5. Given $f(x)=1-(x-a)(x-b)$, and $m, n$ are the roots of the equation $f(x)=0$. Then the possible order of the real numbers $a, b, m, n$ is $(\quad)$.
(A) $m<a<b<n$
(B) $a<m<n<b$
(C) $a<m<b<n$
(D) $m<a<n<b$
5.A.

Notice that $f(a)=$ $f(b)=1$, the parabola $y=$ $-x^{2}+(a+b) x+1-$ $a b$ opens downwards, therefore, the graph can only be as shown in Figure 6 (assuming $m<$ $n$).