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JEE Mains · Maths · STD 12 - 8. Application and integration

If the area of the larger portion bounded between the curves \(x^2+y^2=25\) and \(y=|x-1|\) is \(\frac{1}{4}(b \pi+c), b, c \in N\), then \(b+c\) is equal to

  1. A 55
  2. B 66
  3. C 77
  4. D 88
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Answer & Solution

Correct Answer

(C) 77

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Detailed explanation

\begin{aligned} & \mathrm{x}^2+\mathrm{y}^2=5 \\ & \mathrm{x}^2+(\mathrm{x}-1)^2=25 \Rightarrow \mathrm{x}=4 \\ & \mathrm{x}^2+(-\mathrm{x}+1)^2=5 \Rightarrow \mathrm{x}=-3 \\ & \mathrm{~A}=25 \pi-\int_{-3}^4 \sqrt{25-\mathrm{x}^2} \mathrm{dx}+\frac{1}{2} \times 4 \times…

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