11. (20 points) Through the right focus $F$ of the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$, draw a line $l$ intersecting the ellipse $C$ at two distinct points $A$ and $B$. Let $\overrightarrow{F A}=\lambda \overrightarrow{F B}$, and $T(2,0)$. If $\lambda \in[-2,-1]$, find the range of $|\overrightarrow{T A}+\overrightarrow{T B}|$.
11. It is easy to verify that the slope of line $l$ is not 0.

Let the line $l: x=k y+1$, substituting into the equation of the ellipse $C$ yields
$$
\left(k^{2}+2\right) y^{2}+2 k y-1=0 \text {. }
$$

Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
y_{1}+y_{2}=-\frac{2 k}{k^{2}+2}, y_{1} y_{2}=-\frac{1}{k^{2}+2} \text {. }
$$

From $\overrightarrow{F A}=\lambda \overrightarrow{F B}$, we get $\frac{y_{1}}{y_{2}}=\lambda$, and $\lambda<0$.
$$
\begin{array}{l}
\text { Hence } \frac{\left(y_{1}+y_{2}\right)^{2}}{y_{1} y_{2}}=-\frac{4 k^{2}}{k^{2}+2} \\
\Rightarrow \lambda+\frac{1}{\lambda}+2=-\frac{4 k^{2}}{k^{2}+2} .
\end{array}
$$

Given $\lambda \in[-2,-1]$
$\Rightarrow-\frac{5}{2} \leqslant \lambda+\frac{1}{\lambda} \leqslant-2$
$\Rightarrow-\frac{1}{2} \leqslant \lambda+\frac{1}{\lambda}+2 \leqslant 0$
$\Rightarrow-\frac{1}{2} \leqslant-\frac{4 k^{2}}{k^{2}+2} \leqslant 0$
$\Rightarrow k^{2} \leqslant \frac{2}{7} \Rightarrow 0 \leqslant k^{2} \leqslant \frac{2}{7}$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{T A}=\left(x_{1}-2, y_{1}\right), \overrightarrow{T B}=\left(x_{2}-2, y_{2}\right) \\
\Rightarrow \overrightarrow{T A}+\overrightarrow{T B}=\left(x_{1}+x_{2}-4, y_{1}+y_{2}\right) . \\
\text { Also } x_{1}+x_{2}-4=k\left(y_{1}+y_{2}\right)-2 \\
=-\frac{4\left(k^{2}+1\right)}{k^{2}+2} .
\end{array}
$$

Thus, $|\overrightarrow{T A}+\overrightarrow{T B}|^{2}$
$$
\begin{array}{l}
=\left(x_{1}+x_{2}-4\right)^{2}+\left(y_{1}+y_{2}\right)^{2} \\
=\frac{16\left(k^{2}+1\right)^{2}}{\left(k^{2}+2\right)^{2}}+\frac{4 k^{2}}{\left(k^{2}+2\right)^{2}} \\
=16-\frac{28}{k^{2}+2}+\frac{8}{\left(k^{2}+2\right)^{2}} .
\end{array}
$$

Let $t=\frac{1}{k^{2}+2}$.
From $0 \leqslant k^{2} \leqslant \frac{2}{7} \Rightarrow \frac{7}{16} \leqslant \frac{1}{k^{2}+2} \leqslant \frac{1}{2}$
$\Rightarrow t \in\left[\frac{7}{16}, \frac{1}{2}\right]$.
Then $|\overrightarrow{T A}+\overrightarrow{T B}|^{2}=8 t^{2}-28 t+16$
$$
=8\left(t-\frac{7}{4}\right)^{2}-\frac{17}{2} \text {. }
$$

Therefore, $|\overrightarrow{T A}+\overrightarrow{T B}| \in\left[2, \frac{13 \sqrt{2}}{8}\right]$.