$127 m \in \mathbf{R}$. Planar point set
$$
\begin{array}{c}
M=\left\{(x, y) \mid y^{2}=x-1\right\}, \\
N=\left\{(x, y) \mid y=2 x-2 m^{2}+m-2\right\},
\end{array}
$$

Then the number of elements contained in the set $M \cap N$ is
A. 0 or 1 or 2
B. 0 or 1
C. 0
D. 1 or 2
127 D. Since $y=2 x-2 m^{2}+m-2=2 x-2\left(m-\frac{1}{4}\right)^{2}-\frac{15}{8} \leqslant 2 x-\frac{15}{8}$, the family of lines with a slope of 2, $y=2 x-2 m^{2}+m-2$, is the set of the line $y=2 x-\frac{15}{8}$ and all lines parallel to it and below it.
The line $y=2 x-\frac{15}{8}$ is tangent to the parabola $y^{2}=x-1$.

Therefore, the answer is $\mathrm{D}$.