Given f : Z → Z such that f (x) = ax2 + bx + c, where a, b, c ∊ Q We have to find the value of a + b Now, For x = 0 , f (0) = c = integer [Since f (0) ∊ Z because f : Z → Z] ⇒ c is an integer For x = 1, f (1) = a + b + c = k (say ) [where f (1) = k ∊ Z] ⇒ a + b = k - c ... (i) Since k ∊ c , c ∊ Z ⇒ k - c ∊ Z ⇒ a + b ∊ Z [Using (i)] ⇒ a + b is an integer Hence, option 'B' is correct