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
43√3e−π/6
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C
83√3e−π/6
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D
43
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Solution
The correct option is B43√3e−π/6 Given : x=eθ(sinθ+cosθ),y=eθ(sinθ−cosθ) dxdθ=eθ(sinθ+cosθ)+eθ(cosθ−sinθ)⇒dxdθ=2eθcosθdydθ=eθ(sinθ−cosθ)+eθ(cosθ+sinθ)⇒dydθ=2eθsinθ⇒dydθdxdθ=tanθ⇒dydx=tanθd2ydx2=sec2θ⋅dθdx=sec2θ2eθcosθ∴d2ydx2∣∣∣θ=π6=43√3e−π/6