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If $6 x-x^2+12$ attains its extreme value $\beta$ at $x=\alpha$ then $\beta=$
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Verified Answer
The correct answer is:
$7 \alpha$
We are given that
$$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=-\mathrm{x}^2+6 \mathrm{x}+12 \\
& \mathrm{f}^{\prime}(\mathrm{x})=-2 \mathrm{x}+6 \\
& \mathrm{f}^{\prime}(\mathrm{x})=-2 \\
& \text { Now } \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow-2 \mathrm{x}+6=0 \Rightarrow \mathrm{x}=3 \\
& \therefore \mathrm{f}(\alpha)=\mathrm{f}(3)=-9+18+12=-9+30=21 \\
& \Rightarrow \mathrm{f}(\alpha)=\mathrm{f}(3)=7 \times 3=7 \alpha
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=-\mathrm{x}^2+6 \mathrm{x}+12 \\
& \mathrm{f}^{\prime}(\mathrm{x})=-2 \mathrm{x}+6 \\
& \mathrm{f}^{\prime}(\mathrm{x})=-2 \\
& \text { Now } \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow-2 \mathrm{x}+6=0 \Rightarrow \mathrm{x}=3 \\
& \therefore \mathrm{f}(\alpha)=\mathrm{f}(3)=-9+18+12=-9+30=21 \\
& \Rightarrow \mathrm{f}(\alpha)=\mathrm{f}(3)=7 \times 3=7 \alpha
\end{aligned}
$$
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