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Solution for Let F ( X ) = Tan − 1 ( G ( X ) ) , ,Where G (X) is Monotonically Increasing for 0 < X < π 2 . Then, F(X) is (A) Increasing on (0, π/2) (B) Decreasing on (0, π/2) - CBSE (Commerce) Class 12 - Mathematics

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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
(a) increasing on (0, π/2)
(b) decreasing on (0, π/2)
(c) increasing on (0, π/4) and decreasing on (π/4, π/2)
(d) none of these

Solution

(a) 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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Solution for question: Let F ( X ) = Tan − 1 ( G ( X ) ) , ,Where G (X) is Monotonically Increasing for 0 < X < π 2 . Then, F(X) is (A) Increasing on (0, π/2) (B) Decreasing on (0, π/2) concept: Increasing and Decreasing Functions. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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