w为正实数,f(x)=(1/2)sin(wx/2)cos(wx/2)在[-π/3,π/4]上为增函数,则w的取值范围?
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w为正实数,f(x)=(1/2)sin(wx/2)cos(wx/2)在[-π/3,π/4]上为增函数,则w的取值范围?
w为正实数,f(x)=(1/2)sin(wx/2)cos(wx/2)在[-π/3,π/4]上为增函数,则w的取值范围?
w为正实数,f(x)=(1/2)sin(wx/2)cos(wx/2)在[-π/3,π/4]上为增函数,则w的取值范围?
f(x)=(1/2)sin(wx/2)cos(wx/2)=(1/4)sin(wx)
函数f(x)的增函数区间为wx∈[-π/2+2kπ,π/2+2kπ]
即-π/2+2kπ≤wx≤π/2+2kπ,(k∈Z)
∵w>0,即有 (-π/2+2kπ)/w≤x≤(π/2+2kπ)/w
而由题意,f(x)在[-π/3,π/4]上为增函数
∴有 (-π/2+2kπ)/w≤-π/3≤x≤π/4≤(π/2+2kπ)/w
解两端不等式,分别得:0
即k∈(-1,1) ∴k=0
即不等式的解为0
有原题得:f(x)=(1/4)sin(wx)
则该函数周期T=(2*π)/w........................................(1)
又因函数在[-π/3,π/4]上为增函数,所以w>0.
周期T>2*[(π/4)-(-π/3)]=(7*π)/12..........................(2)
且w*(-π/3)≥(-π/2...
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有原题得:f(x)=(1/4)sin(wx)
则该函数周期T=(2*π)/w........................................(1)
又因函数在[-π/3,π/4]上为增函数,所以w>0.
周期T>2*[(π/4)-(-π/3)]=(7*π)/12..........................(2)
且w*(-π/3)≥(-π/2)................................................(3)
w*(π/4)≤(π/2)......................................................(4)
将(1)带入(2),解得:w<12/7................................(5)
联立(3)(4)(5)得:w∈(0,1.5]。
收起
f(x)=(1/2)sin(wx/2)cos(wx/2)=(1/4)*[2sin(wx/2)cos(wx/2)]=(1/4)sin(wx)
后面的会做了吧!