若抛物线y=-x^+2ax+b的顶点在直线mx-y-2m+1=0上移动,且与抛物线y=x^有公共点.求m的取值范围.

抛物线y=-x^2+2ax+b的顶点(a,a^2+b)在直线mx-y-2m+1=0上移动,∴ma-(a^2+b)-2m+1=0,m(a-2)=a^2+b-1.①它与抛物线y=x^2有公共点,∴2x^2-2ax-b=0有实根,∴△/4=a^2+2b>=0,b>=-a^2/2.由①,a=2,b=-3,m∈R;a>2时m=(a^2+b-1...