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  • Let  <img alt="\mathrm{ f(x)=[\tan x]+\sqrt{\tan x-[\tan x]} \quad \forall 0 \leq x<\frac{\pi}{2}, where [\cdot] }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20f%28x

Let  \mathrm{ f(x)=[\tan x]+\sqrt{\tan x-[\tan x]} \quad \forall 0 \leq x<\frac{\pi}{2}, where [\cdot] } denotes the greatest integer function, then

 

Option: 1

\mathrm{ f(x) is \, \, continuous \, \, in \left[0, \frac{\pi}{2}\right)}
 


Option: 2

\mathrm{f(x) \, is \, \, discontinuous \, \, at \, x=0 \& \frac{\pi}{4}}
 
 


Option: 3

\mathrm{f(x) \, is \, \, discontinuous \, \, at \, x=0 }


Option: 4

\mathrm{ f(x) \, has \, \, infinite \, \, points\, \, of \, \, discontinuity. }


Answers (1)

best_answer

For \mathrm{0 \leq x<\frac{\pi}{2} }, we have
\mathrm{ \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\{[\tan x]+\sqrt{\tan x-[\tan x]}\}=0=f(x) }
\mathrm{\therefore \quad f(x) is\, \, continuous\, \, at\, \, x=0 }
Consider any point in \mathrm{\left[0, \frac{\pi}{2}\right), say x=\frac{\pi}{4} }
\mathrm{\begin{aligned} & \text { Now, } \lim _{x \rightarrow \frac{\pi^{+}}{4}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{4}}\{[\tan x]+\sqrt{\tan x-[\tan x]}\} \\ & =[1]+\sqrt{1-1}=1 \end{aligned} }
So, \mathrm{\lim _{x \rightarrow \frac{\pi}{4}^{+}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=f\left(\frac{\pi}{4}\right)=1 }
\mathrm{\therefore \quad f(x) is\, \, continuous \, \, at \, x=\frac{\pi}{4} \in\left[0, \frac{\pi}{2}\right) }
\mathrm{\therefore f(x) is\, \, continuous \, at,\, x=0 \& each\, \, point\, \, in \left[0, \frac{\pi}{2}\right). }

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shivangi.bhatnagar

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