Question

已知a＞0,函數(shù)f(x)=ax-bx².

(1)當(dāng)b＞0時,若對任意x∈R都有f(x)≤1,證明a≤;

(2)當(dāng)0＜b≤1時,討論:對任意x∈［0,1］,|f(x)|≤1的充要條件.

Answer 1

(1)證明:依題設(shè),對任意x∈R,都有f(x)≤1,

∵f(x)=-b(x-)²+,

∴f()=≤1.

∵a＞0,b＞0,∴a≤.

(2)解析:因為a＞0,0＜b≤1時,對任意x∈［0,1］,

f(x)=ax-bx²≥-b≥-1,即f(x)≥-1;

f(x)≤1f(1)≤1a-b≤1,即a≤b+1,

a≤b+1f(x)≤(b+1)x-bx²≤1,即f(x)≤1.

所以,當(dāng)a＞0,0＜b≤1時,對任意x∈［0,1］,|f(x)|≤1的充要條件是a≤b+1.