Column 7 Let the function $f(x)=-\frac{1}{2} x^{2}+x+a\left(\right.$ constant $\left.a \leqslant \frac{5}{2}\right)$, Question: Does there exist real numbers $m$ and $n(m<n)$, such that the domain and range of the function $f(x)$ are the intervals $[m, n]$ and $[3 m, 3 n]$ respectively?
Solve $f(x)=-\frac{1}{2}(x-1)^{2}+a+\frac{1}{2}$, if $n>1$, then $a+\frac{1}{2} \geqslant 3 n>3$, making $a>\frac{5}{2}$, which is impossible, so $n \leqslant 1$. $f(x)$ is increasing on $[m, n]$, so $\left\{\begin{array}{l}f(m)=3 m \\ f(n)=3 n,\end{array}\right.$, leading to the equation $f(x)=3 x$, i.e., $x^{2}+4 x-2 a=0$ has two distinct real roots in $(-\infty, 1]$. Therefore, let $g(x)=x^{2}+4 x-2 a$, (1) if one root is 1, then $f(1)=0, a=\frac{5}{2}$, the other root is -5; (2) if both roots are in $(-\infty, 1)$, then $\left\{\begin{array}{l}\Delta=16+8 a>0 \\ g(1)=5-2 a>0 \\ -2<1 .\end{array}\right.$ Solving this gives $-2<a<\frac{5}{2}$, so $-2<a \leqslant \frac{5}{2}$. Therefore, when $a \in\left(-2, \frac{5}{2}\right]$, there exist $m=-2-\sqrt{4+2 a}, n=-2+\sqrt{4+2 a}$ that satisfy the conditions.