已知向量a=(coswx-sinwx,sinwx),b=(-coswx-sinwx,2√3coswx).设函数f(x)=
1个回答

a·b=-(coswx-sinwx)(coswx+sinwx)+√3sin(2wx)

=√3sin(2wx)-cos(2wx)

=2sin(2wx-π/6)

故:f(x)=2sin(2wx-π/6)+λ

关于x=π对称,即:2wπ-π/6=kπ+π/2,k∈Z

即:2w=k+1/2+1/6=k+2/3

即:w=k/2+1/3,k∈Z

w∈(1/2,1),当k=1时,w=5/6满足条件

1

故:f(x)=2sin(5x/3-π/6)+λ

最小正周期:2π/(5/3)=6π/5

2

函数点(π/4,0),即:2sin(5π/12-π/6)+λ

=2sin(π/4)+λ=√2+λ=0

即:λ=-√2

即:f(x)=2sin(5x/3-π/6)-√2

x∈[0,3π/5],故:5x/3-π/6∈[-π/6,5π/6]

故:sin(5x/3-π/6)∈[-1/2,1]

故:2sin(5x/3-π/6)-√2∈[-1-√2,2-√2]

以上回答你满意么?