The interval in which the function \( f(x)=\displaystyle\frac{x}{\log_e x} \) is increasing, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (e,\infty) \)
The function is defined for x > 0, x ≠ 1 (log x must exist and be non-zero).
\( \displaystyle f'(x)=\frac{\log_e x\cdot 1-x\cdot\frac{1}{x}}{(\log_e x)^2} \)
\( \displaystyle =\frac{\log_e x-1}{(\log_e x)^2} \)
The denominator is positive, so
\( \displaystyle f'(x)\gt 0\ \Rightarrow\ \log_e x\gt 1\ \Rightarrow\ x\gt e \)
So f is increasing on \( (e,\infty) \).
Why the other options are incorrect
Option (1) \( (-\infty,e) \) — Incorrect. f is not even defined for x ≤ 0 or at x = 1.
Option (3) \( (0,e) \) — Incorrect. f is not defined at x = 1, and on (0, 1) and (1, e), \( \log_e x\lt 1 \) gives f′(x) < 0.
Option (4) \( (0,e)-\{1\} \) — Incorrect. On (0, e) − {1}, f′(x) < 0; this is where f is decreasing.