JEE Advanced 2023 Paper 2, Mathematics Section 3 Q1: Maxima & Minima

By the Leibniz rule,

$f^{\prime }(x)=\frac{e^{x{\tan }^{-1}x-\cos (x{\tan }^{-1}x)}}{1+(x{\tan }^{-1}x{)}^{2023}}\cdot \left({\tan }^{-1}x+\frac{x}{1+x^2}\right).$

The product $x{\tan }^{-1}x\ge 0$ for every real $x$ (it is the product of two quantities with the same sign), so the denominator is positive and the exponential factor is positive. Therefore the sign of $f^{\prime }$ matches the sign of $g(x)={\tan }^{-1}x+\frac{x}{1+x^2}$.

Now $g(0)=0$, and $g^{\prime }(x)=\frac{1}{1+x^2}+\frac{1-x^2}{(1+x^2{)}^2}>0$ at $x=0$ and more generally $g$ is strictly increasing through $0$ (both terms have the sign of $x$). Hence $f^{\prime }(x)<0$ for $x<0$ and $f^{\prime }(x)>0$ for $x>0$.

Therefore $f$ attains its minimum at $x=0$, and $f(0)={\int }_0^0\cdots dt=0.$