Question

The minimum value $4e^{2x}+9e^{-2x}$ is

• A. $11$
• B. $12$
• C. $10$
• D. $14$

Answer 1

The correct option is B

$12$

Explanation for the correct option

Step 1: Solve for the critical point

Given function is, $4e^{2x}+9e^{-2x}$,
Let $f\left(x\right)=4e^{2x}+9e^{-2x}$

A function has local minima at $x$ if $f\text{'}\left(x\right)=0$ and $f\text{'}\text{'}\left(x\right)>0$

We have,
$f\text{'}\left(x\right)=8e^{2x}-18e^{-2x}$ and
$f\text{'}\text{'}\left(x\right)=16e^{2x}+36e^{-2x}$

Setting $f\text{'}\left(x\right)=0$,
$\begin{array}{cc}\Rightarrow & 8e^{2x}-18e^{-2x}=0\\ \Rightarrow & 2\left[\frac{4e^{2x}·e^{2x}}{e^{2x}}-\frac{9}{e^{2x}}\right]=0\\ \Rightarrow & 4e^{4x}=9\\ \Rightarrow & e^{4x}=\frac{9}{4}\\ \Rightarrow & 4x=\ln \left(\frac{9}{4}\right)\\ \Rightarrow & x=\frac{1}{4}\ln \left(\frac{3^2}{2^2}\right)\\ \Rightarrow & x=\frac{1}{2}\ln \left(\frac{3}{4}\right)\left[\because \ln \left(a^n\right)=n\ln \left(a\right)\right]\end{array}$

Step 2: Solve for the minimum value
$\begin{array}{rcl}f\text{'}\text{'}\left(\frac{1}{2}\ln \left(\frac{3}{2}\right)\right)& =& 16e^{\cancel{2}\left[\frac{1}{\cancel{2}}\ln \left(\frac{3}{2}\right)\right]}+36e^{-\cancel{2}\left[\frac{1}{\cancel{2}}\ln \left(\frac{3}{2}\right)\right]}\\ & =& 16·\frac{3}{2}+\frac{36}{{\displaystyle \frac{3}{2}}}\left[\because e^{\ln \left(x\right)}=x\right]\\ & =& 24+24\\ & =& 48>0\end{array}$

Thus, $f\left(x\right)$ has a minima at $x=\frac{1}{2}\ln \left(\frac{3}{2}\right)$

So the minimum value of $f\left(x\right)$ is,
$\begin{array}{rcl}f\left(\frac{1}{2}\ln \left(\frac{3}{2}\right)\right)& =& 4e^{\cancel{2}\left[\frac{1}{\cancel{2}}\ln \left(\frac{3}{2}\right)\right]}+9e^{-\cancel{2}\left[\frac{1}{\cancel{2}}\ln \left(\frac{3}{2}\right)\right]}\\ & =& 4\times \frac{3}{2}+9\times \frac{2}{3}\\ & =& 6+6\\ & =& 12\end{array}$

Therefore the minimum value of $4e^{2x}+9e^{-2x}$ is $12$

Hence, option(B) i.e. $12$ is correct.