If $$x = -1$$ and $$x = 2$$ are extreme points of $$f(x) = \alpha \log|x| + \beta x^2 + x$$, then:

We are given the function $$f(x)=\alpha\,\log|x|+\beta\,x^{2}+x$$ and told that $$x=-1$$ and $$x=2$$ are its extreme points. For a real-valued differentiable function, extreme points occur where its first derivative is zero (provided the derivative exists). Therefore, we begin by differentiating.

The derivative of $$\log|x|$$ with respect to $$x$$ is $$\dfrac{1}{x}$$, the derivative of $$x^{2}$$ is $$2x$$, and the derivative of $$x$$ is $$1$$. Using these standard results, we obtain

$$f'(x)=\alpha\cdot\dfrac{1}{x}+\beta\cdot 2x+1=\dfrac{\alpha}{x}+2\beta x+1.$$

At an extreme point, we set $$f'(x)=0$$. Since both $$x=-1$$ and $$x=2$$ are given as extreme points, we substitute each value into the derivative and equate to zero.

First, for $$x=-1$$:

$$0=f'(-1)=\dfrac{\alpha}{-1}+2\beta(-1)+1=-\alpha-2\beta+1.$$

Simplifying, we have

$$-\alpha-2\beta+1=0 \quad\Longrightarrow\quad \alpha=1-2\beta.$$(1)

Next, for $$x=2$$:

$$0=f'(2)=\dfrac{\alpha}{2}+2\beta(2)+1=\frac{\alpha}{2}+4\beta+1.$$

So we get

$$\frac{\alpha}{2}+4\beta+1=0.$$

Now we substitute the expression for $$\alpha$$ from equation (1) into this second equation.

$$\frac{1-2\beta}{2}+4\beta+1=0.$$

Multiplying out the fraction and collecting like terms:

$$\frac{1}{2}-\beta+4\beta+1=0,$$ $$\left(-\beta+4\beta\right)+\left(\frac{1}{2}+1\right)=0,$$ $$3\beta+\frac{3}{2}=0.$$

Solving for $$\beta$$ gives

$$3\beta=-\frac{3}{2}\quad\Longrightarrow\quad\beta=-\frac{1}{2}.$$

Finally we substitute $$\beta=-\tfrac12$$ back into equation (1) to obtain $$\alpha$$:

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

Thus, the required values are $$\alpha=2$$ and $$\beta=-\dfrac12$$.

Hence, the correct answer is Option A.