作者phs (世故人情情难还...)
看板Math
标题Re: [中学] 不等式范围
时间Wed Apr 27 09:43:02 2011
※ 引述《debdeb ()》之铭言：
: Find the range of values of the constant a for which there exist
: exactly two integers satisfying the two quadratic inequalities in x
: x^2 - 4x - 5 ≧ 0
: x^2 - (3a-2)x -6a ＜ 0
: simultaneously.
: 请问a的范围该怎麽算？
: 谢谢！！
(sol):
x^2 - (3a-2)x - 6a ＜ 0 ----------(1)
=> -x^2 + (3a-2)x + 6a > 0
since it is simultaneous with the inequality
x^2 - 4x - 5 ≧ 0 -----------(2)
-1 3a-2 6a -10 5
=> ---- = ------- = ------ => a = ----- , ---- , 2 --------(*)
1 -4 -5 9 6
Note that in Eq.(1), there is no equal sign,
thus we should be careful.
For Eq.(2) , its solution is x = -1,5 for the equal sign case.
We have x = -1 => a = -1/3 ,
----------------(**)
x = 5 => a = 5/3
if the equal sign holds in Eq.(1).
We can see that the values of "a" for Eqs.(*) and (**)
are different.
Thus the solution is Eq.(*).
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◆ From: 118.168.207.13
※ 编辑: phs 来自: 118.168.207.13 (04/27 09:54)
1F：→ phs :答案有点怪怪的... 04/27 09:54
2F：→ debdeb :谢谢你，答案就是下面那篇那样 04/27 12:57
※ 编辑: phs 来自: 140.112.101.4 (11/04 17:25)