11. (This question is worth 20 points) Let $F_{1}, F_{2}$ be the left and right foci of the ellipse $\frac{x^{2}}{2}+y^{2}=1$, respectively. Suppose a line $l$ that does not pass through the focus $F_{1}$ intersects the ellipse at two distinct points $A, B$. The distance from the focus $F_{2}$ to the line $l$ is $d$. If the slopes of $A F_{1}, l, B F_{1}$ form an arithmetic sequence, find the range of $d$.
Solution: From the given conditions, the coordinates of points $F_{1}$ and $F_{2}$ are $(-1, 0)$ and $(1, 0)$, respectively. Let the equation of line $l$ be $y = kx + m$, and the coordinates of points $A$ and $B$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. Then $x_{1}, x_{2}$ satisfy the equation $\frac{x^{2}}{2} + (kx + m)^{2} = 1$, which simplifies to $\left(2k^{2} + 1\right)x^{2} + 4kmx + \left(2m^{2} - 2\right) = 0$.
Since points $A$ and $B$ are distinct and the slope of line $l$ exists, $x_{1}, x_{2}$ are the two distinct real roots of equation (1). Therefore, the discriminant of equation (1) is $\Delta = (4km)^{2} - 4 \cdot \left(2k^{2} + 1\right) \cdot \left(2m^{2} - 2\right) = 8\left(2k^{2} + 1 - m^{2}\right) > 0$, which implies $2k^{2} + 1 > m^{2}$. (2)
From the slopes of lines $AF_{1}$, $l$, and $BF_{1}$, which are $\frac{y_{1}}{x_{1} + 1}$, $k$, and $\frac{y_{2}}{x_{2} + 1}$, respectively, and form an arithmetic sequence, we have $\frac{y_{1}}{x_{1} + 1} + \frac{y_{2}}{x_{2} + 1} = 2k$. Since $y_{1} = kx_{1} + m$ and $y_{2} = kx_{2} + m$, we get $\left(kx_{1} + m\right)\left(x_{2} + 1\right) + \left(kx_{2} + m\right)\left(x_{1} + 1\right) = 2k\left(x_{1} + 1\right)\left(x_{2} + 1\right)$, which simplifies to $(m - k)\left(x_{1} + x_{2} + 2\right) = 0$.
If $m = k$, then the equation of line $l$ is $y = kx + k$, which passes through point $F_{1}(-1, 0)$, which is not allowed.
Therefore, it must be that $x_{1} + x_{2} + 2 = 0$. From equation (1) and Vieta's formulas, we have $\frac{4km}{2k^{2} + 1} = -\left(x_{1} + x_{2}\right) = 2$, which simplifies to $m = k + \frac{1}{2k}$.
From (2) and (3), we have $2k^{2} + 1 > m^{2} = \left(k + \frac{1}{2k}\right)^{2}$, which simplifies to $k^{2} > \frac{1}{4k^{2}}$, equivalent to $|k| > \frac{\sqrt{2}}{2}$.
Conversely, when $m$ and $k$ satisfy (3) and $|k| > \frac{\sqrt{2}}{2}$, line $l$ does not pass through point $F_{1}$ (otherwise, it would lead to $m = k$, contradicting (3)), and $m$ and $k$ satisfy (2), ensuring that $l$ intersects the ellipse at two distinct points $A$ and $B$, and also ensuring that the slopes of $AF_{1}$ and $BF_{1}$ exist (otherwise, one of $x_{1}, x_{2}$ would be -1, and combining $x_{1} + x_{2} + 2 = 0$ would imply $x_{1} = x_{2} = -1$, contradicting the fact that equation (1) has two distinct real roots). 10 points
The distance from point $F_{2}(1, 0)$ to line $l: y = kx + m$ is
$$
d = \frac{|k + m|}{\sqrt{1 + k^{2}}} = \frac{1}{\sqrt{1 + k^{2}}} \cdot \left|2k + \frac{1}{2k}\right| = \frac{1}{\sqrt{\frac{1}{k^{2}} + 1}} \cdot \left(2 + \frac{1}{2k^{2}}\right).
$$

Noting that $|k| > \frac{\sqrt{2}}{2}$, let $t = \sqrt{\frac{1}{k^{2}} + 1}$, then $t \in (1, \sqrt{3})$, and the above expression can be rewritten as
$$
d = \frac{1}{t} \cdot \left(\frac{t^{2}}{2} + \frac{3}{2}\right) = \frac{1}{2} \cdot \left(t + \frac{3}{t}\right).
$$

Considering the function $f(t) = \frac{1}{2} \cdot \left(t + \frac{3}{t}\right)$ is monotonically decreasing on $[1, \sqrt{3}]$, from (4) we get $f(\sqrt{3}) < d < f(1)$, i.e., $d \in (\sqrt{3}, 2)$. 20 points