Question

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

• A. $f\text{'}\text{'}\left(x\right)$ vanishes at least twice $[0,1]$
• B. $f\text{'}\left(\frac{1}{2}\right)=0$
• C. ${\int }_{-0.5}^{0.5}f\left(x+\frac{1}{2}\right)\sin xdx=0$
• D. ${\int }_0^{0.5}f\left(t\right)e^{\mathrm{sin\pi }t}dt={\int }_{0.5}^{1}f(1-t)e^{\mathrm{sin\pi }t}dt$

Answer 1

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

${\int }_0^{0.5}f\left(t\right)e^{\mathrm{sin\pi }t}dt={\int }_{0.5}^{1}f(1-t)e^{\mathrm{sin\pi }t}dt$

Explanation for the correct options:

Option (B):

Finding whether the given function is symmetric about $x=\frac{1}{2}$:

$$\begin{gathered} f\left(x\right)=f(1-x) \\ \Rightarrow f\left(x+\frac{1}{2}\right)=f\left(\frac{1}{2}-x\right) \end{gathered}$$

$f$ is symmetric about $x=\frac{1}{2}$Therefore $f\text{'}\left(\frac{1}{2}\right)=0$

Hence, option (B) is correct

Option(A):

$f\text{'}\left(\frac{1}{4}\right)=f\text{'}\left(\frac{3}{4}\right)=0$and $f\text{'}\left(\frac{1}{2}\right)=0$

Applying Rolle's theorem on$f\text{'}\left(x\right)$ in the intervals $\left[\frac{1}{4},\frac{1}{2}\right]$and$\left[\frac{1}{2},\frac{3}{4}\right]$

It implies that $f\text{'}\text{'}\left(x\right)$vanishes twice on$\left[0,1\right]$

Hence, option (A) is correct

Option (C):

Determining if ${\int }_{-0.5}^{0.5}f\left(x+\frac{1}{2}\right)\sin xdx=0$ is true:

$f\left(x+\frac{1}{2}\right)$is an even function

So $f\left(x+\frac{1}{2}\right)\sin x$ is an odd function

Therefore ${\int }_{-0.5}^{0.5}f\left(x+\frac{1}{2}\right)\sin xdx=0$

Hence, option (C) is correct

Option (D):

Determining if ${\int }_0^{0.5}f\left(t\right)e^{\mathrm{sin\pi }t}dt={\int }_{0.5}^{1}f(1-t)e^{\mathrm{sin\pi }t}dt$ is true:

Substitute $x=1-t$

This gives

$$\begin{gathered} {\int }_0^{0.5}f\left(x\right)e^{\mathrm{sin\pi }x}dx={\int }_{0.5}^{1}f(1-t)e^{\mathrm{sin\pi }t}dt \\ \Rightarrow {\int }_0^{0.5}f\left(t\right)e^{\mathrm{sin\pi }t}dt={\int }_{0.5}^{1}f(1-t)e^{\mathrm{sin\pi }t}dt \end{gathered}$$

Hence option (D) is correct

Therefore, the correct answers are options (A), (B), (C), and (D).