Indicate all correct alternatives: if `f(x) = x/2-1`, then on the interval `[0, pi]`:

Options

A. (a) `tan (f(x)) and (1)/(f(x))` are both continuous
B. (b) `tan (f(x)) and (1)/(f(x))` are both discontinuous
C. (c) `tan (f(x)) and f^(-1)(x)` are both continuous
D. (d) `tan(f(x))` is continuous but `(1)/(f(x))` is not continuous

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x}{2} - 1 \) and its derived functions over the interval \([0, \pi]\). We will check the continuity of \( \tan(f(x)) \), \( \frac{1}{f(x)} \), and \( f^{-1}(x) \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \frac{x}{2} - 1 \]

2. **Find the range of \( f(x) \) on the interval \([0, \pi]\)**:
- At \( x = 0 \):
\[ f(0) = \frac{0}{2} - 1 = -1 \]
- At \( x = \pi \):
\[ f(\pi) = \frac{\pi}{2} - 1 \]
- Therefore, the range of \( f(x) \) on \([0, \pi]\) is:
\[ f(x) \in [-1, \frac{\pi}{2} - 1] \]

3. **Check for continuity of \( \frac{1}{f(x)} \)**:
- The function \( \frac{1}{f(x)} \) is continuous wherever \( f(x) \neq 0 \).
- Set \( f(x) = 0 \):
\[ \frac{x}{2} - 1 = 0 \implies x = 2 \]
- Since \( 2 \) is within the interval \([0, \pi]\), \( \frac{1}{f(x)} \) is discontinuous at \( x = 2 \).

4. **Check for continuity of \( \tan(f(x)) \)**:
- The function \( \tan(f(x)) \) is continuous wherever \( f(x) \) does not take values that make \( \tan \) undefined (i.e., \( f(x) \neq \frac{\pi}{2} + n\pi \) for any integer \( n \)).
- The maximum value of \( f(x) \) is \( \frac{\pi}{2} - 1 \), which does not equal \( \frac{\pi}{2} \). Thus, \( \tan(f(x)) \) is continuous on the interval \([0, \pi]\).

5. **Find the inverse function \( f^{-1}(x) \)**:
- To find the inverse, we set \( y = f(x) \):
\[ y = \frac{x}{2} - 1 \implies x = 2y + 2 \]
- Therefore, the inverse function is:
\[ f^{-1}(x) = 2x + 2 \]
- Since \( f^{-1}(x) \) is a linear function, it is continuous everywhere, including on the interval \([0, \pi]\).

### Conclusion:
- \( \tan(f(x)) \) is continuous.
- \( \frac{1}{f(x)} \) is discontinuous at \( x = 2 \).
- \( f^{-1}(x) \) is continuous.

Thus, the correct alternatives are:
- **C**: \( \tan(f(x)) \) and \( f^{-1}(x) \) are both continuous.
- **D**: \( \tan(f(x)) \) is continuous, but \( \frac{1}{f(x)} \) is not continuous.