Question

高一数学集合方面的两个题,1.函数f(x)=mx/4x-3(x不等于3/4)在定义域内恒有f{f(x)}=x 则m=2.

Answer 1

1.f[f(x)]=m[mx/(4x-3)]/{4[mx/(4x-3)]-3}

=m²x/[4mx-3(4x-3)]

=m²x/(4mx-12x+9)=x

m²x=(4m-12)x²+9x

4(m-3)x=m²-9=(m-3)(m+3)恒成立

则m=3

2.x²+4x=0 x(x+4)=0

解得x=0或-4

所以A={0,-4}

如果A∪B=A

1.B为空集时,方程无解

判别式=[2(a+1)]²-4(a²-1)