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}