Question

已知集合A={(x,y)|x2+mx－y+2=0},B={(x,y)|x－y+1=0,且0≤x≤2},如果A∩B≠ ,求实数m的取值范围.

A∩B≠空集

Answer 1

A为抛物线，B为直线，交集不为空表明两者有交点
A: y=x^2+mx+2
B: y=x+1
即方程x^2+mx+2=x+1有实数根
mx=-x^2+x-1
m=-x-1/x+1
0<=x<=2,则有2<=x+1/x<=5/2
即有-5/2+1<=-(x+1/x)+1<=-2+1
即有-3/2<=m<=-1

Answer 2

A为抛物线，B为直线，交集不为空表明两者有交点
A: y=x^2+mx+2
B: y=x+1
即方程x^2+mx+2=x+1有实数根
x^2+(m-1)x+1=0
delta=m^2-2m+1-4=(m-3)(m+1)>=0
得： m>=3 或m<=-1

Answer 3

抛物线，B是一条直线，路口之间的交叉
答：Y = X ^ 2 + MX +2

B：Y = X +1

即不为空方程x ^ 2 + MX +2 = X +1实根

MX =-X ^ 2 + X-1

=-X-1 / X +1

0 <= X <= 2时，2 <= 1 / X的<= 5/2

即-5 / 2，1 <= - （1 / x）的1 <= -2 1
-3 / 2 <= M <= -1即