1. Tardigrade
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  4. Let A= (x, y) ∈ R 2: y ≥ 0,2 x ≤ y ≤ √4-(x-1)2 and B= (x, y) ∈ R × R: 0 ≤ y ≤ min 2 x, √4-(x-1)2 . Then the ratio of the area of A to the area of B is

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Q. Let $A=\left\{(x, y) \in R ^2: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^2}\right\}$ and
$B=\left\{(x, y) \in R \times R : 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^2}\right\}\right\} .$
Then the ratio of the area of $A$ to the area of $B$ is

JEE MainJEE Main 2023Application of Integrals

Solution:

$y^2+(x-1)^2=4$
image
shaded portion $=\text { circular }( OABC )$
$ -\operatorname{Ar}(\triangle OAB ) $
$=\frac{\pi(4)}{4}-\frac{1}{2}(2)(1)$
$ A =(\pi-1)$
image
Area $B =\operatorname{Ar}(\triangle AOB )+$ Area of arc of circle $( ABC )$
$ =\frac{1}{2}(1)(2)+\frac{\pi(2)^2}{4}=\pi+1 $
$ \frac{ A }{ B }=\frac{\pi-1}{\pi+1}$