`{:(f(x) = cos x and H_(1)(x) = min{f(t), 0 le t lt x},),(0 le x le (pi)/(2) = (pi)/(2)-x,(pi)/(2) lt x le pi),(f(x) = cos x and H_(2) (x) = max {f(t), o le t le x},),(0 le x le (pi)/(2) = (pi)/(2) - x","(pi)/(2) lt x le pi),(g(x) = sin x and H_(3)(x) = min{g(t),0 le t le x},),(0 le x le (pi)/(2)=(pi)/(2) - x, (pi)/(2) le x le pi),(g(x) = sin x and H_(4)(x) = max{g(t),0 le t le x},),(0 le x le (pi)/(2) = (pi)/(2) - x, (pi)/(2) lt x le pi):}`
Which of the following is true for `H_(3) (x)`?

Options

A. Continuous and derivable in `[0, pi]`
B. Continuous but not derivable at `x = (pi)/(2)`
C. Neither continuous nor derivable at `x = (pi)/(2)`
D. None of the above

Answer

To solve the problem regarding the function \( H_3(x) \) defined as \( H_3(x) = \min \{ g(t), 0 \leq t \leq x \} \) where \( g(x) = \sin x \), we will analyze the function step-by-step.

### Step 1: Define the function \( g(x) \)
Given:
\[ g(x) = \sin x \]

### Step 2: Determine the intervals for \( H_3(x) \)
We need to analyze \( H_3(x) \) for the intervals \( 0 \leq x \leq \frac{\pi}{2} \) and \( \frac{\pi}{2} < x \leq \pi \).

1. **For \( 0 \leq x \leq \frac{\pi}{2} \)**:
\[ H_3(x) = \min \{ g(t) \mid 0 \leq t \leq x \} = \min \{ \sin t \mid 0 \leq t \leq x \} \]
Since \( \sin t \) is increasing in this interval, the minimum occurs at \( t = 0 \):
\[ H_3(x) = \sin(0) = 0 \]

2. **For \( \frac{\pi}{2} < x \leq \pi \)**:
\[ H_3(x) = \min \{ g(t) \mid 0 \leq t \leq x \} = \min \{ \sin t \mid 0 \leq t \leq \frac{\pi}{2} \text{ and } \frac{\pi}{2} < t \leq x \} \]
In the interval \( 0 \leq t \leq \frac{\pi}{2} \), the minimum is still \( 0 \) (at \( t = 0 \)). For \( \frac{\pi}{2} < t \leq x \), \( \sin t \) decreases from \( 1 \) to \( 0 \). Thus, the minimum will still be \( 0 \):
\[ H_3(x) = 0 \]

### Step 3: Conclusion about continuity
Since \( H_3(x) = 0 \) for both intervals \( 0 \leq x \leq \frac{\pi}{2} \) and \( \frac{\pi}{2} < x \leq \pi \), we can conclude that:
- \( H_3(x) \) is continuous at all points in the interval \( [0, \pi] \).

### Final Answer
Thus, \( H_3(x) \) is continuous for \( 0 \leq x \leq \pi \).

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