11. (15 points) Let $A$ and $B$ be the two intersection points of the line $y=x+b$ and the ellipse $3 x^{2}+y^{2}=1$, and $O$ be the origin. If $\triangle O A B$ is an acute triangle, find the range of $b$.

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11. Substituting $y=x+b$ into $3 x^{2}+y^{2}=1$ yields $3 x^{2}+(x+b)^{2}=1$, which simplifies to $4 x^{2}+2 b x+b^{2}-1=0$.
From $\Delta>0$, we get $-\frac{2 \sqrt{3}}{3}<b<\frac{2 \sqrt{3}}{3}$.
In $\triangle O A B$, as $|O A|>0$, when $b (b>0)$ increases to $b^{\prime}$, points $A$ and $B$ become $A^{\prime}$ and $B^{\prime}$, respectively.

From Figure 6, we know that the x-coordinate of point $A$ is greater than that of point $A^{\prime}$, and the x-coordinate of point $B$ is less than that of point $B^{\prime}$. Therefore, $\angle A^{\prime} O B^{\prime}$ is inside $\angle A O B$, so $\angle A O B$ decreases as $b$ increases;

On the other hand, let $O A^{\prime}$ intersect $A B$ at point $P$. Then $\angle B P O$ is an exterior angle of $\triangle A P O$. Hence,
$\angle B^{\prime} A^{\prime} O=\angle B P O>\angle B A O$.
Thus, $\angle B A O$ increases as $b$ increases.
Therefore, when $b>0$, $\triangle O A B$ is an acute triangle if and only if $b \in\left(\frac{\sqrt{2}}{2}, 1\right)$.

Finally, by the symmetry of the ellipse, when $\triangle O A B$ is an acute triangle, the range of real number $b$ is
$$
\left(-1,-\frac{\sqrt{2}}{2}\right) \cup\left(\frac{\sqrt{2}}{2}, 1\right) \text {. }
$$