设a,b是实数,二次方程x^2-ax+b=0的一个根属于区间【-1,1】,另一根属于区间【1,2】.求a-2b的取值范围

根据题意有
f(-1)*f(1)<=0
f(1)*(f2)<=0
0<=a<=3
-2<=b<=2
即(1+a+b)(1-a+b)<=0且(1-a+b)(4-2a+b)<=0且0<=a<=3且-2<=b<=2
1+a+b>=0,1-a+b<=0,4-2a+b>=0
b>=-a-1>=-4,b>=2a-4>=-4
所以0<=a<=3,-2<=b<=2
-4<=-2b<=4
-4<=a-2b<=7