帮忙解道题，谢谢

1、 x^2-(a+2)x+a+1=0
(x-1)[x-(a+1)]=0
x=1或x=a+1
A={1,a+1}
1+a+1=a+2

2、 f(x) =-[x^2-2(a-1)+(a-1)^2]+(a-1)^2+1
=[x-(a-1)]^2+(a-1)^2+1
x∈[a-2,a^2] (x-a+1)∈[-1,a^2-a+1]
a^2-a+1=(a-1/2)^2+3/4>=3/4

若 a^2-a+1∈[3/4,1] a∈[0,1]
则[x-(a-1)]^2∈[0,1]
f(x)max=1+(a-1)^2+1=(a-1)^2+2 此时 x=a

若a^2-a+1>1 a>1或a<0
则[x-(a-1)]^2∈[0,（a^2-a+1)^2]
f(x)max=（a^2-a+1)^2+(a-1)^2+1=a^4+2a^3+3
此时x-a+1=a^2-a+1 x=a^2

1.x^2-(a+2)x+a+1=(x-1)(x-a-1)=0
解为x=1 或 x=a+1 此即A的唯一两个元素 和为 a+2

2.f(x)=-x^2+2(a-1)x+1=-x^2+2(a-1)x-(a-1)^2+(a-1)^2+1
=-(x-a+1)^2+(a-1)^2+1 又x∈[a-2,a^2] 又a-2<a-1<a^2恒成立
所以 x=a-1时 有最大值 a-1)^2+1