The interval in which \( f(x)=\displaystyle\frac{x}{\log x} \) is increasing, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (e,\infty) \)
The function is defined for x > 0, x ≠ 1 (log 1 = 0 would make the denominator zero).
\( \displaystyle f'(x)=\frac{\log x\cdot 1-x\cdot\frac{1}{x}}{(\log x)^2} \)
\( \displaystyle =\frac{\log x-1}{(\log x)^2} \)
The denominator is positive, so the sign of \( f'(x) \) is the sign of \( \log x-1 \):
\( \displaystyle f'(x)\gt 0\ \Rightarrow\ \log x\gt 1\ \Rightarrow\ x\gt e \)
So f is increasing on \( (e,\infty) \).
Why the other options are incorrect
Option (1) \( (0,\infty) \) — Incorrect. f is not even defined at x = 1, and \( f'(x)\lt 0 \) for 0 < x < 1.
Option (2) \( (0,e) \) — Incorrect. On \( (0,e) \), \( \log x\lt 1 \), so \( f'(x)\lt 0 \) and f is decreasing (and undefined at x = 1).
Option (3) \( (1,\infty) \) — Incorrect. For 1 < x < e, \( \log x-1\lt 0 \), so f is decreasing on part of this interval.