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 = π 6 is Options: A . &nbsp π 4 B . &nbsp π 3 C . &nbsp π 2 D . &nbsp 2 π 3

Answer: Option B : B and D 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 d y d x = 4 s i n x c o s x [ d y d x ] x − π 6 = 4 ( 1 2 ) √ 3 2 = √ 3 = m 1 ( s a y ) On differentiating equation (2) w.r.t x, we get d y d x = − 2 s i n 2 x [ d y d x ] x − π 6 = − 2 s i n π 3 = − √ 3 = m 2 ( s a y ) Hence, angle between the two curves is θ = ± t a n − 1 ( m 1 − m 2 1 + m 1 m 2 ) = ± t a n − 1 √ 3 = π 3 o r 2 π 3 Hence (b) is the correct answer.

The angle of intersection of the curves y=2sin2 x and y= c

Discussion Forum : Application Of Derivatives

Question -

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

Options:

A .

\frac{π}{4}

B .

\frac{π}{3}

C .

\frac{π}{2}

D .

\frac{2 π}{3}

Answer: Option B
:
B and D
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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