Properties of Symmetric Twice Differentiable Function and Its Integrals

To determine which of the given statements are correct, we analyze them step by step:

1. Analysis of Option (B):
We are given that $f(x) = f(1-x)$. Differentiating both sides with respect to $x$ using the chain rule, we obtain:
$f'(x) = -f'(1-x)$
Substituting $x = \frac{1}{2}$:
$f'\left(\frac{1}{2}\right) = -f'\left(1 - \frac{1}{2}\right) = -f'\left(\frac{1}{2}\right)$
$2f'\left(\frac{1}{2}\right) = 0 \implies f'\left(\frac{1}{2}\right) = 0$
Thus, Option (B) is correct.

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2. Analysis of Option (A):
From the relation $f'(x) = -f'(1-x)$ and the given condition $f'\left(\frac{1}{4}\right) = 0$, we find:
$f'\left(\frac{3}{4}\right) = f'\left(1 - \frac{1}{4}\right) = -f'\left(\frac{1}{4}\right) = 0$
We now have three points in $[0, 1]$ where the derivative vanishes:
$f'\left(\frac{1}{4}\right) = 0, \quad f'\left(\frac{1}{2}\right) = 0, \quad f'\left(\frac{3}{4}\right) = 0$
Since $f(x)$ is twice differentiable on $(-\infty, \infty)$, $f'(x)$ is continuous and differentiable on $[0, 1]$.
By Rolle's Theorem:

• There exists at least one $c_1 \in \left(\frac{1}{4}, \frac{1}{2}\right)$ such that $f''(c_1) = 0$.
• There exists at least one $c_2 \in \left(\frac{1}{2}, \frac{3}{4}\right)$ such that $f''(c_2) = 0$.

Since $\left(\frac{1}{4}, \frac{1}{2}\right) \subset [0, 1]$ and $\left(\frac{1}{2}, \frac{3}{4}\right) \subset [0, 1]$, $f''(x)$ vanishes at least twice on $[0, 1]$.
Thus, Option (A) is correct.

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3. Analysis of Option (C):
Consider the integral:
$I = \int_{-1/2}^{1/2} f\left(x + \frac{1}{2}\right) \sin x \, dx$
Let the integrand be $g(x) = f\left(x + \frac{1}{2}\right) \sin x$. Testing for symmetry:
$g(-x) = f\left(-x + \frac{1}{2}\right) \sin(-x) = -f\left(\frac{1}{2} - x\right) \sin x$
Using the identity $f(t) = f(1-t)$ with $t = \frac{1}{2} + x$:
$f\left(\frac{1}{2} - x\right) = f\left(1 - \left(\frac{1}{2} + x\right)\right) = f\left(\frac{1}{2} + x\right)$
Therefore:
$g(-x) = -f\left(x + \frac{1}{2}\right) \sin x = -g(x)$
Since $g(x)$ is an odd function integrated over a symmetric interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$, we have:
$I = \int_{-1/2}^{1/2} f\left(x + \frac{1}{2}\right) \sin x \, dx = 0$
Thus, Option (C) is correct.

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4. Analysis of Option (D):
Consider the integral on the right-hand side:
$J = \int_{1/2}^1 f(1-t) e^{\sin \pi t} \, dt$
Using the substitution $u = 1 - t$, we get $dt = -du$. The limits change as follows:

• When $t = \frac{1}{2}$, $u = \frac{1}{2}$.
• When $t = 1$, $u = 0$.

Substituting these into the integral gives:
$J = \int_{1/2}^0 f(u) e^{\sin(\pi(1-u))} (-du) = \int_0^{1/2} f(u) e^{\sin(\pi - \pi u)} \, du$
Using the trigonometric identity $\sin(\pi - \theta) = \sin \theta$:
$J = \int_0^{1/2} f(u) e^{\sin \pi u} \, du$
Changing the dummy variable $u$ back to $t$:
$\int_{1/2}^1 f(1-t) e^{\sin \pi t} \, dt = \int_0^{1/2} f(t) e^{\sin \pi t} \, dt$
Thus, Option (D) is correct.

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Conclusion:
All four statements (A), (B), (C), and (D) are correct.