Given f(θ) = 2^(cos^2 (θ)) + 2^(sin^2 (θ)) find { ((maximum value)),((minimum value)) :}


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

	Answered by Dwaipayan Shikari last updated on 04/Oct/20

	$Minimum$ $\frac{2^{\sin^{2} θ} + 2^{\cos^{2} θ}}{2} ⩾ \sqrt{2^{\sin^{2} θ + \cos^{2} θ}}$ $2^{\sin^{2} θ} + 2^{\cos^{2} θ} ⩾ 2 \sqrt{2}$ $Minimum is 2 \sqrt{2}$
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