The exhaustive set of values of b such that {In(x2−2x+2)}2+bIn(x2−2x+2)+1>0∀x>1 is
A
(−3,∞)
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B
(−1,∞)
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C
(−2,∞)
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D
(2,∞)
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Solution
The correct option is C(−2,∞) Let t=In(x2−2x+2)=In[(x−1)2+1] Then t > 0 ∀ x > 1. So we need to find b such that t2 + bt + 1 > 0 ∀ t > 0 ⇒b>−(t+1t)⇒b>−2