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

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

Case 1: If $x+2\ge 0i.e.x\ge -2$, we get

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

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

But $x\ge -2$

$\therefore xϵ[-2,-\sqrt{2}]\cup (\sqrt{2},\infty )$ ... (i)

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

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

Which is true for all x

$\therefore xϵ(-\infty ,-2)$ ... (ii)

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