题目

已知函数y=g(x)与f(x)=loga(x+1)(a>1)的图象关于原点对称(括号内为真数)
(1)写出y=g(x)的解析式
(2)若函数F(X)=F()x)+G(X)+M为奇函数,试确定实数M的值
(3)当x∈[0,1)时,总有f(x)+g(x)≥n成立,求实数n的取值范围

提问时间：2020-10-09

答案

(1)
图象关于原点对称,则g(x) + f(-x) = 0
g(x) = -f(-x) = -loga(1 - x)
(2)
F(x) = f(x) + g(x) = loga(x + 1) - loga(1 - x) + m = loga[(1+ x)/(1 - x)] + m
F(x)为奇函数,则F(-x) = -F(-x)
loga[(1- x)/(1 + x)] + m = -loga[(1+ x)/(1 - x)] - m
2m = -loga[(1+ x)/(1 - x)] - loga[(1- x)/(1 + x)]
= loga[(1 - x)/(1 + x)] - loga[(1- x)/(1 + x)]
= 0
m = 0
(3)
令G(x) = f(x) + g(x) = loga[(1+ x)/(1 - x)]
G'(x) =[(1- x)/(1 + x)]lna
a > 1,lna > 0
x∈[0,1)时:1-x > 0,1 + x > 0,G'(x) > 0
[0,1)上的最小值为G(0) = 0
n = 0