If x=2sinθ-sin2θ and y=2cosθ-cos2θ, θ∈0,2π, then (d^2y)/(dx^2) at θ=π is:

Correct answer: (B) -3/8. dx/(dθ)=2cosθ-2cos2θ dy/(dθ)=-2sinθ+2sin2θ ∴ dy/dx=(sin2θ-sinθ)/(cosθ-cos2θ) =(2sinθ2.cos3θ)/22sin(θ)/2.sin(3θ)/2=cot(3θ)/2 (d^2y)/(dx^2)=d/(dθ)dy/dx(dθ)/dx=-3/2cosec^2(3θ)/2.(dθ)/dx ⇒(d^2y)/(dx^2)=(-32cosec^23θ)/22cosθ-cos2θ ⇒d^2ydx^2_(θ=π)=3/(4-1-1)=3/8

Q31 / 75 · JEE Main 2020 (09 Jan Shift 2) · Mathematics

If $x = 2 \sin θ - \sin 2 θ$ and $y = 2 \cos θ - \cos 2 θ,$ $θ \in [0, 2 π],$ then $\frac{d^{2} y}{d x^{2}}$ at $θ = π$ is:

A
$\frac{3}{4}$
Correct answer
B
$- \frac{3}{8}$
C
$\frac{3}{2}$
D
$- \frac{3}{4}$

Solution

$\frac{d x}{d θ} = 2 \cos θ - 2 \cos 2 θ$

$\frac{d y}{d θ} = - 2 \sin θ + 2 \sin 2 θ$

$∴ \frac{d y}{d x} = \frac{\sin 2 θ - \sin θ}{\cos θ - \cos 2 θ}$

$= \frac{2 \sin \frac{θ}{2} . \cos \frac{3 θ}{2}}{2 \sin \frac{θ}{2} . \sin \frac{3 θ}{2}} = \cot \frac{3 θ}{2}$

$\frac{d^{2} y}{d x^{2}} = \frac{d}{d θ} (\frac{d y}{d x}) \frac{d θ}{d x} = - \frac{3}{2} {cosec}^{2} \frac{3 θ}{2} . \frac{d θ}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- \frac{3}{2} {cosec}^{2} \frac{3 θ}{2}}{2 (\cos θ - \cos 2 θ)}$

$\Rightarrow {\frac{d^{2} y}{d x^{2}}|}_{θ = π} = \frac{3}{4 (- 1 - 1)} = \frac{3}{8}$

Hence, the correct answer is option: 'B'.

If x=2sinθ-sin2θ and y=2cosθ-cos2θ, θ∈0,2π, then d2ydx2 at θ=π is:

Solution

More Differentiation Questions

If $x = 2 \sin θ - \sin 2 θ$ and $y = 2 \cos θ - \cos 2 θ,$ $θ \in [0, 2 π],$ then $\frac{d^{2} y}{d x^{2}}$ at $θ = π$ is: