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

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

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]2x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=0f(x)=2sin(x+π/3)+1f(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}3a[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

就是这个解释