Let-f-x-ax-2-bx-c-where-a-b-c-are-real-numbers-If-the-numbers-2a-a-b-and-c-are-all-integers-then-the-number-of-integral-values-between-1-and-5-that-f-x-can-take-is-

Question Number 21234 by Tinkutara last updated on 17/Sep/17

Let f (x) = {a x}^{2} + b x + c, where a, b, c

are real numbers . If the numbers 2 a,

a + b, and c are all integers, then the

number of integral values between 1

and 5 that f (x) can take is

Answered by Tinkutara last updated on 17/Sep/17

f (0) = c \in Z

f (1) = a + b + c

f (- 1) = a - b + c

S i n c e 2 a, a + b, c a r e i n t e g e r s, s o

2 a - (a + b) + c = a - b + c \in Z \Rightarrow f (- 1) \in Z

S i m i l a r l y (a + b) + c = f (1) \in Z .

f (1) - f (- 1) = 2 b \in Z

f (2) = 4 a + 2 b + c \in Z

f (3) = 9 a + 3 b + c = a + b + c + 8 a + 2 b \in Z

f (4) = 16 a + 4 b + c \in Z

f (5) = 25 a + 5 b + c = a + b + c + 24 a + 4 b \in Z

∴ f (x) h a s 5 i n t e g r a l v a l u e s i n [1, 5] .