Question

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

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

Answer 1

Answer: (B)

Let $ y=4{{e}^{2x}}+9{{e}^{-2x}} $
$ \therefore $ $ \frac{dy}{dx}=8{{e}^{2x}}-18{{e}^{-2x}} $
$ \Rightarrow $ x $ \frac{{{d}^{2}}y}{d{{x}^{2}}}=16{{e}^{2x}}+36{{e}^{-2x}} $
For minimum, put $ \frac{dy}{dx}=0 $
$ \Rightarrow $ $ 8{{e}^{2x}}-18{{e}^{-2x}}=0 $
$ \Rightarrow $ $ 8{{e}^{2x}}=18{{e}^{-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=4{{e}^{2\left( \frac{1}{4}\log \frac{9}{4} \right)}}+9{{e}^{-2\left( \frac{1}{4}\log \frac{9}{4} \right)}} $
$ =4{{e}^{\log {{\left( \frac{9}{4} \right)}^{1/2}}}}+9{{e}^{\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 $