JEE Advanced 2023 — Paper 2 — Question 8

$\text{Define\; }g(x)=x\arctan x.$

$\text{Note\; that\; } \arctan x \text{ is\; odd,\; so\; } g(-x)=(-x)\arctan(-x)=x\arctan x=g(x), \text{ hence\; } g \text{ is even.}$

\text{Also\; } \arctan x\in\big(-\tfrac{\pi}{2},\tfrac{\pi}{2}\big)\text{ and\; for\; }x>0,\ \arctan x>0,\ so; $g(x)\ge0\ \forall x\in\mathbb{R}$ and $g(0)=0.$

\text{The\; integrand\; }h(t)=\dfrac{e^{,t-\cos t}}{1+t^{2023}} \text{ is\; continuous\; on\; }[0,\infty), {because\; for\; }t\ge0\

we have $1+t^{2023}>0$ and the exponential factor is positive. $

$\text{Hence\; for \; any\; }x\in\mathbb{R},\ f(x)=\int_{0}^{g(x)} h(t),dt\ \text{is\; well-defined\; and\; }f(x)\ge0,$ since $h(t)>0\text{ on }[0,g(x)]\text{ when }g(x)>0.$

$\text{Moreove\;r }f(0)=\int_{0}^{0}h(t),dt=0.$

$\text{Therefore\; the\; minimum\; value\; of\; }f(x)\text{ on\; }\mathbb{R}\text{ is\; }0,$ attained at $x=0.$

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