JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 17)
Let the area enclosed between the curves $|y| = 1 - x^2$ and $x^2 + y^2 = 1$ be $\alpha$. If $9\alpha = \beta \pi + \gamma; \beta, \gamma$ are integers, then the value of $|\beta - \gamma|$ equals:
15
18
33
27
Explanation
$$\begin{aligned} & C_1:|y|=1-x^2 \\ & C_2: x^2+y^2=1 \end{aligned}$$
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$\therefore$ Required Area
$=\alpha=4\left[\right.$ Area of circle in $1^{\text {st }}$ quad. $\left.-\int_0^1\left(1-\mathrm{x}^2\right) \mathrm{dx}\right]$
$$\begin{aligned} & =4\left[\frac{\pi}{4}-\left[\mathrm{x}-\frac{\mathrm{x}^3}{3}\right]_0^1\right] \\ & \alpha=\pi-\frac{8}{3} \\ & \therefore 3 \alpha=3 \pi-8 \\ & \therefore 9 \alpha=9 \pi-24 \\ & \therefore \beta=9, \gamma=-24 \\ & \therefore|\beta-\gamma|=33 \end{aligned}$$
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