A. maths Inequalities (Ergent HW!!!!!!!!!!!!!!) 10pts

-3x^2+6x-k ≤ 0 for all real values of x is equivalent to

3x^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

-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.

∵a= - 3 ＜ 0 and -3x^2+6x-k ≤ 0
∴△≦0
△=36-4(-3)(-k)≦0
36-12k≦0
36≦12k
3≦k