Question

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

Options

• `(-oo, -2) ∪ (2, oo)`

• `(- oo, - sqrt(2)) ∪ (sqrt(2), oo)`

• `(- oo, -1) ∪ (1 , oo)`

• `(sqrt(2), oo)`

Answer 1

`(- oo, - sqrt(2)) ∪ (sqrt(2), oo)`

Explanation:

Case - I: If x ≥ – 2

∴ x² – (x + 2) + x > 0

⇒ x² – x – 2 + x > 0

⇒ x² – 270

⇒ `(x - sqrt(2)) (x + sqrt(2)) > 0`

∴ `x ∈ [-2, - sqrt(2)] ∪ (sqrt(2), oo)` ......(i)

Case - II: If x < – 2

x² + x + 2 + x > 0

⇒ x² + 2x + 2 > 0

⇒ (x + 1)² + 1 > 0

⇒ x ∈ R `(- oo, - 2)`  ......(ii)

From (i) and (ii)

`x ∈ (- oo, - sqrt(2)) ∪ (sqrt(2), oo)`