11. Given the sequence $a_{n+1}=3 a_{n}-a_{n}^{2}-1(n=1,2, \cdots), a_{1}=a$, the following statements:
A. If $a \neq 1$, then $\left\{a_{n}\right\}$ is strictly monotonically decreasing
B. If $a=\frac{3}{2}$, then for any $n \in \mathbf{N}^{*}$, $a_{n}>1$
C. If $a=3$, then there exists a positive integer $k_{0}$ such that $\sum_{i=1}^{k_{0}} \frac{1}{a_{i}-2}<\frac{1}{2}$
D. Given the sequence $a_{n+1}=3 a_{n}-a_{n}^{2}+k(n=1,2, \cdots), a_{1}=1$, if $k \in\left[-\frac{3}{4}, 0\right]$, then the sequence $\left\{a_{n}\right\}$ is bounded. $\qquad$
Answer
As shown in the figure, the fixed point of the sequence $\left\{a_{n}\right\}$ satisfies
$$
x=3 x-x^{2}-1 \Rightarrow x=1 \text {, }
$$

and $3 x-x^{2}-1=1 \Rightarrow x=2$,
that is, when $a=2$, $a_{2}=a_{3}=\cdots=1$.
Option $A$ is incorrect;
From the cobweb diagram, when $11$, option $B$ is correct; $\square$
$a_{n+1}-a_{n}=2 a_{n}-a_{n}^{2}-1=-\left(a_{n}-1\right)^{2}\frac{1}{2}$. Option $C$ is incorrect; For option $D, 3 x-x^{2}+k>x \Rightarrow k>(x-1)^{2}-1$, from the cobweb diagram, the sequence $\left\{a_{n}\right\}$ is bounded. Therefore, the answer is $B D$.