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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Area Under Curves

NDA 2024 · Mathematics · Area Under Curves

The circle x^2+y^2-2 x=0 is partitioned by line y=x in two...

The circle x2+y2−2x=0x^2+y^2-2 x=0 is partitioned by line y=xy=x in two segments. Let A1,A2\mathrm{A}_1, \mathrm{ A}_2 be the areas of major and minor segments respectively. What is the value of 2(A1+A2)A1−3A2\frac{2\left(A_1+A_2\right)}{A_1-3 A_2} ?

Options

  1. Aπ\pi
  2. B1
  3. C-1
  4. D−π-\pi

Correct answer

A. π\pi

Step-by-step solution

 Given, x2+y2−2x=0⇒(x−1)2+y2=1 or  y2=1−(x−1)2\begin{aligned} \text{ Given, } \quad & & x^2+y^2-2 x & =0 \\ \Rightarrow & & (x-1)^2+y^2 & =1 \\ \text{\text{ or } } & & y^2 & =1-(x-1)^2 \end{aligned} Area of minor segment (A2)\left(\mathrm{A}_2\right)=∫011−(x−1)2−(x)dx⇒A2=[(x−1)21−(x−1)2+12sin⁡−1(x−1)−x22]01⇒A2=[(0+12(0)−12)]−[0+12sin⁡−1(−1)−0]⇒A2=π−24\begin{aligned} & =\int_0^1 \sqrt{1-(x-1)^2-(x)} d x \\ \Rightarrow \mathrm{ A}_2 & =\left[\frac{(x-1)}{2} \sqrt{1-(x-1)^2}+\frac{1}{2} \sin ^{-1}(x-1)-\frac{x^2}{2}\right]_0^1 \\ \Rightarrow \mathrm{ A}_2 & =\left[\left(0+\frac{1}{2}(0)-\frac{1}{2}\right)\right]-\left[0+\frac{1}{2} \sin ^{-1}(-1)-0\right] \\ \Rightarrow \mathrm{A}_2 & =\frac{\pi-2}{4} \end{aligned}2(A1+A2)A1−3A2=2(π)(3π+24)−3(π−2)4=π\frac{2\left(\mathrm{ A}_1+\mathrm{A}_2\right)}{\mathrm{A}_1-3 \mathrm{ A}_2}=\frac{2(\pi)}{\left(\frac{3 \pi+2}{4}\right)-\frac{3(\pi-2)}{4}}=\pi
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