1. Determination of personal motivation of students. To continue education, for self-development and intellectual growth, it is necessary to study diligently and consciously and take care of your health. 2. Access to the concept of "parameter". Parameter - a value that characterizes the main properties of a change in a system or phenomenon. ( Dictionary)
In equations (inequalities), the coefficients for unknowns or free terms given not by specific numerical values, but indicated by letters, are called parameters. Example: To solve a problem with a parameter means, for each value of the parameter, to find x values that satisfy the condition of this problem.
X y x y a > 0 a 0, (2 roots) 0 a 0, (2 roots)"> 0 a 0, (2 roots)"> 0 a 0, (2 roots)" title="(!LANG:x y x y a > 0 a 0, (2 roots)"> title="x y x y a > 0 a 0, (2 roots)"> !}
x uuuuuuuuuuh
2. when the equation takes the form, and has a root x \u003d 0. 3. when we find the roots of the equation according to the formula Answer: when there are no roots; with one root x = 0. with two roots 1. the left side of the equation is non-negative for any value of the unknown x,. no solutions. x y 0 y = a "LOOK!" Method 1 (analytical) Method 2 (graphic)
At what values of the parameter a does the equation have one solution? Let's write the equation in the form: x Let's plot the graphs of the functions: Answer: a = 3 and a moving straight line y = a. a
For what values of the parameter a does the equation have no solutions? x y Let's build a graph According to the figure, we see at and a straight line y \u003d a. there are no solutions. a Answer:
(Graphic method for solving problems with a parameter) A problem with a parameter can be considered as a function f (x; a) =0 1. We build a graphic image 2. We intersect the resulting graph with straight lines parallel to the x-axis 3. “Read” the necessary information Solution scheme: !!!
3 Answer: 1 root "title="(!LANG: Specify the number of roots of the equation f(x)= a for all values of parameter a. 1 35-2 1 x a -5 3 1 root, a3 Answer: 1 root" class="link_thumb"> 15 !} Indicate the number of roots of the equation f (x) \u003d a for all values of the parameter a x a root, a3 Answer: 1 root for a 3 2 roots for a \u003d -5, a \u003d 3 3 roots for 1 3 Answer: 1 root "> 3 Answer: 1 root with a 3 2 roots with a \u003d -5, a \u003d 3 3 roots with 1 3 Answer: 1 root "title="(!LANG: Specify the number of roots of the equation f(x)= a for all values of parameter a. 1 35-2 1 x a -5 3 1 root, a3 Answer: 1 root"> title="Indicate the number of roots of the equation f(x)= a for all values of the parameter a. 1 35-2 1 x a -5 3 1 root, a3 Answer: 1 root">!}
X y y For what values of the parameter a does the equation have two roots? x y x
1) When a \u003d 3, the top right angle; Find the sum of integer values of the parameter a for which the equation has three roots. The original equation is equivalent to the set B Expressing the parameter a, we get: It can be seen from the figure that the equation has three roots in 3 cases x a a 1 = 3 a 2 = ? and 3 = ? Then a = = 5. Answer. 8. 2) For x 4, a 2 = 5 a 3 a 3 4, a 2 \u003d 5 a 3 a 3 "\u003e 
For each value of the parameter a a solve the inequality | 2 x + a | ≤ x + 2 |2x+a| \leq x+2 .
Let's solve an auxiliary problem first. Consider this inequality as an inequality with two variables x x and a a and draw on the coordinate plane x O a xOa all points whose coordinates satisfy the inequality.
If 2 x + a ≥ 0 2x+a \geq 0 (i.e., on the line a = - 2 xa=-2x and higher), then we get 2 x + a ≤ x + 2 ⇔ a ≤ 2 - x 2x+ a \leq x+2 \Leftrightarrow a \leq 2-x .
The set is shown in Fig. eleven.
Now let's solve the original problem using this drawing. If we fix a a , then we get a horizontal line a = const a = \textrm(const) . To determine the values x x , you need to find the abscissas of the points of intersection of this line with the solution set of the inequality. For example, if a = 8 a=8 , then the inequality has no solutions (the line does not intersect the set); if a = 1 a=1 , then all x x from the interval [ - 1 ; 1 ] [-1;1], etc. So, there are three options.
1) If $$a>4$$, then there are no solutions.
2) If a = 4 a=4 , then x = - 2 x=-2 .
ANSWER
for $$a
with a = 4 a=4 - x = - 2 x=-2 ;
for $$a>4$$ - there are no solutions.
Find all values of the parameter a a such that the inequality $$3-|x-a| > x^2$$ a) has at least one solution; b) has at least one positive solution.
Let's rewrite the inequality as $$3-x^2 > |x-a)$$. Let's plot the graphs of the left and right parts on the plane x O y xOy . The graph on the left side is a downward-branching parabola with apex at (0 ; 3) (0;3) . The graph crosses the x-axis at points (± 3 ; 0) (\pm \sqrt(3);0) . The graph of the right side is an angle with a vertex on the abscissa axis, the sides of which are directed upwards at an angle of 45 ° 45^(\circ) to the coordinate axes. The abscissa of the vertex is the point x = a x=a .
a) In order for the inequality to have at least one solution, it is necessary and sufficient that at least at one point the parabola is higher than the graph y = | x - a | y=|x-a| . This is done if the corner vertex lies between points A A and B B of the x-axis (see Fig. 12 - points A A and B B are not included). Thus, it is necessary to determine at what position of the vertex one of the branches of the corner touches the parabola.
Consider the case when the vertex of the corner is at the point A A . Then the right branch of the corner touches the parabola. Its slope equal to one. Hence, the derivative of the function y = 3 - x 2 y = 3-x^2 at the point of contact is equal to 1 1 , i.e. - 2 x = 1 -2x=1 , whence x = - 1 2 x = -\frac( 1)(2) . Then the ordinate of the touch point is y = 3 - (1 2) 2 = 11 4 y = 3 - (\frac(1)(2))^2 = \frac(11)(4) . The equation of a straight line with a slope k = 1 k=1 and passing through a point with coordinates (- 1 2 ; 11 4) (-\frac(1)(2); \frac(11)(4)) , the following is * ( \^* : y - 11 4 = 1 · (x + 1 2) y - \frac{11}{4} = 1 \cdot (x+ \frac{1}{2}) , откуда y = x + 13 4 y = x + \frac{13}{4} .!}
This is the equation of the right branch of the corner. The abscissa of the point of intersection with the xx axis is - 13 4 -\frac(13)(4) , i.e. point AA has coordinates A (- 13 4 ; 0) A(-\frac(13)(4); 0) . For reasons of symmetry, the point B B has coordinates: B (13 4 ; 0) B(\frac(13)(4); 0) .
Hence we get that a ∈ (- 13 4 ; 13 4) a\in (-\frac(13)(4); \frac(13)(4)) .
b) The inequality has positive solutions if the vertex of the corner is between the points F F and B B (see Fig. 13). Finding the position of the point FF is not difficult: if the vertex of the corner is at the point FF, then its right branch (the straight line given by the equation y \u003d x - ay \u003d xa passes through the point (0; 3) (0; 3) . From here we find that a \u003d - 3 a=-3 and the point FF has coordinates (- 3 ; 0) (-3; 0) Therefore, a ∈ (- 3 ; 13 4) a \in (-3; \frac(13)(4) ) .
ANSWER
a) a ∈ (- 13 4 ; 13 4) , a\in (-\frac(13)(4); \frac(13)(4)),\:\:\: b) a ∈ (- 3 ; 13 4) a \in (-3; \frac(13)(4)) .
* {\^* Полезные формулы: !}
- \-- a straight line passing through the point (x 0; y 0) (x_0; y_0) and having a slope kk is given by the equation y - y 0 = k (x - x 0) y-y_0=k(x-x_0 ) ;
- \-- slope of the straight line passing through the points (x 0 ; y 0) (x_0; y_0) and (x 1 ; y 1) (x_1; y_1) , where x 0 ≠ x 1 x_0 \neq x_1 , is calculated from formula k = y 1 - y 0 x 1 - x 0 k = \dfrac(y_1-y_0)(x_1-x_0) .
Comment. If you need to find the value of the parameter at which the straight line y = kx + ly=kx+l and the parabola y = ax 2 + bx + cy = ax^2+bx+c touch, then you can write the condition that the equation kx + l = ax 2 + bx + c kx+l = ax^2+bx+c has exactly one solution. Then another way to find the values of the parameter aa, in which the corner vertex is at point A A, is the following: the equation x - a = 3 - x 2 xa = 3-x^2 has exactly one solution ⇔ D = 1 + 4 (a + 3) = 0 ⇔ a = - 13 4 \Leftrightarrow D = 1 + 4(a+3) = 0 \Leftrightarrow a = -\ dfrac(13)(4) .
Please note that in this way it is impossible to write down the condition of touching a straight line with an arbitrary graph. For example, the line y = 3 x - 2 y = 3x - 2 touches the cubic parabola y = x 3 y=x^3 at the point (1 ; 1) (1; 1) and intersects it at the point (- 2 ; - 8) (-2;-8) , i.e. the equation x 3 = 3 x + 2 x^3 = 3x+2 has two solutions.
Find all values of the parameter aa for which the equation (a + 1 - | x + 2 |) (x 2 + 4 x + 1 - a) = 0 (a+1-|x+2|)(x^2 +4x+1-a) = 0 has a) exactly two distinct roots; b) exactly three distinct roots.
Let's do the same as in Example 25. Draw the set of solutions of this equation on the plane x O a xOa . It is equivalent to the combination of two equations:
1) a = | x + 2 | - 1 a = |x+2| -1 is the angle with branches up and apex at (- 2 ; - 1) (-2;-1) .
2) a \u003d x 2 + 4 x + 1 a \u003d x ^ 2 + 4x + 1 is a parabola with branches up and a vertex at the point (- 2; - 3) (-2; -3) . See fig. 14.
Find the intersection points of two graphs. The right branch of the angle is given by the equation y = x + 1 y=x+1 . Solving the Equation
x + 1 = x 2 + 4 x + 1 x+1 = x^2+4x+1
we find that x = 0 x=0 or x = - 3 x=-3 . Only the value x = 0 x=0 is suitable (because for the right branch x + 2 ≥ 0 x+2 \geq 0). Then a = 1 a=1 . Similarly, we find the coordinates of the second intersection point - (- 4 ; 1) (-4; 1) .
We return to the original problem. The equation has exactly two solutions for those a a , for which the horizontal line a = const a=\textrm(const) intersects the solution set of the equation at two points. From the graph, we see that this is true for a ∈ (- 3 ; - 1) ∪ ( 1 ) a\in (-3;-1)\bigcup\(1\) . Exactly three solutions will be in the case of three intersection points, which is possible only for a = - 1 a=-1 .
ANSWER
a) a ∈ (- 3 ; - 1) ∪ ( 1 ) ; a\in (-3;-1)\bigcup\(1\);\:\:\: b) a = - 1 a=-1 .
$$\begin(cases) x^2-x-a \leq 0,\\ x^2+2x-6a \leq 0 \end(cases) $$
has exactly one solution.
Let us represent the solutions of the system of inequalities on the plane x O a xOa . Rewrite the system as $$ \begin(cases) a \leq -x^2+x,\\ a \geq \dfrac(x^2+6x)(6) .\end(cases) $$
The first inequality is satisfied by the points lying on the parabola a = - x 2 + xa = -x^2+x and below it, and the second - by the points lying on the parabola a = x 2 + 6 x 6 a = \dfrac(x^2 +6x)(6) or higher. We find the coordinates of the vertices of the parabolas and their points of intersection, and then build a graph. The vertex of the first parabola is (1 2 ; 1 4) (\dfrac(1)(2);\dfrac(1)(4)) , the second parabola is (- 1 ; - 1 6) (-1; -\dfrac( 1)(6)) , the intersection points are (0 ; 0) (0;0) and (4 7 ; 12 49) (\dfrac(4)(7); \dfrac(12)(49)) . The set of points that satisfy the system is shown in fig. 15. It can be seen that the horizontal line a = const a=\textrm(const) has exactly one common point with this set (which means that the system has exactly one solution) in the cases a = 0 a=0 and a = 1 4 a= \dfrac(1)(4) .
ANSWER
A = 0 , a = 1 4 a=0,\: a=\dfrac(1)(4)
Find the smallest value of the parameter a a , for each of which the system
$$\begin(cases) x^2+y^2 + 3a^2 = 2y + 2\sqrt(3)ax,\\ \sqrt(3)|x|-y=4 \end(cases) $$
It has only decision.
Let's transform the first equation, highlighting full squares:
(x 2 - 2 3 a x + 3 a 2) + (y 2 - 2 y + 1) = 1 ⇔ (x - a 3) 2 + (y - 1) 2 = 1 . 18 (x^2- 2\sqrt(3)ax+3a^2)+(y^2-2y+1)=1 \Leftrightarrow (xa\sqrt(3))^2+(y-1)^2 =1. \:\:\:\left(18\right)
Unlike the previous tasks, it is better to depict a drawing on the x O y xOy plane (a drawing in the “variable - parameter” plane is usually used for problems with one variable and one parameter - as a result, a set is obtained on the plane. In this problem, we are dealing with two (x; y; a) (x; y; a) in three-dimensional space is a difficult task; besides, such a drawing is unlikely to be visual). Equation (18) defines a circle with center (a 3 ; 1) (a\sqrt(3);1) of radius 1. Depending on the value of a a , the center of this circle can be at any point of the line y = 1 y=1 .
The second equation of the system y = 3 | x | - 4 y = \sqrt(3)|x|-4 defines an angle with sides up at an angle of 60 ° 60^(\circ) to the x-axis (the slope of a straight line is the tangent of the slope angle tg 60 ° = 3 \textrm(tg )(60^(\circ)) = \sqrt(3)), ending at (0 ; - 4) (0;-4) .
This system of equations has exactly one solution if the circle touches one of the corner branches. This is possible in four cases (Fig. 16): the center of the circle can be at one of the points A A , B B , C C , D D . Since we need to find the smallest value of the parameter a a , we are interested in the abscissa of the point D D . Consider right triangle D H M DHM . The distance from the point D D to the line H M HM is equal to the radius of the circle, therefore D H = 1 DH=1 . So D M = D H sin 60 ° = 2 3 DM=\dfrac(DH)(\textrm(sin)(60^(\circ))) = \dfrac(2)(\sqrt(3)) . The coordinates of the point M M are found as the coordinates of the point of intersection of two lines y = 1 y=1 and y = - 3 x - 4 y=-\sqrt(3)x-4 ( left-hand side corner).
We get M (- 5 3) M(-\dfrac(5)(\sqrt(3))) . Then the abscissa of the point DD is - 5 3 - 2 3 = - 7 3 -\dfrac(5)(\sqrt(3))-\dfrac(2)(\sqrt(3))=-\dfrac(7)(\ sqrt(3)) .
Since the abscissa of the center of the circle is a 3 a\sqrt(3) , it follows that a = - 7 3 a=-\dfrac(7)(3) .
ANSWER
A = - 7 3 a=-\dfrac(7)(3)
Find all values of the parameter a a , for each of which the system
$$\begin(cases) |4x+3y| \leq 12a,\\ x^2+y^2 \leq 14ax +6ay -57a^2+16a+64 \end(cases) $$
has exactly one solution.
Let us depict the solution sets of each of the inequalities on the plane x O y xOy .
In the second inequality, we select the perfect squares:
x 2 - 14 ax + 49 + y 2 - 6 ay + 9 a 2 ≤ a 2 + 16 a + 64 ⇔ (x - 7 a) 2 + (y - 3 a) 2 ≤ (a + 8) 2 (19 ) x^2-14ax+49 + y^2-6ay + 9a^2 \leq a^2 + 16a + 64 \Leftrightarrow (x-7a)^2+(y-3a)^2 \leq (a+8 )^2 \:\:\:\: (19)
For a + 8 = 0 a+8=0 (a = - 8 a=-8) inequality (19) defines a point with coordinates (7 a ; 3 a) (7a;3a) , i.e. (- 56 ; - 24) (-56; -24) . For all other values of a a (19) defines a circle centered at the point (7 a ; 3 a) (7a;3a) of radius | a + 8 | |a+8| .
Consider the first inequality.
1) For negative a a it has no solutions. Hence, the system also has no solutions.
2) If a = 0 a=0 , then we get a straight line 4 x + 3 y = 0 4x+3y=0 . From the second inequality, we get a circle with center (0 ; 0) (0; 0) of radius 8. Obviously, more than one solution comes out.
3) If $$a>0$$, then this inequality is equivalent to the double inequality - 12 a ≤ 4 x + 3 y ≤ 12 a -12a \leq 4x+3y \leq 12a . It defines a strip between two lines y = ± 4 a - 4 x 3 y=\pm 4a -\dfrac(4x)(3) , each of which is parallel to the line 4 x + 3 y = 0 4x+3y=0 (Fig. 17).
Since we are considering $$a>0$$, the center of the circle is located in the first quadrant on the line y = 3 x 7 y = \dfrac(3x)(7) . Indeed, the coordinates of the center are x = 7 a x=7a , y = 3 a y=3a ; expressing a a and equating, we get x 7 = y 3 \dfrac(x)(7)=\dfrac(y)(3) , whence y = 3 x 7 y = \dfrac(3x)(7) . For the system to have exactly one solution, it is necessary and sufficient that the circle touches the line a 2 a_2 . This happens when the radius of the circle is equal to the distance from the center of the circle to the line a 2 a_2 . According to the formula for the distance from a point to a line * (\^{*} получаем, что расстояние от точки (7 a ; 3 a) (7a;3a) до прямой 4 x + 3 y - 12 a = 0 4x+3y-12a=0 равно | 4 · 7 a + 3 · 3 a - 12 a | 4 2 + 3 2 = 5 a \dfrac{|4\cdot 7a + 3\cdot 3a -12a|}{\sqrt{4^2+3^2}} = 5\left|a\right| . Приравнивая к радиусу круга, получаем 5 a = | a + 8 | 5{a} = |a+8| . Так как $$a>0$$, опускаем модули и находим, что a = 2 a=2 .!}
ANSWER
A=2 a=2
* {\^{*} Пусть даны точка M (x 0 ; y 0) M (x_0;y_0) и прямая l l , !} given by the equation a x + b y + c = 0 ax+by+c=0 . Then the distance from the point M M to the line l l is determined by the formula ρ = | a x 0 + b x 0 + c | a 2 + b 2 \rho = \dfrac(|ax_0+bx_0+c|)(\sqrt(a^2+b^2)) .
For what values of the parameter a a the system
$$\begin(cases) |x|+|y|=1,\\ |x+a|+|y+a|=1 \end(cases)$$ has no solutions?
The first equation of the system defines the square ABCD ABCD on the plane x O y xOy (to construct it, consider x ≥ 0 x\geq 0 and y ≥ 0 y\geq 0 . Then the equation takes the form x + y = 1 x+y=1 . We get a segment - part of the line x + y = 1 x + y \u003d 1, lying in the first quarter.Next, we reflect this segment about the axis O x Ox, and then we reflect the resulting set about the axis O y Oy) (see Fig. 18). The second equation gives the square P Q R S PQRS equal to the square A B C D ABCD , but centered at the point (- a ; - a) (-a;-a) . On fig. 18 for example this square is shown for a = - 2 a=-2 . The system has no solutions if these two squares do not intersect.
It is easy to see that if the segments P Q PQ and B C BC coincide, then the center of the second square is at the point (1 ; 1) (1; 1) . We will use those values of a a , at which the center is located “above” and “to the right”, i.e. $$a1$$.
ANSWER
A ∈ (- ∞ ; - 1) ∪ (1 ; + ∞) a\in (-\infty;-1)\bigcup(1;+\infty) .
Find all values of the parameter b b for which the system
$$\begin(cases) y=|b-x^2|,\\ y=a(x-b) \end(cases) $$
has at least one solution for any value of a a .
Let's consider several cases.
1) If $$b2) If b = 0 b=0 , then the system becomes $$\begin(cases) y=x^2,\\ y=ax .\end(cases) $$
For any a a, the pair of numbers (0 ; 0) (0;0) is a solution to this system, hence b = 0 b=0 fits.
3) Fix some $$b>0$$. The first equation is satisfied by the set of points obtained from the parabola y \u003d x 2 - b y \u003d x ^ 2-b by reflecting a part of this parabola about the O x Ox axis (see Fig. 19a, b). The second equation defines a family of lines (substituting various meanings a a , you can get all possible lines passing through the point (b ; 0) (b; 0) , except for the vertical one) passing through the point (b ; 0) (b; 0) . If the point (b ; 0) (b; 0) lies on the segment [ - b ; b ] [-\sqrt(b);\sqrt(b)] . the x-axis, then the straight line intersects the graph of the first function at any slope (Fig. 19a). Otherwise (Fig. 19b), in any case, there is a straight line that does not intersect this graph. Solving the inequality - b ≤ b ≤ b -\sqrt(b)\leq b \leq \sqrt(b) and taking into account that $$b>0$$, we get that b ∈ (0 ; 1 ] b \in ( 0;1] .
Combine the results: $$b \in $$.
ANSWER
$$b \in $$
Find all values a a , for each of which the function f (x) = x 2 - | x - a 2 | - 3 x f(x) = x^2-|x-a^2|-3x has at least one maximum point.
Expanding the module, we get that
$$f(x) = \begin(cases) x^2-4x+a^2, \:\:\: x\geq a^2 ,\\ x^2-2x-a^2, \:\ :\: x\leq a^2 . \end(cases) $$
On each of the two intervals, the graph of the function y = f (x) y=f(x) is a parabola with branches up.
Since upward-branching parabolas cannot have maximum points, the only possibility is that the maximum point is the boundary point of these gaps - the point x = a 2 x=a^2 . There will be a maximum at this point if the vertex of the parabola y = x 2 - 4 x + a 2 y=x^2-4x+a^2 falls on the interval $$x>a^2$$, and the vertex of the parabola y = x 2 - 2 x - a 2 y=x^2-2x-a^2 - on the interval $$x\lt a^2$$ (see Fig. 20). This condition is given by the inequalities and $$2 \gt a^2$$ and $$1 \lt a^2$$, solving which we find that a ∈ (- 2 ; 1) ∪ (1 ; 2) a\in (-\ sqrt(2);1)\bigcup(1;\sqrt(2)) .
ANSWER
A ∈ (- 2 ; 1) ∪ (1 ; 2) a\in (-\sqrt(2);1)\bigcup(1;\sqrt(2))
Find all values a a , for each of which general solutions inequalities
y + 2 x ≥ a y+2x \geq a and y - x ≥ 2 a (20) y-x \geq 2a \:\:\:\:\:\:\:\: (20)
are solutions to the inequality
$$2y-x>a+3 \:\:\:\:\:\:\:\:\: (21)$$
In order to orientate yourself in a situation, it is sometimes useful to consider a single value of a parameter. Let's make a drawing, for example, for a = 0 a=0 . Inequalities (20) (in fact, we are dealing with a system of inequalities (20)) are satisfied by the points of the angle BAC BAC (see Fig. 21) - points, each of which lies above both lines y = - 2 xy = -2x and y = xy =x (or on these lines). Inequality (21) is satisfied by points lying above the line y = 1 2 x + 3 2 y = \dfrac(1)(2)x + \dfrac(3)(2) . It can be seen that for a = 0 a=0 the condition of the problem is not fulfilled.
What will change if we take a different value of the parameter a a ? Each of the lines will move and go into a line parallel to itself, since the slopes of the lines do not depend on a a . To fulfill the condition of the problem, it is necessary that the entire angle B A C BAC lie above the line l l . Since the modulus of the slopes of the lines A B AB and A C AC is greater than the slope of the line l l , it is necessary and sufficient that the vertex of the angle lies above the line l l .
Solving the system of equations
$$\begin(cases) y+2x=a,\\ y-x=2a, \end(cases)$$
find the coordinates of point A (- a 3 ; 5 a 3) A(-\dfrac(a)(3);\dfrac(5a)(3)) . They must satisfy inequality (21), so $$\dfrac(10a)(3)+\dfrac(a)(3) > a+3$$, whence $$a>\dfrac(9)(8)$$ .
ANSWER
$$a>\dfrac(9)(8)$$
In order to fully reveal the possibilities of this method, we will consider the main types of problems.
Sample tasks for developing knowledge and skills in solving problems with parameters using a graphical method (coordinate plane)
Exercise 1.
At what valuesaequation = has two roots?
Solution.
We pass to the equivalent system:
This system defines a curve on the coordinate plane (;). It is clear that all points of this arc of the parabola (and only they) have coordinates that satisfy the original equation. Therefore, the number of solutions to the equation for each fixed value of the parameter, is equal to the number of points of intersection of the curve with the horizontal line corresponding to this parameter value.
Obviously, for the indicated lines intersect the graph at two points, which is equivalent to the original equation having two roots.
Answer: at.
Task 2.
Find all values of a for which the system has a unique solution.
Solution.
Let's rewrite the original system in this form:
All solutions of this system (view pairs) form the area shown in the figure by hatching. The requirement for the uniqueness of the solution of this system into a graphical language is translated as follows: horizontal lines must have only one common point with the resulting area. It is easy to see that only straight linesand satisfy the stated requirement.
Answer: or.
The two problems just analyzed allow us to give more specific recommendations compared to those given earlier:
try to express the parameter through a variable, i.e. get equalities of the form, then
graph a function on a plane.
Task 3.
At what valuesa does the equation have exactly three roots?
Solution.
We have
The graph of this set is the union of the “corner” and the parabola. Obviously, only the line intersects the resulting union at three points.
Answer: .
Comment: The parameter is usually considered as fixed, but unknown number. Meanwhile, from a formal point of view, a parameter is a variable, moreover, “equal” with others present in the task. With this view of the form parameter, functions are defined not with one, but with two variables.
Task 4.
Find all parameter values, for which the equation has one solution.
Solution.
A fraction is equal to zero if and only if the numerator of the fraction zero, and the denominator is different from zero.
Finding roots square trinomial:
Using the resulting system, it is easy to plot the original equation. It is the presence of "punctures" in this graph that allows for and = to have a unique solution to the equation. This is the determining factor in the decision.
Answer: and.
Task 5.
At what values of the parameter,a the equation has a unique solution.
Solution.
Let us write a system equivalent to the original equation
From here we get
We build a graph and we will draw straight lines perpendicular to the axisa .
The first two inequalities of the system define a set of points shown by hatching, and this set does not include hyperbolas and.
Then a segment and a ray, a segment and a ray, lying respectively on the lines and , are the graph of the original equation. One solution will be if 2< < или < или = .
Answer : 2 < < или < или = .
Task 6.
Find all parameter valuesa , for which the equation
has exactly two various solutions
Solution.
Consider the set of two systems
If , then.
If < , then.
From here
or
Parabolas and a straight line have two common points:A (-2; - 2), V(-1; -1), moreover, V is the vertex of the first parabola,D - top of the second. So, the graph of the original equation is shown in the figure.
There must be exactly two different solutions. This is done with or.
Answer: or.
Task 7.
Find the set of all numbers, for each of which the equation
has only two distinct roots.
Solution.
Let us rewrite this equation in the form
The roots of the equation, provided that.
We build a graph of this equation. In this case, it is convenient to build a graph by assigning the y-axis to the variable. Here we “read” the answer with vertical lines, we get that this equation has only two different roots at = -1 or or.
Dotted lines say that.
Answer: at = -1 or or.
Task 8.
For which the set of solutions of the inequality contains a gap.
Solution.
Let's write down the set of two systems, which is equivalent to the original equation:
or
Since in the solution of the first system neithera segment cannot be included, then we will carry out the necessary studies for the second system.
We have
Denote . Then the second inequality of the system takes the form< - and defines the set shown in the figure on the coordinate plane.
With the help of the figure, we establish that for the resulting set contains all points, the abscissas in which run through all the values of the interval
Then, from here.
Answer : .
Task 9.
Find all non-negative numbers, for which there is a single number that satisfies the system
Solution.
We have
The first equation on the coordinate plane defines a family of vertical lines. The straight lines and divide the planes into four regions. Some of them are solutions to the system's inequality. Specifically, which ones can be established by taking a trial point from each area. The area whose point satisfies the inequality is its solution (this technique is associated with the method of intervals when solving inequalities with one variable). We build straight lines
For example, we take a point and we substitute in Coordinates of a point satisfy an inequality.
We get two areas (I) and ( II), but, given that by condition, we take only the area (I). We build straight lines , k .
So, the original system is satisfied by all points (and only them) lying on the rays and highlighted in the drawing by bold lines (i.e., we build points in a given area).
Now we need to find the only one for fixed. Draw parallel lines intersecting the axis. and find where there will be one point of intersection with the line.
We find from the figure that the requirement of uniqueness of the solution is achieved if (for already 2 points),
where is the ordinate of the point of intersection of the lines and,
where is the ordinate of the intersection point of the lines and.
So we get< .
Answer: < .
Task 10.
At what values of the parameter a does the system have solutions?
Solution.
We factorize the left side of the inequality of the system, we have
We build straight lines and We show in the figure by shading the set of points of the plane that satisfies the inequality of the system.
We build a hyperbola = .
Then the abscissas of the distinguished arcs of the hyperbola are solutions of the original system.M , P , N , Q - nodal points. Let's find their abscissas.
For points P , Q we have
It remains to write down the answer: or.
Answer: or.
Task 11.
Find all values for which any solution to the inequality does not exceed two in absolute value ().
Solution .
Let us rewrite this inequality in this form. We construct graphs of equations and =.
Using the “interval method”, we establish that the shaded areas will be the solution to the original inequality.
Now we build the area and see which part of it falls into the shaded area.
Those. now, if for some fixed value, the line at the intersection with the resulting area gives only points whose abscissas satisfy the condition < 2, then is one of the required values of the parameter.
So we see that.
Answer: .
Task 12.
For what values of the parameter does the set of solutions to the inequality contain at most four integer values?
Solution.
Let us transform this inequality to the form This inequality is equivalent to the combination of two systems
or
Using this set, we depict the solution of the original inequality.
Let's draw straight lines, where. Then the value for which the line intersects the lines at most in four points from the marked set will be the desired one. So we see that either or.
Answer: or or.
Task 13.
At what values of the parametera has a solution system
Solution.
The roots of the square trinomial i.
Then
We build straight lines and
Using the “intervals” method, we find the solution to the system inequality (shaded area).
That part of the circle with the center at the origin and radius 2, which falls into the shaded area and will be the solution of this system. .
Values and find from the system
Values and - from the system.
Answer:
Task 14.
Depending on the parameter valuesa solve inequality > .
Solution.
Let us rewrite this inequality in the form and consider the function, which, expanding the modules, we write as follows:
We build a chart. The graph splits the coordinate plane into two regions. Taking m. (0; 0) and substituting and into the original inequality, we get that 0 > 1, and therefore the original inequality is satisfied in the area lying above the graph.
Directly from the figure we get:
no solutions;
at ;
at.
Answer: no solutions;
at ;
at.
Task 15.
Find all values of the parameter for which the system of inequalities
only one is satisfied.
Solution.
Let's rewrite this system in the following form:
Let us construct the area specified by this system.
1) , is the vertex of the parabola.
2) is a straight line passing through the points and.
The requirement for the uniqueness of the solution is translated into a graphical language as follows: horizontal lines with the resulting area must have only one common point. The above requirement is satisfied by the lines and, where is the ordinate of the point of intersection of the parabola and the line.
Let's find the value:
= (does not fit the meaning of the problem),
We find the ordinate:
Answer: ,
Task 16.
Find all parameter valuesa, under which the system of inequalities
satisfies only for one x.
Solution .
Let us construct parabolas and show the solution of the last system by shading.
1) , .
2) , .
It can be seen from the figure that the condition of the problem is satisfied for or.
Answer: or.
Task 17.
For what values does the equation have exactly three roots?
Solution.
This equation is equivalent to the set
The population plot is the union of the parabola and angle plots.
The lines intersect the resulting union at three points.
Answer: at.
Task 18.
For what values does the equation have exactly three solutions?
Solution.
Let's transform the left side of this equation. We get a quadratic equation for.
We get the equation
which is equivalent to the aggregate
The union of the graphs of parabolas is the solution of the set.
Find the ordinate of the points of intersection of the parabolas:
We read the necessary information from the figure: this equation has three solutions for or
Answer: at or
Task 19.
Depending on the parameter, determine the number of roots of the equation
Solution .
Consider this equation as a quadratic one with respect to a.
,
.
We get the set
We build graphs of the equations of the set and answer the question of the problem.
Answer:: no solutions;
: one solution;
: two solutions;
or: three solutions;
or: four solutions.
Task 20.
How many solutions does the system have
Solution.
It is clear that the number of roots of the second equation of the system is equal to the number of solutions of the system itself.
We have .
Considering this equation as a quadratic one, we obtain the set.
Now referring to the coordinate plane makes the task simple. We find the coordinates of the intersection points by solving the equation
From here
Vertices of parabolas and.
Answer: : four solutions;
: two solutions;
: one solution;
: no solutions.
Task 21.
Find all real values of the parameter for which the equation has only two distinct roots. Write down these roots.
Solution .
Let's find the roots of the square trinomial in brackets:
We depict the set of solutions of this equation in the coordinate plane by plotting graphs provided that
We read the necessary information from the picture. So, this equation has two different roots at (u) and at (u)
Answer: with (and) and
at (and).
Task 2 2 .
Solve the system of inequalities:
Solution.
We build in the plane graphs of a parabola and a straight line.
All points of the shaded area are the solution of the system. Let's divide the constructed area into two parts.
If and, then there are no solutions.
If, then the abscissas of the points of the shaded area will be greater than the abscissas of the points of the straight line, but less than the abscissas (larger root of the equation) of the parabola.
We express through from the equation of a straight line:
Let's find the roots of the equation:
Then.
If, then.
Answer: for and 1 there are no solutions;
at;
at.
Task 23.
Solve the system of inequalities
Solution.
– top of the parabola.
Top of the parabola.
Find the abscissas of the intersection points of the parabolas:
The shaded area is the solution of the system. Let's break it down into two parts.
In the equations of parabolas, we express through:
We write down answer:
if and, then there are no solutions;
if, then< ;
if, then.
Task 24.
At what values, and the equation has no solutions?
Solution.
The equation is equivalent to the system
Let us construct a set of solutions to the system.
Three pieces of a parabola are the solution to this equation.
Let us find under which and exclude it.
So, for there are no solutions;
when there are no solutions;
(note: for the restathere are one or two solutions).
Answer: ; .
Task 25.
For what real values of the parameter there is at least one that satisfies the conditions:
Solution.
Let's graphically solve the inequality in using the "interval method" and build a graph. Let's see what part of the graph falls into the constructed region for solving the inequality, and find the corresponding valuesa.
We build graphs of lines and
They divide the coordinate plane into 4 regions.
Let's graphically solve the last inequality using the "interval method".
The shaded area is its solution. Part of the parabola graph falls into this area. On the interval; (by condition the inequality of the system is strict) exist that satisfy the conditions of the given system.
Answer:
Task 26.
Find all values of the parameter, for each of which the set of solutions to the inequality does not contain any solutions to the inequality.
Solution.
Let us construct a set of solutions to the inequality (“by the method of intervals”). Then we will build a "band" The desired values of the parameterq those for which none of the points of the indicated areas belongs to the "strip"
Answer: or.
Task 27.
For what values of the parameter, the equation has a unique solution.
Solution.
Let's factorize the numerator of the fraction.
This equation is equivalent to the system:
Let's build a population graph in the coordinate plane.
or
– point of intersection of lines and. A population graph is a union of lines.
We “poke out” the points of the graph with abscissas.
We draw straight lines and see where there is one point of intersection with the graph.
It is obvious that only for or this equation has a unique solution.
Answer: or.
Task 28.
For what real values of the parameter the system of inequalities has no solutions.
Solution.
The set of points in the plane of the shaded area satisfies the given system of inequalities.
We build straight lines. From the figure, we determine that for (- the abscissa of the intersection point of the hyperbola and the line), the lines do not intersect the shaded area.
Answer: at.
Task 29.
At what values of the parametera the system has a unique solution.
Solution.
Let's move on to a system equivalent to the given one.
In the coordinate plane, we plot the graphs of the parabolas and Vertices of the parabolas, respectively, the points and.
Calculate the abscissas of the intersection points of the parabolas by solving the equation
The shaded area is the solution of the system of inequalities. Direct and
has one common point with the shaded area.
Answer: at i.
Task 30.
Solve the inequality:
Solution.
Depending on the parameter, we find the value.
We will solve the inequality using the “interval method”.
Let's build parabolas
: .
Calculate the coordinates of the point of intersection of the parabolas:
The points of the shaded area satisfy this inequality. By drawing a straight line, we divide this area into three parts.
1) If, then there are no solutions.
2) If, then in the equation we express through:
Thus, in the areaI we have.
If, then look:
a) area II .
We express in the equation through.
smaller root,
Bigger root.
So, in the area II we have.
b) area III : .
Answer: when there are no solutions;
at
at, .
Literature:
Galitsky M. L., Goldman A. M., Zvavich L. I. Collection of problems in algebra for grades 8 - 9: Tutorial for students of schools and classes with in-depth study of mathematics - 2nd ed. – M.: Enlightenment, 1994.
P. I. Gorshtein, V. B. Polonsky, M. S. Yakir. Tasks with parameters. 3rd edition, enlarged and revised. - M .: Ileksa, Kharkov: Gymnasium, 2003.
Faddeev D. K. Algebra 6 - 8. - M .: Education, 1983 (b - ka teacher of mathematics).
A. H. Shakhmeister. Equations and inequalities with parameters. Edited by B. G. Ziva. C - Petersburg. Moscow. 2004.
V. V. Amelkin, V. L. Rabtsevich. Tasks with parameters Minsk "Asar", 2002.
A. H. Shakhmeister. Tasks with parameters in the exam. Moscow University Publishing House, CheRo on the Neva MCNMO.
TO tasks with parameter include, for example, the search for solutions to linear and quadratic equations v general view, the study of the equation for the number of roots available depending on the value of the parameter.
Without giving detailed definitions, consider the following equations as examples:
y = kx, where x, y are variables, k is a parameter;
y = kx + b, where x, y are variables, k and b are parameters;
ax 2 + bx + c = 0, where x are variables, a, b and c are parameters.
To solve an equation (inequality, system) with a parameter means, as a rule, to solve an infinite set of equations (inequalities, systems).
Tasks with a parameter can be conditionally divided into two types:
a) the condition says: solve the equation (inequality, system) - this means, for all values of the parameter, find all solutions. If at least one case remains unexplored, such a solution cannot be considered satisfactory.
b) required to specify possible values parameter under which the equation (inequality, system) has certain properties. For example, has one solution, has no solutions, has solutions, belonging to the interval etc. In such tasks, it is necessary to clearly indicate at what value of the parameter the required condition is satisfied.
The parameter, being an unknown fixed number, has, as it were, a special duality. First of all, it must be taken into account that the alleged fame suggests that the parameter must be perceived as a number. Secondly, the freedom to handle a parameter is limited by its unknown. So, for example, the operations of dividing by an expression in which there is a parameter or extracting a root of an even degree from a similar expression require preliminary research. Therefore, care must be taken in handling the parameter.
For example, to compare two numbers -6a and 3a, three cases need to be considered:
1) -6a will be greater than 3a if a is a negative number;
2) -6a = 3a in the case when a = 0;
3) -6a will be less than 3a if a is a positive number 0.
The decision will be the answer.
Let the equation kx = b be given. This equation is shorthand for an infinite set of equations in one variable.
When solving such equations, there may be cases:
1. Let k be any real number non-zero and b is any number from R, then x = b/k.
2. Let k = 0 and b ≠ 0, the original equation will take the form 0 · x = b. Obviously, this equation has no solutions.
3. Let k and b be numbers equal to zero, then we have the equality 0 · x = 0. Its solution is any real number.
The algorithm for solving this type of equations:
1. Determine the "control" values of the parameter.
2. Solve the original equation for x with the values of the parameter that were determined in the first paragraph.
3. Solve the original equation for x with parameter values that differ from those selected in the first paragraph.
4. You can write down the answer in the following form:
1) when ... (parameter value), the equation has roots ...;
2) when ... (parameter value), there are no roots in the equation.
Example 1
Solve the equation with the parameter |6 – x| = a.
Solution.
It is easy to see that here a ≥ 0.
By the rule of modulo 6 – x = ±a, we express x:
Answer: x = 6 ± a, where a ≥ 0.
Example 2
Solve the equation a(x - 1) + 2(x - 1) = 0 with respect to the variable x.
Solution.
Let's open the brackets: ax - a + 2x - 2 \u003d 0
Let's write the equation in standard form: x(a + 2) = a + 2.
If the expression a + 2 is not zero, i.e. if a ≠ -2, we have the solution x = (a + 2) / (a + 2), i.e. x = 1.
If a + 2 is equal to zero, i.e. a \u003d -2, then we have the correct equality 0 x \u003d 0, therefore x is any real number.
Answer: x \u003d 1 for a ≠ -2 and x € R for a \u003d -2.
Example 3
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a \u003d 0, then we transform the equation to the form a + x \u003d a 2 + ax or (a - 1) x \u003d -a (a - 1). The last equation for a = 1 has the form 0 · x = 0, therefore, x is any number.
If a ≠ 1, then the last equation will take the form x = -a.
This solution can be illustrated on the coordinate line (Fig. 1)
Answer: there are no solutions for a = 0; x - any number at a = 1; x \u003d -a with a ≠ 0 and a ≠ 1.
Consider another way to solve equations with a parameter - graphical. This method is used quite often.
Example 4
How many roots, depending on the parameter a, does the equation ||x| – 2| = a?
Solution.
To solve by a graphical method, we construct graphs of functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows the possible cases of the location of the line y = a and the number of roots in each of them.
Answer: the equation will have no roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case a = 2; four roots - at 0< a < 2.
Example 5
For which a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let's draw graphs of functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x - 1|, expanding the modules by the gap method, we get:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x – 1, for x > 1.
On the Figure 3 it is clearly seen that the equation will have a unique root only when a = 1.
Answer: a = 1.
Example 6
Determine the number of solutions of the equation |x + 1| + |x + 2| = a depending on the parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at the points (-2; 1) and (-1; 1) (picture 4).
Answer: if the parameter a is less than one, then the equation will have no roots; if a = 1, then the solution of the equation is an infinite set of numbers from the interval [-2; -one]; if the values of the parameter a are greater than one, then the equation will have two roots.
Do you have any questions? Don't know how to solve equations with a parameter?
To get the help of a tutor - register.
The first lesson is free!
site, with full or partial copying of the material, a link to the source is required.