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Question

Let f(x)=x0{(a1)(t2+t+1)2(a+1)(t4+t2+1)}dt. Then the set of values of a for which f(x)=0 has two distinct real roots is:

A
(,0)
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B
(,1)(1,)
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C
(,2)(2,)
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D
(,2)(1,)
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Solution

The correct option is C (,2)(2,)
Differentitating the given equation using Leibnitz theorem, we get
f(x)=(a1)(x2+x+1)2(a+1)(x2+x+1)(x2x+1).
Now, f(x)=0(a1)(x2+x+1)(a+1)(x2x+1)=0x2ax+1=0
For distinct real roots
D>0a24>0a2>4a(,2)(2,)

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