Question

向量a=(√3six,cosx),b=(cosx,cosx),f(x)=2a*b+2m-1,求f(x)的表达式最小正周期

Answer 1

解析：f(X)=2(√3sinxcosx+cos^2x)+2m-1=√3sin2x+2cos^2x-1+2m=√3sin2x+cos2x+2m

=2(sin2x*√3/2+cos2x*1/2)+2m=2sin(2x+π/6)+2m,

∴T=2π/2=π,

当x∈[0,π/2]时,π/6≤2x+π/6≤7π/6,

-1/2≤sin(2x+π/6)≤1,

∴f(x)min=2*(-1/2)+2m=2m-1=5,

m=3