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If $x=-1$ and $x=2$ are extreme points of $\mathrm{f}(x)=\alpha \log x+\beta x^2+x, \alpha$ and $\beta$ are constants, then the value of $\alpha^2+2 \beta$ is
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The correct answer is:
$3$
According to the given condition, $\mathrm{f}^{\prime}(1)=0$ and $\mathrm{f}^{\prime}(2)=0$
$$
\mathrm{f}(x)=\alpha \log x+\beta x^2+x
$$
$$
\begin{array}{ll}
\therefore & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \\
\therefore & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 \\
& \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2
\end{array}
$$
$\therefore \quad$ From (i) and (ii), we get
$$
\begin{aligned}
& \beta=\frac{-1}{2} \text { and } \alpha=2 \\
\therefore \quad & \alpha^2+2 \beta=4-1=3
\end{aligned}
$$
$$
\mathrm{f}(x)=\alpha \log x+\beta x^2+x
$$
$$
\begin{array}{ll}
\therefore & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \\
\therefore & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 \\
& \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2
\end{array}
$$
$\therefore \quad$ From (i) and (ii), we get
$$
\begin{aligned}
& \beta=\frac{-1}{2} \text { and } \alpha=2 \\
\therefore \quad & \alpha^2+2 \beta=4-1=3
\end{aligned}
$$
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