If `x in (0,pi/2)`, then the function `f(x)= x sin x +cosx +cos^(2)x` is (a) Increasing (b) Decreasing (c) Neither increasing nor decreasing (d) None of these

Options

A. increasing
B. Decreasing
C. Neither increasing nor decreasing
D. None of these

Explanation

To determine whether the function \( f(x) = x \sin x + \cos x + \cos^2 x \) is increasing or decreasing in the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Find the derivative of the function
We need to compute the derivative \( f'(x) \) to analyze the monotonicity of the function.

\[ f'(x) = \frac{d}{dx}(x \sin x) + \frac{d}{dx}(\cos x) + \frac{d}{dx}(\cos^2 x) \]

Using the product rule on \( x \sin x \) and the chain rule on \( \cos^2 x \):

1. Derivative of \( x \sin x \):
\[ \frac{d}{dx}(x \sin x) = x \cos x + \sin x \]

2. Derivative of \( \cos x \):
\[ \frac{d}{dx}(\cos x) = -\sin x \]

3. Derivative of \( \cos^2 x \) using the chain rule:
\[ \frac{d}{dx}(\cos^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x \]

Combining these results:
\[ f'(x) = (x \cos x + \sin x) - \sin x - 2 \sin x \cos x \]

### Step 2: Simplify the derivative
Now, we simplify \( f'(x) \):

\[ f'(x) = x \cos x + \sin x - \sin x - 2 \sin x \cos x \]
\[ f'(x) = x \cos x - 2 \sin x \cos x \]
\[ f'(x) = \cos x (x - 2 \sin x) \]

### Step 3: Analyze the sign of \( f'(x) \)
To determine where \( f'(x) \) is positive or negative, we need to analyze \( \cos x \) and \( (x - 2 \sin x) \) in the interval \( (0, \frac{\pi}{2}) \).

1. \( \cos x \) is positive in the interval \( (0, \frac{\pi}{2}) \).
2. We need to check the sign of \( (x - 2 \sin x) \).

### Step 4: Check the inequality \( x < 2 \sin x \)
We will check if \( x < 2 \sin x \) for \( x \in (0, \frac{\pi}{2}) \).

- At \( x = 0 \):
\[ 0 < 2 \sin(0) \quad \text{(true)} \]

- At \( x = \frac{\pi}{2} \):
\[ \frac{\pi}{2} < 2 \sin\left(\frac{\pi}{2}\right) = 2 \quad \text{(false)} \]

To find where \( x = 2 \sin x \), we can analyze the function \( g(x) = 2 \sin x - x \). We can see that \( g(0) > 0 \) and \( g\left(\frac{\pi}{2}\right) < 0 \), indicating that there is a root in \( (0, \frac{\pi}{2}) \).

### Conclusion
Since \( f'(x) < 0 \) in the interval \( (0, \frac{\pi}{2}) \), the function \( f(x) \) is decreasing in this interval.

Thus, the answer is:
**(b) Decreasing**