Question

已知向量a=(sinx,-cosx),b=(cosx,根号3cosx)，函数f(x)=a*b+根号3/2

1.求f(x)的最小正周期，并求其图像对称中心的坐标

2.当0<=x<=π/2时，求函数f(x)的值域

Answer 1

1.f(x)=sinxcosx-√3cos²x+√3/2=1/2sin2x-√3/2cos2x-√3/2+√3/2=sin(2x-π/3)
最小正周期为π，对称中心坐标为2x-π/3=kπ,x=kπ/2+π/6,即(kπ/2+π/6,0)
2.f(x)单调递增区间为：-π/2+2kπ≤2x-π/3≤π/2+2kπ,即-π/12+kπ≤x≤5π/12+kπ
单调递减区间为：5π/12+kπ≤x≤11π/12+kπ
[0,π/2]分为两个区间：递增区间[0,5π/12],递减区间(5π/12,π/2]
f(0)=-√3/2,f(5π/12)=1,f(π/2)=√3/2
f(x)在[0,π/2]值域为[-√3/2,1]

Answer 2

1. f(x)=sinxcosx-√3cos^2x+√3/2
=1/2sin2x-√3/2cos2x
=sin(2x-π/3)
f(x)的最小正周期T=π
对称中心
2x-π/3=kπ x=kπ/2+π/6
图像对称中心的坐标 (kπ/2+π/6，0) k∈Z
2. 0<=x<=π/2
-π/3<=2x-π/3<=2π/3
-1/2<=sin(2x-π/3)<=1
当0<=x<=π/2时，函数f(x)的值域 [-1/2,1]

Answer 3

1. f(x)=sinxcosx-√3cos^2x+√3/2
=1/2sin2x-√3/2cos2x
=sin(2x-π/3)
f(x)的最小正周期T=π
对称中心
2x-π/3=kπ x=kπ/2+π/6
图像对称中心的坐标 (kπ/2+π/6，0) k∈Z
2. 0<=x<=π/2
-π/3<=2x-π/3<=2π/3
-√3/2<=sin(2x-π/3)<=1
当0<=x<=π/2时，函数f(x)的值域 [-√3/2,1]

Answer 4

1
f(x)=sinxcosx-√3cos²x+√3/2
=(1/2)sin2x-(√3/2)cos2x-√3/2+√3/2
=sin(2x-π/3)
T=2π/2=π
对称中心2x-π/3=kπ+π/2
x=kπ/2+5π/12
2
0≤x≤π/2
-π/3≤2x-π/3-≤2π/3
f(x)min=-√3/2
f(x)max=1

Answer 5

已知向量a=(sinx,-cosx)，向量b=(cosx,(√3)cosx),函数f(x)=a•b+((√3)/23)/2;(1)求函数的最小正周期，并求起图像对称中心的坐标;(2)当0≤x≤π/2时，求函数的值域.
解:(1)
f(x)=a•b=sinxcosx-(√3)cos²x+(√3)/2=(1/2)sin2x-(√3)(1+cos2x)/2+(√3)/2
=(1/2)sin2x-(√3/2)cos2x=sin2xcos(π/3)-cos2xsin(π/3)=sin(2x-π/3)
故最小正周期T=π;对称中心:令2x-π/3=kπ，x=π/6+kπ/2=π/6+kπ/2，故全部对称中心的的坐标为(π/6+kπ/2，0)。
(2).当0≤x≤π/2时，minf(x)=f(0)=sin(-π/3)=;maxf(x)=f(5π/12)=sin(5π/6-π/3)=sin(π/2)=1
即在0≤x≤π/2时f(x)的值域为[-√3/2，1]。