一、分类讨论思想例1已知在区间[1,2]上函数f(x)=loga1a-2x+1恒为正.求实数a的取值范围.解在区间[1,2]上f(x)恒为正.(1)当a1时,1a-2x+11,即1a-2x0,1a-20,a21,与a1矛盾.(2)当0a1时,令g(x)=1a-2x+1,要使f(x)0,只要0g(x)1.①当a=21时,g(x)=1,此时f(x)=0,不能使f(x)恒为 First, the classification and discussion of thinking Example 1 is known in the interval [1,2] function f (x) = loga1a-2x + 1 is always positive. Find the range of the value of the real number a. Solution in the interval [1,2] f (x) Constant is positive. (1) When a1, 1a-2x+11, ie, 1a-2x0, 1a-20, a21, contradicts a1. (2) When 0a1, let g(x)=1a- 2x+1, to make f(x)0, as long as 0g(x)1.1 when a=21, g(x)=1, then f(x)=0, cannot make f(x) constant