Question

已知函数f(x)=sin(wx+π/3）（w>0),，f(π/6）=f(π/3），且f(x)在区间（π/6，π/3）有最小值，无最大值

求w=

Answer 1

已知函数f(x)=sin(ωx+π/3）（ω>0),，f(π/6）=f(π/3），且f(x)在区间（π/6，π/3）有最小值，无最大值，求ω=?
解：(1/2)(π/6+π/3)=π/4，∵f(π/6）=f(π/3），∴x=π/4是其对称轴，又因为f(x)有最小值-1，
故有f(π/4)=sin(ωπ/4+π/3)=-1，考虑到ω>0，故有ωπ/4+π/3=3π/2，ωπ/4=3π/2-π/3=7π/6，
∴ω=14/3。
事实上，f(π/6)=sin[(14/3)×(π/6)+π/3)]=sin(10π/9)=sin(π+π/9)=-sin(π/9)
f(π/3)=sin[(14/3)×(π/3)+π/3)]=sin(17π/9)=sin(2π-π/9)=-sin(π/9)=f(π/6)
f(π/4)=sin[(14/3)×(π/4)+π/3]=sin(9π/6)=sin(π+π/2)=-sin(π/2)=-1.
π/6<π/4<π/3.即在区间(π/6，π/3)上确有最小值-1，但无最大值(因为是开区间).

Answer 2

f(x)在区间（π/6，π/3）有最小值，无最大值
说明该区间小于一个周期
所以w*π/6+π/3=π-(w*π/6+π/3)
w=1