Question

已知向量a=(2cosx,1),b=(根号2（x+π/4),m)令f(x)=a点积b

Answer 1

f(x)=a点积b=2根号2sin（x+π/4)cosx+m=2根号2(sinxcosπ/4+cosxsinπ/4)cosx+m=sin 2x+2cos^2x+m=sin2x+cos2x+1+m

所以若x属于R,求f(x)最小正周期T=π

df=2cos2x-2sin2x,在[0,π/4],df(π/8)=0,f(π/8)=根号2+1+m

f(0)=2+m=f(π/4)

f(x)在[0,π/4]上最大值与最小值之和为6,即为根号2+1+m+2+m=6,m=(3-根号2)/2