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UPSC NDA I 2024 Mathematics Paper

Consider the following for the next two (02) items that follow :

The circle

x^2+y^2-2 x=0

is partitioned by line

y=x

in two segments. Let

A_1, A_2

be the areas of major and minor segments respectively.

Question : 98 of 120

Marks: +1, -0

What is the value of

\;\frac{2 (A_1+A_2)}{A_1-3 A_2}

Solution:

Given,

\;\; x^2+y^2-2x=0

\Rightarrow \;\; (x-1)^2+y^2=1

\Rightarrow y^2=1-(x-1)^2

Area of minor segment

(A_2) = \int\limits_{0}^{1} \left( \sqrt{1-(x-1)^2} - x \right) dx

\Rightarrow 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 A_2= \left[ \left(0 + \frac{1}{2}(0) - \frac{1}{2}\right) - \left(0 + \frac{1}{2} \sin^{-1}(-1) - 0\right) \right]

\Rightarrow A_2 = \frac{\pi-2}{4}

Since, area of given circle

= \pi(1)^2 = \pi

\Rightarrow A_1 + A_2 = \pi

\Rightarrow A_1 = \pi - \frac{\pi-2}{4}

= \frac{3\pi+2}{4}

\;\frac{2 (A_1+A_2)}{A_1-3 A_2}\; = \; \frac{2\pi}{\left( \frac{3\pi+2}{4} \right) - \frac{3(\pi-2)}{4}}

\; = \pi

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