The solution set of inequality x + 2 + 2x-1 > A of X is a, and set B = {X - 1 ≤ x ≤ 3}. If a ∩ B ≠ &;, then the value range of real number a is

The solution set of inequality x + 2 + 2x-1 > A of X is a, and set B = {X - 1 ≤ x ≤ 3}. If a ∩ B ≠ &;, then the value range of real number a is

A
If the set M = {x ax2-2 (a + 1) X-1 > 0}, m ≠ & # 8709;, M is contained in (0, positive infinity), then the value range of real number a is obtained
The inequality ax ^ 2-2 (a + 1) X-1 > 0 has solutions and only has positive solutions
When a = 0, the inequality is - 2x-1 > 0, then: x0, that is A0
A
Given the real number a > 0, the function f (x) = ax (X-2) 2 (x ∈ R) has a maximum of 32. (1) find the value of real number a; (2) find the monotone interval of function f (x)
(1) ∵ f (x) = ax (X-2) 2 = ax3-4ax2 + 4ax, ∵ f ′ (x) = 3ax2-8ax + 4a. From F ′ (x) = 0, we get 3ax2-8ax + 4A = 0. ∵ a ≠ 0, ∵ 3x2-8x + 4 = 0. The solution is x = 2 or X = 23. ∵ a ＞ 0, ∵ x ＜ 23 or X ＞ 2, f ′ (x) ＞ 0; 23 ＜ x ＜ 2, f ′ (x) ＜ 0
When the value of real number a is, the equation AX = ㏑ X has no solution,
Finding the tangent of y = LNX passing through the origin (0,0)
Let the tangent point be (x0, Y0)
y'=1/x k=1/x0=y0/x0 y0=1 y0=lnx0 x0=e
The tangent point is (E, 1)
When a = 1, ax = ㏑ X has a solution
When 0
a=LnX/X;
Let f (x) = LNX / X;
F'(X)=(1-LnX)/X;
So f (x) increases in (0, e) and decreases in (E, infinity);
F(e)=1/e；
When x approaches zero, f (x) approaches negative infinity;
When x approaches infinity, f (x) approaches 0;
It can be seen from the image that when a has two solutions at (0,1 / E), a has no solution at (1 / E, positive infinity); when a = 1 / E, a has one solution at [0, negative infinity]. ... unfold
a=LnX/X;
Let f (x) = LNX / X;
F'(X)=(1-LnX)/X;
So f (x) increases in (0, e) and decreases in (E, infinity);
F(e)=1/e；
When x approaches zero, f (x) approaches negative infinity;
When x approaches infinity, f (x) approaches 0;
It can be seen from the image that when a has two solutions at (0,1 / E), a has no solution at (1 / E, positive infinity); when a = 1 / E, a has one solution at [0, negative infinity]. Put it away
When a = 0, there is a solution, x = 1
When a > 0, there is no solution
When a
Given that a, B and C are in equal proportion sequence, then the number of intersections between the image and X axis of quadratic function f (x) = AX2 + BX + C is ()
A. 0b. 0 or 1C. 1D. 2
From a, B, C into an equal ratio sequence, we get B2 = AC, and AC > 0, let AX2 + BX + C = 0 (a ≠ 0), then △ = b2-4ac = ac-4ac = - 3aC < 0, so the number of intersections between the image of function f (x) = AX2 + BX + C and X axis is 0
If the equation LG (AX-2) - LG (x + 1) = 1 has a solution, then the value range of a is 0
lg(ax-2)-lg(x+1)=1
Logarithmically significant, AX-2 > 0 ax > 2 x + 1 > 0 x > - 1
lg[(ax-2)/(x+1)]=1
(ax-2)/(x+1)=10
ax-2=10x+10
(a-10)x=12
There is no solution when a = 10
When a is not equal to 10, x = 12 / (A-10) > - 1
A
The quadratic function FX = ax ^ 2 + BX + C (a > 0 and C > 0) has two different intersections between the image and the X axis, one of which is (C, 0) when 0 < x < C, there is always FX > 0
The area of the triangle formed by the intersection of the image of quadratic function and the coordinate axis is 8
Find the value range of a
According to the meaning of the question, the intersection of the function and the x-axis is on the right side of the y-axis, and the point (C, 0) can only be the left side of the intersection of the function and the x-axis, which is denoted as x1, and the right side is denoted as X2, then: X1 + x2 = - B / A, that is, C + x2 = - B / A, so the intersection of the left side of the x2-c = - B / a-2c function and the x-axis is ((- b-sqrt (b ^ 2-4ac)) / 2a, 0), so C = (- b-sqrt (b
Let me tell you the general idea: Method 1: first, the quadratic function a > 0, opening upward, has two different intersections with the X axis, then the third intersection is on the Y axis, and the coordinates are (0, c), The triangle formed can be regarded as half of the parallelogram, and the height is C. when 0 < x < C and FX > 0, the two intersections of the function are on the same side of the y-axis. Compare the size of C and another intersection (which should be in the positive direction of the x-axis). The short side of the parallelogram is the difference between the two intersections, and the long side is the coordinate of the larger value of the intersection
Method 2: using Veda's theorem to derive (x1-x2) expansion about
Let me tell you the general idea: Method 1: first, the quadratic function a > 0, opening upward, has two different intersections with the X axis, then the third intersection is on the Y axis, and the coordinates are (0, c), The triangle formed can be regarded as half of the parallelogram, and the height is C. when 0 < x < C and FX > 0, the two intersections of the function are on the same side of the y-axis. Compare the size of C and another intersection (which should be in the positive direction of the x-axis). The short side of the parallelogram is the difference between the two intersections, and the long side is the coordinate of the larger value of the intersection
Method 2: the relation of (x1-x2) with respect to a, B and C is derived by using Weida's theorem;
Method 3: using the area, using two right triangle area difference of only 8 can be calculated
If all roots of equation LG (AX) times LG (a times the square of x) = 4 are greater than 1, the value range of a is obtained
LG (AX) * LG (AX ^ 2) = 4 (LGA + lgx) * (LGA + lgx ^ 2) = 4 a > 0 (LGA + lgx) * (LGA + 2lgx) = 42 (lgx) ^ 2 + 3lga * lgx + (LGA) ^ 2-4 = 0 and lgx = t because all solutions of LG (AX) * LG (AX ^ 2) = 4 are greater than one, so x > 1, so t = lgx > LG1 = 0, that is, equation 2T ^ 2 + 3lga * t + (LGA) ^ 2 -
Given the quadratic function f (x) = ax + BX + C, the image is symmetric about the y-axis, f (x) ≤ 1 is constant for X ∈ R, and f (x) = 0, find the analytic expression of F (x)
Because the image is symmetric about the Y axis, B = 0,
Because f (x)
From the quadratic function f (x) = AX2 + BX + C image on the y-axis symmetry, we can get b = 0, f (x) ≤ 1 is constant in X ∈ R, we can get C = 1, a < 0.
If all solutions of the equation LG (AX) * LG (a ^ 2x) = 2 about X are greater than 1, the value range of a is obtained
The original equation of LG (AX) = LGA + lgx, LG (a ^ 2x) = 2lga + lgx can be reduced to: [LGA + lgx] [2lga + lgx] = 2, because x > 1, so lgx > 0, let t = lgx, then the meaning of the problem is equivalent to: equation [T + LGA] [T + 2lga] = 2