If both f (x) = - x2 + 2aX & nbsp; and G (x) = ax + 1 & nbsp; are decreasing functions in the interval [1,2], then the value range of a is () A. （-1，0）∪（0，1）B. （-1，0）∪（0，1]C. （0，1]D. （0，1）

If both f (x) = - x2 + 2aX & nbsp; and G (x) = ax + 1 & nbsp; are decreasing functions in the interval [1,2], then the value range of a is () A. （-1，0）∪（0，1）B. （-1，0）∪（0，1]C. （0，1]D. （0，1）

If f (x) = - x2 + 2aX is a decreasing function in the interval [1,2], then the image of a ≤ 1 function g (x) = ax + 1 & nbsp; is a hyperbola with (- 1,0) as the symmetry center. If G (x) = ax + 1 & nbsp; is a decreasing function in the interval [1,2], then a > 0. In conclusion, the value range of a is (0,1], so C is selected

Given a ∈ R, the function f (x) = x ^ 2-2ax + 5
If the inequality f (x) is greater than 0 and holds for any x ∈ (0, positive infinity), the value range of a is obtained

F (x) = x ^ 2-2ax + 5 > 0 holds for (0, + ∞)
∴x^2-2ax+5>0
x^2+5>2ax
∵x>0
∴x+5/x>2a
X + 5 / X ≥ 2 √ 5.. mean inequality
∴2√5>2a
a

It is known that the length of the major axis of the ellipse R: X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a > b > 0) is 4
It is known that the length of the major axis of the ellipse R: X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a > b > 0) is 4, and it passes through the point (∨ 3,1 / 2)
(1) Let a, B and m be the three points on the ellipse. If the vector om = 3 / 5, the vector OA + 4 / 5, the vector ob, the point n is the midpoint of the line AB, and the coordinates of C and D are (- ∨ 6 / 2,0) and (∨ 6 / 2,0), respectively, then we prove | NC | + | nd | = 2 ∨ 2
QAQ is really in a hurry. I need your help in an hour The key is to write the second question,

"\ /" is it the legendary √ bar (1) x ^ 2 / 4 + y ^ 2 = 1 (2) (vector om = 3 / 5, vector OA + 4 / 5, vector OB) that seems to tell you ∠ AOB = 90 ° from the topic to prove that the locus of point n in a and B is on the ellipse of a = √ 2, C = √ 6 / 2, that is, x ^ / 2 + y ^ 2 / (1 / 2) = 1, and the number is special, so it is estimated that

Given the set M = {(x, y)} | x-3 ≤ y ≤ X-1}, n = {P | PA ≥ 2PB, a (- 1,0), B (1,0)}, then what is the graphic area of M ∩ n?

1. Drawing: establish the coordinate system: m is the set of points (including points on the line) between the straight line y = X-1 and y = x-3. There is the distance between two points. Formula: root [(x1-x2) ^ 2 + (y1-y2) ^ 2] if the coordinate of P point is (x, y), PA ≥ 2PB can be expressed as: root [(x + 1) ^ 2 + y ^ 2] ≥ 2 * root [(x-1) ^ 2 + y ^ 2]

Inequality | x-3 | + | Y-3|

The area shown is the area of the diamond with the center at (3,3). The diagonal lines are perpendicular to each other, and the length is 6. Therefore, it is also a square, and the side length is 3 √ 2, so the area is 18

If the curve y = a [x] and the line y = x + a (a > 0) have two common points, then the value range of a is

The curve y = a | x | is divided into two parts
x> When y = 0, y = ax
x1

The area of the area enclosed by the curve and the two coordinate axes represented by equation (X-Y + 1) (x + Y-1) = 0 is

X-Y + 1 = 0, y = x + 1, x + Y-1 = 0, y = - x + 1, y = x + 1, y = - x + 1, intersection with (0,1) and X axis intersection with (0,1) respectively
(- 1,0) (X-Y + 1) (x + Y-1) = 0 indicates that the area enclosed by the curve and two coordinate axes is [1 - (- 1)] * 1 / 2 = 2 / 2 = 1

In triangle ABC, point D is on BC, and BC vector = 4bd vector. If Ad vector = m * AB vector + n * AC vector, find real numbers m and n
m=3/4 n=1/4

My vector is not marked with an arrow, you should also be able to understand
AD=AB+BD=AB+(1/4)*BC=AB+(1/4)*(AC-AB)=(3/4)*AB+(1/4)*AC
Do you understand? Vector is a word. It's a word. It's a word. It's a word. It's a word. It's a word

If vector a is parallel to vector B, then vector a equals vector B and vector a equals negative vector B
This sentence is right or wrong

Obviously wrong
"Vector a equals vector B, vector a equals negative vector B"
Vector a is equal to K times of vector b (k is any real number that is not zero)
Because vector a is parallel to vector B, it can only show that the directions of vector a and vector B are the same or opposite, but it does not show the quantitative relationship

For AP / Pb = 1 / 2, find the equation of line L at this time!
It is known that the circle C: x2 + (Y-1) 2 = 5, the line L: mx-y + 1-m = 0, and the line L intersect the circle at two points a and B. if the fixed point P (1,1) chord AB is AP / Pb = 1 / 2, the equation of the line l at this time can be solved!

A straight line L: mx-y + 1-m = 0, Y-1 = m (x-1) passes through a fixed point (1,1). Make a straight line parallel to the x-axis and passing through the center of the circle (0,1) and intersect at two points Mn. Obviously, PM = (√ 5) - 1, PN = (√ 5) + 1. According to the chord secant theorem, PM / PA = Pb / PN, PA · Pb = PA · 2PA = 2PA & # 178; = PM · PN = [(√ 5) - 1] ·