Question

The angle of intersection of the curves y = 2 sin² x and y = cos 2 x at \[x = \frac{\pi}{6}\] is ____________ .

Options

• π/4

• π/2

• π/3

• none of these

Answer 1

π/3

 

\[\text { Given }:\]

\[x = \frac{\pi}{6}\]

\[\text { Now }, \]

\[y = 2 \sin^2 x\]

\[ \Rightarrow \frac{dy}{dx} = 4\sin x \cos x\]

\[ \Rightarrow \frac{dy}{dx} = 2 \sin 2x\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{6}} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}\]

\[\text { Also }, \]

\[y = \cos 2x\]

\[ \Rightarrow \frac{dy}{dx} = - 2 \sin2x\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{6}} = - 2 \times \frac{\sqrt{3}}{2} = - \sqrt{3}\]

\[ \therefore \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\sqrt{3} + \sqrt{3}}{1 - \sqrt{3}\sqrt{3}} \right| = \left| \frac{2\sqrt{3}}{- 2} \right| = \sqrt{3}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \sqrt{3} \right) = \frac{\pi}{3}\]