Which of the following functions is strictly increasing on (0, ∞) ?

The correct answer is Option (3) → \(f(x) = \log\left(1 + x^{2}\right)\)

A function is strictly increasing on an interval if \(f'(x) \gt 0\) at every point of it. For \(f(x) = \log(1 + x^{2})\), by the chain rule,

    \( \displaystyle f'(x) = \frac{2x}{1 + x^{2}} \)

For x > 0 both 2x and \(1 + x^{2}\) are positive, so \(f'(x) \gt 0\) on all of \((0, \infty)\).

Why the other options are incorrect

• Option (1) \(f(x) = x^{2} - 2x\) — Incorrect. \(f'(x) = 2x - 2 \lt 0\) for 0 < x < 1, so it decreases there.

• Option (2) \(f(x) = x + \displaystyle\frac{4}{x}\) — Incorrect. \(f'(x) = 1 - \frac{4}{x^{2}} \lt 0\) for 0 < x < 2, so it decreases there.

• Option (4) \(f(x) = e^{-x^{2}}\) — Incorrect. \(f'(x) = -2x\,e^{-x^{2}} \lt 0\) for every x > 0, so it is strictly decreasing on \((0, \infty)\).