作者yueayase (scrya)
看板Math
标题Re: [中学] 多项式 谢谢大大～～
时间Thu Jan 13 22:24:19 2022
※ 引述《LKJX (LKJX)》之铭言：
: 各位大大您们好，
: 请问：一题多项式（不知答案）
: https://imgur.com/685CfRk
: 谢谢大大~~
解为x=5或6≦x≦7
=>5-5≦f(5)≦0
=>f(5)=0且由因式定理 x-5|f(x)
2
由题目f(x)=-x +ax+b，可假设f(x)=-(x-5)(x-k) for some k
则原不等式可改成: 5-x≦-(x-5)(x-k)≦0
5-x≦-(x-5)(x-k) 且 -(x-5)(x-k)≦0
=> 0≦-(x-5)(x-k)+(x-5)=-(x-5)(x-k-1) 且 -(x-5)(x-k)≦0
讨论(1) k≧5 => 不等式为 5≦x≦k+1 且 (x≧k或x≦5) => k≦x≦ k+1或x=5
(2) 4≦k＜5 => 不等式为 5≦x≦k+1 且 (x≧5或x≦k) => 5≦x≦k+1
(3) k＜4 => 不等式为 k+1≦x≦5 且(x≧5或x≦k) => x=5
只有(1)合理，所以k=6
2 2
=>f(x) = -(x-5)(x-6)=-(x -11x+30) = -x +11x-30
-----------------------------------------------------
(O)(1)y=f(x)与x轴交於5和6两点
(X)(2)5-x=-(x-5)(x-6) => -(x-5)(x-6)+(x-5) = -(x-5)(x-7)=0 => x=5 or 7
交於2点
(X)(3) -(x-5)(x-6) < 5-x => -(x-5)(x-6)+(x-5)=-(x-5)(x-7) < 0 => (x-5)(x-7)>0
=> x＞7 或 x＜5
(O)(4) -(x-5)(x-6)>0 => (x-5)(x-6)<0 => 5＜x＜6
(O)(5) y=11x-30
x截距30/11 y截距-30
作图可得
|
| (30/11,0)
|-- /------
| /
| /
(0,-30) |/
直线不过第二象限
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1F：推 LTOTE : 谢谢 yueayase 大大 01/13 22:35