Question

1/3（x^3）-a^2x满足,对任意x1,x2∈[0,1]｜f(x1)-f(x2)｜≤1恒成立则a的取值范围

Answer 1

f'(x) = x^2 - a^2
若a<0或者a>1时, 则[0,1]
上, f'(x)>0, 为f(x)增区间 单
调区间
|f(x1)-f(x2)| ≤ |f(1) - f(0)
| = |1/3 - a^2| ≤ 1
解得 -2/√3 ≤ a ≤ 0 或 1 ≤ a
≤ 2/√3
若0≤a≤1, 令f'(x) = x^2 -
a^2 = 0 得 x=a时, 有最值
x<a时, f'(x) < 0 ; x>a时, f
'(x) > 0 , 则x=a为最小值,此
时
|f(x1)-f(x2)|≤1, 等价于
|f(0) - f(a)| ≤ 1, 且 |f(1) -
f(a)| ≤ 1
即
|-1/3*a^3 + a^3| ≤ 1, 且
|1/3 - a^2 - 1/3*a^3 +
a^3| ≤ 1
对于|-1/3*a^3 + a^3| ≤ 1,
0≤a≤1恒满足.
对于|1/3 - a^2 - 1/3*a^3 +
a^3| ≤ 1
整理得 |( 2a + 1 )( a - 1 )
^2| ≤ 3, 0≤a≤1恒满足.
综上, a∈[-2/√3 , 2/√3] 化
简即[-2√3/3，2√3/3]