急.已知集合A={(x,y)│x的平方+mx-y+2=0} ,B={(x,y)│x-y+1=0 ,0≤x≤2},已知集合A={(x,y)│x的平方+mx-y+2=0} ,B={(x,y)│x-y+1=0 ,0≤x≤2},如果A∩B≠空集,求实数m的取值范围!

Question

急.已知集合A={(x,y)│x的平方+mx-y+2=0} ,B={(x,y)│x-y+1=0 ,0≤x≤2},
已知集合A={(x,y)│x的平方+mx-y+2=0} ,B={(x,y)│x-y+1=0 ,0≤x≤2},如果A∩B≠空集,求实数m的取值范围!

Answer 1

x^2+mx-y+2=0与x-y+1=0 ,0≤x≤2
联立得：x^2+(m-1)x+1=0
要使A∩B≠空集,则方程x^2+(m-1)x+1=0在[0,2]有解；
令f(x)=x^2+(m-1)x+1
当f(x)在[0,2]上与x轴有交点时,则
f(0)*f(2)≤0 或△=(m-1)^2-4≥0
0≤-1/[2(m-1)]≤2
f(0)≥0且f(2)≥0
解得：m≤-2/3 或-2/3≤m≤3/4
所以实数m的取值范围是 (-∞,3/4]