Question 67, Given real numbers $x, y$ satisfy $x^{2}+(y-2)^{2} \leq 1$, find the range of $\frac{x+\sqrt{3} y}{\sqrt{x^{2}+y^{2}}}$.
Question 67, Solution: Notice that $\left(\frac{x}{\sqrt{x^{2}+y^{2}}}\right)^{2}+\left(\frac{y}{\sqrt{x^{2}+y^{2}}}\right)^{2}=1$. Let $\frac{x}{\sqrt{x^{2}+y^{2}}}=\cos \theta, \frac{y}{\sqrt{x^{2}+y^{2}}}=\sin \theta$, then $\frac{x+\sqrt{3} y}{\sqrt{x^{2}+y^{2}}}=\cos \theta+\sqrt{3} \cdot \sin \theta=2 \sin \left(\theta+\frac{\pi}{6}\right)$.

Notice that the point $(x, y)$ lies within the unit disk $x^{2}+(y-2)^{2}=1$, hence $\frac{y}{x} \in\left(-\infty,-\frac{\sqrt{3}}{2}\right] \cup\left[\frac{\sqrt{3}}{2},+\infty\right)$, which means $\theta \in\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right]$. Therefore, the range of $\frac{x+\sqrt{3} y}{\sqrt{x^{2}+y^{2}}}=2 \sin \left(\theta+\frac{\pi}{6}\right)$ is $[1,2]$.