Let `f : [ 0, oo) to [ 0,3]` be a function defined by
`f(x) = {{:(max { sin t : 0 le t le x}" , " 0 le x le pi),( 2 + cos x", " x gt pi ):}`
Then which of the following is true ?

Options

A. ƒ is continuous everywhere but not differentiable exactly at one point in ` (0, oo)`
B. ƒ is differentiable everywhere in `(0, oo)`
C. ƒ is not continuous exactly at two points in `(0,oo)`
D. ƒ is continuous everywhere but not differentiable exactly at two points in `(0,oo)`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise:

\[ f(x) = \begin{cases} \max \{ \sin t : 0 \leq t \leq x \} & \text{for } 0 \leq x \leq \pi \\ 2 + \cos x & \text{for } x > \pi \end{cases} \]

### Step 1: Analyze the function for \( 0 \leq x \leq \pi \)

For \( 0 \leq x \leq \pi \), we need to find the maximum value of \( \sin t \) for \( t \) in the interval \([0, x]\).

- The function \( \sin t \) increases from \( 0 \) to \( \frac{\sqrt{2}}{2} \) at \( t = \frac{\pi}{4} \) and reaches its maximum value of \( 1 \) at \( t = \frac{\pi}{2} \).
- Therefore, for \( 0 \leq x \leq \frac{\pi}{2} \), \( f(x) = \sin x \) since \( \sin t \) is increasing.
- For \( \frac{\pi}{2} < x \leq \pi \), the maximum value of \( \sin t \) is \( 1 \) (which occurs at \( t = \frac{\pi}{2} \)). Thus, \( f(x) = 1 \).

So, we can summarize:
\[ f(x) = \begin{cases} \sin x & \text{for } 0 \leq x \leq \frac{\pi}{2} \\ 1 & \text{for } \frac{\pi}{2} < x \leq \pi \end{cases} \]

### Step 2: Analyze the function for \( x > \pi \)

For \( x > \pi \), we have:
\[ f(x) = 2 + \cos x \]
This function oscillates between \( 1 \) and \( 3 \) as \( \cos x \) varies between \(-1\) and \(1\).

### Step 3: Check continuity at the critical points

The critical points to check for continuity are \( x = \pi \) and \( x = \frac{\pi}{2} \).

1. **At \( x = \frac{\pi}{2} \)**:
- \( \lim_{x \to \frac{\pi}{2}^-} f(x) = \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \lim_{x \to \frac{\pi}{2}^+} f(x) = 1 \)
- Thus, \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

2. **At \( x = \pi \)**:
- \( \lim_{x \to \pi^-} f(x) = 1 \)
- \( \lim_{x \to \pi^+} f(x) = 2 + \cos(\pi) = 2 - 1 = 1 \)
- Thus, \( f(x) \) is continuous at \( x = \pi \).

### Step 4: Check differentiability

To check differentiability, we need to compute the left-hand and right-hand derivatives at the critical points.

1. **At \( x = \frac{\pi}{2} \)**:
- Left-hand derivative: \( f'(x) = \cos x \) at \( x = \frac{\pi}{2} \) gives \( 0 \).
- Right-hand derivative: \( f'(x) = 0 \) for \( \frac{\pi}{2} < x < \pi \).
- Both derivatives are equal, so \( f(x) \) is differentiable at \( x = \frac{\pi}{2} \).

2. **At \( x = \pi \)**:
- Left-hand derivative: \( f'(x) = 0 \) for \( x \) approaching \( \pi \).
- Right-hand derivative: \( f'(x) = -\sin x \) at \( x = \pi \) gives \( 0 \).
- Both derivatives are equal, so \( f(x) \) is differentiable at \( x = \pi \).

### Conclusion

Since \( f(x) \) is continuous and differentiable everywhere in the interval \( [0, \infty) \), we conclude that the correct option is:

**The function \( f(x) \) is differentiable everywhere in the interval \( [0, \infty) \).**