JEE 2008

Let f(x) be a non constant twice differentiable function defined on (-∞,∞) 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) ∫(frm -1/2 to 1/2)f(x+1/2)sinxdx=0
d) ∫(frm 0 to 1/2)f(t)esinπtdt=∫(1/2 to 1)f(1-t)esinπtdt

2 Answers

Sushovan Halder ·


Manish Shankar ·

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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