If f (x) = x ^ 2 + (K-4) x-2k + 4 is always greater than 0 for any k ∈ [- 1,1], the value range of X is? A. - 1 / 3 b.x > 4 c.x < 1 or X > 3 D.X < 1 Item a is changed to x < 0

If f (x) = x ^ 2 + (K-4) x-2k + 4 is always greater than 0 for any k ∈ [- 1,1], the value range of X is? A. - 1 / 3 b.x > 4 c.x < 1 or X > 3 D.X < 1 Item a is changed to x < 0


x²-4x+kx-2k+4>0
x²-4x+4>(2-x)k
k∈[-1,1],(2-x)k ≤|x-2|
(x-2)²≥|x-2|
|x-2|≥1
Choose answer C



Given the function f (x) = x ^ 2 + 2x + A / x, X belongs to {1, positive infinity) for any x belongs to 1 to positive infinity, f (x) > 0 is constant, the value range of a is obtained
eleven thousand one hundred and eleven


f(x)=x^2+2x+a/x>0
x^2+2x+a>0
Y = x ^ 2 + 2x + A, X is an increasing function from 1 to positive infinity
Satisfy x = 1, Y > 0
1+2*1+a>0
a>-3



For any k ∈ [- 1,1], the value of function f (x) = x & sup2; + (K-4) x-2k + 4 is always greater than 0, and the value range of X is obtained
rt
Is the value range of X!


The value of F (x) = x ^ 2 + (K-4) x-2k + 4 is always greater than 0
With the opening upward, the axis of symmetry x = - (K-4) / 2 = 2-k / 2
∵ f (k) = 2-k / 2 (K ∈ [- 1,1]) is a decreasing function
When k = - 1, the axis of symmetry is on the far right; when k = 1, the axis of symmetry is on the far left
In order to make the value of function f (x) = x & sup2; + (K-4) x-2k + 4 be always greater than 0, so:
When k = - 1, X must be greater than the right intersection of the graph and X axis;
When k = 1, X must be less than the left intersection of the graph and X axis
(1) When k = - 1, f (x) = x ^ 2 + (- 1-4) X-2 * (- 1) + 4 = x ^ 2-5x + 6 = (X-2) (x-3)
When k = - 1, X must be greater than the right intersection of the graph and X axis
∴x>3
(2) When k = 1, f (x) = x ^ 2 + (1-4) X-2 * 1 + 4 = x ^ 2-3x + 2 = (x-1) (X-2)
When k = 1, X must be less than the left intersection of the graph and X axis
∴x<1
In conclusion, K ∈ (- ∞, 1), (3, + ∞)



Function f (x) = x ^ 2 + 2x-3a, X ∈ [- 2,2] Q: if f (x) + 2A ≥ 0 holds, find the value range of A


f(x)=x²+2x-3a
f(x)+2a≥0
Namely:
x²+2x-a≥0
A ≤ X & # 178; + 2x is constant for all x ∈ [- 2,2]. Constant small means that a on the left is smaller than the minimum on the right,
The minimum value of X & # 178; + 2x = (x + 1) &# 178; - 1 is - 1,
a≤-1