Question

Let $f\left(x\right)$ be a non-constant twice differentiable function definied 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. ${\int }_{\frac{-1}{2}}^{\frac{1}{2}}f(x+\frac{1}{2})\sin xdx=0$
• D. ${\int }_0^{\frac{1}{2}}f\left(t\right)e^{\sin \pi t}dt={\int }_{\frac{1}{2}}^{1}f(1-t)e^{\sin \pi t}dt$

Answer 1

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

The correct options are
A $f^{\prime \prime }\left(x\right)$ vanishes at least twice on $[0,1]$
B $f^{\prime }\left(\frac{1}{2}\right)=0$

C ${\int }_{\frac{-1}{2}}^{\frac{1}{2}}f(x+\frac{1}{2})\sin xdx=0$
D ${\int }_0^{\frac{1}{2}}f\left(t\right)e^{\sin \pi t}dt={\int }_{\frac{1}{2}}^{1}f(1-t)e^{\sin \pi t}dt$

$\therefore f\left(x\right)$ is a non constant twice differentiable function that $f\left(x\right)=f(1-x)\Rightarrow f^{\prime }\left(x\right)=-f^{\prime }(1-x)$
For $x=\frac{1}{2},\text{we get}{f}^{\prime }\left(\frac{1}{2}\right)=-f^{\prime }(1-\frac{1}{2})$
$\Rightarrow f^{\prime }\left(\frac{1}{2}\right)+f^{\prime }\left(\frac{1}{2}\right)=0\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=0$
$\Rightarrow$ (b) is correct.

For $x=\frac{1}{4},\text{we get}{f}^{\prime }\left(\frac{1}{4}\right)=f^{\prime }\left(\frac{3}{4}\right)$
but given that $f^{\prime }\left(\frac{1}{4}\right)=0\therefore f^{\prime }\left(\frac{1}{4}\right)=f^{\prime }\left(\frac{3}{4}\right)=0$
Hence, f'(x) satisfies all conditions of Rolle's theorem for $x\in [\frac{1}{4},\frac{1}{2}]$ and $[\frac{1}{2},\frac{3}{4}],$ So there exists at least one point $C_1\in (\frac{1}{4},\frac{1}{2})$ and at least one point $C_2\in (\frac{1}{2},\frac{3}{4})$ such that $f"\left(C_1\right)=0\text{and}f"\left(C_2\right)=0$
$\therefore f"\left(x\right)$varishes at least twice on [0,1] $\Rightarrow$ (a)is correct.
Also using f(x)=f(1-x)
$\Rightarrow f(x+\frac{1}{2})=f(1-x-\frac{1}{2})=f(-x+\frac{1}{2})\Rightarrow f(x+\frac{1}{2})\text{is an odd function.}\Rightarrow {\int }_{-\frac{1}{2}}^{\frac{1}{2}}f(x+\frac{1}{2})sinxdx=0\therefore$ (c)is correct.
Substitute $t=1-x\Rightarrow dt=-dx$
$\therefore {\int }_0^{\frac{1}{2}}f\left(t\right)e^{\sin \pi t}dt={\int }_0^{\frac{1}{2}}f-(1-x)e^{\sin (\pi -\pi t)}dx={\int }_{\frac{1}{2}}^{1}f(1-t)e^{\sin \pi t}dt$