Q107 / 180 · COMEDK 2018 · Mathematics
The maximum value of $\left(\frac{1}{x}\right)^{2 x^{2}}$ is
- Correct answerA$e^{1 / e}$
- B$\sqrt{e}$
- C1
- D$e^{2}$
Solution
Let $f(x)=\left(\frac{1}{x}\right)^{2 x^{2}}=x^{-2 x^{2}}$
$$
\begin{array}{ll}
\therefore & \log f(x)=-2 x^{2} \log x \\
\Rightarrow \quad & \frac{1}{f(x)} f^{\prime}(x)=-4 x \log x-\frac{2 x^{2}}{x} \\
\Rightarrow \quad & f^{\prime}(x)=x^{-2 x^{2}} \quad[-4 x \log x-2 x]
\end{array}
$$
For maximum,
$f^{\prime}(x)=0$
$\Rightarrow \quad x^{-2 x^{2}}[-4 x \log x-2 x]=0$
$\Rightarrow \quad-4 x \log x-2 x=0$
$\Rightarrow \quad \log x=-\frac{1}{2}$
$\Rightarrow \quad x=e^{-\frac{1}{2}}$
Again, $f^{\prime \prime}(x)=x^{-2 x^{2}}[-4 x \log x-2 x]^{2}$
$+x^{-2 x^{2}}[-4 \log x-4-2]$
$\therefore \quad f^{\prime \prime}\left(e^{-\frac{1}{2}}\right) < 0$
So, $f(x)$ has maximum value at $x=e^{-\frac{1}{2}}$
$$
\therefore \quad f_{\max }=f\left(e^{-\frac{1}{2}}\right)=\left(\frac{1}{\frac{1}{\sqrt{e}}}\right)^{2\left(\frac{1}{\sqrt{e}}\right)^{2}}=(\sqrt{e})^{2\left(\frac{1}{e}\right)}=e^{1 / e}
$$
Hence, the correct answer is option: 'A'.