Question

If the maximum value of the function f(x) = $\frac{\log_e x}{x}$, x > 0 occurs at x = a, then a2f''(a) is equal to

• A. $-\frac{5}{e}$
• B. $-\frac{1}{e}$
• C. $-\frac{1}{e^3}$
• D. $-5e^3$

Hint:

When finding maxima/minima, always find the first derivative and set it to zero. For complex expressions, be systematic with differentiation rules (product, quotient, chain). Double-check the final expression the question asks for.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

To find the maximum of a function, we find its critical points by setting the first derivative to zero. The second derivative test can confirm if a critical point is a maximum. We then evaluate the required expression.

Step 2: Key Formula or Approach:

Use the quotient rule for differentiation: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. \] For a maximum at \( x = a \), we must have \( f'(a) = 0 \) and \( f''(a) < 0 \).

Step 3: Detailed Explanation:

Given function: \( f(x) = \frac{\ln x}{x} \).

First Derivative:

Using the quotient rule with \( u = \ln x \) and \( v = x \):

\[ f'(x) = \frac{\left(\frac{1}{x}\right) \cdot x - (\ln x) \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}. \] To find the maximum, set \( f'(x) = 0 \): \[ \frac{1 - \ln x}{x^2} = 0 \implies 1 - \ln x = 0 \implies \ln x = 1 \implies x = e. \] So, the maximum occurs at \( a = e \).

Second Derivative:

Now, differentiate \( f'(x) \) using the quotient rule with \( u = 1 - \ln x \) and \( v = x^2 \): \[ f''(x) = \frac{\left(-\frac{1}{x}\right) \cdot x^2 - (1 - \ln x) \cdot (2x)}{(x^2)^2} \] \[ f''(x) = \frac{-x - 2x + 2x \ln x}{x^4} = \frac{-3x + 2x \ln x}{x^4} = \frac{2 \ln x - 3}{x^3}. \] Now evaluate \( f''(a) = f''(e) \): \[ f''(e) = \frac{2 \ln e - 3}{e^3} = \frac{2(1) - 3}{e^3} = \frac{-1}{e^3}. \]

Calculate the final expression:

We need to find \( a^2 f''(a) \). With \( a = e \): \[ a^2 f''(a) = e^2 \cdot \left(\frac{-1}{e^3}\right) = \frac{-e^2}{e^3} = -\frac{1}{e}. \]

Step 4: Final Answer:

The value of \( a^2 f''(a) \) is \(-\frac{1}{e}\).