Question

设函数f（x）=x平方+px+q,A={x|f(x)=x},B={x|f(x-1)=x+1},若A={2},求B

Answer 1

f(x)=x²+px+q=x

x²+(p-1)x+q=0

x1=x2=2

p=-3,q=4

f(x)=x²-3x+4;

f(x-1)=x+1

(x-1)²-3(x-1)+4=x+1

整理得：x²-6x+7=0;

x=3±√2

B={3+√2,3-√2}