Topic 04 Computation of Sets

Topic 04 Computation of Sets (Knowledge combing + 7 point-to-point training + basic clearance + expansion and improvement) learning target 1. Understand the meaning of union and intersection of two sets, and find the union and intersection of two simple sets. (Important, difficult) 2. Be able to use Venn diagrams to express the relationship and operation of sets, and experience the role of diagrams in understanding abstract concepts. (Difficulty) 3. Understand the meaning of the complete work and its symbolic representation. (Easy to mix) 4. Understand the meaning of the complement of a subset in a given set, and will find the complement of a given subset. (Important, difficult) New knowledge points Knowledge point 01: Intersection and its nature Intersection: A A set consisting of a set and all the common elements B of the set is called A the B intersection of and, denoted as " A  B " and pronounced as "A intersects B", i.e A  B   x | x  A且 x  B  ① A  B  B  A ； ② A  A  A ， A     ； ③ A  (B  C )  ( A  B)  C ； ④ A  B  A, A  B  B ； (5) if A  B  A , then A  B ; A ∩ B can be visualized with a Venn chart (1) It means that sets A and B have both common elements and non-common elements, and the shadow part A ∩B is both a true subset of A and a true subset of B; (2) Indicates that the set A is a subset of B, where A∩B=A; Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 1 / 44 ( 3 ) indicates that there is no common element between sets A and B, where A∩B=∅ . Knowledge point 02: Union and its properties Union: A set consisting of all elements A that belong to a set B or belong to a set A is called a union of set i.e A  B A  B  4 and is denoted as " B " and is pronounced "A and B",  x | x  A或 x  B  A  B  B  A ； ② A  A  A ， A    A ； ③ A  (B  C )  ( A  B)  C ； ④ A  A  B, B  A  B ； (5) if A  B  B , then A  B ; Several different cases of A∪B can be visually reflected in the Venn chart, as shown in the diagram where the shaded part represents A∪B. (1) Indicates that sets A and B have both common and non-common elements, and both A and B are true subsets of A∪B (2) indicates that set A is a subset of B, where A∪B=B (3) Indicates that there are no common elements in sets A and B Knowledge point 03: Complete works, supplements and their properties 1. Complete works (1) Definition: If a set contains all the elements involved in the research problem, then the set is called a complete set (2) Notation: The complete works are usually written as U 2. Patches For a set A, the set consisting of all the elements in the set U that do not Written language Symbolic language Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd belong to set A is called the complement of set A with respect to set U and is denoted as A A ={x|x∈U,且 x∉ A} 2 / 44 Graphical A language In the For practice Point-to-point training 1: The concept and operation of intersection Typical example questions Example 1. Known Set A  1, 0,1, 2 , B  1, 2,3 then A  B  【Answer】 1, 2 【Knowledge Points】The concept and operation of intersection 【Analysis】The solution can be obtained according to the concept of intersection. [Detailed explanation] can be obtained from the meaning of A  B  1, 2 the title So the answer is: 1, 2 . Example 2. set a set A  0,1, 2,3, 4 , B   x 0  x  3 then A  B  [Answer] 1, 2  /  2,1 【Knowledge Points】The concept and operation of intersection 【Analysis】Calculated according to the definition of intersection. 【Explanation】Because A  0,1, 2,3, 4 , B   x 0  x  3 , So A  B  1, 2 ... So the answer is: 1, 2 Example 3 is known A  { x | x  N,1  x  10} , B   x | x is a multiple of 3 } , then can A  B be expressed by enumeration as 【Answer】 3, 6,9 【Knowledge Points】The enumeration method represents the concepts and operations of sets and intersections 【Analysis】According to the meaning A  1, 2,3, 4,5, 6, 7,8,9,10 of the problem, it can be solved by combining the intersection operation Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 3 / 44 【Detailed explanation】From the meaning of the title, it can be known: A  {x | x  N,1  x  10}  1, 2,3, 4,5, 6, 7,8,9,10 , And B   x | x it's a multiple of 3 } , so A  B  3, 6,9 ... So the answer is: 3, 6,9 . Example 4 if A  x∣ 2  x  3 , B  {x∣x  a} ,seek A  B . 【Knowledge Points】The concept and operation of intersection 【Analysis】For the a classification discussion, the number axis is used to find the intersection of the two sets 【Details】Mark the set on the number line A , as shown in the figure When a  2 , A  B  A  x∣ 2  x  3 ； When 2  a  3 , A  B  {x∣a  x  3} ； When a  3 , A B   ． To sum up, when a  2 , A  B   x∣ 2  x  3 ; When 2  a  3 , A  B  {x∣a  x  3} ; When a  3 , A  B   ... concise 1. known set A  (2,  ), B  1, 2,3, 4,5  , then A  B  【Answer】 {3, 4,5} Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 4 / 44 【Knowledge Points】The concept and operation of intersection 【Analysis】According to the given conditions, the answer is directly obtained by using the definition of intersection. 【Detailed explanation】According to the meaning of the topic, A  B  {3, 4,5} So the answer is: {3, 4,5} 2. Collection A  (1,1), B  Z , then A  B  【Answer】 0 【Knowledge Points】The concept and operation of intersection 【Analysis】Calculate according to the intersection of sets. Because A  (1,1), B  Z of the set, then A  B  0 So the answer is: 0 . 3. A   x 1  x  3 B  {x x  2 or x  2} , then A  B  【Answer】 { x | 2  x  3} 【Knowledge Points】The concept and operation of intersection 【Analysis】Solve according to the operation of the intersection. 【Explanation】Because A   x 1  x  3 , B  {x x  2 or x  2} , So A  B  {x | 2  x  3} , So the answer is: {x | 2  x  3} 4.   ) y  x2  3 then A  B  ) y  x  3} , B  ( x, y∣ known set A  {( x , y∣ 【Answer】  0, 3  , 1, 4  【Knowledge Points】The concept and operation of intersection y  x  3 【Analysis】Find  the solution of the system of equations, and according to the meaning 2 y  x  3 of the intersection of sets, you can get the answer x  0 x  1 y  x  3 【Detailed explanation】Solution  , get  or  , 2 y  4 y  x  3 y  3  y  x  3  Therefore A  B   x, y  |     0,3 , 1, 4  , 2  y  x  3  So the answer is:  0,3 , 1, 4    5. Known sets A  x∣x  5k  1, k  N , B  x∣x  6, x Q , find A  B 【Answer】 A  B  1, 4, 6 Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 5 / 44 【Knowledge Points】The descriptive method represents the set, and the enumeration method represents the concept and operation of the set and intersection 【Analysis】The A elements in the are first represented from small to large by enumeration (until there are elements greater than or equal to 6), and the solution can be obtained by combining the concept of intersection     x x  5 k 1, k N  1, 6, 11, 4, 21, 26, 31, 6, 41, , 【Detailed explanation】∵ A  ∣ B   x∣x  6, x Q ，∴ A  B  1, 4, 6 ． 2 6. Known sets A  {( x, y )∣y  4 x  1} , sets B  ( x , y )∣y  x  2  , and sets are represented by enumeration A  B 【Answer】 {(1,3), (3,11)} 【Knowledge Points】The enumeration method represents the concepts and operations of sets and intersections 【Analysis】The elements in the set A and B are all points on the function image, so the intersection of A and B is y  4 x  1 the set of the intersection points of y  x2  2 the and  x 1  y  4x  1  x3 【Detailed explanation】Lianli  , solution:  or  , therefore A  B  {(1,3), (3,11)} 2 y  x  2  y  11 y  3 Point-to-point training 2: Find the set or parameters according to the intersection results Typical example questions Example 1.Known set A  1, 2 , B  a,3 if A  B  2 , then a  【Answer】 2 【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】The answer can be obtained from the definition of intersection. 【Detailed explanation】Because A  1, 2 , B  a,3 , A  B  2 , 2  B Then,So a  2 . So the answer is: 2 Example 2.(23-24 Gao Yi Shang Shanghai Putuo Midterm) Known A   x x  2 , B   x x  a if , then A  B   the value range of a the real number is 【Answer】 a a  2 【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】According to A  B   the range of values of a the real numbers that can be obtained Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 6 / 44 【Explanation】Because A   x x  2 , B   x x  a , A  B   And,Then a  2 . So the answer is: a a  2 . Example 3 x 2  px  4  0 The solution set of the equation is M , and x 2  x  q  0 the N value of the M  N  4 equation is p  q equal to [Answer] -25 【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】4 is the common solution of the two equations, which can be substituted. 42  4 p  4  0  p  5  M  N  4   so  2 [Detailed explanation], , therefore  , at this time M  1, 4 ,  q  20 4  4  q  0 N  5, 4 satisfied M  N  4 So p  q  25 ... So the answer is: -25   2 Example 4. A known set M  x x  x  6  0 , N   x mx  1  0 if M  N  N , then m the set of values of real numbers is  1 1 【Answer】   , 0,   3 2 【Knowledge Points】Find the parameters according to the inclusion relationship of the set, and find the set or parameters according to the intersection results 【Analysis】The set can be obtained by solving the equation M , and the number of elements in the set can be discussed N according to the calculation results between the sets   2 【Detailed explanation】From M  x x  x  6  0  3, 2 , M  N  N , Then N  M , Again N   x mx  1  0 , So N there's so much one element in the 1 set When N there is 0 an element, i.e N   ., time, mx  1  0 there is no solution, m  0 , When N there 1is an element in the collection, i.e., N  3 or N  2 1 if N  3 , then 3m  1  0 , m   , 3 if N  2 , then 2m  1  0 , m  1 , 2  1 1 To sum up, the m set of values is   , 0,  ,  3 2 Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 7 / 44  1 1 So the answer is:   , 0,  .  3 2 Example 5. Known 2 2 2 2 Collection A   x∣x  3 x  2  0 , B   x∣x  ax  a  1  0 , C   x∣x  2  m  1 x  m  5  0 (1) If A  B  B , find a the value of the real number; (2) If A  C  C so, find m the range of values of real numbers 【Answer】(1) a  2 or 3 ; (2) (, 3] 【Knowledge Points】Find the parameters according to the inclusion relationship of the set, and find the set or parameters according to the intersection results 【Analysis】(1) If x 2  3 x  2  0 the root of the equation is solved first, then the set A can be known, and then x 2  ax  a  1  0 the root of the solution can be determined, B and the value can be found according to A  B  B the obtainable value that B  A a  1 can be solved a ; (2) According to A  C  C the obtained C  A , the C value is determined by considering the a situation of empty set, single element set, and double element set respectively 【Explanation】(1) By x 2  3 x  2  0 the or x  1 2 , so, A  {1, 2} By x 2  ax  a  1  0 or so x  1 a  1 , 1  B, a  1  B Because A  B  B , so B  A , So a  1  1 or 2 , so a  2 or 3 ; (2) Because A  C  C , so C  A , 2 2 When C   , Δ  4( m  1)  4  m  5   0 ,Solution m  3 , Δ  4( m  1) 2  4 m 2  5   0 C  1  ,  When , 2  1  2( m  1)  m  5  0   Δ  4  m  12  4 m 2  5  0 When C  2 ,  ,Solution m  3 , 2  4  4  m  1   m  5  0 1  2  2( m  1) When C  1, 2 ,  , 2  1 2  m  5 In summary, the range of values for the real number m is (, 3] concise 1. set up A   x  2  x  5 , B   x x  a , and A  B  A , then the value range of a the real number is 【Answer】 a  2 Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 8 / 44 【Knowledge Points】Find the set or parameters according to the intersection results, and find the parameters according to the inclusion relationship of the set 【Analysis】According to the intersection result, the inclusion relationship is obtained, so as to obtain a the value range of the number 【Explanation】 A  B  A Because,So A  B , Therefore a  2 So the answer is: a  2 2 2．Konw set A   x∣x  x  6  0 ， B   y∣ay  3  0 ，if A  B  B ，real number a  ． 3 Answer, 0 or 1or  , 2 【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】It can be solved according to the computational properties of the set intersection. 2 [Detailed explanation] A   x∣x  x  6  0  2, 3 , When a  0 , B   ,Apparently in line with A  B  B , 3 When a  0 , B    ,Because A  B  B , a So there is 3 3 3  3 , or, solution  2 a  1 , or a   , a 2 a 3 In summary: a  0 , or 1 or  , 2 3 So the answer is: 0 , or 1 or  , 2 3. 2 2 Known set A   x x  ax  a  19  0 , B  2,3 ,, C   4, 2 if A  B   , and A  C   , then a the value of the real number is 【Answer】 2 【Knowledge Points】Find sets or parameters according to the intersection results, and find parameters according to the relationship between elements and sets 【Analysis】According to the meaning of 4, 2  A,3  A the question, you can substitute 3 to x 2  ax  a 2  19  0 obtain the value of a. The answer can be determined after verification [Detailed explanation] by the meaning A  B   of the title, and A  C   , Available 4, 2  A,3  A , Therefore 32  3a  a 2  19  0 , the solution a  5 is or a  2 , 2 When a  5 , A   x x  5 x  6  0  2, 3 not satisfied A  C   ; 2 When a  2 , A   x x  2 x  15  0  5, 3 ,In line with the topic, Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 9 / 44 Therefore, the value of the a real number 2 is , So the answer is: 2   2 4. set A   x∣ 3  x  3 , B  y∣y   x  t if A  B   , the t range of values of real numbers 【Answer】  , 3  【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】The set is determined by the quadratic function properties B , and the conclusion of the intersection is concluded to obtain the parameter range   2 【Explanation】∵ A   x∣ 3  x  3 , B  y∣y   x  t , ∴ B   y∣y  t , and A  B   , ∴ t  3 , The value range of t  , 3 ∴ real numbers is 5. (24-25 High School Upper Shanghai Stage Exercise) Known 2 set A  2, 1, x  x  1, B  y , 2, x  1 , A B  1,3  , find x, y the value of real numbers 【Answer】 x  2, y  1 【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】First obtained x 2  x  1  3 , the value can be solved x and then classified and discussed 【Detailed explanation】From the meaning of the question x 2  x  1  3 , the solution or x  1 2, 当 x  1 时， B   y, 2, 0 ，不满足 A  B  1,3 ，故舍去； x  2 When, A  2, 1, 3 , B   y , 2, 3 , A  B  1,3 because,then y  1 , In summary, x  2, y  1 . 6. 5m   2  6  0    , set B  { x | 1  x  4 and x  Z} , known set A   x nx  mx  2   C  {x | ax  3  0} if A  B  A , let the set of values of m be D , if C  D   , find the value of m and the value range of corresponding a 【Knowledge Points】Find the set or parameters according to the intersection results, and find the parameters according to the inclusion relationship of the set [Analysis] points n  0 , n  0 and   0 , n  0 and   0 three cases, respectively, to obtain the value of m, and then according to the C  D   solution of the range corresponding a 【Detailed explanation】Because B  { x | 1  x  4 of and x  Z} , therefore, B  2, 3 When a  0 , C  R ,At this time C  D  D   ,,It doesn't fit the topic,So a  0 , Due to A  B  A the has, A  B , (1) If n  0 , if m  0 , then A   , does not fit the topic, so m  0 , then x  Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 10 / 44 5m  12 , 2m 5m  12  2 When, the solution m  12 , at this time D  {12} , 2m And because C  D   , therefore a 12  3  0 , it can be solved a≤  1 ; 4 5m  12  3 When, the solution m  12 , at this time D  {12} , 2m And because C  D   , therefore a  (12)  3  0 , it can be solved a  2 (2) if n  0 ,   0 when, i.e m  4n( At that 2  A time 4n  2m  1 ; 4 5m  6)  0 2 n  1 5m  6  0 , the joint solution was obtained  , at this time D  {4} , 2 m  4 3 And because C  D   , then, the 4a  3  0 solution a   ; 4 At that 3  A time 9n  3m  n  1 5m  6  0 , the joint solution was obtained  , at this time D  {6} , 2 m  6 1 And because C  D   , therefore 6a  3  0 , it can be solved a   ; 2 (3) if n  0 ,   0 when, i.e m 2  4n( 5m  6)  0 2 From A  B  A , obtain, A  B from the relationship between the root and the m  12 n 5  n  13  coefficient,  5 , and the solution is obtained   m  60 2 m6 6   13  n 60 At this time D  { } ,   ( 60 ) 2  4  12  ( 13 13 13 And because C  D   , therefore 5 2 60 13  6)  0 in line with the topic, 13 60 a  3  0 , the solution a   , 20 13 1 4 To sum up, when n  0 , m  12 then a  (,  ] ; 1 When n  0 , m  12 ,then a  [ ,  ) ; 4 3 4 When n  1 , m  4 ,then a  (,  ] ; 1 When n  1 , m  6 ,then a  ( ,  ] ; 2 When n  Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 12 60 13 ,m  , then a  (,  ] . 20 13 13 11 / 44 Peer Training 3: The Concept and Operation of Union Typical example questions Example 2 set up a set A  1, 2 , B  1, 2,3 then A  B  【Answer】 1,2,3 【Knowledge Points】The concept and operation of union 【Analysis】You can solve it directly according to the definition of union 【Explanation】Because A  1, 2 , B  1, 2,3 , So A  B  1,2,3 ... So the answer is: 1,2,3 Example 3. Known Set A   x 2  x  3 , B   x x  0 then A  B  【Answer】  x x  2 【Knowledge Points】The concept and operation of union 【Analysis】It can be solved by union operation. 【Explanation】Because A   x 2  x  3 , B   x x  0 , So A  B   x x  2 So the answer is:  x x  2 2 2 Example 4. Known set A  x∣x  px  6  0 , set B   x∣x  5 x  q  0 , and set A  B  {3} , find the value of the real number p , q and A  B 【Answer】 p  1, q  6, A  B  {2, 2,3} 【Knowledge Points】The concept and operation of intersection, the concept and operation of finding sets or parameters according to the results of intersection, and the concept and operation of union 【Analysis】According to the definition of intersection and the root of the unary quadratic equation, it is solved. 【Detailed explanation】Substituting both equations into the equation x  3 yields: 9  3 p  6  0,9  15  q  0 , Solution: p  1, q  6, x 2  x  6  0  ( x  2)( x  3)  0  x  2 x 2  5 x  6  0  ( x  2)( x  3)  0  x  2or 3, So A  2,3 , B  2,3 , Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 12 / 44 or 3,  A  B  {2, 2,3} . concise  1 1. A  x | 0  x  1 B   x∣  x   2  2 then A  B   【Answer】  x | 0  x  2 【Knowledge Points】The concept and operation of union 【Analysis】According to the problem, it can be solved by combining union operations.  1  【Explanation】Because A   x | 0  x  1 , B   x∣  x  2 ,  2  So A  B   x | 0  x  2 ... So the answer is:  x | 0  x  2 . 2 2. Known sets A  1, 0,1 , B   x x  2 x  0 , then A  B  【Answer】 1, 0,1, 2 【Knowledge Points】The concept and operation of union 【Analysis】Find the set B first, and then apply the union definition calculation. [Detailed explanation]. A  1, 0,1 , B  0, 2 , A  B  1, 0,1, 2 So the answer is: 1, 0,1, 2 . 3. known set A  {1, 2}, B  {1, a,3} , and A  B  B , then a  【Answer】2 【Knowledge Points】The concept and operation of union 【Analysis】From A  B the meaning of the topic, it can be obtained a 【Explanation】 A  B  B Because,So A  B , And because A  {1, 2}, B  {1, a,3} , so a  2 So the answer is: 2 . 4. 2 2 set the set A  3, a  1, a  , B  2a  1, a  1 if A I B  3 , try to find the real number a and the set A  B 【Answer】 a  1 , A  B  3, 0,1, 2 【Knowledge Points】Finding parameters according to the relationship between elements and sets, finding sets or parameters according to the intersection results, and the concept and operation of union 【Analysis】From the meaning of 3  B the question, it is discussed 2a  1  3 and solved in two a 2  1  3 and two cases, and then solved according to the definition of union A  B Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 13 / 44 【Explanation】 A I B  3 Because,So 3  B , When 2a  1  3 , i.e., a  1 , A  3, 0,1 , B  3, 2 , satisfies the topic; At that time a 2  1  3 , the equation has no solution, and it is rounded In summary, a  1 , A  B  3, 0,1, 2 Point-to-point training 4: Find the set or parameters according to the union results Typical example questions Example 1. If the set A    ,1 , B  a ,   and A  B  R , then the a value range is 【Answer】  ,1 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】The answer is obtained directly from the concept of union. 【Detailed Explanation】Collection A    ,1 , B  a ,   , A  B  R And,Then a  1 . So the answer is:  ,1 . set A  1  1, 2, 3 , then the set A that satisfies the condition has a total of Example 2 【Answer】2 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】Find the set of all the conditions that satisfies A it If A  1  1, 2, 3 so, then A  2,3 or A  1, 2,3 , there are 2 of them So the answer is:2. Example 3. set A   x, y  2 y  4 x  1 , B   x, y  x  2ay  b if A  B  A  B , then ab the value is [Answer]/ 1 0.0625 16 【Knowledge Points】Find the set or parameter according to the union result, find the set or parameter according to the intersection result, and find the parameter according to the equality of two sets 【Analysis】It is obtained by using A  B  A  B the available and A  B then combining the constraints of the two sets a, b 【Detailed explanation】From A  B  A  B , obtained A  B , so the equation x  2ay  b is the 4 x  2 y  1 same as the equation, Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 14 / 44 Then 1 1 2a b   , the solution a  b  , 4 4 2 1 So ab  1 ... 16 So the answer is: Example 4 1 16 Let the set A  {x | ( x  3)( x  a )  0} , B  {x | x 2  5x  4  0} , set A  B  {1,3, 4} , then a the value of the real number is 【Answer】1 or 3 or 4. 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】Solve the quadratic equation to find the set A, B , and find the value of the a real number according to the union result [Detailed explanation] by x 2  5 x  4  0 the solution x  1 or 4 , so B  {1, 4} , By ( x  3)( x  a )  0 the solution x  3 or a , (i) if a  3 , then A  {3} , satisfied A  B  {1, 3, 4} ; (ii) if a  3 , then A  {3, a} , because A  B  {1,3, 4} , So a  1 or 4 , The a value of the real number is 1 or 3 or 4 So the answer is: 1 or 3 or 4. Example 5 is known A  {x |1  x  7}, B  {x | m  1  x  2m  1} , and B   , if A  B  A , find m the range of values for real numbers 【Answer】  2, 3 【Knowledge Points】Find the set or parameters according to the union result, and find the parameters according to the relationship between the elements and the set m  1  1 【Analysis】From B   the get m  1  2m  1 , by A  B  A the get  , the solution can be  2m  1  7 m  1  1 [Detailed explanation] by B   the de m  1  2m  1 , that is. m  2 by A  B  A the de  ,  2m  1  7 by the solution 2  m  3 Therefore, the value range of m real numbers is  2, 3 concise 1. known set A   x x  1 , B   x x  a and A  B  R , then the value range of the real number a is 【Answer】 a  1 【Knowledge Points】Find the set or parameters according to the union result Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 15 / 44 【Analysis】Using the number axis, draw a set on the number axis, and the combination of numbers and shapes can obtain the solution. [Detailed explanation] A   x x  1 , B  {x | x  a} , And A  B  R , as shown in the figure, so at that a  1 time, the proposition is true So the answer is: a  1 . 2. set M  {0,1} , then M  N  {0,1, 2} the number of N sets satisfied is 【Answer】4 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】According to the question, the set N contains at least element 2, and then all the possible results can be enumerated 【Explanation】Because M  0,1 , M  N  0,1, 2 , So N at least element 2 is included So the set N may be {2},{0, 2},{1, 2},{0,1, 2} , a total of 4 So the answer is:4. 3. 2 2 known set A   x x  8 x  15  0 , B   x x  ax  b  0 if A  B  3 , A  B  2,3,5 then ab  【Answer】 30 【Knowledge Points】Find the set or parameters according to the union result, and find the set or parameter according to the intersection result 【Analysis】First find the set A , and then according to the results of the intersection and union to obtain the root of the equation, you can find the parameters, so as to obtain the solution, pay attention to verify whether the set obtained by the parameters meets the problem 2 [Detailed explanation], A   x x  8 x  15  0  3, 5 because A  B  3 , A  B  2,3,5 , So 2  B,3  B , that is, 2,3 the solution of the x 2  ax  b  0 equation So 2  3   a, 2  3  b , the solution a  5, b  6 , When a  5, b  6 , x 2  5 x  6  0 the root of the equation is 2,3 , at this time B  2, 3 , Satisfied A  B  3 , A  B  2,3,5 ,In line with the topic, So ab   5  6  30 ... So the answer is: 30 Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 16 / 44 4. 2 If the known set A  {1, 2,3, m}, B  m ,3 satisfies A  B  {1, 2,3, m} the value of the m real number, it is Answer: 1 or  2 or 0 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】According to the union result, the equation is obtained, and the solution is solved in turn to determine whether it meets the requirements. 【Explanation】Because A  B  {1, 2,3, m} , so m2  1 or m 2  2 or m 2  m , If m2  1, the solution is m  1 or m  1 , when there m  1 A are two 1s, contradictions; Comply m  1 with the requirements at the time; If m 2  2 the solution m  2 is or m   2 , it is verified that the requirements are met; If m 2  m the solution m  1 is or m  0 , it is not conformed by the knowledge m  1 above, and m  0 it is verified that it is compliant So or m  1 m  2 or or m   2 m  0 So the answer is: 1 or  2 or 0 2 2 5. Known A  x | x  mx  3  0, x  R, B  x | x  x  n  0, x  R  , if A  B  {3,0,1} , find m, n the value of the real number 【Answer】 m  2, n  0 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】The product of the two roots x 2  mx  3  0 that can be known from Veda's theorem 3 is, so that the A  3,1 sum of the two roots can be found by using the sum of the two roots m m , and A  B  {3,0,1} therefore, the 0  B meaning can be obtained by solving the equation n [Detailed explanation] because x 2  mx  3  0 of the middle   m2  12  0 , and the product of the two roots is 3 , and A  B  {3,0,1} , 2 Therefore A   x | x  mx  3  0, x  R  3,1 , therefore  m   3  1   2 , then m  2 , From the above: 0  B , therefore n  0 , substitution B  {0,1} , obviously satisfied A  B  {3,0,1} So m  2, n  0 ... Point-to-point training 5: The concept and operation of complement set Typical example questions Example 1. Known Complete Works U  1, 2,3, 4,5, 6 , Collection A  1,3, 4, 6 , then A = 【Answer】 2,5 【Knowledge Points】The concept and operation of supplements 【Analysis】The set is obtained from the concept of complement. Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 17 / 44 [Detailed explanation]. A  2,5 So the answer is: 2,5 Example 2 is known to a be a real number, the whole set U  1,2,4, B  a ,2  if, B  4 then a  【Answer】 1 【Knowledge Points】The concept and operation of supplements 【Analysis】It can be found according to the properties of the complement operation. Because U  1,2,4, B  a ,2  , B  4 , So a  1 ... So the answer is: 1 Example 3: If A  {x∣x  2} the collection, when the complete set U takes the following sets respectively, find A (1) U  R ； (2) U  x∣x  0 ； (3) U  x∣x  2 ． 【Answer】(1) A   x x  2 (2) A   x 0  x  2 (3) A   x∣x  2 【Knowledge Points】The concept and operation of supplements 【Analysis】According to the definition of the complement, combined with different complete sets, the solution can be solved. 【Explanation】(1) Because U  R , A   x x  2 ,So A   x x  2 . (2) Because U   x x  0 , A   x x  2 so A   x 0  x  2 (3) Because U   x x  2 , A   x x  2 ,so A   x x  2 . Example 4. Known Complete z 2  z  5} ,Seeking A  B , B  A , A  B Works U   x∣x  2 , A  { y∣3  y  4} , B  {∣ 【Answer】 A  B   2, 3   4, 5  , B  A   , A  B   2,   【Knowledge Points】The concept and operation of intersection, the concept and operation of union, the concept and operation of complementation 【Analysis】Find it first A, B , and then combine the intersection and complement operation to solve 【Explanation】Because A   2,3   4,   , B  5,   , Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 18 / 44 So A  B   2,3   4,5 , B  A   , A  B   2,   . concise 1. Known Complete Works U  1, 0,1 , A   x | 1  x  1, x  N , then A  【Answer】 1 ; 【Knowledge Points】The concept and operation of supplements 【Analysis】Calculate according to the complement definition of the set. Because of the complete works U  1, 0,1 , A   x | 1  x  1, x  N  0,1 , So A   1 ... So the answer is: 1 2. Known Complete Works U   x x  9, x  N , A  1, 2, 4,5, 6,8 , B  2, 4,5, 6, 7,9 , A  B  【Answer】 {1,8} 【Knowledge Points】The concept and operation of intersection, the concept and operation of supplement 【Analysis】The whole set is represented by the enumeration method U , and then solved according to the inter-set operation 【Detailed explanation】From the meaning of the title, U  0,1, 2,3, 4,5, 6, 7,8,9  , ∵ B  2, 4, 5, 6, 7, 9 ∴ B = {0,1,3,8}， ∴ A  B = {1,8} . So the answer is: {1,8} . 3. known set A  (1, ) , set B  ( , a) , and A  B  B , then the value range of the real number a is 【Answer】 (1,  ) 【Knowledge Points】Finding parameters according to the inclusion relationship of sets, finding sets or parameters according to union results, and the concept and operation of complements 【Analysis】According to the conclusion of union, the inclusion relationship of the set is obtained, and then the conclusion is obtained from the inclusion relation. 【Explanation】 A  B  B Because,So A  B , Again A  ( ,1] , so a  1 , So the answer is: (1, ) . 4. (22-23 High School One Upper Shanghai Pudong New Stage Practice) Known Complete Works U  {1, 2, 3, 4, 5} , Collection A  {1, 2,3} , Collection B  {3, 4} , Finding: A  B, A  B, A  B, A  B Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 19 / 44 【Answer】 {3} , {5} ,, {1, 2} . {3, 4,5} 【Knowledge Points】The concept and operation of intersection, the concept and operation of union, the concept and operation of complementation 【Analysis】It can be solved according to the intersection, union and complement of the set. 【Explanation】Because U  {1, 2, 3, 4, 5} , A  {1, 2,3} ,, B  {3, 4} So A  B  {3} , A  B  {1, 2,3, 4) ,, A  {4,5} B  {1, 2,5} Therefore A  B  5 , A  B  {1,2} , A  B  {3,4,5} Point-to-point training 6: Mixed operations of intersection and complementation Typical example questions Example 1.Known Complete Works U  1, 2,3, 4,5, 6, 7 , Collection M  1, 2,5 , N  3,5, 6 , Then M  N  【Answer】 4, 7 【Knowledge points】Mixed operation of intersection and complementation 【Analysis】It can be solved according to the union and complement set operations. 【Explanation】Because M  1, 2,5 , N  3,5, 6 , So M  N  1, 2,3,5, 6 , M  N  4, 7 , So the answer is: 4, 7 Example 2. set set A  1, 2, 6 , B  2, 4 , C  1, 2,3, 4 ,then  A  B   C  . 【Answer】 1, 2, 4 【Knowledge points】Mixed operation of intersection and complementation 【Analysis】It can be calculated directly by using the calculation law of the set. [Detailed explanation] can be known from the question, A  B  1, 2, 4, 6 so  A  B   C  1, 2, 4 ... So the answer is: 1, 2, 4 Example 3. If the whole set U  1,2,3,4,5,6, 7,8,9 , A, B is a U subset, and A  B  1,9 , A  B  2 , A  B  4,6,8 then A  【Answer】 2,3,5, 7 【Knowledge points】Mixed operation of intersection and complementation 【Analysis】Draw a Venn diagram according to the meaning of the title. [Detailed explanation], A  B  1,9 , A  B  2 A  B  4,6,8 make a Wayne diagram, as shown in the figure: Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 20 / 44 Then A  2,3,5, 7 So the answer is: 2,3,5, 7 Example 4. Knowing the complete works U  1, 2,3, 4 , and A  1, 2,3  2 ,, A  1, 2,3  U Find the set A 【Answer】 A  2, 4 【Knowledge points】Mixed operation of intersection and complementation 【Analysis】According to A  1, 2,3  2 the element 2 , there is none 1, 3 , and the A  1, 2,3  U element can 4 be solved [Detailed explanation] Solution: ∵ complete works U  1, 2,3, 4 , Satisfied A  1, 2,3  2 , A  1, 2,3  U ,  A There are elements 2 , there are none 1, 3 , there are 4 , ∴ A  2, 4 ． 2 2 Example 5, A   x x  ax  2  0 B   x x  3 x  b  0 , A  B  1 C  2, 4 , (1) Find the values of a and b and A and B; (2) Seeking  A  B   C 【Answer】(1) a  1 , b  2 ,, A  2,1 ; B  1, 2 (2)  A  B   C  2,1 . 【Knowledge Points】According to the intersection results, the concept and operation of the set or parameters, the concept and operation of union, and the mixed operation of intersection and complementation 【Analysis】(1) According to A  B  1 the obtained 1  A,1  B , the equation will x  1 be substituted and calculated a  1 , b  2 so as to find A and B; (2) Find A  B  2,1, 2 , so as to obtain  A  B   C  2,1 [Explanation] (1) because A  B  1 , therefore, 1  A,1  B So 1  a  2  0 , 1  3  b  0 , Solution: a  1 , b  2 , 2 2 Therefore A   x x  x  2  0  2,1 , B   x x  3 x  2  0  1, 2 ; Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 21 / 44 （2） A  B  2,1, 2 ，  A  B   C  2,1 . concise 1. Set the whole set U  1, 2,3, 4,5, 6, 7 , the set A  1, 4, 7, B  2, 4,5, 7  , then the set represented by the shaded part shown in the figure is ( ). A． 4, 7 B． 1, 2, 4,5, 7 C． 3, 6 D． 1, 2,3,5, 6 【Answer】C 【Knowledge Points】Intersection and complement mixed operation, using Venn graph to find the set 【Analysis】According to the intersection and complementation operation of the set, it can be solved in combination with the graph. The set represented by the shaded part is: ðU  A  B  By A  1, 4, 7, B  2, 4,5, 7  that A  B  1, 4, 7  2, 4,5, 7   1, 2, 4,5, 7  , So ðU  A U B   3, 6 , Therefore, C 2. known complete works U  1, 2,3, 4,5, 6, 7,8,9 , if A  B  1,3 , A  B  5, 7 ,, A  B  2, 4, 6,8 ,then A  B  【Answer】 {9} 【Knowledge Points】The concept and operation of intersection, the mixed operation of intersection and complementation, and the use of Venn graph to find the set 【Analysis】Make a Venn diagram from the question, and the answer can be obtained from the diagram. 【Detailed explanation】Make a Venn diagram from the title, as shown in the figure, Fill in the elements in the corresponding set, as can be seen from the diagram A  B  {9} So the answer is: {9} Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 22 / 44 3. Let the complete set U={1,2,3,4,5},A={1,3,5}, then the difference between the maximum and minimum values of the number of true subsets of the set represented by the shaded part in the figure is 【Answer】 28 【Knowledge points】Determine the number of subsets of the set (true subset), intersect and complement the mixed operation 【Analysis】Categorize and discuss according to the shaded part, and find the correct answer. 【Explanation】The shaded part is indicated 痧 U  A  B    U A   痧 U B   2, 4   U B  , If 2, 4   ðU B   2, 4 so, If 2, 4   ðU B   1, 2,3, 4,5 so, the true subset has 22  1  3 a the true subset has 25  1  31 a So the difference between the maximum and minimum values of the number of true subsets is 31  3  28 So the answer is: 28 4. known complete works U  R , set A  x∣x  a  1 , B  x∣x  a  2 ,C  {x∣x  0 or x  4} , and, then A U B  C the value range of a real numbers is ., 【Answer】  , 2  U 5,   【Knowledge Points】Parameters are calculated according to the inclusion relationship of the set, and mixed operations are intersected and complemented 【Analysis】Using the definition of union A  B , it is obtained that it is A  B=ðU  A  B  solved according to the set containment of the series inequality 【Detailed explanation】Complete U  R works,Collection A   x x  a  1 , B   x x  a  2 , So A  B   x x  a  2 or x  a  1 ,  So ðU  A  B   x a 1  x  a  2 ... Collection C   x x  0 or x  4 , and ðU  A  B   C , So a  2  0 or a  1  4 , Solution or a  2 a  5 , i.e a . the  , 2  U 5,   range is So the answer is:  , 2  U 5,   . 5. Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd if the whole collection U  1, 2,3, 4,5 , set A  1, 2,3, 4 , B  1, 2,5 then A  B = 23 / 44 【Answer】  【Knowledge points】Mixed operation of intersection and complementation 【Analysis】First, find the union set, and then find the complement set. [Detailed explanation], A U B  1, 2,3, 4,5 so A  B   So the answer is:  6. known complete works U  {x∣x  10, x  N} , B  A  1,9 , A  B  3 ,, A  B  4, 6, 7 ,Seek the set A, B 【Answer】 A  1,3,9 , B  0, 2,3,5,8 【Knowledge Points】Intersection and complement mixed operation, using Venn graph to find the set 【Analysis】Draw a graph according to the result of the set operation Venn , and determine the result according to the Venn graph. or use the definition method of set operation to solve the answer; Method 1 ( Venn Diagram Method): Make a diagram according to the meaning of the topic, Venn as shown in the figure As can be seen from the figure A  1,3,9 , B  0, 2,3,5,8 Method 2 (Definition): B  A  1,9 , A  B  4, 6, 7 , ∴ B  1, 4, 6, 7,9 Again U  0,1, 2,3, 4,5, 6, 7,8,9 , ∴ B  0, 2,3,5,8 ∵ B  A  1,9 ， A  B  3 ，∴ A  1,3,9 ． 7. 5 set the complete works as R , set A   x 4  x  2 , B   x 1  x  3 P  {x x  0 or x  } , 2 seek A  B , B  P ,  A  B   P ., .,   5 Answer, A  B   x 1  x  2 B P  {x x  0 or x  } ,  A  B   P  x 0  x  2 2 【Knowledge Points】The concept and operation of intersection, and the mixed operation of intersection and complementation 【Analysis】According to the definition of intersection, union, and complement, combined with the number axis, it can be solved. 【Explanation】The sets A, B, P are represented separately on the number line, as shown in the figure Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 24 / 44 ∵ A   x 4  x  2 ， B   x 1  x  3 ，  A  B   x 1  x  2 , B  {x x  1 or x  3} , 5 5 Or P  {x x  0 x  } ,  B  P  {x x  0 or x  } , 2 2  Again P   x 0  x   5  ,  A  B   P   x 0  x  2 . 2 Point-to-point training 7: Determine the set or parameters according to the mixed operation of intersection and complementation Typical example questions Example 1If A∩{-1,0,1}={0,1} and A∪{-2,0,2}={-2,0,1,2}, then there are a total of sets A that meet the above conditions. 【Answer】4 【Knowledge Points】Determine the set or parameters according to the mixed operation of intersection and complementation 【Analysis】Using the properties of intersection and intersection, the set A that satisfies the conditions is enumerated, and the result can be obtained. 【详解】解:∵A∩{-1,0,1}={0,1},且 A∪{-2,0,2}={-2,0,1,2} ∴ set A that satisfies the conditions are: A＝{0，1}，A＝{﹣2，0，1}，A＝{0，1，2}，A＝{﹣2，0，1，2} There are 4 sets A ∴ meet the above conditions. So the answer is:4. Example 2.Known Collection A  {x | 3  x ≤ 6} , B  x | b  3  x  b  7  , M   x | 4 ≤ x  5 ,Complete Works U  R . A  M  【Answer】  x | 3  x  5 ； If B  M  R so, then the value range of the real number b is . 2  b  1 【Knowledge Points】Find the set or parameter according to the intersection result, and determine the set or parameter according to the intersection and complement mixed operation 【Analysis】The definition of the intersection is used to solve it directly A  M , and then the M complement of the set is obtained, and then the B  M  R value range of the real number b can be obtained from the column inequality group Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 25 / 44 【Explanation】Because A   x | 3  x ≤ 6 , M   x | 4 ≤ x  5 ,so A  M   x | 3  x  5 ; Because M   x | 4 ≤ x  5 , so M    ,  4   5,   ; b  3   4 Because B  M  R , so  , b  7  5 So 2  b  1 ... So the answer is:  x | 3  x  5 ; 2  b  1 ． Example 3 set set A   x∣x  m  0 , B  {x∣ 2  x  4} complete set U  R , and A  B   , find m the value range of real numbers 【Answer】  2,   【Knowledge Points】Determine the set or parameters according to the mixed operation of intersection and complementation 【Analysis】According to the set A  x∣x  m , the complete set U  R , obtained A  {x∣x   m} , and then based on A  B   , the m scope of the solution 【Detailed explanation】Collection A  x∣x  m  0  x∣x  m , Complete Works U  R , ∴ A  {x∣x   m} Again B  {x∣ 2  x  4} , A  B   , then  m  2, m  2 , i.e m .,  2,   the scope is    2 2 2 Example 4 set A  x x  3x  2  0 , B  x x  2 a  1x  a  5  0  (1) If A  B  2 , find a the value of the real number; (2) If the complete set is R , A  B  A , find a the range of values of real numbers Answer: (1) a  1 or 3 (2) a a  3 and a    3  1 a  1 and a  3  1 【Knowledge Points】Calculate the parameters according to the inclusion relationship of the set, find the set or parameters according to the intersection result, and determine the set or parameters according to the intersect and complement mixed operation 2 2 【Analysis】(1) Solve the set A , according to A  B  2 x  2 x  2  a  1 x  a  5  0 the two elements of the equation, the a value is obtained, and then the A  B  2 solution can be obtained by combining the test; (2) the analysis is available A  B , which is discussed in two situations: B   according to the range   0 that can be obtained a ; B   , analysis shows that , 1 is 2 not Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 26 / 44 x 2  2  a  1 x  a 2  5  0 the root of the equation, can be derived about the group of inequalities about the real numbers a , and the synthesis can be derived from a the range of values of the real numbers     2 2 2 【Explanation】(1) Because A  x x  3 x  2  0  1, 2 , B  x x  2 a  1x  a  5  0 ,and A  B  2 , 2 2 is x  2 the root of the x  2  a  1 x  a  5  0 equation 2 2 So, 4  4  a  1  a  5  a  4a  3  0 ,Solve a  1 or 3 ,   2 When a  1 , B  x x  4  0   2, 2  at this time, at this time, A  B  2 is in line with the topic;   2 When a  3 , B  x x  4x  4  0  2  at this time, at this time, A  B  2 is in line with the topic In summary, a  1 or 3 .   2 2 2 (2) For the equation x  2  a  1 x  a  5  0 , Δ  4  a  1  4 a  5  8  a  3 , 2 Because the complete collection is R , A  B  A , then A  B , it is discussed in the following situations: At that B   time, then Δ  8  a  3  0 , can be obtained a  3 , at this time, A  B  R it is in line with the topic; When B   the time comes, then Δ  8  a  3  0 , available a  3 2 2 Because A  B , then 1 , 2 is not x  2  a  1 x  a  5  0 the root of the equation 2 2 1  2  a  1  a  5  a  2a  2  0 So,  , 2 2  4  4  a  1  a  5  a  4a  3  0 Solve it and a  1 a  3 a  1  3 and a  3  1 ,   At this point, a  a 3  a  1  3 or 1  3  a  1 or 1  a  3  1or or a  3  1  To sum up, a the range of values for a a  3 real numbers is and a   3  1 and a  1 a  3  1 and concise 1. As shown in the figure, the shaded part of the figure is ( ) Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 27 / 44 A．  A  C    B  C  B．  A  B    A  C  C．  A  B    B  C  D．  A B   C 【Answer】B 【Knowledge Points】Determine the set or parameters according to the intersecting and complementing mixed operation, and use the Venn graph to find the set 【Analysis】From the Venn diagram, it can be seen that the elements in the shaded part satisfy the "elements of B and the elements of C, or the elements of A", and the answer can be obtained from the relationship between the Venn diagram and the set. The shaded part in the diagram indicates that the element satisfies: it is the element in A, or the common element of B and C Therefore it can be expressed as A   B  C  or as:  A  B    A  C  Therefore, B. 2. known complete works U  1, 2,3, 4,5, 6, 7,8 , if A  B  2,3 , B  A  4,8 ,, A  B  1, 7 then set A  【Answer】 2, 3,5, 6 【Knowledge Points】Determine the set or parameters according to the mixed operation of intersection and complementation 【Analysis】Find it first A  B , then find it A  B , so that you can find A it 【Detailed explanation】 A  B  1, 7 Because,Therefore A  B  {2,3, 4,5, 6,8} , And A  B   A  B    B  A    A  B  A  B, B  A, A  B the intersection of two pairs is an empty set, Therefore A  B  5, 6 , therefore, A   A  B    A  B   2, 3, 5, 6 So the answer is: 2,3,5, 6 3. If the complete set U contains 10 elements, P､Q all U are subsets, P  Q containing 2 elements, P  Q containing 4 elements, and P  Q containing 3 elements, then P it contains elements 【Answer】3 【Knowledge Points】Determine the set or parameters according to the intersecting and complementing mixed operation, and use the Venn graph to find the set 【Analysis】A Venn diagram is made according to the meaning of the topic, and the result can be obtained from the Venn diagram. Because the complete set U contains 10 elements, P､Q they are all U subsets, P  Q containing 2 elements, P  Q containing 4 elements, and P  Q containing 3 elements Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 28 / 44 So make a Venn diagram as shown in the figure, then 4  x  2  3  10 , get x  1 , So P the number of elements in the set is 2  1  3 one So the answer is: 3 4. known complete works U  R , set A   1,4  , B   x | 2t  3  x  t  1 , if, find A  B   the value range of the real number t Answer: 1  t  3 or t  4 【Knowledge Points】Calculate the parameters according to the inclusion relationship of the set, and determine the set or parameters according to the mixed operation of intersection and complementation 【Analysis】By A  B   de, B  A then classify and discuss B   and B   discuss, so as to get the solution 【Explanation】 A  B   Because,So B  A , Because A   1,4  , B   x | 2t  3  x  t  1 , When B   , 2t  3  t  1 then t  4 , satisfies at this time B  A ; 2t  3  1 When B   , t  4 then  , solution 1  t  3 ; t  1  4 In summary, 1  t  3 or t  4 5. Known Complete 2 2 Works U  1, 2,3, 4,5 , A   x∣x  5 x  m  0} , B   x∣x  nx  12  0 A  B  1,3, 4,5 , m  n And,Value 【Answer】 m  n  1 【Knowledge Points】Intersect and complement mixed operation, determine the set or parameters according to the intersection and complement mixed operation 【Analysis】From the meaning 2  A of the question, 2 is x a root x 2  5x  m  0 of the equation m  6 about A  {2,3} , and so it A  1, 4,5 is obtained 3  B , and x 3 must be a root x 2  nx  12  0 of the equation n about, and m  n the value obtained can be obtained Solution: ∵ U  1, 2,3, 4,5 , A  B  1,3, 4,5 ,   2 ∴ 2  A , again A  x x  5x  m  0 , ∴2 is about x a root x 2  5x  m  0 of the equation Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 29 / 44 ∴ 4  10  m  0 ， ∴ m  6 and A  2,3 , ∴ A  1, 4,5 , while A  B  1,3, 4,5 ,   2 ∴ 3  B , again B  x x  nx  12  0 , ∴3 must be about x a root x 2  nx  12  0 of the equation ∴ 9  3n  12  0 ， ∴ n  7 and B  3, 4 , ∴ m  n  1 ．   2 known set A  x x  4x  3  0 , set B   x 2m  x  1  m 6. (1) If m  1 , ask A  B (2) If A  B   so, find m the range of values of real numbers 【Answer】(1)  x 1  x  2 (2) m  0 【Knowledge Points】The concept and operation of intersection, and the determination of sets or parameters according to the mixed operation of intersection and complementation 【Analysis】(1) List the inequality groups according to the set inclusion relationship, and find the value range of the real number m; (2) Divide B   and B   discuss, list the unequal relations, and find the range of values of the real number m 【Explanation】(1) Because m  1 , so B   x 2  x  2 , Again A   x 1  x  3 , so A  B  x 1  x  2 .... (2), A   x 1  x  3 because A  B   , Therefore B   , when the time comes 2m ³ 1- m , the solution is in m  1 line with the topic; 3  2m  1  m  2m  1  m 1 When B   the time comes, then  or  , solution 0  m  3 1  m  1  2m  3 To sum up, the range of values for the real number m is m  0 Let A, B be two subsets of the U whole work, A  B then the following formula is true ( ). 1. A． A  B B． A  B  U C． A  B   D． A  B   【Answer】C Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 30 / 44 【Knowledge Points】Intersect and complement mixed operations, and judge the inclusion relationship between two sets 【Analysis】Obtain the correct answer according to the knowledge of subset, complement, union and intersection. According to the title, A, B it is U two subsets of the A  B whole work,, Option A, A  B so option A is wrong Option B, A  B  A  U so option B is wrong C option, A  B   so C option is correct D option, A  B   so D option is wrong Therefore, C set the set P  2.  x , y  y  x 2   1, x  R , Q   x , y  y  2x A． (1, 0)  ( 1, 0) B． {(1,0),( 1,0)} C． { x  1, y  0 or x  1, y  0} D．  1, 2  2   2, x  R then P  Q  ( ). 【Answer】B 【Knowledge Points】The concept and operation of intersection 【Analysis】According to the given conditions, the solution is obtained by using the definition of the intersection.  y  x2 1 x  1 【Detailed explanation】According to the meaning of the topic,  ,Solve or  2 y  0  y  2 x  2  x  1 ,  y  0 So P  Q  {(1, 0), ( 1, 0)} ... Therefore, B The sum of the elements of the known set A and set B is m, and A  B there are n elements, 3. and if A  B   , then A  B the number of elements is ( ). A．mn B． n  m C． m  n D． m  n 【Answer】D 【Knowledge points】Application of the principle of repulsion [Analysis] can card  A  B   card  A  card  B   card  A  B  be obtained from the formula 【Detailed explanation】From the question, card  A  B   n, card  A  card  B   m So card  A  B   card  A  card  B   card  A  B   m  n ... Therefore, D Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 31 / 44 If the complete set U , the non-empty sets M and N are satisfied M  N , then the following 4. sets represent the empty set is ( ). A． M I N B． M  N D． M  N C． M U N 【Answer】A 【Knowledge Points】Intersection and complement mixed operation, using Venn graph to find the set 【Analysis】You can use Venn a graph to represent a set M , N and you U can find out the option to represent an empty set by combining it with a graph 【Explanation】You can use Venn a graph to represent the set M , N as U follows: Observing the graph, yes M  N   , M  N   , M  N  M   M I N  M   ,,A is, BCD is not Therefore, A As shown in the figure, it is known that U it is a complete set, A and , B is a C subset of 5. it, then the set represented by the shaded part is ( ). A． A  B  C B．  B  A   C C．  A  B   C D． A  B  C 【Answer】D 【Knowledge Points】Intersection and complement mixed operation, using Venn graph to find the set 【Analysis】It can be solved by combining the known with the Venn diagram. Solution: Combined with the Venn diagram, it can be seen that it D  A  B is the set represented by the following shaded parts then the shaded part of the stem represents the set D  C Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 32 / 44 That is, the set is A  B  C Selection: D. 6. known complete works U  1, 2,3, 4,5 , collections A  1, 2 , collections B  2, 4, 5 , then A  B  【Answer】 1, 2, 4,5 【Knowledge Points】The concept and operation of union 【Analysis】Apply the union operation of the set to find the set. [Detailed explanation] is set by the question A  B  1, 2  2, 4, 5  {1, 2, 4, 5} So the answer is: 1, 2, 4,5 7. known set A  1, 0,1, 2, B  x∣x  1 x  2  0  , then A  B  【Answer】 1 【Knowledge Points】The concept and operation of intersection 【Analysis】According to the simplification of the set of one-dimensional quadratic equations B , it can be solved by intersecting operations [Detailed explanation] by B   x∣ x  1 x  2   0 the de B  1, 2 , So A  B   1 , So the answer is: 1 8. known complete works U  1, 2,3, 4,5 , collection A  1,3,5 , then A  . 【Answer】 2, 4 【Knowledge Points】The concept and operation of supplements 【Analysis】The solution can be obtained by using the complement operation of the set. 【Explanation】Because U  1, 2,3, 4,5 , A  1,3,5 , So A  2, 4 ... So the answer is: 2, 4 . 2 9. Let the complete set be U  R , if U the subset of the set A   x | x  5 x  14  0 , the subset B   x | 3  2 x  5  5 , then A  B = 【Answer】 7 【Knowledge points】Mixed operation of intersection and complementation Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 33 / 44 【Analysis】According to the unary quadratic equation and the group of unary inequalities, the set can be obtained, and the answer can be obtained according to the operation of the complement and the intersection. [Detailed explanation] by x 2  5 x  14  0 ,  x  7  x  2   0 , solution x  7 or 2 , then A  7, 2 ; By 3  2 x  5  5 , solved 1  x  5 , then B   x 1  x  5 , can be obtained B   x x  1 or x  5 ; So A  B  7 ... So the answer is: 7 . 10. Known Collection A   x | a  x  a  2 , Collection B  {x | x  1 or x  5} , Complete Works U  R (1) If a  1 , ask A  B ; (2) If A  B so, find a the range of values of real numbers 【Answer】(1) {x | x  1 or x  3} ; (2) a  3 or a  5 【Knowledge Points】The concept and operation of complement, the concept and operation of union, and the calculation of parameters according to the inclusion relationship of the set 【Analysis】(1) Apply the parallel operation of the set to find the set; (2) The range of parameters can be found according to the inequality of the containment series. 【Explanation】(1) By the question A   x |1  x  3 , then A  { x | x  1 or x  3} , So A  B  {x | x  1 or x  3} (2) If it A  B is established by and a  2  a constantly, it A is a non-empty set So a  2  1 or a  5 , i.e a  3 . or a  5 . 1.If the non-empty set A, B satisfies A  B , and U is the complete set, then the empty set in the following set is ( ). A． A  B ； B． A  B ； C． A  B ； D． A  B ． 【Answer】D 【Knowledge Points】Use Venn graphs to find sets 【Analysis】According to venn the diagram, analyze each option one by one to find the correct answer [Detailed explanation] can be seen according to venn the following Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 34 / 44 Right A, A  B  A   ; To B, A  B  B   ; To C, A  B   ; To D, A  B   . . Therefore, D 2. known complete works U  R , sets A   x 2  x  5 , B   x m  1  x  2m  1 , if A  B   , then m the value range of real numbers is ( ). A．  3, 3 B．  2, 3 C．  , 2  D．  ,3 【Answer】D 【Knowledge Points】The concept and operation of complement set, and the set or parameter is calculated according to the intersection result 【Analysis】First A , find the complement of the set, and then discuss the empty B and non-empty sets of the set, and establish the inequality relationship, and then solve it [Explanation] from the known available A  {x x  5 or x  2} , and A  B   , At B   that time, m  1  2m  1 the solution is satisfied m  2 , and the topic is satisfied at this time;  m  1  2m  1  At that B   time, to meet the topic, you only need to  m  1  2 solve it 2  m  3  2m  1  5  To sum up, the range of real numbers m is  , 3 Therefore, D 3. non-empty number set A  R , which satisfies the following two properties at the same time: 1 (1) if a, b  A , then ab  A ; (2) If a  A then, then  A is called A a "closed set", as described a below: (1) If A it is a "closed set", then 1 A ; a (2) if A it is a "closed set" and a, b  A ;  A b (3) if A, B they are all "closed sets", then the A  B sufficient and necessary conditions for "closed sets" are A  B or B  A ; (4) If A, B they are all "closed sets", then A  B the sufficient and necessary conditions for "closed sets" are A  B or B  A Correct is ( ) A．①③④ Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd B．①②③④ C．①②③ 35 / 44 D．①②④ 【Answer】D 【Knowledge Points】Set new definitions 【Analysis】From the definition of the closed set, it is enough to judge item by item, and at the same time, (3) use examples, and (4) use the counter-proof method. For (1), because it A is a "closed set", it can be known by definition a  A 1 1  A , then a   1  A , a a correct; 1 a For (2), because A for a "closed set, a, b  A so  A , so  A , correct; b b 1 1 For (3), A  { ,1, 2} ,, B  { ,1,3} A  B  {1} are both closed sets, apparently A  B or B  A not, 2 3 false For (4), sufficiency: A, B are all "closed sets", if A  B or B  A , it is easy to know A  B that they are "closed sets". Necessity: If A  B it is a "closed set", order A  B  C , Assume A  B and B  A ... then there is a  A, a  B , b  B, b  A ,, a  C , b  C , Because A  B  C it is a "closed set", So ab  C , 1  C it is discussed in two categories ab If ab  A , then a  A, 1 1  A, so ab   b  A , this contradicts the hypothesis; a a If ab  A , then b  B, 1 1  B, so ab   a  B , this contradicts the hypothesis; b b Therefore, the assumption is not valid, and the original conclusion A  B is that the necessity of the "closed set" is A  B B  A valid, so it is correct; Therefore, D 4. known complete 2 2 works U  4, 3, 1, 2 , A  a , a  1, 3 , B  a  3, 2a  1, a  1 , A I B  3 if,then A  B  . 【Answer】 3, 1, 0,1 【Knowledge points】According to the intersection results, find the set or parameters, and the intersection and complement the mixed operation 【Analysis】Divide a  3  3 and 2a  1  3 sum a 2  1  3 three cases, and then you can get the set A, B , and a the value you find can get the result 2 2 【Explanation】Because A  a , a  1, 3 , B  a  3, 2a  1, a  1 ,, A  B   3 Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 36 / 44 When a  3  3 the solution is obtained a  0 , the set A  0,1, 3 is B  3, 1,1 not satisfied A  B   3 ; When 2a  1  3 the time comes, the solution is obtained a  1 , the collection A  1, 0, 3 is B  4, 3, 2 satisfied A  B   3 , and the satisfaction is satisfied; Apparently a 2  1  3 not; In summary: a  1 , A  1, 0, 3 ,, B  4, 3, 2  B  1 ， A  B  3, 1, 0,1 ， So the answer is: 3, 1, 0,1.     2 2 6. A  x x  px  q  0 , B  x qx  px  1 0 A  B   ,,, A  B  2 ,then p  q  Answer: 1 or 5 【Knowledge Points】Find the set or parameter according to the intersection result 【Analysis】The rules are determined by m  A 1  B the meaning of the question, and then m passed A  B   , obtained, m and solved [Details] design m  A , if m  0 , then q  0 , again A  B  2 , so 2  A ,  1 So p  2 , at this time, you can get B     it doesn't match A  B   , so m  0  2 2 1 1 Then m2  pm  q  0 , both sides are divided together m2 , and you can get q    p    1  0 , m m so, 1 B m Because A  B   , so there must be m  1 , so m2  1 , ie m  1 , m m  1 When, again A  B  2 , so 2  A , so A  2,1 ,  1 From Veda's theorem, it follows: p  1, q  2 , which B  1,   is in line with this time; 2  m  1 When, again A  B  2 , so 2  A , so A  2, 1 ,  1 From Veda's theorem, it follows: p = 3, q = 2 , which B  1,   is in line with this time; 2  So p  q  1 or p  q  5 So the answer is: 1 or 5 7. Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 2 2 known set A  x | x  3x  2  0 , B  x | x  (m  1)x  m  0  , if A  B  A , then m  37 / 44 Answer: 1 or 2 【Knowledge Points】Find the set or parameters according to the union result 【Analysis】Discuss the elements in set B, according to the A  B  A available solution 2 【Explanation】Because A   x | x  3 x  2  0  1, 2 ,   B  x | x 2  ( m  1) x  m  0   x | ( x  m)( x  1)  0 ， When m  1 , B  1 ,At this time, A  B  A ,,Meet the topic, When m  1 , B  1, m by A  B  A the available m  2 , i.e m  2 . . . . In summary, m  1 or 2 So the answer is: 1 or 2. 8. Known Sets A   t ,t  1  t  2,t  4 ,0  A . If there is a positive number such  that it a  A is true  a  A for any t of the Answer: 2 or  8 5 【Knowledge Points】The concept and operation of union, and the parameters are calculated according to the relationship between elements and sets 【Analysis】According to t the different ranges in which the a  t , t  1 sum is a  t  2, t  4  obtained, the  range in which it is located; Then use the upper and lower limits of the set A to obtain  a t the equiquantitative relationship with , so as to construct the equation and t obtain the value Because 0  A , you only need to consider the following three situations: 1  1 1   1 1 , U , , (1) When t  0 , because a  t , t  1  t  2, t  4  , then   a  t  4 t  2   t  1 t  And   0 , available          ,  , , a  t  4 t  2   t  1 t      t t 2    t  4 t 1  A And because , then  and  , a    t 1   t  4 t  2   t t  t  4     t  t  4  Available at:  ,  t  1 t  2      t  1 t  2  Then   t  t  4   t  1 t  2  , the solution is obtained t  2 ; (2) When it t  4  0 is imminent t  4 , it is the same as (1) the structural equation, that is, t  2 it does not fit the topic and is discarded; Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 38 / 44     t t 2   t  1  0 t 1 t  4 (3) When  , i.e., 2  t  1 obtain:  and  , t  2  0   t 4   t 1   t t  2 8 Available   t  t  1  t  2 t  4  , solved t   ; 5 To sum up: t  2 or  8 5 So the answer is: 2 or  8 5 Key points: This question examines the use of the relationship between the set and the elements to solve the value of the parameters, the key is to t be able to obtain the range of the relevant a through the different value ranges  , so as to use the upper and lower limits of the set to obtain a  the equivalent relationship about, so as to construct t the equation about; The challenge is to be able to t classify the ranges accurately 2 2 2 2 9. set A   x | x  x  6  0 , B   x | x  2 x  8  0 , C   x | x  ax  2a  0 , (1) Try to find the range of the value of the real number a, so that  A  B   C ; (2) If Z it is an integer set, is there a positive number a , satisfied B  C  Z  3, 4, 5, 6, 7,8, 9,10,11 ? If it exists, the a value range is obtained, and if it does not exist, explain the reason 【Answer】(1) a  3 or a  3 ; 2 (2) does not exist a , the reason is explained in the analysis; 【Knowledge Points】Find the set or parameters according to the intersection results, and find the parameters according to the inclusion relationship of the set 【Analysis】(1) The set can be obtained by solving the inequality A, B , and then the value range of the real number a can be classified and discussed. (2) The results can be obtained from the conclusions in (1) and classified and discussed according to the intersection results. [Detailed explanation] (1) The solution of inequality x 2  x  6  0 can be obtained A   x | 2  x  3 The solution to the inequality x 2  2 x  8  0 can be obtained B   x x 2 or x  4 , Therefore available A  B   x | 2  x  3 ; At a  0 the time, C   it is not on topic; When a  0 the time comes, the solution is obtained C   x |  a  x  2a Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 39 / 44 a  2 3 if  A  B   C ，then  ，then a  ； 2 a  3 2  When a  0 the time comes, the solution is obtained C   x | 2a  x   a  2a  2 if  A  B   C ，then  ，then a  3 ； a  3 To sum up, the value range of the real number a is a  3 or a  3 ; 2 (2) From (1) B   x x 2 can be known or x  4 , Obviously 3, 4,5, 6, 7,8,9,10,11  B , and 3, 4, 5, 6, 7,8, 9,10,11  Z ; So just be satisfied 3, 4,5, 6, 7,8,9,10,11  C , and because a is a positive number, Knowable a  0 time, C   x |  a  x  2a because B  C  Z  3, 4, 5, 6, 7,8, 9,10,11 11 11  2 a  12  a6 It can be obtained  , it can be solved  2 , and there is no solution at this time;  a   5   a  5 Therefore, there is no such thing as satisfying the topic a 10. known y  x2  2ax  a (1) If the equation y  0 has two real roots x1 , x2 , , and x1 x2 3   6 find a the value of the x2 x1 x1 x2 real number; 2 2 2 (2) If A   x x  4 x  0 the set, B  {x | x  2ax  a    4a  2  x  a  a  1} if, the A  B  A range of values a to be found 【Answer】(1) 3 2 (2) ( , 1]  1 【Knowledge Points】The solution set of a quadratic equation and the relationship between its roots and coefficients, the set or parameters according to the union result, and the parameters according to the inclusion relationship of the set 【Analysis】(1) Obtained by Vedic theorem x1  x2  2a , x1 x2  a , according to the meaning of the 2 question, transformed ( x1  x2 )  8x1x2  3  0 , substituted to obtain 4a 2  8a  3  0 , the a obtained value, combined,   0 you can get the answer (2) Obtain A  {4, 0} , B  {x | x 2  2(a  1) x  a 2  1  0} by A  B  A , obtain B  A , divide B   , B  0 B  4 and, discuss the B  4,0 four cases, and solve it Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 40 / 44 [Detailed explanation] (1) Solution: by the function y  x2  2ax  a , order y  0 , i.e x 2  2ax  a  0 ., Because the equation y  0 has two real roots x1 , x2 , it can be obtained x1  x2  2a , x1 x2  a And because x1 x2 3 ( x  x ) 2  2 x1 x2 6 x1 x2 3   6  , it is available 1 2 x2 x1 x1 x2 x1 x2 x1 x2 2 2 Namely ( x1  x2 )  2 x1 x2  6 x1 x2  3, namely, ( x1  x2 )  8x1x2  3  0 namely Will x1  x2  2a , x1 x2  a be substituted, obtainable 4a 2  8a  3  0 , solved a  1 3 or a  , 2 2 Again   (2 a) 2  4 a  0 , i.e a( a  1)  0 ., the solution or a  0 a  1 , So a  3 3 , i.e., a the value of the real number is 2 2 2 (2) Solution: by set A   x x  4 x  0  4, 0 , 2 2 gather B  {x | x  2ax  a    4a  2  x  a  a  1}  {x | x 2  2( a  1) x  a 2  1 0} Because A  B  A , it's available B  A If B   so, i.e., the equation x 2  2( a  1) x  a 2  1  0 has no real roots then satisfied   4( a  1) 2  4( a 2 1)  0 , solved a  1 ; If B  0 so, substitute x  0 into the equation x 2  2( a  1) x  a 2  1  0 and solve it a  1 When a  1 , the equation x 2  4 x  0 , the solution x  0 or x  4 , at this time B  {4,0} , the round; At that a  1 time, the equation x 2  0 is solved x  0 , and at this time, B  {0} it conforms to the topic; If B  4 so, substitute x  4 into the equation x 2  2( a  1) x  a 2  1  0 and solve or a  1 a  7 , When a  1 , the equation x 2  4 x  0 , the solution x  0 or x  4 , at this time B  {4,0} , the round; When a  7 , the equation x 2  16 x  48  0 , the solution x  4 or x  12 , at this time B  {4, 12} , the round; a 2  1  0 If B  4,0 the time, it can be obtained  2 , solved a  1 , and in line with the topic a  8a  7  0 In summary, it can be concluded that the real number a satisfies a  1 a  1 or, that is, the a value range of the ( , 1]  1 real number is 11. 2 2 2 set set A   x x  3 x  2  0 , B   x x  2  a  1 x  a  5  0 (1) If A  B  2 , find the value of the real number a; (2) If there are two elements in the set B x1 , x2 find the range of the value of the real number a, and express it by the algebraic formula containing a x1  x2 ; Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 41 / 44 (3) If the whole set U  R is complete A  B   , find the value range of the real number a Answer: (1) a  3 or a  1 (2) | x1  x2 | 8a  24 (3) {a | a  3} 【Knowledge Points】Find the set or parameters according to the intersection results, and find the parameters according to the number of elements in the set 【Analysis】(1) From 2  B , substitute the equation and test whether it satisfies the problem; (2) According to Vedica's theorem and the squared formula of complete difference, the evaluation can be simplified; (3) Solve according to B the classification of the set elements 2 【Detailed explanation】(1) From the meaning of the title A  {x | x  3x  2  0}  1, 2 , because A B  {2} , so 2  B , So 22  4(a  1)  a 2  5  0that is 4  4 a  4  a 2  5  0 , Simplification a 2  4a  3  0 , i.e., (a  3)(a  1)  0solution a  3 or a  1 , Inspection: When a  3 , B  { x | x2  4 x  4  0}  {2},Satisfied A B  {2} , a  1 When, B  {x | x 2  4  0}  {2 , 2} ,Satisfied A B  {2} , So a  3 or a  1 (2) Because B there are two elements in the set x1 , x2 the equation x 2  2(a  1) x  (a 2  5)  0 has two roots 2 So   4( a  1) 2  4( a 2  5)  8a  24  0 and x1  x2   2(a  1) , x1 x2  a  5 , So | x1  x2 | ( x1  x2 )2  4 x1 x2  4(a  1)2  4(a 2  5)  8a  24 ... (3) Because A  1, 2 , and U  R, B  A   , At B   that time,   4( a  1) 2  4( a 2  5)  8a  24  0 the solution is in a  3 line with the topic; Δ  4(a  1)2  4(a 2  5)  8a  24  0 When B  {1} the time comes, there is  2 no solution; 2 1  2(a  1)  (a  5)  0 Δ  4(a  1)2  4(a 2  5)  8a  24  0 B  {2} When, then  2 , so a  3 ; 2 2  4(a  1)  a  5  0 Δ  4( a  1) 2  4( a 2  5)  8a  24  0  When B  {1 , 2} then 1  2  2( a  1) , then, there is no solution, 2  a 2  5  In summary, the a range is {a | a  3} 12. Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd  The known set P is a set of non-empty numbers, defined P  a  b a, b  P , 42 / 44    P  a  b a ,b  P ., (1) If it is a set P  2, 2024 , please write the sum of the set directly  P  P ; * (2) if 1  n  2023 and n  N* the set P   x n  x  2023, x  N  satisfies  P   P   the n minimum value that is obtained; (3) If the set P   x1 , x2 , x3 , x4 , x5  , x1  x2  x3  x4  x5 and  P  P , verify: x1  x5  x2  x4   【Answer】(1), P  4, 2026, 4048 P  0, 2022 (2) 675 (3) See analysis for proof 【Knowledge Points】New definition of sets, and finding sets or parameters according to the intersection results 【Analysis】(1) The results can be obtained directly according to the definition of  P the basis and the sum  P ; (2) First explain that 1  n  674 the conditions are not met at the time, and then explain that the conditions are met at the time n  675 , and the minimum value n 675 that can be obtained is (3) The  P value of the property is determined first x1  0 , and then x2  x1, x3  x2 , x4  x2 , x5  x2the value is repeatedly discussed, and the conclusion to be proved can be obtained [Explanation] (1) According to  P  P the definition of and, there     are P  a  b a, b  2, 2024  4, 2026, 4048 , P  a  b a ,b  2, 2024   0, 2022  (2) When 1  n  674 , due to n  3n  3  674  2022  2023 , therefore n,3n  P  So 2n  n  n   P 2n  3n  n  P , this  P   P   contradicts this; n  675 When a, b  P , to arbitrary n  a, b  2023 , due to a  b  2n  2  675  1350 , therefore a  b  2023  n  2023  675  1348 ,  This means P  1350,   ,  P   ,1348 so...  P   P   In summary, n the minimum value is 675 (3) Due to P   x1 , x2 , x3 , x4 , x5  , x1  x2  x3  x4  x5 .  Therefore x3  x2 , x4  x2 , x5  x2  P , x3  x2  x4  x2  x5  x2 . . Obviously  P , there are no negative numbers, and they must be contained 0 , so it is  P  P known x1  0 And then by  P  P , x1  0  x3  x2  x3 , know x3  x2  x2 , i.e x3  2x2 .. . Further there are x2  x3  x2  x4  x2  x4, therefore, x4  x2  x3 i.e x4  x2  x3  x2  2x2  3x2 . . Further there is x3  x4  x2  x5  x2  x5, therefore, x5  x2  x4 i.e x5  x2  x4  x2  3x2  4x2 . So x1  x5  0  4 x2  x2  3 x2  x2  x4... Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 43 / 44 Key points: The key to this question is to understand  P  P the definition of and, only by understanding the definition, can the corresponding problem be solved Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd Discipl ine Network (Beijin g) Co., Ltd 44 / 44