The function y = f (x) is an even function on R and an increasing function on (- ∞, 0]. If f (a) ≤ f (2), the value range of real number a is () A. a≤2 B. a≥-2 C. -2≤a≤2 D. A ≤ - 2 or a ≥ 2

The function y = f (x) is an even function on R and an increasing function on (- ∞, 0]. If f (a) ≤ f (2), the value range of real number a is () A. a≤2 B. a≥-2 C. -2≤a≤2 D. A ≤ - 2 or a ≥ 2

F (x) is a monotone decreasing function on (0, + ∞),
Thus there
a＜0
a≤−2 or
a＞0
a≥2 ，
The solution is a ≤ - 2 or a ≥ 2,
Therefore, D

If the even function f (x) is an increasing function on (- ∞, 0], the value range of real number a satisfying f (1) ≤ f (a) is __

∵ even function f (x) is an increasing function on (- ∞, 0],
‡ f (x) is a subtractive function on [0, + ∞),
When a ≥ 0, 0 ≤ a ≤ 1 is obtained from F (1) ≤ f (a);
When a < 0, the inequality f (1) ≤ f (a), that is, f (- 1) ≤ f (a), can be - 1 ≤ a < 0
To sum up, the value range of real number a satisfying f (1) ≤ f (a) is [- 1,1]
So the answer is: [- 1, 1]

Given that the definition field of function f (x) is [- 1,1], and the definition field of function f (x) = f (x + m) - f (x-m) exists, the value range of real number m is __

∵ the definition field of function f (x) is [- 1, 1],
The definition field of - 1 ≤ x ≤ 1, f (x) = f (x + m) - f (x-m) exists
∴-1≤x+m≤1，-1≤x-m≤1①，
And - 1 ≤ - x-m ≤ 1 ②,
① + ②,
-2≤-2m≤2，
∴-1≤m≤1，
Therefore, the answer is: - 1 ≤ m ≤ 1;

The odd function f (x) is a subtractive function in the definition field (- 1,1), and f (1-A) + F (1-A ^ 2) < 0 to find the value range of the real number a

First, consider the domain, - 1 < 1-A < 1, - 1 < 1-A ²＜ one
Get 0 < a < radical 2
Now look at the topic
f(1-a)+f(1-a ²)＜ 0
f(1-a)＜-f(1-a ²)
And f (x) is an odd function
Then - f (x) = f (- x)
Then - f (1-A ²)= f(a ²- 1)
Then f (1-A) ＜ f (a) ²- 1)
Since f (x) is a subtractive function on (- 1,1)
Then it is known from the above that 1-A > a ²- one
Get a ∈ (- 2,1)
A ∈ (0,1) is known from the comprehensive definition field
Then, in the face of this problem, we should first consider the problem of defining domain, because the smaller things are
The more easily people forget, they often lose points due to negligence, which is very uneconomical
Also, in the face of this problem, we should see the trick. First, we use the monotonicity of the function to solve the monotonicity of the function
Inequality is very common. If you move a function past and find that it cannot be solved, there must be parity
Other things like sex or periodicity help
Finally, I wish you a happy study of mathematics. You can ask me if you have any questions~

Please help me solve the problem in detail. It's hard. I'm in the exam The even function f (x) of the definition field R, when x > 0, f (x) = LNX ax (a belongs to R), equation f (x) = 0, there are exactly five different real number solutions on R. one, find X

The title is incomplete
I guess the first question is: ask for X

Let the definition domain of even function f (x) be r, when x ∈ [0, + ∞), f (x) is an increasing function, then the size relationship of F (- 2), f (π), f (- 3) is ___

From the relationship between even function and monotonicity, if f (x) is an increasing function when x ∈ [0, + ∞), then f (x) is a decreasing function when x ∈ (- ∞, 0),
  Therefore, the geometric feature of the image is that the smaller the absolute value of the independent variable, the smaller the function value,
∵|-2|＜|-3|＜π
∴f（π）＞f（-3）＞f（-2）
  Therefore, the answer is f (π) > F (- 3) > F (- 2)