Let f(x) = ax2 + bx + c where a, b and care rational, and f:Z→Z where Zis the set of integer. Then a+b is :a)A negative integerb)An integerc)Non-integral rational numberd)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Let f(x) = ax2 + bx + c where a, b and care rational, and f:Z&rarr;Z where Zis the set of integer. Then a+b is :a)A negative integerb)An integerc)Non-integral rational numberd)None of theseCorrect answer is option 'B'. Can you explain this answer?

Answer

f:Z →Z such that f(x) = ax² + bx + c where a,b and c ∈ Q
We have to find the value of a+b

Now for x = 0,f(0) = c = integer

⇒ c is and integer

Answer

Explanation:

Given function:
Let f(x) = ax^2 + bx + c where a, b, and c are rational numbers, and f:Z→Z where Z is the set of integers.

Key Points:
- For f(x) to be a function from integers to integers, a, b, and c must be integers.
- The sum of two integers is always an integer.

Solution:

Step 1: Since f(x) is a function from integers to integers, a, b, and c must be integers.

Step 2: In the given function f(x) = ax^2 + bx + c, a, b, and c are rational numbers, so they can be expressed as fractions.

Step 3: Let's assume a = p/q, b = r/s, and c = m/n, where p, q, r, s, m, n are integers.

Step 4: The sum of a and b is (p/q) + (r/s) = (ps + rq) / qs.

Step 5: Since ps, rq, and qs are integers, their sum (ps + rq) / qs is also an integer.

Step 6: Therefore, the sum of a and b, i.e., a + b, is an integer.

Conclusion:
Hence, the correct answer is option 'B' - An integer.

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