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≥

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 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 #