If the maximum value of the function \(f(x) = \displaystyle\frac{\log_e x}{x},\ x \gt 0\) occurs at \(x = a\), then \(a^2f''(a)\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{1}{e}\)
By the quotient rule:
\( \displaystyle f'(x) = \frac{\frac{1}{x}\cdot x - \log_e x}{x^2} = \frac{1 - \log_e x}{x^2} \)
At the maximum, \(f'(a) = 0\):
\(1 - \log_e a = 0\)
\(a = e\)
Differentiating f′(x) again by the quotient rule:
\( \displaystyle f''(x) = \frac{-\frac{1}{x}\cdot x^2 - (1 - \log_e x)\cdot 2x}{x^4} \)
\( \displaystyle f''(x) = \frac{-x - 2x + 2x\log_e x}{x^4} = \frac{2\log_e x - 3}{x^3} \)
At x = e:
\( \displaystyle f''(e) = \frac{2 - 3}{e^3} = -\frac{1}{e^3} \)
Hence:
\( \displaystyle a^2f''(a) = e^2\left(-\frac{1}{e^3}\right) = -\frac{1}{e} \)
Why the other options are incorrect
Option (1) \(-\displaystyle\frac{5}{e}\) — Incorrect. The value of \(a^2f''(a)\) is \(-\frac{1}{e}\), not \(-\frac{5}{e}\).
Option (3) \(-\displaystyle\frac{1}{e^3}\) — Incorrect. This is \(f''(a)\) alone; the question asks for \(f''(a)\) multiplied by \(a^2 = e^2\).
Option (4) \(-5e^3\) — Incorrect. The value of \(a^2f''(a)\) is \(-\frac{1}{e}\), not \(-5e^3\).