Question

1.设f(x)=x的平方+ax+b,A={x|f(x)=x}={a},求a,b的值 2.已知集合A={x|x≤-3或x≥-1}.......

Answer 1

f(x)=x^2+ax+b = x
=> x^2+（a-1）x+b = 0

for x=a
a^2+(a-1)a+b=0
2a^2-a+b=0

for double root(a)
1-8b=0
b= 1/8

2a^2-a+b=0
16a^2-8a +1 =0
(4a-1)^2=0
a = 1/4

a=1/4 and b=1/8 #

A={x|x≤-3 or x≥-1},
B={x|2m<x<m-1, m∈R}
if A∪B=A => B is subset of A
case 1 if B=ø => B is subset of A
m-1 < 2m
m > -1

case 2, B≠ø
=> m-1≤-3 or 2m≥-1
=> m≤-2 or m≥-1/2

case 1 or case 2
m > -1 or (m≤-2 or m≥-1/2)

ie m>-1 or m≤-2 #