JEE MAIN - Mathematics (2007 - No. 15)
The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an incresing function in
$$\left( {0,{\pi \over 2}} \right)$$
$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$
$$\left( { {\pi \over 4},{\pi \over 2}} \right)$$
$$\left( { - {\pi \over 2},{\pi \over 4}} \right)$$
Explanation
Given $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$
$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$
$$ = {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
$$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.
Hence $$f(x)$$ is increasing, if $$ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$
$$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$
Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$
$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$
$$ = {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
$$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.
Hence $$f(x)$$ is increasing, if $$ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$
$$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$
Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$
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