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Let f(x) be a non constant twice differentiable function defined on (âˆž,âˆž) such that f(x)=f(1x) and f'(1/4)=0.Then
a) f''(x) vanishes at least twice on [0,1]
b) f'(1/2)=0
c) âˆ«(frm 1/2 to 1/2)f(x+1/2)sinxdx=0
d) âˆ«(frm 0 to 1/2)f(t)e^{sinÏ€t}dt=âˆ«(1/2 to 1)f(1t)e^{sinÏ€t}dt

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2 Answers
Manish Shankar
·20140805 05:48:15
We have f(1/2+x)=f(1/2x)
So graph is symmetrical about x=1/2
So f'(1/2)=0 as it is differentiable at x=1/2,
Also we have f'(3/4)=0 as it is symmetrical about x=12