✔ 最佳答案
1.a. f(x) = 2x^2 - kx + 12
= 2(x^2 - kx/2 + (k/4)^2) - k^2/8 + 12
= 2(x - k/4)^2 + 12 - (k^2/8)
For x = 5/2, f(x) attains the minimum
This happens when (x - k/4)^2 = 0
So, 5/2 - k/4 = 0
k = 10
b. For the greatest value of 4 - f(x), f(x) is at minimum value
min f(x) = 12 - (10)^2/8 = -1/2
So, the maximum value of 4 - f(x) = 4 - (-1/2) = 9/2
2.a. 2x^2 + 6x + 9 = a(x + b)^2 + c
= a(x^2 + 2bx + b^2) + c
= ax^2 + 2abx + (ab^2 + c)
Comparing coefficients, a = 2, 2ab = 6, so, b = 6/(2 X 2) = 3/2
(2)(3/2)^2 + c = 9
c = 9/2
b. 2x^2 + 6x + 9 = 2(x + 3/2)^2 + 9/2
The minimum value of this function is 9/2.
c. As minimum value of the equation 2x^2 + 6x + 9 = 9/2
So, 2/(2x^2 + 6x + 9) attains maximum when 2x^2 + 6x + 9 = 9/2
So, 2/(2x^2 + 6x + 9) <= 2/(9/2) = 4/9
And the denominator will converge to +ve infinity as x tends to infinity, so
2/(2x^2 + 6x + 9) > 0
Therefore,
0 < 2/(2x^2 + 6x + 9) <= 4/9
