ω是正实数,函数f(x)=2sinωx在[-π/3,π/4]上递增,那么ω的范围是A(0,3/2)B(0,2] C(0,24/7]D([2,+∞)
问题描述:
ω是正实数,函数f(x)=2sinωx在[-π/3,π/4]上递增,那么ω的范围是
A(0,3/2)
B(0,2]
C(0,24/7]
D([2,+∞)
答
A
x在[-π/3,π/4]上递增,则ωx在[-ωπ/3,ωπ/4]上递增,又ω是正实数(ω>0),则
[-ωπ/3,ωπ/4]包含于[-π/2,π/2],于是有-ωπ/3>-π/2,ωπ/4
答
用w代替
sinx在(-π/2,π/2)递增
所以-π/2