O asks one between a and B; PA =] Pb =] two; does there exist a point P on the number axis so that PA + Pb = 5? 3、 Point P moves to the right from point o at the speed of one unit length / min, while point a moves to the left at the speed of 5 units / min, and point B moves to the right at the speed of 20 units / min. in the process of movement, m and N are the midpoint of AP and ob respectively. Please explain why the value of AB op / Mn changes

Your question is wrong: (AB OP) / Mn if wrong, the value does not change, the ratio is 2, because the direction is different, we first set the positive direction to the right, let the speed of p be x (unit), then the speed of a is - 5x (unit), the speed of B is 20x (unit), when the time is t, Op = XT; OA = - 5xt; ob = 20XT
It's too simple
PA=|-1-x| PB=|3-x|
2. Still doing
3. Someone answered
Given that point a is - 4 on the number axis and point B is 1, let the number corresponding to point P on the number axis be X. when the absolute value of pa - the absolute value of Pb is equal to 2, find the value of X
|x+4|-|x-1|=2
x> 1 x + 4-x + 1 = 2 does not hold
-4
-1/2
A -- m -- P -- N -- B as shown in the figure, line AB = 10, P is a moving point on line AB, M is the midpoint of PA, n is the midpoint of Pb
Q: when point P moves, does the length of line segment Mn change? If not, please find out the length of line segment Mn. If not, please explain the reason
Let AP = x, then BP = 10-x
∵ m is the midpoint of AP and N is the midpoint of BP
∴MP=1/2AP=1/2X,NP=1/2BP=5-1/2X
∴MN=MP+NP=1/2X+5-1/2X=5
Therefore, the length of Mn is constant at 5 and does not change
unchanged
MN=5
AB = 10, MP = AP / 2, NP = BP / 2, Mn = (AP + BP) / 2 = AB / 2 = 10 / 2 = 5
Find the range of function f (x) = 2x ^ 2-4x + 5 (x ∈ [0,3])
The axis of symmetry is x = 1
So when x = 1, take the minimum value of 3
When x = 3, the maximum value is 11
That is, the value range is [3,11]
f(x)=2x^2-4x+5=2（x-1)^2+3
Due to 0
The derivative function f (x) = 1 / 3x3 + 1 / 2ax2 + 2bx + C on R. if x belongs to (0,1) to get the maximum and X belongs to (1,2) to get the minimum, then the value range of (b-2) / (A-1) is
If f (x) = 1 / 3x ^ 3 + 1 / 2aX ^ 2 + 2bx + C, then
f'(x)=x^2+ax+2b,
Let x ^ 2 + ax + 2B = (x-x1) (x-x2), (x1)
Let the image of the function f (x) = AX2 + BX + 3A + B be symmetric with respect to the y-axis, and its domain of definition is [A-1, 2A] (a, B ∈ R)
It can be seen from the meaning that the function must be a quadratic function, that is, a ≠ 0, while the image is symmetric about the y-axis, B = 0 can be judged, that is, the analytic expression of the function is reduced to f (x) = AX2 + 3a. From the definition field [A-1, 2A] which is symmetric about the y-axis, A-1 + 2A = 0, a = 13 is obtained, that is, the analytic expression of the function is reduced to f (x) = 13x2 + 1, and the value field of X ∈ [- 23, 23] f (x) is [13127]
Given the function f (x) = 13x3 + 12ax2 + 2bx + C (a, B, C ∈ R), and the function f (x) obtains the maximum value in the interval (0,1) and the minimum value in the interval (1,2), then the value range of Z = (a + 3) 2 + B2 ()
A. （22，2）B. （12，4）C. （1，2）D. （1，4）
∵f（x）=13x3+12a X2 + 2bx + C  f ′ (x) = x2 + ax + 2B ∵ function f (x) obtains the maximum value in the interval (0,1) and the minimum value in the interval (1,2). F ′ (x) = x2 + ax + 2B = 0. In (0,1) and (1,2), there is a root f ′ (0) > 0, f ′ (1) < 0, f ′ (2) > 0, that is, b > 0A + 2B + 1 < A + B + 2 > 00 (a + 3) 2 + B2 represents the square of the distance from point (a, b) to point (- 3,0), It is known from the graph that the distance from (- 3,0) to the straight line a + B + 2 = 0 is 22, and the square is 12, which is the minimum value. From a + 2B + 1 = 0A + B + 2 = 0, the distance between (- 3,1) (- 3,0) and (- 3,1) is 1, and the distance between (- 3,0) and (- 1,0) is 2, so the value range of Z = (a + 3) 2 + B2 is (12,4), so the option is B
Let the image of the function f (x) = AX2 + BX + 3A + B be symmetric with respect to the y-axis, and its domain of definition is [A-1, 2A] (a, B ∈ R)
It can be seen from the meaning that the function must be a quadratic function, that is, a ≠ 0, while the image is symmetric about the y-axis, B = 0 can be judged, that is, the analytic expression of the function is reduced to f (x) = AX2 + 3a. From the definition field [A-1, 2A] which is symmetric about the y-axis, A-1 + 2A = 0, a = 13 is obtained, that is, the analytic expression of the function is reduced to f (x) = 13x2 + 1, and the value field of X ∈ [- 23, 23] f (x) is [13127]
Let function = - 1 / 3x3 + 2ax2-3a2x + 1,0
(1) In this paper, we let y = 0-x = a, x = a, x = a, x = a, x = a, x = a, x = a, x = a, let A0, f (x) be a 0, f (x) let: x = a, f (a) is the maximum, and we get: F (a) = 1-4a & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\inthis paper, the author introduces the concept of
(I) f ′ (x) = - x2 + 4ax-3a2, and 0 ＜ a ＜ 1, (1 point)
When f ′ (x) > 0, a < x < 3a is obtained;
When f ′ (x) ＜ 0, X ＜ a or X ＞ 3a is obtained;
The monotone increasing interval of F (x) is (a, 3a);
The monotone decreasing intervals of F (x) are (- ∞, a) and (3a, + ∞)
So when x = 3A, f (x) has a maximum, and its maximum is f (3a) = 1
(II) f ′ (x) = - x2 + 4A... Expansion
(I) f ′ (x) = - x2 + 4ax-3a2, and 0 ＜ a ＜ 1, (1 point)
When f ′ (x) > 0, a < x < 3a is obtained;
When f ′ (x) ＜ 0, X ＜ a or X ＞ 3a is obtained;
The monotone increasing interval of F (x) is (a, 3a);
The monotone decreasing intervals of F (x) are (- ∞, a) and (3a, + ∞)
So when x = 3A, f (x) has a maximum, and its maximum is f (3a) = 1
（Ⅱ）f′（x）=-x2+4ax-3a2=-（x-2a）2+a2，
I) when 2A ≤ 1-A, i.e. 0 < a ≤
Thirteen
F ′ (x) decreases monotonically in the interval [1-A, 1 + a]
∴[f′（x）]max=f′（1-a）=-8a2+6a-1，[f′（x）]min=f′（1+a）=2a-1．
∵-a≤f′（x）≤a，∴
-8a2+6a-1≤a2a-1≥-a
Qi
a∈Ra≥13
∴a≥
Thirteen
.
At this point, a=
Thirteen
(9 points)
II) when 2A > 1-A and 2A < A + 1, i.e
Thirteen
＜a＜1，[f′（x）]max=f′（2a）=a2．
∵-a≤f′（x）≤a，∴
f′(1+a)≥-af′(1-a)≥-af′(2a)≤a
Namely
2a-1≥-a-8a2+6a-1≥-aa2≤a
Qi
a≥137-1716≤a≤7+17160≤a≤1.
Qi
Thirteen
≤a≤
7+1716
.
At this point,
Thirteen
＜a≤
7+1716
(12 points)
(3) when 2A ≥ 1 + A, the contradiction between a ≥ 1 and known 0 < a < 1 was obtained. (13 points)
To sum up, the value range of real number a is[
Thirteen
,
7+1716
]14 points
Let the image of the function f (x) = AX2 + BX + 3x + B be symmetric about the y-axis, and its domain of definition be [a-1,2a] (a, B ∈ R), and find the range of the function f (x)
The image of function f (x) = AX2 + BX + 3x + B is symmetric about y axis, that is, even function
So, f (- x) = ax ^ 2 - (B + 3) x + B = f (x)
So: B + 3 = 0, B = - 3
The domain of definition is [a-1,2a] (a, B ∈ R). Since the domain of definition is symmetric about the origin, A-1 + 2A = 0, a = 1 / 3
f(x)=x^2/3-3,-2/3
The image of function f (x) = ax ^ 2 + (B + 3) x + B is symmetric about y axis
2 a = - 3, B = - 3) / X;
Because a = 2A + 1-1, a = 1-1;
So ^ 2 [- 2 / 3] = x / 3
f(x)∈[-3,-77/27].

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