2001年全国高中数学联赛第11题是这样的:函数 的值域是_. 本题解法较多,在不同的数学思想指导下,可以得到不同的解题思路,现归纳如下. 一、运用方程的思想,可以得到如下的解法 解法1(反函数法)由 得 由(2)得x=(2-y2)/(3-2y),代入(1)得y≥(2-y2)/(3-2y),∴1≤y≤3/2,或y≥2,故函数的值域是{y|1≤y≤3,或y≥2}. The 11th question of the National High School Mathematics League in 2001 was as follows: The range of the function is _. This problem has many solutions. Under the guidance of different mathematical thoughts, different ideas can be obtained. These are summarized as follows. First, the use of equations The idea can be obtained as follows: Solution 1 (inverse function) From (2) get x=(2-y2)/(3-2y), Substitute (1) get y≥(2-y2)/(3 -2y), ∴ 1 ≤ y ≤ 3/2, or y ≥ 2, so the range of the function is {y|1 ≤ y ≤ 3, or y ≥ 2}.