The number of points where `f(x)={{:([cos pi x],"," 0 le x le 1),(|2x-3|[x-2],","1 lt x le 2):}`
([x] is the greatest integer less than or equal to x) is discontinuous is

Options

A. 2
B. 4
C. 3
D. 1

Answer

To determine the number of points where the function \( f(x) \) is discontinuous, we will analyze the function piece by piece.

The function is defined as:
\[ f(x) = \begin{cases} \lfloor \cos(\pi x) \rfloor & \text{for } 0 \leq x \leq 1 \\ |2x - 3| \cdot \lfloor x - 2 \rfloor & \text{for } 1 < x \leq 2 \end{cases} \]

### Step 1: Analyze \( f(x) \) for \( 0 \leq x \leq 1 \)

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

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

3. **Check continuity at \( x = 0 \)**:
- Left-hand limit as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lfloor \cos(0) \rfloor = 1 \]
- Right-hand limit as \( x \to 0^- \):
\[ f(0) = 1 \]
- Since \( f(0) \neq \lim_{x \to 0^+} f(x) \), there is a discontinuity at \( x = 0 \).

4. **Check continuity at \( x = 1 \)**:
- Left-hand limit as \( x \to 1^- \):
\[ \lim_{x \to 1^-} f(x) = \lfloor \cos(\pi) \rfloor = -1 \]
- Right-hand limit as \( x \to 1^+ \):
- For \( x = 1 \), we switch to the second part of the function.
\[ f(1) = -1 \]
- Since both limits are equal, \( f(x) \) is continuous at \( x = 1 \).

5. **Check continuity at \( x = \frac{1}{2} \)**:
- Left-hand limit as \( x \to \frac{1}{2}^- \):
\[ \lim_{x \to \frac{1}{2}^-} f(x) = \lfloor \cos(\frac{\pi}{2}) \rfloor = \lfloor 0 \rfloor = 0 \]
- Right-hand limit as \( x \to \frac{1}{2}^+ \):
\[ \lim_{x \to \frac{1}{2}^+} f(x) = \lfloor \cos(\frac{\pi}{2}) \rfloor = 0 \]
- Since both limits are equal, \( f(x) \) is continuous at \( x = \frac{1}{2} \).

### Step 2: Analyze \( f(x) \) for \( 1 < x \leq 2 \)

1. **Identify the point where the modulus breaks**:
- The expression \( |2x - 3| \) changes at \( x = \frac{3}{2} \).

2. **Check continuity at \( x = \frac{3}{2} \)**:
- Left-hand limit as \( x \to \frac{3}{2}^- \):
\[ f\left(\frac{3}{2}\right) = |2 \cdot \frac{3}{2} - 3| \cdot \lfloor \frac{3}{2} - 2 \rfloor = |0| \cdot \lfloor -\frac{1}{2} \rfloor = 0 \]
- Right-hand limit as \( x \to \frac{3}{2}^+ \):
\[ f\left(\frac{3}{2}\right) = (2 \cdot \frac{3}{2} - 3) \cdot \lfloor \frac{3}{2} - 2 \rfloor = 0 \cdot \lfloor -\frac{1}{2} \rfloor = 0 \]
- Since both limits are equal, \( f(x) \) is continuous at \( x = \frac{3}{2} \).

3. **Check continuity at \( x = 2 \)**:
- Left-hand limit as \( x \to 2^- \):
\[ f(2) = |2 \cdot 2 - 3| \cdot \lfloor 2 - 2 \rfloor = |1| \cdot \lfloor 0 \rfloor = 0 \]
- Right-hand limit as \( x \to 2^+ \):
\[ f(2) = 0 \]
- Since both limits are equal, \( f(x) \) is continuous at \( x = 2 \).

### Conclusion

The points of discontinuity are:
- \( x = 0 \)
- \( x = \frac{1}{2} \)
- \( x = 2 \)

Thus, the number of points where \( f(x) \) is discontinuous is **3**.