If $x=-1$ and $\cdot x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$, then
If $x=-1$ and $\cdot x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$, then
$\alpha=-6, \beta=\frac{1}{2}$
$\alpha=-6, \beta=-\frac{1}{2}$
$\alpha=2, \beta=-\frac{1}{2}$
$\alpha=2, \beta=\frac{1}{2}$
Solution
According to the given condition, $\mathrm{f}^{\prime}(-1)=0$ and $\mathrm{f}^{\prime}(2)=0$ $\mathrm{f}(x)=\alpha \log |x|+\beta x^2+x$
$\begin{array}{ll}
\therefore & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \\
\therefore & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 ...(i)\\
& \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2...(ii)
\end{array}$ From (i) and (ii), we get
$\alpha=2 \text { and } \beta=\frac{-1}{2}$