Question

已知向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1),f(x）=向量a•向量b+m

1、求f(x)在[0,2派]上的单调区间

2、当x属于[0，派\2]时，f（x）的最小值为2，求f（x）大于等于2成立的x的取值集合
3、若存在实数a，b，c，使得a[f（x）-m]+b[f（x-c）-m]=1，对任意x属于R恒成立，求bcosc\a的值

Answer 1

向量a=(2sinx\2,根号下3+1),向量b=(cosx\2-根号下3sinx\2,1)
f(x)=a●b+m
=2sinx/2(cosx/2-√3sinx/2)+√3+1+m
=2sinx/2cox/2-2√3sin²x/2+√3+1+m
=sinx-√3(1-cosx)+√3+1+m
=2(1/2sinx+√3/2cosx)+1+m
=2sin(x+π/3)+1+m
∵x∈[0,2π]
∴x+π/3∈[π/3,7π/3]
∴x+π/3∈[π/3,π/2]或x+π/3∈[3π/2,7π/3],
即x∈[0,π/6],或x∈[7π/6,2π]
函数f(x)递增
当x+π/3∈[π/2,3π/2]，即x∈[π/6,7π/6]
函数递减
∴f(x)在[0,2派]上的单调递增区间为
[0,π/6],[7π/6,2π]
单调递增区间为[π/6,7π/6]
2
x属于[0，派\2]时，x+π/3∈[π/3,5π/6]
∴x+π/3=5π/6时，f(x)min=2+m=2
∴m=0
f(x)=2sin(x+π/3)+1
f(x)≥2即 sin(x+π/3)≥1/2
2kπ+π/6≤x+π/3≤ 2kπ+5π/6
2kπ-π/6≤x≤ 2kπ+π/2
∴f(x)≥2成立的x的取值集合
为｛x| 2kπ-π/6≤x≤ 2kπ+π/2,k∈Z}
3
a[f（x）-m]+b[f（x-c）-m]=1
即a[2sin(x+π/3)+1]+b[2sin(x-c+π/3)+1]=1
2[asin(x+π/3)+bsin(x+π/3-c)]+a+b=1对任意x属于R恒成立
x=-π/3时， 2bsin(-c)+a+b=1
x=2π/3时，2bsinc+a+b=1
x=π/6时，2a+2bcosc+a+b=1
∴a+b=1,sinc=0, cosc=±1
若cosc=1,那么a+b=0矛盾
∴cosc=-1,2a-2b=0,a=b
∴a=b=1/2,c=2kπ+π,k∈Z
此时2[asin(x+π/3)+bsin(x+π/3-c)]+a+b
=sin(x+π/3)+sin(x+π/3-2kπ-π)+a+b
=sin(x+π/3)-sin(x+π/3)+a+b
=a+b=1恒成立
∴ bcos(c/a)=1/2cos(4kπ+2π)=1/2