The function `f(x)=[x]^(2)+[-x^(2)]`, where [.] denotes the greatest integer function, is

Options

A. Continuous as well as differentiable at x = 0
B. Neither continuous nor differentiable at x = 2
C. Continuous but not differentiable at x = 2
D. None of these

Answer

To analyze the function \( f(x) = [x^2] - [-x^2] \), where \([.]\) denotes the greatest integer function (also known as the floor function), we need to check its continuity and differentiability at the points \( x = 0 \) and \( x = 2 \).

### Step 1: Evaluate \( f(x) \) at \( x = 0 \)

1. **Left-hand limit as \( x \) approaches 0**:
\[ f(0 - h) = [(-h)^2] - [-(-h)^2] = [h^2] - [-h^2] \]
As \( h \) approaches 0 from the left, \( h^2 \) approaches 0, and thus:
\[ [h^2] = 0 \quad \text{and} \quad [-h^2] = -1 \]
Therefore,
\[ f(0 - h) = 0 - (-1) = 1 \]

2. **Right-hand limit as \( x \) approaches 0**:
\[ f(0 + h) = [h^2] - [-h^2] \]
As \( h \) approaches 0 from the right, we have:
\[ [h^2] = 0 \quad \text{and} \quad [-h^2] = -1 \]
Therefore,
\[ f(0 + h) = 0 - (-1) = 1 \]

3. **Value of the function at \( x = 0 \)**:
\[ f(0) = [0^2] - [-0^2] = [0] - [0] = 0 - 0 = 0 \]

### Conclusion for \( x = 0 \):
- Left-hand limit: \( 1 \)
- Right-hand limit: \( 1 \)
- \( f(0) = 0 \)

Since the left-hand limit and right-hand limit are equal but not equal to \( f(0) \), the function is **discontinuous** at \( x = 0 \).

### Step 2: Evaluate \( f(x) \) at \( x = 2 \)

1. **Left-hand limit as \( x \) approaches 2**:
\[ f(2 - h) = [(2 - h)^2] - [-(2 - h)^2] \]
As \( h \) approaches 0 from the left:
\[ (2 - h)^2 = 4 - 4h + h^2 \]
For small \( h \), this is slightly less than 4, thus:
\[ [4 - 4h + h^2] = 3 \quad \text{and} \quad [-(2 - h)^2] = -4 \]
Therefore,
\[ f(2 - h) = 3 - (-4) = 7 \]

2. **Right-hand limit as \( x \) approaches 2**:
\[ f(2 + h) = [(2 + h)^2] - [-(2 + h)^2] \]
As \( h \) approaches 0 from the right:
\[ (2 + h)^2 = 4 + 4h + h^2 \]
For small \( h \), this is slightly greater than 4, thus:
\[ [4 + 4h + h^2] = 4 \quad \text{and} \quad [-(2 + h)^2] = -5 \]
Therefore,
\[ f(2 + h) = 4 - (-5) = 9 \]

3. **Value of the function at \( x = 2 \)**:
\[ f(2) = [2^2] - [-2^2] = [4] - [-4] = 4 - (-4) = 8 \]

### Conclusion for \( x = 2 \):
- Left-hand limit: \( 7 \)
- Right-hand limit: \( 9 \)
- \( f(2) = 8 \)

Since the left-hand limit and right-hand limit are not equal, the function is **discontinuous** at \( x = 2 \).

### Final Summary:
- The function \( f(x) \) is **discontinuous** at both \( x = 0 \) and \( x = 2 \), hence it is not differentiable at these points.