Question

The local maximum value of the function $f\left(x\right)={\left(\frac{2}{x}\right)}^{x^2},x>0,$ is

• A. ${\left(2\sqrt{e}\right)}^{\frac{1}{e}}$
• B. ${\left(\frac{4}{\sqrt{e}}\right)}^{\frac{e}{4}}$
• C. $1$
• D. $(e{)}^{\frac{2}{e}}$

Answer 1

The correct option is D $(e{)}^{\frac{2}{e}}$

Let $y={\left(\frac{2}{x}\right)}^{x^2},(x>0)$
Taking ${\log }_e$ both sides, we get
$\ln y=x^2\ln \left(\frac{2}{x}\right)=x^2(\ln 2-\ln x)$
Differentiate both sides
$\frac{1}{y}\cdot y^{\prime }=2x(\ln 2-\ln x)+x^2\left(\frac{-1}{x}\right)$
$=x\left[\right(2\ln 2-1)-2\ln x]$
$y^{\prime }={\left(\frac{2}{x}\right)}^{x^2}x(\ln \frac{4}{e}-\ln x^2)$
For local maxima,
$y^{\prime }=0$
$\Rightarrow \ln \left(\frac{4}{e}\right)=\ln x^2$
$\Rightarrow x^2=\frac{4}{e}$
$\Rightarrow x=\frac{2}{\sqrt{e}}[\because x>0]$
$y$ is maximum at $x=\frac{2}{\sqrt{e}}$ as can be seen from sign change of $y^{\prime }$ across $x=\frac{2}{\sqrt{e}}$.
$y_{max}=y\left(\frac{2}{\sqrt{e}}\right)={\left(\sqrt{e}\right)}^{\frac{4}{e}}=e^{\frac{1}{2}\times \frac{4}{e}}=(e{)}^{\frac{2}{e}}$