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Question

The function f(x)=3+2(a+1)x+(a2+1)x2x3 has a local minimum at x=x1 and local maximum at x=x2 such that x1<2<x2, then a belongs to interval(s).

A
(,32)
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B
(32,1)
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C
(0,)
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D
(1,)
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Solution

The correct options are
A (,32)
D (1,)
Since, x1 & x2 are local minima & local maxima of f(x)=3+2(a+1)x+(a2+1)x2x3
Therefore, x1 & x2 are the roots of the equation f(x)=0
3x2+2(a2+1)x+2(a+1)=0
Since, 2 lies between the roots of the above quadratic equation
Therefore, f(2)>0
12+4(a2+1)+2(a+1)>0
Therefore, a belongs to (,32) & (1,)
Ans: A,D

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