8. If the roots of the equation $4 x^{2}-2 m x+n=0$ with respect to $x$ are both greater than 0 and less than 1, then the values of the natural numbers $m, n$ are ( ).
(A) $m=1, \bar{n}=1$
(B) $m=1, n=2$
(C) $m=2, n=1$
(D) $m=2, n=2$
8. C.

Solution Let the two roots of the original equation be $x_{1}, x_{2}$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{m}{2}, x_{1} x_{2}=\frac{n}{4} . \\
\left(x_{1}-1\right)\left(x_{2}-1\right) \\
=x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1 \\
=\frac{n}{4}-\frac{m}{2}+1 . \\
\text { By }\left\{\begin{array} { l } 
{ n ^ { 2 } \geq 4 \ldots , } \\
{ 0 < m < 4, } \\
{ \frac { n } { 4 } > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=2, \\
n=1 .
\end{array}\right.\right.
\end{array}
$$