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Question
Let \[f\left( x \right) = \tan^{- 1} \left( g\left( x \right) \right),\],where g (x) is monotonically increasing for 0 < x < \[\frac{\pi}{2} .\] Then, f(x) is
Options
increasing on (0, π/2)
decreasing on (0, π/2)
increasing on (0, π/4) and decreasing on (π/4, π/2)
none of these
Solution
increasing on (0, \[\pi\]/2)
\[\text { Given:}g\left( x \right) \text { is increasing on }\left( 0, \frac{\pi}{2} \right). \text { Then, }\]
\[ x_1 < x_2 , \forall x_1 , x_2 \in \left( 0, \frac{\pi}{2} \right)\]
\[ \Rightarrow g\left( x_1 \right) < g\left( x_2 \right)\]
\[ {\text { Taking } tan}^{- 1} \text { on both the sides, we get } \]
\[ \tan^{- 1} \left( g\left( x_1 \right) \right) < \tan^{- 1} \left( g\left( x_2 \right) \right)\]
\[ \Rightarrow f\left( x_1 \right) < f\left( x_2 \right), \forall x_1 , x_2 \in \left( 0, \frac{\pi}{2} \right)\]
\[\text { So,}f\left( x \right)\text { is increasing on }\left( 0, \frac{\pi}{2} \right).\]
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