The function \( f(x)=\displaystyle\frac{x}{\log x} \) is increasing in the interval
(consider \( \log_e x=\log x \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (e,\infty) \)
The function is defined for x > 0, x ≠ 1. By the quotient rule,
\( \displaystyle f'(x)=\frac{\log x\cdot 1-x\cdot\frac{1}{x}}{(\log x)^2} \)
\( \displaystyle =\frac{\log x-1}{(\log x)^2} \)
The denominator is positive, so the sign of f′(x) is the sign of \( \log x-1 \).
\( \displaystyle f'(x)\gt 0\ \Rightarrow\ \log x\gt 1 \)
\( \displaystyle x\gt e \)
So f is increasing in \( (e,\infty) \).
Why the other options are incorrect
Option (1) \( (1,\infty) \) — Incorrect. On (1, e), \( \log x\lt 1 \), so f′(x) < 0 and f is decreasing there; f is increasing only from x = e onwards.
Option (3) \( (-\infty,e) \) — Incorrect. \( \log x \) is not defined for x ≤ 0, and on (0, e) the function is decreasing (except that it is not defined at x = 1).
Option (4) \( (-\infty,-1) \) — Incorrect. \( \log x \) is not defined for negative x, so the function does not exist on this interval.