Question

已知a>0,函數(shù)f(x)=ax-bx^2.

(1)當(dāng)b>0時(shí),若對(duì)任意x∈R都有f(x)≤1,證明a≤2;

(2)當(dāng)b>1時(shí),證明對(duì)任意x∈［0,1］,|f(x)|≤1的充要條件是b-1≤a≤2.

Answer 1

證明:（1）依題意設(shè)對(duì)任意x∈R都有f(x)≤1,

∵=-b(x-)²+,

∴f()=≤1.

∵a>0,b>0,∴a≤2.

(2)必要性:對(duì)任意x∈［0,1］,|f(x)|≤1f(x)≥-1,

∴f(1)≥-1,即a-b≥-1.

∴a≥b-1.

對(duì)任意x∈［0,1］,|f(x)|≤1f(x)≤1,

∵b>1,可以推出f()≤1,

即a·-1≤1.

∴a≤.∴b-1≤a≤.

充分性:∵b>1,a≥b-1,對(duì)任意x∈［0,1］,可以推出ax-bx²≥b(x-x²)-x≥-x≥-1,即ax-bx²≥-1.

∵b>1,a≤2,

對(duì)任意x∈［0,1］,可以推出ax-bx²≤2bx-bx²≤1,即ax-bx²≤1.

∴-1≤f(x)≤1.

綜上,當(dāng)b>1時(shí),對(duì)任意x∈［0,1］,|f(x)|≤1的充要條件是b-1≤a≤2.