Mathematics - Functions Question with Solution | TestHub

$af\left(x\right)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta }{x} \dots \left(1\right)$

replace $x$ by $\frac{1}{x}$

$af\left(\frac{1}{x}\right)+\alpha f\left(x\right)=\frac{b}{x}+\beta x \dots \left(2\right)$

$\left(1\right)+\left(2\right)$

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

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