Let `f (x) =cos x, g (x)={{:(min {f (t):0 le t le x}",", x in[0","pi]),((sin x)-1"," , x gtpi):}` Then

Options

A. g (x) is discontinuous at `x =pi`
B. ` g (x) ` is continous for `x in [0,oo)`
C. `g (x)` is differentiable at `x = pi`
D. `g (x)` is differentiable for `x in [0,oo)`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) as defined in the question.

1. **Define the functions**:
- \( f(x) = \cos x \)
- For \( g(x) \):
\[ g(x) = \begin{cases} \min \{ f(t) : 0 \leq t \leq x \} & \text{if } x \in [0, \pi] \\ \sin x - 1 & \text{if } x > \pi \end{cases} \]

2. **Analyzing \( g(x) \) for \( x \in [0, \pi] \)**:
- Since \( f(t) = \cos t \) is a decreasing function in the interval \([0, \pi]\), the minimum value of \( f(t) \) occurs at \( t = x \).
- Therefore, for \( x \in [0, \pi] \):
\[ g(x) = \cos x \]

3. **Analyzing \( g(x) \) for \( x > \pi \)**:
- For \( x > \pi \):
\[ g(x) = \sin x - 1 \]

4. **Finding continuity at \( x = \pi \)**:
- We need to check if \( g(x) \) is continuous at \( x = \pi \):
- \( g(\pi) = \cos(\pi) = -1 \)
- \( \lim_{x \to \pi^-} g(x) = \lim_{x \to \pi^-} \cos x = -1 \)
- \( \lim_{x \to \pi^+} g(x) = \lim_{x \to \pi^+} (\sin x - 1) = \sin(\pi) - 1 = -1 \)
- Since both limits and the function value at \( x = \pi \) are equal, \( g(x) \) is continuous at \( x = \pi \).

5. **Finding differentiability at \( x = \pi \)**:
- We need to check if \( g(x) \) is differentiable at \( x = \pi \):
- For \( x < \pi \):
\[ g'(x) = -\sin x \]
- For \( x > \pi \):
\[ g'(x) = \cos x \]
- Evaluating at \( x = \pi \):
- \( g'(\pi^-) = -\sin(\pi) = 0 \)
- \( g'(\pi^+) = \cos(\pi) = -1 \)
- Since the left-hand derivative and right-hand derivative at \( x = \pi \) are not equal, \( g(x) \) is not differentiable at \( x = \pi \).

6. **Conclusion**:
- \( g(x) \) is continuous at \( x = \pi \) but not differentiable at \( x = \pi \).

### Final Answer:
- The correct option is that \( g(x) \) is continuous at \( x = \pi \) but not differentiable at \( x = \pi \).