Question

Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$ such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$ Then

• A. $f''( x )$ vanishes at least twice on $[0,1]$
• B. $f'\left(\frac{1}{2}\right)=0$
• C. $\int\limits_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
• D. $\int\limits_{0}^{1 / 2} f(t) e^{\sin \pi t} d t=\int\limits_{1 / 2}^{1} f(1-t) e^{\sin \pi t} d t$

Answer 1

Answer: (A), (B), (C), (D)

$( A , B , C , D )$
$f ( x )= f (1- x )$
Put $x =1 / 2+ x$

$f \left(\frac{1}{2}+ x \right)= f \left(\frac{1}{2}- x \right)$
Hence $f(x+1 / 2)$ is an even function or $f(x+1 / 2) \sin x$ is an odd function.
Also, $f'(x)=-f'(1-x)$ and for $x=1 / 2$, we have $f'(1 / 2)=0$.
Also, $\int\limits_{1 / 2}^{1} f (1- t ) e ^{\sin \pi t} d t =-\int\limits_{1 / 2}^{0} f ( y ) e ^{\sin \pi y } dy$
(obtained by putting, $\left.1- t = y \right)$.
Since $f '(1 / 4)=0, f '(3 / 4)=0$.
Also $f '(1 / 2)=0$
$\Rightarrow f ''( x )=0$ atleast twice in $[0,1]$ (Rolle's Theorem)