题文
(本小题满分10分)已知函数
解析

解得0 ≤ x ≤ 1;·········································································· 6分
(3) 当0≤ x

≤ 1时,g(x) -f (x) =

-log 2(x+1)

,
令

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

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

时,x=

∈[0,1],
∴当x=

时,[g(x)-f (x)]max=

.········································· 10分
考点
据考高分专家说,试题“(本小题满分10分) 已知函数,当点(x.....”主要考查你对 [对数函数的解析式及定义(定义域、值域) ]考点的理解。 对数函数的解析式及定义(定义域、值域)对数函数的定义:
一般地,我们把函数y=logax(a>0,且a≠1)叫做对数函数,其中x是自变量,函数的定义域是(0,+∞),值域是R。
对数函数的解析式:
y=logax(a>0,且a≠1)
在解有关对数函数的解析式时注意:
在涉及到对数函数时,一定要注意定义域,即满足真数大于零;求值域时,还要考虑底数的取值范围。


