The area (in sq. units) of the region in the first quadrant bounded by the x-axis, the line y = mx, where m > 0, and the circle \(x^{2}+y^{2}=4\) is \(\displaystyle\frac{\pi}{3}\). Then m is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{\sqrt{3}}\)
Let the line y = mx make an angle \(\theta\) with the x-axis, so \(m = \tan\theta\). The region lies between the x-axis and this line inside the circle of radius 2, so it is a sector of angle \(\theta\). Its area is the fraction \(\frac{\theta}{2\pi}\) of the disc area \(4\pi\):
\( \displaystyle \frac{\theta}{2\pi}\times4\pi = 2\theta \)
\( \displaystyle 2\theta = \frac{\pi}{3},\ \ \theta = \frac{\pi}{6} \)
\( \displaystyle m = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \)
Why the other options are incorrect
Option (1) \(\sqrt{3}\) — Incorrect. This line makes \(60^{\circ}\) with the x-axis and gives a sector of area \(\frac{2\pi}{3}\), not \(\frac{\pi}{3}\).
Option (2) 1 — Incorrect. The line y = x gives a sector of angle \(\frac{\pi}{4}\) and area \(\frac{\pi}{2}\).
Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. \(\frac{1}{2}\) is \(\sin\frac{\pi}{6}\), not \(\tan\frac{\pi}{6}\); the slope of a line is the tangent of its angle.