已知f(x)=sin²wx+2倍根号3 sin(wx+π/4)cos(wx-π/4)-cos²wx-根号3,w>0(1)若函数f(x+π/6w)在(-π/3,2π/3)上是单调递增增函数,求w的取值范围(2)若函数f(x)图象的一个对称中心到相邻的对称轴距离为π/4,求f
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![已知f(x)=sin²wx+2倍根号3 sin(wx+π/4)cos(wx-π/4)-cos²wx-根号3,w>0(1)若函数f(x+π/6w)在(-π/3,2π/3)上是单调递增增函数,求w的取值范围(2)若函数f(x)图象的一个对称中心到相邻的对称轴距离为π/4,求f](/uploads/image/z/572339-11-9.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dsin%26%23178%3Bwx%2B2%E5%80%8D%E6%A0%B9%E5%8F%B73+sin%28wx%2B%CF%80%2F4%29cos%28wx-%CF%80%2F4%29-cos%26%23178%3Bwx-%E6%A0%B9%E5%8F%B73%2Cw%3E0%281%29%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%2B%CF%80%2F6w%29%E5%9C%A8%28-%CF%80%2F3%2C2%CF%80%2F3%29%E4%B8%8A%E6%98%AF%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E6%B1%82w%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%282%29%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%E5%9B%BE%E8%B1%A1%E7%9A%84%E4%B8%80%E4%B8%AA%E5%AF%B9%E7%A7%B0%E4%B8%AD%E5%BF%83%E5%88%B0%E7%9B%B8%E9%82%BB%E7%9A%84%E5%AF%B9%E7%A7%B0%E8%BD%B4%E8%B7%9D%E7%A6%BB%E4%B8%BA%CF%80%2F4%2C%E6%B1%82f)
已知f(x)=sin²wx+2倍根号3 sin(wx+π/4)cos(wx-π/4)-cos²wx-根号3,w>0(1)若函数f(x+π/6w)在(-π/3,2π/3)上是单调递增增函数,求w的取值范围(2)若函数f(x)图象的一个对称中心到相邻的对称轴距离为π/4,求f
已知f(x)=sin²wx+2倍根号3 sin(wx+π/4)cos(wx-π/4)-cos²wx-根号3,w>0
(1)若函数f(x+π/6w)在(-π/3,2π/3)上是单调递增增函数,求w的取值范围
(2)若函数f(x)图象的一个对称中心到相邻的对称轴距离为π/4,求f(x)在[-π/12,25/36π]上的最大值和最小值,并指出相应的x的值
已知f(x)=sin²wx+2倍根号3 sin(wx+π/4)cos(wx-π/4)-cos²wx-根号3,w>0(1)若函数f(x+π/6w)在(-π/3,2π/3)上是单调递增增函数,求w的取值范围(2)若函数f(x)图象的一个对称中心到相邻的对称轴距离为π/4,求f
(1) f(x)=sin²ωx+2√3 sin(ωx+π/4)cos(ωx-π/4)-cos²ωx-√3
=2√3 ·√2/2(sinωx+cosωx)·√2/2(sinωx+cosωx)-(cos²ωx-sin²ωx)-√3
=√3(1+sin2ωx)-cos2ωx-√3
=√3sin2ωx-cos2ωx
=2sin(2ωx-π/6)
f[x+π/(6ω)]=2sin{2ω[x+π/(6ω)]-π/6}=2sin(2ωx+π/6)
一个单调增区间满足 -π/2≤2ωx+π/6≦π/2
即 -π/(3ω)≤x≦π/(6ω)
所以 -π/(3ω)≦-π/3 且 π/(6ω)≧2π/3
解得 0<ω≦1/4
(2)设f(x)最小正周期为T,由题意,T/4=π/4, T=π
所以 2π/2ω=π, ω=1
所以 f(x)=2sin(2x-π/6)
x∈[-π/12,25π/36]时,2x-π/6∈[-π/3,11π/9]
而 11π/9<4π/3
所以 f(x)∈[-√3,2]
当 2x-π/6=-π/3 即 x=-π/12 时,f(x)有最小值-√3;
当 2x-π/6=π/2 即 x=π/3 时,f(x)有最大值2.
(1) f(x)=sin²ωx+2√3 sin(ωx+π/4)cos(ωx-π/4)-cos²ωx-√3
=2√3 ·√2/2(sinωx+cosωx)·√2/2(sinωx+cosωx)-(cos²ωx-sin²ωx)-√3
=√3(1+sin2ωx)-cos2ωx-√3
=√3sin2ωx-cos2ωx
=...
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(1) f(x)=sin²ωx+2√3 sin(ωx+π/4)cos(ωx-π/4)-cos²ωx-√3
=2√3 ·√2/2(sinωx+cosωx)·√2/2(sinωx+cosωx)-(cos²ωx-sin²ωx)-√3
=√3(1+sin2ωx)-cos2ωx-√3
=√3sin2ωx-cos2ωx
=2sin(2ωx-π/6)
f[x+π/(6ω)]=2sin{2ω[x+π/(6ω)]-π/6}=2sin(2ωx+π/6)
一个单调增区间满足 -π/2≤2ωx+π/6≦π/2
即 -π/(3ω)≤x≦π/(6ω)
所以 -π/(3ω)≦-π/3 且 π/(6ω)≧2π/3
解得 0<ω≦1/4
(2)设f(x)最小正周期为T,由题意,T/4=π/4, T=π
所以 2π/2ω=π, ω=1
所以 f(x)=2sin(2x-π/6)
x∈[-π/12,25π/36]时,2x-π/6∈[-π/3,11π/9]
而 11π/9<4π/3
所以 f(x)∈[-√3,2]
当 2x-π/6=-π/3 即 x=-π/12 时,f(x)有最小值-√3;
当 2x-π/6=π/2 即 x=π/3 时,f(x)有最大值2.
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