When x ∈ [- 2,0] and f (x) = X-1, then the inequality XF (x) > 0 is an even function with period 4 Solution set on [- 1,3]

Drawing can be obtained as follows:
[-2,0]：f(x)=x-1；
[0,2]：f(x)=-x-1；
[2,4]：f(x)=x-5；
[- 1,0]: XF (x) = x (x-1) > 0; the result is: [- 1,0];
[0,2]: XF (x) = x (- x-1) > 0;
[2,4]: XF (x) = x (X-5) > 0;
To sum up, the solution set is: [- 1,0]

Let f (x) be an even function defined on R. when x > 0, f (x) + XF '(x) > 0, and f (1) = 0, then the solution set of inequality XF (x) > 0 is () A. （-1，0）∪（1，+∞） B. （-1，0）∪（0，1） C. （-∞，-1）∪（1，+∞） D. （-∞，-1）∪（0，1）

Let g (x) = XF (x), then G '(x) = [XF (x)]' = x'f (x) + XF '(x) = XF' (x) + F (x) > 0,
The function g (x) is an increasing function on the interval (0, + ∞),
∵ f (x) is an even function defined on R,
/ / g (x) = XF (x) is an odd function on R,
The function g (x) is an increasing function on the interval (- ∞, 0),
∵f（1）=0，
∴f（-1）=0；
That is, G (- 1) = 0, G (1) = 0
﹥ XF (x) ﹥ 0 becomes g (x) ﹥ 0,
Let x > 0, so the inequality is g (x) > G (1), i.e. 1 < X;
Let x < 0, so the inequality is g (x) > G (- 1), that is - 1 < x < 0
So the solution set is (- 1,0) ∪ (1, + ∞)
Therefore, a

It is known that the function f (x) is an even function on R, and there is f '(x) > 0 on (0, + ∞). If f (- 1) = 0, then the solution set of the inequality x f (x) < 0 about X is______ .

The function f (x) is an even function on R, and there is f '(x) > 0 on (0, + ∞),
Therefore, the function is an increasing function on (0, + ∞) and a decreasing function on (- ∞, 0),
F (- 1) = 0, so f (1) = 0, the function graph is shown in the figure
From the graph, we know that the solution set of the inequality x f (x) < 0 is (- ∞, - 1) ∪ (0, 1)
So the answer is (- ∞, - 1) ∪ (0, 1)

F (x) is an even function defined on R. when x < 0, f (x) + X · f ′ (x) < 0, and f (- 4) = 0, then the solution set of the inequality XF (x) > 0 is______ .

When x is less than 0, f (x) + X · f ′ (x) ＜ 0, i.e. [XF (x)] ′ < 0, so the function y = XF (x) is a minus function on (- ∞, 0). According to the fact that f (x) is an even function, we can get that the function y = XF (x) is an odd function and a decreasing function on (0, + ∞)

Let f (x) be an even function defined on R and a periodic function with period 4. When x belongs to [0,2], f (x) = 2x cosx, then a = f (- 3 / 2) and f (x) = 2x cosx The size relation of B = f (5 / 12)

When x belongs to [0,2], f (x) = 2x cosx, then a = f (- 3 / 2) = f (3 / 2) = 3-cos1.5
b=5/6-cos（5/12）
A>b

3. The function f (x) = 2sinx cosx is (a) odd function with minimum positive period 2 (b) even function (c) with minimum positive period 2

f(x)=sin2x
T=π
Odd functions,

Let f (x) = log4 (4 ^ x + 1) + KX be even functions, let g (x) = log4 (a * 2 ^ x-4a / 3) If the function y = f (x) - G (x) has and only one zero point, find the value range of real number a

1. Find K first, according to f (x) = log4 (4 ^ x + 1) + KX is even function, we get f (x) = f (- x)
That is log4 (4 ^ x + 1) + KX = log4 [1 / (4 ^ x) + 1] - KX, k = - 1 / 2
2. Find the value range of real number a
If y = f (x) - G (x) has and only one zero point, then log4 (4 ^ x + 1) + KX = log4 (a * 2 ^ x-4a / 3)
First, the domain defined by G (x) has a * (2 ^ x-4 / 3) > 0. When x > log2 (4 / 3), a > 0
When x < log2 (4 / 3), a < 0
3. Verify whether there is only one solution and find the
In order to make f (x) = g (x) = > to simplify writing, let 2 ^ x = t
That is (A-1) T ^ 2-4a / 3t-1 = 0
In order for t = 2 ^ x to have one and only one solution, it must be △ = B ^ 2-4ac = 0, where the unique solution of F (x) = g (x) is t = 2 ^ x = - B / (2a)
When 16 / 9A ^ 2 + 4 (A-1) = 0, f (x) = g (x) has a unique solution
The unique solution of A1 = - 3 is t = 2 ^ x = {4A / 3} / {2 (A-1)} = 1 / 2, that is, x = - 1
Or the unique solution of A2 = 3 / 4 is t = 2 ^ x = {4A / 3} / {2 (A-1)} = - 2, that is, x = log2 (- 2) should be omitted
Conclusion: if and only if a = - 3, there is and only one zero point, and the solution is x = - 1

Let f (x) = log4 (4 ^ x + 1) + KX (K ∈ R) be even functions Let f (x) = log4 (4 ^ x + 1) + KX (K ∈ R) be even functions. Let H (x) = log4 (a times the x power of two minus four-thirds a), if the functions f (x) and H (x) are equal to each other Let f (x) = log4 (4 ^ x + 1) + KX (K ∈ R) be even functions. Let H (x) = log4 (a times the x power of two minus four-thirds a), if the image of function f (x) and H (x) have and only one common point, the value range of real number a is calculated

From F (x) = f (- x), it is obtained that f (- 1) = f (1) {log4 (4-1 + 1) - k = log4 (4 + 1) + k = - 1 / 2, that is, the graph of F (x) = log4 (4 ^ x + 1) - 1 / 2x function f (x) and G (x) have and only one common point, that is, the equation log4 (4 ^ x + 1) - 1 / 2x = log4 (a · 2 ^ x-4 / 3a) has and only one common point

Given that the function f (x) = log4 (4 ^ x + 1) + X / 2 is even function, if the equation f (x) - M ＜ 0 has a solution, find the value range of M

If f (x) = log4 (4 ^ x + 1) - X / 2, it is even function
It is easy to prove that [0, + ∞) is monotonically increasing
The minimum value of F (x) is obtained when x = 0
That is, f (x) min = f (0) = 1 / 2
So the equation f (x) - M 〈 0 has a solution as long as m ≥ 1 / 2
That is, the value range of M is m ≥ 1 / 2

Let f (x) = log4 (4 ^ x + 1) + KX (K ∈ R) be even functions (1) Find the value of K: (2) if the equation f (x) - M = 0 has a solution, find the value range of M Explain the second question in detail

(1) Log (x) = log4 (4 ^ x + 1) + KX (K ∈ R) is even function, and (- x) = f (x), that is, log [4 ^ (- x) + 1] + K (- x) = log (4 ^ x + 1) + K (- x) = log (4 ^ x + 1) + KX, log {[4 ^ (- x) + 1] / (4 ^ x + 1)} = 2kx, - x = 2kx, k = - 1 / 2. (2) f (x) = log4 (4 ^ x + 1) - X / 2-m = 0m = log4 (4 ^ x + 1) - X / 2 = log4 (4 ^ x + 1) - X / 2 = log4 (4 ^ 4 ^ x + 1) - X / 2 = log4 (4 ^ + 1) - X / 2 = log4 (4 x + 1) - log4 [4

When x ∈ [- 2,0] and f (x) = X-1, then the inequality XF (x) > 0 is an even function with period 4 Solution set on [- 1,3]