题目

已知w>0,a向量=(2sinwx+coswx,2sinwx-coswx)b向量=(sinwx,coswx),f(x)=a向量*b向量,
f(x)=a向量*b向量,f(x)图像上相邻的两条对称轴的距离为π/2.
求w的值
求函数f(x)在[0,π/2]上的单调区间及最值

提问时间：2020-12-17

答案

1
f(x)=a·b=(2sin(wx)+cos(wx),2sin(wx)-cos(wx))·(sin(wx),cos(wx))
=(2sin(wx)+cos(wx))sin(wx)+(2sin(wx)-cos(wx))cos(wx)
=2sin(wx)^2+sin(wx)cos(wx)+2sin(wx)cos(wx)-cos(wx)^2
=3sin(2wx)/2+(1-cos(2wx))-(1+cos(2wx))/2
=3sin(2wx)/2-3cos(2wx)/2+1/2
=(3√2/2)sin(2wx-π/4)+1/2
f(x)图像相邻的两条对称轴的距离为π/2
即f(x)的最小正周期：T=π
即：2π/(2w)=π
即：w=1
即：f(x)=(3√2/2)sin(2x-π/4)+1/2
2
增区间：2x-π/4∈[2kπ-π/2,2kπ+π/2]
即：x∈[kπ-π/8,kπ+3π/8],k∈Z
减区间：2x-π/4∈[2kπ+π/2,2kπ+3π/2]
即：x∈[kπ+3π/8,kπ+7π/8],k∈Z
x∈[0,π/2],故当k=0时,增区间：x∈[0,3π/8]
减区间：x∈[3π/8,π/2]
x∈[0,π/2],2x-π/4∈[-π/4,3π/4]
sin(2x-π/4)∈[-√2/2,1]
故：(3√2/2)sin(2x-π/4)+1/2∈[-1,(3√2+1)/2]
即：fmax=(3√2+1)/2
fmin=-1