Task 1 #6329
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the system \[\begin(cases) (x-2a-2)^2+(y-a)^2=1\\ y^2=x^2\end(cases)\]
has exactly four solutions.
(USE 2018, main wave)
The second equation of the system can be rewritten as \(y=\pm x\) . Therefore, consider two cases: when \(y=x\) and when \(y=-x\) . Then the number of solutions of the system will be equal to the sum of the number of solutions in the first and second cases.
1) \(y=x\) . Substitute in the first equation and get: \
(note that in the case of \(y=-x\) we will do the same and also get a quadratic equation)
For the original system to have 4 various solutions, you need to get 2 solutions in each of the two cases.
A quadratic equation has two roots when its \(D>0\) . Let's find the discriminant of equation (1):
\(D=-4(a^2+4a+2)\) .
Discriminant greater than zero: \(a^2+4a+2<0\)
, откуда \(a\in (-2-\sqrt2; -2+\sqrt2)\).
2) \(y=-x\) . We get a quadratic equation: \
The discriminant is greater than zero: \(D=-4(9a^2+12a+2)>0\) , whence \(a\in \left(\frac(-2-\sqrt2)3; \frac(-2+\sqrt2)3\right)\).
It is necessary to check whether the solutions in the first case are the same as the solutions in the second case.
Let \(x_0\) be the general solution of equations (1) and (2), then \
From here we get that either \(x_0=0\) or \(a=0\) .
If \(a=0\) , then equations (1) and (2) turn out to be the same, therefore, they have identical roots. This case does not suit us.
If \(x_0=0\) is their common root, then \(2x_0^2-2(3a+2)x_0+(2a+2)^2+a^2-1=0\), whence \((2a+2)^2+a^2-1=0\) , whence \(a=-1\) or \(a=-0,6\) . Then the whole original system will have 3 different solutions, which does not suit us.
Given all this, the answer will be:
Answer:
\(a\in\left(\frac(-2-\sqrt2)3; -1\right)\cup\left(-1; -0.6\right)\cup\left(-0.6; - 2+\sqrt2\right)\)
Task 2 #4032
Task level: Equal to the Unified State Examination
Find all values \(a\) , for each of which the system \[\begin(cases) (a-1)x^2+2ax+a+4\leqslant 0\\ ax^2+2(a+1)x+a+1\geqslant 0 \end(cases)\ ]
It has only decision.
Let's rewrite the system as: \[\begin(cases) ax^2+2ax+a\leqslant x^2-4\\ ax^2+2ax+a\geqslant -2x-1 \end(cases)\] Consider three functions: \(y=ax^2+2ax+a=a(x+1)^2\) , \(g=x^2-4\) , \(h=-2x-1\) . It follows from the system that \(y\leqslant g\) , but \(y\geqslant h\) . Therefore, for the system to have solutions, the graph \(y\) must be in the area, which is given by the conditions: “above” the graph \(h\) , but “below” the graph \(g\) :
(we will call the “left” region region I, the “right” region - region II)
Note that for every fixed \(a\ne 0\) graph \(y\) is a parabola whose vertex is at the point \((-1;0)\) , and whose branches are either up or down. If \(a=0\) , then the equation looks like \(y=0\) and the graph is a straight line coinciding with the x-axis.
Note that in order for the original system to have a unique solution, it is necessary that the graph \(y\) has exactly one common point with region I or region II (this means that the graph \(y\) must have a single common point with the border of one of these regions).
Let's consider several cases separately.
1) \(a>0\) . Then the branches of the parabola \(y\) are turned upwards. In order for the original system to have a unique solution, it is necessary that the parabola \(y\) touches the boundary of region I or the boundary of region II, that is, it touches the parabola \(g\) , and the abscissa of the point of contact should be \(\leqslant -3\) or \(\geqslant 2\) (that is, the parabola \(y\) must touch the border of one of the regions that is above the x-axis, since the parabola \(y\) lies above the x-axis).
\(y"=2a(x+1)\) , \(g"=2x\) . Conditions for graphs \(y\) and \(g\) to touch at the point with abscissa \(x_0\leqslant -3\) or \(x_0\geqslant 2\) : \[\begin(cases) 2a(x_0+1)=2x_0\\ a(x_0+1)^2=x_0^2-4 \\ \left[\begin(gathered)\begin(aligned) &x_0\leqslant - 3\\ &x_0\geqslant 2 \end(aligned)\end(gathered)\right. \end(cases) \quad\Leftrightarrow\quad \begin(cases) \left[\begin(gathered)\begin(aligned) &x_0\leqslant -3\\ &x_0\geqslant 2 \end(aligned)\end(gathered) \right.\\ a=\dfrac(x_0)(x_0+1)\\ x_0^2+5x_0+4=0 \end(cases)\] From the given system \(x_0=-4\) , \(a=\frac43\) .
We got the first value of the parameter \(a\) .
2) \(a=0\) . Then \(y=0\) and it is clear that the line has an infinite number of points in common with region II. Therefore, this parameter value does not suit us.
3) \(a<0\) . Тогда ветви параболы \(y\) обращены вниз. Чтобы у исходной системы было единственное решение, нужно, чтобы парабола \(y\) имела одну общую точку с границей области II, лежащей ниже оси абсцисс. Следовательно, она должна проходить через точку \(B\) , причем, если парабола \(y\) будет иметь еще одну общую точку с прямой \(h\) , то эта общая точка должна быть “выше” точки \(B\) (то есть абсцисса второй точки должна быть \(<1\) ).
Find \(a\) for which the parabola \(y\) passes through the point \(B\) : \[-3=a(1+1)^2\quad\Rightarrow\quad a=-\dfrac34\] We make sure that with this value of the parameter, the second point of intersection of the parabola \(y=-\frac34(x+1)^2\) with the line \(h=-2x-1\) is a point with coordinates \(\left(-\frac13; -\frac13\right)\).
Thus, we got one more parameter value.
Since we have considered all possible cases for \(a\) , the final answer is: \
Answer:
\(\left\(-\frac34; \frac43\right\)\)
Task 3 #4013
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the system of equations \[\begin(cases) 2x^2+2y^2=5xy\\ (x-a)^2+(y-a)^2=5a^4 \end(cases)\]
has exactly two solutions.
1) Consider the first equation of the system as quadratic with respect to \(x\) : \ The discriminant is equal to \(D=9y^2\) , therefore, \
Then the equation can be rewritten as \[(x-2y)\cdot (2x-y)=0\] Therefore, the whole system can be rewritten as \[\begin(cases) \left[\begin(gathered)\begin(aligned) &y=2x\\ &y=0.5x\end(aligned)\end(gathered)\right.\\ (x-a)^2 +(y-a)^2=5a^4\end(cases)\] The set defines two straight lines, the second equation of the system defines a circle with center \((a;a)\) and radius \(R=\sqrt5a^2\) . For the original equation to have two solutions, the circle must intersect the population graph at exactly two points. Here is the drawing when, for example, \(a=1\) : 
Note that since the coordinates of the center of the circle are equal, the center of the circle “runs” along the straight line \(y=x\) .
2) Since the line \(y \u003d kx\) has the tangent of the angle of inclination of this line to the positive direction of the axis \(Ox\) is \(k\), then the tangent of the slope of the line \(y=0.5x\) is equal to \ (0,5\) (let's call it \(\mathrm(tg)\,\alpha\) ), the straight line \(y=2x\) is equal to \(2\) (let's call it \(\mathrm(tg)\ ,\beta\) ). notice, that \(\mathrm(tg)\,\alpha\cdot \mathrm(tg)\,\beta=1\), Consequently, \(\mathrm(tg)\,\alpha=\mathrm(ctg)\,\beta=\mathrm(tg)\,(90^\circ-\beta)\). Hence \(\alpha=90^\circ-\beta\) , whence \(\alpha+\beta=90^\circ\) . This means that the angle between \(y=2x\) and the positive direction \(Oy\) is equal to the angle between \(y=0.5x\) and the positive direction \(Ox\) : 
And since the line \(y=x\) is the bisector of the I coordinate angle (that is, the angles between it and the positive directions \(Ox\) and \(Oy\) are equal in \(45^\circ\) ), then the angles between \(y=x\) and lines \(y=2x\) and \(y=0.5x\) are equal.
We needed all this in order to say that the lines \(y=2x\) and \(y=0.5x\) are symmetrical to each other with respect to \(y=x\) , therefore, if the circle touches one of them , then it necessarily touches the second line.
Note that if \(a=0\) , then the circle degenerates into the point \((0;0)\) and has only one point of intersection with both lines. That is, this case does not suit us.
Thus, in order for the circle to have 2 points of intersection with the lines, it must be tangent to these lines: 
We see that the case when the circle is located in the third quarter is symmetrical (with respect to the origin of coordinates) to the case when it is located in the first quarter. That is, in the first quarter \(a>0\) , and in the third \(a<0\)
(но такие же по модулю).
Therefore, we will consider only the first quarter. 
notice, that \(OQ=\sqrt((a-0)^2+(a-0)^2)=\sqrt2a\), \(QK=R=\sqrt5a^2\) . Then \ Then \[\mathrm(tg)\,\angle QOK=\dfrac(\sqrt5a^2)(\sqrt(2a^2-5a^4))\] But on the other side, \[\mathrm(tg)\,\angle QOK=\mathrm(tg)\,(45^\circ-\alpha)=\dfrac(\mathrm(tg)\, 45^\circ-\mathrm(tg) \,\alpha)(1+\mathrm(tg)\,45^\circ\cdot \mathrm(tg)\,\alpha)\] Consequently, \[\dfrac(1-0.5)(1+1\cdot 0.5)=\dfrac(\sqrt5a^2)(\sqrt(2a^2-5a^4)) \quad\Leftrightarrow\quad a =\pm \dfrac15\] Thus, we already immediately got both a positive and a negative value for \(a\) . Therefore, the answer is: \
Answer:
\(\{-0,2;0,2\}\)
Task 4 #3278
Task level: Equal to the Unified State Examination
Find all values \(a\) , for each of which the equation \
has a unique solution.
(USE 2017, official trial 04/21/2017)
Let's make the replacement \(t=5^x, t>0\) and move all terms into one part: \
We have obtained a quadratic equation whose roots, according to the Vieta theorem, are \(t_1=a+6\) and \(t_2=5+3|a|\) . In order for the original equation to have one root, it is enough that the resulting equation with \(t\) also has one (positive!) root.
We note at once that \(t_2\) for all \(a\) will be positive. Thus, we get two cases:
1) \(t_1=t_2\) : \ &a=-\dfrac14 \end(aligned) \end(gathered) \right.\]
2) Since \(t_2\) is always positive, \(t_1\) must be \(\leqslant 0\) : \
Answer:
\((-\infty;-6]\cup\left\(-\frac14;\frac12\right\)\)
Task 5 #3252
Task level: Equal to the Unified State Examination
\[\sqrt(x^2-a^2)=\sqrt(3x^2-(3a+1)x+a)\]
has exactly one root on the interval \(\) .
(Unified State Examination 2017, reserve day)
The equation can be rewritten as: \[\sqrt((x-a)(x+a))=\sqrt((3x-1)(x-a))\] Thus, note that \(x=a\) is the root of the equation for any \(a\) , since the equation becomes \(0=0\) . In order for this root to belong to the segment \(\) , you need \(0\leqslant a\leqslant 1\) .
The second root of the equation is found from \(x+a=3x-1\) , i.e. \(x=\frac(a+1)2\) . In order for this number to be the root of the equation, it must satisfy the ODZ of the equation, that is: \[\left(\dfrac(a+1)2-a\right)\cdot \left(\dfrac(a+1)2+a\right)\geqslant 0\quad\Rightarrow\quad -\dfrac13\leqslant a\leqslant 1\] In order for this root to belong to the segment \(\) , it is necessary that \
Thus, for the root \(x=\frac(a+1)2\) to exist and belong to the segment \(\) , it is necessary that \(-\frac13\leqslant a\leqslant 1\).
Note that then for \(0\leqslant a\leqslant 1\) both roots \(x=a\) and \(x=\frac(a+1)2\) belong to the segment \(\) (that is, the equation has two roots on this segment), except for the case when they coincide: \
So we fit \(a\in \left[-\frac13; 0\right)\) and \(a=1\) .
Answer:
\(a\in \left[-\frac13;0\right)\cup\(1\)\)
Task 6 #3238
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the equation \
has a single root on the segment \(.\)
(Unified State Examination 2017, reserve day)
The equation is equivalent: \ odz equation: \[\begin(cases) x\geqslant 0\\ x-a\geqslant 0\\3a(1-x) \geqslant 0\end(cases)\] On the ODZ, the equation will be rewritten in the form: \
1) Let \(a<0\) . Тогда ОДЗ уравнения: \(x\geqslant 1\) . Следовательно, для того, чтобы уравнение имело единственный корень на отрезке \(\) , этот корень должен быть равен \(1\) . Проверим: \ Doesn't match \(a<0\) . Следовательно, эти значения \(a\) не подходят.
2) Let \(a=0\) . Then the ODZ equation is: \(x\geqslant 0\) . The equation will be rewritten as: \ The resulting root fits under the ODZ and is included in the segment \(\) . Therefore, \(a=0\) is suitable.
3) Let \(a>0\) . Then ODZ: \(x\geqslant a\) and \(x\leqslant 1\) . Therefore, if \(a>1\) , then the ODZ is an empty set. Thus, \(0
The derivative is \(y"=3x^2-2ax+3a\) . Let's determine what sign the derivative can be. To do this, find the discriminant of the equation \(3x^2-2ax+3a=0\) : \(D=4a( a-9)\) Therefore, for \(a\in (0;1]\) the discriminant \(D<0\)
. Значит, выражение \(3x^2-2ax+3a\)
положительно при всех \(x\)
. Следовательно, при \(a\in (0;1]\)
производная \(y">0\) . Hence \(y\) is increasing. Thus, by the property of an increasing function, the equation \(y(x)=0\) can have at most one root.
Therefore, in order for the root of the equation (the intersection point of the graph \(y\) with the x-axis) to be on the segment \(\) , it is necessary that \[\begin(cases) y(1)\geqslant 0\\ y(a)\leqslant 0 \end(cases)\quad\Rightarrow\quad a\in \] Considering that initially in the case under consideration \(a\in (0;1]\) , then the answer is \(a\in (0;1]\) . Note that the root \(x_1\) satisfies \((1) \) , the roots \(x_2\) and \(x_3\) satisfy \((2)\) . Also note that the root \(x_1\) belongs to the segment \(\) .
Consider three cases:
1) \(a>0\) . Then \(x_2>3\) , \(x_3<3\)
, следовательно, \(x_2\notin .\)
Тогда уравнение будет иметь один корень на \(\)
в одном из двух случаях:
- \(x_1\) satisfies \((2)\) , \(x_3\) does not satisfy \((1)\) , or matches \(x_1\) , or satisfies \((1)\) , but not included in the segment \(\) (that is, less than \(0\) );
- \(x_1\) does not satisfy \((2)\) , \(x_3\) satisfies \((1)\) and is not equal to \(x_1\) .
Note that \(x_3\) cannot be both less than zero and satisfy \((1)\) (i.e. greater than \(\frac35\) ). Given this remark, the cases are recorded in the following set: \[\left[ \begin(gathered)\begin(aligned) &\begin(cases) \dfrac9(25)-6\cdot \dfrac35+10-a^2>0\\ 3-a\leqslant \dfrac35\ end(cases)\\ &\begin(cases) \dfrac9(25)-6\cdot \dfrac35+10-a^2\leqslant 0\\ 3-a> Solving this collection and taking into account that \(a>0\) , we get: \
2) \(a=0\) . Then \(x_2=x_3=3\in .\) Note that in this case \(x_1\) satisfies \((2)\) and \(x_2=3\) satisfies \((1)\) , then there is an equation that has two roots at \(\) . This value \(a\) does not suit us.
3) \(a<0\) . Тогда \(x_2<3\) , \(x_3>3\) and \(x_3\notin \) . Arguing similarly to paragraph 1), you need to solve the set: \[\left[ \begin(gathered)\begin(aligned) &\begin(cases) \dfrac9(25)-6\cdot \dfrac35+10-a^2>0\\ 3+a\leqslant \dfrac35\ end(cases)\\ &\begin(cases) \dfrac9(25)-6\cdot \dfrac35+10-a^2\leqslant 0\\ 3+a> \dfrac35\end(cases) \end(aligned) \end(gathered)\right.\] Solving this collection and taking into account that \(a<0\) , получим: \\]
Answer:
\(\left(-\frac(13)5;-\frac(12)5\right] \cup\left[\frac(12)5;\frac(13)5\right)\)
1. Systems of linear equations with a parameter
Systems of linear equations with a parameter are solved by the same basic methods as conventional systems of equations: the substitution method, the method of adding equations, and the graphical method. Knowing the graphical interpretation of linear systems makes it easy to answer the question about the number of roots and their existence.
Example 1
Find all values for the parameter a for which the system of equations has no solutions.
(x + (a 2 - 3) y \u003d a,
(x + y = 2.
Solution.
Let's look at several ways to solve this problem.
1 way. We use the property: the system has no solutions if the ratio of the coefficients in front of x is equal to the ratio of the coefficients in front of y, but not equal to the ratio of free terms (a/a 1 = b/b 1 ≠ c/c 1). Then we have:
1/1 \u003d (a 2 - 3) / 1 ≠ a / 2 or a system
(and 2 - 3 = 1,
(a ≠ 2.
From the first equation a 2 \u003d 4, therefore, taking into account the condition that a ≠ 2, we get the answer.
Answer: a = -2.
2 way. We solve by the substitution method.
(2 - y + (a 2 - 3) y \u003d a,
(x = 2 - y,
((a 2 - 3) y - y \u003d a - 2,
(x = 2 - y.
After taking the common factor y out of brackets in the first equation, we get:
((a 2 - 4) y \u003d a - 2,
(x = 2 - y.
The system has no solutions if the first equation has no solutions, that is
(and 2 - 4 = 0,
(a - 2 ≠ 0.
It is obvious that a = ±2, but taking into account the second condition, only the answer with a minus is given.
Answer: a = -2.
Example 2
Find all values for the parameter a for which the system of equations has an infinite number of solutions.
(8x + ay = 2,
(ax + 2y = 1.
Solution.
By property, if the ratio of the coefficients at x and y is the same, and is equal to the ratio of the free members of the system, then it has an infinite set of solutions (i.e., a / a 1 \u003d b / b 1 \u003d c / c 1). Hence 8/a = a/2 = 2/1. Solving each of the equations obtained, we find that a \u003d 4 is the answer in this example.
Answer: a = 4.
2. Systems of rational equations with a parameter
Example 3
(3|x| + y = 2,
(|x| + 2y = a.
Solution.
Multiply the first equation of the system by 2:
(6|x| + 2y = 4,
(|x| + 2y = a.
Subtract the second equation from the first, we get 5|x| = 4 – a. This equation will have a unique solution for a = 4. In other cases, this equation will have two solutions (for a< 4) или ни одного (при а > 4).
Answer: a = 4.
Example 4
Find all values of the parameter a for which the system of equations has a unique solution.
(x + y = a,
(y - x 2 \u003d 1.
Solution.
We will solve this system using the graphical method. So, the graph of the second equation of the system is a parabola, lifted up along the Oy axis by one unit segment. The first equation defines the set of lines parallel to the line y = -x (picture 1). The figure clearly shows that the system has a solution if the straight line y \u003d -x + a is tangent to the parabola at the point with coordinates (-0.5; 1.25). Substituting these coordinates into the equation of a straight line instead of x and y, we find the value of the parameter a: 
1.25 = 0.5 + a;
Answer: a = 0.75.
Example 5
Using the substitution method, find out at what value of the parameter a, the system has a unique solution.
(ax - y \u003d a + 1,
(ax + (a + 2)y = 2.
Solution.
Express y from the first equation and substitute it into the second:
(y \u003d ah - a - 1,
(ax + (a + 2) (ax - a - 1) = 2.
We bring the second equation to the form kx = b, which will have a unique solution for k ≠ 0. We have:
ax + a 2 x - a 2 - a + 2ax - 2a - 2 \u003d 2;
a 2 x + 3ax \u003d 2 + a 2 + 3a + 2.
The square trinomial a 2 + 3a + 2 can be represented as a product of brackets
(a + 2)(a + 1), and on the left we take x out of brackets:
(a 2 + 3a) x \u003d 2 + (a + 2) (a + 1).
Obviously, a 2 + 3a must not be equal to zero, therefore,
a 2 + 3a ≠ 0, a(a + 3) ≠ 0, which means a ≠ 0 and ≠ -3.
Answer: a ≠ 0; ≠ -3.
Example 6
Using the graphical solution method, determine at what value of the parameter a, the system has a unique solution.
(x 2 + y 2 = 9,
(y - |x| = a.
Solution.
Based on the condition, we build a circle with a center at the origin of coordinates and a radius of 3 unit segments, it is this circle that sets the first equation of the system
x 2 + y 2 = 9. The second equation of the system (y = |x| + a) is a broken line. By using figure 2 we consider all possible cases of its location relative to the circle. It is easy to see that a = 3. 
Answer: a = 3.
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The purpose of this work is to study various ways of solving problems with parameters. The ability and ability to solve problems with parameters demonstrate mastery of methods for solving equations and inequalities, a meaningful understanding of theoretical information, the level of logical thinking, and stimulate cognitive activity. For the development of these skills, more efforts are needed, which is why in profile grades 10-11 with an in-depth study of the exact sciences, a course was introduced: “Mathematical Practicum”, part of which is the solution of equations and inequalities with parameters. The course is one of the disciplines included in the curriculum component of the school.
The successful study of methods for solving problems with parameters can be helped by an elective or optional course, or a component behind a grid on the topic: “Problems with parameters”.
Consider four large classes of problems with parameters:
- Equations, inequalities and their systems that need to be solved for any parameter value, or for parameter values that belong to a certain set.
- Equations, inequalities and their systems, for which it is required to determine the number of solutions depending on the value of the parameter.
- Equations, inequalities and their systems, for which it is required to find all those values of the parameter for which the specified equations (systems, inequalities) have a given number of solutions.
- Equations, inequalities and their systems, for which, for the desired values of the parameter, the set of solutions satisfies the given conditions in the domain of definition.
Methods for solving problems with parameters.
1. Analytical method.
This is a direct solution method that repeats the standard procedures for finding an answer in problems without a parameter.
Example 1: Find all parameter values a, for which the equation:
(2a – 1)x 2 + ax + (2a – 3) =0 has at most one root.
At 2 a– 1 = 0 this equation is not quadratic, so the case a=1/2 are analyzed separately.
If a a= 1/2, then the equation becomes 1/2 x– 2 = 0, it has one root.
If a a≠ 1/2, then the equation is quadratic; for it to have at most one root it is necessary and sufficient that the discriminant be nonpositive:
D= a 2 – 4(2a – 1)(2a – 3) = -15a 2 + 32a – 12;
In order to write down the final answer, it is necessary to understand
2. Graphical method.
Depending on the task (with a variable x and parameter a) graphs are considered in the coordinate plane ( x;y) or in the plane ( x;a).
Example 2. For each parameter value a quantify the solutions of the equation 
.
Note that the number of solutions to the equation equal to the number of intersection points of function graphs 
and y = a.
Function Graph 
shown in Fig.1.
y=a is a horizontal line. According to the graph, it is easy to establish the number of intersection points depending on a(for example, when a= 11 – two points of intersection; at a= 2 – eight points of intersection).
Answer: when a < 0 – решений нет; при a= 0 and a= 25/4 – four solutions; at 0< a < 6 – восемь решений; при a= 6 – seven solutions; at
6 < a < 25/4 – шесть решений; при a> 25/4 - two solutions.
3. Method of decision regarding the parameter.
When solving this way, the variables X and a are taken equal, and the variable with respect to which the analytical solution becomes simpler is selected. After simplifications, you need to return to the original meaning of the variables X and a and complete the solution.
Example 3: Find all parameter values a, for each of which the equation = - ax +3a+2 has a unique solution.
We will solve this equation by change of variables. Let = t , t≥ 0 then x = t 2 + 8 and the equation becomes at 2 +t + 5a– 2 = 0 . Now the task is to find all a, for which the equation at 2 +t + 5a– 2 = 0 has a unique non-negative solution. This takes place in the following cases.
1) If a= 0, then the equation has a unique solution t = 2.
Solution of some types of equations and inequalities with parameters.
Tasks with parameters help in the formation of logical thinking, in acquiring research skills.
The solution of each problem is unique and requires an individual, non-standard approach, since there is no single way to solve such problems.
. Linear equations.Task number 1. For what values of the parameter b equation has no roots?
Task number 2. Find all parameter values a, for which the set of solutions to the inequality:
contains the number 6, and also contains two segments of length 6 that do not have common points.
Let us transform both sides of the inequality.
In order for the set of solutions to the inequality to contain the number 6, it is necessary and sufficient that the following condition be satisfied: 
Fig.4
At a> 6 set of solutions to the inequality: 
.
The interval (0;5) cannot contain any segment of length 6. Hence, two non-intersecting segments of length 6 must be contained in the interval (5; a).
. Exponential equations, inequalities and systems.Task number 3. In the domain of the function definition 
Take all positive integers and add them up. Find all values for which such a sum is greater than 5 but less than 10.
1) Graph of a linear-fractional function 
is a hyperbole. By condition x> 0. With an unlimited increase X fraction decreases monotonically and approaches zero, and the values of the function z increase and approach 5. In addition, z(0) = 1.
2) By definition of the degree, the domain of definition D(y) consists of solutions to the inequality . At a= 1 we get an inequality that has no solutions. Therefore, the function at nowhere defined.
3) At 0< a< 1 показательная функция с основанием a decreases and the inequality is equivalent to the inequality . Because x> 0 , then z(x) > z(0) = 1 . So every positive value X is a solution to the inequality . Therefore, for such a the amount specified in the condition cannot be found.
4) When a> 1 exponential function with base a increases and the inequality is equivalent to the inequality . If a a≥ 5, then any positive number is its solution, and the sum specified in the condition cannot be found. If 1< a < 5, то множество положительных решений – это интервал (0;x 0) , where a = z(x 0) .
5) Integers are located in this interval in a row, starting from 1. Let's calculate the sums of consecutive natural numbers, starting from 1: 1; 1+2 = 3; 1+2+3 = 6; 1+2+3+4 = 10;… Therefore, the indicated sum will be greater than 5 and less than 10 only if the number 3 lies in the interval (0; x 0), and the number 4 does not lie in this interval. So 3< x 0 ≤ 4 . Since it increases by , then z(3) < z(x 0) ≤ z(4) .
The solution of irrational equations and inequalities, as well as equations, inequalities and systems containing modules are considered in Annex 1.
Problems with parameters are complex because there is no single algorithm for solving them. The specificity of such problems is that, along with unknown quantities, they include parameters whose numerical values are not specified specifically, but are considered known and given on a certain numerical set. At the same time, the values of the parameters significantly affect the logical and technical course of solving the problem and the form of the answer.
According to statistics, many of the graduates do not start solving problems with parameters for the USE. According to FIPI, only 10% of graduates start solving such problems, and the percentage of their correct solution is low: 2–3%, so the acquisition of skills for solving difficult, non-standard tasks, including tasks with parameters, by school students is still relevant.
MKOU "Lodeynopolskaya secondary comprehensive school No. 68"
_________________________________________________________________________________________________________________________________
Speech at a meeting of the Moscow Region
Problem solving methods
with parameters
Prokusheva Natalya Gennadievna
Lodeynoye Pole
2013-2014
Tasks with parameters
Problems with parameters are among the most difficult of the problems offered both in the Unified State Examination and in additional competitive exams for universities.
They play an important role in shaping logical thinking and mathematical culture. The difficulties that arise in solving them are related to the fact that each problem with parameters is a whole class of ordinary problems, for each of which a solution must be obtained.
If in the equation (inequality) some coefficients are not specified by specific numerical values, but are indicated by letters, then they are called parameters, and the equation (inequality) is parametric.
As a rule, unknowns are denoted by the last letters of the Latin alphabet: x, y, z, ..., and parameters - by the first: a, b, c, ...
To solve an equation (inequality) with parameters means to indicate at what values of the parameters solutions exist and what they are. Two equations (inequalities) containing the same parameters are called equivalent if:
a) they make sense for the same values of the parameters;
b) each solution to the first equation (inequality) is a solution to the second and vice versa.
Naturally, such a small class of problems does not allow many to grasp the main thing: the parameter, being fixed, but unknown number, has a dual nature. Firstly, the assumed fame allows you to “communicate” with the parameter as with a number, and secondly, the degree of freedom of communication is limited by its unknownness. So, division by an expression containing a parameter, extracting a root of an even degree from such expressions require preliminary research. As a rule, the results of these studies influence both the decision and the answer.
How to start solving such problems? Do not be afraid of tasks with parameters. First of all, you need to do what is done when solving any equation or inequality - bring given equation(inequality) to more plain sight, if possible: factorize a rational expression, factorize a trigonometric polynomial, get rid of modules, logarithms, etc.. then you need to carefully read the task again and again.
When solving problems containing a parameter, there are problems that can be conditionally divided into two large classes. The first class includes problems in which it is necessary to solve an inequality or an equation for all possible values parameter. The second class includes tasks in which it is necessary to find not all possible solutions, but only those that satisfy some additional conditions.
The most understandable way for schoolchildren to solve such problems is that they first find all the solutions, and then select those that satisfy additional conditions. But this is not always possible. There are a large number of problems in which it is impossible to find the entire set of solutions, and we are not asked about this. Therefore, one has to look for a way to solve the problem without having the entire set of solutions to a given equation or inequality, for example, to look for properties of the functions included in the equation that will allow one to judge the existence of a certain set of solutions.
Main task types with parameters
Type 1. Equations, inequalities, their systems and sets, which must be solved either for any value of the parameter (parameters), or for parameter values that belong to a predetermined set.
This type of problem is basic when mastering the topic "Problems with parameters", since the invested work predetermines success in solving problems of all other basic types.
Type 2. Equations, inequalities, their systems and sets, for which it is required to determine the number of solutions depending on the value of the parameter (parameters).
We draw attention to the fact that when solving problems of this type, there is no need to either solve the given equations, inequalities, their systems and combinations, etc., or give these solutions; such extra work in most cases is a tactical mistake, leading to unjustified expenditure of time. However, this should not be taken as an absolute, since sometimes a direct solution in accordance with type 1 is the only reasonable way to get an answer when solving a type 2 problem.
Type 3. Equations, inequalities, their systems and collections, for which it is required to find all those values of the parameter for which the indicated equations, inequalities, their systems and collections have a given number of solutions (in particular, they do not have or have an infinite number of solutions).
It is easy to see that problems of type 3 are in some sense the inverse of problems of type 2.
Type 4. Equations, inequalities, their systems and collections, for which, for the desired values of the parameter, the set of solutions satisfies the given conditions in the domain of definition.
For example, find the parameter values for which:
1) the equation is fulfilled for any value of the variable from the given interval;
2) the set of solutions of the first equation is a subset of the set of solutions of the second equation, and so on.
Comment. The variety of problems with a parameter covers the entire course of school mathematics (both algebra and geometry), but the vast majority of them in the final and entrance exams belong to one of the four listed types, which for this reason are called basic.
The most popular class of problems with a parameter is problems with one unknown and one parameter. The next paragraph indicates the main ways of solving problems of this particular class.
Basic methods for solving problems with a parameter
Method I(analytical). This is a method of the so-called direct solution, which repeats the standard procedures for finding an answer in problems without a parameter. Sometimes they say that this is a way of force, in good sense"brazen" decision.
Method II(graphic). Depending on the task (with a variable x and parameter a) are considered graphs or in the coordinate plane ( x; y), or in the coordinate plane ( x; a).
Comment. Exceptional visibility and beauty graphic way solving problems with a parameter is so captivating for students of the topic “Problems with a parameter” that they begin to ignore other methods of solving, forgetting the well-known fact: for any class of problems, their authors can formulate one that is brilliantly solved by this method and with enormous difficulties by other methods. Therefore, on initial stage study is dangerous to start with graphic techniques solving problems with a parameter.
Method III(parameter decision). When solving this way, the variables x and a are taken equal, and the variable is chosen, with respect to which the analytical solution is recognized as simpler. After natural simplifications, we return to the original meaning of the variables x and a and finish the solution.
Let us now proceed to demonstrate the indicated methods for solving problems with a parameter.
1. Linear equations and inequalities with parameters
Linear function: - equation of a straight line with a slope . The slope is equal to the tangent of the angle of inclination of the straight line to the positive direction of the axis .
Linear equations with parameters of the form
If a , equation has the only thing solution.
If a , then the equation has no solutions, when , and the equation has infinitely many solutions, when .
Example 1 solve the equation | x | = a .
Solution:
a > 0, => x 1.2 = ± a
a = 0, => x = 0
a < 0, =>there are no solutions.
Answer: x 1.2 = ± a at a > 0; x= 0 at a= 0; no solutions for a < 0.
Example 2 Solve equation |3 - x | = a .
Solution:
a > 0, => 3 – x = ± a , => x= 3 ± a
a = 0, => 3 – x = 0. => x = 3
a < 0, =>there are no solutions.
Answer: x 1.2=3± a at a > 0; x= 3 at a= 0; no solutions for a < 0.
Example 3 solve the equation m ² x – m = x + 1.
Solution:
m ² x – m = x + 1
m ² x – x = m + 1
(m² - 1)x = m + 1
Answer: 
at m
≠ ±
1; x
Є
R at m= –1; no solutions for m
= 1.
Example 4 a solve the equation: ( a 2 – 4) x = a + 2 .
Solution: Let us decompose the coefficient at into factors. .
If a , equation has the only thing solution: .
If a , the equation has no solutions.
If a , then the equation has infinitely many solutions .
Example 6 For all parameter values a
solve the equation: 
.
Solution: ODZ: .
Under this condition, the equation is equivalent to the following: 
.
Let's check the belonging to the ODZ: ,
if .
If ,
then the equation has no solutions.
Example 7 For all parameter values a solve the equation: | X + 3| – a | x – 1| = 4.
Solution: We divide the number line into 3 parts by points at which the expressions under the module sign vanish and solve 3 systems:
1) 
,
if .
Found will be the solution if .
2) 
,
if .
Found satisfies the desired inequality, therefore, is a solution for .
If ,
then the solution is any .
3) 
,
if .
Found not satisfies the desired inequality, therefore, not is a solution for .
If ,
then the solution is any x > 1.
Answer: at ; at ;
P ri ; is also a solution for all .
Example 8 Find all a, for each of which at least one of the solutions of Equation 15 x – 7a = 2 – 3ax + 6a less 2 .
Solution: Let us find solutions to the equation for each .
, if .
Let's solve the inequality: 
.
For , the equation has no solutions.
Answer : aÎ (–5 , 4) .
Linear inequalities with parameters
For example: Solve the inequality: kx < b .
If a k> 0, then 
. If a k
< 0, то 
. If a k= 0, then b> 0 the solution is any x
Є R, and when 
there are no solutions.
Solve the remaining inequalities in the box in the same way.
Example 1 For all values of the parameter a, solve the inequality 
.
Solution:
. If the parenthesis before x is positive, i.e. at 
, then 
. If the parenthesis before x is negative, i.e. at 
, then 
. If a= 0 or a = , then there are no solutions.
Answer: at ; at ;
no solutions for a= 0 or a = .
Example 2. For all parameter values a solve inequality | X– a| – | x + a| < 2a .
Solution:
At a=0 we have the wrong inequality 0< 0, т.е. решений нет. Пусть a >0, then for x< –a both modules are expanded with a minus and we get the wrong inequality 2 a < 2a, i.e. there are no solutions. If a x Є [– a ; a] , then the first module is expanded with a minus, and the second with a plus, and we get the inequality –2 x < 2a, i.e. x > –a, i.e., the solution is any x Є (– a ; a]. If a x > a both modules are expanded with a plus and we obtain the correct inequality –2 a < 2a, i.e. , the solution is any x Є ( a; +∞). Combining both answers, we get that a > 0 x Є (– a ; +∞).
Let a < 0, тогда первое слагаемое больше, чем второе, поэтому разность в левой части неравенства положительна и, следовательно, не может быть меньше negative number 2a. Thus, at a < 0 решений нет.
Answer: x
Є (– a; +∞) at a> 0, there are no solutions for 
.
Comment. The solution to this problem is faster and easier if you use the geometric interpretation of the modulus of the difference of two numbers, as the distance between points. Then the expression on the left side can be interpreted as the difference in distances from the point X to points a and - a .
Example 3 Find all a, for each of which all solutions of the inequality 
satisfy inequality 2 x
– a² + 5< 0.
Solution:
By solving the inequality |x | ≤ 2 is the set A=[–2; 2], and the solution of inequality 2 x
– a² + 5< 0 является множество B
= (–∞; 
) . To satisfy the condition of the problem, it is necessary that the set A is included in the set B (). This condition is satisfied if and only if .
Answer: a Є (–∞; –3)U (3; +∞).
Example 4 Find all values of a for which the inequality 
performed for everyone x from the cut.
Solution:
The fraction is less than zero between the roots, so you need to find out which root is greater.
–3a
+ 2 < 2a
+ 4 
and -3 a
+ 2 > 2a
+ 4 
. Thus, at 
xЄ (–3 a
+ 2; 2a+ 4) and in order for the inequality to hold for all x from the segment , it is necessary that
At 
xЄ (2 a
+ 4; –3a+ 2) and that the inequality holds for all x from the segment , you need to
For a = – (when the roots coincide) there are no solutions, because in this case, the inequality takes the form: .
Answer: 
.
Example 5 a the inequality is valid for all negative values X?
Solution:
The function increases monotonically if the coefficient at x is non-negative, and it decreases monotonically if the coefficient at x negative.
Find out the sign of the coefficient at 
a ≤ –3,
a ≥ 1; (a² + 2 a – 3) < 0 <=> –3 < a < 1.
a ≤ –3,
Let a≥ 1. Then the function f
(x
)
does not decrease monotonically, and the condition of the problem will be satisfied if f
(x
)
≤ 0 <=> 3a
² – a
– 14 ≤ 0 <=> 
.
a ≤ –3,
Together with conditions a≥ 1; we get: 
Let -3< a < 1. Тогда функция f (x ) decreases monotonically, and the condition of the problem can never be satisfied.
Answer:
.
2. Quadratic equations and inequalities with parameters
.
In the set of real numbers, this equation is studied according to the following scheme.
Example 1. At what values a the equationx ² – ax + 1 = 0 has no real roots?
Solution:
x ² – ax + 1 = 0
D = a ² – 4 1 =a ² - 4
a ² - 4< 0 + – +
( a – 2)( a + 2) < 0 –2 2
Answer: ata Є (–2; 2)
Example 2For what values of a does the equation a (X ² – X + 1) = 3 X + 5 has two distinct real roots?
Solution:
a (X ² – X + 1) = 3 X + 5, a ≠ 0
Oh ² – ah + a – 3 X – 5 = 0
Oh ² – ( a + 3) X + a – 5 = 0
D = ( a +3)² - 4a ( a – 5) = a ² +6a + 9 – 4 a ² + 20a = –3 a ² + 26a + 9
–3 a ² + 26 a + 9 > 0
3 a ² - 26a – 9 < 0
D \u003d 26² - 4 3 (-9) \u003d 784
a
1
=
;
a
2
=
+ –
+
0 9
Answer:ataЄ (–1/3; 0)U (0; 9)
Example 3. Solve the equation 
.
Solution:
ODZ: x ≠1, x ≠ a
x
– 1 +
x
–
a
= 2, 2
x
= 3 +
a
, 
1)
; 3 +
a
≠ 2;
a
≠ –1
2) 
; 3 +
a
≠ 2
a
;
a
≠ 3
Answer:
ata
Є (–∞; –1)U
(–1; 3)
U
(3; +∞);
no solutions fora = -1; 3.
Example4 . solve the equation | x ²–2 x –3 | = a .
Solution:
Consider the functions y = | x ²–2 x –3 | andy = a .
At a
<
0
no solutions;
at a
=
0 and a> 4 two solutions;
at 0< a
< 4 – четыре решения;
at a= 4 – three solutions.
Answer:
at a
< 0 нет решений;
at a= 0 and a> 4 two solutions;
at 0< a
< 4 – четыре решения;
at a= 4 – three solutions.
Example 5Find all values
a
, for each of which the equation
|
x
²–(
a
+2)
x
+2
a
| = |
3
x
–6
|
has exactly two roots. If such values
a
more than one, indicate their product in your answer.
Solution:
Let's decompose square trinomial x
²–(
a
+2)
x
+2
a
for multipliers.
;
;
;
Get |
(
x
–2)(
x
–
a
)
| =
3
|
x
–2
|.
This equation is equivalent to the set
Therefore, this equation has exactly two roots, if a+ 3 = 2 and a
– 3 = 2.
Hence we find that the desired values a are a
1
= –1; a
2
= 5; a
1
·
a
2
= –5.
Answer: –5.
Example 6Find all values a , for which the roots of the equation ax ² – 2( a + 1) x – a + 5 = 0 positive.
Solution:
Check Point a= 0, because changes the essence of the equation.
1. a = 0 –2x + = 0;
Answer: a Є U .
Example 7Atwhat parameter values a the equation | x ² - 4 x + 3 | = ax has 3 roots.
Solution:
Let's build graphs of functions y = | x ² - 4 x + 3 | and y = ax .
The graph of the function is plotted on the segment 
.
This equation will have three roots if the graph of the function y
=
ax will be tangent to the graph y
= x
²+ 4
x
– 3
on the
segment .
The tangent equation has the form y
=
f
(x
0
) +
f
’(x
0
)(x
–
x
0
),
Because tangent equation y
=
a, we obtain a system of equations 
Because x 0 Є ,
Answer: at a
= 4 – 2
.
Square inequalities with parameters
Example.Find all parameter values
a
, for each of which among the solutions of the inequality 
there is no cutoff point.
Solution:
First, we solve the inequality for all values of the parameter, and then we find those for which there is not a single point of the segment among the solutions .
Let 
, ax
= t
²
t ≥ 0
With such a change of variables in the DPV, the inequalities are satisfied automatically. x can be expressed through t, if a≠ 0. Therefore, the case when a
= 0, we will consider separately.
1.Let a
= 0, then X> 0, and the given segment is a solution.
2. Let a≠ 0, then 
and inequality will take the form 
, 
The solution of the inequality depends on the values a, so we have to consider two cases.
1) If a>0, then 
at 
, or in old variables,
The solution does not contain any point of the given segment if and only if the conditions are met a ≤ 7,
16a≥ 96. Hence, a
Є .
2). If a a< 0, то ; 
; tЄ (4 a
; a). Because t≥ 0, then there are no solutions.
Answer: .
Irrational equations with parameters
When solving irrational equations and inequalities with a parameter, first, one should take into account the range of admissible values. Secondly, if both parts of the inequality are non-negative expressions, then such an inequality can be squared with the inequality sign preserved.
In many cases, irrational equations and inequalities reduce to quadratic equations after a change of variables.
Example 1 solve the equation 
.
Solution:
ODZ: x + 1 ≥ 0, x ≥ –1, a ≥ 0.
x + 1 = a ².
If a x = a² - 1, then the condition is met.
Answer: x = a² - 1 at a≥ 0; no solutions for a < 0.
Example 2. Solve the equation 
.
Solution:
ODZ: x + 3 ≥ 0, x ≥ –3,
a-x ≥ 0; x ≤ a;
x + 3 = a-x,
2x = a – 3,
<=>
<=>
<=>
a
≥ –3.
Answer: at a≥ -3; no solutions for a
< –3.
Example 3 How many roots does the equation have 
depending on parameter values a?
Solution:
The range of admissible values of the equation: x Є [–2; 2]
Let's build graphs of functions. The graph of the first function is the upper half of the circle x² + y² = 4. The graph of the second function is the bisector of the first and second coordinate angles. Subtract the graph of the second function from the graph of the first function and get the graph of the function 
. If replace at on the a, then the last graph of the function is the set of points (x; a) that satisfy the original equation. 
We see the answer on the graph.
Answer: at aЄ (–∞; –2) U (1; +∞), no roots;
at aЄ [–2; 2), two roots;
at a= 1, one root.
Example 4 At what values of the parameter a the equation 
has a unique solution?
Solution:
1 way (analytical):
Answer:
2 way (graphical):
Answer: for a ≥ –2, the equation has a unique solution
Example 5 For what values of the parameter a the equation = 2 + x has a unique solution.
Solution:
Consider a graphical version of the solution of this equation, that is, we will build two functions:
at 1 = 2 + X and at 2 =
The first function is linear and passes through the points (0; 2) and (–2; 0).
The graph of the second function contains a parameter. Consider first the graph of this function for a= 0 (Fig. 1). When changing the value of the parameter, the graph will move along the axis OH to the corresponding value to the left (with positive a) or to the right (with negative a) (Fig. 2)
It can be seen from the figure that at a < –2 графики не пересекают друг друга, а следовательно не имеют common solutions. If the value of the parameter a is greater than or equal to -2, then the graphs have one intersection point, and therefore one solution.
Answer: at a≥ –2 the equation has a unique solution.
Trigonometric equations with parameters.
Example 1Solve the Equation sin (– x + 2 x – 1) = b + 1.
Solution:
Given the oddness of the function 
, this equation can be reduced to the equivalent 
.
1. b = –1
3. b =–2
4. | b + 1| > 1
There are no solutions.
5. bЄ(–1; 0)
6. bЄ(–2; –1)
Example 2Find all values of the parameter p for which the equation
has no solutions.
Solution:
Express cos 2 x through sinx.
Let 
then the task was reduced to finding all the values p, for which the equation has no solutions on [–1; one]. The equation is not algorithmically solved, so we will solve the problem using the graph. We write the equation in the form , and now the sketch of the graph of the left side 
easy to build.
The equation has no solutions if the line y
=
p+ 9 does not intersect the graph on the segment [–1; 1], i.e. 
Answer:p Є (–∞; –9) U (17; +∞).
Systems of equations with parameters
Systems of two linear equations with parameters
System of equations 
The solutions of a system of two linear equations are the points of intersection of two lines: and .
3 cases are possible:
1. Lines are not parallel . Then their normal vectors are not parallel either, i.e. . In this case, the system has only decision.
2. The lines are parallel and do not coincide. Then their normal vectors are also parallel, but the shifts are different, i.e. .
In this case no decision system .
3. Straight lines coincide. Then their normal vectors are parallel and the shifts coincide, i.e. . In this case, the system has infinite number of solutions all points of the line .
Type equation f(x; a) = 0 is called variable equation X and parameter a.
Solve an equation with a parameter a This means that for every value a find values X satisfying this equation.
Example 1 Oh= 0
Example 2 Oh = a
Example 3
x + 2 = ax
x - ax \u003d -2
x (1 - a) \u003d -2
If 1 - a= 0, i.e. a= 1, then X 0 = -2 no roots
If 1 - a 0, i.e. a 1, then X =
Example 4
(a 2 – 1) X = 2a 2 + a – 3
(a – 1)(a + 1)X = 2(a – 1)(a – 1,5)
(a – 1)(a + 1)X = (1a – 3)(a – 1)
If a a= 1, then 0 X = 0
X- any real number
If a a= -1, then 0 X = -2
no roots
If a a 1, a-1 then X= (the only solution).
This means that each valid value a matches a single value X.
For example:
if a= 5, then X = = ;
if a= 0, then X= 3 etc.
Didactic material
1. Oh = X + 3
2. 4 + Oh = 3X – 1
3. a = +
at a= 1 there are no roots.
at a= 3 no roots.
at a = 1 X any real number except X = 1
at a = -1, a= 0 there are no solutions.
at a = 0, a= 2 no solutions.
at a = -3, a = 0, 5, a= -2 no solutions
at a = -With, With= 0 there are no solutions.
Quadratic equations with a parameter
Example 1 solve the equation
(a – 1)X 2 = 2(2a + 1)X + 4a + 3 = 0
At a = 1 6X + 7 = 0
When a 1 select those values of the parameter for which D goes to zero.
D = (2(2 a + 1)) 2 – 4(a – 1)(4a + 30 = 16a 2 + 16a + 4 – 4(4a 2 + 3a – 4a – 3) = 16a 2 + 16a + 4 – 16a 2 + 4a + 12 = 20a + 16
20a + 16 = 0
20a = -16
If a a < -4/5, то D < 0, уравнение имеет действительный корень.
If a a> -4/5 and a 1, then D > 0,
X = 
If a a= 4/5, then D = 0,
Example 2 At what values of the parameter a the equation
x 2 + 2( a + 1)X + 9a– 5 = 0 has 2 different negative roots?
D = 4( a + 1) 2 – 4(9a – 5) = 4a 2 – 28a + 24 = 4(a – 1)(a – 6)
4(a – 1)(a – 6) > 0
according to t. Vieta: X 1 + X 2 = -2(a + 1)
X 1 X 2 = 9a – 5
By condition X 1 < 0, X 2 < 0 то –2(a + 1) < 0 и 9a – 5 > 0
| Eventually | 4(a – 1)(a – 6) > 0 - 2(a + 1) < 0 9a – 5 > 0 |
a < 1: а > 6 a > - 1 a > 5/9 |
 (Rice. one) < a < 1, либо a > 6 |
Example 3 Find values a for which this equation has a solution.
x 2 - 2( a – 1)X + 2a + 1 = 0
D = 4( a – 1) 2 – 4(2a + 10 = 4a 2 – 8a + 4 – 8a – 4 = 4a 2 – 16a
4a 2 – 16 0
4a(a – 4) 0
a( a – 4)) 0
a( a – 4) = 0
a = 0 or a – 4 = 0
a = 4
(Rice. 2)
Answer: a 0 and a 4
Didactic material
1. At what value a the equation Oh 2 – (a + 1) X + 2a– 1 = 0 has one root?
2. At what value a the equation ( a + 2) X 2 + 2(a + 2)X+ 2 = 0 has one root?
3. For what values of a is the equation ( a 2 – 6a + 8) X 2 + (a 2 – 4) X + (10 – 3a – a 2) = 0 has more than two roots?
4. For what values of a equation 2 X 2 + X – a= 0 has at least one common root with equation 2 X 2 – 7X + 6 = 0?
5. For what values of a do the equations X 2 +Oh+ 1 = 0 and X 2 + X + a= 0 have at least one common root?
1. When a = - 1/7, a = 0, a = 1
2. When a = 0
3. When a = 2
4. When a = 10
5. When a = - 2
Exponential Equations with a Parameter
Example 1.Find all values a, for which the equation
9 x - ( a+ 2) * 3 x-1 / x +2 a*3 -2/x = 0 (1) has exactly two roots.
Solution. Multiplying both sides of equation (1) by 3 2/x, we obtain an equivalent equation
3 2(x+1/x) – ( a+ 2) * 3 x + 1 / x + 2 a = 0 (2)
Let 3 x+1/x = at, then equation (2) takes the form at 2 – (a + 2)at + 2a= 0, or
(at – 2)(at – a) = 0, whence at 1 =2, at 2 = a.
If a at= 2, i.e. 3 x + 1/x = 2 then X + 1/X= log 3 2 , or X 2 – X log 3 2 + 1 = 0.
This equation has no real roots because it D= log 2 3 2 – 4< 0.
If a at = a, i.e. 3 x+1/x = a then X + 1/X= log 3 a, or X 2 –X log 3 a + 1 = 0. (3)
Equation (3) has exactly two roots if and only if
D = log 2 3 2 – 4 > 0, or |log 3 a| > 2.
If log 3 a > 2, then a> 9, and if log 3 a< -2, то 0 < a < 1/9.
Answer: 0< a < 1/9, a > 9.
Example 2. At what values of a equation 2 2x - ( a - 3) 2 x - 3 a= 0 has solutions?
For a given equation to have solutions, it is necessary and sufficient that the equation t 2 – (a - 3) t – 3a= 0 has at least one positive root. Let's find the roots using Vieta's theorem: X 1 = -3, X 2 = a = >
a is a positive number.
Answer: when a > 0
Didactic material
1. Find all values of a for which the equation
25 x - (2 a+ 5) * 5 x-1 / x + 10 a* 5 -2/x = 0 has exactly 2 solutions.
2. For what values of a does the equation
2 (a-1) x? + 2 (a + 3) x + a \u003d 1/4 has a single root?
3. For what values of the parameter a the equation
4 x - (5 a-3) 2 x +4 a 2 – 3a= 0 has a unique solution?
Logarithmic Equations with a Parameter
Example 1 Find all values a, for which the equation
log 4x (1 + Oh) = 1/2 (1)
has a unique solution.
Solution. Equation (1) is equivalent to the equation
1 + Oh = 2X at X > 0, X 1/4 (3)
X = at
au 2 - at + 1 = 0 (4)
The (2) condition from (3) is not satisfied.
Let a 0, then au 2 – 2at+ 1 = 0 has real roots if and only if D = 4 – 4a 0, i.e. at a 1. To solve inequality (3), we construct graphs of functions Galitsky M.L., Moshkovich M.M., Shvartsburd S.I. In-depth study of the course of algebra and mathematical analysis. - M.: Enlightenment, 1990