2. (2004 National College Entrance Examination, Fujian Province) Given $f(x)=\frac{2 x-a}{x^{2}+2}(x \in \mathbf{R})$ is an increasing function on the interval $[-1,1]$.
(I) Find the set $A$ of real numbers $a$;
(II) Suppose the equation $f(x)=\frac{1}{x}$ has two roots $x_{1} 、 x_{2}$. Is there a real number $m$ such that the inequality $m^{2}+t m+1 \geqslant\left|x_{1}-x_{2}\right|$ holds for any $a \in A$ and $t \in[-1,1]$? If it exists, find the range of $m$; if not, explain the reason.
2. ( I ) $f^{\prime}(x)=\frac{4+2 a x-2 x^{2}}{\left(x^{2}+2\right)^{2}}=\frac{-2\left(x^{2}-a x-2\right)}{\left(x^{2}+2\right)^{2}}$.

Since $f(x)$ is an increasing function on $[-1,1]$, then $f^{\prime}(x) \geqslant 0$ for all $x \in[-1,1]$.
Let $g(x)=x^{2}-a x-2$, then
$\left\{\begin{array}{l}g(1)=1-a-2 \leqslant 0, \\ g(-1)=1+a-2 \leqslant 0 .\end{array}\right.$ This gives $-1 \leqslant a \leqslant 1$.
Furthermore, for $x \in[-1,1]$, $f(x)$ is a continuous function, and only when $a=1$, $f^{\prime}(-1)=0$, and when $a=-1$, $f^{\prime}(1)=$
0.
Thus, $A=\{a \mid-1 \leqslant a \leqslant 1\}$.
( II ) From $\frac{2 x-a}{x^{2}+2}=\frac{1}{x}$, we get $x^{2}-a x-2=0$.
Since $\Delta=a^{2}+8>0$, then $x_{1} 、 x_{2}$ are the two real roots of $x^{2}-a x-2=0$.
Thus, $x_{1}+x_{2}=a, x_{1} x_{2}=-2$.
Therefore, $\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{a^{2}+8}$.
Since $-1 \leqslant a \leqslant 1$, then $\left|x_{1}-x_{2}\right|=\sqrt{a^{2}+8} \leqslant 3$.
To make the inequality $m^{2}+t m+1 \geqslant\left|x_{1}-x_{2}\right|$ hold for any $t \in[-1,1]$ and $a \in A$, it is necessary and sufficient that $m^{2}+t m+1 \geqslant 3$ for any $t \in[-1,1]$, i.e., $m^{2}+t m-2 \geqslant 0$ for any $t \in[-1,1]$.
Let $g(t)=m^{2}+t m-2=m t+\left(m^{2}-2\right)$, then
$$
\left\{\begin{array}{l}
g(-1)=m^{2}-m-2 \geqslant 0, \\
g(1)=m^{2}+m-2 \geqslant 0 .
\end{array}\right.
$$

Solving this gives $m \leqslant-2$ or $m \geqslant 2$.
Therefore, there exist real numbers $m$ such that the inequality $m^{2}+m t+1 \geqslant\left|x_{1}-x_{2}\right|$ holds for any $a \in A$ and $t \in[-1,1]$, and the range of $m$ is $\{m \mid m \leqslant-2$ or $m \geqslant 2\}$.