Question

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

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

Answer 1

Answer: (B)

Let $y=4e^{2x}+9e^{-2x}$
$\therefore$ $\frac{dy}{dx}=8e^{2x}-18e^{-2x}$
$\Rightarrow$ x $\frac{d^{2}y}{d{x}^2}=16e^{2x}+36e^{-2x}$
For minimum, put $\frac{dy}{dx}=0$
$\Rightarrow$ $8e^{2x}-18e^{-2x}=0$
$\Rightarrow$ $8e^{2x}=18e^{-2x}$
$\Rightarrow$ $e^{4x}=\frac{18}{8}=\frac{9}{4}$
$\Rightarrow$ $4x=\log \frac{9}{4}$
$\Rightarrow$ $x=\frac{1}{4}\log \frac{9}{4}$
At $x=\frac{1}{4}\log \frac{9}{4}$
$\frac{d^{2}y}{d{x}^2}>0$
$\therefore$ Minimum value at $x=\frac{1}{4}\log \frac{9}{4}$
is $y=4e^{2\left(\frac{1}{4}\log \frac{9}{4}\right)}+9e^{-2\left(\frac{1}{4}\log \frac{9}{4}\right)}$
$=4e^{\log {\left(\frac{9}{4}\right)}^{1/2}}+9e^{\log {\left(\frac{9}{4}\right)}^{-1/2}}$
$=4{\left(\frac{9}{4}\right)}^{1/2}+9{\left(\frac{9}{4}\right)}^{-1/2}$
$=4.\frac{3}{2}+9.\frac{2}{3}=6+6=12$