Question 11: Given the mapping f: $\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ is both injective and surjective, and satisfies $f(x)+f(f(x))=6$, then $f(1)=$ $\qquad$ _.
Question 11, Solution: Let $S=\{1,2,3,4,5\}$, for any $\mathrm{a} \in S$, according to the conditions:
$$
\left\{\begin{array}{c}
f(a)+f(f(a))=6 \\
f(f(a))+f(f(f(a)))=6
\end{array}\right.
$$

Thus, $\mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{a})))$.
Since $f$ is a bijection, $\{f(a) \mid a \in S\}=S$, hence for any $a \in S$, we have $a=f(f(a))$. Therefore, $6=f(a)+f(f(a))=f(a)+a$, which means $f(a)=6-a$.
Based on the above analysis, we know $f(1)=5$.