11. (20 points) Let the function $f(x)=a x^{3}+b x^{2}+c x+d$ have an image $\Gamma$ with two extremum points $P$ and $Q$, where $P$ is the origin.
(1) When point $Q(1,2)$, find the analytical expression of $f(x)$;
(2) When point $Q$ is on the circle $C:(x-2)^{2}+(y-3)^{2}=1$, find the maximum value of the slope of the tangent line to the curve $\Gamma$.
11. Since $f(x)=a x^{3}+b x^{2}+c x+d$, we have
$$
f^{\prime}(x)=3 a x^{2}+2 b x+c \text {. }
$$

Because the graph $\Gamma$ has an extremum point $P$ at the origin, we have
$$
f^{\prime}(0)=0 \text {, and } f(0)=0 \text {. }
$$

Thus, $c=d=0$.
(1) When point $Q(1,2)$, from $f^{\prime}(1)=0$ and $f(1)=2$, we get
$$
3 a+2 b=0, a+b=2 \text {. }
$$

Solving these, we find $a=-4, b=6$.
Thus, $f(x)=-4 x^{3}+6 x^{2}$.
(2) Since $f^{\prime}(x)=3 a x^{2}+2 b x$, and given that point $Q$ lies on the circle $C$ and $a<0$, the maximum value of the slope $k$ of the tangent line to the curve $\Gamma$ is the maximum value of $f^{\prime}(x)$, i.e.,
$$
k_{\max }=-\frac{b^{2}}{3 a} \text {. }
$$

Let point $Q(m, n)$. Then
$$
\begin{array}{l}
f^{\prime}(m)=0, \text { and } f(m)=n \\
\Rightarrow 3 a m^{2}+2 b m=0, \text { and } a m^{3}+b m^{2}=n \\
\Rightarrow \frac{b}{a}=-\frac{3 m}{2}, b=\frac{3 n}{m^{2}} \\
\Rightarrow k_{\max }=-\frac{b^{2}}{3 a}=\frac{3}{2} \cdot \frac{n}{m} .
\end{array}
$$

Since $\frac{n}{m}$ represents the slope of a line passing through the origin and intersecting the circle $C$, and the maximum slope of such a line is $2+\frac{2 \sqrt{3}}{3}$, we have
$$
\begin{array}{l}
k_{\max }=-\frac{b^{2}}{3 a}=\frac{3}{2} \cdot \frac{n}{m} \\
\leqslant \frac{3}{2}\left(2+\frac{2 \sqrt{3}}{3}\right)=3+\sqrt{3} .
\end{array}
$$

Therefore, the maximum value of the slope of the tangent line to the curve $\Gamma$ is $3+\sqrt{3}$.