If f (1-A) + F (1-a2) > 0, the range of a is obtained

If f (1-A) + F (1-a2) > 0, the range of a is obtained

∵ f (x) is an odd function, ∵ f (1-A) + F (1-a2) > 0 can be transformed into f (1-A) > - f (1-a2) = f (A2-1), and f (x) is increasing in the domain (- 1, 1), ∵ 1 − a > A2 − 1 − a < 1 − a < 1 − A2 < 1, that is − 2 < a < 10 < a < 2 − 2 < a < 0 or 0 < a < 2, the solution is 0 < a < 1
It is known that the odd function f (x) increases monotonically over the domain (- 1.1) and has f (1-A) + F [(1 / 2) - 2A]
So f (x) = - f (- x),
So f (1-A) + F [(1 / 2) - 2A] 0.5
And 2a-0.5, 1-A are within (- 1.1)
75 > a > 0.5
It is known that the parabola y2 = 2px (P > 0) intersects with the straight line y = - x + 1 at two points a and B. the circle with chord length AB as diameter just passes through the origin. The equation of the parabola is obtained
Let a (x1, Y1), B (X2, Y2), y = - x + 1, x = 1-y, then: y2 = 2p (1-y), Y2 + 2py-2p = 0, Y1 + y2 = - 2p, y1y2 = - 2p, x1x2 = (1-y1) (1-y2) = 1 - (Y1 + Y2) + y1y2 = 1 + 2p-2p = 1. ∵ the circle with the diameter of string AB just passes through the origin, ∵ OA ⊥ ob, ∵ x1x2 + y1y2 = 0 ∵ x1x2 + y1y2 = 1-2p = 0 ∵ P = 12. ∵ the equation of parabola is: y2 = X
Given the set a = {x | (x, y) | y = | x-a |}, B = {(x, y) | (x-1) 2 + (Y-2) 2 = 2}, if there are only two elements in a ∩ B, then the range of real number a is
My answer is - 3
The line L passing through P (0,1) and parabola y ^ 2 = 2x intersect at two points m, N, O as the origin. If Kom + Kon = 1, then the equation of line L is
Let m (x1, Y1), n (X2, Y2)
The linear equation is y = KX + 1
From Kom + Kon = 1, Y2 / x2 + Y1 / X1 = 1
That is, (kx2 + 1) / x2 + (kx1 + 1) / X1 = 1
It is reduced to 2K + (x1 + x2) / x1x2 = 1
From the linear equation and parabolic equation
k^2*x^2+(2k-2)*x+1=0
The K is calculated by the Veda theorem
What's the meaning of Kom + Kon = 1?
We know that the real number set a satisfies: if x ∈ a, and X is not equal to positive and negative 1 and 0, then 1-x / 1 + X ∈ a, let positive and negative 1 and 0 not belong to a, non empty set a
It is known that the real number set a satisfies: if x ∈ a, and X is not equal to positive and negative 1 and 0, then 1-x / 1 + X ∈ a, let positive and negative 1 and 0 not belong to a, how many elements are there in the nonempty set a?
One, ∵ when 1 + X / 1-x = 0, it does not hold, and it is an empty set. When 1 + X / 1-x = - 1, 0 = 0, then x takes any real number that is not ± 1,0. At this time, there is only one element in A. when 1 + X / 1-x = 1, that is, 1 + x = 1-x, that is, 2x = 0, x = 0 does not hold
The line L with slope 1 and parabola y ^ 2 = 2x intersect at two points a and B, and the circle with diameter AB passes through the origin to find the equation of line L
If x belongs to a, and X is not equal to 1 and 0, then (x + 1) / (1-x) belongs to A. when x belongs to a, set a also has elements ()
X∈A ,(X+1)/(1-X) ∈A ,T=(X+1)/(1-X),（T+1）/（T-1）= -1/X∈A〈（-1/X）+1〉/〈1-（-1/X）〉=（X-1）/（X+1）∈AS=（X-1）/（X+1）,(S+1)/(1-S)=X∈A∴X∈A ,(X+1)/(1-X) ∈A ,-1/X∈A,（X-1）/（X+1）∈A...
Let me give you a hint, (1 + TaNx) / (1-1 * TaNx) = Tan (45 + x) can know up to 4 elements. You just count four, and the fifth will loop (and it does) x, (1 + x) / (1-x), - 1 / x, (x-1) / (1 + x)
Given the parabola y ^ 2 = 4x, the straight line passing through the point (0, - 2) intersects the parabola at two points a and B, and O is the origin. If the vertical bisector of line AB intersects the X axis at the point (n, 0), the value range of N can be obtained
The linear equation y + 2 = KX, let a (x1, Y1), B (X2, Y2) be substituted into the parabolic equation, then by subtraction, Y1 ^ 2-y2 ^ 2 = 4 (x1-x2), (y1-y2) / (x1-x2) = 4 / (Y1 + Y2) = k, let m (x0, Y0), Y0 = (Y1 + Y2) / 2 = 2 / K be substituted into the linear equation, then x0 = 2 / K / K + 2 / K,
The vertical equation Y-2 / k = - 1 / K (X-2 / K / K-2 / k), then n = 2 / K / K + 2 / K + 2 = 2 (1 / K + 1 / 2) ^ 2 + 1.5
The linear equation is replaced by the parabolic equation, (4 / k) ^ 2-4 (- 8 / k) > 0, (1 / K + 2) (1 / k) > 0,
So 1 / K is smaller than - 2 or larger than 0, and substituting n, we get
N is larger than 2
N>6
Given the function f (x) = asin (π x + α) + bcos (π x + β), where a, B, α and β are all non-zero real numbers. If f (2008) = - 1, what is f (2009) equal to
OK,
F (x) = asin (π x + α) + bcos (π x + β) f (2005) = asin (2008 π + α) + bcos (2008 π + β) = asin (α) + bcos (β) = - 1asin (α) + bcos (β) = - 1, then - asin (α) - bcos (β) = 1F (2009) = asin (2009 π + α) + bcos (2009 π + β) = asin (π + α) + bcos (π + β) = - ASI