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

$f \text{ twice differentiable on } \mathbb{R},\ \text{non-constant}$

$f(x) = f(1-x)\ (\text{symmetric about } x=\tfrac12),\qquad f'\!\left(\tfrac14\right) = 0$

$\text{To find: the correct statement(s) among } (1)\text{–}(4)$

$f'(x) = -f'(1-x)$

$x=\tfrac12:\quad f'\!\left(\tfrac12\right) = -f'\!\left(\tfrac12\right) \Rightarrow f'\!\left(\tfrac12\right)=0$

$x=\tfrac14:\quad f'\!\left(\tfrac14\right) = -f'\!\left(\tfrac34\right) \Rightarrow f'\!\left(\tfrac34\right)=0$

$f''(x) = f''(1-x)$

$\text{Rolle on } f':\ f''=0 \text{ in } \left(\tfrac14,\tfrac12\right) \text{ and in } \left(\tfrac12,\tfrac34\right)$

$\Rightarrow f'' \text{ vanishes at least twice on } [0,1] \Rightarrow (1)\ \text{TRUE}$

$u = x+\tfrac12:\quad \int_{-1/2}^{1/2} f(x+\tfrac12)\sin x\,dx = \int_0^1 f(u)\sin(u-\tfrac12)\,du$

$f(u) \text{ even, } \sin(u-\tfrac12) \text{ odd about } \tfrac12 \Rightarrow \text{integrand odd}$

$\Rightarrow (3)\ \text{TRUE}$

$u = 1-t:\quad \int_{1/2}^{1} f(1-t)e^{\sin\pi t}\,dt = \int_0^{1/2} f(u)e^{\sin\pi(1-u)}\,du$

$\sin\pi(1-u) = \sin\pi u \Rightarrow (4)\ \text{TRUE}$

$\boxed{\text{Correct: } (1),\ (2),\ (3),\ (4)}$