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Question

If x=eθ(sinθcosθ),yeθ(sinθ+cosθ), then dydx at θ=π4 is

A
1
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B
0
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C
12
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D
2
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Solution

The correct option is A 1
We have,
dxdθ=eθ(cosθ+sinθ)+eθ(sinθcosθ)

dxdθ=2eθsinθ
Also,
dydθ=eθ(cosθsinθ)+eθ(sinθ+cosθ)dydθ=2eθcosθ

So,
dydx=cotθ
and at θ=π/4

dydx=cot(π/4)=1

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