JEE 2008

VITEEE Papers › Calculus questions › JEE 2008

Let f(x) be a non constant twice differentiable function defined on (-âˆž,âˆž) such that f(x)=f(1-x) 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€tdt=âˆ«(1/2 to 1)f(1-t)e^sinÏ€tdt

#Mathematics #Calculus #Relation & Functions

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2 Answers

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Sushovan Halder ·2014-08-05 05:27:47

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Manish Shankar ·2014-08-05 05:48:15

We have f(1/2+x)=f(1/2-x)

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

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