4 math problems in the first year of high school, you will get points

f(x)=x -4ax+2a+6=(x-2a) -4a +2a+6 0 is established.

So -4a +2a +6 0 constant is established.

2a²-a-3≤0

a+1)(2a-3)≤0

1≤a≤3/2

So A+3>0

g(a)=2-a(a+3)=-[a+(3/2)]²17/4)

When a=-1, g(a) is the maximum, and g(-1)=4

When a=3 2, g(a) takes the minimum value, and g(3 2) = -19 4

The range of g(a) is [-19 4,4].

a<0, so the images of f(x) and g(x) are both parabolas with openings facing down and the axis of symmetry x=1.

f(x)=a(x-1) +b, the maximum value f(1)=b in [0,2].

g(x)=a(x-1) +b decreases monotonically in [2,3], and the maximum value f(2)=a+b can be obtained

The axis of symmetry of the image of the quadratic function f(x) is x=2

Again, the distance between the image and the intersection of the x-axis is 8

So the abscissa of the intersection is x1=2-(8 2)=-2, x2=2+(8 2)=6

i.e. x1 = -2 and x2 = 6 are the two roots of the equation f(x) = 0.

Obviously, at -31, the function increases monotonically.

1) If t 0, then t x t + 1

When x=t+1, f(x) is the minimum value, and g(t)=f(t+1)=t-2

2) When 01,x [t,t+1], so x can take 1

When x=1, f(x) is the minimum and g(t)=f(1)=-2

3) When t 1, 1 t x t + 1

When x=t, f(x) takes the minimum value, and g(t)=f(t)=t-2t-1

The expression for g(t) is:

g(t)=t²-2，t≤0

g(t)=-2，0g(t)=t²-2t-1，t≥1

The range of f(x) is a non-negative number.

So f(x) does not have a part below the x-axis.

Therefore, the discriminant formula is less than or equal to 0

So 16a 2-4(2a+6)<=0

2a^2-a-3<=0

a+1)(2a-3)<=0

1<=a<=3/2

So A+3>0a+3|=a+3

So g(a)=2-a 2-3a=-(a+3 2) 2+17 41<=a<=3 2

The axis of symmetry a=-3 2, no longer defines the domain, the opening is downward, and the axis of symmetry decreases to the right.

So a=-1 is the maximum, and a=3 2 is the minimum.

g(3/2)=-19/4,g(-1)=4

So the value range [-19 4,4].

f(x)=a(x-1) 2+b(x belongs to [0,2])a<0, the opening is downward.

The axis of symmetry x=1, within the defined domain.

So when x=1 there is a maximum value =b

g(x)=a(x-1) 2+b([x belongs to 2,3])a<0, the opening is downward.

The axis of symmetry x=1 is not in the defined domain.

The defined domain is on the right side of the axis of symmetry, so it decreases, so there is a maximum value of =a*(2-1) 2+b=a+b at x=2

When x=2 there is a maximum value of 16

So y=a(x-2) 2+16, a<0

The length of the line segment obtained by the truncation of the x-axis of the image is 8

So the two of equations a(x-2) 2+16=0 satisfy |x1-x2|=8ax^2-4ax+4a+16=0

x1+x2=4,x1*x2=(4a+16) a, so (x1-x2) 2=(x1+x2) 2-4x1x2=16-4(4a+16) a=8 2

a+4)/a=-3

a=-1, so f(x)=-x 2+4x+12x2+4x+12=0

x-6)(x+2)=0

x=6,x=-2

So the two are in (-3, -1) and (5, 7) respectively f(x)=x 2-2x-1=(x-1) 2-2 axis of symmetry x=1, with the opening pointing upwards.

If t<=1<=t+1, the axis of symmetry is in the defined domain.

In this case, 0<=t<=1

then when x=1, the minimum value is =-2

If the symmetry axis is to the left of the definition domain, then 1t+1, i.e., t<0, the definition domain is to the left of the symmetry axis, and the function decreases.

So when x=t+1, there is a minimum value f(t+1)=(t+1-1) 2-2=t 2-2

In summary. t<0,g(t)=t^2-2

0<=t<=1,g(t)=-2

t>1,g(t)=t^2-2t-1

First question. Because b = [-1,4].

So cub=(-infinity, -1) (4,+infinity) so according to the title.

Get. A is greater than 4

Second question. cua = [a, + infinity].

According to the question. Available.

A is greater than 4

Hope. Xie Shihuai Xie.

Do you have any questions about searching for suspects**q423237840.Hope.

Get x4 or a<-1 from the problem feast

From question 2 x>=a is not in [Nebi state, so a>4.

You are sure that the answer to the second question is correct, when a<-1, cua=-1 to positive infinity, and the intersection of liquid silver and b is not an empty set.

These two problems are the same thing (the first set on both sides of the formula will get the second formula), and you represent the sets a and b on the number line.

From the first equation we can get the set b should.

Contains in parsley.

Set A, so A 4

The result of the second equation is also a 4

The first question is to use sine to calculate AC and then cosine to calculate CD.

The first formula of the second question, a=105°=60°+45°, uses the formula of two angles to find tana, sina

Use the sinusoidal area formula to find the area.

Let's calculate the details of the landlord himself

Exam, so many people are too lazy to do it! (1) Easy to get sina = 3 10 10, sinc = 2 5 5, from the sine theorem to get ab = 2 2, from the cosine theorem to get 8 = 3 2 + ac 2-2* 5 5 * 3 ac, the solution to ac = 5 or 5 5, because abc is an acute triangle, so ac = 5, the triangle abc is complemented into a parallelogram, and the opposite angle of c is d, from the cosine theorem to get cd 2 = ac 2 + ad 2-2cos cad*ca*ad=5+9+2* 5 5 * 5 * 3 = 20, so cd = 2 5, so the midline length is cd 2 = 5

2) Because (sina + cosa) 2-1 = sin2a = -1 2, so 2a = 210, a = 105, so tan105 = tan60 + 45 = - (2 + 3).

It is easy to get sin105=( 6 + 2) 4, so s = 1 2 * sin105 * 3 * 2 = 3 ( 6 + 2) 4

The criteria for determining victory and defeat are only based on m-points, so these distances are not important.

Then look at the critical situation, that is, both sides reach M at the same time, as long as this analysis is clear, then the attacking side has a way to win.

Depending on the conditions, for any AD direction, am = 2bm, so that the trajectory of the M point is a circle, and this circle is called the Apollonius circle. This is as you prove it yourself, and it's simple for first-year high school students.

Then it's easy to do, if AD and the Apollonius circle have an intersection, then the defender wins, otherwise the attacker wins, so the attacking player only needs to choose the direction that does not intersect with the Apollonius circle and is towards the safety line, and from the perspective of saving physical strength, the tangent line of the Apollonius circle should be selected and then a little outward.

1. The equation of the straight line is set to y-3=k(x-1), and the distance d from the center of the circle to the straight line is calculated by using the distance formula from the straight line to the point, and d 2+(2 times the root number 3 2) 2=radius 2, you can find k

2. Let the straight line be y=kx+b, and solve two points (x1,y1) and (x2,y2) respectively in m and n, and the midpoint of these two points is (5,2), so there is x1+x2=5*2, y1+y2=2*2, and k,b can be solved immediately by coupling

3. It is easy to know that L3: 2X-5Y+1=0 to L1 and L2 are equal distances, so the midpoint of AB must be on the straight line 2X-5Y+1=0, which is coupled with X-4Y-1=0, and the coordinates of this midpoint can be solved, and then the midpoint coordinates and (2,3) are used to find the straight line L

4. Bring x=2y+2k into 2x-3y-k=0 to get y=-3k, then x=2y+2k=-4k, that is, 3x+4y-7=0 pass the point (-4k, -3k), and find k.

I didn't bother to count, I just said the method, and I won't ask me again.

L1: 2x-5y+9=0 L2: 2x-5y-7=0, C1 of L1 is 9, C2 of L2 is -7, C3=(C1+C2) 2=1, i.e., L3: 2x-5y+1=0

4. The intersection point of two straight lines is (-4k, -3k), and the point is brought into the straight line 3x+4y-7=0, and k=-7 24 is obtained

2. Let the straight line be y=ax+k, the intersection point with m is (4k+1) (4a+3), and the intersection point of (8ak+a+3k) (4a+3) and n is (3k+10) (4-3a), (4k+10a) (4-3a).

Because the midpoint is (5,2), there are (4k+1) (4a+3)+(3k+10) (4-3a)=10 and (8ak+a+3k) (4a+3)+(4k+10a) (4-3a)=4, and the two equations are solved to obtain a and k, and this line is obtained.

I'll give me a **! I'll tell you when I come to my house!

The solution set is r because it is satisfied for any real number.

2。Get. χ5|<15 Get: -154. This is a quadratic function, the vertex coordinates (-1, -2) open upward, so the minimum value of y is -2, and the maximum value is x=6, and y=47 so the range is [-2,47].

6."Each price increase of 1 yuan, its sales will be reduced by 20 "whether it is 80 yuan or 90 yuan, if it is 90 yuan."

y=(x-80)=(x-80)(220-20x)

It is a real number.

2.-103.Sufficient condition.

5.It's a senior year, and I have long forgotten this question in my first year of high school.

6.（80+x）*（600-20x）

The last one simplifies it for itself.

1 All real numbers.

3 p:1q:0p is a sufficient and unnecessary condition for q.

5 Forgot. 6 The sales volume is: 400-20(x-90)y=x[400-20(x-90)]-80x=-20x 2+2120x

1 x less than 1

2 x is less than 20 or x is greater than -10

3 is obtained by p, 14 y=(x+1) 2-2, the axis of symmetry is x=-1, then the range is -2,47 5 2a+b 1+b-a

6 y=(x-80)(2200-20x)

1.Solution: Define the domain as r f(x)=(1 3) [x-1) 2-1] For the function h(x)=(x-1) 2-1, we can know that it is a single decrease on (- 1) and a single increase on [1,+].

And (1 3) x is a subtraction function, then the increase interval of f(x) is determined by h(x), i.e., a single increase on (- 1). Then its value range is the maximum value at 1. That is, f(1)=1, and the range is (- 1].

The definition proves that the single increase interval is to take two numbers on (- 1] as x1 and x2, and let x10 define the domain: -10 is loga(1+x) (1-x)>0=loga 1 because a>1, so it is single-incremented, so (1+x) (1-x) >1 so 0

Don't answer, it's not easy to look at the one in front of you

a^3+b^3+ab-a^2-b^2=(a+b)(a^2-ab+b^2)-(a^2-ab+b^2)=(a+b-1)(a^2-ab+b^2)=0

Because ab is not equal to 0, a 2>0, b 2>0

So a+b=1

It is proved below that a+b=1 is a sufficient condition for a3+b 3+ab-a2-b 2=0.

a+b=1, so a 3+b 3+ab-a 2-b 2=(a+b-1)(a 2-ab+b 2)=0