The interval in which `f(x)=cot^(-1)x+x` increases , is

Options

A. a) `R`
B. b) `(0,oo)`
C. c) `R-{npi}`
D. d) None of these

Answer

To determine the interval in which the function \( f(x) = \cot^{-1}(x) + x \) is increasing, we will follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = \cot^{-1}(x) + x \]

### Step 2: Find the derivative
Next, we need to find the derivative \( f'(x) \) to analyze the monotonicity of the function. The derivative of \( \cot^{-1}(x) \) is given by:
\[ \frac{d}{dx}(\cot^{-1}(x)) = -\frac{1}{1 + x^2} \]
The derivative of \( x \) is simply \( 1 \). Therefore, we can write:
\[ f'(x) = -\frac{1}{1 + x^2} + 1 \]

### Step 3: Set the derivative greater than zero
For the function to be increasing, we need:
\[ f'(x) > 0 \]
This leads us to:
\[ -\frac{1}{1 + x^2} + 1 > 0 \]

### Step 4: Simplify the inequality
Rearranging the inequality gives:
\[ 1 - \frac{1}{1 + x^2} > 0 \]
This can be rewritten as:
\[ \frac{1 + x^2 - 1}{1 + x^2} > 0 \]
which simplifies to:
\[ \frac{x^2}{1 + x^2} > 0 \]

### Step 5: Analyze the inequality
The expression \( x^2 \) is always non-negative, and \( 1 + x^2 \) is always positive for all real \( x \). Therefore, the inequality \( \frac{x^2}{1 + x^2} > 0 \) holds true for all \( x \neq 0 \).

### Step 6: Consider the domain of \( f(x) \)
However, we must also consider the domain of \( f(x) \). The function \( \cot^{-1}(x) \) is not defined for \( x = n\pi \) where \( n \) is an integer, as it approaches infinity at these points.

### Conclusion
Thus, the function \( f(x) \) is increasing for:
\[ x \in \mathbb{R} \setminus \{ n\pi \} \]
This means the correct interval in which \( f(x) \) increases is:
\[ \text{Option C: } \mathbb{R} - \{ n\pi \} \]