1. We are given $f(x)=\alpha \log|x|+\beta x^2+x$ and told that $x=-1$ and $x=2$ are extreme points.

2. For an extreme point, we must have $f'(x)=0.$

First, differentiate: $\frac{d}{dx}(\log|x|)=\frac{1}{x}\quad (x\ne 0).$ So, $f'(x)=\frac{\alpha}{x}+2\beta x+1.$

3. Since $x=-1$ is an extreme point, $f'(-1)=0.$ Thus, $\frac{\alpha}{-1}+2\beta(-1)+1=0$ $-\alpha-2\beta+1=0$ $\alpha+2\beta=1 \quad ...(1)$

4. Since $x=2$ is an extreme point, $f'(2)=0.$ Thus, $\frac{\alpha}{2}+2\beta(2)+1=0$ $\frac{\alpha}{2}+4\beta+1=0 \quad ...(2)$

5. Solve the system: From (1), $\alpha=1-2\beta.$ Substitute into (2): $\frac{1-2\beta}{2}+4\beta+1=0$ $\frac{1}{2}-\beta+4\beta+1=0$ $\frac{3}{2}+3\beta=0$ $3\beta=-\frac{3}{2}$ $\beta=-\frac{1}{2}.$

Now from (1), $\alpha+2\left(-\frac{1}{2}\right)=1$ $\alpha-1=1$ $\alpha=2.$

6. Hence, $\alpha=2,\quad \beta=-\frac{1}{2}.$ So the correct option is A.

7. Verification of extreme points: $f''(x)=-\frac{\alpha}{x^2}+2\beta.$ Substitute $\alpha=2,\beta=-\frac12$: $f''(x)=-\frac{2}{x^2}-1<0 \quad (x\ne 0).$ Therefore both $x=-1$ and $x=2$ are indeed points of local maximum, so they are extreme points.

8. Comparison with stored answer: Stored correct answer = A, which matches our result.