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JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 16)

Let $$A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$$ and

$$ B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. } $$.

Then the ratio of the area of A to the area of B is

$$\frac{\pi}{\pi+1}$$
$$\frac{\pi-1}{\pi+1}$$
$$\frac{\pi}{\pi-1}$$
$$\frac{\pi+1}{\pi-1}$$

Explanation

$y^{2}+(x-1)^{2}=4$

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 49 English Explanation 1

shaded portion $=$ circular $(\mathrm{OABC})$

$$ \begin{aligned} & -\operatorname{Ar}(\Delta \mathrm{OAB}) \\\\ & =\frac{\pi(4)}{4}-\frac{1}{2}(2)(1) \\\\ & \mathrm{A}=(\pi-1) \end{aligned} $$

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 49 English Explanation 2
Area $B=\operatorname{Ar}(\triangle \mathrm{AOB})+$ Area of arc of circle $(\mathrm{ABC})$

$$ =\frac{1}{2}(1)(2)+\frac{\pi(2)^{2}}{4}=\pi+1 $$

$$ \frac{\mathrm{A}}{\mathrm{B}}=\frac{\pi-1}{\pi+1} $$

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