→ phs :答案有點怪怪的... 04/27 09:54
→ debdeb :謝謝你，答案就是下面那篇那樣 04/27 12:57
※ 編輯: phs 來自: 140.112.101.4 (11/04 17:25)
※ 引述《debdeb ()》之銘言：
: 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.
: 請問a的範圍該怎麼算？
: 謝謝！！
(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
※ 編輯: phs 來自: 118.168.207.13 (04/27 09:54)