If `F(x)=Ax^(2)+Bx+C,and f(x) is integer when x is integer then

Options

A. (A+B+C) is an integer
B. (A+B+C) is an integer
C. (8A+2C)is an integer
D. A,BC are integers

Answer

To solve the problem, we need to analyze the function \( F(x) = Ax^2 + Bx + C \) and determine the conditions under which \( F(x) \) is an integer when \( x \) is an integer.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( F(x) = Ax^2 + Bx + C \) is a quadratic function where \( A \), \( B \), and \( C \) are constants.

2. **Condition for Integer Outputs**:
For \( F(x) \) to be an integer for all integer values of \( x \), the coefficients \( A \), \( B \), and \( C \) must be such that the output is always an integer. This means that \( A \), \( B \), and \( C \) should be integers.

3. **Setting Coefficients**:
Let \( A \), \( B \), and \( C \) be integers. This ensures that for any integer \( x \), \( Ax^2 \), \( Bx \), and \( C \) will also be integers.

4. **Evaluating Integer Values**:
We can evaluate \( F(x) \) for a few integer values of \( x \):
- For \( x = 0 \): \( F(0) = C \)
- For \( x = 1 \): \( F(1) = A + B + C \)
- For \( x = -1 \): \( F(-1) = A - B + C \)
- For \( x = 2 \): \( F(2) = 4A + 2B + C \)
- For \( x = -2 \): \( F(-2) = 4A - 2B + C \)

5. **Conclusion**:
Since \( A \), \( B \), and \( C \) are integers, all the evaluated values of \( F(x) \) for integer \( x \) will also be integers. Thus, the function \( F(x) \) will yield integer outputs for all integer inputs.

### Final Result:
Therefore, if \( A \), \( B \), and \( C \) are integers, then \( F(x) \) will be an integer for all integer values of \( x \).