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MathematicsDifferentiationParametric/Determinant/Logarithmic InverseMedium2 minPYQ_2020
MathematicsMediumsingle choice

Ifx=2sinθ-sin2θandy=2cosθ-cos2θθ0,2π,thend2ydx2atθ=πis:

Options:

Answer:
B
Solution:

dxdθ=2cosθ-2cos2θ

dydθ=-2sinθ+2sin2θ

 dydx=sin2θ-sinθcosθ-cos2θ

=2sinθ2.cos3θ22sinθ2.sin3θ2=cot3θ2

d2ydx2=ddθdydxdθdx=-32cosec23θ2.dθdx

d2ydx2=-32cosec23θ22cosθ-cos2θ

d2ydx2θ=π=34-1-1=38

Stream:JEESubject:MathematicsTopic:DifferentiationSubtopic:Parametric/Determinant/Logarithmic Inverse
2mℹ️ Source: PYQ_2020

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