发布时间：2022-05-06 点击量：1322次

独立思考:

若函数 $f%28x%29%3D%5Csin%20%5Cvarpi%20x$ (ω>0)在区间 $%5Cleft%5B%200%2C%5Cfrac%7B%5Cpi%20%7D%7B3%7D%20%5Cright%5D$ 上单调递增，在区间 $%5Cleft%5B%20%5Cfrac%7B%5Cpi%20%7D%7B3%7D%2C%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%5D$ 上单调递减，则ω=

A $%5Cfrac%7B2%7D%7B3%7D$ B $%5Cfrac%7B3%7D%7B2%7D$ C 2 D 3

动态解析: 点击查看

教师讲解: 点击查看

由- $%5Cfrac%7B%CF%80%7D%7B2%7D$ +2kπ≤ωx≤ $%5Cfrac%7B%CF%80%7D%7B2%7D$ +2kπ，k∈Z，ω>0得

- $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ ≤x≤ $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ ，k∈Z

所以函数 $f%28x%29%3D%5Csin%20%5Cvarpi%20x$ 的单增区间是[- $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ , $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ ]，k∈Z

同理可得f(x)的单减区间是[ $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ , $%5Cfrac%7B3%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ ]，k∈Z

又f(x)在区间 $%5Cleft%5B%200%2C%5Cfrac%7B%5Cpi%20%7D%7B3%7D%20%5Cright%5D$ 上单调递增

所以 $%5Cleft%5B%200%2C%5Cfrac%7B%5Cpi%20%7D%7B3%7D%20%5Cright%5D$ ⊆[- $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ , $%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D$ + $%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D$ ]

则 $%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D-%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D%2B%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D%5Cleqslant%200%5C%5C%5Cfrac%7B%CF%80%7D%7B2%CF%89%7D%2B%5Cfrac%7B2k%CF%80%7D%7B%CF%89%7D%5Cgeqslant%20%5Cfrac%7B%CF%80%7D%7B3%7D%5Cend%7Barray%7D%5Cright.$ ，解得 $%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7Dk%5Cleqslant%20%5Cfrac%7B1%7D%7B4%7D%5C%5C%CF%89%5Cleqslant%20%5Cfrac%7B3%7D%7B2%7D%2B6k%5Cend%7Barray%7D%5Cright.$ ，k∈Z