Test Index

UPSC NDA I 2024 Mathematics Paper

Question Numbers: 97-98

Consider the following for the items that follow :

The circle

x^{2}+y^{2}-2x=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\left(A_1+A_2\right)}{A_1-3A_2}

Solution:

Concept:

The expression simplifies to

\pi

after substituting the areas

A_1

and

A_2

derived from a given circle equation.

Explanation:

The equation

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

can be rewritten as

(x - 1)^2 + y^2 = 1

This represents a circle with centre at

(1,0)

and radius

1

Given

A_1 = \frac{3\pi + 2}{4}

and

A_2 = \pi - \frac{3\pi + 2}{4} = \frac{\pi - 2}{4}

Now compute

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

A_1 + A_2 = \frac{3\pi + 2}{4} + \frac{\pi - 2}{4} = \frac{4\pi}{4} = \pi

A_1 - 3A_2 = \frac{3\pi + 2}{4} - 3 \cdot \frac{\pi - 2}{4} = \frac{3\pi + 2 - 3\pi + 6}{4} = \frac{8}{4} = 2

Thus the expression becomes

\frac{2 \cdot \pi}{2} = \pi

Answer:

\pi

(Option A).

Go to Question:

Prev Question

Next Question