It is known that the image of positive scale function y = 2x and the image of linear function y = 4x-3 intersect at point P (see supplement for specific topic) It is known that the image of positive scale function y = 2x and the image of primary function y = 4x-3 intersect at point P, and there is a point a on the x-axis, so that s △ AOP = 4, the coordinates of point a can be obtained

It is known that the image of positive scale function y = 2x and the image of linear function y = 4x-3 intersect at point P (see supplement for specific topic) It is known that the image of positive scale function y = 2x and the image of primary function y = 4x-3 intersect at point P, and there is a point a on the x-axis, so that s △ AOP = 4, the coordinates of point a can be obtained

Let the coordinates of point a be (a, 0)
Y = 2x, y = 4x-3
Point P coordinates (3 / 2,3)
So the high of △ AOP is 3
S△AOP=1/2*｜a｜*3=4
The solution is a = 8 / 3 or - 8 / 3
What is the number of solutions for lgx + LG (x-1) = LG a (a > 0)?
LGX（X－1）＝LGA
So x ＾ 2-x-a = 0
B ＾ 2-4ac = 1 4A greater than 0 (because a is greater than 0)
And X1 * x2 = - A is less than 0
There are two different signs
Consider the definition field of X, X is greater than 0, X-1 is greater than 0, so x is required to be greater than 1
So the negative root has to be discarded, so there is only one
Given the intersection of a positive proportion function and a first-order function with point A: (1,4), and the intersection of the image of a first-order function and Y axis with point B: (0,6), find the expressions of two functions
Let the positive proportion function be y = KX, and the primary function be y = ax + B. take point a into the two functions and get k = 4
So the positive proportion function y = 4x can also get a + B = 4, and then (0,6) is brought into y = ax + B
B = 6 and a = - 2, so y = - 2x + 6
Process of solving equation (lgx) ^ 2-lgx ^ 2 = 3
（lgx）^2-lgx^2=3
Let lgx = t, T ^ 2-2t-3 = 0; t = 3 or T = - 1;
So x = 1000, or 0.1
（lgx）^2-lgx^2=3
（lgx）^2-2lgx-3=0
（lgx-3)(lgx+1)=0
lgx=3
x=100
Or lgx = - 1
x=0.1
The image of the first-order function passing through point a (0,3) intersects with the image of the positive scale function y = 2x in B (1,2). The equation that can represent the image of the first-order function is ()
A.2x-y+3=0 B.x-y-3=0
C.2y-x+3=0 D.x+y-3=0
For answers, explain why
Supplement: this one-time function has passed (0,3)
y=kx+b
Over ab
Then 3 = 0 + B
2=k+b
So B = 3, k = 2-B = - 1
y=-x+3
That is, x + Y-3 = 0
Choose D
F (x) is a piecewise function. When x is greater than or equal to 3 / 2, f (x) = lgx. When x is less than 3 / 2, f (x) = LG (3-x),
What is the title for? Is it for K range?
If it is
The solution is as follows:
From F (x) = k, the range of K is the range of F (x)
When x ≥ 3 / 2, f (x) = lgx ≥ LG (3 / 2)
When x < 3 / 2, 3-x > 3 / 2, f (x) = LG (3-x) > LG (3 / 2)
Because the definition field of function is x ≥ 3 / 2 and x < 3 / 2, it is r
Therefore, the value range is also the combination of the above two ranges, i.e. [LG (3 / 2), + ∞]
That is, the range of K [LG (3 / 2), + ∞)
Can there be only one intersection point between positive scale function and inverse scale function?
A mathematical problem. → known positive proportion function y = 4x, inverse proportion function y = x / K. These two functions can only have one intersection? All request coordinates. If not, please explain the reason
First of all, the inverse proportion function should be y = K / X. if so, it should be divided into two cases. Because the positive proportion function y = 4x must pass through one or three quadrants, it depends on whether the K in the inverse proportion function y = K / X is positive or negative. 1. If K is positive, then the inverse proportion function passes through one or three quadrants
F (x) is a piecewise function. When x is greater than or equal to 3 / 2, f (x) = lgx. When x is less than 3 / 2, f (x) = LG (3-x)
If x is greater than or equal to 3 / 2, f (x) = lgx. If x is less than 3 / 2, f (x) = LG (3-x)
If f (x) = k has a real solution, find the value range of K
k≥lg3/2
Given an intersection of the inverse and positive proportional functions, find the analytic expression and another intersection
It is known that the abscissa of an intersection of the image with the inverse scale function y = K / X (k is not equal to 0) and the image with the positive scale function y = 2x is 2. Find the analytic expression of the inverse scale function and the coordinate of another intersection
K is 8, and the other intersection is (- 2, - 4). Take x = 2 into y = 2x to get y = 4, so the intersection point is (2,4). This point satisfies both the positive proportion function and the inverse proportion function. Take y = 8 into y = K / X to get y = 8, and then find another intersection. First, solve the equation 8 / x = 2x, and the other intersection is (- 2, - 4) (of course, you can also look at the image. It's easy to look at the image. This method is special for lazy people,
Solving the equation lgx = - x + 1
Detailed process
Draw f (x) = LG x, f (x) = - x + 1 respectively
It can be seen that only the unique solution x = 1
F (x) = LG x is an increasing function
F (x) = - x + 1 is a decreasing function
There can be no more intersections
It's like dichotomy, isn't it???