Discuss the continuity of f(x) in [0, 2], where `f(x) = {{:([cos pi x]",",x le 1),(|2x - 3|[x - 2]",",x gt 1):}` where [.] denotes the greatest integral function.

Answer

To discuss the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \lfloor \cos(\pi x) \rfloor & \text{for } x \leq 1 \\ |2x - 3|(x - 2) & \text{for } x > 1 \end{cases} \]

on the interval \([0, 2]\), we will analyze the function at critical points and check for continuity.

### Step 1: Determine the function values at critical points

1. **At \( x = 0 \)**:
\[ f(0) = \lfloor \cos(0) \rfloor = \lfloor 1 \rfloor = 1 \]

2. **At \( x = 1 \)**:
\[ f(1) = \lfloor \cos(\pi) \rfloor = \lfloor -1 \rfloor = -1 \]

3. **At \( x = 2 \)**:
\[ f(2) = |2(2) - 3|(2 - 2) = |4 - 3| \cdot 0 = 1 \cdot 0 = 0 \]

### Step 2: Analyze the intervals

- For \( x \in [0, 1] \):
- The function \( f(x) = \lfloor \cos(\pi x) \rfloor \) is continuous as \( \cos(\pi x) \) is continuous. However, we need to check the values at the endpoints:
- At \( x = 0 \), \( f(0) = 1 \).
- At \( x = 1 \), \( f(1) = -1 \).

- For \( x \in (1, 2] \):
- The function \( f(x) = |2x - 3|(x - 2) \) is continuous as both \( |2x - 3| \) and \( (x - 2) \) are continuous functions.

### Step 3: Check continuity at critical points

1. **At \( x = 1 \)**:
- Left-hand limit as \( x \to 1^- \):
\[ \lim_{x \to 1^-} f(x) = f(1) = -1 \]
- Right-hand limit as \( x \to 1^+ \):
\[ \lim_{x \to 1^+} f(x) = |2(1) - 3|(1 - 2) = |2 - 3|(-1) = 1(-1) = -1 \]
- Since both limits are equal and equal to \( f(1) \), \( f(x) \) is continuous at \( x = 1 \).

2. **At \( x = 2 \)**:
- Left-hand limit as \( x \to 2^- \):
\[ \lim_{x \to 2^-} f(x) = |2(2) - 3|(2 - 2) = |4 - 3| \cdot 0 = 1 \cdot 0 = 0 \]
- Right-hand limit as \( x \to 2^+ \):
\[ f(2) = 0 \]
- Since both limits are equal and equal to \( f(2) \), \( f(x) \) is continuous at \( x = 2 \).

### Conclusion

The function \( f(x) \) is continuous on the interval \([0, 2]\).