JEE Advanced 2023 Paper 2 · Q08 · Maxima & Minima

Question

For $x \in \mathbb{R}$, let $	an^{-1}(x) \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}ight)$. Then the minimum value of the function $f: \mathbb{R} 	o \mathbb{R}$ defined by $f(x) = \displaystyle\int_0^{x	an^{-1}x} \dfrac{e^{(t - \cos t)}}{1 + t^{2023}}\,dt$ is

SolveKar AI · Accepted Answer

By the Fundamental Theorem of Calculus and the chain rule: $f'(x) = \dfrac{e^{(x	an^{-1}x - \cos(x	an^{-1}x))}}{1 + (x	an^{-1}x)^{2023}} \cdot \dfrac{d}{dx}(x	an^{-1}x)$. The factor $\dfrac{d}{dx}(x	an^{-1}x) = 	an^{-1}x + \dfrac{x}{1+x^2}$ vanishes only at $x = 0$. Sign analysis: it's negative for $x < 0$ and positive for $x > 0$, so $f$ has a global minimum at $x = 0$. $f(0) = \int_0^0 = 0$.

JEE Advanced 2023 Paper 2 · Q08 · Maxima & Minima

Solution