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. $f^{\prime \prime }(x)$ vanishes at least twice on $[0,1]$
• B. $f^{\prime }\left(\frac{1}{2}\right)=0$
• C. $\underset{-1/2}{\overset{1/2}{\int }}f\left(x+\frac{1}{2}\right)\sin xdx=0$
• D. $\underset{0}{\overset{1/2}{\int }}f(t)e^{\sin \pi t}dt=\underset{1/2}{\overset{1}{\int }}f(1-t)e^{\sin \pi t}dt$

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, $f^{\prime }(x)=-f^{\prime }(1-x)$ and for $x=1/2$, we have $f^{\prime }(1/2)=0$.
Also, $\underset{1/2}{\overset{1}{\int }}f(1-t)e^{\sin \pi t}dt=-\underset{1/2}{\overset{0}{\int }}f(y)e^{\sin \pi y}dy$
(obtained by putting, $1-t=y)$.
Since $f^{\prime }(1/4)=0,f^{\prime }(3/4)=0$.
Also $f^{\prime }(1/2)=0$
$\Rightarrow f^{\prime \prime }(x)=0$ atleast twice in $[0,1]$ (Rolle's Theorem)