Question

已知函数fx=ax^2-1(a,x属于R),设集合A=｛x/fx=x},集合B=｛x/f[f(x)] =x},且A=B不

Answer 1

重点化简集合B.f[f(x)] =a(ax^2-1)^2-1=x a(ax^2-1)^2=x+1

a(ax^2-1)^2-ax^2=x+1-ax^2

a(ax^2-1+x)(ax^2-1-x)+(ax^2-x-1)=0

(ax^2-x-1)(a^2*x^2+ax-a+1)=0

集合A.ax^2-x-1=0

A=B不等于空集,所以ax^2-x-1=0有解,且a^2*x^2+ax-a+1=0无解

1）a=0成立

2)a不等于0,则1+4a>=0,且a^2-4a^2(1-a)