Question

Let $f\left(x\right)=\frac{x}{x-1}\mathrm{and}\frac{f\left(\alpha \right)}{f\left(\alpha +1\right)}=f\left({\alpha }^k\right),$ then k = ______________.

Answer 1

Given: $f\left(x\right)=\frac{x}{x-1}\mathrm{and}\frac{f\left(\alpha \right)}{f\left(\alpha +1\right)}=f\left({\alpha }^k\right)$

$$\begin{gathered} \frac{f\left(a\right)}{f\left(a+1\right)}=\frac{{\displaystyle \frac{a}{a-1}}}{{\displaystyle \frac{a+1}{a+1-1}}} \\ =\frac{{\displaystyle \frac{a}{a-1}}}{{\displaystyle \frac{a+1}{a}}} \\ =\frac{a^2}{\left(a-1\right)\left(a+1\right)} \\ =\frac{a^2}{a^2-1}...\left(1\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}, \\ \frac{f\left(a\right)}{f\left(a+1\right)}=f\left(a^k\right) \\ \Rightarrow \frac{a^2}{a^2-1}=\frac{a^k}{a^k-1} \\ \Rightarrow k=2 \end{gathered}$$

Hence, k = 2.