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Question Number 116433 by bemath last updated on 04/Oct/20
Given f(θ) = 2^(cos^2 (θ)) + 2^(sin^2 (θ)) find { ((maximum value)),((minimum value)) :}
$$\mathrm{Given}\:\mathrm{f}\left(\theta\right)\:=\:\mathrm{2}^{\mathrm{cos}\:^{\mathrm{2}} \left(\theta\right)} \:+\:\mathrm{2}^{\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right)} \\ $$$$\mathrm{find}\:\begin{cases}{\mathrm{maximum}\:\mathrm{value}}\\{\mathrm{minimum}\:\mathrm{value}}\end{cases} \\ $$
Answered by bobhans last updated on 04/Oct/20
⇒ f(θ) = 2^(1−sin^2 (θ)) + 2^(sin^2 (θ)) let 2^(sin^2 (θ)) = t ⇒f(t)= (2/t)+t first step →ln t = sin^2 (θ).ln (2) (1/t) (dt/dθ) = sin (2θ).ln (2) (dt/dθ) = (sin 2θ. ln (2)).t Now f ′(t) = {1−(2/t^2 ) }.{sin (2θ).ln (2))t taking f ′(t) = 0 we get { ((2t^2 −2=0→(t−1)(t+1)=0)),((t = 0 (rejected),because t >0)),((sin 2θ=0⇒θ=(k+1).(π/2))) :} case(1) for t = 1 →f(θ) = 3 (max) case(2) for θ=(π/2)→f(θ)=2^0 +2^1 = 3 (max) minimum value we get when 2^(cos^2 (θ)) = 2^(cos^2 (θ)) ⇒sin^2 (θ)=cos^2 (θ) or tan^2 (θ)=1 ⇒ θ = (π/4), ((3π)/4),... . Thus minimum value of f(θ) = 2^(sin^2 ((π/4))) + 2^(cos^2 ((π/4))) = 2(√2)
$$\Rightarrow\:\mathrm{f}\left(\theta\right)\:=\:\mathrm{2}^{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right)} \:+\:\mathrm{2}^{\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right)} \\ $$$$\mathrm{let}\:\mathrm{2}^{\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right)} \:=\:\mathrm{t}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\:\frac{\mathrm{2}}{\mathrm{t}}+\mathrm{t} \\ $$$$\mathrm{first}\:\mathrm{step}\:\rightarrow\mathrm{ln}\:\mathrm{t}\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right).\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{t}}\:\frac{\mathrm{dt}}{\mathrm{d}\theta}\:=\:\mathrm{sin}\:\left(\mathrm{2}\theta\right).\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{dt}}{\mathrm{d}\theta}\:=\:\left(\mathrm{sin}\:\mathrm{2}\theta.\:\mathrm{ln}\:\left(\mathrm{2}\right)\right).\mathrm{t} \\ $$$$\mathrm{Now}\:\mathrm{f}\:'\left(\mathrm{t}\right)\:=\:\left\{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{2}} }\:\right\}.\left\{\mathrm{sin}\:\left(\mathrm{2}\theta\right).\mathrm{ln}\:\left(\mathrm{2}\right)\right)\mathrm{t} \\ $$$$\mathrm{taking}\:\mathrm{f}\:'\left(\mathrm{t}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{2t}^{\mathrm{2}} −\mathrm{2}=\mathrm{0}\rightarrow\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}+\mathrm{1}\right)=\mathrm{0}}\\{\mathrm{t}\:=\:\mathrm{0}\:\left(\mathrm{rejected}\right),\mathrm{because}\:\mathrm{t}\:>\mathrm{0}}\\{\mathrm{sin}\:\mathrm{2}\theta=\mathrm{0}\Rightarrow\theta=\left(\mathrm{k}+\mathrm{1}\right).\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{for}\:\mathrm{t}\:=\:\mathrm{1}\:\rightarrow\mathrm{f}\left(\theta\right)\:=\:\mathrm{3}\:\left(\mathrm{max}\right) \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{for}\:\theta=\frac{\pi}{\mathrm{2}}\rightarrow\mathrm{f}\left(\theta\right)=\mathrm{2}^{\mathrm{0}} +\mathrm{2}^{\mathrm{1}} =\:\mathrm{3}\:\left(\mathrm{max}\right) \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{we}\:\mathrm{get}\:\mathrm{when}\:\mathrm{2}^{\mathrm{cos}\:^{\mathrm{2}} \left(\theta\right)} \:=\:\mathrm{2}^{\mathrm{cos}\:^{\mathrm{2}} \left(\theta\right)} \\ $$$$\Rightarrow\mathrm{sin}\:^{\mathrm{2}} \left(\theta\right)=\mathrm{cos}\:^{\mathrm{2}} \left(\theta\right)\:\mathrm{or}\:\mathrm{tan}\:^{\mathrm{2}} \left(\theta\right)=\mathrm{1} \\ $$$$\Rightarrow\:\theta\:=\:\frac{\pi}{\mathrm{4}},\:\frac{\mathrm{3}\pi}{\mathrm{4}},…\:.\:\mathrm{Thus}\:\mathrm{minimum}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\theta\right)\:=\:\mathrm{2}^{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)} \:+\:\mathrm{2}^{\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)} \:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Commented by bemath last updated on 04/Oct/20
Answered by Dwaipayan Shikari last updated on 04/Oct/20
Minimum ((2^(sin^2 θ) +2^(cos^2 θ) )/2)≥(√2^(sin^2 θ+cos^2 θ) ) 2^(sin^2 θ) +2^(cos^2 θ) ≥2(√2) Minimum is 2(√2)
$$\mathrm{Minimum} \\ $$$$\frac{\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta} +\mathrm{2}^{\mathrm{cos}^{\mathrm{2}} \theta} }{\mathrm{2}}\geqslant\sqrt{\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta} } \\ $$$$\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta} +\mathrm{2}^{\mathrm{cos}^{\mathrm{2}} \theta} \geqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Minimum}\:\mathrm{is}\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$

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