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Question

Let M=[sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1,
Where α=α(θ) and β=β(θ) are real numbers, and I is the 2×2 identity matrix. If α is the minimum of set {α(θ):θ[0,2π)} and
β is the minimum of set {β(θ):θ[0,2π]}, then the value of α+β is

A
3716
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B
3116
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C
2916
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D
1716
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Solution

The correct option is C 2916
M=[sin4θ1sin2θ1+cos2θcos4θ]=αI+βM1,M2=αM+βI
[sin4θ1sin2θ1+cos2θcos4θ][sin4θ1sin2θ1+cos2θcos4θ]=α[sin4θ1sin2θ1+cos2θcos4θ]+β[1001]
On comparing both sides we get,
sin8θ1sin2θcos2θsin2θcos2θ=αsin4θ+β
sin8θsin2θcos2θ2=αsin4θ+β ...(i)
Also sin4θ(1+cos2θ)+cos4θ(1+cos2θ)=α(1+cos2θ)
sin4θ(1+cos2θ)+cos4θ(1+cos2θ)(1+cos2θ)=α
α=sin4θ+cos4θ ...(ii)
α=112(sin2θ)2
α is minimum whensin2θ1
α=αmin=112=12
solving (i) and (ii) we get,
β=[sin4θ.cos4θ+sin2θ.cos2θ+2]β=(t2+t+2), where t=sin22θ4[0,14]βmin=β=3716
α+β=123716=2916


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