The inequality 2sinθ+2cosθ≥21−(1/√2) holds for all real values of θ.
A
True
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B
False
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Solution
The correct option is A True Ans. True. We have 12[2sinθ+2cosθ]≥√2sinθ.2sinθ [∵A.M≥G.M.] ⇒2sinθ+2cosθ≥2.2(sinθ+cosθ)/2(1) Now (sinθ+cosθ)=√2sin(θ+π/4)≥−√2 for all real θ. ∴2sinθ+2cosθ≥2.2(sinθ+cosθ)/2>2.2−√2/2=21−(1/√2) Thus 2sinθ+2cosθ≥21−(1/√2).