How to define an even or odd online. Properties of functions

How to insert mathematical formulas on the site?

If you need to ever add one or two mathematical formulas on a web page, it is easiest to do this, as described in the article: Mathematical formulas are easily inserted into the site in the form of pictures that automatically generates alpha tungsten. In addition to simplicity, this universal way will help improve the visibility of the site in search engines. It works for a long time (and, I think, will work forever), but morally outdated.

If you are constantly using mathematical formulas on your site, then I recommend that you use MathJax - a special JavaScript library that displays mathematical designations in web browsers using Mathml, LaTEX or ASCIIMATHML markup.

There are two ways to start using MathJax: (1) With the help of a simple code, you can quickly connect the MathJax script to your site, which will be in the right moment automatically boost from a remote server (server list); (2) Download MathJax script from a remote server to your server and connect to all pages of your site. The second method is more complicated and long - will speed up the download of the pages of your site, and if the MathJax parent server for some reason becomes temporarily unavailable, it will not affect your own website. Despite these advantages, I chose the first way as a simpler, fast and not requiring technical skills. Follow my example, and after 5 minutes you can use all the features of MathJax on your website.

You can connect the MathJax library script from a remote server using two code options taken on MathJax's main website or on the documentation page:

One of these code options must be copied and insert into the code of your webpage, preferably between the tags and or immediately after the tag . According to the first version, MathJax is loaded faster and slows down the page. But the second option automatically tracks and loads the latest MathJax versions. If you insert the first code, it will need to be periodically updated. If you insert the second code, the pages will be loaded slower, but you will not need to constantly monitor MathJax updates.

Connect MathJax is the easiest way to Blogger or WordPress: Add a widget for inserting a third-party JavaScript code to insert the first or second version of the download code presented above and place the widget closer to the beginning of the template (by the way, it is not at all necessary Since the MathJax script is loaded asynchronously). That's all. Now read Mathml, Latex and Asciimathml markup syntax, and you are ready to insert mathematical formulas on the web pages of your site.

Any fractal is based on a specific rule that is consistently applied to an unlimited number of times. Everyone is called iteration.

The iterative algorithm for constructing the menger sponge is quite simple: the source cube with a side 1 is divided by planes parallel to its faces, on 27 equal cubes. One central cube and 6 adjacent cubes are removed from it. A set consisting of 20 remaining smaller cubes is obtained. By doing the same with each of these cubes, we obtain a set, consisting already from 400 smaller cubes. Continuing this process infinitely, we get a sponge of Menger.

To do this, use a millimeter or graphic calculator. Choose a few any numeric values \u200b\u200bof an independent variable. X (\\ DisplayStyle X) and substitute them to function to calculate the values \u200b\u200bof the dependent variable Y (\\ DisplayStyle Y). Found coordinates of points Apply to the coordinate plane, and then connect these points to build a function graph.

• Function Positive numeric values X (\\ DisplayStyle X) and corresponding negative numeric values. For example, a function is given. Substitute the following values \u200b\u200binto it. X (\\ DisplayStyle X):
  • f (1) \u003d 2 (1) 2 + 1 \u003d 2 + 1 \u003d 3 (\\ displaystyle f (1) \u003d 2 (1) ^ (2) + 1 \u003d 2 + 1 \u003d 3) (1, 3) (\\ DisplayStyle (1,3)).
  • F (2) \u003d 2 (2) 2 + 1 \u003d 2 (4) + 1 \u003d 8 + 1 \u003d 9 (\\ displaystyle f (2) \u003d 2 (2) ^ (2) + 1 \u003d 2 (4) +1 \u003d 8 + 1 \u003d 9). Received a point with coordinates (2, 9) (\\ DisplayStyle (2,9)).
  • F (- 1) \u003d 2 (- 1) 2 + 1 \u003d 2 + 1 \u003d 3 (\\ displaystyle f (-1) \u003d 2 (-1) ^ (2) + 1 \u003d 2 + 1 \u003d 3). Received a point with coordinates (- 1, 3) (\\ DisplayStyle (-1.3)).
  • f (- 2) \u003d 2 (- 2) 2 + 1 \u003d 2 (4) + 1 \u003d 8 + 1 \u003d 9 (\\ displaystyle f (-2) \u003d 2 (-2) ^ (2) + 1 \u003d 2 ( 4) + 1 \u003d 8 + 1 \u003d 9). Received a point with coordinates (- 2, 9) (\\ DisplayStyle (-2.9)).

The parity and oddness of the function are one of its main properties, and in parity occupies an impressive part of the school course in mathematics. It determines the nature of the behavior of the function and greatly facilitates the construction of an appropriate schedule.

Determine the parity of the function. Generally speaking, the function under study is considered even if for opposite values \u200b\u200bof an independent variable (X), which are in its definition area, the corresponding values \u200b\u200bof y (functions) will be equal.

We will give a more stringent definition. Consider some function f (x), which is set in the region D. It will be even if for any point x in the field of definition:

• -x (opposite point) also lies in this area of \u200b\u200bdefinition,

• f (-x) \u003d f (x).

From the given definition, the condition necessary for the determination of such a function is followed, namely, symmetry relative to the point of the onset of coordinates, since if some point b is contained in the field of determining the even function, the corresponding point is also lies in this area. Of the foregoing, therefore, it follows: even function It has symmetrical in relation to the axis of the ordinate (OY) view.

How to practice the parity of the function?

Let be given using the formula H (x) \u003d 11 ^ x + 11 ^ (- x). Following the algorithm that arises directly from the definition, we examine primarily its definition area. Obviously, it is defined for all the values \u200b\u200bof the argument, that is, the first condition is fulfilled.

The next step is to substitute instead of the argument (x) its opposite value (-X).
We get:
h (-x) \u003d 11 ^ (- x) + 11 ^ x.
Since the addition satisfies the commutative (transitional) law, then obviously h (-x) \u003d h (x) and the specified functional dependence is even.

Check the parity of the function h (x) \u003d 11 ^ x-11 ^ (- x). Following the same algorithm, we obtain that H (-x) \u003d 11 ^ (- x) -11 ^ x. I will make minus, as a result, we have
h (-X) \u003d - (11 ^ x-11 ^ (- x)) \u003d - h (x). Consequently, H (x) is odd.

By the way, you should recall that there are functions that cannot be classified according to these features, they are called neither even or odd.

Even functions have a number of interesting properties:

• as a result of the addition of such functions, it is also obtained;
• as a result, the subtraction of such functions receive even;
• even, also even;
• as a result of multiplying two such functions, it is also obtained;
• as a result of multiplication of odd and even functions, the odd is obtained;
• as a result of the division of odd and even functions, they receive odd;
• the derivative of such a function is odd;
• if we build an odd function in a square, we get even.

The readiness of the function can be used in solving equations.

To solve the equation of type G (x) \u003d 0, where the left part of the equation is an even function, it will be enough to find it solutions for non-negative values \u200b\u200bof the variable. The obtained roots of the equation must be combined with opposite numbers. One of them is subject to verification.

This is successfully used to solve non-standard tasks with a parameter.

For example, is there any value of the parameter A, in which the equation 2x ^ 6-X ^ 4-AX ^ 2 \u003d 1 will have three roots?

If we consider that the variable is included in the equation in even degrees, it is clear that the replacement of X is x the specified equation will not change. It follows that if a certain number is its root, then it is also the opposite number. The conclusion is obvious: the roots of the equation other than zero are included in the many of his "couples" solutions.

It is clear that the number 0 itself is not, that is, the number of roots of a similar equation can only be even even and, naturally, none of the parameter value it cannot have three roots.

But the number of roots of the equation 2 ^ x + 2 ^ (- x) \u003d AX ^ 4 + 2x ^ 2 + 2 may be odd, and for any value of the parameter. Indeed, it is easy to check that many roots of this equation Contains solutions "couples". Check whether 0 is the root. When substituting it into the equation, we obtain 2 \u003d 2. Thus, in addition to the "paired" 0, it is also the root, which proves their odd amount.

evenif at all \\ (x \\) from its definition area is true: \\ (f (-x) \u003d f (x) \\).

An even function graph is symmetrical with respect to the axis \\ (Y \\):

Example: the function \\ (f (x) \u003d x ^ 2 + \\ cos X \\) is even, because \\ (f (-X) \u003d (- x) ^ 2 + \\ cos ((- x)) \u003d x ^ 2 + \\ cos x \u003d f (x) \\).

\\ (\\ BLACKTRIANGLERIGHT \\) The function \\ (f (x) \\) is called oddif at all \\ (x \\) from its definition area is true: \\ (f (-x) \u003d - f (x) \\).

The schedule of an odd function is symmetrical on the start of the coordinates:

Example: the function \\ (f (x) \u003d x ^ 3 + x \\) is an odd, because \\ (f (-x) \u003d (- x) ^ 3 + (- x) \u003d - x ^ 3 - x \u003d - (x ^ 3 + x) \u003d - f (x) \\).

\\ (\\ blacktriangleright \\) Functions that are neither even nor odd, are called functions general view. Such a function can always be unique in the form of an even and odd function as the sum.

For example, the function \\ (f (x) \u003d x ^ 2-x \\) is the sum of the even function \\ (F_1 \u003d x ^ 2 \\) and the odd \\ (F_2 \u003d -x \\).

\\ (\\ blacktriangleright \\) Some properties:

1) The product and private two functions of the same parity are an even function.

2) The product and private two functions of different parity are an odd function.

3) The sum and difference of even functions is an even function.

4) The sum and the difference of odd functions is an odd function.

5) If \\ (F (x) \\) is an even function, the equation \\ (f (x) \u003d c \\ (C \\ in \\ mathbb (r) \\)) has the only root then and only when when \\ (x \u003d 0 \\).

6) If \\ (f (x) \\) is an even or odd function, and the equation \\ (f (x) \u003d 0 \\) has a root \\ (x \u003d b \\), then this equation will necessarily have a second root \\ (x \u003d -B \\).

\\ (\\ BLACKTRIANGLERIGHT \\) The function \\ (F (x) \\) is called periodic to \\ (x \\), if for a certain number \\ (t \\ ne 0 \\) is made \\ (F (x) \u003d f (x + t) \\), where \\ (x, x + t \\ in x \\). The smallest \\ (t \\), for which this equality is satisfied, is called the main (basic) function period.

In the periodic function, any number of species \\ (nt \\), where \\ (n \\ in \\ mathbb (z) \\) will also be a period.

Example: anyone trigonometric function is periodic;
functions \\ (f (x) \u003d \\ sin x \\) and \\ (f (x) \u003d \\ cos x \\) the main period is \\ (2 \\ pi \\), for functions \\ (f (x) \u003d \\ mathrm ( TG) \\, x \\) and \\ (f (x) \u003d \\ mathrm (CTG) \\, X \\) The main period is \\ (\\ pi \\).

In order to build a graph of a periodic function, you can construct its schedule on any segment length \\ (T \\) (the main period); Then the schedule of the entire function is completed with a shift of the constructed part by an integer number of periods to the right and left:

\\ (\\ (\\ blacktriangleright \\) The definition area \\ (D (F) \\) functions \\ (F (x) \\) is a set consisting of all the values \u200b\u200bof the argument \\ (x \\), in which the function makes sense (defined).

Example: function \\ (F (X) \u003d \\ SQRT X + 1 \\) Definition area: \\ (X \\ IN

Task 1 # 6364

Task level: equal to ege

Under what values \u200b\u200bof the parameter \\ (a \\) equation

it has only decision?

Note that since \\ (x ^ 2 \\) and \\ (\\ cos x \\) are even functions, if the equation has a root \\ (x_0 \\), it will also have a root \\ (- x_0 \\).
Indeed, let \\ (x_0 \\) - the root, that is, equality \\ (2x_0 ^ 2 + a \\ mathrm (Tg) \\, (\\ cos x_0) + a ^ 2 \u003d 0 \\) right. Substitute \\ (- x_0 \\): \\ (2 (-x_0) ^ 2 + A \\ MathRM (TG) \\, (\\ cos (-x_0)) + a ^ 2 \u003d 2x_0 ^ 2 + A \\ MathRM (TG) \\, (\\ cos x_0) + a ^ 2 \u003d 0 \\).

Thus, if \\ (x_0 \\ ne 0 \\), the equation will already have at least two roots. Consequently, \\ (x_0 \u003d 0 \\). Then:

We obtained two values \u200b\u200bof the parameter \\ (a \\). Note that we used what \\ (x \u003d 0 \\) is precisely the root of the original equation. But we have not used anywhere that it is the only one. Therefore, it is necessary to substitute the resulting values \u200b\u200bof the parameter \\ (a \\) to the initial equation and check at which \\ (a \\) root \\ (x \u003d 0 \\) will indeed be the only one.

1) If \\ (a \u003d 0 \\), the equation will take the form \\ (2x ^ 2 \u003d 0 \\). Obviously, this equation has only one root \\ (x \u003d 0 \\). Consequently, the value \\ (a \u003d 0 \\) is suitable for us.

2) if \\ (a \u003d - \\ mathrm (Tg) \\, 1 \\), the equation will take the form \ Rewrite the equation in the form \ As \\ (- 1 \\ leqslant \\ cos x \\ leqslant 1 \\)T. \\ (- \\ MathRM (TG) \\, 1 \\ Leqslant \\ Mathrm (TG) \\, (\\ COS X) \\ Leqslant \\ Mathrm (TG) \\, 1 \\). Consequently, the values \u200b\u200bof the right part of the equation (*) belong to the segment \\ ([- \\ MathRM (TG) ^ 2 \\, 1; \\ MathRM (TG) ^ 2 \\, 1] \\).

Since \\ (x ^ 2 \\ geqslant 0 \\), the left part of the equation (*) is greater than or equal to \\ (0+ \\ MathRM (TG) ^ 2 \\, 1 \\).

Thus, equality (*) can be performed only when both parts of the equation are \\ (\\ MathRM (TG) ^ 2 \\, 1 \\). And this means that \\ [\\ Begin (Cases) 2x ^ 2 + \\ MathRM (TG) ^ 2 \\, 1 \u003d \\ MathRM (TG) ^ 2 \\, 1 \\\\\\\\\\ Mathrm (TG) \\, 1 \\ Cdot \\ Mathrm (TG) \\ \u003d \\ MathRM (TG) \\, 1 \\ END (CASES) \\ quad \\ leftrightarrow \\ quad x \u003d 0 \\] Consequently, the value \\ (a \u003d - \\ mathrm (TG) \\, 1 \\) is suitable for us.

Answer:

\\ (A \\ in \\ (- \\ MathRM (TG) \\, 1; 0 \\) \\)

Task 2 # 3923

Task level: equal to ege

Find all parameter values \u200b\u200b\\ (a \\), each you are a function graph \

symmetrical on the start of the coordinates.

If the graph of the function is symmetrical relative to the start of the coordinates, this function is odd, that is, it is made \\ (f (-x) \u003d - f (x) \\) for any \\ (x \\) from the function of determining the function. Thus, it is required to find those values \u200b\u200bof the parameter at which it is made \\ (f (-x) \u003d - f (x). \\)

\\ [\\ begin (aligned) & 3 \\ mathrm (tg) \\, \\ left (- \\ dfrac (AX) 5 \\ RIGHT) +2 \\ sin \\ dfrac (8 \\ pi a + 3x) 4 \u003d - \\ left (3 \\ , \\ dfrac (AX) 5 + 2 \\ sin \\ dfrac (8 \\ pi a + 3x) 4 \u003d - \\ left (3 \\ MathRM (TG) \\, \\ left (\\ dfrac (AX) 5 \\ RIGHT) +2 \\ 3X) 4 \u003d 0 \\ quad \\ rightarrow \\ quad2 \\ sin \\ dfrac12 \\ left (\\ dfrac (8 \\ pi a + 3x) 4+ \\ dfrac (8 \\ PI A-3X) 4 \\ Right) \\ CDOT \\ COS \\ DFRAC12 \\ left (\\ dfrac (8 \\ pi a + 3x) 4- \\ dfrac (8 \\ pi a-3x) 4 \\ right) \u003d 0 \\ quad \\ rightarrow \\ quad \\ sin (2 \\ pi a) \\ cdot \\ cos \\ The latter equation must be performed for all \\ (x \\) from the definition area \\ (F (x) \\), therefore,

\\ (\\ sin (2 \\ pi a) \u003d 0 \\ rightarrow a \u003d \\ dfrac n2, n \\ in \\ mathbb (z) \\) \\ (\\ dfrac n2, n \\ in \\ mathbb (z) \\).

Answer:

Task 3 # 3069

Find all the parameter values \u200b\u200b\\ (a \\), each you each of which the equation \\ has 4 solutions, where \\ (F \\) is an even periodic with a period \\ (T \u003d \\ DFRAC (16) 3 \\) a function defined on the entire numeric direct , moreover, \\ (f (x) \u003d ax ^ 2 \\) when

Task level: equal to ege

\\ (0 \\ leqslant x \\ leqslant \\ dfrac83. \\) (Task from subscribers)

Since \\ (f (x) \\) is an even function, then its graph is symmetrical relative to the ordinate axis, therefore, when

\\ (- \\ dfrac83 \\ leqslant x \\ leqslant 0 \\) \\ (F (x) \u003d AX ^ 2 \\). Thus, when \\ (- \\ dfrac83 \\ leqslant x \\ leqslant \\ dfrac83 \\) , and this is a length of length \\ (\\ dfrac (16) 3 \\), the function \\ (F (x) \u003d AX ^ 2 \\).1) Let \\ (a\u003e 0 \\). Then the function graph (F (X) \\) will look like this:

Then so that the equation has 4 solutions, it is necessary that the graph \\ (G (x) \u003d | A + 2 | \\ CDOT \\ SQRTX \\) passed through the point \\ (A \\):


Hence,


\\ [\\ dfrac (64) 9a \u003d | a + 2 | \\ Cdot \\ sqrt8 \\ quad \\ leftrightarrow \\ quad \\ left [\\ begin (gathered) \\ begin (aligned) & 9 (a + 2) \u003d 32a \\\\ & 9 (a +2) \u003d - 32A \\ END (Aligned) \\ END (Gathered) \\ Right. \\ quad \\ leftrightarrow \\ quad \\ left [\\ begin (gathered) \\ begin (aligned) & a \u003d \\ dfrac (18) (23) \\\\ & A \u003d - \\ DFRAC (18) (41) \\ END (Aligned) \\ END ( Gathered) \\ Right. \\] Since \\ (a\u003e 0 \\), it is suitable \\ (a \u003d \\ dfrac (18) (23) \\). 2) Let \\ (a

<0\) . Тогда картинка окажется симметричной относительно начала координат:


It is necessary that the graph \\ (G (x) \\) passed through the point \\ (B \\): \\ [\\ dfrac (64) 9A \u003d | A + 2 | \\ CDOT \\ SQRT (-8) \\ quad \\ leftrightarrow \\ quad \\ left [\\ begin (gathered) \\ begin (aligned) & a \u003d \\ dfrac (18) (23 ) \\\\ & A \u003d - \\ DFRAC (18) (41) \\ END (ALIGNED) \\ END (Gathered) \\ Right. \\] As \\ (a<0\) , то подходит \(a=-\dfrac{18}{41}\) .

3) the case when \\ (a \u003d 0 \\) is not suitable, since then \\ (f (x) \u003d 0 \\) for all \\ (x \\), \\ (g (x) \u003d 2 \\ sqrtx \\) and The equation will have only 1 root.

Answer:

\\ (a \\ in \\ left \\ (- \\ dfrac (18) (41); \\ DFRAC (18) (23) \\ Right \\) \\)

Task 4 # 3072

Task level: equal to ege

Find all values \u200b\u200b\\ (a \\), each you are \

it has at least one root.

Since \\ (f (x) \\) is an even function, then its graph is symmetrical relative to the ordinate axis, therefore, when

Rewrite the equation in the form \ and consider two functions: \\ (G (x) \u003d 7 \\ sqrt (2x ^ 2 + 49) \\) and \\ (f (x) \u003d 3 | x-7a | -6 | x | -a ^ 2 + 7a \\ The function \\ (G (x) \\) is even, has a point of minimum \\ (x \u003d 0 \\) (and \\ (G (0) \u003d 49 \\)).
The function \\ (F (x) \\) with \\ (x\u003e 0 \\) is decreasing, and with \\ (x
Indeed, with \\ (x\u003e 0 \\), the second module will reveal positively (\\ (| x | \u003d x \\)), therefore, regardless of how the first module is revealed, \\ (F (x) \\) will be equal to \\ ( KX + A \\), where \\ (A \\) is the expression from \\ (a \\), and \\ (k \\) is equal to either \\ (- 9 \\), or \\ (- 3 \\). With \\ (x<0\) – возрастающей, следовательно, \(x=0\) – точка максимума.
Find the value \\ (F \\) at the maximum point: \\<0\) наоборот: второй модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(3\) , либо \(9\) .
In order for the equation to have at least one solution, it is necessary that the graphs of functions \\ (F \\) and \\ (G \\) have at least one point of intersection. Therefore, you need:

\\ (A \\ In \\ (- 7 \\) \\ CUP \\) \ \\]

Answer:

Task 5 # 3912

Find all the parameter values \u200b\u200b\\ (a \\), each you are

Task level: equal to ege

Has six different solutions. \

we will replace \\ ((\\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) \u003d T \\), \\ (T\u003e 0 \\). Then the equation will take the form

We will gradually write out the conditions under which the initial equation will have six solutions. \ Note that the square equation \\ ((*) \\) can maximize two solutions. Any cubic equation \\ (AX ^ 3 + BX ^ 2 + CX + D \u003d 0 \\) may have no more than three solutions. Therefore, if the equation \\ ((*) \\) has two different solutions (positive!, Since \\ (t \\) must be larger than zero) \\ (T_1 \\) and \\ (t_2 \\), then by making a replacement, we We get:
\\ [\\ left [\\ begin (gathered) \\ begin (aligned) & (\\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) \u003d t_1 \\\\ \\ (\\ sqrt2) ^ (x ^ 3-3x ^ 2 +4) \u003d T_2 \\ END (Aligned) \\ END (Gathered) \\ Right. \\] Since any positive number can be represented as \\ (\\ sqrt2 \\) to some extent, for example, \\ (t_1 \u003d (\\ sqrt2) ^ (\\ log _ (\\ sqrt2) t_1) \\) , the first equation of the aggregate will rewrite in the form ofAs we have already spoken, any cubic equation has no more than three solutions, therefore, each equation from the aggregate will have no more than three solutions. So, the entire totality will have no more than six decisions. \
It means that the initial equation has six solutions, the square equation \\ ((*) \\) must have two different solutions, and each obtained cubic equation (from the aggregate) should have three different solutions (no solution of one equation should coincide with what -Lo decision of the second!)
Obviously, if the square equation \\ ((*) \\) will have one solution, we will not get six solutions in the original equation.

Thus, the solution plan becomes clear. Let's repel the conditions that must be performed.

1) To the equation \\ ((*) \\) had two different solutions, its discriminant must be positive: \

2) It is also necessary that both roots are positive (since \\ (T\u003e 0 \\)). If the product of the two roots is positive and the amount is positive, then the roots themselves will be positive. Therefore, you need: \\ [\\ Begin (Cases) 12-A\u003e 0 \\\\ - (A-10)\u003e 0 \\ End (Cases) \\ Quad \\ Leftrightarrow \\ Quad A<10\]

Thus, we have already provided two different positive roots \\ (t_1 \\) and \\ (t_2 \\).

3) Let's look at such an equation \ At what \\ (t \\) will it have three different solutions?
Consider the function \\ (f (x) \u003d x ^ 3-3x ^ 2 + 4 \\).
You can decompose on multipliers: \ Consequently, its zeros: \\ (x \u003d -1; 2 \\).
If you find the derivative \\ (f "(x) \u003d 3x ^ 2-6x \\), then we obtain two extremum points \\ (x_ (max) \u003d 0, x_ (min) \u003d 2 \\).
Therefore, the schedule looks like this:


We see that any horizontal straight line \\ (y \u003d k \\), where \\ (0 \\ (x ^ 3-3x ^ 2 + 4 \u003d \\ log _ (\\ sqrt2) t \\) had three different solutions, you need to \\ (0<\log_ {\sqrt2}t<4\) .
Thus, you need: \\ [\\ Begin (Cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\] Let us immediately note that if the numbers \\ (t_1 \\) and \\ (t_2 \\) are different, then the numbers \\ (\\ log _ (\\ sqrt2) t_1 \\) and \\ (\\ log _ (\\ sqrt2) t_2 \\) will be different, So the equations \\ (x ^ 3-3x ^ 2 + 4 \u003d \\ log _ (\\ sqrt2) t_1 \\) and \\ (x ^ 3-3x ^ 2 + 4 \u003d \\ log _ (\\ sqrt2) t_2 \\) Will have the roots at all.
The system \\ ((**) \\) can be rewritten so: \\ [\\ Begin (Cases) 1

Thus, we determined that both roots of the equation \\ ((*) \\) must lie in the interval \\ ((1; 4) \\). How to write this condition?
In explicit form, write the roots we will not.
Consider the function \\ (G (T) \u003d T ^ 2 + (A-10) T + 12-A \\). Its graph is a parabola with branches up, which has two intersection points with the abscissa axis (we recorded this condition in paragraph 1)). How should its schedule look like that the intersection points with the abscissa axis were in the interval \\ ((1; 4) \\)? So:


First, the values \u200b\u200b\\ (g (1) \\) and \\ (g (g) \\) functions at points \\ (1 \\) and \\ (4 \\) should be positive, secondly, the pearabol vertex \\ (T_0 \\ Therefore, you can write the system: \\ [\\ begin (Cases) 1 + A-10 + 12-A\u003e 0 \\\\ 4 ^ 2 + (A-10) \\ CDOT 4 + 12-A\u003e 0 \\\\ 1<\dfrac{-(a-10)}2<4\end{cases}\quad\Leftrightarrow\quad 4\\ (a \\) always has at least one root \\ (x \u003d 0 \\). It means to fulfill the condition of the task you need to equate \

there had four different roots other than zero, representing with \\ (x \u003d 0 \\) arithmetic progression.

Note that the function \\ (y \u003d 25x ^ 4 + 25 (a - 1) x ^ 2-4 (A-7) \\) is even, it means that \\ (x_0 \\) is the root of the equation \\ ((*) \\ Then it is necessary that the roots of this equation are ordered by increasing numbers: \\ (- 2d, -d, d, 2d \\) (then \\ (d\u003e 0 \\)). It was then that the data five numbers will form an arithmetic progression (with a difference \\ (D \\)).

So that these roots are numbers \\ (- 2d, -d, d, 2d \\), it is necessary that the numbers \\ (d ^ (\\, 2), 4d ^ (\\, 2) \\) are roots of the equation \\ (25t ^ 2 +25 (A-1) T-4 (A-7) \u003d 0 \\). Then the Vieta Theorem:

Rewrite the equation in the form \ and consider two functions: \\ (G (x) \u003d 20a-a ^ 2-2 ^ (x ^ 2 + 2) \\) and \\ (f (x) \u003d 13 | x | -2 | 5x + 12a | \\) .
The function \\ (G (x) \\) has a maximum point \\ (x \u003d 0 \\) (and \\ (G _ (\\ Text (versh)) \u003d g (0) \u003d - a ^ 2 + 20a-4 \\)):
\\ (G "(x) \u003d - 2 ^ (x ^ 2 + 2) \\ CDOT \\ LN 2 \\ CDOT 2x \\). Zero derivative: \\ (x \u003d 0 \\). With \\ (x<0\) имеем: \(g">0 \\), with \\ (x\u003e 0 \\): \\ (G "<0\) .
The function \\ (f (x) \\) with \\ (x\u003e 0 \\) is increasing, and with \\ (x<0\) – убывающей, следовательно, \(x=0\) – точка минимума.
Indeed, with \\ (x\u003e 0 \\), the first module will reveal positively (\\ (| x | \u003d x \\)), therefore, regardless of how the second module is revealed, \\ (F (x) \\) will be equal to \\ ( KX + A \\), where \\ (a \\) is the expression from \\ (a \\), and \\ (k \\) is equal to either \\ (13-10 \u003d 3 \\), or \\ (13 + 10 \u003d 23 \\). With \\ (x<0\) наоборот: первый модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(-3\) , либо \(-23\) .
We find the value \\ (F \\) at the point of the minimum: \

\\ (A \\ In \\ (- 7 \\) \\ CUP \\) \ Resolving this set of systems, we will receive the answer: \\]

Answer:

\\ (a \\ in \\ (- 2 \\) \\ CUP \\)