Discussion
The maximum value of \(\dfrac{\log x}{x}\), \(x \gt 0\) is :
Answer: B
The maximum value is \(\dfrac{1}{e}\).
Let \(g(x) = \dfrac{\log x}{x}\). Then
\[ g'(x) = \frac{1 - \log x}{x^{2}} \]
which is positive for \(x \lt e\) and negative for \(x \gt e\). So g rises to a maximum at \(x = e\) and falls after it, and the maximum value is \(g(e) = \dfrac{\log e}{e} = \dfrac{1}{e}\).
\(e\) is where the maximum occurs, not its value. \(e + 1/e\) and \(e - 1/e\) are not values of g at all: since \(\log x \lt x\), \(g(x) \lt 1\) everywhere, while both those numbers exceed 2.
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