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The extremum values of the function f(x)=\frac{1}{\sin x+4}-\frac{1}{\cos x-4} where, \mathrm{x} \in R is 

Option: 1

\frac{4}{8-\sqrt{2}}


Option: 2

\frac{2 \sqrt{2}}{8-\sqrt{2}}


Option: 3

\frac{2 \sqrt{2}}{4 \sqrt{2}+1}


Option: 4

\frac{4 \sqrt{2}}{8+\sqrt{2}}


Answers (1)

best_answer

f^{\prime}(x)=0

\Rightarrow \quad(\sin x+\cos x)(\text { non-zero quantity) }=0

\begin{array}{ll} \Rightarrow & \tan x=-1 \\ \\\Rightarrow & x=\frac{3 \pi}{4} \text { or } \frac{7 \pi}{4} \end{array}

Global minimum =x=2 n \pi+\left(\frac{3 \pi}{4}\right)

Global maximum =x=2 n \pi+\left(\frac{7 \pi}{4}\right)

M=\frac{4}{8-\sqrt{2}} ; m=\frac{4}{8+\sqrt{2}}

Posted by

Ritika Kankaria

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