The function f(x)=cot^-1x+x increases in the interval:

The correct answer is Option (3) → \( (-\infty,\infty) \)

Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Options (1), (2), (3) and (4) are all correct, as explained below.

Using \( \displaystyle\frac{d}{dx}\cot^{-1}x = -\displaystyle\frac{1}{1+x^2} \):

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

\( f'(x) > 0 \) for every x ≠ 0, and \( f'(0) = 0 \) only at the single point 0. So f is increasing on the whole real line, and therefore also on every interval inside it.

Option (1) \( (1,\infty) \) — Should also be correct, though not the answer given by NTA. It lies inside \( (-\infty,\infty) \).

Option (2) \( (-\infty,0) \) — Should also be correct, though not the answer given by NTA. It lies inside \( (-\infty,\infty) \).

Option (3) \( (-\infty,\infty) \) — Correct. As shown above.

Option (4) \( (0,\infty) \) — Should also be correct, though not the answer given by NTA. It lies inside \( (-\infty,\infty) \).

Advice for the exam: When more than one option is correct, mark the largest interval, here \( (-\infty,\infty) \), which contains all the others. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.