11. Let $x, y \in[0,1]$, find the range of $f(x, y)=\sqrt{\frac{1+x y}{1+x^{2}}}+\sqrt{\frac{1-x y}{1+y^{2}}}$.
When $x \geqslant y$, $\frac{1+x y}{1+x^{2}}, \frac{1-x y}{1+y^{2}} \in[0,1]$, thus
$1 \leqslant f(x, y) \leqslant \frac{1+x y}{1+x^{2}}+\frac{1-x y}{1+y^{2}} \leqslant \frac{1+x y}{1+x^{2}}+\frac{1-x y}{1+x^{2}}=\frac{2}{1+x^{2}} \leqslant 2 ;$
$xy^{-1}+y \Rightarrow \frac{y-x}{x^{-1}+x}<\frac{y-x}{y^{-1}+y}$ and $1-\frac{x+y}{y^{-1}+y}<1-\frac{y-x}{y^{-1}+y}$,
therefore $1<f(x, y)<\sqrt{1+\frac{y-x}{y^{-1}+y}}+\sqrt{1-\frac{y-x}{y^{-1}+y}}$
$<\sqrt{(1+1)\left(1+\frac{y-x}{y^{-1}+y}+1-\frac{y-x}{y^{-1}+y}\right)}=2$. Also, $f(0,0)=2, f(1,1)=1$,
so $f(x, y)=\sqrt{\frac{1+x y}{1+x^{2}}}+\sqrt{\frac{1-x y}{1+y^{2}}} \in[1,2]$.