If `x in (0,pi//2)`, then the function `f(x)= x sin x +cosx +cos^(2)x` is

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 on the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

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

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

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

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

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

### Step 3: Analyze the sign of the derivative
Next, we need to analyze the sign of \( f'(x) \) on the interval \( (0, \frac{\pi}{2}) \).

We need to check when \( f'(x) < 0 \):

\[ f'(x) < 0 \implies \cos x (x - 2 \sin x) < 0 \]

Since \( \cos x > 0 \) in the interval \( (0, \frac{\pi}{2}) \), we focus on the term \( (x - 2 \sin x) \):

\[ x - 2 \sin x < 0 \implies x < 2 \sin x \]

### Step 4: Show that \( x < 2 \sin x \)
To show that \( x < 2 \sin x \) for \( x \in (0, \frac{\pi}{2}) \), we can analyze the function \( g(x) = 2 \sin x - x \).

1. At \( x = 0 \):
\[ g(0) = 2 \sin(0) - 0 = 0 \]

2. The derivative \( g'(x) = 2 \cos x - 1 \).
- \( g'(x) > 0 \) when \( \cos x > \frac{1}{2} \) (which is true for \( x < \frac{\pi}{3} \)).
- \( g'(x) < 0 \) when \( \cos x < \frac{1}{2} \) (which is true for \( x > \frac{\pi}{3} \)).

3. Therefore, \( g(x) \) increases on \( (0, \frac{\pi}{3}) \) and decreases on \( (\frac{\pi}{3}, \frac{\pi}{2}) \).

4. Since \( g(0) = 0 \) and \( g(x) \) is increasing initially, \( g(x) > 0 \) for \( x \in (0, \frac{\pi}{2}) \).

Thus, \( x < 2 \sin x \) holds true for \( x \in (0, \frac{\pi}{2}) \).

### Step 5: Conclusion
Since \( f'(x) < 0 \) for \( x \in (0, \frac{\pi}{2}) \), the function \( f(x) \) is decreasing on this interval.

Therefore, the function \( f(x) = x \sin x + \cos x + \cos^2 x \) is **decreasing** on \( (0, \frac{\pi}{2}) \).

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