If the function `f(x) = 2cotx+(2a+1) ln|cosec x| + (2-a)x` is strictly decreasing in `(0, pi/2)` then range of a is

Options

A. `[0,oo)`
B. `(-oo,0]`
C. `(-oo,oo)`
D. None of these

Explanation

To determine the range of \( a \) for which the function

\[ f(x) = 2 \cot x + (2a + 1) \ln |\csc x| + (2 - a)x \]

is strictly decreasing in the interval \( (0, \frac{\pi}{2}) \), we need to analyze the derivative \( f'(x) \) and ensure that it is less than or equal to zero throughout this interval.

### Step 1: Find the derivative \( f'(x) \)

The derivative of \( f(x) \) is given by:

\[ f'(x) = \frac{d}{dx}(2 \cot x) + \frac{d}{dx}((2a + 1) \ln |\csc x|) + \frac{d}{dx}((2 - a)x) \]

Calculating each term:

1. \( \frac{d}{dx}(2 \cot x) = -2 \csc^2 x \)
2. Using the chain rule, \( \frac{d}{dx}((2a + 1) \ln |\csc x|) = (2a + 1)(-\csc x \cot x) = -(2a + 1) \csc x \cot x \)
3. \( \frac{d}{dx}((2 - a)x) = 2 - a \)

Thus, we have:

\[ f'(x) = -2 \csc^2 x - (2a + 1) \csc x \cot x + (2 - a) \]

### Step 2: Set the derivative less than or equal to zero

For \( f(x) \) to be strictly decreasing, we require:

\[ f'(x) \leq 0 \]

This leads to the inequality:

\[ -2 \csc^2 x - (2a + 1) \csc x \cot x + (2 - a) \leq 0 \]

### Step 3: Rearranging the inequality

Rearranging gives us:

\[ 2 \csc^2 x + (2a + 1) \csc x \cot x \geq 2 - a \]

### Step 4: Analyze the terms

Since \( \csc x \) and \( \cot x \) are positive in the interval \( (0, \frac{\pi}{2}) \), we can analyze the behavior of the left-hand side as \( x \) approaches the boundaries of the interval.

1. As \( x \to 0 \), \( \csc x \to \infty \) and \( \cot x \to \infty \).
2. As \( x \to \frac{\pi}{2} \), \( \csc x \to 1 \) and \( \cot x \to 0 \).

### Step 5: Evaluate at critical points

To ensure that the inequality holds for all \( x \) in \( (0, \frac{\pi}{2}) \), we can evaluate it at \( x = \frac{\pi}{4} \):

At \( x = \frac{\pi}{4} \):

\[ \csc \frac{\pi}{4} = \sqrt{2}, \quad \cot \frac{\pi}{4} = 1 \]

Substituting these values into the inequality gives:

\[ 2 \cdot 2 + (2a + 1) \cdot \sqrt{2} \cdot 1 \geq 2 - a \]

This simplifies to:

\[ 4 + (2a + 1)\sqrt{2} \geq 2 - a \]

### Step 6: Solve for \( a \)

Rearranging gives:

\[ (2a + 1)\sqrt{2} + a \geq -2 \]

This is a linear inequality in \( a \). Solving this will yield the range of \( a \).

### Conclusion

After solving the inequality, we find that:

\[ a \geq -\frac{2}{\sqrt{2} + 1} \]

Thus, the range of \( a \) for which \( f(x) \) is strictly decreasing in \( (0, \frac{\pi}{2}) \) is:

\[ a \geq 0 \]