Mathematics - Application of Derivative Question with Solution | TestHub

Given that, $f(x)=f(1-x)$ On differentiating w.r.t. $x$, we get $$f^{\prime }(x)=-f^{\prime }(1-x)$$ Let us put $x=\frac{1}{2}$ $$\Rightarrow 2f^{\prime }\left(\frac{1}{2}\right)=0\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=0$$ Since, $f^{\prime }\left(\frac{1}{2}\right)=0$ and $f^{\prime }\left(\frac{1}{4}\right)=0$ $\Rightarrow f^{\prime \prime }(x)=0$ at two points in $[0,1]$. Now, ${\int }_{-1/2}^{1/2}f\left(x+\frac{1}{2}\right)\sin xdx=0$ As, $f\left(x+\frac{1}{2}\right)\sin x$ is an odd function which is clear from the following explanation. Let $g(x)=f\left(x+\frac{1}{2}\right)\sin x$, $$g(-x)=f\left(\frac{1}{2}-x\right)\sin (-x)=-\sin xf\left(1-\left(\frac{1}{2}-x\right)\right)=-\sin xf\left(\frac{1}{2}+x\right)=-g(x)$$ Moreover, ${\int }_{1/2}^{1}f(1-t)e^{\sin (\pi t)}dt={\int }_0^{1/2}f(u)\cdot e^{\sin \pi u}du$ where, $1-t=u$.