Maximum and minimum value of f(x) = max (sin t), `0 lt t lt x le 0 x le 2pi` are

Options

A. `1, 0`
B. `1 , -1`
C. `0 , -1`
D. None of these

Answer

To find the maximum and minimum values of the function \( f(x) = \max(\sin t) \) for \( 0 < t < x \) and \( 0 < x \leq 2\pi \), we can analyze the behavior of the sine function over the specified intervals.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) \) is defined as the maximum value of \( \sin t \) for \( t \) ranging from 0 to \( x \). The sine function oscillates between -1 and 1.

2. **Analyzing the Interval**:
- For \( 0 < x \leq \frac{\pi}{2} \):
- In this interval, \( \sin t \) is increasing from \( \sin(0) = 0 \) to \( \sin\left(\frac{\pi}{2}\right) = 1 \).
- Therefore, the maximum value of \( \sin t \) in this range is \( \sin x \) since \( x \) is less than or equal to \( \frac{\pi}{2} \).
- Thus, \( f(x) = \sin x \).

3. **Continuing the Analysis**:
- For \( \frac{\pi}{2} < x \leq \frac{3\pi}{2} \):
- Here, \( \sin t \) reaches its maximum at \( \sin\left(\frac{\pi}{2}\right) = 1 \) and then decreases to \( \sin\left(\frac{3\pi}{2}\right) = -1 \).
- Since \( \sin t \) has already reached its maximum value of 1 at \( t = \frac{\pi}{2} \), we have \( f(x) = 1 \).

4. **Final Interval**:
- For \( \frac{3\pi}{2} < x \leq 2\pi \):
- In this range, \( \sin t \) starts from \( -1 \) at \( t = \frac{3\pi}{2} \) and increases back to \( \sin(2\pi) = 0 \).
- The maximum value of \( \sin t \) in this interval is still 1 (achieved at \( t = \frac{\pi}{2} \)).
- Thus, \( f(x) = 1 \).

5. **Conclusion**:
- For \( 0 < x \leq \frac{\pi}{2} \), \( f(x) = \sin x \).
- For \( \frac{\pi}{2} < x \leq 2\pi \), \( f(x) = 1 \).
- Therefore, the maximum value of \( f(x) \) is 1, and the minimum value occurs when \( x \) is in the interval \( (0, \frac{\pi}{2}) \) where \( f(x) \) can take values from 0 to 1.

### Maximum and Minimum Values:
- **Maximum Value**: \( 1 \)
- **Minimum Value**: \( 0 \) (when \( x \) approaches 0)