设函数f(x)=sin(2x+μ)(-π
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设函数f(x)=sin(2x+μ)(-π设函数f(x)=sin(2x+μ)(-π设函数f(x)=sin(2x+μ)(-π解析:已知函数f(x)图像的一条对称轴是直线x=π/8,则可得:当x=π/8时,
设函数f(x)=sin(2x+μ)(-π
设函数f(x)=sin(2x+μ)(-π
设函数f(x)=sin(2x+μ)(-π
解析:
已知函数f(x)图像的一条对称轴是直线x=π/8,则可得:
当x=π/8时,函数f(x)取得最值
即有:2×(π/8)+φ=kπ+π/2,k∈Z
所以:φ=kπ+π/4
又-π<φ<0,则符合题意的φ=-3π/4,此时k=-1
所以函数:f(x)=sin(2x-3π/4)
易知当2kπ-π/2≤2x-3π/4≤2kπ+π/2即kπ-π/8≤x≤kπ+7π/8,k∈Z时,函数f(x)是增函数
所以函数y=f(x)的单调递增区间为:[kπ-π/8,kπ+7π/8],k∈Z
φ=-3π/4
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