The function `f(x) =[x] cos{(2x-1)//2} pi` denotes the greatest integer function, is discontinuous at:

Options

A. all x
B. all integer points
C. no x
D. x whixh is not an integer

Answer

To determine the points of discontinuity of the function \( f(x) = [x] \cos\left(\frac{2x-1}{2} \pi\right) \), where \([x]\) denotes the greatest integer function, we will analyze the function step by step.

### Step 1: Understand the Function
The function \( f(x) \) is defined as the product of the greatest integer function \([x]\) and the cosine function. The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\).

### Step 2: Identify Points of Discontinuity
The greatest integer function \([x]\) is discontinuous at integer values of \(x\). Therefore, we will check the points where \(x\) is an integer, specifically \(x = n\) where \(n\) is any integer.

### Step 3: Check Discontinuity at Integer Points
Let's check the behavior of \(f(x)\) around an integer point, say \(x = 1\).

1. **Left-hand limit as \(x\) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} [x] \cos\left(\frac{2x-1}{2} \pi\right) \]
For \(x\) just less than 1, \([x] = 0\):
\[ \lim_{x \to 1^-} f(x) = 0 \cdot \cos\left(\frac{2(1)-1}{2} \pi\right) = 0 \cdot \cos\left(\frac{\pi}{2}\right) = 0 \]

2. **Right-hand limit as \(x\) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} [x] \cos\left(\frac{2x-1}{2} \pi\right) \]
For \(x\) just greater than 1, \([x] = 1\):
\[ \lim_{x \to 1^+} f(x) = 1 \cdot \cos\left(\frac{2(1)-1}{2} \pi\right) = 1 \cdot \cos\left(\frac{\pi}{2}\right) = 1 \cdot 0 = 0 \]

3. **Value of the function at \(x = 1\)**:
\[ f(1) = [1] \cos\left(\frac{2(1)-1}{2} \pi\right) = 1 \cdot \cos\left(\frac{\pi}{2}\right) = 1 \cdot 0 = 0 \]

### Step 4: Conclusion
Since the left-hand limit, right-hand limit, and the value of the function at \(x = 1\) are all equal to 0, the function \(f(x)\) is continuous at \(x = 1\).

### Step 5: Generalization
This reasoning can be applied to any integer \(n\):
- For \(x = n\), the function is continuous because the limits from both sides will yield the same result as the function value at that point.

### Final Answer
The function \(f(x)\) is continuous for all \(x\) and thus is discontinuous at **no points**.