5. (1994 Shanghai Competition Question) Given the function $y=\sqrt{1994-x}+\sqrt{x-1993}$, find the range of $y$.


5. Since $1994-x \geqslant 0$ and $x-1993 \geqslant 0$, then $1993 \leqslant x \leqslant 1994$, we can know that $y \geqslant 1$. Let $\vec{a}=(\sqrt{1994-x}, \sqrt{x-1993}), \vec{b}=(1,1)$,
$$
\begin{aligned}
1 & =1994-x+x-1993=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}} \\
& =\frac{(\sqrt{1994-x}+\sqrt{x-1993})^{2}}{1^{2}+1^{2}}=\frac{(\sqrt{1994-x}+\sqrt{x-1993})^{2}}{2},
\end{aligned}
$$

Thus, $(\sqrt{1994-x}+\sqrt{x-1993})^{2} \leqslant 2, \sqrt{1994-x}+\sqrt{x-1993} \leqslant \sqrt{2}$. 
Since $y \geqslant 1$,
the range of the function $y=\sqrt{1994-x}+\sqrt{x-1993}$ is $1 \leqslant y \leqslant \sqrt{2}$.