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    (苏皖学校高三第三次月考数学试卷) 已知二次函数f(x)=ax2+bx+c(a,b,c均为实数),满足a-b+c=0.对于任意实数x 都有f (x)-x≥0.并且当x∈(0.2)时,有f (x)≤. (1)求f (1)的值, (2)证明:ac≥, (3)当x∈[-2.2]且a+c取得最小值时.函数F(x)=f (x)-mx (m为实数)是单调的.求证:m≤或m≥. 解:(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 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 【查看更多】

     

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