
解得0 ≤
x ≤ 1;·········································································· 6分
(3) 当0≤
x
≤ 1时,
g(
x) -
f (
x) =

-log
2(
x+1)

,
令

,
则
kx2 + (2
k-3)
x + (
k-1) = 0,························································· 7分
∵
x的取值存在,∴D = (2
k-3)
2-4
k(
k-1) ≥ 0,
解得:

,·········································································· 9分
当
k=

时,
x=

∈[0,1],
∴当
x=

时,[
g(
x)-
f (
x)]
max=

.········································· 10分