JEE Main 2025 Jan 29 Shift 2, Mathematics Q3: Area as Definite Integral

Both curves are symmetric about the $x$- and $y$-axes, so compute the first-quadrant area and quadruple it. In the first quadrant the parabola is $y=1-x^2$ and the circle is $y=\sqrt{1-x^2}$, with $\sqrt{1-x^2}\ge 1-x^2$ on $[0,1]$.

First-quadrant slice between the two:

$A_1={\int }_0^1\left[\sqrt{1-x^2}-(1-x^2)\right]dx=\frac{\pi }{4}-{\left[x-\frac{x^3}{3}\right]}_0^1=\frac{\pi }{4}-\frac{2}{3}$.

Hence $\alpha =4A_1=\pi -\frac{8}{3}$, so $9\alpha =9\pi -24$.

Comparing with $9\alpha =\beta \pi +\gamma$ gives $\beta =9$, $\gamma =-24$, and $|\beta -\gamma |=|9-(-24)|=33$.