7.102 Given the linear fractional transformation of $x$ as $f_{1}(x)=\frac{2 x-1}{x+1}$. For $n=1$, $2,3, \cdots$, if we define $f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$. Assuming $f_{35}(x)=f_{5}(x)$, then $f_{28}(x)$ equals
(A) $x$.
(B) $\frac{1}{x}$.
(C) $\frac{x-1}{x}$.
(D) $\frac{1}{1-x}$.
(E) None of these.
(22nd American High School Mathematics Examination, 1971)
[Solution] If $g(x)$ is the inverse function of the transformation $f_{1}(x)$, then
$$
g\left[f_{n+1}(x)\right]=f_{n}(x) .
$$

Since $f_{35}(x)=f_{5}(x)$, applying this formula repeatedly, we get
$$
\begin{aligned}
g\left[f_{35}(x)\right]=f_{34}(x) & =g\left[f_{5}(x)\right]=f_{4}(x), \\
f_{34}(x) & =f_{4}(x) .
\end{aligned}
$$

Similarly, we have $f_{33}(x)=f_{3}(x), \cdots, f_{31}(x)=f_{1}(x)$.
Thus, $f_{30}(x)=g\left[f_{1}(x)\right]=x$, and $f_{29}(x)=g(x)$, so $f_{28}(x)=g[g(x)]$. Given that $g(x)=\frac{x+1}{2-x}$, we have $f_{28}(x)=g[g(x)]=\frac{1}{1-x}$. Therefore, the answer is $(D)$.