Question

The minimum value of ${27}^{\cos \left(x\right)}·{81}^{\sin \left(2x\right)}$ is

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

Answer 1

The correct option is C

$\frac{1}{243}$

Explanation for the correct option

Given function is ${27}^{\cos \left(x\right)}·{81}^{\sin \left(2x\right)}$
Let, $f\left(x\right)={27}^{\cos \left(x\right)}·{81}^{\sin \left(2x\right)}$
$\begin{array}{rcl}& \Rightarrow & f\left(x\right)={\left(3^3\right)}^{\cos \left(2x\right)}·{\left(3^4\right)}^{\sin \left(2x\right)}\\ & =& 3^{3\cos \left(2x\right)+4\sin \left(2x\right)}\end{array}$

We observe that the function will be minimum if the exponent is minimum.

We also know that,
$-\sqrt{a^2+b^2}\le a\sin \theta +b\cos \theta \le \sqrt{a^2+b^2}$

Here, $a=3$, $b=4$, and $\theta =2x$

Thus,
$\begin{array}{cc}& -\sqrt{3^2+4^2}\le 4\sin \left(2x\right)+3\cos \left(2x\right)\le \sqrt{3^2+4^2}\\ \Rightarrow & -\sqrt{25}\le 4\sin \left(2x\right)+3\cos \left(2x\right)\le \sqrt{25}\\ \Rightarrow & -5\le 4\sin \left(2x\right)+3\cos \left(2x\right)\le 5\end{array}$

Therefore, the minimum value of $4\sin \left(2x\right)+3\cos \left(2x\right)$ is $-5$

Therefore, the minimum value of $3^{4\sin \left(2x\right)+3\cos \left(2x\right)}$ is $3^{-5}=\frac{1}{243}$

Hence, option(C) i.e. $\frac{1}{243}$ is correct.