Let `f(x)=cot^-1g(x)]` where g(x) is an increasing function on the interval `(0,pi)` Then f(x) is

Options

A. increasing on `(0,pi)`
B. decreasing on `(0,pi)`
C. increasing on `(0,pi//2)` and decreasing on `(pi//2 , pi)`
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) = \cot^{-1}(g(x)) \), where \( g(x) \) is given to be an increasing function on the interval \( (0, \pi) \). We will determine whether \( f(x) \) is increasing or decreasing in this interval.

### Step-by-step Solution:

1. **Understanding the Function**:
We have \( f(x) = \cot^{-1}(g(x)) \). The cotangent inverse function, \( \cot^{-1}(x) \), is a decreasing function. This means that if \( g(x) \) increases, \( f(x) \) will decrease.

2. **Finding the Derivative**:
To analyze whether \( f(x) \) is increasing or decreasing, we need to find the derivative \( f'(x) \).
\[ f'(x) = \frac{d}{dx} \left( \cot^{-1}(g(x)) \right) \]
Using the chain rule, we have:
\[ f'(x) = -\frac{1}{1 + (g(x))^2} \cdot g'(x) \]

3. **Analyzing the Derivative**:
- The term \( 1 + (g(x))^2 \) is always positive since \( g(x) \) is a real-valued function.
- Since \( g(x) \) is an increasing function, its derivative \( g'(x) \) is greater than 0 for \( x \in (0, \pi) \).

4. **Conclusion about \( f'(x) \)**:
Since \( f'(x) = -\frac{g'(x)}{1 + (g(x))^2} \):
- The numerator \( g'(x) > 0 \) (because \( g(x) \) is increasing).
- The denominator \( 1 + (g(x))^2 > 0 \).
- Therefore, \( f'(x) < 0 \) (since it is negative).

5. **Final Result**:
Since \( f'(x) < 0 \) for all \( x \in (0, \pi) \), we conclude that \( f(x) \) is a decreasing function on the interval \( (0, \pi) \).

### Answer:
Thus, \( f(x) \) is a decreasing function on the interval \( (0, \pi) \).