2. The quadratic function $f(x)=x^{2}-a x+b$ has one root in $[-1,1]$ and the other root in $[1,2]$, then the minimum value of $a-2 b$ is
A. 0
B. -1
C. -2
D. 1
2. B Hint: According to the problem, $f(-1)=1+a+b \geqslant 0, f(1)=1-a+b \leqslant 0, f(2)=4-2 a+b \geqslant 0$. Therefore, $a-b-1 \geqslant 0, 3 a-3 b-3 \geqslant 0$, so $(4-2 a+b)+(3 a-3 b-3)=a-2 b+1 \geqslant 0, a-2 b \geqslant-1$.