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The sum of the absolute maximum and absolute minimum values of the function \mathrm{f(x)=\tan ^{-1}(\sin x-\cos x)} in the interval \mathrm{[0, \pi]} is :

Option: 1

0


Option: 2

\mathrm{\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}}


Option: 3

\mathrm{\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}}


Option: 4

\mathrm{\frac{-\pi}{12}}


Answers (1)

best_answer

\begin{aligned} \mathrm{f^{\prime}(x)} &=\mathrm{\frac{1}{1+(\sin x-\cos x)^2}[\cos x+\sin x] }\\ &=\mathrm{\frac{1}{1+(\sin x-\cos x)^2} \cdot \frac{1}{\sqrt{2}} \sin \left(x+\frac{\pi}{4}\right) }\\ &\mathrm{\therefore f^{\prime}(x)>0 \text { for }\left(0, \frac{3 \pi}{4}\right) }\text { and } \\ &\mathrm{f^{\prime}(x)<0} \text{ for } \left( \frac{3\pi}{4}, \pi\right ) \\ \mathrm{f(0)} &=\mathrm{\tan ^{-1}(-1)=-\frac{\pi}{4} }\\ \mathrm{f\left(\frac{3 \pi}{4}\right) }&=\mathrm{\tan ^{-1}(\sqrt{2})} \\ \mathrm{f(\pi)} &=\mathrm{\tan ^{-1}(1)=\frac{\pi}{4}} \\ \text { Sum } &=\mathrm{\tan ^{-1}(\sqrt{2})-\frac{\pi}{4}} \\ &=\mathrm{\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}} \\ &\therefore Option (c) \end{aligned}

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Kuldeep Maurya

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