Question

已知集合A={x| |x-a|=4}，集合B={1，2，b}.

（1）是否存在实数a，使得对于任意实数b都有A包含于B？若存在，求出a=？，若不存在，说明理由。
（2）若A包含于B成立，求出对应的实数（a，b）.

Answer 1

A={x| |x-a|=4}
={ a+4 ， a-4 }

B={1，2，b}

if B is subset of A, for all b

then
case 1
a+4 =1 and a-4 =2
a = -3 and a = 6 (rejected)

or
case 2
a+4 = 2 and a-4 = 1
a = -2 and a = 5 ( rejected )

不局数存在a， 对于任意实简腊改数b都有A包拦判含于B

if A is subset of B, then
case 1
a+4 =1 and a-4=b
=> a= -3 and b = -7

or
case 2
a+4=2 and a-4=b
=>a = -2 and b= -6
or
case 3
a-4= 1 and a+4 =b
=>a = 5 and b = 9

or
case 4
a-4 = 2 and a+4 =b
=> a =6 and b=10

(a,b) = (-3,-7) or (-2,-6) or (5,9) or (6,10) #