Question

Let $f(x)=\min (\cot x, \tan x)$,
$x \in(0,2 \pi)-\left\{\frac{\pi}{2}, \pi, \frac{3 \pi}{2}\right\}$
If $m:$ number of points in the domain where $f$ is not differentiable
$n$ : number of points in the domain where $f$ is not continuous
and the fundamental period of function $f$ is $p \pi$, then evaluate $m+n-p$

Answer 1

$f(x)=\min (\cot x, \tan x), x \in(0,2 \pi)-\left\{\frac{\pi}{2}, \pi, \frac{3 \pi}{2}\right\}$


Continuity at $x=\frac{\pi}{4}$ :
$f\left(\frac{\pi}{4}\right)=\displaystyle\lim _{x \rightarrow \frac{\pi}{4}^{-}} f(x)=\displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{4}} f(x)=1$
$\Rightarrow f$ is continuous at $x=\frac{\pi}{4}$
Similarly, it can be shown that $f$ is continuous at
$x=\frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} .$
Differentiability at $x=\frac{\pi}{4}$ :
L.h.lim. $=\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}}\left(\frac{f(x)-f\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\right)$
$=\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}}\left(\frac{\tan x-1}{x-\frac{\pi}{4}}\right)$
let $x-\frac{\pi}{4}=t$
$=\displaystyle\lim _{t \rightarrow 0^{-}}\left(\frac{\tan \left(\frac{\pi}{4}+t\right)-1}{t}\right)$
$=\displaystyle\lim _{t \rightarrow 0^{-}}\left(\frac{\frac{1+\tan t}{1-\tan t}-1}{t}\right)$
$=\displaystyle\lim _{t \rightarrow 0^{-}}\left(\frac{2 \tan t}{t} \times \frac{1}{1-\tan t}\right)$
$=2$
R.h.lim. $=\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}}\left(\frac{f(x)-f\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\right)$
$=\displaystyle\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}}\left(\frac{\cot x-1}{x-\frac{\pi}{4}}\right)$
$\operatorname{let} x-\frac{\pi}{4}=t$
$=\displaystyle\lim _{t \rightarrow 0^{+}}\left(\frac{\cot \left(\frac{\pi}{4}+t\right)-1}{t}\right)$
$=\displaystyle\lim _{t \rightarrow 0^{+}} \frac{\left(\frac{\cot \frac{\pi}{4} \cot t-1}{\cot \frac{\pi}{4}+\cot t}-1\right)}{t}$
$=\displaystyle\lim _{t \rightarrow 0} \frac{-2}{t(1+\cot t)}$
$=\displaystyle\lim _{t \rightarrow 0} \frac{-2 \sin t}{t(\cos t+\sin t)}=-2$
$\Rightarrow f$ is not differentiable at $x=\frac{\pi}{4}$.
Similarly it can be shown that $f$ is not differentiable at
$x=\frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$.
From figure, it is clear that $f$ is periodic with fundamental period $=\pi$.
$\Rightarrow m=0, n=4, p=1 $
$\Rightarrow m+n-p=0+4-1=3$