已知两集合A={x|x=t的平方+(a+1)t=b,t属于R},B={x|x=-t的平方-(a-1)t-b,t属于R},且A∩B={x|-1≤x≤2} 求常数a,b的值

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已知两集合A={x|x=t的平方+(a+1)t=b,t属于R},B={x|x=-t的平方-(a-1)t-b,t属于R},
且A∩B={x|-1≤x≤2} 求常数a,b的值

A={x|x=t^2+(a+1)t+b,t∈R},B={x|x=-t^2-(a-1)t-b,t∈R},
A∩B={x|-1≤x≤2}
consider
f(x)= x^2+(a+1)x+b
f'(x) = 2x + (a+1) = 0
x = -(a+1)/2
f''(x) = 2 (min)
f( -(a+1)/2 ) = ((a+1)/2)^2 -(a+1)(a+1)/2 + b
= - (a+1)^2/4 + b = -1 (1)
let
g(x) = -x^2-(a-1)x-b
g'(x) = -2x - (a-1) = 0
x = -(a-1)/2
g''(x) = -2 ( max )
g( - (a-1)/2 ) = (a-1)^2/4- b = 2 (2)
(1) + (2)
- (a+1)^2/4 + (a-1)^2/4 = 1
-a^2-2a-1+a^2-2a+1 = 4
-4a=4
a =-1
b = -1