Answer: $\textbf {B}$
Solution:
Given: $\mathrm {f(x) = \frac{x-1}{x+1}}$
$\begin {align} \Rightarrow \mathrm {f^2(x)} &= \mathrm{f(f(x))} \\&= \mathrm{\frac{f(x)-1}{f(x) + 1}} \\&= \mathrm{\dfrac{\frac{x-1}{x+1} + 1}{\frac{x-1}{x+1} + 1}} \\&= \mathrm {\frac{-1}{x}}\end {align}$
$\begin {align} \mathrm {f^3(x)} &= \mathrm {f(f^2(x))}\\&=\mathrm {f\bigg(\frac{-1}{x}\bigg)}\\&=\mathrm{\frac{-x-1}{x-1}} \end {align}$
$\begin{align} \mathrm {f^4(x)} &= \mathrm{f(f^3(x))} \\&= \mathrm{\dfrac{\frac{-x-1}{x-1}-1}{\frac{-x-1}{x-1}+1}} \\&=\mathrm x \end {align}$
$\therefore \mathrm {f^4(x) = x}\tag{1}$
Similarly,
$\mathrm {f^5(x) = f(x)}$
$\mathrm {f^6(x) = f^2(x)}$
$\mathrm {f^7(x) = f^3(x)}$
$\mathrm {f^8(x) = f^4(x) =x}$ $\tag{[From (1)]}$
Now the pattern goes like:
$\begin {align}\mathrm {f^4(x) = f^8(x) = f^{12}(x)= \cdots \cdots=f^{4k}(x) = x}~ \text{where $\mathrm k$ $\in$ $\mathrm {I^+}$} \end {align}$
$\therefore \mathrm {f^{100}(x) = x} \\ \Rightarrow \mathrm {f^{100}(10) = 10}$
$\therefore \textbf B$ is the correct option.