15. Given that $f$ is a real-valued function on the set of all real numbers such that for any real numbers $a$ and $b$,
$$
\mathrm{f}(a \mathrm{f}(b))=a b
$$

Find the value of $\mathrm{f}(2011)$.
15. Answer: 2011
From the recurrence relation, $\mathrm{f}(\mathrm{f}(1) \mathrm{f}(b))=\mathrm{f}(1) b$ and $\mathrm{f}(\mathrm{f}(b) \mathrm{f}(1))=\mathrm{f}(b) \cdot 1$. Hence, $\mathrm{f}(b)=$ $\mathrm{f}(1) b$. By letting $b=f(1)$, we obtain $\mathrm{f}(\mathrm{f}(1))=(\mathrm{f}(1))^{2}$. Also, from the given functional equation, we have $\mathrm{f}(\mathrm{f}(1))=1$, hence $(f(1))^{2}=1$, following that $f(1)$ is either 1 or -1 . Hence $\mathrm{f}(2011)=2011$.