10 Let the real-coefficient quadratic equation $x^{2}+a x+2 b-2=0$ have two distinct real roots, one of which lies in the interval $(0,1)$ and the other in the interval $(1,2)$. Then the range of $\frac{b-4}{a-1}$ is $\qquad$
According to the problem, let the two distinct real roots be $x_{1}$ and $x_{2}$, and
$$
0<x_{1}<1<x_{2}<2,
$$

then
$$
\begin{array}{l}
1<x_{1}+x_{2}=-a<3, \\
0<x_{1} x_{2}=2 b-2<2 .
\end{array}
$$

Thus, we have $-3<a<-1,1<b<2$, which means
$$
\begin{array}{l}
-\frac{1}{2}<\frac{1}{a-1}<-\frac{1}{4} \\
-3<b-4<-2 .
\end{array}
$$

Therefore, $\frac{1}{2}<\frac{b-4}{a-1}<\frac{3}{2}$.