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Question

If x=eθ(sinθ+cosθ) and y=eθ(sinθ cosθ), where θ is a real parameter, then d2ydx2 at θ=π6 is

A
32
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B
433eπ/6
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C
833eπ/6
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D
43
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Solution

The correct option is B 433eπ/6
Given : x=eθ(sinθ+cosθ), y=eθ(sinθcosθ)
dxdθ=eθ(sinθ+cosθ)+eθ(cosθsinθ)dxdθ=2eθcosθdydθ=eθ(sinθcosθ)+eθ(cosθ+sinθ)dydθ=2eθsinθdydθdxdθ=tanθdydx=tanθd2ydx2=sec2θdθdx=sec2θ2eθcosθd2ydx2θ=π6=433eπ/6

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