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The set of all real numbers x for which x2 - |x + 2| + x > 0 is 1. (-∞, -2) ∪ (2, ∞) 2. (-∞, -√2) ∪ (√2, ∞) 3. (-∞, -1) ∪ (1, ∞) 4. (√2, ∞)
Real numbers are the combination of rational numbers as well as irrational numbers.
Answer: Option (2) , x ∈ (-∞, -√2) ∪ (√2, ∞)
Let us proceed step by step
Explanation:
From the given data, x2 – |x + 2| + x > 0
We can discuss this problem by observing two different cases.
Case 1: When (x + 2) ≥ 0.
Therefore, x2 – x – 2 + x > 0
Hence, x2 – 2 > 0
x2 > 2
⇒ x < -√2 or x > √2
Hence, x ∈ (-∞, -√2) ∪ (√2, ∞) ……. (1)
Case 2: When (x + 2) < 0
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 (Because (x + 1)2 ≥ 0 ⇒(x + 1)2 + 1 ≥ 1 > 0)
Hence, x ∈ (-∞, ∞) ……… (2)
From equations (1) and (2), we take the intersection, hence we get x ∈ (-∞, -√2) ∪ (√2, ∞).
Therefore, x ∈ (-∞, -√2) ∪ (√2, ∞) is the required answer.
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