The angle of intersection of the curves $y = 2 s i n^{2} x a n d y = c o s 2 x a t x = \frac{π}{6}$ is

$\frac{π}{4}$

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$\frac{π}{3}$

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$\frac{π}{2}$

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$\frac{2 π}{3}$

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Solution

The correct option is D
$\frac{2 π}{3}$

We have,
$y = s i n^{2} x . . . (1) y = c o s 2 x . . . (2)$ And
On differentiating equation (1) w.r.t x, we get
$\frac{d y}{d x} = 4 s i n x c o s x {[\frac{d y}{d x}]}_{x - \frac{π}{6}} = 4 (\frac{1}{2}) \frac{\sqrt{3}}{2} = \sqrt{3} = m_{1} (s a y)$
On differentiating equation (2) w.r.t x, we get
$\frac{d y}{d x} = - 2 s i n 2 x {[\frac{d y}{d x}]}_{x - \frac{π}{6}} = - 2 s i n \frac{π}{3} = - \sqrt{3} = m_{2} (s a y)$
Hence, angle between the two curves is
$θ = \pm t a n^{- 1} (\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}) = \pm t a n^{- 1} \sqrt{3} = \frac{π}{3} o r \frac{2 π}{3}$
Hence (b) is the correct answer.

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