Question

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

• A. $f^{\prime \prime }\left(x\right)$ vanishes at least twice on $[0,1]$
• B. $f^{\prime }\left(\frac{1}{2}\right)=0$
• C. ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}f(x+\frac{1}{2})\cdot \sin xdx=0$
• D. ${\int }_0^{\frac{1}{2}}f\left(t\right)\cdot e^{\sin \pi t}dt={\int }_{\frac{1}{\sqrt{2}}}^{1}f(1-t)\cdot e^{\sin \pi t}dt$

Answer 1

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

The correct options are
A $f^{\prime \prime }\left(x\right)$ vanishes at least twice on $[0,1]$
B ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}f(x+\frac{1}{2})\cdot \sin xdx=0$
C $f^{\prime }\left(\frac{1}{2}\right)=0$
D ${\int }_0^{\frac{1}{2}}f\left(t\right)\cdot e^{\sin \pi t}dt={\int }_{\frac{1}{\sqrt{2}}}^{1}f(1-t)\cdot e^{\sin \pi t}dt$

Given that $f\left(x\right)=f(1-x)$
On differentiating w.r..t $x$, we get
$f^{\prime }\left(x\right)=-f^{\prime }(1-x)$
Put $x=\frac{1}{2}$

$\Rightarrow 2f^{\prime }\left(\frac{1}{2}\right)=0\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=0$
Since $f^{\prime }\left(\frac{1}{2}\right)=0$ and $f^{\prime }\left(\frac{1}{4}\right)=0$
$\Rightarrow f^{\prime \prime }\left(x\right)=0$ at two points $[0,1]$
Now, ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}f(x+\frac{1}{x})\sin xdx=0$
Since, $f(x+\frac{1}{2})sinx$ is an odd function which is clear from the following explanation.
Let $g\left(x\right)=f(x+\frac{1}{2})\sin x$, then
$g(-x)=f(\frac{1}{2}-x)\sin (-x)=-\sin xf(1-(\frac{1}{2}-x))$
$=-\sin xf(\frac{1}{2}+x)=-g\left(x\right)$
Moreover ${\int }_{\frac{1}{2}}^{1}f{(1-t)e}^{\sin \left(\pi t\right)}dt={\int }_0^{\frac{1}{2}}f\left(u\right).e^{\sin \pi u}du$, where $1-t=u$