f ( x ) = ( e 2 x + 1 e 2 x − 1 ) is

∵ f ( x ) = ( e 2 x + 1 e 2 x − 1 ) ∴ f ( − x ) = e − 2 x + 1 e − 2 x − 1 = 1 + e 2 x 1 − e 2 x ⇒ f ( − x ) = e 2 x + 1 − ( e 2 x − 1 ) = − f ( x ) ∴ f ( x ) is an odd function. Again, f ′ ( x ) = ( 1 + e 2 x ) 2 4 e 2 x > 0 , ∀ x ∈ R ⇒ f ( x ) is an increasing function.

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-Increasing-DecreasingMedium2 minPYQ_2021

MathematicsMediumsingle choice

$f (x) = (\frac{e ^{2 x} - 1}{e ^{2 x} + 1})$ is

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Solution:

$∵ f (x) = (\frac{e ^{2 x} - 1}{e ^{2 x} + 1}) ∴ f (- x) = \frac{e ^{- 2 x} - 1}{e ^{- 2 x} + 1} = \frac{1 - e ^{2 x}}{1 + e ^{2 x}} \Rightarrow f (- x) = \frac{- ( e ^{2 x} - 1 )}{e ^{2 x} + 1} = - f (x)$ $∴ f (x)$ is an odd function. Again, $f^{'} (x) = \frac{4 e ^{2 x}}{( 1 + e ^{2 x} ) ^{2}} > 0, \forall x \in R$ $\Rightarrow f (x)$ is an increasing function.

Stream:BITSATSubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-Increasing-Decreasing

⏱ 2mℹ️ Source: PYQ_2021

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