China Mathematical Competition

Let $x<2\Rightarrow 2-x>0$. Then $$\begin{array}{rl}f(x)& =\frac{1+(4-4x+x^2)}{2-x}=\frac{1}{2-x}+(2-x)\\ & \ge 2\times \sqrt{\frac{1}{2-x}\times (2-x)}=2.\end{array}$$ The equality holds if and only if $\frac{1}{2-x}=2-x$, and it is so when $x=1\in (-\infty ,2)$. This means that $f$ reaches the minimum value 2 at $x=1$.

Answer: C