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

The correct option is B

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

Given, $x^2-|x+2|+x>\mathrm{0....}\left(i\right)$
Case I When $x+2\ge 0$
$\therefore x^2-x-2+x>0\Rightarrow x^2-2>0\Rightarrow x<-\sqrt{2}orx>\sqrt{2}\Rightarrow x\in [-2,-\sqrt{2})\cup (\sqrt{2},\infty )....\left(ii\right)$
Case II When x + 2 < 0
$\therefore x^2+x+2+x>0\Rightarrow x^2+2x+2>0\Rightarrow (x+1{)}^2+1>0$
Which is true for all x.
$\therefore x\le -2orx\in (-\infty ,-2)...\left(iii\right)$
From Eqs. (ii) and (iii), we get
$x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$