1. We are given $a+\alpha=1,\qquad b+\beta=2$ and $a f(x)+\alpha f\left(\frac1x\right)=bx+\frac{\beta}{x}, \qquad x\ne 0.$

We need to find $\frac{f(x)+f\left(\frac1x\right)}{x+\frac1x}.$

2. Write the given relation again, and then replace $x$ by $\dfrac1x$.

From the given equation: $a f(x)+\alpha f\left(\frac1x\right)=bx+\frac{\beta}{x} \quad \cdots (1)$

Replacing $x$ by $\dfrac1x$: $a f\left(\frac1x\right)+\alpha f(x)=\frac{b}{x}+\beta x \quad \cdots (2)$

3. Add equations (1) and (2).

Left-hand side: $a f(x)+\alpha f\left(\frac1x\right)+a f\left(\frac1x\right)+\alpha f(x)$ $=(a+\alpha)\left(f(x)+f\left(\frac1x\right)\right).$

Since $a+\alpha=1$, this becomes $f(x)+f\left(\frac1x\right).$

Right-hand side: $bx+\frac{\beta}{x}+\frac{b}{x}+\beta x$ $=(b+\beta)\left(x+\frac1x\right).$

Since $b+\beta=2$, this becomes $2\left(x+\frac1x\right).$

Hence, $f(x)+f\left(\frac1x\right)=2\left(x+\frac1x\right).$

4. Therefore,