14.42 In the region $x>0, y>0, x y=1$, find the range of the function
$$
f(x, y)=\frac{x+y}{[x][y]+[x]+[y]+1}
$$

(Austrian Mathematical Competition, 1988)
[Solution] By symmetry, let $x \geqslant 1$.
(1) When $x=1, y=1$.
$$
f(1,1)=\frac{1}{2} \text {. }
$$
(2) When $x>1$, let $x=n+\alpha$, where $n \geqslant 1, n \in N, 0 \leqslant \alpha<1$.
The function $y=x+\frac{1}{x}$ is an increasing function when $x>1$.
For this, let $x_{1}>x_{2}>1$, then
$$
\begin{aligned}
y_{1}-y_{2} & =\left(x_{1}-x_{2}\right)+\left(\frac{1}{x_{1}}-\frac{1}{x_{2}}\right) \\
& =\left(x_{1}-x_{2}\right)\left(1-\frac{1}{x_{1} x_{2}}\right) .
\end{aligned}
$$

Since $x_{1}>x_{2}>1$, then $x_{1}-x_{2}>0,1-\frac{1}{x_{1} x_{2}}>0$, thus
$$
y_{1}-y_{2}>0, \quad y_{1}>y_{2} \text {. }
$$

That is, the function $y=x+\frac{1}{x}$ is an increasing function when $x>1$.
Considering function (1), since

then
$$
n \leqslant n+\alpha < n+1 .
$$

Thus, when $n \geqslant 1$, we have
$$
b_{1}>b_{2}>b_{3}>b_{4}>b_{5}>\cdots \text {, }
$$

which means $b_{1}$ is the largest.
Therefore, when $x>1$, the range of $f(x, y)$ is
$$
\left[a_{2}, b_{1}\right)=\left[\frac{5}{6}, \frac{5}{4}\right) .
$$

Thus, the range of $f(x, y)$ is
$$
\left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right) .
$$