Given the function f (x) = 4x ^ 3-4ax, X ∈ [0,1], the solution set of inequality f (x) ^ 2 ＞ 1 about X is an empty set, then the value of real number a satisfying the condition

Given the function f (x) = 4x ^ 3-4ax, X ∈ [0,1], the solution set of inequality f (x) ^ 2 ＞ 1 about X is an empty set, then the value of real number a satisfying the condition

If the solution set of F (x) ^ 2 > 1 is an empty set, then f (x) ^ 2

The function f (x) = x 2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant, and the value range of real number a is obtained

∵ function f (x) = x2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant, ∵ (x-1) a ≥ - x2-3, when x ∈ [- 2,2], f (x) ≥ - x2 − & nbsp; when x ∈ [- 2,2]; 3x − 1 is constant in X ∈ (1,2] such that G (x) = − x2 − 3x − 1, X ∈ (1,2], i.e., a ≥ g (x) max ∵ g ′ (x) = − (x − 3) (x + 1) (x − 1) 2, ■ (1,2] is the increasing interval, G (2) is the largest, and is - 7  a ≥ - 7; ② when x ∈ [- 2,1], a ≤ − x2 − & nbsp; 3x − 1 is constant in X ∈ [- 2,1), so that G (x) = − x2 − 3x − 1, X ∈ [- 2,1), that is, a ≤ g (x) min, and the minimum value of G (x) = − x2 − 3x − 1 in ∈ [- 2,1) is g (- 1) = 2, ｜ a ≤ 2; in conclusion, the value range of real number a is: [- 7,2]

Given that the function FX is a decreasing function over the domain R, the inequality f (1 / x) is larger than the range of X of F 1

（-∞,0）∪（1,+∞）

Given the function f (x) = x2 + 1 & nbsp; (x ≥ 0) 1 & nbsp; (x < 0), then the value range of X satisfying the inequality f (1-x2) > F (2x) is ()
A. （-1，0）B. （0，1）C. （-1，2-1）D. （-2-1，2-1）

From the meaning of the question, draw the image of function f (x), as shown in the figure: ∵ f (1-x2) > F (2x) ∵ 1 − x2 > 02x < 0 or 1 − x2 > 02x ≥ 01 − x2 > 2x solution: - 1 ∵ x < 0 or 0 ≤ x < 2 − 1 ∵ x < 2 − 1, so choose C