11. A. For the function $y=x^{2}+(2 k-1) x+k^{2}$, are the two intersection points with the $x$-axis both to the left of the line $x=1$? If so, please explain the reason; if not, find the range of values for $k$ when both intersection points are to the right of the line $x=1$.
Three, 11. A. Not necessarily.
For example, when $k=0$, the intersections of the function's graph with the $x$-axis are at $(0,0)$ and $(1,0)$, not both to the right of the line $x=1$.

Let the $x$-coordinates of the two intersections of the function with the $x$-axis be $x_{1}$ and $x_{2}$. Then $x_{1}+x_{2}=-(2 k-1), x_{1} x_{2}=k^{2}$.
The parabola intersects the $x$-axis at two points to the right of the line $x=1$ if and only if the following conditions are satisfied:
$$
\left\{\begin{array}{l}
\Delta \geqslant 0, \\
\left(x_{1}-1\right)+\left(x_{2}-1\right)>0, \\
\left(x_{1}-1\right)\left(x_{2}-1\right)>0
\end{array}\right.
$$

By
$$
\text { } \left\{\begin{array} { l } 
{ ( 2 k - 1 ) ^ { 2 } - 4 k ^ { 2 } \geqslant 0 , } \\
{ - 2 k - 1 > 0 , } \\
{ k ^ { 2 } + 2 k > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k \leqslant \frac{1}{4}, \\
k < 0 .
\end{array}\right.\right.
$$

Therefore, when $k<-2$, the parabola intersects the $x$-axis at two points to the right of the line $x=1$.