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Question

Let f(x) be a non-constant twice differentiable function definied on (,) such that f(x)=f(1x) and
f(14)=0. Then

A
f′′(x) vanishes at least twice on [0,1]
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B
f(12)=0
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C
1212f(x+12)sinxdx=0
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D
120f(t)esinπtdt=112f(1t)esinπt dt
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Solution

The correct option is D 120f(t)esinπtdt=112f(1t)esinπt dt
f(x) is a non constant twice differentiable function that f(x)=f(1x)f(x)=f(1x)
For x=12, we get f(12)=f(112)
f(12)+f(12)=0f(12)=0
(b) is correct.

For x=14, we get f(14)=f(34)
but given that f(14)=0f(14)=f(34)=0
Hence, f'(x) satisfies all conditions of Rolle's theorem for x[14,12] and [12,34], So there exists at least one point C1(14,12) and at least one point C2(12,34) such that f"(C1)=0 and f"(C2)=0
f"(x)varishes at least twice on [0,1] (a)is correct.
Also using f(x)=f(1-x)
f(x+12)=f(1x12)=f(x+12)f(x+12) is an odd function.1212f(x+12)sinxdx=0 (c)is correct.
Substitute t=1xdt=dx
120f(t)esinπtdt=120f(1x)esin(ππt)dx=112f(1t)esinπt dt

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