Question

向量a=(√3sinx,cosx),向量b=(cosx,cosx)

1·向量a*向量b=1 且x∈【-π/4，π/4】求x的值
2·设f（x）=向量a*向量b，求f（x）的周期及单调减区间

Answer 1

解：(1)
∵a•b=√3sinx cosx+ cos^2x
=(√3/2)sin2x+[(2 cos^2x-1)+1]/2
=(√3/2)sin2x+(1/2)cos2x+(1/2)
= sin[2x+(π/6)]+ (1/2)
=1
则，sin[2x+(π/6)]= 1/2=sin(π/6)
又∵x∈[-π/4，π/4]
∴2x+(π/6)= π/6
∴x=0
（2)
又(1)得，f（x）=a•b=sin[2x+(π/6)]+ (1/2)
则，最小正周期T =π
令（π/2）《2x+(π/6)《(3π/2)
解得，(π/6)《x《(2π/2)
所以f(x)在[kπ+π/6,kπ+2π/3]（k∈Z）上单调递减

Answer 2

向量a*向量b
=√3sinxcosx+cosxcosx
=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2

1 向量a*向量b=1 所以sin(2x+π/6)=1/2 x=0
2 f（x）=sin(2x+π/6)+1/2
周期为π
单调减区间为kπ+π/6,kπ+2π/3

Answer 3

1.
向量a*向量b=√3sinxcosx+(cosx)^2
=(√3/2)sin2x+(1/2)cos2x+(1/2)
=sin(2x+π/6)+(1/2)

sin(2x+π/6)+(1/2)=1
sin(2x+π/6)=(1/2)
而：-π/4<=x<=π/4
-π/3<=2x+π/6<=2π/3

所以：2x+π/6=π/6
x=0

2.
f(x)=sin(2x+π/6)+(1/2)
最小正周期:2π/2=π

单调减区间:
2kπ+(π/2)<=2x+π/6<=2kπ+(3π/2)
kπ+(π/3)<=x<=kπ+(2π/3) 其中：k为整数
即：[kπ+(π/3),kπ+(2π/3)]