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, ∞)
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, ∞)
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, ∞)
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, ∞)
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, ∞).
