The function \(f(x) = x - \log x\), \(x \gt 0\), is strictly increasing in the interval
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((1, \infty)\)
Differentiating:
\( \displaystyle f'(x) = 1 - \frac{1}{x} \)
\( \displaystyle = \frac{x - 1}{x} \)
Since x > 0, the sign of f’(x) is the sign of x − 1. So f’(x) > 0 for x > 1 and f’(x) < 0 for 0 < x < 1.
Hence f is strictly increasing in \((1, \infty)\) and strictly decreasing in (0, 1).
Why the other options are incorrect
Option (1) \((0, 1)\) — Incorrect. Here f’(x) < 0, so f is decreasing.
Option (3) \((0, \infty)\) — Incorrect. This includes (0, 1), where f is decreasing.
Option (4) \((0, e)\) — Incorrect. This also includes (0, 1), where f is decreasing.