1.ω>0,若函數(shù)F(x)=2sinωx在[-π/3,π/4]上單調(diào)遞增,求ω的范圍
1.ω>0,若函數(shù)F(x)=2sinωx在[-π/3,π/4]上單調(diào)遞增,求ω的范圍
2.已知函數(shù)y=tanωx在 (﹣π/2,π/2)內(nèi)是減函數(shù),則ω的取值范圍 兩道題可不可以用同一種解法?
優(yōu)質(zhì)解答
1、F(x)=2sinwx在[-π/3,π/4]內(nèi)遞增,則只要函數(shù)F(x)的半個周期大于等于π/3即可,得:
T=2π/w,(1/2)T≥π/3
得:w≤3
則:0要使得函數(shù)y=sin(wx)在[-π/3,π/4]上遞增,則必須在[-π/3,π/3]上遞增,考慮到這個函數(shù)在半個周期內(nèi)是遞增的,則這半個周期必須大于等于2π/3即:(1/2)T≥2π/3π/w≥2π/3w≤3/2則:0