(1) The minimum value of the function $f(x)=\frac{5-4 x+x^{2}}{2-x}$ on $(-\infty, 2)$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
(1) When $x<2$, therefore
$$
\begin{aligned}
f(x) & =\frac{1+\left(4-4 x+x^{2}\right)}{2-x} \\
& =\frac{1}{2-x}+(2-x) \\
& \geqslant 2 \cdot \sqrt{\frac{1}{2-x} \cdot(2-x)} \\
& =2,
\end{aligned}
$$

Equality holds if and only if $\frac{1}{2-x}=2-x$. This equation has a solution $x=1 \in(-\infty, 2)$, so the minimum value of $f(x)$ on $(-\infty, 2)$ is 2. Therefore, the answer is C.