對(duì)任意x1∈[0,1],存在x2∈[0,1],使得g(x2)=f(x1)成立
那么當(dāng)定義域?yàn)閇0,1]時(shí),
f(x)的值域是g(x)值域的子集
g(x)=10x-1 為增函數(shù)
∵x∈[0,1],
∴g(x)值域?yàn)閇-1,9]
f'(x)=3x^2-3a=3(x^2-a)
∵a>1
∴f'(x)=3(x+√a)(x-√a)
當(dāng)x∈[0,1],f'(x)≤0恒成立,
∴f(x)在[0,1]上為減函數(shù)
f(x)max=f(0)=a^2,
f(x)min=f(1)=a^2-3a+1
∴{a^2≤9
{a^2-3a+1≥-1
解得{-3≤a≤3
{a≤1或a≥2
{a>1
∴2≤a≤3