f(x+1)=f(x)+2xf(0)=11=cf(1)=f(0)a+b+1=1==>b=-af(x)=ax^2-ax+1在f(x+1)=f(x)+2x中,令x= - 1f(0)=f(-1)-2f(-1)=32a+1=3a=1f(x)=x^2-x+1mm而右边y=x^2-3x+1这条抛物线的开口向上,对称轴:x=3/2,所以函数,y=x^2-3x+1在[-1,1] 上是减函数,y(min)=y(1)=-1所以,m<-1