For x ∈ R, let tan^-1(x) ∈ (-(π)/2, (π)/2 ). Then the minimum value of the function f : R → R defined by f(x) = displaystyle∫limits_0^xtan^-1x dfrace^(t-cos t)1+t^2023 dt is

For x ∈ R, let tan^-1(x) ∈ (-(π)/2, (π)/2 ). Then the… — JEE Advanced 2023 Mathematics PYQ

JEE Advanced 2023 · Paper 2 · numerical answer

For

x \in \mathbb{R}

, let

\tan^{-1}(x) \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)

. Then the minimum value of the function

f : \mathbb{R} \to \mathbb{R}

defined by

f(x) = \displaystyle\int\limits_{0}^{x\tan^{-1}x} \dfrac{e^{(t-\cos t)}}{1+t^{2023}}\, dt

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[JEE Advanced 2023 · Paper 2]

Practise Definite Integrals and Areas the way it is asked

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