Question

The maximum value of \( \frac{\log x}{x}, \; 0 < x < \infty \) is

• A. $\infty$
• B. $e$
• C. 1
• D. $e^{-1}$

Hint:

The function $x^{1/x}$ and $(\log x)/x$ both reach their maximum at $x = e$.

Answer 1

The Correct Option is D

Step 1: Differentiation
Let $y = \frac{\log x}{x}$. $\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x}{x^{2}} = \frac{1 - \log x}{x^{2}}$.
Step 2: Critical Points
Put $\frac{dy}{dx} = 0 \implies \log x = 1 \implies x = e$.
Step 3: Verification
The second derivative $\frac{d^{2}y}{dx^{2}} = -\frac{(3 - 2 \log x)}{x^{3}}$. At $x = e$, $\frac{d^{2}y}{dx^{2}} = -\frac{1}{e^{3}}<0$, which indicates a maximum.
Step 4: Calculation
Maximum value at $x = e$ is $\frac{\log e}{e} = \frac{1}{e} = e^{-1}$.
Final Answer: (d)