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If x^2 + ax + b is an integer for every ...

If `x^2 + ax + b` is an integer for every integer x then

A

'a' is always an integer but 'b' need not be an integer.

B

'b' is always an integer but 'a' need not be an integer.

C

a + b is always an integer.

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the quadratic expression \( x^2 + ax + b \) and determine the conditions under which it remains an integer for every integer \( x \). ### Step-by-step Solution: 1. **Substituting \( x = 0 \)**: - When we substitute \( x = 0 \) into the expression, we get: \[ 0^2 + a(0) + b = b \] - Since the expression must be an integer for every integer \( x \), it follows that \( b \) must be an integer. 2. **Substituting \( x = 1 \)**: - Now, substituting \( x = 1 \) into the expression, we have: \[ 1^2 + a(1) + b = 1 + a + b \] - This expression must also be an integer. Since \( 1 \) is an integer, it follows that \( a + b \) must also be an integer (because the sum of integers is an integer). 3. **Substituting \( x = -1 \)**: - Next, substituting \( x = -1 \) into the expression gives us: \[ (-1)^2 + a(-1) + b = 1 - a + b \] - This must also be an integer, which reinforces that \( b \) is an integer and \( -a + b \) must also be an integer. 4. **Conclusion**: - From the above substitutions, we have established that: - \( b \) is an integer. - \( a + b \) is an integer. - Since \( b \) is an integer, it follows that \( a \) must also be an integer (because the sum of an integer and another integer is an integer). 5. **Final Result**: - Therefore, we conclude that both \( a \) and \( b \) are integers, and thus \( a + b \) is also an integer. ### Answer: The correct option is **C: \( a + b \) is always an integer**.
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