11. (15 points) Given the function $f(x)=-2 x^{2}+b x+c$ has a maximum value of 1 at $x=1$, and $0<m<n$. When $x \in [m, n]$, the range of $f(x)$ is $\left[\frac{1}{n}, \frac{1}{m}\right]$. Find the values of $m$ and $n$.
Three, 11. From the problem, we have
$$
f(x)=-2(x-1)^{2}+1 \text {. }
$$

Then $f(x) \leqslant 1 \Rightarrow \frac{1}{m} \leqslant 1 \Rightarrow m \geqslant 1$.
Thus, $f(x)$ is monotonically decreasing on $[m, n]$. Therefore,
$$
f(m)=-2(m-1)^{2}+1=\frac{1}{m} \text {, }
$$

and $f(n)=-2(n-1)^{2}+1=\frac{1}{n}$.
Hence, $m$ and $n$ are the two solutions of the equation
$$
f(x)=-2(x-1)^{2}+1=\frac{1}{x}
$$

which can be rewritten as
$$
(x-1)\left(2 x^{2}-2 x-1\right)=0 .
$$

Solving the equation, we get $x=1, \frac{1+\sqrt{3}}{2}, \frac{1-\sqrt{3}}{2}$.
Since $1<m<n$, then $m=1, n=\frac{1+\sqrt{3}}{2}$.