Question

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

• A. $\frac{1}{243}$
• B. $\frac{1}{27}$
• C. $-5$
• D. $\frac{1}{5}$

Answer 1

The correct option is A

$\frac{1}{243}$

Explanation for the correct option:

Find the minimum value of the given trigonometric function.

A trigonometric function ${27}^{\cos 2x}·{81}^{\sin 2x}$ is given.

Rewrite the function as follows:

$$\begin{gathered} {27}^{\cos 2x}·{81}^{\sin 2x}={\left(3\right)}^{3\cos 2x}·{\left(3\right)}^{4\sin 2x} \\ \Rightarrow {27}^{\cos 2x}·{81}^{\sin 2x}={\left(3\right)}^{3\cos 2x+4\sin 2x} \end{gathered}$$

We know that, the minimum value of the function $a\cos 2x+b\sin 2x$ is given by $-\sqrt{a^2+b^2}$.

Therefore, the minimum value of the given function is given by:

$$\begin{gathered} {\left(3\right)}^{-\sqrt{3^2+4^2}}={\left(3\right)}^{-\sqrt{9+16}} \\ \Rightarrow {\left(3\right)}^{-\sqrt{3^2+4^2}}={\left(3\right)}^{-\sqrt{25}} \\ \Rightarrow {\left(3\right)}^{-\sqrt{3^2+4^2}}={\left(3\right)}^{-5} \\ \Rightarrow {\left(3\right)}^{-\sqrt{3^2+4^2}}=\frac{1}{243} \end{gathered}$$

Therefore, the minimum value of ${27}^{\cos 2x}·{81}^{\sin 2x}$ is $\frac{1}{243}$.

Hence, the option $A$ is the correct answer.