If `f: R to R` is a function defined by `f(x) = [x] cos((2x-1)/2)pi` where [x] denotes the greatest integer function, then f is:

Options

A. continous for every real R
B. discontinous only at x=0
C. discontinuous only at non-zero integral values of k
D. continuous only at x = 0

Answer

To determine the continuity 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 behavior of the function at integer points and in between.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is defined as \( f(x) = [x] \cos\left(\frac{2x-1}{2}\pi\right) \). Here, \([x]\) is the greatest integer less than or equal to \(x\).

2. **Analyzing the Cosine Term**:
The cosine term can be rewritten for clarity:
\[ \cos\left(\frac{2x-1}{2}\pi\right) = \cos\left(\pi x - \frac{\pi}{2}\right) = \sin(\pi x) \]
This is because \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta) \).

3. **Rewriting the Function**:
Thus, we can express \( f(x) \) as:
\[ f(x) = [x] \sin(\pi x) \]

4. **Checking Continuity at Integer Points**:
We need to check the limit of \( f(x) \) as \( x \) approaches an integer \( n \):
- For \( x \to n^+ \) (approaching from the right):
\[ f(n^+) = [n] \sin(n\pi) = n \cdot 0 = 0 \]
- For \( x \to n^- \) (approaching from the left):
\[ f(n^-) = [n-0.0001] \sin(n\pi) = (n-1) \cdot 0 = 0 \]

5. **Conclusion on Continuity**:
Since both the left-hand limit and right-hand limit at \( n \) are equal to 0, and \( f(n) = 0 \), we conclude that:
\[ \lim_{x \to n} f(x) = f(n) = 0 \]
Therefore, \( f(x) \) is continuous at every integer \( n \).

6. **Checking Continuity at Non-Integer Points**:
For non-integer values of \( x \), the function \( [x] \) remains constant, and since \( \sin(\pi x) \) is continuous, \( f(x) \) is continuous for all non-integer values as well.

### Final Conclusion:
The function \( f(x) \) is continuous for every real number \( x \).