JEE Mains · Maths · STD 12 - 8. Application and integration
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
- A \(\frac{\pi-1}{\pi+1}\)
- B \(\frac{\pi}{\pi-1}\)
- C \(\frac{\pi}{\pi+1}\)
- D \(\frac{\pi+1}{\pi-1}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi-1}{\pi+1}\)
Step-by-step Solution
Detailed explanation
\(y^2+(x-1)^2=4\) \(\text { shaded portion }=\text { circular }( OABC )\) \(-\operatorname{Ar}(\triangle OAB )\) \(=\frac{\pi(4)}{4}-\frac{1}{2}(2)(1)\) \(A =(\pi-1)\) \(\text { Area } B=\operatorname{Ar}(\triangle AOB )+\text { Area of arc of circle }( ABC )\)…
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