a·b=-(coswx-sinwx)(coswx+sinwx)+√3sin(2wx)
=√3sin(2wx)-cos(2wx)
=2sin(2wx-π/6)
故：f(x)=2sin(2wx-π/6)+λ
關(guān)于x=π對(duì)稱,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),當(dāng)k=1時(shí),w=5/6滿足條件
1
故：f(x)=2sin(5x/3-π/6)+λ
最小正周期：2π/(5/3)=6π/5
2
函數(shù)點(diǎn)(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]
以上回答你滿意么?