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Grade: 11

                        

The set of all real numbers x for which x 2 - |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 x 2 - |x + 2| + x > 0 Two cases are possible: Case 1: When (x+2) ≥ 0. Therefore, x 2 - x - 2 + x > 0 Hence, x 2 – 2 > 0 So, either x √2. Hence, x ∈ [-2, -√2) ∪ (√2, ∞) ……. (1) Case 2: When (x+2) Then x 2 + x + 2 + x > 0 So, x 2 + 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 x 2 - |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 x 2 - |x + 2| + x > 0 Two cases are possible: Case 1: When (x+2) ≥ 0. Therefore, x 2 - x - 2 + x > 0 Hence, x 2 – 2 > 0 So, either x √2. Hence, x ∈ [-2, -√2) ∪ (√2, ∞) ……. (1) Case 2: When (x+2) Then x 2 + x + 2 + x > 0 So, x 2 + 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, ∞).

2 years ago

Answers : (1)

Umang goyal
23 Points
							X^2 -|x+2|+X >0Two cases arises Case 1 when |X+2| >0So.X^2-X-2+X >0X^2>2(X-√2)(X+√2)>0---------(1)Case 2 when|X+2|0X^2 +x+2+x>0---------(2)SOLVING both equation we get the answers....
						
2 years ago
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