Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Case 1: When x + 2 ≥ 0 i.e/ x ≥ -2,

Then given inequality becomes

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

⇒ therefore, in this case the part of the solution set is

[-2,-$\sqrt{2}$) ∪ ($\sqrt{2}$, ∞).

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

Then given inequality becomes $x^2+(x+2)+x>0$

⇒ $x^2+2x+2>0$ ⇒ $(x+1{)}^2+1>0$, which is true for all real x

Hence, the part of the solution set in this case is (-∞, -2). Combining the two cases, the solution set is (-∞, -2)∪[-2, -$\sqrt{2}$)∪($\sqrt{2},\infty )=(-\infty ,-\sqrt{2}$)∪($\sqrt{2}$, ∞)