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}

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}
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}

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}
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

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