If x=a(cosθ+θsinθ),y=f(θ),f(2π)=0, dydx=tanθθ,θ≠0 and θ≠(2n+1)π2, then f(π3)=

x=a(cosθ+θsinθ), y=f(θ) dxdθ=aθcosθ, dydθ=f1(θ) ⇒dydx=dydθdxdθ=f1(θ)aθcosθ=tanθθ⇒f1(θ)=asinθIntegrate and apply condition.

If x=a(cosθ+θsinθ),y=f(θ),f(2π)=0, dydx=tanθθ,θ≠0 and θ≠(2n+1)π2, then f(π3)=

x=a(cosθ+θsinθ), y=f(θ) dxdθ=aθcosθ, dydθ=f1(θ) ⇒dydx=dydθdxdθ=f1(θ)aθcosθ=tanθθ⇒f1(θ)=asinθIntegrate and apply condition.

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If $x = a (\cos θ + θ \sin θ), y = f (θ), f (2 π) = 0, \frac{d y}{d x} = \frac{\tan θ}{θ}, θ \neq 0$ and $θ \neq (2 n + 1) \frac{π}{2},$ then $f (\frac{π}{3}) =$

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$2 a π$

$\frac{π}{2} a$

$\frac{a}{2}$

$- 2 a$

answer is C.

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Detailed Solution

$x = a (\cos θ + θ \sin θ), y = f (θ)$
$\frac{d x}{d θ} = a θ \cos θ, \frac{d y}{d θ} = f^{1} (θ)$
$\Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{f^{1} (θ)}{a θ \cos θ} = \frac{\tan θ}{θ} \Rightarrow f^{1} (θ) = a \sin θ$
Integrate and apply condition.