设A={x|-2≤x≤a},B={y|y=2x+3,x∈A}，C={z|z=x^2,x∈A},且C包含于B，求a的取值范围。要详细解题过程
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A={x|-2≤x≤a},B={y|y=2x+3,x∈A},C={z|z=x^2,x∈A},
所以
B={y|-1<=y<=2a+3} (a>=-2)
1.C={z|0<=z<=4} (0<=a<=2)
2.C={z|0<=z<=a^2} (a>2)
3.C={z|a^2<=z<=4} (a<0)
因为C包含于B,
所以
1.
4<=2a+3
a>=1/2
取1/2<=a<=2
2.
a^2<=2a+3
a^2-2a-3<=0
(a+1)(a-3)<=0
-1<=a<=3
取2<a<=3
3.
4<=2a+3
a>=1/2矛盾!
所以
a的取值范围:1/2<=a<=3.
所以
B={y|-1<=y<=2a+3} (a>=-2)
1.C={z|0<=z<=4} (0<=a<=2)
2.C={z|0<=z<=a^2} (a>2)
3.C={z|a^2<=z<=4} (a<0)
因为C包含于B,
所以
1.
4<=2a+3
a>=1/2
取1/2<=a<=2
2.
a^2<=2a+3
a^2-2a-3<=0
(a+1)(a-3)<=0
-1<=a<=3
取2<a<=3
3.
4<=2a+3
a>=1/2矛盾!
所以
a的取值范围:1/2<=a<=3.
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