2. (The 15th "Hope Cup" Mathematics Invitational Test) When $x \in [0,1]$, the absolute value of the function $f(x) = ax^2 + bx + c (a \neq 0)$ does not exceed 1, then the maximum value of $|a| + |b| + |c|$ is
A. 3
B. 9
C. 17
D. 19
2. C. Using the three-point form of a quadratic function: From $f(0)=c, f\left(\frac{1}{2}\right)=\frac{1}{4} a+\frac{1}{2} b+c, f(1)=a+b+c$, we get $a=2 f(1)-4 f\left(\frac{1}{2}\right)+2 f(0), b=-f(1)+4 f\left(\frac{1}{2}\right)-3 f(0), c=f(0)$. Thus, $|a| \leqslant 2|f(1)|+4\left|f\left(\frac{1}{2}\right)\right|+2|f(0)| \leqslant 2+4+2=8,|b| \leqslant|f(1)|+4\left|f\left(\frac{1}{2}\right)\right|+3|f(0)| \leqslant 1+4+3=8,|c|=|f(0)| \leqslant$ 1, hence $|a|+|b|+|c| \leqslant 17$. The equality in the above expression can hold, for example, by taking $f(x)=8 x^{2}-8 x+1, x \in[0,1]$.