The function \(f(x) = \log(1 + x) - \displaystyle\frac{x}{1 + x}\), \(x \gt -1\), is strictly increasing in the interval
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \((0, \infty)\)
Differentiating, using the quotient rule for the second term:
\( \displaystyle f'(x) = \frac{1}{1 + x} - \frac{(1 + x) - x}{(1 + x)^{2}} \)
\( \displaystyle = \frac{1}{1 + x} - \frac{1}{(1 + x)^{2}} \)
\( \displaystyle = \frac{(1 + x) - 1}{(1 + x)^{2}} \)
\( \displaystyle = \frac{x}{(1 + x)^{2}} \)
The denominator is positive for \(x \gt -1\), so f′(x) has the sign of x:
\( f'(x) \lt 0 \) for \( -1 \lt x \lt 0 \)
\( f'(x) \gt 0 \) for \( x \gt 0 \)
So f is strictly increasing in \((0, \infty)\).
Why the other options are incorrect
Option (1) \((-1, \infty)\) — Incorrect. It includes (−1, 0), where f′(x) < 0.
Option (2) \((-1, 0)\) — Incorrect. On (−1, 0) the numerator x is negative, so f is decreasing there.
Option (3) \((-1, 1)\) — Incorrect. f decreases on (−1, 0) and increases only from 0 onwards.