To solve the first kind of parameter problems, we usually adopt the method of "classified discussion", that is, according to the conditions of the problem and the concepts involved; Use theorems, formulas, properties, operational needs, graphic positions, etc. Carry out scientific and reasonable classification, then discuss them one by one, explore their respective results, and finally draw the conclusion of the proposition to achieve the purpose of solving the problem. In fact, it is a kind of hard work. Strategies and methods of solving problems from complex to simple.
First, scientific and reasonable classification.
A set A is divided into several non-empty proper subset Ai(i= 1, 2,3 n) (n ≥ 2, n∈N), so that each element in set A belongs to only one subset. that is
①A 1∪A2∪A3∪An = A
② AI ∩ AJ = φ (i，j ∈ n，i≠j)。
It is called the scientific classification (or logical division) of set A.
Scientific classification meets two conditions: condition ① ensures that classification is not omitted; Condition ② ensures that the classification is not repeated. On this basis, according to the conditions and nature of the problem, try to reduce the classification.
Second, determine the classification criteria.
After determining the object of discussion, the most difficult thing is to determine the classification standard. Generally speaking, there are usually three classification criteria:
(1) Determine the classification standard according to mathematical concepts.
For example, the absolute value is defined as:
Therefore, when solving the inequality | logx |+| log (3-x) | ≥ 1 with absolute values, it is necessary to divide the domain (0,3) into three sections for discussion according to the x value 1 and 2 that determine the sign of logx and log (3-x), that is, 0 < x
For example 1, it is known that the distance from the moving point m to the origin o is m, the distance from the straight line L: x = 2 is n, and m+n = 4.
(1) Find the trajectory equation of point M. ..
(2) The straight line with the inclination angle α passing through the origin O intersects the trajectory curve of point M at two points P and Q, and the maximum chord length ||| PQ||| and the corresponding inclination angle α are found.
Solution: (1) Let the coordinate of point M be (x, y), which can be obtained according to the meaning of the question: += 4.
According to the concept of absolute value, the trajectory equation depends on whether x > 2 or x≤2, so the classification discussion based on 2 can be
The trajectory equation is: y =
Solution (2) As shown in figure 1, due to the position change of p and q,
The expressions of chord length ||| pq ||| are different, so it is necessary to divide points P and Q into two parts, that is, they are both on curve y2=4(x+ 1), one part on curve y2=4(x+ 1) and the other part on curve Y2 =-12 (x).
So as to know when or when,
(2) Determine the classification standard according to the theorems, formulas and properties in mathematics.
Some formulas, theorems and properties in mathematics have different conclusions under different conditions. When it is used, it should be classified and discussed, and the classification is based on the conditions in the formula.
For example, the monotonicity of logarithmic function y = logax is given in two cases: 0 < a < 1 and a > 1, so the solution base contains letter inequality; For example, logx & gt- 1 should be classified and discussed according to cardinality X > 1 and 0 < X < 1, that is, when X > 1, when 0 < X < 1,
For another example, the first few terms and formulas of geometric series are given respectively:
Therefore, when solving this kind of problems, if Q is a variable, we should use Q as the standard for classification and discussion.
Example 2, let the first term be 1, the sum of the first n terms in geometric progression be Sn, let Tn =, n = 1, 2 ... find TN.
Solution: when Q = 1, Sn = N, TN =
When q≠ 1, sn =
All in all,
(3) Determine the classification standard according to the operation needs.
For example, solving inequalities.
Obviously, A should be divided into three cases: 1 < A ≤ 3, 3 < A ≤ 4 and based on 3 4.
Example 3, Solving Inequalities about X
Where a > 0 and a≠ 1.
Solution, because all inequalities contain parameter a, and the situation of its solution depends on A > 1 or A < 1, so 1 is the standard of classification.
(I) when 0 < a < 1, the following solution can be obtained:
(2) when a > 1, the solution can be: whether the inequality group has a solution depends on the size relationship with 2, so a = 3 is taken as the standard of the second classification.
(1) When 1 < a ≤ 3, the solution set is φ.
(2) when a > 3, the solution set is
To sum up: when 0 < a < 1, the original inequality solution set is (2,; When 1 < a ≤ 3, the solution set is φ;
When a > 3, the solution set is (2,.
Third, the methods and steps of classified discussion
(1) Determine whether classified discussion is needed, and if so, the object and its value range;
(2) Determine the classification standard and make scientific and reasonable classification;
(3) Discuss one by one and get various results;
(4) Summarize various conclusions.
Example 4: If the image passing points of the function f (x) = a+bcosx+csinx (0, 1) and (1), x ∈ [0, 0], | f (x) | ≤ 2 hold, try to find the value range of a.
Solution: from F (0) = A+B = 1, F () = A+C = 1, we can get b = c =1-a.
f(x)= a+( 1-a)(sinx+cosx)= a+( 1-a)sin(x+)
∵
① when a≤ 1,1≤ f (x) ≤ a+(1-a) ∞| f (x) |≤ 2 ∴ as long as a+( 1-a) ≤ 2 is a ≥ ② when
Example 5, the function f (x) = sim2x-asim2 is known.
Try to find the maximum value b of f(x) represented by a.
Solution: The original function is f(x)= 1
Let t = cosx, then-1 ≤ T ≤ 1.
Remember that g (t) =- (. t∈[- 1， 1]
Because the acquisition of the maximum value of quadratic function g(t) is closely related to the position of the vertex abscissa of the image of quadratic function y=g(t) relative to the definition domain [- 1, 1], it is discussed in three cases with the position relative to the interval [- 1, 1]:
(1) When-1 ≤ 1, that is, -4 ≤ A ≤ 4, b = g (t) max =, t =；;
(2) When
(3) when > 1, that is, A > 4 and B = 0, at this time, t= 1.
To sum up: b =
Example 6, arithmetic progression {an} tolerance d < 0, Sn is the sum of the first n terms, if sp = sq, (p, q∈N, p≠q) try d, p and q represent the maximum value of Sn.
Slight solution: It can be obtained from SP = SQP ≠ Q.
∵ d < 0, ∴ a 1 > 0, and Sn is maximum if and only if.
From an≥0, n≤, from an+ 1≤0, n≥
∴≤n≤, ∫n∈n, ∴ We should discuss it in two categories according to whether it is a positive integer, that is, whether p+q is odd or even.
(1) when p+q is even, n =, Sn is the largest, and (Sn)max=
(2) when p+q is odd, n = or n =, Sn is the largest, and (Sn)max= =
Example 7, (Beijing 20 1 1 year college entrance examination science 18)
Known function.
(i) Monotone interval of the solution;
(2) If there is, there is a range of ≤
Solution: (1)
Order, take it.
When k>0, the situation is as follows.
x
()
(，k)
k
+
—
+
↗
↘
↗
Therefore, the monotone decreasing interval of is () and; A single high interval is when K.
x
()
(，k)
k
—
+
—
↘
↗
↘
Therefore, the monotone decreasing interval of is () and; The single high-level interval is
(2) when k > 0, because, so there will beno..
When k < 0, we know from (I) that the maximum value on (0,+) is
So this is equivalent to
Solve.
So the range of k at that time was
The idea of classified discussion is an important problem-solving strategy, which undoubtedly helps to cultivate the rigor, rigor and flexibility of students' thinking and improve their ability to analyze and solve problems. However, it is not necessary to discuss the problems with parameters in categories as soon as they appear. If we can combine the idea of combining numbers and shapes with the idea of functions, we can avoid or simplify the classified discussion, so as to achieve a fast and accurate problem-solving effect.
Example 7. Solve the inequality about x: ≥ a-x.
Simple solution: use the idea of combining numbers and shapes to solve problems, as shown in the figure:
Make images with y = and y = a-x in the same coordinate system, and take the intercept of L 1, L2 and L3 on the Y axis as the classification standard.
Know: when a ≤- 1; - 1≤x≤3
When-1 < a ≤ 3; ≤x≤3
When 3 < A 1+2;
When a > 1+2, the inequality has no solution.
Example 8. What is the value of the real number k? Does the equation kx2+2|x|+k=0 have a real number solution?
Simple solution: solving problems with the idea of function;
K = can be obtained from the equation.
So when the equation has a solution, the value range of k is the function f (x) =, obviously-1 ≤ f (x) ≤ 0.
Therefore,-1 ≤ K ≤ 0 is the demand.
In short, in the teaching of each module, mathematical ideas such as classified discussion are gradually infiltrated to solve problems. Classified discussion covers more knowledge points, which is conducive to examining students' knowledge. Classified thinking is diverse, logical and comprehensive. To establish classified discussion thinking, we should pay attention to understanding and mastering the principles, methods and skills of classification, so as to determine all objects, clarify classification standards, analyze and discuss at different levels, do not repeat or omit, and achieve phased results; Finally, summarize and draw a comprehensive conclusion.