Ask a few high school math questions, ask a few high school math questions

y=x -2ax+3=(x-a) +3-a) The axis of symmetry of the function image is x=a

Because the value of a is unknown, it is discussed.

1) A<-2.

The x=-2 function takes the minimum value, the f(-2)=7+4ax=2 function takes the maximum value, and the f(2)=7-4a function

The function range is: [7+4a,7-4a].

2)-2 a<0.

The x=a function takes the minimum value, and f(a)=3-a

The function of x=2 takes the maximum value, and f(2)=7-4a

The range of the function is: [3-a ,7-4a].

3) When a=0.

The x=0 function takes the minimum value, and f(0)=3

The x=2 function takes the maximum value, and f(2)=7

The range of the function is: [3,7].

4) 02 o'clock.

The x=2 function is the minimum value, and f(2)=7-4a

The maximum value of the x=-2 function is taken, and the range of the f(-2)=7+4a function is: [7-4a,7+4a].

4) 2x2-9x+a>0 constant establishment.

Then the equation 2x2-9x+a=0 has no real solution.

then =81-8a<0

a>81/8

5) When x (2,3), the inequality 2x -9x+a<0 is constant.

i.e. a<-(2x, 2-9x) is constant.

f(x)=-2x 2+9x=-2(x-9 4) 2+81 8 axis of symmetry x=9 4, opening downward.

2So, the range of a is a<=10

3. A needs to be classified and discussed.

The axis of symmetry of this function is x=a

When a is greater than or equal to 2, x=2 obtains the minimum value of the function, 7-4, ax=-2 obtains the maximum value of the function, 7+4a

When a is less than or equal to -2, x=2 gives the maximum value of the function 7-4, and ax=-2 gives the minimum value of the function 7+4a

When a is between -2 and 2, x=a gives the minimum value of the function.

The maximum value needs to be further subdivided into a.

When a is greater than or equal to -2 and less than or equal to 0, x=2 is the maximum value.

When a is greater than 0 and less than or equal to 2, x=-2 obtains the maximum value.

4. The unary quadratic function corresponding to this inequality is .

f（x）=2*x^2-9x+a

When it is less than 0, the function has no intersection point with the x-axis, and because the coefficient of the quadratic term is greater than 0, the inequality is constant at this time.

Solution a 81 8

5. The unary quadratic function corresponding to this inequality is.

f（x）=2*x^2-9x+a

Symmetrical pumping is x=9 4 between 2 and 3.

Therefore, the condition for the original inequality to be constant is.

f(2) is less than 0, f(3) is less than 0 and f(9 4) is also less than 0, and the solution a is less than 9

Question 3: The upstairs idea is right, and I didn't see the process.

There are other ways to do the last two questions.

4.2x 2-9x+a>0 then a>9x-2x 2 let y=9x-2x 2

then y=-2(x-9 4) 2+81 8, and there is a maximum value when x=9 4.

So a>81 8, the inequality is constant.

5.2x 2-9x+a>0 then a>9x-2x 2 let y=9x-2x 2

then y=-2(x-9 4) 2+81 8

Because the curve of the equation is open downward, it can be seen from the curve graph that this time is when.

The minimum value can be taken when x = 2 (y = -10) or x = 3 (y = -9).

Therefore, when a<-10, the inequality is constant.

First, the trim y=x-2ax+3=(x-a) +3-a) shows that the axis of symmetry of the function image is x=a

1) A<-2.

The x=-2 function takes the minimum value, the f(-2)=7+4ax=2 function takes the maximum value, and the f(2)=7-4a function

The function range is: [7+4a,7-4a].

2) - 22 hours.

The x=2 function is the minimum value, and f(2)=7-4a

The maximum value of the x=-2 function is taken, and the range of the f(-2)=7+4a function is: [7-4a,7+4a].

2x2-9x+a>0 is established.

Because it represents a parabola with an opening upward, the question is that the equation 2x2-9x+a=0 has no real solution.

then =81-8a<0

a>81/8

When x (2,3), the inequality 2x -9x + a<0 is constant.

That is to say, a<-(2x, 2-9x) is constant.

Let f(x)=-2x 2+9x=-2(x-9 4) 2+81 8, then it is OK to make a less than the maximum value of f(x).

The axis of symmetry of f(x) x = 9 4, with the opening pointing downward.

2So, the range of a is a<=10

Solution:1Let the difference number be listed first, A1 first, and the difference value D.

The first 4 = a1 + a2 + a3 + a4 = 4a1 + d + 2d + 3d = 4a1 + 6d = 124

After s, 4=an+an-1+an-2+an-3=4a1+(n-1)d+(n-2)d+(n-3)d+(n-4)d=4a1+4nd-10d=156

sn=na+[n(n-1)/2]d=210

From the above three equations, a system of ternary linear equations is formed, and n is found.

2.(1) Sufficiency: "For any n n*, the point pn(n,an) is on the straight line y=2x+1" can get an=2n+1, then an-an-1=2. It can be proven: "{an} is a series of equal differences".

2) Necessity: "{an} is the difference series", (let the first term be a1, and the difference value is d), an-an-1=d, when d is not equal to 2, pn(n,an) will not always be on the straight line y=2x+1. When d=2, the points pn(n,an) may all be on the straight line y=2x+1.

an-an-1=-3, the resulting series is a series of equal differences with 101 as the first term and -3 as the equal difference.

Find the value from the first term less than 0: a1+md=101-3m<0 to find m>101 3>33

Values from 34 items onwards are less than 0. Sequence {| an|The sum of the first n terms of } is tn=(a1+....+a33)-(a34+…+an)

4.In the series of equal difference {an}, when n is odd, sn=n*[a(n+1) 2].

s9/s5=9a5/5a3=1

5.In the equation series {an}, let the difference be a17=a1+16d=-12 to obtain d=4.

Find the value from the first term greater than or equal to 0 : a1+md=-60+4m>=0 to find m>=15

Values from 15 terms can be greater than or equal to 0: Sequence {| an|The sum of the first n terms of } is tn=(a1+....+a14)+(a15+…+an)

There is no solution to the first problem, there are eight numbers in total, the sum of the first four terms of a difference series is 124, and the sum of the last four terms is 156, the sum of the eight numbers should be 124 + 156 280, how can it be 210

1 16 (x y) 81, 1 8 1 xy 1 3, multiply the two to give 2 x 3 y 4 27.

2 The inequality is simplified to obtain (a+b) x -(a-b) 0, and a ≠ b, so that x -x 0 is obtained, and the solution is 0 x 1

3 f(x) is an odd function, so f(0)=0,1) takes b=-b in the inequality, then there is f(a)+f(-b) a-b 0, i.e., f(a)-f(b) a-b 0, from the question a b, so f(a) f(b).

2) From (1) it is known that f(x) is a monotonic increasing function in the defined domain, so the inequality can be reduced to (x-1) 2 (x-1) 4, and -1 (x-1) 2 1, -1 (x-1) 4 1, and the group of inequalities above -1 x 1

4 From the question, we know that a=(a+b) 2, g= (ab), ag= (a+b)(ab) 2, and a, b are both negative, then there is (a+b) 2=-[(a)+(b)] 2 -(ab), since a≠b, ag -(ab) (ab) =-ab, or ag ab chooses d.

1, c (2 3 4 but not 4) 3, d

4. c (am+bm+cm is a zero vector and any vector is collinear) 5、①op=2/3oa+1/3ob

Left and right multiplied by 3 3OP=2OA+OB

Adding 3ao to the left and right gives 3(ao+op)=3ao+2oa+ob=ao+ob

i.e., 3ap=ab ap ab two vectors collinear proves a p b three-point collinear.

op=aoa+bob Substituting b=1-a gives op=aoa+(1-a)ob

op=aoa-aob+ob

Translate bo+op=a(oa-ob) oa-ob=oa+bo=ba

bp=aba

BP Ba two vectors are collinear.

That is, a p b is three points collinear.

1、c

3、d4、c

abxap = (ob-oa)x(op-oa)= (ob-oa)x(1/3ob-1/3oa)= 0;(x is the cross product).

So a, b, p are collinear.

2) Same as above.

abxap = (ob-oa)x(op-oa)= (ob-oa)x(b*ob-b*oa)= 0;(x is the cross product).

So a, b, p are collinear.

1.From a2, a5, the common ratio q=1 2 can be obtained

So a1 = so ana(n+1)=(1 2) (n-5)=16(1 2) (n-1) is obviously also a proportional series, just sum the formula.

2.The general term an=3 (n-1) of the proportional series, use the formula to calculate s10 and s3, and subtract the two.

Let me start with the third question.

1／cos2a+tan2a=1/((cosa)^2-(sina)^2)+2sinacosa/((cosa)^2-(sina)^2)

sina + cosa) 2 ((cosa) 2-(sina) 2) because the m vector is parallel to the n vector, so.

cosa+sina = 2006 (cosa-sina) Because cosa+sina is not 0, both sides of the equation are multiplied by cosa+sina.

sina+cosa）^2=2006((cosa)^2-(sina)^2)

Original = 2006