A中
xx - 2x -3<=0
=>
-1<=x<=3
而A∩B={x│-1≤x<2},说明B中不等式较大的一解为2,另外一个解不比-1小
=>
设另外一解为a
=>-1<=a<2
由伟达定理:
a + 2 = -p
2a = q
=>
p = -2-a
-4q = 2a
-2<=q<4
已知集合A={x│x^2-2x-3≤0},B={x│x^2+px+q
已知集合A={x│x^2-2x-3≤0},B={x│x^2+px+q<0},若A∩B={x│-1≤x<2},求p.q的范围.
数学人气:895 ℃时间:2020-03-21 08:33:27
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