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$\mathrm{A}(1,15), \mathrm{B}(3,-12), \mathrm{C}(6,12)$ are three consecutive turning points of a continuous curve $y=f(x)$. If $f(x)=0$ only for $x=\alpha$ and $x=\beta$, then $|\beta-\alpha| < $
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The correct answer is:
$5$
Given that $\mathrm{A}(1,15), \mathrm{B}(3,-12), \mathrm{C}(6,12)$ are threeconsecutive turning points of continuous curve $y=f(x)$ and intersect $x$-axis at $x=\alpha$, and $x=\beta$.
It is clear from graph is
$\begin{aligned}
& 1 < \alpha < 3 \text { and } 3 < \beta < 6 \Rightarrow-\alpha < -1 \text { and } \beta < 6 \\
& \Rightarrow|\beta-\alpha| < 6-1 \Rightarrow|\beta-\alpha| < 5 .
\end{aligned}$
It is clear from graph is
$\begin{aligned}
& 1 < \alpha < 3 \text { and } 3 < \beta < 6 \Rightarrow-\alpha < -1 \text { and } \beta < 6 \\
& \Rightarrow|\beta-\alpha| < 6-1 \Rightarrow|\beta-\alpha| < 5 .
\end{aligned}$
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