Q1 Consider the equation
$$(x-1)^2 = |3x-5|.$$
(1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$.
(2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.