已知a>0,函数f(x)=ax-bx`2.
(1）当b>0时，若对任意x∈R都有f(x)<=1,证明：a<=2√b;
(2)当b>1时，证明：对任意x∈[0,1]都有│f(x)│<=1的充要条件是b-1<=a<=2√b。

已知a>0,函数f(x)=ax-bx`2.
(1）当b>0时，若对任意x∈R都有f(x)<=1,证明：a<=2√b;
(2)当b>1时，证明：对任意x∈[0,1]都有│f(x)│<=1的充要条件是b-1<=a<=2√b

i）∵a<=2√b==>a^2<=4b==>a^2/4b<=1,f(x)<=1
∴f(x)=ax-bx^2
==>-b(x^2-xa/b)
==>-b(x^2-xa/b+a^2/4b^2)+a^2/4b
==>-b(x-a/2b)^2+a^2/4b<=a^2/4b<=1
∴a<=2√b

ii）必要条件：
)当b>1时，对任意x∈[0,1]都有│f(x)│<=1
∵│f(x)│<=1==>f(x)>=-1
==>f(1)>=-1
==>a-b>=-1
==>a>=b-1
│f(x)│<=1==>f(x)<=1,(b>1)
==>f(1/√b)<=1
==>(a/√b)-1<=1
==>a<=2√b
所以：b-1<=a<=2√b

充分条件：
b>1,a>=b-1,x∈[0,1].
f(x)=ax-bx^2>=(b-1)x-bx^2=b(x-x^2)-x
==>b(x-x^2)-x>=-x>=-1
即：f(x)>=-1

b>1,a<=2√b,x∈[0,1]
f(x)=ax-bx^2<=x2√b-bx^2<=1

∴-1=<f(x)<=1
即：对任意x∈[0,1]都有│f(x)│<=1的充要条件是b-1<=a<=2√b。