In which of the following interval, the function \(f(x) = \displaystyle\frac{x}{\log x}\) is decreasing?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((0, e) - \{1\}\)
The function is defined only for \(x \gt 0\) and \(x \ne 1\), since \(\log x\) must exist and must not be 0.
By the quotient rule,
\( \displaystyle f'(x) = \frac{(\log x)(1) - x \cdot \frac{1}{x}}{(\log x)^2} \)
\( \displaystyle = \frac{\log x - 1}{(\log x)^2} \)
The denominator is positive, so \(f'(x) \lt 0\) exactly when
\(\log x - 1 \lt 0\)
\(\log x \lt 1\)
\(x \lt e\)
Together with the domain, \(f'(x) \lt 0\) for \(0 \lt x \lt e,\ x \ne 1\), that is, on \((0, e) - \{1\}\).
Strictly, f is decreasing on each part, (0, 1) and (1, e), separately: f is negative on (0, 1) and positive on (1, e), so it does not keep decreasing across x = 1.
Why the other options are incorrect
Option (1) \((-\infty, e)\) — Incorrect. \(\log x\) is not defined for \(x \le 0\).
Option (2) \((0, e)\) — Incorrect. It contains x = 1, where \(\log x = 0\) and f is not defined.
Option (4) \((e, \infty)\) — Incorrect. For \(x \gt e\), \(\log x \gt 1\), so \(f'(x) \gt 0\) and f is increasing there.