Question

求解题，谢谢。

Answer 1

A = { x| -2≤x≤a }, A≠Φ
B={ y| y=2x+3, x∈A}
C = { z| z=x^2 ,x∈A}
C is subset of B

Solution:
A≠Φ
=> a ≥-2 (1)

C = { z| z=x^2 ,x∈A}

case 1: -2≤a<0
A = { x| -2≤x≤a }, A≠Φ
C = { z| z=x^2 ,x∈A}
= { z| a^2≤z≤4 }
B={ y| y=2x+3, x∈A}
2(-2)+3 ≤ a^2 and 2(a) + 3 ≥ 4
-1≤ a^2 and a≥ 1/2
a≥ 1/2
case 1: 舍去

case 2: 0≤a≤2
A = { x| -2≤x≤a }, A≠Φ
C = { z| z=x^2 ,x∈A}
= { z| 0≤z≤a^2 }
B={ y| y=2x+3, x∈A}
2(-2)+3 ≤ 0 and 2(a) + 3 ≥ a^2
-1≤ 0 and a^2-2a-3≤0
a^2-2a-3≤0
(a-3)(a-1)≤0
1≤a≤3
solution for case 2: 1≤a≤2

case 3: a>2
A = { x| -2≤x≤a }, A≠Φ
C = { z| z=x^2 ,x∈A}
= { z| 4≤z≤a^2 }
B={ y| y=2x+3, x∈A}
2(-2)+3 ≤ 4 and 2(a) + 3 ≥ a^2
-1≤ 4 and a^2-2a-3≤0
a^2-2a-3≤0
(a-3)(a-1)≤0
1≤a≤3
solution for case 3: 2<a≤3

C is subset of B
case 1 or case 2 or case 3
(1≤a≤2) or (2<a≤3 )
1≤a≤3