(1)∵对于任意x∈R,都有f (x)-x≥0,且当x∈(0,2)时, 有f (x) ≤.令x=1 ∴1≤f (1) ≤. 即f (1)="1.·······················" 5分 (2) 由a-b+c=0及f (1)=1. 有,可得b=a+c=.·············· 7分 又对任意x,f(x)-x≥0,即ax2-x+c≥0. ∴a>0且△≤0. 即-4ac≤0,解得ac≥.················ 9分 (3) 由(2)可知a>0,c>0. a+c≥2≥2·=.················ 10分 当且仅当时等号成立.此时 a=c=.························ ∴f (x)= x2+x+, F (x)=f (x)-mx=[x2+(2-4m)x+1].············· 12分 当x∈[-2,2]时,f (x)是单调的,所以F (x)的顶点一定在[-2,2]的外边. ∴≥2.····················· 13分 解得m≤-或m≥. …………………………………………………………..14分 |