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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

ConceptIncreasing and Decreasing Functions

#### 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 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.
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