Question

若实数x,y,z满足x²+y²+z²=1,则xy+yz+zx的取值范围是

A.［-1,1］              B.［-,1］           C.［-1,］            D.［-,］

Answer 1

B

解:∵xy+yz+zx≤++=x²+y²+z²=1.

又∵2(xy+yz+zx)=(x+y+z)²-(x²+y²+z²)≥0-1=-1,

∴xy+yz+zx≥-,故选择B.