11. (22 points) Let $x, y \in [0,1]$. Find the range of
$$
f(x, y)=\sqrt{\frac{1+x y}{1+x^{2}}}+\sqrt{\frac{1-x y}{1+y^{2}}}
$$
11. When $x \geqslant y$,
$$
\begin{array}{l}
\frac{1+x y}{1+x^{2}}, \frac{1-x y}{1+y^{2}} \in[0,1] \\
\Rightarrow 2 \geqslant f(x, y) \geqslant \frac{1+x y}{1+x^{2}}+\frac{1-x y}{1+y^{2}} \\
\quad \geqslant \frac{1+x y}{1+x^{2}}+\frac{1-x y}{1+x^{2}}=\frac{2}{1+x^{2}} \geqslant 1 .
\end{array}
$$

When $x \leqslant y$, it is obvious that
$$
\begin{array}{l}
f(x, y) \geqslant \sqrt{\frac{1+x y}{1+x^{2}}} \geqslant 1, x+x^{-1} \geqslant y+y^{-1} \\
\Rightarrow 1 \leqslant f(x, y) \\
\quad=\sqrt{1+\frac{y-x}{x+x^{-1}}}+\sqrt{1-\frac{x+y}{y+y^{-1}}} \\
\quad \leqslant \sqrt{1+\frac{y-x}{y+y^{-1}}}+\sqrt{1-\frac{y-x}{y+y^{-1}}} \leqslant 2 .
\end{array}
$$

The last inequality above uses:
When $0 \leqslant t \leqslant 1$, $\sqrt{1+t}+\sqrt{1-t} \leqslant 2$.
By the continuity of $f(x, y)$ and
$$
f(1,1)=1, f(0,0)=2 \text {, }
$$

the range of $f(x, y)$ is $[1,2]$.