数学天才请进!题六!解释越详细越好!

y2-y1=x^2-2x+1=(x-1)^2>=0
y2>=y1
2.
假设存在y3
由于y1和y2相交于(1,2)
故y2也要经过(1,2)
可设y3=a(x+5)(x-1)+2
a(x+5)(x-1)+2<=x^2+1
a(x+5)(x-1)+2>=2x
(a-1)x^2+4ax-5a+1<=0
a<1
(a-1)(-2a/(a-1))^2-4a*2a/(a-1)-5a+1<=0
4a^2-8a^2-5a(a-1)+(a-1)>=0
-4a^2-5a^2+5a+a-1=-9a^2+6a-1=-(3a-1)^2
-(3a-1)^2>=0
a=1/3
ax^2+(4a-2)x-5a+2>=0
x^2/3-2x/3-1/3>=0
x^2-2x-1>=0
此时必有两个根，既不恒大于等于0
不存在y3