Given that f (x) = x2-2x-3, the monotonicity of function f (5-x2) is discussed

Given that f (x) = x2-2x-3, the monotonicity of function f (5-x2) is discussed

Let t = 5-x2, obviously its axis of symmetry is x = 0, and t ≤ 5. ∵ it is known that the axis of symmetry of F (x) = x2-2x-3 is x = 1. ① when t = 5-x2 ≥ 1, f (T) is an increasing function, and the solution is - 2 ≤ x ≤ 2. In the interval [- 2, 0), function T is an increasing function, so function f (5-x2) is an increasing function; in the interval [0, 2], function T is a decreasing function, so function f (5-x2) is a decreasing function. ② when t = 5-x2 < 1, function f (5-x2) is an increasing function In the interval (- ∞, - 2), the function T is an increasing function, so the function f (5-x2) is a decreasing function. In the interval (2, + ∞), the function T is a decreasing function, so the function f (5-x2) is an increasing function. In conclusion, the increasing interval of the function f (5-x2) is [- 2,0), (2, + ∞), and the decreasing interval is [0,2], (- ∞, - 2)
It is known that the solution of the system of equations {ax + 3By = C; 2aX by = 5C} about XY is {x = 1; y = 2} to find a: B: C
By introducing x, y, a + 6B = C; 2a-2b = 5C; 2A + 12b = 2C; 2a-2b = 5C; 14b = - 3C; - > b: C = - 3:14; a + 6B = C; 6a-6b = 15C; 7a = 16C; - > A: C = 16:7 = 32:14; a: B: C = 32: - 3:14
The number of solutions of equation | X's square-2x | = A's square + 1 (a belongs to positive real number)
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When solving the equations ax + by = 2,2ax-by = - 2, a misread a certain coefficient in the first equation and got x = 1, y = - 1
If you misread the coefficient of the first equation, you should not misread the coefficient of the second equation
Substituting (1, - 1): 2A + B = - 2
Similarly, - 2a-6b = 7
The solution is: a = - 1 / 2, B = - 1
Then ax + by = 7,2ax-by = - 2 becomes: - X / 2-y = 7, - x + y = - 2
x=-10/3
y= -16/3
A
The number of real number solutions of the equation cos (2x + π 6) = 13 (x ∈ [0, 2 π]) is ()
A. 1B. 2C. 3D. 4
∵ x ∈ [0, 2 π], ∵ 2x + π 6 ∈ [π 6, 25 π 6], in the same coordinate system, make the image of y = cos (2x + π 6) of cosine function and y = 13 of straight line. From the figure, we can see that y = cos (2x + π 6) of cosine function and y = 13 of straight line have four intersections in X ∈ [0, 2 π], that is, the number of real number solutions of equation cos (2x + π 6) = 13 (x ∈ [0, 2 π]) is four
When the following equations ax + BX = 1 and 2aX + 3By = - 1 are solved as x = 2, y = 1 by misinterpreting the positive and negative coefficients of the unknown number y in equation 2, try to find the value of a / b
Let x = 2, y = 1 generation ax + BX = 1,2ax-3by = - 1 get a = 1, B = - 1
Hope to adopt
If the graph of function y = | x | - 1 and the curve of equation x2 + λ y2 = 1 have exactly two different common points, then the value range of real number λ is ()
A. （-∞，-1]∪[0，1）B. [-1，1）C. {-1，0}D. [-1，0）∪（1，+∞）
From y = | x | - 1, we can get that when x ≥ 0, y = X-1; when x < 0, y = - X-1, the image of function y = | x | - 1 and the curve of equation x2 + λ y2 = 1 must intersect (± 1, 0). So in order to make the image of function y = | x | - 1 and the curve of equation x2 + λ y2 = 1 have exactly two different common points, then y = X-1 can be substituted into equation x2 + λ y2 = 1, and (1 + λ) x2-2 λ x + λ - 1 = 0 can be obtained When y = - X-1 is substituted into equation x2 + λ y2 = 1, we can get (1 + λ) x2 + 2 λ x + λ - 1 = 0. When λ = - 1, x = - 1 satisfies the meaning of the problem. Because △ 0, - 1 is the root of the equation, when - 1 ＜ λ ＜ 1, two different signs satisfy the meaning of the problem[- 1, 1) so B
When someone solves the equations ax + by = 1 (1) and 2aX + 3By = - 1 (2), because he misinterprets the positive and negative sign of Y in 2, the solution of x = 2, y = 1 can find the value of a and B
When x = 2, y = 1, 2A + B = 1, 4a-3b = - 1, the solution is a = 1 / 5, B = 3 / 5
It is known that the equation (x + 1) 2 + (X-B) 2 = 2 about X has a unique real solution, and the image of inverse scale function y = 1 + BX increases with the increase of X in each quadrant, then the relation of inverse scale function is ()
A. y＝−3xB. y＝1xC. y＝2xD. y＝−2x
The general form of the equation (x + 1) 2 + (X-B) 2 = 2 of X is: 2x2 + (2-2b) x + (b2-1) = 0, △ = (2-2b) 2-8 (b2-1) = - 4 (B + 3) (B-1) = 0. The solution is: B = - 3 or 1. ∵ the image of inverse scale function y = 1 + BX increases with the increase of X in each quadrant, ∵ 1 + B < 0 ∵ B < - 1, ∵ B = - 3
When solving the equations ax + by = 7,2ax-by = - 2, a misread a certain coefficient in the first equation and got x = 1, y = - 1
B misread some coefficient in the second equation, and the solution is x = - 2, y = - 6
This problem is not like the previous one. It's not "wrong one, the other equation must be satisfied". It's "wrong some coefficient". I can't do it. Who can teach me!
I'll go after the minute~
My time is not long
If you misread the coefficient of the first equation, you should not misread the coefficient of the second equation
Substituting (1, - 1): 2A + B = - 2
Similarly, - 2a-6b = 7
The solution is: a = - 1 / 2, B = - 1
Then ax + by = 7,2ax-by = - 2 becomes: - X / 2-y = 7, - x + y = - 2
x=-10/3
y= -16/3