Question:
If $a+\alpha=1, b+\beta=2$ and $a f(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}, x \neq 0$, then the value of the expression $\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}$ is
Solution:
$\operatorname{af}(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}$..............
$x \rightarrow \frac{1}{x}$
af $\left(\frac{1}{x}\right)+\operatorname{af}(x)=\frac{b}{x}+\beta x$
...........
$(i)+(i i)$
$(a+\alpha)\left[f(x)+f\left(\frac{1}{x}\right)\right]=\left(x+\frac{1}{x}\right)(b+\beta)$
$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}=\frac{2}{1}=2$
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