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Let $f(x)=a x^{2}+b x+c$, where $a, b, c$ are integers. Suppose $f(1)=0,40 < f(6) < 50,60 < f(7) < 70$, and $1000 \mathrm{t} < \mathrm{f}(50) < 1000(\mathrm{t}+1)$ for some integer $\mathrm{t}$. Then the value of $\mathrm{t}$ is
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4
$\begin{array}{l}
\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c} \\
\text { given } \mathrm{f}(1)=0 \\
\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\
\text { and } 40 < \mathrm{f}(6) < 50 \\
\Rightarrow 40 < 36 \mathrm{a}+6 \mathrm{~b}+\mathrm{c} < 50 \\
\Rightarrow 40 < 35 \mathrm{a}+5 \mathrm{~b} < 50 \\
\Rightarrow 8 < 7 \mathrm{a}+\mathrm{b} < 10 \\
7 \mathrm{a}+\mathrm{b}=\text { integer }=9 \\
\text { and } 60 < \mathrm{f}(7) < 70 \\
\Rightarrow 60 < 49 \mathrm{a}+7 \mathrm{~b}+\mathrm{c} < 70 \\
\Rightarrow 60 < 48 \mathrm{a}+6 \mathrm{~b} < 70 \\
\Rightarrow 10 < 8 \mathrm{a}+\mathrm{b} < 11.6 \\
8 \mathrm{a}+\mathrm{b}=\text { integer }=11
\end{array}$
Solving $(1) \&(2)$
$\begin{array}{l}
a=2, b=-5, c=3 \\
\therefore \quad f(x)=2 x^{2}-5 x+3 \\
\quad f(50)=4753 \\
1000 t < f(50) < 1000(t+1) \\
(1000 \times 4) < 4753 < 1000(4+1) \\
\therefore \quad t=4
\end{array}$
\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c} \\
\text { given } \mathrm{f}(1)=0 \\
\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\
\text { and } 40 < \mathrm{f}(6) < 50 \\
\Rightarrow 40 < 36 \mathrm{a}+6 \mathrm{~b}+\mathrm{c} < 50 \\
\Rightarrow 40 < 35 \mathrm{a}+5 \mathrm{~b} < 50 \\
\Rightarrow 8 < 7 \mathrm{a}+\mathrm{b} < 10 \\
7 \mathrm{a}+\mathrm{b}=\text { integer }=9 \\
\text { and } 60 < \mathrm{f}(7) < 70 \\
\Rightarrow 60 < 49 \mathrm{a}+7 \mathrm{~b}+\mathrm{c} < 70 \\
\Rightarrow 60 < 48 \mathrm{a}+6 \mathrm{~b} < 70 \\
\Rightarrow 10 < 8 \mathrm{a}+\mathrm{b} < 11.6 \\
8 \mathrm{a}+\mathrm{b}=\text { integer }=11
\end{array}$
Solving $(1) \&(2)$
$\begin{array}{l}
a=2, b=-5, c=3 \\
\therefore \quad f(x)=2 x^{2}-5 x+3 \\
\quad f(50)=4753 \\
1000 t < f(50) < 1000(t+1) \\
(1000 \times 4) < 4753 < 1000(4+1) \\
\therefore \quad t=4
\end{array}$
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