The minimum value of 27^{cos 2x} 81^{sin 2x} | vedclass

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Question

(A) Given expression is $f(x) = 27^{\cos 2x} 81^{\sin 2x}$.We can write this as $f(x) = (3^3)^{\cos 2x} (3^4)^{\sin 2x} = 3^{3 \cos 2x + 4 \sin 2x}$.T

Vedclass · Accepted Answer

$1/243$

Vedclass · Answer

(A) Given expression is $f(x) = 27^{\cos 2x} 81^{\sin 2x}$.We can write this as $f(x) = (3^3)^{\cos 2x} (3^4)^{\sin 2x} = 3^{3 \cos 2x + 4 \sin 2x}$.To find the minimum value of $f(x)$,we need to find the minimum value of the exponent $g(x) = 3 \cos 2x + 4 \sin 2x$.The range of the function $a \cos 	heta + b \sin 	heta$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = 3$ and $b = 4$,so the minimum value of $g(x) = -\sqrt{3^2 + 4^2} = -\sqrt{9 + 16} = -\sqrt{25} = -5$.Therefore,the minimum value of $f(x) = 3^{-5} = \frac{1}{3^5} = \frac{1}{243}$.

The minimum value of $27^{\cos 2x} 81^{\sin 2x}$ is.......

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