It is known that the quadratic function f x ax2 bx c satisfies for any real number, there is f x x

1.For any x, f(x) x is satisfied, so there is f(2) 2;

And 2 is in the interval (1,3), so there is f(2) (2+2) 8=2

So there is f(2)=2

f(2)=4a+2b+c=2, subtract the two formulas to obtain 4b=2, b=1 2

f(5/2)=(25/4)a+(5/2)b+c=(25/4)a+c+5/4

And according to f(x) (x+2) 8, we get f(5 2) (5 2 +2) 8=81 32, so that we have (25 4)a+c+(5 4) 81 32

(25/4)a+c≤41/32

And by f(-2)=4a-2b+c=0, we can get 4a+c=2b=1, c=1-4a

Bring in the above formula: 25 4)a+c=(9 4)a+(4a+c)+5a=1+(9 4)a 41 32

a 1 8 and f(3 2) at f(3 2) 8=49 32

and f(3 2)=(9 4)a+(3 2)b+c=(9 4)a+c+3 4=(-7 4a)+4a+c+3 4=(-7 4)a+1+3 4=(-7a 4)+7 4

So there is (-7a 4) +7 4 49 32

a 1 8 then a = 1 8 gives c = 1 2

So f(x)=x 8 +x 2+1 2

3.The condition of the original problem is equivalent to the fact that there is always a real root in the equation f(x)-g(x)=0 when x [-2,2] exists.

and f(x)-g(x)=x 8-x 2+(1 2 -m).

Simplifying the equation yields: x -4x + (4-8m) = 0

And let the parabola h(x)=x-4x+(4-8m).

The problem then becomes that h(x) must intersect the x-axis in the interval of x[-2,2].

The axis of symmetry of h(x) can easily be obtained as x=2, the vertex (2,-8m) falls exactly at the right end of the interval [-2,2], and the opening of h(x) is still upward, so it can be determined that h(x) is monotonically decreasing on [-2,2], and its vertex (2,-8m) falls exactly at the right end of the interval.

For this parabola to have a must have an intersection with the x-axis on [-2,2], then all that is required is that the vertex function is less than or equal to 0 and that f(-2) 0:

Vertex function values -8m 0, m 0;f(-2)=16-8m≥0,m≤2

Therefore, the value range of m is [0,2].

1) Since x (1,3), f(x)<=(x+2) 2 8, substituting x=2 gives f(2)<=2

And f(2)>=2, so f(2)=2.

2) Since f(x)-x>=0 is constant, and f(2)-2=0, f(x)=x+a(x-2) 2, and f(-2)=16a-2=0, a=1 8

f(x)=x+(x-2)^2/8=(x+2)^2/8=x^2/8+x/2+1/2

3) From the problem, we know that f(x)-g(x)=(x-2) 2 8-m has a solution in [-2,2], and in x [-2,2], 0<=(x-2) 2 8<=2, so 0<=m<=2

(1) f(x)=ax2+bx+c satisfies For any real number, there is f(x) x,f(2)>=2, when x (1,3), f(x) (x+2)2 8 is constant, f(2)<=2, f(2)=2

2)f(2)=4a+2b+c=2,f(-2)=4a-2b+c=0,b=1/2,4a+c=1,……

（1）f(2)≥2

2 (1,3) has f(2) 2

So f(2)=2

2) f(2)=0 obtains: 4a+2b+c=2

f(-2)=0: 4a-2b+c=0

So b=1 2

2,0) is the vertex coordinate of f(x).

b/2a=-2

So a=1 8

c=1/2f(x)=1/8*x^2+1/2*x+1/2(3)g(x)=1/8*x^2+1/2*x+1/2-mx/2g'(x)=1/4*x+1/2-m/2

When x 0, g(x) must be a single increase, that is, 1 4*x+1 2-m 2>0 and x=0, g(0)>1 4

So they are solved separately: m<1

I hope it can help you, and I wish you progress in learning, and don't forget to adopt!

(1) Proof: f(2)=2

For for any real number x, there is f(x) x, f(x) (1 8)(x+2) 2

So: f(2) 2, and f(2) (1 8)*(2+2) 2=2

i.e., 2 f(2) 2

So f(2)=2

2) From f(2)=2, f(-2)=0, 4a+2b+c=0......①4a-2b+c=0……Get b = 1 23C=1-4a, substituting f(x) x, we get ax 2-(x 2)+1-4a 0 for any real number x constantly, a>0 and 0, i.e., a>0 and (8a-1) 2 0, but (8a-1) 2 0, a=1 8, c=1 2, it is verified that for any real number x, there is f(x) 1 8(x+2) 2 constant, f(x)=(1 8)x 2+(x 2)+(1 2).

3) g(x)=(1 8)x 2+(1-m)x 2)+(1 2) 1 4(x 0), i.e. x 0,x 2+(4m-1)x+2 0 has a solution on [0,+. ∴=8[(m-1)^2-1]≥0……①2（n-1）≥0……Solution m 1 + ( 2 2 ).

(1) It is known that there are: f(2)>=2.

Because 2 (1,3), f(2)<=(1 8)(2+2) 2=2

In summary, we get: f(2)=2

2) From f(2)=0 and f(-2)=0, three conclusions can be obtained:

4a+2b+c=0

4a-2b+c=0

f(0)=c>=0 (this is obtained from f(0)0): c=-4a>=0, i.e. a<=0

And there is a theorem (this should be the conclusion that the teacher gave you):

For any real number x, there is f(x) x, then: a>0, and <=0. (f(x)=f(x)-x, is the discriminant of f(x)).

So it follows that the expression for such f(x) does not exist.

3) From f(2)=0: 4a+2b+c=0, it is estimated that there is no way to get the result if there is no conclusion combined with the second question.

I remember encountering a similar problem 3 years ago.

Considering that x-1 and the function x 2-3x+3 are cut to (2,1) f(2)=1

f'(2)=(x-1)'|x=2=1

It is then coupled with f(-1)=0.

The solution is a = 2 9

b=1/9c=-1/9

f(x)=(2 9) x 2+(1 9)x-1 9 second sub-question.

Shift the term, using the discriminant 0 of the root

Solve the third sub-problem.

Observe n [-3,3], which crosses the second sub-question area, so when x1 = x2 |x1-x2|=0 to get the minimum value.

m^2+tm+1≤0

t is optional, and t=0 is obtained.

m^2+1≤0

It cannot be established, so there is no such m making the inequality.

m 2+TM+1 x1-x2 for any n [-3,3] constant.

Answer: Solution: By the condition of any real number x, there is f(x) x, and f(2) 2 is true.

When x(1,3), there is f(x).

x+2)2 is established, take x=2 when the morning brigade, f(2) 2+2)2=2 into a land bend bench, f(2)=2

4a+2b+c=2①

f(-2)=0

4a-2b+c=0②

From available, 4a+c=2b=1,b=

Hence B

Your conjecture is correct.

There is a>0, a+c=1 2

According to the basic inequality, there is indeed AC 1 16

Your thinking is very good, and this question is not rigorous.

The solution is from 1 4-4ac 0, and the solution is ac 1 16 and then from a 0, there must be c 0

The reason ac is positive.

i.e. c 0

(1),1/2[f(x1)+f(x2)]

ax1²+ax2²+bx1+bx2+2c]/2=[a(x1²+x2²)/2+b(x1+x2)/2+cf(x1+x2)/2)

a(x1+x2)/2)²+b(x1+x2)/2)+c=a(x1+x2)²/4+b(x1+x2)/2+c∵2(x1x2)≤(x1²+x2²)

2(x1x2)+x1²+x2²≤2(x1²+x2²)(x1+x2)²≤2(x1²+x2²)

x1+x2) 4 (x1 +x2 ) 2 when a>0.

a(x1+x2)²/4≤a(x1²+x2²)/2∴a(x1+x2)²/4+b(x1+x2)/2+c≤a(x1²+x2²)/2+b(x1+x2)/2+c

Therefore f((x1+x2) 2) 2

2) When x belongs to [-1,1], f(x) <=1, is there a,b,c such that f2) >36 5 is true? If so, write a set of values a, b, and c that meet the conditions; If not, please explain why.

It is known that the quadratic function f(x) = ax +bx+c

When x belongs to [-1,1], |f(x)|≤1

Let x=1,-1,0, respectively, get.

f(-1)|=|a-b+c|≤1

f(1)|=|a+b+c|≤1

f(0)|=|c|≤1

f2)|=|4a+2b+c|

3(a+b+c) +a-b+c) -3c| ≤3|a+b+c | a-b+c| +3|c|

f2)|≤3|a+b+c | a-b+c| +3|c| ≤7|f2)|7<36 5 does not exist.

1. Utilize (x+y) 2 2(x 2+y 2)1 2[f(x1)+f(x2)] = 1 2a(x1 2+x2 2) +1 2b(x1+x2) +c

a(x1+x2)^2 /4 + 1/2b(x1+x2) +c=f[(x1+x2)/2]

2,x belongs to [-1,1], |f(x)|<=1 gives x=1, -1, 0, respectively.

1 ≤ a+b+c ≤ 1

1 ≤ a-b+c ≤ 1

1 ≤ c ≤ 1

f(2) = 4a+2b+c = 3(a+b+c) +a-b+c) -3c

So-7 f(2) 7

So |f(2)|7<36 5 does not exist.

f(2)≥2

2 (1,3) has f(2) 2

So f(2)=2

f(2)=2: 4a+2b+c=2

f(-2)=0: 4a-2b+c=0

So b=1 2

2,0) is the vertex coordinate of f(x).

b/2a=-2

So a=1 8

Because there is always f(x)>x, a>=0 and the parabola is upward. And because when x belongs to (1,3), there is f(x)<=1 8(x+2) 2 is true, so a is not equal to 0, and f(1)=1 8(1+2) 2,f(3)=1 8(3+2) 2,f(-2)=0The system of simultaneous equations yields a, b, c...

I haven't done this kind of question for many years, and I miss it.

Substituting x=1 into , we get f(1)=1

b=1 23, the formula in can be reduced to .

ax²＋1／2x＋1／2﹣a≤1／4x²﹢1／2x﹢1／4﹙a－1／4﹚﹙x²－1﹚≤0

a=1／4，c＝1／4

The third question is just fine.