The function `f(x)=[x]^2-[x^2]` is discontinuous at (where `[gamma]` is the greatest integer less than or equal to `gamma`), is discontinuous at

Options

A. all integers
B. all integers except 0 and 1
C. all integers except 0
D. all integers except 1

Answer

To determine where the function \( f(x) = [x]^2 - [x^2] \) is discontinuous, we need to analyze the behavior of the function around integer values.

### Step-by-Step Solution

1. **Understanding the Function**:
The function \( f(x) = [x]^2 - [x^2] \) involves the greatest integer function, denoted by \( [ \cdot ] \). This function returns the largest integer less than or equal to the input.

2. **Case 1: When \( x \) is an integer**:
Let \( x = n \) where \( n \) is an integer.
- Then, \( [x] = n \) and \( [x^2] = [n^2] = n^2 \).
- Therefore, \( f(n) = n^2 - n^2 = 0 \).

3. **Case 2: When \( x \) is slightly less than an integer**:
Let \( x = n - h \) where \( h \) is a small positive number (i.e., \( h \to 0^+ \)).
- Here, \( [x] = n - 1 \) and \( [x^2] = [(n-h)^2] \).
- Since \( (n-h)^2 = n^2 - 2nh + h^2 \), for small \( h \), \( [x^2] = n^2 - 1 \) (as long as \( n > 1 \)).
- Thus, \( f(n - h) = (n - 1)^2 - (n^2 - 1) = n^2 - 2n + 1 - n^2 + 1 = -2n + 2 \).

4. **Case 3: When \( x \) is slightly more than an integer**:
Let \( x = n + h \) where \( h \) is a small positive number (i.e., \( h \to 0^+ \)).
- Here, \( [x] = n \) and \( [x^2] = [(n+h)^2] \).
- Since \( (n+h)^2 = n^2 + 2nh + h^2 \), for small \( h \), \( [x^2] = n^2 \).
- Thus, \( f(n + h) = n^2 - n^2 = 0 \).

5. **Finding Limits**:
- **Right-hand limit (RHL)** at \( x = n \):
\[ \lim_{h \to 0^+} f(n + h) = 0 \]
- **Left-hand limit (LHL)** at \( x = n \):
\[ \lim_{h \to 0^+} f(n - h) = -2n + 2 \]

6. **Discontinuity Condition**:
The function \( f(x) \) is discontinuous at \( x = n \) if \( \text{LHL} \neq f(n) \) or \( \text{RHL} \neq f(n) \).
- For integers \( n \):
- \( f(n) = 0 \)
- LHL = \( -2n + 2 \)
- RHL = \( 0 \)

Setting LHL equal to \( f(n) \):
\[ -2n + 2 = 0 \implies n = 1 \]
Thus, the function is continuous at \( x = 1 \).

7. **Conclusion**:
The function \( f(x) \) is discontinuous at all integers except \( x = 1 \).

### Final Answer:
The function \( f(x) \) is discontinuous at all integers except \( 1 \).