Question

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

Answer 1

$af\left(x\right)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta }{x}\cdots \left(1\right)$
Replace $x$ by $\frac{1}{x}$
$af\left(\frac{1}{x}\right)+\alpha f\left(x\right)=\frac{b}{x}+\beta x\cdots \left(2\right)$
$\left(1\right)+\left(2\right),$ we get
$(a+\alpha )\left[f\right(x)+f\left(\frac{1}{x}\right)]=(x+\frac{1}{x})(b+\beta )$
$\Rightarrow \frac{f\left(x\right)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{2}{1}=2$