Question

The angle of intersection of the curves y = 2 sin² x and y = cos 2 x at $x=\frac{\mathrm{\pi }}{6}$ is

(a) π/4
(b) π/2
(c) π/3
(d) none of these

Answer 1

(c) π/3

$$\begin{gathered} \text{Given:} \\ x=\frac{\mathrm{\pi }}{6} \\ \mathrm{Now}, \\ y=2{\sin }^{2}x \\ \Rightarrow \frac{dy}{dx}=4\sin x\cos x \\ \Rightarrow \frac{dy}{dx}=2\sin 2x \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{6}}=2\times \frac{\sqrt{3}}{2}=\sqrt{3} \\ \mathrm{Also}, \\ y=\cos 2x \\ \Rightarrow \frac{dy}{dx}=-2\sin 2x \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{6}}=-2\times \frac{\sqrt{3}}{2}=-\sqrt{3} \\ \therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{3}+\sqrt{3}}{1-\sqrt{3}\sqrt{3}}\right|=\left|\frac{2\sqrt{3}}{-2}\right|=\sqrt{3} \\ \Rightarrow \theta ={\tan }^{-1}\left(\sqrt{3}\right)=\frac{\mathrm{\pi }}{3} \end{gathered}$$