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

Answer: 3

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


Continuity at $x=\frac{\pi }{4}$ :
$f\left(\frac{\pi }{4}\right)=\underset{x\to {\frac{\pi }{4}}^{-}}{\lim }f(x)=\underset{x\to \frac{{\pi }^{+}}{4}}{\lim }f(x)=1$
$\Rightarrow f$ 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. $=\underset{x\to {\left(\frac{\pi }{4}\right)}^{-}}{\lim }\left(\frac{f(x)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}\right)$
$=\underset{x\to {\left(\frac{\pi }{4}\right)}^{-}}{\lim }\left(\frac{\tan x-1}{x-\frac{\pi }{4}}\right)$
let $x-\frac{\pi }{4}=t$
$=\underset{t\to 0^{-}}{\lim }\left(\frac{\tan \left(\frac{\pi }{4}+t\right)-1}{t}\right)$
$=\underset{t\to 0^{-}}{\lim }\left(\frac{\frac{1+\tan t}{1-\tan t}-1}{t}\right)$
$=\underset{t\to 0^{-}}{\lim }\left(\frac{2\tan t}{t}\times \frac{1}{1-\tan t}\right)$
$=2$
R.h.lim. $=\underset{x\to {\left(\frac{\pi }{4}\right)}^{+}}{\lim }\left(\frac{f(x)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}\right)$
$=\underset{x\to {\left(\frac{\pi }{4}\right)}^{+}}{\lim }\left(\frac{\cot x-1}{x-\frac{\pi }{4}}\right)$
$\mathrm{let}x-\frac{\pi }{4}=t$
$=\underset{t\to 0^{+}}{\lim }\left(\frac{\cot \left(\frac{\pi }{4}+t\right)-1}{t}\right)$
$=\underset{t\to 0^{+}}{\lim }\frac{\left(\frac{\cot \frac{\pi }{4}\cot t-1}{\cot \frac{\pi }{4}+\cot t}-1\right)}{t}$
$=\underset{t\to 0}{\lim }\frac{-2}{t(1+\cot t)}$
$=\underset{t\to 0}{\lim }\frac{-2\sin t}{t(\cos t+\sin t)}=-2$
$\Rightarrow f$ 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 $=\pi$.
$\Rightarrow m=0,n=4,p=1$
$\Rightarrow m+n-p=0+4-1=3$