x=θsin2θ, y=θcos2θ then dydxθ=π2=

x=θ sin2θ, y=θ·cos2θ dxdθ=θ(cos2θ 2)+(sin2θ) dydθ=θ(-2sin2θ)+cos2θ(1) dydx=dy/dθdx/dθ=cos2θ-2θ sin2θsin2θ+2θ cos2θ dydxθ=π2=cos π-π(0)0+π(-1)=1π

x=θsin2θ, y=θcos2θ then dydxθ=π2=

x=θ sin2θ, y=θ·cos2θ dxdθ=θ(cos2θ 2)+(sin2θ) dydθ=θ(-2sin2θ)+cos2θ(1) dydx=dy/dθdx/dθ=cos2θ-2θ sin2θsin2θ+2θ cos2θ dydxθ=π2=cos π-π(0)0+π(-1)=1π

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x= $θ \sin 2 θ$ , y= $θ \cos 2 θ$ then ${[\frac{d y}{d x}]}_{θ = \frac{π}{2}} =$

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$\frac{π}{2}$

$\frac{1}{π}$

$\frac{1}{2}$

$- \frac{1}{2}$

answer is D.

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Detailed Solution

$x = θ \sin 2 θ, y = θ \cdot \cos 2 θ \frac{dx}{dθ} = θ (\cos 2 θ 2) + (\sin 2 θ) \frac{dy}{dθ} = θ (- 2 \sin 2 θ) + \cos 2 θ (1) \frac{dy}{dx} = \frac{dy / dθ}{dx / dθ} = \frac{\cos 2 θ - 2 θ \sin 2 θ}{\sin 2 θ + 2 θ \cos 2 θ} {(\frac{dy}{dx})}_{θ = \frac{π}{2}} = \frac{\cos π - π (0)}{0 + π (- 1)} = \frac{1}{π}$