Question

A function $f$ is defined by $f\left(x\right)={\int }_0^{x}costcos(x-t)dt,0\le x\le 2\pi$ 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 $\pi$
• C. There exists atleast one $cϵ(0,2\pi )s.t.f^{\prime }\left(c\right)=0$.
• D. Minimum value of $f$ is $-\pi /2$

Answer 1

Answer: (C), (D)

The correct options are
C There exists atleast one $cϵ(0,2\pi )s.t.f^{\prime }\left(c\right)=0$.
D Minimum value of $f$ is $-\pi /2$

Multiplying and dividing by 2, we get
$\frac{1}{2}\left(2cos\right(t).cos(x-t)$
$=\frac{1}{2}\left[cos\right(x)+cos(2t-x\left)\right]$
Now
$\frac{1}{2}{\int }^{x}cos\left(x\right)+cos(2t-x).dt$
$=\frac{1}{2}[t.cos(x)+\frac{sin(2t-x)}{2}{]}_{t=0}^{t=x}$
$=\frac{1}{2}\left[xcos\right(x)+\frac{sinx}{2}-\frac{sin(-x)}{2}]$
$f\left(x\right)=\frac{1}{2}[xcosx+sin(x\left)\right]$
Now
$f^{\prime }\left(x\right)=\frac{1}{2}\left[cos\right(x)-xsin(x)+cos(x\left)\right]$
$=cos\left(x\right)-\frac{x}{2}sin\left(x\right)$
$=0$
Implies
$cos\left(x\right)=sin\left(x\right).\frac{x}{2}$
Or
$cot\left(x\right)=\frac{x}{2}$
Or
$2cot\left(x\right)-x=0$
Minimum value of $f\left(x\right)$ is at $x=\pi$.
$f\left(\pi \right)$
$=\frac{1}{2}[\pi .cos(\pi )-0]$
$=-\frac{\pi }{2}$.