The number of point `f(x) ={{:([ cos pix],0le x lt1),( |2x-3|[x-2],1lt xle2):}` is discontinuous at Is ([.] denotes the greatest intgreal function )

Options

A. two
B. three
C. four
D. zero

Answer

To determine the number of points of discontinuity for the function \( f(x) \) defined as:

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

we will analyze the function at the critical points and intervals.

### Step 1: Analyze the function on the interval \( [0, 1) \)

1. **At \( x = 0 \)**:
- \( f(0) = \lfloor \cos(0) \rfloor = \lfloor 1 \rfloor = 1 \)
- **Right-hand limit** as \( x \to 0^+ \):
- For \( 0 < x < 1 \), \( \cos(\pi x) \) is continuous and \( \cos(\pi x) \) decreases from 1 to 0.
- Thus, \( \lfloor \cos(\pi x) \rfloor = 0 \) for \( x \) close to 1.
- **Conclusion**: \( f(0) = 1 \) and \( \lim_{x \to 0^+} f(x) = 0 \) → Discontinuity at \( x = 0 \).

### Step 2: Analyze the function at \( x = \frac{1}{2} \)

2. **At \( x = \frac{1}{2} \)**:
- \( f\left(\frac{1}{2}\right) = \lfloor \cos(\frac{\pi}{2}) \rfloor = \lfloor 0 \rfloor = 0 \)
- **Right-hand limit** as \( x \to \frac{1}{2}^+ \):
- For \( x \) just greater than \( \frac{1}{2} \), \( f(x) = \lfloor \cos(\pi x) \rfloor = -1 \) (as \( \cos(\pi x) < 0 \)).
- **Conclusion**: \( f\left(\frac{1}{2}\right) = 0 \) and \( \lim_{x \to \frac{1}{2}^+} f(x) = -1 \) → Discontinuity at \( x = \frac{1}{2} \).

### Step 3: Analyze the function at \( x = 1 \)

3. **At \( x = 1 \)**:
- \( f(1) = \lfloor \cos(\pi) \rfloor = \lfloor -1 \rfloor = -1 \)
- **Right-hand limit** as \( x \to 1^+ \):
- For \( x \) just greater than 1, we use the second piece of the function: \( f(x) = |2x - 3| \lfloor x - 2 \rfloor \).
- At \( x = 1 \), \( \lfloor 1 - 2 \rfloor = -1 \) and \( |2(1) - 3| = 1 \) → \( f(1) = 1 \cdot (-1) = -1 \).
- **Conclusion**: Both limits match → Continuous at \( x = 1 \).

### Step 4: Analyze the function at \( x = \frac{3}{2} \)

4. **At \( x = \frac{3}{2} \)**:
- \( f\left(\frac{3}{2}\right) = |2(\frac{3}{2}) - 3| \lfloor \frac{3}{2} - 2 \rfloor = |0| \cdot (-1) = 0 \)
- **Right-hand limit** as \( x \to \frac{3}{2}^+ \):
- For \( x \) just greater than \( \frac{3}{2} \), \( f(x) = |2x - 3| \lfloor x - 2 \rfloor \) → \( f(x) = 0 \).
- **Conclusion**: Both limits match → Continuous at \( x = \frac{3}{2} \).

### Step 5: Analyze the function at \( x = 2 \)

5. **At \( x = 2 \)**:
- \( f(2) = |2(2) - 3| \lfloor 2 - 2 \rfloor = |1| \cdot 0 = 0 \)
- **Left-hand limit** as \( x \to 2^- \):
- For \( x \) just less than 2, \( f(x) = |2x - 3| \lfloor x - 2 \rfloor \) → \( f(x) = 1 \cdot (-1) = -1 \).
- **Conclusion**: \( f(2) = 0 \) and \( \lim_{x \to 2^-} f(x) = -1 \) → Discontinuity at \( x = 2 \).

### Summary of Discontinuities

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

Thus, the total number of points of discontinuity is **3**.

### Final Answer
The number of points where \( f(x) \) is discontinuous is **3**.

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