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  • Let <img alt="\mathrm{A=\left\{(x, y) \in \mathbb{R}^2: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^2}\right\}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%3D%5Cleft%5C%7B%28x%2C

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

Option: 1

\frac{\pi+1}{\pi-1}
 


Option: 2

\frac{\pi}{\pi-1}
 


Option: 3

\frac{\pi-1}{\pi+1}
 


Option: 4

\frac{\pi}{\pi+1}


Answers (1)

best_answer

\mathrm{A=}

\mathrm{B=}

\mathrm{\begin{aligned} & y^2=4-(x-1)^2 \\ \end{aligned}}

\mathrm{\begin{aligned} & (x-1)^2+y^2=2^2 \\ \end{aligned}}

\mathrm{\begin{aligned} & \mathrm{y}=2 \mathrm{x} \\ \end{aligned}}

\mathrm{\begin{aligned} & (\mathrm{x}-1)^2+4 \mathrm{x}^2=4 \\ \end{aligned}}

\mathrm{\begin{aligned} & \mathrm{x}^2+1-2 \mathrm{x}+4 \mathrm{x}^2=4 \\ \end{aligned}}

\mathrm{\begin{aligned} & 5 x^2-2 \mathrm{x}-3=0 \\ \end{aligned}}

\mathrm{\begin{aligned} & 5 \mathrm{x}^2-5 \mathrm{x}+3 \mathrm{x}-3=0 \\ \end{aligned}}

\mathrm{\begin{aligned} & 5 \mathrm{x}(\mathrm{x}-1)+3(\mathrm{x}-1)=0 \\ & \mathrm{x}=1,-3 / 5 \\ \end{aligned}}

\mathrm{\begin{aligned} & \text { For B : req. area }=\operatorname{ar}(\triangle \mathrm{DRQ})+\operatorname{ar}(\mathrm{RPQ}) \\ \end{aligned}}

\mathrm{\begin{aligned} & =\frac{1}{2} \times 1 \times 2+\int_1^3 \sqrt{4-(x-1)^2} \mathrm{dx} \\ \end{aligned}}

\mathrm{\begin{aligned} & =1+\left[\left(\frac{x-1}{2}\right) \sqrt{4-(x-1)^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x-1}{2}\right)\right]_1^3 \\ \end{aligned}}

\mathrm{\begin{aligned} & =1+2 \sin ^{-1} 1=1+\pi \\ \end{aligned}}........................(1)

\mathrm{\begin{aligned} & \text { For A : req. area }=\text { area of semi circle }- \text { shaded area of } \mathrm{B} \\ \end{aligned}}

\mathrm{\begin{aligned} & =\frac{\pi r^2}{2}-(1+\pi) \\ & =\frac{\pi \times 4}{2}-(1+\pi) \quad\{\because r=2\} \\ \end{aligned}}

\mathrm{\begin{aligned} & A=\pi-1 \\ & \end{aligned}} ........................(2)

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

Posted by

Pankaj Sanodiya

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