Let f(x) `= [x]^(2) + [x+1] - 3`, where [.] denotes the greatest integer function. Then

Options

A. `f(x) != 0` for all real values of x
B. f(x) = 0 for only two real value of x
C. f(x) = 0 for infinite values of x.
D. f(x) = 0 for no real value of x

Answer

To solve the problem, we need to analyze the function \( f(x) = [x]^2 + [x + 1] - 3 \), where \([.]\) denotes the greatest integer function (also known as the floor function).

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). For example, \([2.5] = 2\) and \([3] = 3\).

2. **Case 1: \(0 \leq x < 1\)**:
- Here, \([x] = 0\) because \(x\) is less than 1.
- \([x + 1] = [0 + 1] = [1] = 1\).
- Therefore, \(f(x) = 0^2 + 1 - 3 = 1 - 3 = -2\).

3. **Case 2: \(1 \leq x < 2\)**:
- Here, \([x] = 1\) because \(x\) is at least 1 but less than 2.
- \([x + 1] = [1 + 1] = [2] = 2\).
- Therefore, \(f(x) = 1^2 + 2 - 3 = 1 + 2 - 3 = 0\).

4. **Case 3: \(2 \leq x < 3\)**:
- Here, \([x] = 2\) because \(x\) is at least 2 but less than 3.
- \([x + 1] = [2 + 1] = [3] = 3\).
- Therefore, \(f(x) = 2^2 + 3 - 3 = 4 + 3 - 3 = 4\).

5. **Case 4: \(x < 0\)**:
- If \(x\) is negative, say \(x = -0.5\), then \([x] = -1\).
- \([x + 1] = [0.5] = 0\).
- Therefore, \(f(x) = (-1)^2 + 0 - 3 = 1 + 0 - 3 = -2\).

6. **Summary of Results**:
- For \(0 \leq x < 1\), \(f(x) = -2\).
- For \(1 \leq x < 2\), \(f(x) = 0\).
- For \(2 \leq x < 3\), \(f(x) = 4\).
- For \(x < 0\), \(f(x) = -2\).

7. **Conclusion**:
- The function \(f(x)\) takes the values \(-2\) (for \(x < 1\) and \(x < 0\)), \(0\) (for \(1 \leq x < 2\)), and \(4\) (for \(2 \leq x < 3\)).
- Therefore, \(f(x)\) is not onto as it does not cover all integers, but it is an into function.