(2)易知f(x)>=0,∴a>=1/2,f(x)={2x+a,x>=0;.......{a,-a.......{-2x-a,x<=-a.f(x)<=2a-1化为3个不等式组：1)x>=0,2x+a<=2a-1;2)-a3)x<=-a,-2x-a<=2a-1.由1)0<=x<=(a-1)/2;由2),-a=1;由3),(1-3a)/2<=x<=-a.综上，a>=1,(1-3a)/2<=x<=(a-1)/2,依题意(a-1)/2-(1-3a)/2=2a-1=3,a=2,b=(1-3a)/2=-5/2.
