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一、选择题1.(卷Ⅰ,理1)函数y=x(x-1)~(1/2)+x~(1/2)的定义域为().A.{x|x≥0} B.{x|x≥1}C.{x|x≥1}U{0} D.{x|0≤x≤1}解答途径:∵x=0符合题意要求,排除B;∵x=1/2不符合题意要求,排除 A、D,故选 C.解题感悟:本题主要通过求函数定义域的形式,考查简单不等式的解法以及集合的交集、并集运算.解答途径抓住四个选择支的结构特征,取了两个特殊值否定了三个选择支,从而确定了正确的选择支,这
1. Multiple Choice 1. (Volume I, Theory 1) The domain of the function y=x(x-1)~(1/2)+x~(1/2) is ().A.{x|x≥ 0} B.{x|x≥1}C.{x|x≥1}U{0} D.{x|0≤x≤1} Solution route: ∵x=0 meets the question intention requirement, excludes B; ∵x=1/2 does not meet the requirements of the question, exclude A, D, so choose C. problem solving sentiment: This question is mainly through the form of the function definition domain, to examine the solution of the simple inequality and the set of intersection, union operation. The way to seize the structural characteristics of the four alternative branches, taking two special values negating the three alternative branches, thus determining the correct selection branch,