If a + α = 1, b + β = 2 and af(x) + ((1)/(x)) = bx + (β)/(x), x ≠...

$af\left(x\right) + \alphaf\left(\dfrac{1}{x}\right) = bx + \dfrac{\beta}{x} \ldots \left(1\right)$

replace $x$ by $\dfrac{1}{x}$

$af\left(\dfrac{1}{x}\right) + \alphaf\left(x\right) = \dfrac{b}{x} + \betax \ldots \left(2\right)$

\left(1\right) + \left(2\right)$</p><p>\left(a + \alpha\right)f\left(x\right) + \left(a + \alpha\right)f\left(\dfrac{1}{x}\right) = x\left(b + \beta\right) + \left(b + \beta\right)\dfrac{1}{x}$

$\dfrac{f\left(x\right) + f\left(\dfrac{1}{x}\right)}{x + \dfrac{1}{x}} = \dfrac{b + \beta}{a + \alpha} = \dfrac{2}{1} = 2$