设函数 f（x）=x²+bx+c,A=﹛x |f（x）=x﹜,B=﹛x |f（x-1）=x+1﹜,若A=﹛2﹜,求集合B
问题描述：
设函数 f（x）=x²+bx+c,A=﹛x |f（x）=x﹜,B=﹛x |f（x-1）=x+1﹜,若A=﹛2﹜,求集合B
答
A={xIf(x)=x},得到x^2+bx+c=x，所以x^2+(b-1)x+c=0，因为A=﹛2﹜，所以
2*2=-（b-1)，2+2=c解得c=4，b=-3；所以
B=﹛x |f（x-1）=x+1﹜得到（x-1)^2-3(x-1)+4=x+1，解得x=3±根号2，所以
集合B={3＋根号2,3-根号2}
答
B={2,5}
答
f(x)=x^2+bx+c
f(x)=x
=>x^2+(b-1)x+c = 0
A={2}
=> 4+2(b-1)+c =0
c = -2-2b ---(1)
for double roots(a)
=>△ =0
(b-1)^2-4c=0
(b-1)^2 +8(b+1) =0
b^2+6b+9=0
(b+3)^2=0
b= -3
c = 4
f(x)= x^2-3x+4
for B,f(x-1)=x+1
=> (x-1)^2 -3(x-1)+4 = x+1
x^2-6x+9=0
(x-3)^2=0
x=3
=> B ={3}