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Question

If x=sinθ,y=cos3θ, then (1x2)y2xy1 is equal to

A
y
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B
3y
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C
3y
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D
9y
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Solution

The correct option is D 9y
The given information is: x=sinθ, y=cos3θ

Differentiating x and y w.r.t θ we get,

dxdθ=cosθ

dydθ=3sinθ

dydx=3sinθcosθ

d2ydx2=ddθ(dydx).dθdx

d2ydx2=ddθ(3sinθcosθ).dθdx

d2ydx2=3(3cos3θcosθsinθsin3θ)cos2θ.1cosθ

(1x2)y2=(1sin2θ)3(3cos3θcosθsinθsin3θ)cos2θ.1cosθ

(1x2)y2=cos2θ.3(3cos3θcosθsinθsin3θ)cos2θ.1cosθ

(1x2)y2=(9cos3θcosθ+3sinθsin3θ)cosθ

xy1=3sinθsin3θcosθ

(1x2)y2+xy1=9cos3θcosθ+3sinθsin3θ3sinθsin3θcosθ

(1x2)y2+xy1=9cos3θ

(1x2)y2+xy1=9y

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