QJEEAdv23Mad9c7: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Mad9c7 Solution: For , let . Then the minimum value of the function defined by…

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$\begin{aligned} f(x) & =\int\limits_0^{x \tan ^{-1} x} \frac{e^{t-\cos t}}{1+t^{2023}} d t \\\\ f^{\prime}(x) & =\frac{e^{x \tan ^{-1} x-\cos \left(x \tan ^{-1} x\right)}}{1+\left(x \tan ^{-1} x\right)^{2023}}\left(\frac{x}{1-x^2}+\tan ^{-1} x\right)\end{aligned}$

For max/min put $f^{\prime}(x)=0$

$\begin{aligned}& \Rightarrow \frac{x}{1+x^2}+\tan ^{-1} x=0 \\\\& \Rightarrow x=0\end{aligned}$


$\therefore f(x)$ has minimum at $x=0$

And $f(x)_{\min }=f(0)=0$