Question
The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an increasing function in
A.
$$\left( {0,\frac{\pi }{2}} \right)$$
B.
$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
C.
$$\left( {\frac{\pi }{4},\frac{\pi }{2}} \right)$$
D.
$$\left( { - \frac{\pi }{2},\frac{\pi }{4}} \right)$$
Answer :
$$\left( { - \frac{\pi }{2},\frac{\pi }{4}} \right)$$
Solution :
$$\eqalign{
& {\text{Given}}\,f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right) \cr
& f'\left( x \right) = \frac{1}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cdot \left( {\cos x - \sin x} \right) \cr
& = \frac{{\sqrt 2 \cdot \left( {\frac{1}{{\sqrt 2 }}\cos x - \frac{1}{{\sqrt 2 }}\sin x} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cr
& = \frac{{\left( {\cos \frac{\pi }{4} \cdot \cos x - \sin \frac{\pi }{4} \cdot \sin x} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cr
& \therefore f'\left( x \right) = \frac{{\sqrt 2 \cos \left( {x + \frac{\pi }{4}} \right)}}{{1 + {{\left( {\sin x + \cos x} \right)}^2}}} \cr
& {\text{if}}\,f'\left( x \right) > O\,{\text{then}}\,f\left( x \right)\,{\text{is increasing function}}{\text{.}} \cr
& {\text{Hence}}\,f\left( x \right)\,{\text{is increasing,}}\,{\text{if}}\, - \frac{\pi }{2} < x + \frac{\pi }{4} < \frac{\pi }{2} \cr
& \Rightarrow - \frac{{3\pi }}{4} < x < \frac{\pi }{4} \cr
& {\text{Hence, }}f\left( x \right)\,{\text{is increasing when}}\,n \in \left( { - \frac{\pi }{2},\frac{\pi }{4}} \right) \cr} $$