Let` f (x) = px^2 + qx +r,` where p,q,r are rational and `f: Z to Z` where Z is the set of integers. Then p+q is

Options

A. negative integer
B. an integer
C. non-integral rational
D. none

Answer

To solve the problem, we need to analyze the function \( f(x) = px^2 + qx + r \), where \( p, q, r \) are rational numbers, and the function is defined from the set of integers \( \mathbb{Z} \) to \( \mathbb{Z} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = px^2 + qx + r \) is a quadratic polynomial where \( p, q, r \) are rational coefficients. Since the domain is \( \mathbb{Z} \) (the set of integers), we will be substituting integer values for \( x \).

2. **Evaluating the Output**:
The output of the function \( f(x) \) must also be an integer since the range is also \( \mathbb{Z} \). This means that for every integer input \( x \), the result \( f(x) \) must yield an integer.

3. **Analyzing the Coefficients**:
If \( p \) and \( q \) are rational numbers, they can be expressed as fractions. For \( f(x) \) to yield an integer for all integer \( x \), the coefficients \( p \) and \( q \) must be such that the entire expression \( px^2 + qx + r \) remains an integer.

4. **Conditions for Integer Outputs**:
- If \( p \) is a rational number expressed as \( \frac{a}{b} \) (where \( a \) and \( b \) are integers), then \( px^2 \) will be an integer if \( b \) divides \( a \cdot x^2 \).
- Similarly, for \( qx \) to be an integer for all integer \( x \), \( q \) must also be a rational number such that \( b \) divides \( a \cdot x \).

5. **Conclusion on \( p \) and \( q \)**:
Given that \( f(x) \) must yield integers for all integer inputs, both \( p \) and \( q \) must be integers themselves. If \( p \) and \( q \) are integers, then their sum \( p + q \) will also be an integer.

6. **Final Answer**:
Therefore, \( p + q \) is an integer.