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a
−1+sin2α, 1+sin2α
b
−1+cos2α, 1+cos2α
c
[sinα, cosα]
d
no solution
answer is A.
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Detailed Solution
Let fθ=cos θ (sin θ +sin2θ+sin2α)⇒f(θ)secθ−sinθ2=sin2θ+sin2α⇒f2(θ)1+tan2θ−2f(θ)tanθ−sin2α=0⇒f2(θ)tan2θ−2f(θ)tanθ+(f2(θ)−sin2α)=0∵ tanθ is real, we have b2−4ac≥0⇒4f2(θ)−4f2(θ)(f2(θ)−sin2α)≥0⇒4f2(θ)1−f2(θ)+sin2α≥0⇒1−f2(θ)+sin2α≥0⇒f2(θ)≤1+sin2α⇒−1+sin2α≤f(θ)≤1+sin2α