QJEEAdv08Ma1d59: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma1d59 Solution: Let be a non-constant twice differentiable function defined on…

$f(x)$ 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

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