The function f (x) = x ^ 3 - (k ^ 2-k + 1) x ^ 2 + 5x-2, G (x) = k ^ 2x ^ 2 + KX + 1, where k ∈ (1) Let P (x) = f (x) + G (x), if P (x) is not monotone in the interval (0,3), then the range of K is obtained? p'(0)*p'(3)

The function f (x) = x ^ 3 - (k ^ 2-k + 1) x ^ 2 + 5x-2, G (x) = k ^ 2x ^ 2 + KX + 1, where k ∈ (1) Let P (x) = f (x) + G (x), if P (x) is not monotone in the interval (0,3), then the range of K is obtained? p'(0)*p'(3)

(1)p(x)=f(x)+g(x)=x^3-(k^2-k+1)x^2+5x-2+k^2x^2+kx+1=x^3+(k-1)x^2+(k+5)x-1
Obviously, P (x) is differentiable in the interval (0,3) and P (x) is not monotone in the interval (0,3),
∧ there must be a ∈ (0,3) such that p '(a) = 3A ^ 2 + 2 (k-1) a + (K + 5) = 0,
Solve k = u (a) and note 0
It is known that a (√ 3 + 1,1), B (1,1) and C (1,2) are vectors a = CB →, B = ab → and C = Ca →
It is known that a (√ 3 + 1,1), B (1,1) and C (1,2) are vectors a = CB →, B = ab → and C = Ca →
(1) The coordinates of a, B and C
(2) The coordinates of a + 2b-3c
(3) The angle between a and C
a(1-1,2-1),a(0,1)
b(1-(√3+1),1-1),b(-√3,0)
c(√3+1-1,1-2),c(√3,-1)
A + 2b-3c coordinate (- 5 √ 3,4)
a.c=(0*√3)+(1*(-1))=-1
|a|=√(0^2+1^2)=1
|c|=√((√3)^2+(-1)^2)=2
Cosx = A.C / | a | C | = - 1 / 2, the angle is - 60 degrees
What is the relationship between the root discriminant and the coefficient of quadratic equation with one variable?
The quadratic equation of one variable is ax & # 178; + by + C, and the discriminant is Δ = B & # 178; + 4ac,
y=ax^2+bx+c，a≠0
△=b^2-4ac
a. B and C are coefficients respectively
ax² bx c=0.a≠0
Then the discriminant △ = B & # 178; - 4ac
If △ 0, the equation has two real roots
If △ = 0, the equation has two equal real roots
If △ 0, the equation has no real root
y=ax^2+bx+c，a≠0
△=b^2-4ac
Is X-1 / 4 = 2x + 3 / 5 a linear equation with one variable
yes
Because there is only one unknown X
And X is linear
Is 2 / x = 1 a linear equation with one variable
2/X=2*x^(-1)=1
no
Yes, there is only one element x, and X is only once
Given m = 1 - √ 2, n = 1 + √ 2, calculate √ (M & # 178; + n & # 178; - 2Mn + 1)
Solution;
m-n=-2√2
√(m²+n²-2mn+1)
=√(m-n)²+1
=√8+1
=3
√(m²+n²-2mn+1)
= √[(m-n)²+1]
=√(2² +1)
=√5
Qiufeng Yanyan answers question o (∩)_ ∩)O
If you don't understand, you can continue to ask this question
If you are satisfied, please choose it as the satisfactory answer in time. Thank you
√(m²+n²-2mn+1)=√(m-n）²+1)=√[（（1-√2）-（1+√2））²+1]=√1=1
I agree with 334455665.. Qiufengyan seems to be wrong. ok I'm a little late. I can only help you to judge right and wrong.
The function f (x) defined on R satisfies that f (0) = 0, f (x) + F (1-x) = 1, f (x / 5) = 1 / 2F (x), and if 0 ≤ x1
f(0)=0, f(x)+f(1-x)=1 ==> f(1/2)=1/2, f(1)=1
f(1)=1, f(x/5)=1/2f(x), f(x)+f(1-x)=1 ==> f(1/5)=1/2, f(4/5)=1/2
F (1 / 5) = 1 / 2, f (1 / 2) = 1 / 2, f (4 / 5) = 1 / 2, f (x1) ≤ f (x2) {0 ≤ X1 [1 / 5,4 / 5] interval, f (x) = 1 / 2
F (x / 5) = 1 / 2F (x) = > [1 / 5 ^ n, 4 / 5 ^ n] interval f (x) = 1 / 2 ^ n
1 / 2020 belongs to [1 / 5 ^ 5, 4 / 5 ^ 5] = > F (1 / 2020) = 1 / 2 ^ 5 = 1 / 32
Why is vector a = λ B (λ is a real number) a sufficient and unnecessary condition for vector AB to be collinear
"Vector a = λ B" can get "a, B collinear"
But "a, B" collinear can not get "vector a = λ B"
For example: a = (1,1), B = (0,0)
a. B is collinear (0 vector and any non-0 vector are collinear)
When there is no real number λ, a = λ B holds
Therefore, the vector a = λ B (λ is a real number) is a sufficient and unnecessary condition for vector AB to be collinear
x^2-2x-3
Given that x is a real number and T = ----, find the possible value range of real number t
2x^2+2x+1
The title is not good. Here is the formula
Given that x is a real number and T = x ^ 2-2x-3 / 2x ^ 2 + 2x + 1, find the possible value range of real number t
Topic reduction: (2t-1) x ^ + (2t + 2) x + 4 = 0
When 2t-1 = 0, i.e. t = 1 / 2, X has a solution
When 2t-1 is not equal to 0, then △ > = 0, that is, (2t + 2) ^ - 16 (2t-1) > = 0, then t = 5
So t = 1 / 2 or T = 5
The first floor is right
How to calculate (2x-1) / 5 - (x + 1) / 2 = 3 equation of first degree with one variable?
Multiply the left and right sides of the equation by 10 at the same time
2 (2x-1) - 5 (x + 1) = 30
We get 4x-2-5x-5 = 30
5x-4x = - 30-5-2
Combining, we get x = - 37
1. Given 2m-4n = 0, there are two more problems under the value of 3M & # 178; - N & # 178; △ M & # 178; + 2Mn
2. Given 1 / A-1 / b = 5, find the value of 2A + 19ab-2b / b-3ab-a. 3. If a / b (a = a △ B of B), then are a + B / B and C + D / D equal? Try to prove the reason!
1.2m-4n = 0, so m = 2n3m & # 178; - N & # 178; / (fractional line) M & # 178; + 2Mn = (12n & # 178; - N & # 178;) / (4N & # 178; + 4N & # 178;) = 11 / 82.1 / A-1 / b = 5 (B-A) / AB = 5b-a = 5ab (2a + 19ab-2b) / (b-3ab-a) = (19ab-10ab) / (5ab-3ab) = 9 / 23. A + B / b = 1 + A / BC + D / D = 1 + C /