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Question

If y=cotθ(sin2θ+sinθcosθ), then
(where θnπ,nZ)

A
ymin=124
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B
ymin=122
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C
ymax=1+24
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D
ymax=1+22
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Solution

The correct option is D ymax=1+22
y=cotθ(sin2θ+sinθcosθ) =cosθsinθ(sin2θ+sinθcosθ) =cosθ(sinθ+cosθ) =cosθsinθ+cos2θ =sin2θ2+1+cos2θ2 =12+sin2θ+cos2θ2

Now, 2sin2θ+cos2θ2
12212+sin2θ+cos2θ21+22

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