The maximum value of {left( {frac{1}{x}} righ | vedclass

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(B) Let $y = (\frac{1}{x})^{2x^2} = x^{-2x^2}$.Taking the natural logarithm on both sides,we get $\ln y = -2x^2 \ln x$.Differentiating with respect to

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$\sqrt[e]{e}$

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(B) Let $y = (\frac{1}{x})^{2x^2} = x^{-2x^2}$.Taking the natural logarithm on both sides,we get $\ln y = -2x^2 \ln x$.Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = -4x \ln x - 2x^2 (\frac{1}{x}) = -4x \ln x - 2x = -2x(2 \ln x + 1)$.Thus,$\frac{dy}{dx} = -2x(2 \ln x + 1) y$.For critical points,set $\frac{dy}{dx} = 0$. Since $x > 0$ and $y > 0$,we have $2 \ln x + 1 = 0$,which implies $\ln x = -\frac{1}{2}$,so $x = e^{-1/2} = \frac{1}{\sqrt{e}}$.At $x = \frac{1}{\sqrt{e}}$,the function attains its maximum value.Substituting $x = e^{-1/2}$ into the original expression:$y_{max} = (\frac{1}{e^{-1/2}})^{2(e^{-1/2})^2} = (e^{1/2})^{2(e^{-1})} = e^{(1/2) 	imes (2/e)} = e^{1/e} = \sqrt[e]{e}$.

The maximum value of ${\left( {\frac{1}{x}} \right)^{2{x^2}}}$ is

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