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  • <img alt="\mathrm{\operatorname{Let} f(x)=[\tan x \mid \cot x] \mid . x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Coperator

\mathrm{\operatorname{Let} f(x)=[\tan x \mid \cot x] \mid . x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right) } ,( where \mathrm{[\cdot] \text { denotes }} the greatest integer less than or equal to \mathrm{x}). Then the number of proints, where \mathrm{f(x)} is discontinuous is

Option: 1

one


Option: 2

zero


Option: 3

 three


Option: 4

 infinite


Answers (1)

best_answer

When \mathrm{x \in I}.

\mathrm{[\cot x]=\cot x}
\mathrm{\Rightarrow \quad \tan x[\cot x]=1}
\mathrm{\Rightarrow \quad[\tan x[\cot x]]=1}

\mathrm{When x \notin I}
\mathrm{\Rightarrow \quad\left[\begin{array}{ll} & {[\cot x]<\cot x} \\ & 0<\tan x[\cot x]<1\end{array}\right.}

\mathrm{\Rightarrow \quad[\tan x[\cot x]]=0}
\mathrm{\Rightarrow \quad f(x)=[\tan x[\cot x]]= \begin{cases}1, & \cot x \in I \\ 0, & \cot x \notin I\end{cases}}
So, \mathrm{f(x)} is discontinuous when \mathrm{\cot x \in I}

\mathrm{Now \, \frac{\pi}{12} \leq x<\frac{\pi}{2}}
\mathrm{\Rightarrow \quad 0<\cot x \leq 2+\sqrt{3}}

Hence, number of points of discontinuity are
\mathrm{x=\cot ^{-1} 3, \cot ^{-1} 2 \: and \: \cot ^{-1} 1=\frac{\pi}{4}}.
Thus three points of discontinuity.

 

Posted by

Pankaj

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