Let f(x) = ax^2 + bx + c,where a, b, c are in | vedclass

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(C) Given $f(x) = ax^2 + bx + c$ with $a, b, c \in \mathbb{Z}$.Since $f(1) = 0$,we have $a + b + c = 0$,which implies $c = -a - b$.Substituting $c$ in

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$4$

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(C) Given $f(x) = ax^2 + bx + c$ with $a, b, c \in \mathbb{Z}$.Since $f(1) = 0$,we have $a + b + c = 0$,which implies $c = -a - b$.Substituting $c$ into $f(x)$,we get $f(x) = ax^2 + bx - a - b = a(x^2 - 1) + b(x - 1) = (x - 1)(ax + a + b)$.Given $40 Since $a, b$ are integers,$7a + b = 9$.Given $60 Since $a, b$ are integers,$8a + b = 11$.Subtracting the two equations: $(8a + b) - (7a + b) = 11 - 9 \implies a = 2$.Substituting $a = 2$ into $7a + b = 9$,we get $14 + b = 9 \implies b = -5$.Then $c = -a - b = -2 - (-5) = 3$.Thus,$f(x) = 2x^2 - 5x + 3$.Calculating $f(50) = 2(50)^2 - 5(50) + 3 = 5000 - 250 + 3 = 4753$.Given $1000t < 4753 < 1000(t + 1)$,we find $t = 4$.

Let $f(x) = ax^2 + bx + c$,where $a, b, c$ are integers. Suppose $f(1) = 0$,$40 < f(6) < 50$,$60 < f(7) < 70$,and $1000t < f(50) < 1000(t+1)$ for some integer $t$. Then,the value of $t$ is

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