The function `f(x)=1+x(sinx)[cosx], 0ltxlepi//2`, where `[.]` denotes greatest integer function

Options

A. is discontinuous is `(0,pi//2)`
B. is strictly decreasing in `(0,pi//2)`
C. is strictly increasing in `(0,pi//2)`
D. has global maximum value 1

Explanation

To solve the problem, we need to analyze the function \( f(x) = 1 + x \sin x \cdot [\cos x] \), where \( [\cdot] \) denotes the greatest integer function and \( x \) is in the interval \( [0, \frac{\pi}{2}] \).

### Step 1: Determine the range of \( \cos x \)
The function \( \cos x \) decreases from 1 to 0 as \( x \) goes from 0 to \( \frac{\pi}{2} \). Therefore, we have:
\[ \cos(0) = 1 \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) = 0 \]
Thus, the range of \( \cos x \) for \( x \in [0, \frac{\pi}{2}] \) is:
\[ 0 \leq \cos x \leq 1 \]

### Step 2: Apply the greatest integer function
Since \( \cos x \) takes values from 0 to 1, the greatest integer function \( [\cos x] \) will take the following values:
- For \( 0 < \cos x < 1 \), \( [\cos x] = 0 \)
- For \( \cos x = 1 \) (which occurs at \( x = 0 \)), \( [\cos x] = 1 \)

However, since we are interested in the interval \( (0, \frac{\pi}{2}] \), we can conclude that:
\[ [\cos x] = 0 \quad \text{for } x \in (0, \frac{\pi}{2}] \]

### Step 3: Substitute into the function
Now, substituting \( [\cos x] = 0 \) into the function \( f(x) \):
\[ f(x) = 1 + x \sin x \cdot [\cos x] = 1 + x \sin x \cdot 0 = 1 \]
Thus, for \( x \in (0, \frac{\pi}{2}] \), we have:
\[ f(x) = 1 \]

### Step 4: Analyze the function
Since \( f(x) = 1 \) is constant for all \( x \) in the interval \( [0, \frac{\pi}{2}] \):
- The function is neither increasing nor decreasing.
- The function is continuous.

### Step 5: Conclusion
Since \( f(x) \) is constant and equal to 1, we can conclude that:
- The global maximum value of \( f(x) \) is 1.