If f (A-1) + F (3a-1) > 0, find the range of real number a

If f (A-1) + F (3a-1) > 0, find the range of real number a

Because f (x) is an odd function, and it is a decreasing function, it must pass through the origin; and if the condition result is greater than 0, then A-1 + 3a-1 must be less than 0, and A-1 and 3a-1 should be in the domain of definition, so the result of the three constraints is - 1 / 3
If f (A-1) > F (1-3a), we can find the value range of A
From the condition
-1
It is known that the function f (x) is a decreasing function in the domain [- 1.1], and f (x) is an odd function, and f (1-A) + F (1-2a)
f(1-a)+f(1-2a)>0
f(1-a)>-f(1-2a)
F (x) is an odd function
So - f (1-2a) = f [- (1-2a)] = f (2a-1)
So f (1-A) > F (2a-1)
F (x) is a decreasing function
So 1-a2
a>2/3
Another domain
-1
Given that the intersection parabola y ^ 2 = 4x passing through the point m (4,0) is at two points a and B, the position relationship between the circle with the diameter of line AB and the origin is
The origin is on the circle
When ab ⊥ X axis
AM=BM=4=1/2AB
∴∠AOB=90°
O on a circle with the diameter of line ab
Select, just judge the special position
If it's a calculation
It is also necessary to prove that when AB is not perpendicular to the x-axis
Let AB: y = K (x-4)
Substitute y ^ 2 = 4x to get
k^2x^2-(8k^2+4)x+16k^2=0
Let a (x1, Y1), B (X2, Y2)
Vector OA · vector ob
x1x2+y1y1
=x1x2-2√x1*2√x2
=x1x2-4√(x1x2)
Weida theorem
=16-4*4
=0
Ψ vector OA ⊥ vector ob
∴∠AOB=90°
O on a circle with the diameter of line ab
Let s be a nonempty set of real numbers satisfying the following conditions: (1) 1 does not belong to s; (2) if a ∈ s, then 1 / (1-A) ∈ s
Let s be a nonempty set of real numbers satisfying the following conditions: (1) 1 does not belong to s; (2) if a ∈ s, then 1 / (1-A) ∈ s
(1) : 0 is an element in set S. why?
(2) If 2 ∈ s, try to determine a consistent set s
(3) How many elements are there in set s at least? Prove your conclusion
1) 0 is not an element in the set s, because if it is, then: 1 / (1-0) = 1 ∈ s and: 1 are not s contradictions. 2) 2 ∈ S1 / (1-2) = - 1 ∈ S1 / (1 - (- 1)) = 1 / 2 ∈ S1 / (1-1 / 2) = 2 ∈ S. therefore, a consistent set s = {2, - 1,1 / 2} 3) from 2) we can see that three elements can form S0 elements
If a straight line with an inclination angle of π 4 crosses the parabola y2 = 2x at two points m and N through point a (1,0), then | Mn | = n___ .
The linear equation is y = X-1, the simultaneous equation is y = X-1, y2 = 2x, the solution is m (2-3, 1-3), n (2 + 3, & nbsp; 1 + 3), so Mn = 26, so the answer is 26
Let s be a set of real numbers satisfying the following two conditions: 1.1 does not belong to s; 2. A belongs to s, then (1 / 1-A) belongs to s
1. If all the items in the sequence {2 * (- 1) ^ n are in s, find the set s * with the least number of elements in S;
2. Take any three elements a, B, C in s * and find the probability of ABC = - 1;
3. Must the number of elements in s be 3N (belonging to n *)? If so, please give proof; if not, try to explain the reason
1(2 ,1/2,-1)
2 100%
3 is a (1 / 1-A) (A-1 / a)
If the line passing through the focus of parabola y square = - 2x and perpendicular to the line y = 2x intersects the parabola at two points m and N, then the midpoint coordinates of line segment Mn are
Let y = ax + B be the focus of the parabola y square = - 2x and the line perpendicular to the line y = 2x, then y = - X / 2-1 / 4 ------ > 2x + 4Y + 1 = 0 2x + 4Y + 1 = 0y & # 178; = - 2x solve the equations: X1 = - (9 + 4 √ 5) / 2
The answer is (4.5, - 2)
The focus of Y & # 178; = - 2x is (- 1 / 2,0),
The slope of the line perpendicular to y = 2x is - 1 / 2,
Then the straight line is y = - 1 / 2x-1 / 4, and the simultaneous equations of Y & # 178; = - 2x,
The solution is m (- 9 / 2 + 2 √ 5, 2 - √ 5), n (- 9 / 2-2 √ 5, 2 + √ 5),
Then the midpoint of Mn is (- 9 / 2, 2).
(- 9 / 2, 2)
Let a be a set of real numbers and satisfy the following conditions: if a belongs to a, then 1 / 1-A belongs to a (a is not equal to 1). Proof: set a cannot be a single element set
If a belongs to a, then 1 / (1-A) belongs to a
If a and 1 / (1-A) are not equal, there will be at least two elements in a
Let a = 1 / (1-A) be a & # 178; - A + 1 = 0
Because △ = 1-4 = - 3
It can be proved by the counter evidence
Let the line y = KX + 1 and the circle x2 + Y2 + KX + my-4 = 0 intersect at two points m and N, and m and N are symmetric with respect to the line x + y = 0. The inequality system KX − y + 1 ≥ 0kx − my ≤ 0y ≥ 0 denotes the area of the plane region
Because m and N are symmetric with respect to x + y = 0, and the line y = KX + 1 is perpendicular to the line x + y = 0 and bisected by the line, the equation of the line Mn is y = x + 1. According to the property of the intersection of the line and the circle, x + y = 0 passes through the center of the circle x2 + Y2 + KX + my-4 = 0, so substituting k = 1, M = - 1 into the system of inequalities, we get x − y + 1 ≥ 0x + y