If the function f (x) = ax-x-a (a > 0 and a ≠ 1) has two zeros, then the value range of real number a is () A. 0＜a＜1B. 0＜a＜12C. a＞2D. a＞1

If the function f (x) = ax-x-a (a > 0 and a ≠ 1) has two zeros, then the value range of real number a is () A. 0＜a＜1B. 0＜a＜12C. a＞2D. a＞1

If the function f (x) = ax-x-a (a > 0 and a ≠ 1) has two zeros, then the function y = ax and y = x + a have two intersections. When 0 < a < 1, the function y = ax and y = x + a have only one intersection, which does not meet the condition. When a > 1, the function y = ax and y = x + a have two intersections, as shown in the figure: therefore, the value range of real number a is a > 1. Therefore, select D
If the image of function f (x) = (x + a) (| x-a | + | X-2 |) is a centrosymmetric figure, then the value of real number a is______ .
∵ f (x) = (x + a) (| x-a | + | X-2 |). First of all, we notice that the image of function f (x) is a parabola, a line segment and a parabola from left to right. Therefore, the center of symmetry of the graph must be the midpoint of the line segment
If the function f (x) = a ^ x-x-a (a > 0, a ≠ 1) has two zeros, find the value range of A
It can only be solved by drawing the function Y1 = a ^ x, y2 = x + a
When A1,
A ^ x at 0
For any real number x, if f (x) satisfies f (- x) + F (2 + x) + 2 = 0, then what is the symmetry of function image
Point (1, - 1)
If f (x) = − 12x2 + bln (x + 2) is a decreasing function on (- 1, + ∞), find the value range of B
F ′ (x) = - x + BX + 2, so f (x) is a decreasing function on (- 1, + ∞), equivalent to - x + BX + 2 ≤ 0, constant on (- 1, + ∞), x + 2 > 0 from X ＞ - 1, equivalent to B ≤ x2 + 2x, constant on (- 1, + ∞), and y = x2 + 2x monotonically increasing on (- 1, + ∞), X2 + 2x ＞ - 1, so B ≤ - 1, so the value range of B is (- ∞, - 1]
The real root x0 of the equation f (x) = f '(x) is defined as the "new stationary point" of the function f (x). For example, if the "new stationary points" of the function g (x) = x, H (x) ln (x + 1), φ (x) = cosx (x ∈ (π / 2, π)) are respectively α, β, γ, then the size relationship of α, β, γ is as follows
The "new stationary point" of G '(x) = 1 = g (x) = x g (x) is α = 1
h'(x)=1/(x+1)=h(x)=ln(x+1)
H '(x) decreasing and H (x) increasing
When x = 1, H '(x) = 1 / 2, H (x) = LN2 > 1 / 2, that is, H (x) > H' (x)
When h (x) = H '(x), X
If f (x) = − 12x2 + bln (x + 2) is a decreasing function on (- 1, + ∞), find the value range of B
F ′ (x) = - x + BX + 2, so f (x) is a decreasing function on (- 1, + ∞), equivalent to - x + BX + 2 ≤ 0, constant on (- 1, + ∞), x + 2 > 0 from X ＞ - 1, equivalent to B ≤ x2 + 2x, constant on (- 1, + ∞), and y = x2 + 2x monotonically increasing on (- 1, + ∞), X2 + 2x ＞ - 1, so B ≤ - 1, so the value range of B is (- ∞, - 1]
The function f (x) defined on (0, + ∞) satisfies: F (2x) = 2F (x), and when x ∈ (1,2], f (x) = 2-x. if X1 and X2 are two real roots of the equation f (x) = a (0 < a ≤ 1), then x1-x2 cannot be ()
A. 24B. 72C. 96D. 120
Because f (2x) = 2F (x) holds for any x ∈ (0, + ∞), and if x ∈ (1,2], f (x) = 2-x; so f (x) = - x + 2B, X ∈ (B, 2b], B ∈ n *. From the two real roots of the problem equation f (x) = a (0 < a ≤ 1), we obtain that there are two intersections of the function y = f (x) image and the line y = a
If f (x) = − 12x2 + bln (x + 2) is a decreasing function on (- 1, + ∞), find the value range of B
F ′ (x) = - x + BX + 2, so f (x) is a decreasing function on (- 1, + ∞), equivalent to - x + BX + 2 ≤ 0, constant on (- 1, + ∞), x + 2 > 0 from X ＞ - 1, equivalent to B ≤ x2 + 2x, constant on (- 1, + ∞), and y = x2 + 2x monotonically increasing on (- 1, + ∞), X2 + 2x ＞ - 1, so B ≤ - 1, so the value range of B is (- ∞, - 1]
It is known that the set M is the whole set of functions f (x) satisfying the following properties. There exists X1 in the domain D such that f (x1 + 1) = f (x1) + F (1) holds
① Does the function f (x) = 1 / X belong to the set M
② If the function f (x) = KX + B belongs to the set M, try to find the value range of real numbers K and B
③ Let f (x) = ㏒ 10 (A / X & # 178; + 1) belong to the set M, and find the value range of real number a
(1) Let f (x + 1) = f (x) + F (1)
I.e. 1 / (1 + x) = 1 / x + 1, x ^ 2 + X + 1 = 0, no real root, in R, f (x) does not belong to set M
(2) Let f (x + 1) = f (x) + F (1)
That is, KX + B + k = KX + B + K + B has a solution, B = 0, K can be any real number
(3) Easy to know a > 0
Let f (x + 1) = f (x) + F (1)
That is, LG (A / (x + 1) ^ 2 + 1) = LG (A / x ^ 2 + 1) + LG (A / 2) has a solution
That is, a / [(x + 1) ^ 2 + 1] = a ^ 2 / 2 (x ^ 2 + 1)
It is found that (A-2) x ^ 2 + 2aX + 2a-2 = 0 has a solution
If a = 2, there is a solution
When a ≠ 2, Δ = (2a) ^ 2-4 (A-2) (2a-2) ≥ 0, the solution is 3 - √ 5 ≤ a ≤ 3 + √ 5
To sum up, 3 - √ 5 ≤ a ≤ 3 + √ 5