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

Answer 1

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

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

Given, $f\left(x\right)=f(1-x)$
Put $x=\frac{1}{2}+x$
$f(\frac{1}{2}+x)=f(\frac{1}{2}-x)$
Hence, $f(x+\frac{1}{2})$ is an even function or $f(x+\frac{1}{2})\sin x$ is an odd function.
Also, $f\left(x\right)=f(1-x)$ and for $x=\frac{1}{2}$, we have $f\left(\frac{1}{2}\right)=0$
Also, ${\int }_{1/2}^{1}f(1-t)e^{\sin \pi t}dt=-{\int }_{1/2}^{0}f\left(y\right)e^{\sin \pi y}dy$ (obtained by putting, $1-t=y$).
Since $f^{\prime }\left(1/4\right)=0,f^{\prime }\left(\frac{3}{4}\right)=0$.

Also $f\left(\frac{1}{2}\right)=0$
$\Rightarrow f^{\prime }\left(x\right)=0$ atleast twice in $[0,1]$ (Rolle's Theorem)