2. Suppose $a+b\sqrt{11} > 1$. Also, $(a+b\sqrt{11})(a-b\sqrt{11}) = 1$. Let $c := a+b \sqrt{11}$. Then $a-b \sqrt{11} = \frac{1}{c}$. Then $ca-cb \sqrt{11} = 1$. Or $\sqrt{11} = - \frac{\frac{1}{c}-a}{b}$. Or $\frac{1}{\sqrt{11}} = -\frac{b}{\frac{1}{c}-a}$.
We know that $0 < \frac{1}{c} < 1$. So $a > 0$. Also $b > 0$, which forces the quotient to be positive. If $b < 0$ then $a < 0 \implies a+b \sqrt{11} < 1$. So $a >0$ so that the quotient is negative.