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-by-Step Solution:

1. **Understanding the Range of \( \cos x \)**:
- For \( x \) in the interval \( [0, \frac{\pi}{2}] \), the value of \( \cos x \) ranges from \( 1 \) (when \( x = 0 \)) to \( 0 \) (when \( x = \frac{\pi}{2} \)).
- Therefore, \( 0 \leq \cos x \leq 1 \).

**Hint**: Remember that the cosine function decreases from 1 to 0 in this interval.

2. **Applying the Greatest Integer Function**:
- Since \( \cos x \) is always less than or equal to 1 in this interval, the greatest integer function \( [\cos x] \) will take the value \( 0 \) for all \( x \) in \( (0, \frac{\pi}{2}) \) and will be \( 1 \) at \( x = 0 \).
- Thus, \( [\cos x] = 0 \) for \( 0 < x < \frac{\pi}{2} \).

**Hint**: The greatest integer function \( [y] \) gives the largest integer less than or equal to \( y \).

3. **Simplifying the Function \( f(x) \)**:
- For \( 0 < x < \frac{\pi}{2} \), we have:
\[ f(x) = 1 + x \sin x \cdot [\cos x] = 1 + x \sin x \cdot 0 = 1 \]
- At \( x = 0 \), \( f(0) = 1 + 0 \cdot \sin(0) \cdot [\cos(0)] = 1 + 0 = 1 \).
- At \( x = \frac{\pi}{2} \), \( f(\frac{\pi}{2}) = 1 + \frac{\pi}{2} \cdot \sin(\frac{\pi}{2}) \cdot [\cos(\frac{\pi}{2})] = 1 + \frac{\pi}{2} \cdot 1 \cdot 0 = 1 \).

**Hint**: Check the function values at the endpoints and within the interval.

4. **Conclusion About the Function**:
- Since \( f(x) = 1 \) for all \( x \) in \( [0, \frac{\pi}{2}] \), the function is constant.
- A constant function is continuous, neither strictly increasing nor strictly decreasing, and has a global maximum value of \( 1 \).

**Hint**: A constant function has the same value across its entire domain.

### Final Answer:
The function \( f(x) \) is continuous on \( [0, \frac{\pi}{2}] \) and has a global maximum value of \( 1 \).

### Options:
- Discontinuous in \( [0, \frac{\pi}{2}] \) - **False**
- Strictly decreasing in \( [0, \frac{\pi}{2}] \) - **False**
- Strictly increasing in \( [0, \frac{\pi}{2}] \) - **False**
- Has a global maximum value of \( 1 \) - **True** (Correct option)