Re: [中學] 多項式 謝謝大大～～

[yueayase (scrya)]

※ 引述《LKJX (LKJX)》之銘言：
: 各位大大您們好，
: 請問：一題多項式（不知答案）
: https://imgur.com/685CfRk
: 謝謝大大~~

解為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
由題目f(x)=-x +ax+b，可假設f(x)=-(x-5)(x-k) for some k
則原不等式可改成: 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

討論(1) k≧5 => 不等式為 5≦x≦k+1 且 (x≧k或x≦5) => k≦x≦ k+1或x=5

    (2) 4≦k＜5 => 不等式為 5≦x≦k+1 且 (x≧5或x≦k) => 5≦x≦k+1

    (3) k＜4 => 不等式為 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)與x軸交於5和6兩點
(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點
(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
       作圖可得

        |
        | (30/11,0)
        |-- /------
        |  /
        | /
(0,-30) |/

     直線不過第二象限


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