`f(x)=[x^(2)]-{x}^(2),` where [.] and {.} denote the greatest integer function and the fractional part function , respectively , is

Options

A. continuous at x=1,-1
B. continuous at x=-1 but not at x=1
C. continuous at x=1 but not at x=1
D. discontinuous at x=1 and x=-1

Answer

To determine the continuity of the function \( f(x) = [x^2] - \{x^2\} \), where \([.]\) denotes the greatest integer function and \(\{.\}\) denotes the fractional part function, we need to check the continuity at the points \( x = 1 \) and \( x = -1 \).

### Step 1: Check continuity at \( x = 1 \)

1. **Calculate \( f(1) \)**:
\[ f(1) = [1^2] - \{1^2\} = [1] - \{1\} = 1 - 0 = 1 \]

2. **Calculate \( f(1^+) \)** (approaching from the right):
\[ f(1^+) = [1^2] - \{1^2\} = [1^2] - \{1^2\} = [1] - \{1\} = 1 - 0 = 1 \]

3. **Calculate \( f(1^-) \)** (approaching from the left):
\[ f(1^-) = [1^2] - \{1^2\} = [1^2] - \{1^2\} = [1] - \{1\} = 1 - 0 = 1 \]

4. **Compare the values**:
\[ f(1) = 1, \quad f(1^+) = 1, \quad f(1^-) = 1 \]
Since \( f(1) = f(1^+) = f(1^-) \), the function is continuous at \( x = 1 \).

### Step 2: Check continuity at \( x = -1 \)

1. **Calculate \( f(-1) \)**:
\[ f(-1) = [-1^2] - \{-1^2\} = [1] - \{1\} = 1 - 0 = 1 \]

2. **Calculate \( f(-1^+) \)** (approaching from the right):
\[ f(-1^+) = [-1^+] - \{-1^+\} = [(-1 + \epsilon)^2] - \{(-1 + \epsilon)^2\} \]
As \( \epsilon \) approaches 0, \( (-1 + \epsilon)^2 \) approaches \( 1 \), so:
\[ f(-1^+) = [1] - \{1\} = 1 - 0 = 1 \]

3. **Calculate \( f(-1^-) \)** (approaching from the left):
\[ f(-1^-) = [-1^-] - \{-1^-\} = [(-1 - \epsilon)^2] - \{(-1 - \epsilon)^2\} \]
As \( \epsilon \) approaches 0, \( (-1 - \epsilon)^2 \) also approaches \( 1 \), so:
\[ f(-1^-) = [1] - \{1\} = 1 - 0 = 1 \]

4. **Compare the values**:
\[ f(-1) = 1, \quad f(-1^+) = 1, \quad f(-1^-) = 1 \]
Since \( f(-1) = f(-1^+) = f(-1^-) \), the function is continuous at \( x = -1 \).

### Conclusion
The function \( f(x) = [x^2] - \{x^2\} \) is continuous at both \( x = 1 \) and \( x = -1 \).