Let `a,b,c` be rational numbers and `f:Z->Z` be a function given by `f(x)=a x^2+b x+c.` Then, `a+b` is

Options

A. a negative integer
B. an integer
C. non-integral rational number
D. none of these

Answer

B

Explanation

Since `f(x) Z to Z` is given by
`f(x)=ax^(2)+bx+c" for all x "inZ.`
`therefore f(x)=ax^(2)+bc+c` takes integral values for all `x in Z`.
`therefore f(x)` is an integer for all `x in Z`.
`therefore f(0)` and f(1) are integers
`therefore f(1)-f(0)` is an integer.
`therefore (a+b+c)-c` is an integer `[therefore f(0)=c and f(1)=a+b+c]`
`therefore` a+b is an integer.