MCQ
The maximum value of ${\left( {\frac{1}{x}} \right)^{2{x^2}}}$ is
- A$e$
- ✓$\sqrt[e]{e}$
- C$1$
- DNone of these
✓
Answer
Correct option: B.
$\sqrt[e]{e}$
b
Let $y=\left(\frac{1}{x}\right)^{2 x^{2}}$
Let $y=\left(\frac{1}{x}\right)^{2 x^{2}}$
$\Rightarrow \log y=-2 x^{2} \log x$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=-4 x \log x-2 x$
$\Rightarrow \frac{d y}{d x}=-2 x(2 \log x+1) y$
$\Rightarrow \frac{d y}{d x}=-2 x\left(\log x^{2}+1\right)\left(\frac{1}{x}\right)^{2 x^{2}}$
Now, the value of $y$ is maximum or minimum when, $\frac{d y}{d x}=0$ $-2 x\left(\log x^{2}+1\right)\left(\frac{1}{x}\right)^{2 x^{2}}=0$
$\Rightarrow \frac{4 x \log x+2 x}{x^{2 x^{2}}}=0$
$\Rightarrow 2 x=-4 x \log x$
$\Rightarrow \log x=\frac{-1}{2}$
$\Rightarrow x=\frac{1}{\sqrt{e}}$
Its double derivative will be negative there hence value will be maximum
Thus, the maximum value of $y$ is,
$\Rightarrow\left(\frac{1}{\frac{1}{\sqrt{e}}}\right)^{2\left(\frac{1}{\sqrt{e}}\right)^{2}}$
$\Rightarrow(\sqrt{e})^{\frac{2}{e}}$
$\Rightarrow e^{\frac{1}{2} \times \frac{2}{e}}$
$\Rightarrow e^{\frac{1}{e}}$
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