Given-f-2-cos-2-2-sin-2-find-maximum-value-minimum-value-

Question Number 116433 by bemath last updated on 04/Oct/20

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

\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}, \dots . 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}