Answers

-x^2+(a-2)x+a-4<y<x^2-(a-4)x+3
任意のxに対して-x^2+(a-2)x+a-4<x^2-(a-4)x+3を満たせばyも決まりAが成り立つ。
0<2x^2-(2a-6)x+7-a
0<x^2-(a-3)x+7-a/2
0<(x-(a-3/2))^2+7-a/2-(a-3/2)^2
0<(x-(a-3/2))^2+14-2a/4-(a^2-6a+9/4)
0<(x-(a-3/2))^2-a^2+4a+5/4
()^2≧0より-a^2+4a+5/4>0
a^2-4a-5<0 (a+1)(a-5)<0 -1<a<5

Post A Comment
Were you able to resolve your confusion?