Question

若不论k为何值,直线y=k(x-2)+b与曲线x²-y²=1总有公共点,则b的取值范围是(    )

A.(-,)                                B.［-,］

C.(-2,2)                                        D.［-2,2］

Answer 1

答案：B直线过(2,b)点,当x=2时,y²=x²-1=3,∴y=±,故b∈［-,］.