Given that: t is a constant, the maximum value of function y = | x2-2x + t| in the interval [0, 3] is 3, then the real number t=______ .

Given that: t is a constant, the maximum value of function y = | x2-2x + t| in the interval [0, 3] is 3, then the real number t=______ .

Let g (x) = x2-2x + T, X ∈ [0,3],
Then y = f (x) = | g (x) |, X ∈ [0,3]
The f (x) image is obtained by folding the part of the function g (x) image below the x-axis to the upper part of the x-axis,
If its symmetry axis is x = 1, then the maximum value of F (x) must be obtained at x = 3 or x = 1
(1) When the maximum value is obtained at x = 3, f (3) = | 32-2 × 3 + T | = 3,
It is found that t = 0 or - 6. When t = - 6, f (0) = 6 > 3 does not match, and T = 0 is consistent
(2) When the maximum value is obtained at x = 1, f (1) = | 12-2 × 1 + T | = 3, the solution is t = 4 or - 2. When t = 4, f (0) = 4 ﹥ 2 is inconsistent, t = - 2 is consistent
In short, t = 0 or - 2 is consistent
So the answer is: 0 or - 2

Given that t is a constant and the maximum value of the function y = | x? - 2x| on the interval [0,3] is 2, then t =? My answer is this: y=|x²-2x-t|=|(x-1)²-(t+1)| ∵| x | 2x| has a maximum value of 2 on the interval [0,3] The (x-1) 2 is the largest, (T + 1) is the smallest When x = 3, (x-1) 2 max = 4 ∴t+1=2,t=1 After looking at the answer, the result is right, but I always think there are some problems in the process. Please help me to have a look, |Title Modification of X? - 2x-t |

It should be discussed in two ways
When (x-1) ^ 2 > (T + 1), y = (x-1) ^ 2-t-1, and the maximum value of Y in a given interval is 3-T = 2, so t = 1
When (x-1) ^ 2

Let t be a constant and the maximum value of function y = | x2-2x-t| on the interval [0,3] is 2, then t is a constant=______ .

Let g (x) = x2-2x-t, X ∈ [0,3],
Then y = f (x) = | g (x) |, X ∈ [0,3]
The f (x) image is obtained by folding the part of the function g (x) image below the x-axis to the upper part of the x-axis,
If its symmetry axis is x = 1, then the maximum value of F (x) must be obtained at x = 3 or x = 1
(1) When the maximum value is obtained at x = 3, f (3) = | 32-2 × 3-T | = 2,
T = 1 or 5,
When t = 5, f (0) = 5 > 2 does not meet the condition,
When t = 1, f (0) = 1, f (1) = 2
(2) When the maximum value is obtained at x = 1, f (1) = | 12-2 × 1-T | = 2,
T = 1 or - 3,
When t = - 3, f (0) = 3 > 2 does not meet the condition,
When t = 1, f (3) = 2, f (1) = 2
To sum up, when t = 1
So the answer is: 1

Let t be a constant and the maximum value of function y = | x2-2x-t| on the interval [0,3] is 2, then t is a constant=______ .

Let g (x) = x2-2x-t, X ∈ [0,3],
Then y = f (x) = | g (x) |, X ∈ [0,3]
The f (x) image is obtained by folding the part of the function g (x) image below the x-axis to the upper part of the x-axis,
If its symmetry axis is x = 1, then the maximum value of F (x) must be obtained at x = 3 or x = 1
(1) When the maximum value is obtained at x = 3, f (3) = | 32-2 × 3-T | = 2,
T = 1 or 5,
When t = 5, f (0) = 5 > 2 does not meet the condition,
When t = 1, f (0) = 1, f (1) = 2
(2) When the maximum value is obtained at x = 1, f (1) = | 12-2 × 1-T | = 2,
T = 1 or - 3,
When t = - 3, f (0) = 3 > 2 does not meet the condition,
When t = 1, f (3) = 2, f (1) = 2
To sum up, when t = 1
So the answer is: 1

If the polynomial x2-2x + 3 = a (x + 1) 2 + B (x + 1) + C, where a, B, C are constants, then the value of a + B + C is______ .

A (x + 1) 2 + B (x + 1) + C, = a (x2 + 2x + 1) + BX + C, = AX2 + (2a + b) x + A + C, ∵ x2-2x + 3 = a (x + 1) 2 + B (x + 1) + C,  x2-2x + 3 = a (x + 1) 2 + B (x + 1) + C, x22x + 2x + 3 = AX2 + (2a + b) x + A + C,,, \\ = 1, 2A + B = - 2, a + C = 3, a = 1, B = - 4, C = 2, \\\\\\\theanswer is - 1

(1) 1-m = - M + 1 / 2 (2) 1-2x - 2x = 2x? - x (3) AB fraction a? - 3AB + B? = ab fraction-3 hurry up.

Solution 11) 1-m = - M + 1 / 1 (1-m) square
(2) 1-2x - 2x = 2x 2 - x times the square of X
(3) AB fraction a 2 - 3AB + B 2 = the square of part a of AB plus the square of B - 3

The known function f (x) = BX + 1 2X + A, a, B are constants, and ab ≠ 2, if there is f (x) f (1) for all X x) = K (k is a constant) then K=______ .

∵f（x）=bx+1
2x+a，∴f（1
x）=b1
X+1
Twenty-one
x+a=b+x
2+ax，
Then f (x) f (1)
x）=bx+1
2x+a•b+x
2+ax
Then f (x) f (1)
x）-k=(bx+1)(b+x)−k(2x+a)(2+ax)
(2x+a)(2+ax)
=(b−2ak)x2+(b2+1−4k−ka2)x+b−2ak
(2x + a) (2 + ax) = 0 is constant
be
b−2ak＝0
B 2 + 1 − 4K − a2k = 0 can be obtained by eliminating B
4a2k2-（4+a2）k+1=0，
K = 1
4，k=1
a2．
If k = 1
A2, then B = 2ak = 2
a. Then AB = 2,
It is inconsistent with ab ≠ 2
Then k = 1
4．
So the answer is 1
4．

If a? + AB = 4, AB + B? = - 1, find a?? - B? If the value of 3x? 2x + 6 is 8, find the value of 3 / 2x? - x + 1

(1) Subtract the two formulas given: a 2 + AB = 4, AB + B 2 = - 1,
In other words, (a 2 + AB) - (AB + B 2) = 4 - (- 1)
So a 2 - B 2 = 5;
(2) 3x? 2x + 6 = 8, that is, 3x? - 2x-2 = 0,
That is, 3x? - 2x = 2, divide 2 on both sides of the equation, and get: 3 / 2x? - x = 1,
So: 3 / 2x? - x + 1 = 2;
Hope to help you, if you don't understand, please hi me, I wish learning progress!

The value of C can be (.)

Center (1, - 2)
Distance from center of circle to straight line:
|2-2+c|/√(2²+1²)=1
∴c=±√5

How many points are there on the circle x 2 + y 2 + 2 x + 4 y - 3 = 0 whose distance from the straight line x + y + 1 = 0 is equal to 3 √ 2

X 2 + y 2 + 2x + 4y-3 = 0, i.e. (x + 1) 2 + (y + 2) 2 = 8, and the radius is 2 √ 2
The coordinates of the circle point are (- 1, - 2), and the distance from the circle point to the straight line x + y + 1 = 0 is | - 1-2 + 1 | / √ 2 = √ 2 < 2 √ 2
Therefore, if a line intersects a circle, 2 √ 2 + √ 2 = 3 √ 2. Therefore, there is one point on the circle where the distance from the line x ＋ y ＋ 2x + 4y-3 = 0 to the line x + y + 1 = 0 is equal to 3 √ 2