95 Let the function $f(x)$ be defined for $x \in \mathbf{R}$, $f(x) \neq 0 (x \in \mathbf{R})$, and there exists a positive number $a$ such that $f(x-a) = \frac{1}{f(x)} (x \in \mathbf{R})$, then $f(x)=$
A. is definitely not a periodic function
B. is definitely a periodic function
C. may be a periodic function, or may not be a periodic function
D. can only be constantly equal to 1 or -1
95 B. From the given condition, we have
$$
\begin{aligned}
f(x) & =f[(x+a)-a]=\frac{1}{f(x+a)}=\frac{1}{f[(x+2 a)-a]} \\
& =\frac{1}{\frac{1}{f(x+2 a)}}=f(x+2 a) \quad(x \in \mathbf{R}) .
\end{aligned}
$$

That is,
$$
f(x)=f(x+2 a) \quad(x \in \mathbf{R}) .
$$

Therefore, $f(x)$ is a periodic function.