Question

设w＞0为常数,函数y=sinwX在[-π/3,π/4]上单调递增,则实数W的取值范围是

Answer 1

-π/2+2kπ≤wx≤π/2+2kπ(k为整数) (y=sinwx单调递增时wx的范围)-π/2w+2kπ/w≤x≤π/2w+2kπ/w (函数递增时x的范围)∵y=sinwx在[-π/3,π/4]上单调递增∴-π/3≥-π/2w+2kπ/w且π/4≤π/2w+2kπ/w ([-π/3,π/4]为以上求得x范围的子区间)即w≤3/2-6k且w≤2+8k(k为整数)又∵w>0∴k=0 (确保w有解)∴0