Function F X X X 8 X. It is proved that the equation F X F a with respect to X has three real soluti

Solution: (1) F1(x) is a quadratic function, y1=ax2+bx+c passes through vertices (0,0) and points (1,1).

Substituting vertices (0,0) and points (1,1) into y1=ax2+bx+c respectively yield:

c=01=a+b ∴b=1-a

The vertex coordinates of the quadratic function y=ax2+bx+c are (-b 2a, (4ac-b2) 4a).

b 2a = 0 and (4ac-b2) 4a = 0 (substituting c = 0 and b = 1-a into the equation).

A=1, B=0

f1(x)=x2

f2(x) is an inversely proportional function and has two intersections with the line y=x, let the two intersections be (x,x),(x,-x).

Use the two-point distance formula = [(x1-x2)2+(y1-y2)2], and get:

The two intersection points of x=2 2 are points (2 2, 2 2) and points (-2 2, -2 2).

Substituting the points (2 2, 2 2) into y=k x gives k=8

f2(x)=8/x

f(x)=f1(x)+f2(x)=x2+(8/x)（x≠0）

2) Transfer.

x2+8/x-a2-8/a=0

x-a)(ax2+a2x-8)/(ax)=0

i.e. (x-a)(ax2+a2x-8)=0

x=a is a solution.

Let's look at ax2+a2x-8=0

Discriminant δ = a 4+32a=a(a3+32).

Because a>0 so a(a3+32)>0 is δ>0

So ax2+a2x-8=0 has two solutions to real numbers.

In summary, when a>3, the equation for x2+8 x=a2+8 a has three real solutions.

Let h(x)=f x - f a =x 8 x- a -8 ah'(x)=2x-2a-8 x 2-8 a 2=0a 2*x 2(x-a)=4(a 2+x 2)a 2) x 3-(a 3+4) x 2-4a 2=0, let g(x) = (a 2) x 3-(a 3+4) x 2-4a 2g'(x)=3(a 2)x 2-2(a 3+4)x has two roots, and x=(2(a 3+4)) 3(a 2), g(x) is greater than 0

h(x)=f x - f a = x 8 x- a -8 a has three roots.

The equation f x f a about x has three real solutions.

Because f''(x) Continuous.

then there is f''(x) =3x-2f'(x)) x in Nasheng Zen let x=0 continuously.

That attack the old lim(x->0)f''(x)=3-2f''(x) Use the Lobida Rule.

There is f''(x)=1>0

So it's a minima.

by f(-x)=-fx), so there is the equation: (x+1)(x+a) x=-[1-x)(a-x) (x)], unraveling a=-1

1), x>=0, f(x)=x(x-a)=(x-a/2)^2-a^2/4

x<0, f(x)=-x(x-a)=-x-a 2) 2+a 2 4 When a>=0, the monotonic decreasing interval is: x>=a 2 or x<=0 The monotonic decreasing interval is: 0==0 or x<=a 2 The monotonic decreasing interval is:

A 2==0 or x<=A 2 monotonic reduction interval is: A 2==A 2 4---A 2+2A-1<=0-->1- 2=Shouting Filial Piety -5 2<=A

So when -5 2=when a<-5 2, the maximum value is f(-1)=-1-a

1) When a=0, f(x)=|x|*x, which is defined by f(-x)=|-x|*(x)= -|x|*x, hence the odd function;

When a≠0, it is determined by f(a)=0 and f(-a)= -2a*|a|≠0, so the function is neither odd nor even.

2) When a 0, f(x)={-x 2+ax(x<0) ; x 2-ax(x>=0), which is given by -x 2+ax=-(x+a 2) 2+a 2 4 ,x 2-ax=(x-a 2) 2-a 2 4 , the function is an increasing function on (- a 2), a subtraction function on (a 2,0), and an increasing function on (0,+).

3) When a<=0, it is known from 2) that if a<-2, the function is a subtraction function on [-1,0] and an increasing function on [0, and if -2<=a<=0, the function is an increasing function on [-1,a 2], a decreasing function on [a 2 ,0], and an increasing function on [0, because f(-1)= -1-a ,f( ,f(a 2)=a 2 4 , making -1-a >1 4-a 2 , then a<-5 2 , in summary, when a<=-5 2, the maximum value of the function on [-1, is f(-1)= -1-a ;

When -5 2

When a=0, odd function; When a≠0, it is not odd or even.

When x 0, f(x) = x(x-a).

When x 0, f(x) = x(a-x).

Draw the piecewise function image easily f(x) in (- a 2) increments, in (a 2,0) decrement, in (0,+ increments.

When a 2 -1 is a -2, f( f(-1)=-(a+1) f( compare the difference with the size of zero.

When -2 a 0, max=f(x)maxNote: Analogous to the idea of categorical discussion of the axial dynamic interval of the quadratic function, which is a monotonic interval dynamic interval determination. (It is the promotion and understanding of the former, just look at the picture and discuss it, it's easy.) ）

(1) When a=0, this function is odd; When a is not equal to 0, this function is non-odd and non-even.

First, when x=0, f(0)=0;

Then, according to the definition of odd function and even function, it is assumed that the function is even or odd function respectively, that is, f(x)=f(-x) or f(x)+f(-x)=0, respectively.

2) When x is less than 0, f(x)=-x 2+ax=-(x-a 2) 2+(a 2) 4, then, when x is less than a 2, the function increases monotonically, and when x is greater than a 2 and less than 0, the function decreases monotonically.

When x is greater than 0, f(x)=x 2-ax=(x-a 2) 2-(a 2) 4, then, because a is less than or equal to 0, when x is greater than 0, the function increases monotonically.

In summary, when x is less than a 2, the function increases monotonically, when x is greater than a 2 and less than 0, the function decreases monotonically, and when x is greater than 0, the function increases monotonically.

3) When x is greater than 0, the maximum value a is taken at x=, which is a=;

When x is less than 0, it is discussed by category.

When a 2 is greater than or equal to -1 and less than 0, that is, when a is greater than or equal to -2 and less than 0, the maximum value b is taken at x=a 2 and is b=(a 2) 4

When a 2 is less than -1, that is, when a is less than -2, the maximum value is taken when x = -1, and c = -1-a

a-b=-(a+1) 2 4+1 2, because a is greater than or equal to -2 and less than 0, so when a=-2 or 0, a-b has a minimum value of 1 4, and when a=-1, a-b has a maximum value of 1 2It can be seen that a is greater than b, and a is the maximum;

a-c=, because a is less than -2, a-c=0, a=;

Then, when a is less than, a-c is less than 0, a is less than c, and c is the maximum; When a is greater than and less than -2, a is greater than c, and a is the maximum;

In summary, when a is greater than, the maximum value is; When a is less, the maximum value is -1-a

<> Solution: From the meaning of the title, it can be seen that the image of the number f(x) is as follows:

by the equation f2 with respect to x

x)-af(x)=0 has exactly three real solutions that are not stuffy, and we can see that the equation a=f(x) has three different real solutions, that is, the image of the function y=a and the function y=f(x) have three different intersection points

From the image, it is easy to see that the value range of the real number a is (0, Ant Yu 1] so the answer is:

Column inequality groups are not required.

But the big questions of the function class all require derivatives.

f'（x）=3ax ²-2x

x(3a-2)

In general, the abscissa corresponding to the extreme value is 0 and 2 3a (but it is not known which is the maximum value and which is the minimum value).

Use f(0)*f(2 3a) 0

Get (-2, 3) 9, and many problems are like this: One extreme value is equal to a +1 and is greater than 0, just make the other value less than 0.

It will make many calculations simple, and the judgment of symbols requires a gradual understanding of the problem of solving the real numbers of cubic functions.

Three different real solutions are b -4ac 0 and maximum*minima <0, two different real solutions are b -4ac 0 and maximum*minimum=0, and one real number solution is b -4ac 0, note the equal sign.

For this kind of judgment, it is important to have a general picture of the quadratic and cubic functions of certain formulas, especially the images of the cubic functions, etc., etc., engraved in the mind!

a≠01

If a>0, f'(x)=3ax 2-2x order 3a x2-2x=0==".

x1=0; x2=2/3a

Since the function image is capitalized n, that is, it first increases, then decreases, and then increases, so there are three different conditions for the solution of real numbers.

f0)>0

f2/3a)<0

1>0a^2>4/27==>

a>2/3√3

So a>2 3 3

When 2a<0, the function decreases first and then increases and then decreases.

x=0 is the big root of the derivative, 2 3a is the small root, and f(0) > 0

The expression of f(2 3a) <0 is the same as 1), and in a 2>4 27 take a<-2 3 3

So a< -2 3 3 or a> 2 3 3

This function is cubic and is generally solved by derivation.

Using the derivative, the maximum and minimum values of f(x) are determined, and since the equation f(x)=0 needs to have three roots, it is required:

1) the maximum value is greater than 0, and (2) the minimum value is less than 0

Solve this group of inequalities and you're good to go.

Answer: <>

Solution: Let t=f(x), then f(t)=0, if a 0, when x 0, f(x)=a?2x

by f(t)=0, i.e. log

t 0 at this time t=1, when t = 1 to get f(x) = 1, then x=1 has a unique solution, then the condition is satisfied

If a=0, then when x 0, f(x)=a?2x0, at which point the function has an infinite number of points and does not meet the condition

If a 0, when x 0, f(x)=a?2x

0，a]．<

At this time, the maximum value of f(x) is a, so that if the equation f(f(x)) = 0 about x has and only one real number solution, then a 1, at this time 0 a 1, and the value range of the real number a is (- 0) (0, 1), so choose b