Let A = { ( x , y ) 2 : y 0 , 2 x y 4 - ( x - 1 ) 2 ? } and B = { ( x , y ): 0 y min { 2 x , ? 4 - ( x - 1 ) 2 ? } } . Then the ratio of the area of A to the area of B is [2023]

Question

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

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

Then the ratio of the area of $A$ to the area of $B$ is [2023]

Smart Achievers · Accepted Answer

(2)

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

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

We have the following diagram $y^{2} + {(x - 1)}^{2} = 4$

$Shaded portion A$

$Area A = area of quadrant of circle - area of (O A B)$

$= \frac{π (4)}{4} - \frac{1}{2} (2) (1) = (π - 1)$ ...(i)

Now, for finding the area of portion B, we have

$Area B = area of (∆ A O B) + area of quadrant of circle$

$= \frac{1}{2} \cdot (1) (2) + π \frac{{(2)}^{2}}{4} = (π + 1)$ ...(ii)

Thus, according to the question, ratio of area A to area

$B = \frac{π - 1}{π + 1}$

Solution

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

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

Then the ratio of the area of $A$ to the area of $B$ is [2023]

1 $\frac{π}{π + 1}$

2 $\frac{π - 1}{π + 1}$

3 $\frac{π + 1}{π - 1}$

4 $\frac{π}{π - 1}$

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Solution

Q. Let A={(x,y)∈ℝ2:y≥0, 2x≤y≤ 4-(x-1)2 } and B={(x,y)∈ℝ×ℝ:0≤y≤min{2x, 4-(x-1)2 }}. Then the ratio of the area of A to the area of B is [2023]

1 ππ+1

2 π-1π+1

3 π+1π-1