3. (2006 National High School Mathematics League Competition Jiangxi Province Preliminary Test) The set $M$ consists of functions $f(x)$ that satisfy the following condition: when $x_{1}, x_{2} \in[-1,1]$, there is $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leqslant 4\left|x_{1}-x_{2}\right|$. For the two functions $f_{1}(x)=$ $x^{2}-2 x+5, f_{2}(x)=\sqrt{|x|}$, the following relationship that holds is
A. $f_{1} \in M, f_{2} \in M$
B. $f_{1} \notin M, f_{2} \notin M$
C. $f_{1} \notin M, f_{2} \in M$
D. $f_{1} \in M, f_{2} \notin M$
3. D Note that
$$
\begin{array}{l}
\left|f_{1}\left(x_{1}\right)-f_{1}\left(x_{2}\right)\right|=\left|x_{1}^{2}-x_{2}^{2}-2\left(x_{1}-x_{2}\right)\right|=\left|x_{1}-x_{2}\right|\left|x_{1}+x_{2}-2\right| \leqslant 4\left|x_{1}-x_{2}\right|, \\
\left|f_{2}\left(x_{1}\right)-f_{2}\left(x_{2}\right)\right|=\left|\sqrt{\left|x_{1}\right|}-\sqrt{\mid x_{2} T}\right|=\frac{|| x_{1}|-| x_{2}||}{\sqrt{\left|x_{1}\right|}+\sqrt{\left|x_{2}\right|}} .
\end{array}
$$

Let $x_{1}=\frac{1}{900}, x_{2}=\frac{1}{3600}$, then
$$
\left|f_{2}\left(x_{1}\right)-f_{2}\left(x_{2}\right)\right|=\frac{|| x_{1}|-| x_{2}||}{\sqrt{\mid x_{1} T}+\sqrt{\left|x_{2}\right|}}=\frac{\left|x_{1}-x_{2}\right|}{\frac{1}{30}+\frac{1}{60}}=20\left|x_{1}-x_{2}\right|>4\left|x_{1}-x_{2}\right| .
$$