For $\mathrm{x} \in \mathbb{R}$, let $\tan ^{-1}(\mathrm{x}) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the minimum value of the function $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x} \tan ^{-1} \mathrm{x}} \frac{e^{(\mathrm{t}-\cos t)}}{1+\mathrm{t}^{2023}} \mathrm{dt}$ is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Definite Integrals

For

x \in R

, let

tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})

. Then the minimum value of the function

f : R \to R

defined by

f (x) = \int_{0}^{x t a n^{- 1} x} \frac{e ^{(t - c o s t)}}{1 + t ^{2023}} dt

Answer: 0

Differentiate:

f^{'} (x) = \frac{e ^{x t a n^{- 1} x - c o s (x t a n^{- 1} x)}}{1 + ( x t a n ^{- 1} x ) ^{2023}} \cdot (\frac{x}{1 + x ^{2}} + tan^{- 1} x)

Note that

x tan^{- 1} x \geq 0

for all

x \in R

(both factors have the same sign).

Also,

\frac{x}{1 + x ^{2}} + tan^{- 1} x

changes sign at

x = 0

Therefore

f

has a minimum at

x = 0

f (0) = \int_{0}^{0} = 0

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer