高一集合题，急！！

B={y|-1<=y<=2a+3}
1,-2<=a<=0
C={z|a^2<=z<=4}
当C真包含于B时4<=2a+3 a>=1/2,矛盾
2,0<a<2,C={z|0<=z<=4} 4<=2a+3,a>1=/2得到1/2<=a<=2
3,a>=2,C={z|0<=z<=a^2}
当C真包含于B时a^2<=2a+3 ,-1<=a<=3,得到2<=a<=3
所以,实数a的取值范围是1/2<=a<=3

B.-1<=y<=2a+3
画出C的图像 可知
C的取值范围要看a的取值
1.-2<=a<0
C a^2<=z<=4
C真包含于B
2a+3>=4 a>1/2 不成立
2.0<=a<=2
C 0<=z<=4
C真包含于B
2a+3>=4 a>=1/2
1/2<=a<=2
3. a>2
C 0<=z<=a^2
C真包含于B
2a+3>=a^2
a^2-2a-3<=0
(a-3）(a+1)<=0
-1 <=a<=3
所以2<a<=3