已知A={x | x=t²+(a+1)t+b},B={y | y=-t²-(a-1)t-b}.求常数
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已知A={x | x=t²+(a+1)t+b},B={y | y=-t²-(a-1)t-b}.求常数a、b.使得
A∩B={x | -1≤x≤2}
A∩B={x | -1≤x≤2}
![已知A={x | x=t²+(a+1)t+b},B={y | y=-t²-(a-1)t-b}.求常数](/uploads/image/z/19483817-41-7.jpg?t=%E5%B7%B2%E7%9F%A5A%3D%EF%BD%9Bx+%7C+x%3Dt%26%23178%3B%2B%EF%BC%88a%2B1%EF%BC%89t%2Bb%EF%BD%9D%2CB%3D%EF%BD%9By+%7C+y%3D-t%26%23178%3B-%EF%BC%88a-1%EF%BC%89t-b%EF%BD%9D.%E6%B1%82%E5%B8%B8%E6%95%B0)
x=t²+(a+1)t+b=[t+(a+1)/2]²+b-(a+1)²/4≥b-(a+1)²/4
A={x|x=t²+(a+1)t+b}={x|x≥b-(a+1)²/4}
y=-t²-(a-1)t-b=-[t+(a-1)/2]²-b+(a-1)²/4≤-b+(a-1)²/4
B={y|y=-t²-(a-1)t-b}={y|y≤-b+(a-1)²/4}
因为A∩B={x|-1≤x≤2}
所以b-(a+1)²/4=-1,-b+(a-1)²/4=2
解得a=-1,b=-1
A={x|x=t²+(a+1)t+b}={x|x≥b-(a+1)²/4}
y=-t²-(a-1)t-b=-[t+(a-1)/2]²-b+(a-1)²/4≤-b+(a-1)²/4
B={y|y=-t²-(a-1)t-b}={y|y≤-b+(a-1)²/4}
因为A∩B={x|-1≤x≤2}
所以b-(a+1)²/4=-1,-b+(a-1)²/4=2
解得a=-1,b=-1
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