Question
The set of all real numbers x for which x2−|x+2|+x>0, is
Answer: Option B
:
B
Case 1: When x + 2≥ 0 i.e/ x≥ -2,
Then given inequality becomes
x2−(x+2)+x>0 ⇒ x2−2>0 ⇒ |x|>√2
⇒therefore, in this case the part of the solution set is
[-2,-√2) ∪ (√2,∞).
Case 2: When x + 2 <0 i.e. x <-2,
Then given inequality becomes x2+(x+2)+x>0
⇒ x2+2x+2>0⇒ (x+1)2+1>0, which is true for all real x
Hence, the part of the solution set in this case is (-∞, -2). Combining the two cases, the solution set is(-∞, -2)∪[-2, -√2)∪(√2,∞)=(−∞,−√2)∪(√2, ∞)
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:
B
Case 1: When x + 2≥ 0 i.e/ x≥ -2,
Then given inequality becomes
x2−(x+2)+x>0 ⇒ x2−2>0 ⇒ |x|>√2
⇒therefore, in this case the part of the solution set is
[-2,-√2) ∪ (√2,∞).
Case 2: When x + 2 <0 i.e. x <-2,
Then given inequality becomes x2+(x+2)+x>0
⇒ x2+2x+2>0⇒ (x+1)2+1>0, which is true for all real x
Hence, the part of the solution set in this case is (-∞, -2). Combining the two cases, the solution set is(-∞, -2)∪[-2, -√2)∪(√2,∞)=(−∞,−√2)∪(√2, ∞)
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