If `f''(x) gt 0 AA x in R, f'(4) = 0 ` and `g(x) = f ( cot^2 x - 2 cot x + 5) , 0 lt x lt (pi)/(2)` , then

Options

A. g(x) is increasing in `( 0 , (pi)/(2))`
B. g(x) is decreasing in `(0 , (pi)/(2))`
C. g(x) is increasing in `((pi)/(4) , (pi)/(2))`
D. g(x) is decreasing in `(0 , (pi)/(4))`

Answer

To solve the problem step by step, we need to analyze the function \( g(x) = f(\cot^2 x - 2 \cot x + 5) \) and determine where it is increasing or decreasing based on the given conditions.

### Step 1: Find the derivative of \( g(x) \)

To find the increasing or decreasing nature of \( g(x) \), we first need to compute its derivative \( g'(x) \).

Using the chain rule:
\[ g'(x) = f'(\cot^2 x - 2 \cot x + 5) \cdot \frac{d}{dx}(\cot^2 x - 2 \cot x + 5) \]

### Step 2: Differentiate the inner function

Now we differentiate the inner function \( \cot^2 x - 2 \cot x + 5 \):
\[ \frac{d}{dx}(\cot^2 x) = 2 \cot x (-\csc^2 x) = -2 \cot x \csc^2 x \]
\[ \frac{d}{dx}(-2 \cot x) = 2 \csc^2 x \]
Combining these, we get:
\[ \frac{d}{dx}(\cot^2 x - 2 \cot x + 5) = -2 \cot x \csc^2 x + 2 \csc^2 x = 2 \csc^2 x (1 - \cot x) \]

### Step 3: Substitute back into \( g'(x) \)

Now substituting back into the expression for \( g'(x) \):
\[ g'(x) = f'(\cot^2 x - 2 \cot x + 5) \cdot 2 \csc^2 x (1 - \cot x) \]

### Step 4: Analyze the sign of \( g'(x) \)

Given that \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), we know that \( f'(x) \) is increasing. Since \( f'(4) = 0 \), it means that \( f'(x) < 0 \) for \( x < 4 \) and \( f'(x) > 0 \) for \( x > 4 \).

Now we need to analyze \( \cot^2 x - 2 \cot x + 5 \):
- The expression \( \cot^2 x - 2 \cot x + 5 \) can be rewritten as \( (\cot x - 1)^2 + 4 \), which is always greater than or equal to 4 for all \( x \) in \( (0, \frac{\pi}{2}) \).

### Step 5: Determine the behavior of \( \cot x \)

- As \( x \) approaches 0, \( \cot x \) approaches \( +\infty \).
- At \( x = \frac{\pi}{4} \), \( \cot x = 1 \).
- As \( x \) approaches \( \frac{\pi}{2} \), \( \cot x \) approaches 0.

### Step 6: Analyze \( g'(x) \)

- For \( 0 < x < \frac{\pi}{4} \), \( \cot x > 1 \) implies \( 1 - \cot x < 0 \) and thus \( g'(x) < 0 \) (decreasing).
- For \( x = \frac{\pi}{4} \), \( \cot x = 1 \) implies \( g'(x) = 0 \).
- For \( \frac{\pi}{4} < x < \frac{\pi}{2} \), \( \cot x < 1 \) implies \( 1 - \cot x > 0 \) and thus \( g'(x) > 0 \) (increasing).

### Conclusion

Thus, \( g(x) \) is:
- Decreasing on \( (0, \frac{\pi}{4}) \)
- Constant at \( x = \frac{\pi}{4} \)
- Increasing on \( (\frac{\pi}{4}, \frac{\pi}{2}) \)