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If  \mathrm{ f(x)=\min \{\tan x, \cot x\}} , then

Option: 1

f(x) is not differentiableat \mathrm{x=0, \pi / 4,5 \pi / 4} 


Option: 2

f(x) is continuous at \mathrm{ x=0, \pi / 2,3 \pi / 2}


Option: 3

\mathrm{\int_0^{\pi / 2} f(x) d x=\ln \sqrt{2}}


Option: 4

f(x) is periodic with period 0


Answers (1)

best_answer

\text { We have, } f(x)=\left\{\begin{array}{l} \tan x, 0 \leq x \leq \pi / 4 \\ \cot x, \pi / 4 \leq x \leq \pi / 2 \\ \tan x, \pi / 2<x \leq 3 \pi / 4 \\ \cot x, 3 \pi / 4 \leq x<\pi \end{array}\right.

Since \mathrm{\tan x}  and  \mathrm{\cot x} are periodic functions with period \mathrm{\pi}. So, f(x) is also periodic with period \mathrm{\pi} 

It is evident from the graph that f(x) is not continuous at\mathrm{ x=\frac{\pi}{2}} . Since f(x) is period with period \mathrm{\pi}. So, it is not continuous at \mathrm{x=0, \pm \pi / 2, \pm \pi, \neq 3 \pi / 2} 
Also, f(x) is not differentiableted  \mathrm{x=\pi / 4,3 \pi / 4,5 \pi / 4 etc.} 

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SANGALDEEP SINGH

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