Question

The function $f(x) = \cot^{-1} x + x$ is increasing in the interval.

• A. $(-\infty, \infty)$
• B. $(0, 3)$
• C. $(1, \infty)$
• D. $(-1, \infty)$

Hint:

If $f'(x) \ge 0$ and $f'(x) = 0$ only at isolated points, the function is considered strictly increasing on the entire interval.

Answer 1

The Correct Option is A

Step 1: Differentiate the function.
For $f(x) = \cot^{-1}x + x$, using $\tfrac{d}{dx}\cot^{-1}x = -\tfrac{1}{1+x^2}$: \[ f'(x) = -\frac{1}{1+x^2} + 1. \]

Step 2: Combine into one fraction.
\[ f'(x) = \frac{-1 + (1+x^2)}{1+x^2} = \frac{x^2}{1+x^2}. \]

Step 3: Check the sign.
The top $x^2$ is never negative and the bottom $1+x^2$ is always positive, so $f'(x) \ge 0$ for every real $x$. It is zero only at $x=0$ and positive elsewhere.

Step 4: Conclude.
So the function keeps rising on the whole real line. \[ \boxed{(-\infty,\ \infty)} \]