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在高三总复习中接触一道试题:已知a为常数,设f(x)=lg[2/(1-x)+a]是奇函数,则使f(x)<0的x的取值范围是( ).(A)(-1,0)(B)(0,1)(C))(-∞,0)(D)(-∞,0)∪(1,+∞)这是一道常见的题目,主要考察函数奇偶性的定义以及对数不等式的运算,对于该题的解法,笔者同本组教师及部分学生作了一些交流,同时也查阅了一些数学辅导书上给出的答案,作为一道客观题,流行的解法都是这样的:
In a total review of the senior high school entrance examination contact with a test: known as a constant, let f (x) = lg [2 / (1-x) + a] is an odd function, The range is () (-) (- 1, 0) (B) (0,1) (C)) (- ∞, 0) (D) (- ∞, 0) ∪ (1, + ∞) A common topic, the main study of the definition of function parity and logarithmic inequality of the operation, the solution for the problem, the author made some exchanges with the teachers and some students, but also access to a number of mathematical counseling book given Answer, as an objective question, the popular solution is this: