Question

已知a＞0,函数f(x)=ax-bx².

(1)当b＞0时,若对任意x∈R都有f(x)≤1,证明a≤;

(2)当0＜b≤1时,讨论:对任意x∈［0,1］,|f(x)|≤1的充要条件.

Answer 1

(1)证明:依题设,对任意x∈R,都有f(x)≤1,

∵f(x)=-b(x-)²+,

∴f()=≤1.

∵a＞0,b＞0,∴a≤.

(2)解析:因为a＞0,0＜b≤1时,对任意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,当a＞0,0＜b≤1时,对任意x∈［0,1］,|f(x)|≤1的充要条件是a≤b+1.