1. (2002 Hunan Province High School Mathematics Competition) If the quadratic function $f(x)=a x^{2}+b x+c$, where $a \in \mathbf{N}^{*}, c \geqslant$ $1, a+b+c \geqslant 1$, the equation $a x^{2}+b x+c=0$ has two distinct positive roots less than 1, then the minimum value of $a$ is
A. 5
B. 4
C. 3
D. 2
1. A. Let $f(x)=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, then $a+b+c=f(1)=a\left(1-x_{1}\right)\left(1-x_{2}\right) \geqslant 1, c=f(0)=$ $a x_{1} x_{2} \geqslant 1$. Multiplying the two inequalities, we get $a^{2} x_{1}\left(1-x_{1}\right) \cdot x_{2}\left(1-x_{2}\right) \geqslant 1$. Since $x_{1}\left(1-x_{1}\right) \cdot x_{2}\left(1-x_{2}\right) \leqslant \frac{1}{16}$. And $a \in \mathbf{N}^{*}$, so $a \geqslant 5$.