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