Let f(x) be a non-constant twice differentiable function defined on `(oo, oo)` 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) `∫_{-1/2}^{1/2}f(x+1/2)sinxdx=0`
(d) `∫_{0}^{1/2}f(t)e^sinxtdt=∫_{1/2}^{1}f(1−t)e^sinπtdt`

Options

A. `f'(x)` vanishes at least twice on `[0,1]`
B. `f'(1/2)=0`
C. `int_(-1//2)^(1//2)f(x+1/2)sinxdx=0`
D. `int_(0)^(1//2)f(t)e^(sinxt)dt=int_(t//2)^(1)f(1-t)e^(sinpit)dt`

Answer

A, B, C, D

Explanation

`f(x)=f(1-x)`
Replacing `x` by `1/2+x`, we get
`f(1/2+x)=f(1/2-x)`…………..1
Henc `f(x+1//2)` is an even function or `f(x+1//2)sinx` is an odd function.
Also `f'(x)=f'(1-x)`……………2
and for `x=1//2` we have `f'(1//2)=0`
Also `int_(1//2)^(1)f(1-t)e^(sinpit)dt=-int_(1//2)^(0)f(y)e^(sinpiy) dy`
(by putting `1-t=y`)
Since `f'(1//4)=0, f'(3//4)=0` [from equation (2)].
Also `f'(1//2)=0` [from equation (2)]
Thus `f'(x)=0` at least in `[0,1]` (by Rolle's theorem).