Subject: Mathematics — Definite Integration
Question: Let $f(x)$ be a non-constant twice differentiable
function definied on $( - \infty,\infty)$ such that
$f(x) = f(1 - x)$ and $f^{'}\left( \frac{1}{4} \right) = 0$. Then,
find the value of
$\int_{0}^{1/2}\mspace{2mu} f(t)e^{sin\pi t}dt - \int_{0}^{1/2}\mspace{2mu} f(1 - t)e^{sin\pi t}dt$
Answer: 0
Solution: $\mathbf{\therefore}f(x)$ is a non constant twice
differentiable function such
that $f(x) = f(1 - x) \Rightarrow f^{'}(x) = - f^{'}(1 - x)$ .....(i)\
For $x = \frac{1}{2}$, we get
$f^{'}\left( \frac{1}{2} \right) = - f^{'}\left( 1 - \frac{1}{2} \right)$\
$\Rightarrow f^{'}\left( \frac{1}{2} \right) + f^{'}\left( \frac{1}{2} \right) = 0$
$\lbrack\because f(x) = f(1 - x)\rbrack$
$$\Rightarrow f^{'}\left( \frac{1}{2} \right) = 0$$
For $x = \frac{1}{4}$,
we get
$f^{'}\left( \frac{1}{4} \right) = - f^{'}\left( \frac{3}{4} \right)\lbrack\because f(x) = f(1 - x)\rbrack$
but given that $f^{'}\left( \frac{1}{4} \right) = 0$\
$$\therefore f^{'}\left( \frac{3}{4} \right) = f^{'}\left( \frac{1}{4} \right) = 0
$$Hence, $f^{'}(x)$ satisfies all conditions of Rolle\'s theorem for
$x \in \left\lbrack \frac{1}{4},\frac{1}{2} \right\rbrack$ and
$\left\lbrack \frac{1}{2},\frac{3}{4} \right\rbrack$. So there exists
at least one point
$a_{1} \in \left( \frac{1}{4},\frac{1}{2} \right)$ and at least one
point $b_{2} \in \left( \frac{1}{2},\frac{3}{4} \right)$.
Such that $f^{''}\left( a_{1} \right) = 0$ and
$f^{''}\left( b_{2} \right) = 0$\
$\therefore f^{''}(x)$ varishes at least twice on
$\lbrack 0,1\rbrack$.\
Also using $f(x) = f(1 - x)$\
$$\Rightarrow f\left( x + \frac{1}{2} \right) = f\left( 1 - x - \frac{1}{2} \right) = f\left( - x + \frac{1}{2} \right)
$$$\Rightarrow f\left( x + \frac{1}{2} \right)$ is an even function.\
$\Rightarrow sinx.f\left( x + \frac{1}{2} \right)$ is an odd
function.\
$\Rightarrow \int_{- 1/2}^{1/2}\mspace{2mu} f\left( x + \frac{1}{2} \right)sinxdx = 0,$