Question

1.(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},B={x|2m<x<m-1,m∈R}若A∪

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Answer 1

f(x)=x^2+ax+b, A={x|f(x)=x}={a},

x=f(x)
x=x^2+ax+b
x^2+(a-1)x+b=0
for double roots a
(x-a)^2 =0
x-2ax+a^2=0

compare coefficient of x, constant
=> a-1 =-2a
a= 1/3
b=a^2 = 1/9 #

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, B＝ø
m-1<2m
m > -1
case 2, B≠ø
=> m-1 ≤-3 and m-1≥-1
=> m ≤ -2 and m ≥ 0

case 1 or case 2
m > -1 or (m ≤ -2 and m ≥ 0)
=> m≤ -2 and m >-1 #