11. Define an odd function $f(x)$ on $\mathbf{R}$, such that when $x \geqslant 0$, $f(x)=-x^{2}+2 x$. Another function $y=$ $g(x)$ has a domain of $[a, b]$, and a range of $\left[\frac{1}{b}, \frac{1}{a}\right]$, where $a \neq b, a, b \neq 0$. On $x \in[a, b]$, $f(x)=g(x)$. Question: Does there exist a real number $m$, such that the set $\{(x, y) \mid y=g(x), x \in[a, b]\} \cap$ $\left\{(x, y) \mid y=x^{2}+m\right\}$ contains exactly two elements?
11. Solution: $\left\{(x, y) \mid y=x^{2}+m\right\}$ is a family of parabolas formed by translating the graph of $y=x^{2}$ with the $y$-axis as the axis of symmetry. For a given $m$, it represents a parabola. The condition $\{(x, y) \mid y=g(x), x \in[a, b]\} \cap\left\{(x, y) \mid y=x^{2}+m\right\}$ having exactly two elements means that the graph of the function $y=g(x), x \in[a, b]$ intersects the parabola $y=x^{2}+m$ at exactly two points. First, we need to understand $y=g(x)$, $x \in[a, b]$, and then draw its graph.
It is easy to find that for the odd function $y=f(x)$, when $x>0$, it is better to consider the following three cases: $01$. However, when $x \in(0,1]$, $f(x) \leqslant 1$, which contradicts the right endpoint of the range of $g(x)$, $\left[\frac{1}{b}, \frac{1}{a}\right]$, being greater than 1. Therefore, the case $0<a<b \leqslant 1$ does not occur.
(2) If $1 \leqslant a<b<2$, from the graph, we see that $g(x)$ is a decreasing function, so we have $\left\{\begin{array}{l}\frac{1}{b}=g(b)=-b^{2}+2 b \\ \frac{1}{a}=g(a)=-a^{2}+2 a\end{array}\right.$, which simplifies to $\left\{\begin{array}{l}(a-1)\left(a^{2}-a-1\right)=0 \\ (b-1)\left(b^{2}-b-1\right)=0\end{array}\right.$, considering the condition $1 \leqslant a<b<2$, we solve to get $\left\{\begin{array}{l}a=1 \\ b=\frac{1+\sqrt{5}}{2}\end{array}\right.$
Similarly, considering the three cases $-1 \leqslant a<b<0$, $-2<a<-1<b<0$, and $-2<a<b \leqslant-1$, we can derive from the range condition that $\left\{\begin{array}{l}a=\frac{-1-\sqrt{5}}{2} \\ b=-1\end{array}\right.$ in the case $-2<a<b \leqslant-1$. This gives us the function
$$
g^{\prime}(x)=\left\{\begin{array}{lc}
-x^{2}+2 x & \left(1 \leqslant x \leqslant \frac{1+\sqrt{5}}{2}\right) \\
x^{2}+2 x & \left(\frac{-1-\sqrt{5}}{2} \leqslant x \leqslant-1\right)
\end{array}\right. \text {. }
$$

For some $m$, the parabola intersects the graph of the function $g^{\prime}(x)$ at two points, one in the first quadrant and the other in the third quadrant. Therefore, $m$ should make the equation $x^{2}+m=-x^{2}+2 x$ have exactly one real root in $\left[1, \frac{1+\sqrt{5}}{2}\right]$, and the equation $x^{2}+m=x^{2}+2 x$ have exactly one real root in $\left[\frac{-1-\sqrt{5}}{2},-1\right]$. The problem reduces to finding $m$ such that $\left\{\begin{array}{ll}2 x^{2}-2 x+m=0 & \text { in }\left[1, \frac{1+\sqrt{5}}{2}\right] \text { has exactly one real root, (1) } \\ x=\frac{1}{2} m & \text { in }\left[\frac{-1-\sqrt{5}}{2},-1\right] \text { has exactly one real root, (2) }\end{array}\right.$ From (1), the equation $2 x-2 x^{2}=m$ has exactly one root in $\left[1, \frac{1+\sqrt{5}}{2}\right]$, let $h(x)=2 x-2 x^{2}$, then $h\left(\frac{1+\sqrt{5}}{2}\right) \leqslant m \leqslant h(1)$, i.e., $-2 \leqslant m \leqslant 0$, from (2), we get $\frac{-1-\sqrt{5}}{2} \leqslant \frac{1}{2} m \leqslant-1$, i.e., $-1-\sqrt{5} \leqslant m \leqslant-2$, thus $m=-2$. It is easy to verify that the parabola $y=x^{2}-2$ intersects the graph of the function $g(x)$ at exactly two points $(-1, -1)$ and $\left(\frac{\sqrt{5}+1}{2}, \frac{\sqrt{5}-1}{2}\right)$. In summary: the real number $m=-2$ under the given conditions.