The set of all real numbers for which is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
Consider two cases based on the absolute value.
Case 1: x + 2 ≥ 0 (x ≥ -2)
Inequality becomes x² - (x+2) + x > 0 → x² - 2 > 0 → x < -√2 or x > √2.
Combined with x ≥ -2: gives -2 ≤ x < -√2 or x > √2.
Case 2: x + 2 < 0 (x < -2)
Inequality becomes x² + (x+2) + x > 0 → x² + 2x + 2 > 0 → (x+1)² + 1 > 0, which is always true.
So all x < -2 satisfy the inequality.
Union of both cases: x < -2, or -2 ≤ x < -√2, or x > √2 → simplifies to x < -√2 or x > √2, i.e., (-∞, -√2) ∪ (√2, ∞).
Case 1: x + 2 ≥ 0 (x ≥ -2)
Inequality becomes x² - (x+2) + x > 0 → x² - 2 > 0 → x < -√2 or x > √2.
Combined with x ≥ -2: gives -2 ≤ x < -√2 or x > √2.
Case 2: x + 2 < 0 (x < -2)
Inequality becomes x² + (x+2) + x > 0 → x² + 2x + 2 > 0 → (x+1)² + 1 > 0, which is always true.
So all x < -2 satisfy the inequality.
Union of both cases: x < -2, or -2 ≤ x < -√2, or x > √2 → simplifies to x < -√2 or x > √2, i.e., (-∞, -√2) ∪ (√2, ∞).
[AIEEE 2002]