Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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)

$$\mathrm{Minimum} \\ $$$$\frac{\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta} +\mathrm{2}^{\mathrm{cos}^{\mathrm{2}} \theta} }{\mathrm{2}}\geqslant\sqrt{\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta} } \\ $$$$\mathrm{2}^{\mathrm{sin}^{\mathrm{2}} \theta} +\mathrm{2}^{\mathrm{cos}^{\mathrm{2}} \theta} \geqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Minimum}\:\mathrm{is}\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com