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Question

If x=2sinθsin2θ and y=2cosθcos2θ, θ [0,2π], then d2ydx2 at θ=π is :

A
38
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B
34
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C
32
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D
34
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E
None of these
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Solution

The correct option is E None of these
dxdθ=2cosθ2cos 2θ
dydθ=2sinθ+2sin2θ
dydx=2cos3θ2sinθ22sin3θ2sinθ2
dydx=cot3θ2
d2ydx2=32cosec23θ2dθdx
d2ydx2=(32cosec23θ2)1(2cosθ2cos2θ)
d2ydx2θ=π=38
None of the above option satisfies the answer. This is a BONUS question.

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