Given that the domain of the function f (x) = 2Sin ^ 2x - (2 radical 3) sinxcosx-1 + radical 3 is [0, Pai / 2], then the value range of y = f (x) is?, and the zero point is?

Given that the domain of the function f (x) = 2Sin ^ 2x - (2 radical 3) sinxcosx-1 + radical 3 is [0, Pai / 2], then the value range of y = f (x) is?, and the zero point is?

f(x)=1-cos2x-√3sin2x-1+√3
=√3-2sin(2x+π/6)
Because 0 ≤ x ≤ π / 2
So π / 6 ≤ 2x + π / 6 ≤ 7 π / 6
So - 1 / 2 ≤ sin (2x + π / 6) ≤ 1
So f (x) ∈ [√ 3-2, √ 3 + 1]

If the vertex of the image of the quadratic function y = x2-2 (a + b) x + C2 + 2Ab is on the x-axis, and a, B, C are the three sides of △ ABC, then △ ABC is () A. Acute triangle B. Right triangle C. Obtuse triangle D. Isosceles triangle

The formula of the analytic formula of quadratic function is: y = [x - (a + b)] 2 + C2 + 2Ab - (a + b) 2 = [x - (a + b)] 2 + c2-a2-b2
The vertex is (a + B, c2-a2-b2)
According to the meaning of the title, c2-a2-b2 = 0
The △ ABC is a right triangle
Therefore, B is selected

It is known that the image of the quadratic function y = 2x2-4mx + M2 has two intersections a B and the vertex is C. if the area of the triangle ABC is 4 and the root is 2, then M=___ ... It is known that the image of the quadratic function y = 2x2-4mx + M2 has two intersections a B and the vertex is C. if the area of the triangle ABC is 4 and the root is 2, then M=___ (it is better to have a problem-solving process)

The distance formula between the two intersection points of parabola and X axis is ab = √Δ / a = 2 √ (2m ^ 2) / 2 = √ (2m ^ 2)
Formula y = 2x ^ 2-4mx + m ^ 2 = 2 (x-m) ^ 2-m ^ 2,
So the vertex is (m, - m ^ 2)
So △ ABC area = (1 / 2) * AB * m ^ 2 = 4 √ 2,
M = ± 2

Let the quadratic function y = x ^ 2 + 2aX + A ^ 2 / 2 (a

Vertex (- A, - A ^ 2 / 2)
Y = x ^ 2 + 2aX + A ^ 2 / 2 = 0
X = [- 2 + (radical 2)] A / 2 x '= [- 2 - (radical 2)] A / 2
Then B {[- 2 + (radical 2)] A / 2,0} C {[- 2 - (radical 2)] A / 2,0}, a

Let the vertex of the quadratic function y = f (x) be at (0, - 3), and there is a root sign 3. The image of the inverse scale function y = g (x) is in a three quadrant, and has exactly two common points with the image of the function y = f (x), and find the analytic formula of the function f (x) = f (x) + G (x)

Let the vertex of the quadratic function y = f (x) be at (0, - 3), and one root is the root sign 3, then we can find that the image of the inverse scale function y = g (x) of F (x) = x ^ 2-3 is in a three quadrant, then let g (x) = K / X (k > 0), Let f (x) = g (x), then x ^ 2-3 = K / x, that is, x ^ 3-3x-k = 0, let H (x) = x ^ 3-3x-k, find the first order

Quadratic function y = 1 6（x+2 3) The vertex of the image of 2 is a, which intersects with the y-axis at point B. with ab as the edge, an equilateral triangle ABC is made in the second quadrant (1) Find the expression of line AB and the coordinates of point C (2) Point m (m, 1) is in the second quadrant, and the area of △ ABM is equal to the area of △ ABC (3) Take the point n on the X axis as the center of the circle, 1 as the radius of the circle, and tangent to the circle with the point C as the center and the length of CM as the radius

(1) Quadratic function y=
One
Six
（x+2
Three
）The vertex a (- 2) of the image of 2
Three
The intersection point B (0, 2) with the Y axis,
Let the expression of line AB be y = KX + B (K ≠ 0),
K can be obtained=
Three
Three
So the expression of line AB is y=
Three
Three
x+2．
We can get ∠ Bao = 30 ° and ∵ BAC = 60 °,
∴∠CAO=90°．
In RT △ Bao, ab = 4
/ / AC = 4. Point C (- 2
Three
，4）．
(2) ∵ points c and m are in the second quadrant, and the area of ∵ ABM is equal to the area of ∵ ABC,
∴CM∥AB．
Let the expression of the line cm be y=
Three
Three
X + m, point C (- 2)
Three
4) on the straight line cm,
M = 6
The expression of the line cm is y=
Three
Three
x+6．
The coordinates of M can be obtained: (- 5)
Three
，1）．
(3) By C (- 2
Three
，4）、M（-5
Three
1) we can get the following results: 1
CM=
(−2
Three
+5
Three
)2+(4−1)2
=6．
① When ⊙ C and ⊙ n are circumscribed, CN = cm + 1 = 7;
In RT △ can, an=
CN2−CA2
=
72−42
=
Thirty-three
;
∴ON=AN+OA=
Thirty-three
+2
Three
Or on = an-oa=
Thirty-three
-2
Three
That is, the coordinates of point n are: (-
Thirty-three
-2
Three
，0）、（
Thirty-three
-2
Three
，0）．
② When ⊙ C and ⊙ n are inscribed, CN = cm-1 = 5;
In RT △ can, CN = 5, CA = 4, then an = 3;
∴ON=AN+OA=3+2
Three
Or on = oa-an = 2
Three
-3
That is, the coordinate of point n is: (- 3-2)
Three
，0），（3-2
Three
，0）．
To sum up, we can see that:
Coordinates of point n (- 3-2)
Three
，0），（3-2
Three
，0），（-
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Problem analysis
(1) Given the analytic formula of parabola, the coordinates of its vertex and the intersection point of function image and y-axis are easy to obtain. When calculating the coordinates of point C, we should grasp the particularity of RT △ AOB (including 30 ° angle). Obviously, if △ ABC is an equilateral triangle, then AC is perpendicular to x-axis. It is easy to find the coordinates of point C whether we use Pythagorean theorem to calculate the side length or point B is on the vertical line of AC
(2) "Point m is in the second quadrant" determines the approximate range of point M. if "the area of △ ABM is equal to the area of △ ABC", then the distance from point C and point m to line AB is the same, that is, cm ‖ ab. the analytic formula of line AB is easy to find, and the slope is the same if the two lines are parallel, The conclusion can be drawn by substituting the ordinate of point M
(3) If ⊙ C is tangent to ⊙ n, two cases should be considered: ① circumscribed, CN length is equal to the sum of radius of two circles; ② inscribed, CN length is equal to the radius difference of two circles
In the case of definite CN length, the length of an can be obtained by Pythagorean theorem in RT △ can, and then the coordinates of point n can be determined
Comments of famous teachers
Test points:
Quadratic function synthesis problem
Comments on examination sites:
This problem of quadratic function covers Pythagorean theorem, the method of finding the area of a graph, the position relationship between circles and so on. In the last small problem, we must take both the external tangent and the internal tangent into account to avoid missing the solution
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As shown in the figure, the image of the first-order function y = - the root of three-thirds, x + B and the x-axis and y-axis respectively intersect at two points of A. B. take the line segment AB as the edge, and make the equilateral △ in the first quadrant If there is a point P (a, half) in the second quadrant, the area of the quadrilateral ABPO is expressed by the algebraic expression containing a, and the value of a when the area of triangle ABC is equal to that of triangle ABC is obtained

The area of quadrilateral ABPO: S1 = 1 / 2 * ob * a + 1 / 2 * √ 3 * b = 1 / 2 * B (a + √ 3)
Area of triangle ABC: S2 = √ 3 * B
When S1 = S2, a = √ 3
Note: I change "and find the value of a when the area of triangle ABC is equal to that of triangle ABC" as follows: "find the value of a when the area of triangle ABC is equal to that of quadrilateral ABPO."

As shown in the figure, the image of the first order function y = - 2 / 3x + 2 intersects points a and B with X axis and Y axis respectively, and takes line segment AB as edge in the first quadrant, As an isosceles right angle △ ABC, the angle BAC = 90 degrees The third question: if point D and point B are symmetric about the x-axis, is there a point P on the x-axis, which is the maximum of pc-Pd? If it exists, find the coordinates of P! If not, please explain the reason! Big gate, please! A reward of 100 points

The image of the first order function y = - 2 / 3x + 2 intersects points a and B with X axis and Y axis respectively, and takes line AB as edge in the first quadrant,
A=(3,0) B=(0,2)
Isosceles right angle △ ABC, angle BAC = 90 ° know AC linear equation y = 3x / 2 --- 9 / 2
Slope of straight line BC k = 1 / Tan (oba + CBA) = --- 1 / 5
Linear BC equation y = -- 1 / 5x + 3 points c = (5,3)
Point D and point B are symmetric about the x-axis. Point d = (0, -- 2) line segment CD = √ (5 --- 0) ^ 2 + (-- 2 -- 3) ^ 2 = 5 √ 2
Let P = (x, 0) △ PDC
Pc-Pd = √ (X-5) ^ 2 + 9 - √ x ^ 2 + 4
=(---10x+30)/(√(x-5)^2+9+√x^2+4)
If and only if x = 0, pc-Pd is maximum

As shown in the figure: in the rectangular coordinate plane, the positive proportional function straight line y = radical 3x intersects with an inverse scale function image in the first quadrant, a point ab ⊥ X axis is at B, AB=6 ① Find the analytic expression of inverse proportional function ② In line ab The existence point P is above , make P to positive proportional function line OA The distance of is equal to P Arrive at the point Distance of B? If it exists, find point P If not, please explain the reason

(1) AB = 6, ordinate of a = 6,
Y = radical 3x = 6,
X = 2 root sign 3. Point a (2 radical 3,6) is on hyperbola again,
Let hyperbola y = K / x, substituting a coordinate,
K = 12 root sign 3
The analytic formula of inverse proportional function is y = 12 root sign 3 / X
(2) Take e point in OA, make ob = OE = 2, root number 3
According to the formula of distance between two points, and E is on the straight line, e (x, root sign 3x)
The result shows that 2 root sign 3 = x square under big root sign + square of root 3, then E (root number 3,3) is obtained
It is assumed that there exists a point P (2 radical 3, y) such that Pb = PE
Then the triangle ope congruent triangle OPB
Then the formula of distance between two points is used, PE = Pb,
P ordinate = 2
So the point P, P coordinates are (2 radical, 3, 2)

When x takes what value, a = under the root (8-x), B = under the root (3x-4), C = under the root (x = 2), the triangle ABC is a right triangle?

First of all, make the root sign meaningful, so that the formula in the root is greater than 0 to get 2