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Q.

x=sin⁡θcos⁡θ,y=cos⁡θcos⁡2θ then the value of dydx at θ=π4 is

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a

2

b

0

c

3

d

4

answer is B.

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Detailed Solution

dxdθ=sinθ12cosθ−sinθ+cosθcosθ-----1dydθ=cos⁡θ22cos⁡2θ(−sin⁡2θ)+cos⁡2θ(−sin⁡θ)=−cos⁡θ2sin⁡θcos⁡θcos⁡2θ−sin⁡θcos⁡2θdydθ=−2cos2⁡θsin⁡θ−sin⁡θcos⁡2θcos⁡2θdydx=dy∣dθdx∣dθ=−2cos2⁡θsin⁡θ−sin⁡θcos⁡2θcos⁡2θ−sin2⁡θ2cos⁡θ+cos⁡θcos⁡θθ=π4,dydx=−21212−12(0)=0
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