Question

The set of all real numbers x for which x² - |x + 2| + x > 0, is

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

Answer 1

Answer: (B)

(-∞, - $\sqrt { 2 }$ ) ∪ ( $\sqrt { 2 }$ , ∞)

Answer 2

For x < - 2, |x + 2| = - (x + 2) and the inequality becomes

x² + x + 2 + x > 0 ⇒ (x + 1)² + 1 > 0

which is valid $\forall$ x ∈ R but x < - 2

∴ x ∈ (- ∞, - 2) (i)

For x ≥ - 2 |x + 2| = x + 2

and the inequality becomes

x² – x – 2 + x > 0 ⇒ x² > 2

⇒ x > $\sqrt { 2 }$ or x < - $\sqrt { 2 }$

i.e, x ∈ (- ∞, - $\sqrt { 2 }$ ) ∪ ($\sqrt { 2 }$, ∞)

but x ≥ - 2 ⇒ x ∈ [- 2, - $\sqrt { 2 }$ ) ∪ ( $\sqrt { 2 }$ , ∞] (ii)

From (1) and (2) x ∈ ( - ∞, - 2) ∪ [- 2, - $\sqrt { 2 }$ ) ∪

($\sqrt { 2 }$, ∞)

⇒ x ∈ ( - ∞, -$\sqrt { 2 }$) ∪ ($\sqrt { 2 }$, ∞)