The function \(f(x) = \log(1 + x) - x\), \(x \gt -1\), is strictly decreasing in
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((0, \infty)\)
The derivative is
\( \displaystyle f'(x) = \frac{1}{1 + x} - 1 = \frac{-x}{1 + x} \)
For x > −1 the denominator 1 + x is positive, so the sign of \(f'(x)\) is the sign of −x:
\(f'(x) \gt 0\) for \(-1 \lt x \lt 0\)
\(f'(x) \lt 0\) for \(x \gt 0\)
So f is strictly decreasing in \((0, \infty)\).
Why the other options are incorrect
Option (1) \((-1, 0)\) — Incorrect. Here \(f'(x) \gt 0\), so f is strictly increasing in this interval.
Option (3) \((-1, \infty)\) — Incorrect. This includes (−1, 0), where f is increasing.
Option (4) \((-\infty, 0)\) — Incorrect. f is not defined for \(x \le -1\), and it is increasing on (−1, 0).