Inequality

2007-02-23 2:54 am
-2<=(k+1)/(k-1)<=2 (answer: k<=1/3 or k>=3)

-1/2<=(k-1)/(k+1)<=1/2 (answer: 1/3<=k<=3)

兩條數式計出的答案點解咁唔同.....
詳細解釋please!
更新1:

多謝樓下既解答. 其實我係想問 明明第2條數係=1/第一條數 ,咁點解個答案咁唔同. 定係我個concept錯左=.=

回答 (1)

2007-02-23 6:24 am
✔ 最佳答案
-2<=(k+1)/(k- 1)<=2

Case 1, k < 1
-2 ≦ (k+1)/(k-1) ≦ 2
-2 (k-1) ≧ (k+1) ≧ 2 (k-1)  [全式乘 (k-1), 注意 k-1 為負數]
-2k+2 ≧ k+1 and k+1 ≧ 2k-2
3k ≦ 1 and k ≦ 3
∴ k ≦ 1/3

Case 2, k>1
-2 ≦ (k+1)/(k-1) ≦ 2
-2 (k-1) ≦ (k+1) ≦ 2 (k-1)  [全式乘 k-1]
-2k+2 ≦ k+1 and k+1 ≦ 2k-2
3k ≧ 1 and k ≧ 3
∴ k ≧ 3

Case 3, k = 1
Rejected, as (k+1)/(k-1) is not real number.

Therefore, the answer is k ≦ 1/3 or k ≧ 3


-1/2<=(k-1)/( k+1)<=1/2

Case 1, k<-1
-1/2 ≦ (k-1)/(k+1) ≦ 1/2
-(k+1) ≧ 2(k-1) ≧ (k+1)  [全式乘 2(k+1)]
-k-1 ≧ 2k-2 and 2k-2 ≧ k+1
k ≦ 1/3 and k ≧ 3 (All rejected, as no solutoin.)

Case 2, k > -1
-1/2 ≦ (k-1)/(k+1) ≦ 1/2
-(k+1) ≦ 2(k-1) ≦ (k+1)  [全式乘 2(k+1)]
-k-1 ≦ 2k-2 and 2k-2 ≦ k+1
k ≧ 1/3 and k ≦ 3
∴ 1/3 ≦ k ≦ 3

Case 3, k = -1
Rejected, as (k-1)/(k+1) is not real number.

Therefore, the answer is 1/3 ≦ k ≦ 3

2007-02-22 23:26:05 補充:
其實您個 concept 都冇錯,不過實際應用上好難用到黎解題.始終要用番正式計算先得.不過,都解答左您既疑問先.首先, 如果 1/x > 1/y咁即係 x < y跟住比較番您條題目要拆開兩邊- 第一題其中一個可能性 k≦1/3所以第二題 (1/第一題), 就有其中一個答案係 k≧1/3- 同樣, 第一題另一個可能性 k≧3所以第二題相應亦有一部分答案 k≦3不過,再重複一次,您唔可以用呢個方法當做答案架.


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