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Question

Let the function f:(0,π)R be defined by f(θ)=(sinθ+cosθ)2+(sinθcosθ)4. Suppose the function f has a local minimum at θ precisely when θ{λ1π,,λrπ}, where 0<λ1<<λr<1. Then the value of λ1++λr is

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Solution

f(θ)=(sinθ+cosθ)2+(sinθcosθ)4
=sin2θ+cos2θ+2sinθcosθ+((sinθcosθ)2)2
=1+sin2θ+(sin2θ+cos2θ2sinθcosθ)2
=1+sin2θ+(1sin2θ)2
=1+sin2θ+1+sin22θ2sin2θ
=sin22θsin2θ+2
f(θ)=(sin2θ12)2+74

Since θ(0,π),
2θ(0,2π)
f(θ) is minimum when sin2θ=12
2θ=π6,5π6
θ=π12,5π12
λ1=112, λ2=512
λ1+λ2=112+512=12=0.50

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