2 > 0$ ⇒ $(x - \sqrt{2})(x + \sqrt{2}) > 0$

⇒ $x \in ( - \infty, - \sqrt{2}) \cup (\sqrt{2},\infty)$

But $x \geq - 2$

∴ $x \in \lbrack - 2, - \sqrt{2}) \cup (\sqrt{2}\infty)$ …..(i)

Case II: $x + 2 < 0$ i.e. $x < - 2$, then

$x^{2} + x + 2 + x > 0$ ⇒ $x^{2} + 2x + 2 > 0$ ⇒ $(x + 1)^{2} + 1 > 0$.

Which is true for all x

∴ $x \in ( - \infty, - 2)$ …..(ii)

From (i) and (ii), we get, $x \in ( - \infty, - \sqrt{2}) \cup (\sqrt{2},\infty)$