The function [f( x ) = ^ - 1( x + x )] is an incresing function in

Question

The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an incresing function in

Accepted Answer

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 $$n \in \left( { - {\pi  \over 2},{\pi  \over 4}} \right)$$

Solutions · 1