1. Let $\mathrm{f}$ be a function satisfying the following conditions:
(i) If $x>y$, and $f(y)-y \geqslant v \geqslant f(x)-x$, then for some number $z$ between $x$ and $y$, $f(z)=v+z$;
(ii) The equation $\mathrm{f}(\mathrm{x})=0$ has at least one solution, and among the solutions, there is one that is not less than all other solutions;
(iii) $\mathrm{f}(0)=1$;
(iv) $\mathrm{f}(1987) \leq$ 588;
(v) $f(x) \cdot f(y)=f(x \cdot f(y)+y \cdot f(x)$ $-x y)$.
Find $f(1987)$.
1. Solution
(1) From (ii), we know that $f(x)=0$ has at least one solution. Let's assume $u$ is a solution of $f(x)$.
(2) Assume $u>0$.
(3) From (i) and (iii), we know that there exists a number $z$ between 0 and $u$ such that $f(z)=z$.
(4) From (v), we get:
$$
\begin{array}{l} 
0=f(z) \cdot f(u)=f(z \cdot f(u)+u \cdot f(z)-u z \\
=f(0)=1, \text{ contradiction. }
\end{array}
$$
(5) Therefore, from (iii), we have $u<0$.
(6) From (ii), we know that among the solutions of $\mathrm{f}(\mathrm{x})=0$, there is a solution $\mathbf{x}_{0}$ that is not less than all other solutions.
(7) From (5), we get $x_{0}<0$.
(8) From (v), we get $f(x) \cdot f\left(x_{0}\right)=f\left(x \cdot f\left(x_{0}\right)\right.$
$$
\left.+x_{0} \cdot f(x)-x x_{0}\right),
$$
which means $0=f\left(x_{0} \cdot(f(x)-x)\right)$, and $x$ can be any number.
(9) From (6), we get $x_{0} \geqslant x_{0} \cdot(f(x)-x)$. Since $x < 0$, we can rewrite the inequality as $f(x)-x \geqslant 1$.
(10) From (iv) and (9), we get $1987 \leqslant 1988 \leqslant f(1987)$. Therefore, we have $f(1987)=1988$.