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Question

The maximum value of the expression 39cos2Θ+8sinΘcosΘ+3sin2Θ is

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Solution

Consider the given function

f(x)=39cos2θ+8sinθcosθ+3sin2θ

f(x)=38cos2θ+8sinθcosθ+2sin2θ+sin2θ+cos2θ

f(x)=38cos2θ+8sinθcosθ+2sin2θ+1

f(x)=38cos2θ+4sinθcosθ+4sinθcosθ+2sin2θ+1

f(x)=34cosθ(2cosθ+sinθ)+2sinθ(2cosθ+sinθ)+1

f(x)=3(2cosθ+sinθ)(4cosθ+2sinθ)+1

For the maximum value of expression the denominator must be minimum and for denominator to be minimum,

(2cosθ+sinθ)(4cosθ+2sinθ) can be zero.

When

(2cosθ+sinθ)(4cosθ+2sinθ)=0

2cosθ+sinθ=0,4cosθ+2sinθ=0

2cosθ=sinθ,4cosθ=2sinθ

tanθ=2,tanθ=2

Therefore, the maximum value of expression

=3(2cosθ+sinθ)(4cosθ+2sinθ)+1

=30×0+1

=3

Hence, this is the answer.


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