Question

A function is defined by
$f\left(x\right)=\frac{1}{2}\underset{0}{\overset{\pi }{\int }}\cos t\cdot \cos (x-t)dt,0\le x\le 2\pi$ , then which of the following statement(s) is/are true?

• A. Minimum value of $f$ is $\frac{-\pi }{4}$
• B. $f\left(x\right)$ is continuous but not differentiable in $(0,2\pi )$
• C. Maximum value of $f$ is $\pi$
• D. There exists at least one $c\in (0,2\pi )$ such that $f^{\prime }\left(c\right)=0$

Answer 1

Answer: (A), (D)

There exists at least one $c\in (0,2\pi )$ such that $f^{\prime }\left(c\right)=0$

Here $f\left(x\right)=\frac{1}{4}\underset{0}{\overset{\pi }{\int }}2\cos t\cdot \cos (x-t)dx$
$=\frac{1}{4}\underset{0}{\overset{\pi }{\int }}\left[\cos t\right(t+x-t)+\cos \{t-(x-t\left)\right\}]dt$
$=\frac{\cos x}{4}[t{]}_0^{\pi }+\frac{1}{4}\underset{0}{\overset{\pi }{\int }}\cos (2t-x)dt$
$=\frac{\pi \cos x}{4}+\frac{1}{4}\cdot {\left[\frac{\sin (2t-x)}{2}\right]}_0^{\pi }$
$=\frac{\pi \cos x}{4}+\frac{1}{8}[\sin (2\pi -x)-\sin (0-x)]$
$=\frac{\pi \cos x}{4}+\frac{1}{8}(-\sin x+\sin x)$
$\therefore f\left(x\right)=\frac{\pi \cos x}{4}$ Which is clearly continuous and differentiable is $(0,2\pi )$
$\therefore f^{\prime }\left(x\right)=\frac{-\pi }{4}\sin x$
${f^{\prime }}^{\prime }\left(x\right)=\frac{-\pi }{4}\cos x$
When $f^{\prime }\left(x\right)=0\Rightarrow \frac{-\pi }{4}\sin x=0$
$\Rightarrow \sin x=0$
$\Rightarrow x=0,\pi ,2\pi$
At $x=0,2\pi ,{f^{\prime }}^{\prime }\left(x\right)=\frac{-\pi }{4}\times 1<0$
$\therefore x=0,2\pi$ are the points of local maxima
and at $x=\pi ,{f^{\prime }}^{\prime }\left(x\right)=\frac{-\pi }{4}\times (-1)>0$
$\therefore x=\pi$ is the point of local minima
At $x=\pi ,f^{\prime }\left(x\right)=\frac{-\pi }{4}\sin \pi =0$
$\Rightarrow \exists$ atleast one $c=\pi \in (0,2\pi )$ such that $f^{\prime }\left(c\right)=0$ and maximum value of $f\left(x\right)=\frac{\pi }{4}\cos \pi =\frac{-\pi }{4}$
$\text{Max}.f\left(x\right)=\frac{\pi }{4}\text{at}\text{(}x=0,2\pi \text{)}$