$$
y=4 x^{2}-4 k x+(k-1)(-1 \leqslant k \leqslant 1) .
$$

Let the maximum length of the segment intercepted on the $x$-axis by the graph of this function be $m$, and the minimum length be $n$. Then $m+n=(\quad)$.
(A) $\frac{3 \sqrt{3}}{2}$
$\begin{array}{lll}\text { (B) } 1 & \text { (C) } 2 \sqrt{3}\end{array}$
(D) $\sqrt{3}+\frac{1}{4}$
- 1. A.

Let the coordinates of the points where the graph intersects the $x$-axis be $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$, and the distance between the points be $d$.
By Vieta's formulas, we have $x_{1}+x_{2}=k, x_{1} x_{2}=\frac{k-1}{4}$.
Then $d^{2}=\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$
$=k^{2}-k+1=\left(k-\frac{1}{2}\right)^{2}+\frac{3}{4}$.
Note that, $\left(k-\frac{1}{2}\right)^{2} \geqslant 0,-1 \leqslant k \leqslant 1$.
Therefore, when $k=\frac{1}{2}$, $n=\frac{\sqrt{3}}{2}$; when $k=-1$, $m=\sqrt{3}$.
Thus, $m+n=\frac{3 \sqrt{3}}{2}$.