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UPSC NDA I 2015 Mathematics Solved Paper

For the next two (2) items that follow:

Consider the line

x=\sqrt{3}y

and the circle

x^2+y^2=4

[NDA I 2015]

Question : 77 of 120

Marks: +1, -0

What is the area of the region in the first quadrant enclosed by the x-axis, the line

x=\sqrt{3}y

and the circle ?

[NDA I 2015]

$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{3} - \frac{\sqrt{3}}{2}$
None of the above

Validate

Solution:

From the figure, area bounded by x-axis, the line and the circle in the first quadrant is simply the area of the sector of the circle with 30 at center.

∴Area bounded

= \frac{30^{\circ}}{360^{\circ}} \times \pi \times 2^2 = \frac{\pi}{3}

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