Question

已知函数f(x)=loga(2/(1-x)+b),(a>0且a≠1),f(x)为奇函数,求f(x)的表达式

Answer 1

f(0) = 0 -> f(0) = loga(2 + b) = 0 -> 2 + b = 1 -> b = -1

f(x) = loga [(1 + x)/(1 - x)] (-1 < x 1 时,f(x)为单调增函数

f(x² - 2x + 1) + f(2x - 2) > 0 -> f(x² - 2x + 1) > f(2 - 2x) -> x² - 2x + 1 > 2 - 2x

-> x > 1 或 x < -1

综上

1 < x < 3/2