A. maths Inequalities (Ergent HW!!!!!!!!!!!!!!) 10pts
回答 (3)
2007-10-20 5:23 am
✔ 最佳答案
-3x^2+6x-k ≤ 0 for all real values of x is equivalent to3x^2-6x+k≧0 for all real values of x
So △ ≤ 0,
(-6)^2-4(3)(k)≤0
36-12k≤0
12k≧36
k≧3
參考: My Maths Knowledge
2007-10-26 11:06 pm
-3x^2+6x-k ≤ 0
=> -3(x^2 - 2x) - k ≤ 0
=> -3(x^2 - 2x + 1 - 1) - k ≤ 0
=> -3(x - 1)^2 + 3 - k ≤ 0
=> k >= -3(x - 1)^2 + 3
Since we need to cater for all real values of x and the maximum of -3(x - 1)^2 + 3 is reached when x = 1, k >= 3 should hold in order for -3x^2+6x-k ≤ 0 for all real values of x.
=> -3(x^2 - 2x) - k ≤ 0
=> -3(x^2 - 2x + 1 - 1) - k ≤ 0
=> -3(x - 1)^2 + 3 - k ≤ 0
=> k >= -3(x - 1)^2 + 3
Since we need to cater for all real values of x and the maximum of -3(x - 1)^2 + 3 is reached when x = 1, k >= 3 should hold in order for -3x^2+6x-k ≤ 0 for all real values of x.
2007-10-20 6:03 am
∵a= - 3 ＜ 0 and -3x^2+6x-k ≤ 0
∴△≦0
△=36-4(-3)(-k)≦0
36-12k≦0
36≦12k
3≦k
∴△≦0
△=36-4(-3)(-k)≦0
36-12k≦0
36≦12k
3≦k
收錄日期: 2021-04-13 14:02:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071019000051KK04577