The function f(x) = tan^-1 x - x , x ∈ R , is

The correct answer is Option (2) → strictly decreasing on R

Differentiating:

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

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

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

So \(f'(x) \lt 0\) for every \(x \ne 0\), and f’(x) = 0 only at the single point x = 0.

A derivative that is negative everywhere except at isolated points still makes the function strictly decreasing. Hence f is strictly decreasing on R.

Why the other options are incorrect

• Option (1) strictly increasing on R — Incorrect. f’(x) is never positive.

• Option (3) strictly increasing on \((0, \infty)\) and strictly decreasing on \((-\infty, 0)\) — Incorrect. For x > 0, f’(x) < 0, so f is not increasing there.

• Option (4) strictly decreasing on \((0, \infty)\) and strictly increasing on \((-\infty, 0)\) — Incorrect. For x < 0, f’(x) < 0 as well, so f is not increasing there.