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The maximum value of $\frac{\log x}{x}, 0 < x < \infty$ is
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The correct answer is:
$e^{-1}$
Let $\quad y=\frac{\log x}{x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log x}{x^2}=\frac{1-\log x}{x^2}$
Put $\quad \frac{d y}{d x}=0 \Rightarrow \log x=1$
$\Rightarrow \quad x=e$
Now, $\frac{d^2 y}{d x^2}=\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x) 2 x}{\left(x^2\right)^2}$
$=-\frac{(3-2 \log x)}{x^3}$
$\Rightarrow\left(\frac{d^2 y}{d x^2}\right)_{(x=e)}=\frac{-(3-2)}{e^3}=-\frac{1}{e^3} < 0$, maxima
Hence, maximum value at $x=e$ is $\frac{1}{e}$.
$\Rightarrow \quad \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log x}{x^2}=\frac{1-\log x}{x^2}$
Put $\quad \frac{d y}{d x}=0 \Rightarrow \log x=1$
$\Rightarrow \quad x=e$
Now, $\frac{d^2 y}{d x^2}=\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x) 2 x}{\left(x^2\right)^2}$
$=-\frac{(3-2 \log x)}{x^3}$
$\Rightarrow\left(\frac{d^2 y}{d x^2}\right)_{(x=e)}=\frac{-(3-2)}{e^3}=-\frac{1}{e^3} < 0$, maxima
Hence, maximum value at $x=e$ is $\frac{1}{e}$.
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