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

Question

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

Accepted Answer

Answer: C. \cos ^{-1}\left(\frac{1}{\sqrt{3}}ight)-\frac{\pi}{4} f(x) = {	an ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ] Let g(x) = \sin x - \cos x = \sqrt 2 \sin \left( {x - {\pi \over 4}} ight) and x - {\pi \over 4} \in \left[ {{{ - \pi } \over 4},\,{{3\pi } \over 4}} ight] 	herefore g(x) \in \left[ { - 1,\,\sqrt 2 } ight] and {	an ^{ - 1}}x is an increasing function 	herefore f(x) \in \left[ {{{	an }^{ - 1}}( - 1),\,{{	an }^{ - 1}}\sqrt 2 } ight] \in \left[ { - {\pi \over 4},\,{{	an }^{ - 1}}\sqrt 2 } ight] 	herefore Sum of {f_{\max }} and {f_{\min }} = {	an ^{ - 1}}\sqrt 2 - {\pi \over 4} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} ight) - {\pi \over 4}

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

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Explanation

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