Question

The maximum value of \(\frac{\log_ex}{x}\), if x > 0 is

• A. e
• B. 1
• C. \(\frac{1}{e}\)
• D. \(-\frac{1}{e}\)

Answer 1

The Correct Option is C

We need to find the maximum value of \(\frac{\log_e x}{x}\) for x > 0.

Let \(f(x) = \frac{\ln x}{x}\). To find the maximum value, we need to find the critical points by finding the first derivative and setting it to zero.

\(f'(x) = \frac{d}{dx}(\frac{\ln x}{x})\)

Using the quotient rule: \(\frac{d}{dx} \frac{u}{v} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

\(f'(x) = \frac{x(\frac{1}{x}) - \ln x (1)}{x^2} = \frac{1 - \ln x}{x^2}\)

Set \(f'(x) = 0\):

\(\frac{1 - \ln x}{x^2} = 0\)

\(1 - \ln x = 0\)

\(\ln x = 1\)

\(x = e\)

Now we need to verify that x = e corresponds to a maximum. We can use the second derivative test.

\(f''(x) = \frac{d}{dx} (\frac{1 - \ln x}{x^2})\)

Using the quotient rule:

\(f''(x) = \frac{x^2(-\frac{1}{x}) - (1-\ln x)(2x)}{(x^2)^2} = \frac{-x - 2x + 2x\ln x}{x^4} = \frac{-3x + 2x \ln x}{x^4}\)

\(f''(x) = \frac{x(-3 + 2 \ln x)}{x^4} = \frac{-3 + 2 \ln x}{x^3}\)

Evaluate \(f''(e)\):

\(f''(e) = \frac{-3 + 2 \ln e}{e^3} = \frac{-3 + 2(1)}{e^3} = \frac{-1}{e^3}\)

Since \(f''(e) < 0\), the critical point at x = e corresponds to a maximum.

Now, we find the maximum value of f(x) at x = e:

\(f(e) = \frac{\ln e}{e} = \frac{1}{e}\)

Therefore, the maximum value of \(\frac{\log_e x}{x}\) is \(\frac{1}{e}\).

Thus, the correct option is (C) \(\frac{1}{e}\).