已知函数 $f(x)=\left\{\begin{array}{l}e^{x-1}, x<2, \\ \log _{3}\left(x^{2}-1\right), x \geq 2,\end{array}\right.$ 若 $f(a) \geq 1$, 则 $a$ 的取值范围是 ( )
$\because f(x)=\left\{\begin{array}{c}e^{x-1}, x<2, \\ \log _{3}\left(x^{2}-1\right), x \geq 2, \quad f(a) \geq 1\end{array}\right.$

$\therefore\left\{\begin{array}{c}a<2 \\ e^{a-1} \geq 1 \text { 或 }\end{array}\left\{\begin{array}{c}a \geq 2 \\ \log _{3}\left(a^{2}-1\right) \geq 1\end{array}\right.\right.$

即 $\left\{\begin{array}{c}a<2 \\ a-1 \geq 0 \text { 或 }\end{array}\left\{\begin{array}{c}a \geq 2 \\ a^{2}-1 \geq 3\end{array}\right.\right.$

即 $1 \leq \mathrm{a}<2$ 或 $\mathrm{a} \geq 2$

$\therefore a$ 的取值范围是 $\lfloor 1,+\infty)$

故答案为: B

【分析】依题意, 对 $\mathrm{a}$ 分 $\mathrm{a}<2$, 与 $\mathrm{a} \geq 2$ 讨论, 再解相应的不等式即可.