函数y=根号3sinxcosx+cos^2x-1/2在[0,π/2]的值域是?
问题描述:
函数y=根号3sinxcosx+cos^2x-1/2在[0,π/2]的值域是?
A.[-1,1]
B.[1/2,1]
C.[0,1]
D.[-1/2,1]
根号3
答
y=√3sinxcosx+cos^2x-1/2 =(√3/2)sin2x+(1/2)(cos2x+1)-1/2 =(√3/2)sin2x+(1/2)cos2x =sin(2x+π/6)因为x属于[0,π/2],所以2x+π/6属于[π/6,7π/6] 所以最小值为sin(7π/6)=-0....