Question

A function $f$ is defined by $f\left(x\right)={\int }_0^{\pi }\cos t\cos (x-t)dt,0\le x\le 2\pi$. Then which of the following hold(s) good

• A. $f\left(x\right)$ is continuous but not differentiable in $(0,2\pi )$
• B. Maximum value of $f$ is $\frac{\pi }{2}$
• C. There exists atleast one $c\in (0,2\pi )$ such that $f^{\prime }\left(c\right)=0$
• D. Minimum value of $f$ is $-\frac{\pi }{2}$

Answer 1

Answer: (B), (C), (D)

Minimum value of $f$ is $-\frac{\pi }{2}$

$f\left(x\right)={\int }_0^{\pi }\cos t\cos (x-t)dt\cdots \left(i\right)={\int }_0^{\pi }-\cos t\cos (x-\pi +t)dt\therefore f\left(x\right)={\int }_0^{\pi }\cos t\cdot \cos (x+t)dt\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right),$ we get
$2f\left(x\right)={\int }_0^{\pi }\cos t(2\cos x\cdot \cos t)dt\therefore f\left(x\right)=\cos x{\int }_0^{\pi }{\cos }^{2}tdt=2\cos x{\int }_0^{\pi /2}{\cos }^{2}tdt$
$\therefore f\left(x\right)=\frac{\pi \cos x}{2}$,
which is continuous and differentiable, having maximum value $\frac{\pi }{2}$ and minimum value $-\frac{\pi }{2}$.

$\because f\left(0\right)=f\left(2\pi \right)=\frac{\pi }{2}$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem in $[0,2\pi ]$