It is known that the quadratic function f (x) = 2x ^ 2 - (A-2) x-2a ^ 2-2, if it is in the interval [0,1] If there is at least one real number B such that f (b) > 0, then the value range of real number a is?

It is known that the quadratic function f (x) = 2x ^ 2 - (A-2) x-2a ^ 2-2, if it is in the interval [0,1] If there is at least one real number B such that f (b) > 0, then the value range of real number a is?

solution
Quadratic function f (x) = 2x ^ 2 - (A-2) x - (2a ^ 2 + 2)
The opening is upward, the axis of symmetry is x = (A-2) / 4, and intersects the negative half axis of Y axis
If x ∈ [0,1], there exists f (x) > 0:
1. When (A-2) / 4
The opening of quadratic function f (x) is upward, and there is at least one real number B in the interval [0,1] of quadratic function f (x) such that f (b) > 0
Then f (0) > 0 or F (1) > 0 is 2A & # 178; + A-2 < 0, a ∈ (- 17 / 4-1 / 4, 17 / 4-1 / 4)
When x + y = 1, xy = - 2, 3x + 3y-2xy process
A:
x+y=1,xy=-2
3x+3y-2xy
=3(x+y)-2xy
=3*1-2*(-2)
=3+4
=7
So: 3x + 3y-2xy = 7
3x+3y-2xy
=3(x+y)-2xy
=3×1-2×(-2)
=3+4
=7
3x+3y-2xy
=3(x+y)-2xy
=3×1-2×(-2)
=3+4
=7
The following indefinite integrals are obtained by the method of partial integration: ∫ x times the square of SiNx times DX
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I don't know if you are a beginner of integral. I'll give you a basic method. When you come across such a problem in the future, you should deal with it in this way. When you come across a trigonometric function of higher power, you should first consider reducing the power, use the square of SiNx to be equal to (1-cos2x) / 2, and then divide the term and multiply with X respectively.
As for partial integration, I'll give you a formula. The simple choice for derivation is u, and the easy integration factor is DV. Of course, this problem is to choose x as u, cos2x as DV.
If you want to know more about partial integral, I suggest you read Chen Wendeng's book, which is very detailed. ... unfold
I don't know if you are a beginner of integral. I'll give you a basic method. When you come across such a problem in the future, you should deal with it in this way. When you come across a trigonometric function of higher power, you should first consider reducing the power, use the square of SiNx to be equal to (1-cos2x) / 2, and then divide the term and multiply with X respectively.
As for partial integration, I'll give you a formula. The simple choice for derivation is u, and the easy integration factor is DV. Of course, this problem is to choose x as u, cos2x as DV.
If you want to know more about partial integral, I suggest you read Chen Wendeng's book, which is very detailed. I just finished the postgraduate entrance examination. I went to the postgraduate entrance examination class with him. It's really powerful. Hope to help you, come on. Put it away
x-y+4z=10,x+3y+2z=2,x+2y+3z=11.
X-Y + 4Z = 10 (1) x + 3Y + 2Z = 2 (2) x + 2Y + 3Z = 11 (3) (2) - (1): 4y-2z = - 8, that is, 2Y - z = - 4 (4) (3) - (1): 3Y - Z = 1 (5) (5) - (4): y = 5 substitute (5): z = 14 substitute y and Z into (1): x = - 41 answer: (x, y, z) = (- 41,5,14)
Seeking indefinite integral (SiNx ^ 2cosx ^ 2)
sinx^2cosx^2=[(sin2x)/2]^2=[(sin2x)^2]/4=(1-cos4x)/8.
Indefinite integral (SiNx ^ 2cosx ^ 2) = (1 / 8) [x - (sin4x) / 4] + C = x / 8 - (sin4x) / 32 + C
We should know that 99% of indefinite integrals can not be obtained by elementary method, and this is no exception (we can expand the approximate product product with Taylor formula). It should be reminded that its generalized integral from negative infinity to positive infinity exists and can be changed into Fresnel integral in legend.
Equation: x + 2Y + 3Z = 11 X-Y + 4Z = 10 x + 3Y + 2Z = 2
Elimination ah, into binary linear equations, easy solution
The specific methods are as follows:
x+2y+3z=11 ①
x-y+4z=10 ②
x+3y+2z=2 ③
① - 2 can eliminate X and become 3y-z = 1 (1)
② - 3 can eliminate X and become z-2y = 4 (2)
The new system of linear equations of two variables can be solved
(1) + (2) y = 5
Substituting into (1) results: z = 14
Substituting into ①, we get: x = - 41
Indefinite integral of (SiNx cos x) divided by (SiNx + cosx)
The values of solutions X and y of the system of quadratic equations 4x + 3Y = 7, KX + (k-1) y = 3 are equal
Method 1: 4x + 3Y = 7 ①
kx+（k-1）y=3 ②
Equation 1 × K gives 4kx + 3ky = 7K
Equation 2 × 4 gives 4kx + (4k-4) y = 12
By subtracting the two equations, (K-4) y = 12-7k, y = (12-7k) / (K-4)
By substituting it into the equation (1), 4x = 7-3y = (9k-64) / (K-4), (x = (7k-16) / (K-4)
According to the meaning: (7k-16) / (K-4) = (12-7k) / (K-4)
The solution is k = 2
Method 2: ∵ x = y 4x + 3Y = 7
Then 7x = 7,
∴x=y=1
Substituting KX + (k-1) y = 3, we get
k+k-1=3
∴k=2
Answer: k = 2
Bring x = y in
Get: 7x = 7, get x = 1
Take x = 1 into (2k-1) x = 3 to get k = 2
In order to find the indefinite integral of X (SiNx) ^ 2, we can't find its solution,
x(sinx)^2
=x/2(1-cos2x)
And then we can use the step-by-step integration method
x*(-1/2*cos(x)*sin(x)+1/2*x)-1/4*cos(x)^2-1/4*x^2+C
The solutions X and y of the system of quadratic equations 1.4x + 3Y = 7 KX + (k-1) y = 3 are equal, and K is obtained
4x+3y=7(1)
kx+(k-1)y=3(2)
x=y(3)
Because x = y is substituted into (1)
7x=7y=7
therefore
x=y=1
Substituting (2)
k+k-1=3
2k=4
K=2
4x+3y=7
And then there is x = y
So x = y = 1
KX + (k-1) y = 3 becomes K + k-1 = 3
K=2