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Question

x=sinθ+θcosθ,y=cosθθsinθ,then(dydx)θ=π2


A

-π/2

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B

2/π

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C

π/4

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D

4/π

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Solution

The correct option is D

4/π


x=sin θ+θ cosθdxdθ=2cosθθsinθ y=cosθ+θsinθdydθ=2sinθθcosθdydx=2sinθθ cosθ2 cosθθsinθ


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