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

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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
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

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
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)=-0.5
最大值为sin(π/2)=1
故答案选D

根号根到哪里阿?