Question
Let $$f\left( x \right) = {\tan ^{ - 1}}\left\{ {\phi \left( x \right)} \right\},$$ where $$\phi \left( x \right)$$ is $$m.i.$$ for $$0 < x < \frac{\pi }{2}.$$ Then $$f\left( x \right)$$ is :
A.
increasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
B.
decreasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
C.
increasing in $$\left( {0,\,\frac{\pi }{4}} \right)$$ and decreasing in $$\left( {\frac{\pi }{4},\,\frac{\pi }{2}} \right)$$
D.
none of these
Answer :
increasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
Solution :
$$f'\left( x \right) = \frac{{\phi '\left( x \right)}}{{1 + {{\left\{ {\phi \left( x \right)} \right\}}^2}}} > 0{\text{ for }}0 < x < \frac{\pi }{2}$$ because $$\phi '\left( x \right) > 0,\,\phi \left( x \right)$$ being m.i.