$$ \text{Let } f:\mathbb{R}\setminus\{0\}\to\mathbb{R} \text{ be a function such that } f(x)-6f\!\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}. \text{ If } \lim_{x\to 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta, \; \alpha,\beta\in\mathbb{R}, \text{ then } \alpha+2\beta \text{ is equal to:} $$

We are given $$f(x) - 6f(1/x) = \frac{35}{3x} - \frac{5}{2}$$ and asked to find $$\alpha + 2\beta$$ where $$\lim_{x\to 0}(\frac{1}{\alpha x} + f(x)) = \beta$$.

Starting from the given relation, we have $$f(x) - 6f(1/x) = \frac{35}{3x} - \frac{5}{2}\quad (i)$$

Replacing $$x$$ with $$1/x$$ yields $$f(1/x) - 6f(x) = \frac{35x}{3} - \frac{5}{2}\quad (ii)$$

From equation (i) we solve for $$f(x)$$ to get $$f(x) = 6f(1/x) + \frac{35}{3x} - \frac{5}{2}$$

Substituting this expression into (ii) gives $$f(1/x) - 6[6f(1/x) + \frac{35}{3x} - \frac{5}{2}] = \frac{35x}{3} - \frac{5}{2}$$

Rearranging terms yields $$-35f(1/x) = \frac{35x}{3} - \frac{5}{2} + \frac{70}{x} - 15 = \frac{35x}{3} + \frac{70}{x} - \frac{35}{2}$$

Hence $$f(1/x) = -\frac{x}{3} - \frac{2}{x} + \frac{1}{2}$$

Finally, replacing $$x$$ with $$1/x$$ again yields $$f(x) = -\frac{1}{3x} - 2x + \frac{1}{2}$$

To evaluate the limit $$\lim_{x\to 0}(\frac{1}{\alpha x} + f(x))$$, substitute the expression for $$f(x)$$ to obtain $$\frac{1}{\alpha x} + f(x) = \frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2}$$ and rearrange to $$= \frac{1}{x}\left(\frac{1}{\alpha} - \frac{1}{3}\right) - 2x + \frac{1}{2}$$.

Since the limit exists only if the coefficient of $$1/x$$ vanishes, we require $$\frac{1}{\alpha} - \frac{1}{3} = 0$$ which gives $$\alpha = 3$$. Consequently, the remaining part approaches $$\beta = \lim_{x\to 0}(-2x + \frac{1}{2}) = \frac{1}{2}$$.

Finally, $$\alpha + 2\beta = 3 + 2 \times \frac{1}{2} = 4$$.

The correct answer is Option 3: 4.