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Question

The exhaustive set of values of b such that {In(x22x+2)}2+bIn(x22x+2)+1>0x>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(x22x+2)=In[(x1)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

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