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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,
$$f''(x)$$ vanishes at least twice on [0, 1]
$$f'\left(\frac{1}{2}\right) = 0$$
$$\int_{-\frac{1}{2}}^{\frac{1}{2}}f\left(x + \frac{1}{2}\right)\sin x dx = 0 $$
$$\int_{0}^{\frac{1}{2}}f(t)e^{\sin \pi t} dt = \int_{\frac{1}{2}}^{1}f(1 - t)e^{\sin \pi t} dt$$
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