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

Options

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

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 behavior of the function around integer values.

### Step 1: Understanding the Function
The function consists of two parts:
1. The greatest integer function \([x]\), which is discontinuous at integer values.
2. The cosine function \(\cos\left(\frac{2x-1}{2} \pi\right)\), which is continuous everywhere.

### Step 2: Identify Points of Discontinuity
Since \([x]\) is discontinuous at integers, we need to check the continuity of \(f(x)\) at integer points.

### Step 3: Check Continuity at Integer Points
Let's check the continuity at \(x = n\) (where \(n\) is an integer).

#### Left-Hand Limit (LHL) at \(x = n\):
\[ f(n^-) = \lim_{h \to 0} f(n - h) = [n - h] \cos\left(\frac{2(n - h) - 1}{2} \pi\right) \]
For \(h\) approaching \(0\) from the right, \([n - h] = n - 1\) (since \(n - h\) is still less than \(n\)):
\[ f(n^-) = (n - 1) \cos\left(\frac{2(n - h) - 1}{2} \pi\right) \]
As \(h \to 0\), \(\cos\left(\frac{2(n - h) - 1}{2} \pi\right) \to \cos\left(\frac{2n - 1}{2} \pi\right)\).

#### Right-Hand Limit (RHL) at \(x = n\):
\[ f(n^+) = \lim_{h \to 0} f(n + h) = [n + h] \cos\left(\frac{2(n + h) - 1}{2} \pi\right) \]
For \(h\) approaching \(0\) from the left, \([n + h] = n\):
\[ f(n^+) = n \cos\left(\frac{2(n + h) - 1}{2} \pi\right) \]
As \(h \to 0\), \(\cos\left(\frac{2(n + h) - 1}{2} \pi\right) \to \cos\left(\frac{2n - 1}{2} \pi\right)\).

### Step 4: Value of the Function at Integer Points
\[ f(n) = [n] \cos\left(\frac{2n - 1}{2} \pi\right) = n \cos\left(\frac{2n - 1}{2} \pi\right) \]

### Step 5: Compare Limits and Function Value
Now we compare:
- \(f(n^-) = (n - 1) \cos\left(\frac{2n - 1}{2} \pi\right)\)
- \(f(n^+) = n \cos\left(\frac{2n - 1}{2} \pi\right)\)
- \(f(n) = n \cos\left(\frac{2n - 1}{2} \pi\right)\)

Since \(f(n^-) \neq f(n)\) and \(f(n^+) \neq f(n)\) at integer points, the function is discontinuous at each integer \(n\).

### Conclusion
The function \(f(x)\) is discontinuous at all integer values of \(x\).