Question:
Let $f(x)=\frac{x}{x-1}$ and $\frac{f(\alpha)}{f(\alpha+1)}=f\left(\alpha^{k}\right)$, then $k=$ ______________.
Solution:
Given: $f(x)=\frac{x}{x-1}$ and $\frac{f(\alpha)}{f(\alpha+1)}=f\left(\alpha^{k}\right)$
$\frac{f(a)}{f(a+1)}=\frac{\frac{a}{a-1}}{\frac{a+1}{a+1-1}}$
$=\frac{\frac{a}{a-1}}{\frac{a+1}{a}}$
$=\frac{a^{2}}{(a-1)(a+1)}$
$=\frac{a^{2}}{a^{2}-1}$ ...(1)
It is given that,
$\frac{f(a)}{f(a+1)}=f\left(a^{k}\right)$
$\Rightarrow \frac{a^{2}}{a^{2}-1}=\frac{a^{k}}{a^{k}-1}$
$\Rightarrow k=2$
Hence, k = 2.
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