Question

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

Answer 1

Solution: The condition given in the question is x2 - |x + 2| + x > 0

Two cases are possible:

Case 1: When (x+2) ≥ 0.

Therefore, x2 - x - 2 + x > 0

Hence, x2 – 2 > 0

So, either x √2.

Hence, x ∈ [-2, -√2) ∪ (√2, ∞)     ……. (1)

Case 2: When (x+2)

Then x2 + x + 2 + x > 0

So, x2 + 2x + 2 > 0

This gives (x+1)2 + 1 > 0 and this is true for every x

Hence, x ≤ -2 or x ∈ (-∞, -2)       ...…… (2)

From equations (2) and (3) we get x ∈ (-∞, -√2) ∪ (√2, ∞).

The set of all real numbers x for which x2 - |x + 2| + x > 0 is (2002)

1. (-∞, -2) ∪ (2, ∞)                                          2. (-∞, -√2) ∪ (√2, ∞)

3. (-∞, -1) ∪ (1, ∞)                                          4. (√2, ∞)
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