$$f(x)$$ is a non constant twice differential function such that

$$f(x) = f(1 - x) \Rightarrow f'(x) = - f'(1 - x)$$ .... (i)

For $$x = {1 \over 2}$$

We get $$f'\left( {{1 \over 2}} \right) = - f'\left( {1 - {1 \over 2}} \right)$$

$$ = f'\left( {{1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) = 0 \Rightarrow f'\left( {{1 \over 2}} \right) = 0$$

For $$x = {1 \over 4}$$, we get $$f'\left( {{1 \over 4}} \right) = - f'\left( {{3 \over 4}} \right)$$

But given that $$f'\left( {{1 \over 4}} \right) = 0:f'\left( {{1 \over 4}} \right) = f'\left( {{3 \over 4}} \right) = 0$$

Hence $$f'(x)$$ satisfies all conditions of roller's theorem for $$x \in \left[ {{1 \over 4},{1 \over 2}} \right]$$ and $$\left[ {{1 \over 2},{3 \over 4}} \right]$$, so there exist at least one point $${C_1} \in \left( {{1 \over 4},{1 \over 2}} \right)$$ and at least one point $${C_2} \in \left( {{1 \over 2},{3 \over 4}} \right)$$ such that

$$f''(G) = 0$$ and $$f''{(C)_2} = 0$$

$$\therefore$$ $$f''(x)$$ vanishes at least twice on [0, 1]

Also using $$f(x) = f(1 - x)$$

$$f\left( {x + {1 \over 2}} \right) = f\left( {1 - x - {1 \over 2}} \right) = f\left( { - x + {1 \over 2}} \right)$$

$$f'\left( {x + {1 \over 2}} \right)$$ is an odd function

$$ \Rightarrow \sin x\,f\left( {x + {1 \over x}} \right)$$ is an even function

$$\int\limits_{{{ - 1} \over 2}}^{{1 \over 2}} {f\left( {x + {1 \over x}} \right)\sin x\,dx = 0} $$