(1)
△ = [3(m - 4)]² - 4m×(-9)
= 9(m² - 8m + 16) + 36m
= 9(m² - 4m + 16)
= 9(m - 2)² + 108
> 0
所以f（x）有两个零点
(2)
设两个零点分别是 x1 和 x2 ,则
x1 + x2 = 3(4 - m)/m
x1 * x2 = -9/m
(x1 - x2)²
= (x1 + x2)² - 4x1*x2
= 9(4 - m)²/m² + 36/m
= 9(16/m² - 4/m + 1)
= 9(16/m² - 4/m + 1/4) + 27/4
= 9(4/m - 1/2)² + 27/4
当 4/m = 1/2 ,即 m = 8 时 ,最小值是 27/4
两个零点的距离的最小值
= |x1 - x2|
= √(x1 - x2)²
= √(27/4)
= 3√3/2
(3)
m = 1 时,
f(x)
= x² - 9x - 9
= (x - 9/2)² - 117/4
因为 x∈[0 ,2]
所以当 x = 2 时 ,
f(x)取 最小值 -23
f(x) - a > 0
a < f(x)
所以 a < f(x)的最小值
所以 a < -23