Given that the odd function f (x) defined on (- 1,1) is a decreasing function in the domain of definition, and f (1-A) + F (1-2a) > 0, the value range of real number a is obtained

Given that the odd function f (x) defined on (- 1,1) is a decreasing function in the domain of definition, and f (1-A) + F (1-2a) > 0, the value range of real number a is obtained

From F (1-A) + F (1-2a) > 0, we get that f (1-A) > F (1-2a), and ∵ f (x) is an odd function on (- 1,1); - f (1-2a) = f (2a-1), and - 1 < 1-2a < 1 ① In addition, f (x) is a decreasing function defined on (- 1,1); {1-A ＜ 2a-1 and - 1 ＜ 1-A ＜ 1 ② So the value range of real number a is (23,1)
Given that the odd function f (x) is a decreasing function in the domain (- 1,1), and f (1-A) + F (1-2a) > 0, the value range of real number a can be obtained
What kind of question does this question belong to? How to solve this kind of question?
There's another one
It is known that the function f (x) satisfies f (1) = 1988, f (n + 1) = f (n) + 11, and N belongs to positive integer
(1) Finding the value of F (3)
(2) If f (n) = an + B, find the value of a and B
（2）
f(1-a)>-f(1-2a)=f(2a-1)
-1
Let f (x) defined on [- 2.2] monotonically decrease in the interval [0.2], if f (1-m)
Even function, monotonically decreasing on interval [0,2]
Then it increases monotonically on the interval [- 2,0]
Domain of definition
-2
Can a real number whose cube is equal to itself form a set
Can be set as {0,1, - 1}
Can form a set!
{XIX ^ 3 = x, X is a real number}
If the line y = KX + 1 intersects the circle x ^ 2 + y ^ 2 + KX + my-4 = 0 with m and N, and m and N are symmetric with respect to the line x + y = 0, what is the value of K? What is the value of M?
Because the slope of the intersection of a straight line x ^ k + 1 and a straight line Mn + 1 is symmetric with respect to the reciprocal slope of a straight line x ^ k + y = 0
A set of real numbers greater than or equal to - 3 and less than 11
{x│-3
-2 -1 0 1 2 3 4 5 6 7 8 9 10
[-3,11)
{x|-3≤x
If the line kx-y + 1 = 0 intersects the circle x2 + Y2 + 2x my + 1 = 0 at two points m and N, and m and N are symmetric with respect to the line y = - x, then | Mn|=______ .
It can be seen from the meaning of the question that the line y = - x passes through the center of the circle and is perpendicular to the line kx-y + 1 = 0, k = 1, the center coordinate (- 1, M2) of the circle x2 + Y2 + 2x-my + 1 = 0 is on the line x + y = 0, so the radius of M = 2, the center coordinate (- 1, 1), X2 + Y2 + 2x-2y + 1 = 0 is 1, and the distance from the center of the circle to the line y = x + 1 is | - 1 − 1 + 1 | 2 = 22
It is known that s is a set composed of real numbers and satisfies the following conditions: 1) 1 does not belong to s; 2) if a belongs to s, then 1 (1-A). If s is not equal to an empty set, s contains at least
Why?
2 elements
If 2 ∈ s, then - 1 ∈ s, and then 1 / 2 ∈ s, continue to substitute for 2, forming a cycle
If the line y = KX + 2 intersects the circle x + y + KX + my-4 at two points, and Mn is symmetric with respect to the line x + y = 0, the length of Mn is obtained
Mn is symmetric with respect to the line x + y = 0, and the Mn ⊥ line x + y = 0 is obtained
Straight line y = KX + 2 ⊥ straight line x + y = 0
K=1
Mn is symmetric with respect to the line x + y = 0, and m, n are on the circle x2 + Y2 + KX + my-4 = 0
The line x + y = 0 passing through the center O is obtained
Circle x2 + Y2 + KX + my-4 = 0, center O (- K / 2, - M / 2)
k=1,m=-1
Straight line y = x + 2 and circle x2 + Y2 + X-Y-4 = 0 (x + 1 / 2) 2 + (Y-1 / 2) 2 = 9 / 2
Center O (- 1 / 2,1 / 2), radius r = 3 / √ 2
The distance a from the center O (- 1 / 2,1 / 2) to the straight line y = x + 2 is
(-1/2-1/2+2)/√2=1/√2
mn=2*√(r2-a2)=2*√(9/2-1/2)=4
A set of real numbers greater than 3 and less than or equal to 1
Real numbers greater than 3 and less than or equal to 1 do not exist
So their set is empty
I'm glad to answer for you. The 1900 team will answer for you
Please click the [select as satisfactory answer] button below,