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: Understanding the Question:
We need to determine the interval for which the function $f(x)$ is increasing. A function is increasing if its derivative is positive ($f'(x) > 0$).

Step 2: Key Formula or Approach:
Find $f'(x)$ and determine where it is strictly positive. Recall $\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$.

Step 3: Detailed Explanation:
$f(x) = \cot^{-1} x + x$
$f'(x) = -\frac{1}{1+x^2} + 1$
Simplify the expression: $f'(x) = \frac{-(1) + (1+x^2)}{1+x^2} = \frac{x^2}{1+x^2}$.
Since $x^2 \ge 0$ for all real $x$ and $1+x^2 > 0$, $f'(x) \ge 0$ for all $x \in \mathbb{R}$.
The derivative is zero only at $x=0$ and positive everywhere else, so the function is strictly increasing on $(-\infty, \infty)$.

Step 4: Final Answer:
The function is increasing in the interval $(-\infty, \infty)$, which corresponds to option (A).