11. Given the quadratic function $f(x)=x^{2}+a x+b(a, b \in \mathbf{R})$ has two distinct zeros, if $f\left(x^{2}+2 x-1\right)$ $=0$ has four distinct roots $x_{1}<x_{2}<x_{3}<x_{4}$, and $x_{1}, x_{2}, x_{3}, x_{4}$ form an arithmetic sequence, find the range of $a-b$.
Let $x^{2}+a x+b=0$ have two roots $m, n$, then $m+n=-a, m n=b$. The two roots of $x^{2}+2 x-1=m$ are $x_{1}, x_{4}$, and the two roots of $x^{2}+2 x-1=n$ are $x_{2}, x_{3}$. From $\left|x_{1}-x_{4}\right|=3\left|x_{2}-x_{3}\right| \Rightarrow \sqrt{4 m+8}=3 \sqrt{4 n+8} \Rightarrow m+2=9(n+2)$ $\Rightarrow m=9 n+16$. Therefore, $a-b=-(m+n)-m n=-10 n-16-n(9 n+16)$ $=-9 n^{2}-26 n-16=-9\left(n+\frac{13}{9}\right)^{2}+\frac{169}{9}-16=-9\left(n+\frac{13}{9}\right)^{2}+\frac{25}{9}$, since $n>-2$, when $n=-\frac{13}{9}$, $a-b$ reaches its maximum value $\frac{25}{9}$. Therefore, the range of $a-b$ is $\left(-\infty, \frac{25}{9}\right]$.