For the function \(f(x) = \displaystyle\frac{\log x}{x}\), \(x \gt 0\):
(A) \(f'(x) = \displaystyle\frac{1 - \log x}{x^{2}}\).
(B) f is strictly increasing in \((e, \infty)\).
(C) The maximum value of f is \(\displaystyle\frac{1}{e}\).
(D) \(f(x) \lt 0\) for every x in (0, 1).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
By the quotient rule:
\( \displaystyle f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^{2}} = \frac{1 - \log x}{x^{2}} \)
Since \(x^{2} \gt 0\), the sign of f′(x) is the sign of \(1 - \log x\):
\( f'(x) \gt 0 \) for \( 0 \lt x \lt e \)
\( f'(x) \lt 0 \) for \( x \gt e \)
So f rises up to x = e and falls after it, and the maximum value is
\( \displaystyle f(e) = \frac{\log e}{e} = \frac{1}{e} \)
So f′(x) = (1 − log x)/x2, f increases in (0, e), decreases in (e, ∞), has maximum value 1/e, and is negative wherever log x is negative.
Checking the statements
(A) \(f'(x) = \displaystyle\frac{1 - \log x}{x^{2}}\). — Correct. Quotient rule with \((\log x)' = \frac{1}{x}\).
(B) f is strictly increasing in \((e, \infty)\). — Incorrect. For x > e, log x > 1, so f′(x) < 0 and f is decreasing.
(C) The maximum value of f is \(\displaystyle\frac{1}{e}\). — Correct. f′ changes from + to − at x = e, and f(e) = 1/e.
(D) \(f(x) \lt 0\) for every x in (0, 1). — Correct. For 0 < x < 1, log x < 0 while x > 0.