The maximum value of y=x(log x)^2 is | vedclass

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(D) Given $y = x(\log x)^2$. To find the critical points,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = x \cdot (2 \log x \cdot \frac{1}{x

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$4 e^{-2}$

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(D) Given $y = x(\log x)^2$. To find the critical points,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = x \cdot (2 \log x \cdot \frac{1}{x}) + (\log x)^2 \cdot 1 = 2 \log x + (\log x)^2$. Setting $\frac{dy}{dx} = 0$: $\log x (2 + \log x) = 0$. This gives $\log x = 0$ or $\log x = -2$. Thus,$x = e^0 = 1$ or $x = e^{-2}$. Now,we use the second derivative test: $\frac{d^2y}{dx^2} = \frac{2}{x} + 2 \log x \cdot \frac{1}{x} = \frac{2 + 2 \log x}{x}$. At $x = 1$,$\frac{d^2y}{dx^2} = \frac{2 + 0}{1} = 2 > 0$,so $x = 1$ is a point of local minima. At $x = e^{-2}$,$\frac{d^2y}{dx^2} = \frac{2 + 2(-2)}{e^{-2}} = \frac{-2}{e^{-2}} The maximum value is $y(e^{-2}) = e^{-2}(\log e^{-2})^2 = e^{-2}(-2)^2 = 4e^{-2}$.

The maximum value of $y=x(\log x)^2$ is

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