If $$a + \alpha = 1, b + \beta = 2$$ and $$af(x) + \alpha f\left(\frac{1}{x}\right) = bx + \frac{\beta}{x}, x \neq 0$$, then the value of the expression $$\frac{f(x) + f\left(\frac{1}{x}\right)}{x + \frac{1}{x}}$$ is ______.

We are given $$a + \alpha = 1$$, $$b + \beta = 2$$, and the functional equation $$af(x) + \alpha f\!\left(\frac{1}{x}\right) = bx + \frac{\beta}{x}$$ for $$x \neq 0$$.

Replacing $$x$$ with $$\frac{1}{x}$$ in the functional equation gives $$af\!\left(\frac{1}{x}\right) + \alpha f(x) = \frac{b}{x} + \beta x$$.

Adding the original equation and this new equation: $$(a + \alpha)f(x) + (a + \alpha)f\!\left(\frac{1}{x}\right) = (b + \beta)x + (b + \beta)\frac{1}{x}$$, which simplifies to $$(a + \alpha)\left[f(x) + f\!\left(\frac{1}{x}\right)\right] = (b + \beta)\left(x + \frac{1}{x}\right)$$.

Since $$a + \alpha = 1$$ and $$b + \beta = 2$$, we get $$f(x) + f\!\left(\frac{1}{x}\right) = 2\left(x + \frac{1}{x}\right)$$.

Therefore, $$\frac{f(x) + f\!\left(\frac{1}{x}\right)}{x + \frac{1}{x}} = 2$$.

The correct answer is $$2$$.