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Given the proposition p: "equation x2 + y2m = 1 represents the ellipse with focus on Y axis"; proposition q: "equation 2x2-4x + M = 0 has no real root". If P ∧ q is false and P ∨ q is true, the value range of real number m is obtained
If equation 2x2-4x + M = 0 has no real root, then △ = 16-8m ＜ 0, the solution is m ＞ 2, i.e. Q: m ＞ 2. Because P ∧ q is false and P ∨ q is true, then p, q is true and false. If P is true and Q is false, then 1 ＜ m ≤ 2. If P is false and Q is true, then M has no solution
It is known that the ellipse equation is that the square of quarter x plus the square of third y equals one, which is to determine the value range of M, so that for the straight line y = 4x + m, the ellipse is different
3x²+4y²=12
y=4x+m
So 67x & sup2; + 32mx + 4m & sup2; - 12 = 0
The discriminant of two different intersections is greater than 0
1024m²-1072m²+3216>0
m²
According to the elliptic equation: X & sup2 / 4 + Y & sup2 / 3 = 1, that is, 3x & sup2; + 4y2 = 12, ① linear equation y = 4x + m, ② simultaneous equations need to be considered in order to ensure that there are two different intersections between ellipse and straight line. Therefore, substituting formula ② into formula ①, there must be univariate quadratic equation: 67x & sup2; + 32mx + 4m & sup2; - 12 = 0, M & sup2; < 67
According to the elliptic equation: X & sup2 / 4 + Y & sup2 / 3 = 1, that is, 3x & sup2; + 4y2 = 12, ① linear equation y = 4x + m, ② simultaneous equations need to be considered in order to ensure that there are two different intersections between ellipse and straight line. Therefore, substituting formula ② into formula ①, there must be univariate quadratic equation: 67x & sup2; + 32mx + 4m & sup2; - 12 = 0, Then we can find out the value range of M, which is the open interval: (- √ 67, √ 67). Put it away
Why sin2a + sin2b = sin2c leads to sin (a + b) cos (a-b) = 2sinccosc
Sin2a + sin2b = sin2c, sin (a + b) cos (a-b) = 2sinccosc answer: because sin (a + b) = sinacosb + sinbcosa cos (a-b) = cosacosb + sinasinb, sin (a + b) cos (a-b) = (sinacosb + sinbcosa) (cosacosb + sinasinb) finishing = sin2a + sin2b
If the image of quadratic function y = (M + 5) x2 + 2 (M + 1) x + m is all above the x-axis, then the value range of M is___ .
∵ the images of quadratic function y = (M + 5) x2 + 2 (M + 1) x + m are all above the x-axis, ∵ (M + 5) > 0, ∵ 0, ∵ m ＞ - 5, 4 (M + 1) 2-4 (M + 5) × m ＜ 0, so m ＞ 13
This formula proves that sin (a + b) sin (a-b) = sin2a-sin2b 2 represents the square
sin(a+b)=sinacosb+sinbcosa
sin(a+b)=sinacosb-sinbcosa
sin(a+b)sin(a-b)=(sin²acos²b-sin²bcos²a)
=[sin²a(1-sin²b)-sin²b(1-sin²a)]
=sin²a-sin²b
When m is a value, the function y = - (m-2) x ^ m & # 178; - 2 is a quadratic function
m-2≠0
m²=2
It is a quadratic function when m = ± √ 2
The proof of sin (a + b) sin (a-b) = sin2a-sin2b
It is proved that: because 2A = (a + b) + (a-b) B = (a + b) - (a-b) is substituted by the left formula sin2a = sin [(a + b) + (a-b)] = sin (a + b) cos (a-b) + cos (a + B) sin (a-b) sin2b = sin [(a + b) - (a-b)] = sin (a + b) cos (a-b) - cos (a + b) sin (a-b), we can get sin2a + sin2b = 2Sin (a + b) cos (a-b)
It is known that the positive and negative scale function images intersect at the point (- 2,1). The analytic expressions of these two functions and the coordinates of their other intersection are obtained
Method 1: let y = ax, y = B / X ∵ two images intersect at (- 2,1) ∵ when x = - 2, y = 1; 1 = - 2A; 1 = B / - 2 ∵ a = - 0.5, B = - 2 ∵ y = - 0.5x, y = - 2 / X ∵ when two images intersect, y = - 0.5x, y = - 2 / X ∵ X1 = - 2, Y1 = 1 (rounding); x2 = 2, y2 = - 1
Let y = KX, substitute the point (- 2,1), and get the analytic formula y = - 1 / 2x
Let y = K / x, substitute the point (- 2,1), and get y = - 2 / x, simultaneous equations, and get the coordinates of another intersection (2, - 1)
Let y = KX, substitute the point (- 2,1), get y = - 1 / 2 times x, let y = K / x, substitute the point (- 2,1), get y = - 2 / x, simultaneous equations, get the coordinates of another intersection (2, - 1)
Why sin2a + sin2b = 2Sin (a + b) cos (a + b)
sin2a+sin2b
=sin[(a+b)+(a-b)]+sin[(a+b)-(a-b)]
=sin(a+b)cos(a-b)+sin(a-b)cos(a+b)+sin(a+b)cos(a-b)-sin(a-b)cos(a+b)
=2sin(a+b)cos(a+b)
Sin2a + sin2b = 2Sin (a + b) cos (a-b)
There is 2sinxcosx = sin2x. Then change x to a + B
How to list the inverse scale function image according to the analytic expression
For example, if y = 1 / x, I don't know how to list and what numbers to look for
In the inverse proportion function, X and y can take any number and satisfy the relation
For example: x = 1, y = 1
X = 2, y = 1 / 2
X = negative 1, y = negative 1
X = minus 2, y = minus 1 / 2
wait.
RELATED INFORMATIONS
- 1. It is known that the square of ellipse x plus the square of 4Y is equal to 1 and the line y is equal to x plus M. when the line and ellipse have a common point, the value range of real number m is calculated? Don't you understand?
- 2. It is known that the ellipse G: X6 ^ 2 / 4 + y ^ 2 = 1. Through the point (m, 0), make the tangent l of the circle x ^ 2 + y ^ 2 = 1 and intersect the ellipse g at two points a and B It is known that the ellipse G: X6 ^ 2 / 4 + y ^ 2 = 1. The tangent l of circle x ^ 2 + y ^ 2 = 1 intersects the ellipse g at two points a and B through the point (m, 0). (1) find out the focus coordinate and eccentricity of the ellipse g. (2) express labl as a function of M, and find out the maximum value of labl
- 3. From the point a (- 1,4), make the tangent l of the square of the circle (X-2) + (Y-3) = 1, and find the tangent length From point a (- 1,4) make the tangent l of the square of the circle (X-2) + (Y-3) = 1
- 4. Ellipse X & # 178 / M + Y & # 178 / 10 = 1 has the same focus as hyperbola Y & # 178; - X & # 178 / b = 1
- 5. It is known that ellipse C1 and hyperbola C2 have the same focus F1 (- 3,0) and F2 (3,0), and both pass through D (2, radical 2) Let a (x, y) be the point on the ellipse C1. If the minimum distance between a and B (x0,0) (x0 > 0) is not less than 2 root sign 3 / 3, the value range of x0 is obtained
- 6. Log 3 is the logarithm of the bottom 2 / 3. Can we calculate the specific number?
- 7. The function y = 3sin (2x + π / 6) is the closest axis of symmetry to the Y axis
- 8. Given that the maximum value of function f (x) = asin ^ 2 (Wx + α) (a > 0) is 2 and the distance between two adjacent symmetrical axes is 2, then f (1) + F (2) +... + F (100)= RT,
- 9. Let f (x) = asin (ω x + φ), X ∈ R (where a > 0, ω > 0,0)
- 10. The known function f (x) = asin (ω x + φ) (a > 0, ω > 0, - π / 2)
- 11. The eccentricity of the ellipse with focus on the x-axis is two-thirds root sign three, and the ellipse intersects with (X-2) ^ + (Y-1) ^ = 5 / 2 at two points a and B. the length of line AB is equal to the diameter of the circle 1. Find the equation of line ab 2. Find the equation of ellipse
- 12. If the point (m, n) is on the graph of function y = 2x + 1, then the value of 2m-n is () A. 2B. -2C. 1D. -1
- 13. Find the coordinates of the intersection point of the image of the function y = 2x + 1 and y = - 3x-4
- 14. When the value of the first-order function y = (4 + 2m) x + M-4 is known, the image passes through one three four quadrants
- 15. The first-order function y = (2m-1) + X + (3 + n) is used to find the value range of m.n through one two three quadrants
- 16. If the image of the linear function y = (2-m) x + m passes through the first, second and fourth quadrants, then the value range of M is______ .
- 17. As shown in the figure, in the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length of 1 are on the Y-axis and X-axis respectively, As shown in the figure, in the plane rectangular coordinate system, the two vertices a and C of the square oabc with side length of 1 are respectively on the positive half axis of Y axis and X axis, and point O is at the origin. Rotate the square oabc clockwise around point O, and stop rotating when point a falls on the line y = x for the first time. In the process of rotation, intersect the straight line y = x at point m, and BC intersect the X axis at point n ）In the process of rotation, when Mn and AC are parallel, calculate the rotation degree of oabc ）&If the circumference of △ BMN is p, does the value of P change during the rotation of oabc? Please prove your conclusion; ）Let Mn = m, what is the minimum area of △ mon when m is, and what is the minimum value? Then the radius of the inscribed circle of △ BMN is obtained
- 18. In the plane rectangular coordinate system, the vertex o of the rectangular oacb is at the origin of the coordinate, the vertices a and B are respectively on the positive half axis of the x-axis, OA = 3, OB = 4, and D is the center of the edge ob In the plane rectangular coordinate system, the vertex o of the rectangular oacb is at the origin of the coordinate, the vertices a and B are respectively on the positive half axis of the X axis, OA = 3, OB = 4, and D is the midpoint of the edge ob. If e and F are two moving points on the edge OA and EF = 2, then the minimum perimeter of the quadrilateral cdef is the coordinates of points E and F
- 19. If the two focuses and the two vertices of the minor axis of the ellipse are the four vertices of a diamond with a 60 ° angle, then the eccentricity of the ellipse is 0 A. 1 / 2 b. √ 3 / 2 C. √ 3 / 3 d.1/2 or √ 3 / 2
- 20. It is known that the eccentricity of ellipse C: x2 / A2 + Y2 / B2 = 1 is root sign 3 / 2, the straight line X-Y + 1 = 0 passes through the upper vertex of ellipse C, the straight line x = - 1 intersects the ellipse at two points a and B, P is different from the ellipse at any point a and B, the straight line AP BP intersects the straight line L: x = - 4 at two points Q, R, 1 respectively, to solve the equation of ellipse C, 2 to prove that the vector OQ × vector or is a fixed value