Basic Operations of Sets With Answer

Lecture 03 Basic Operations on Sets directory Module 1: Induction of new knowledge Module 2: Draw inferences from the explanation of the test center Test 1: Operations between sets Test point 2: Operations between point sets Test 3: The use of Wayne diagrams in sets Test point 4: Calculate the parameters according to the operation of the set Test 5: The application of set operations in real life Module 4: Customs Clearance Detection Question type is practiced in groups Consolidate and improve comprehensive training Module 1 Induction of new knowledge 【Knowledge point 1】Union 1．The concept of union natural In general, a set consisting of all elements belonging to set A or belonging to set B is called the union of A and B, denoted as A∪B and pronounced "A and B". language Symbolic A∪B={x|x∈A or x∈B} language Graphical language 2．The computational nature of union quality A B=B A A= A A = A definition Satisfy the commutative law The union of any set with itself is equal to the set itself A= A The union of any set with an empty set is equal to the set itself -1Discipline Network (Beijing) Co ., Ltd (A B) C=A (B C) The union of multiple sets satisfies the associative law B) ， B  ( A  B) A  (A Any collection is a subset of that collection that is in union with another collection A B  A B = B Any union of a set with its subsets is itself, and vice versa 【Knowledge point 2】Intersection 1．The concept of intersection natural A set consisting of all the elements that belong to set A and belong to set B, called the intersection language of set A and B, is denoted A∩B and is pronounced "A intersection B". Symbolic A∩B＝{x|x∈A and x∈B} language Graphical language 2．The computational nature of the intersection quality definition A B = B  A Satisfy the commutative law A  = The intersection of an empty set with any collection is an empty set A A= A The intersection of the set and the set itself is still the set itself (A B) C=A (B C) The intersection of multiple sets satisfies the associative law ( A  B)  C = ( A  C )  (B  C ) The synthesis of multiple sets satisfies the distributive property (A If A (A B) C = (A C) (B B = A so, then A  B B )  A, ( A B)  B C) Transformation of intersection and subset relationships The intersection of two sets is a subset of either set 【Knowledge point 3】Complete works and supplements -2Discipline Network (Beijing) Co ., Ltd 1．The concept of the complete works natural In general, if a set contains all the elements involved in the problem under study, then the set is called language a complete set and is usually denoted as U. Symbolic If so A  U , B  U , C  U , , it U is a complete work language Graphical language 2．The concept of complements If set A is a subset of set U, the set consisting of all the elements in set U that do not belong to set A is natural called the complement of set A relative to set U, denoted as C U A language Symbolic ðU A =  x x  U 且 x  A language Graphical language 3．The computational nature of the complement quality definition ( ðU A )  A = U The union of any set and its complement is a complete set ( ðU A ) The intersection of any set with its complement is an empty set A= CU (CU A ) =A 痧 UU = , U = U The complement of any set complement is the set itself The supplement of the complete set is an empty set, and the supplement of an empty set is the complete set 【Knowledge point 4】The principle of repulsion 1．Principle of repulsion: In some finite sets, we often encounter the problem of the number of elements in the set, -3Discipline Network (Beijing) Co ., Ltd and the Venn diagram is often used to represent the intersection, merger, and complement of two sets. If card is used to denote the number of elements of a finite set, i.e., card(A) denotes the number of elements of a finite set A, then the following conclusions are made: （1） card ( A B ) = card ( A ) + card ( B ) − card ( A B) （2） card ( A B C ) = card ( A ) + card ( B ) + card ( C ) − card ( A B ) − card ( B C ) − card ( A C ) + card ( A B C) 【Knowledge point 5】Intervals and related concepts (in order to better solve the problem, explain this part of the content in advance). 1．Representation of the general interval: Let a and b be two real numbers, and a < b, we stipulate that the real numbers here a ,b are called the endpoints of the interval When using intervals to represent a continuous set of numbers, the end that contains the endpoint is indicated by parentheses, and the end that does not contain the endpoint is indicated by parentheses definition name symbol  x | a  x  b closed interval  x | a  x  b open interval ( a,b )  x | a  x  b Half-open and half-closed intervals  a ,b )  x | a  x  b Half-open and half-closed intervals ( a ,b  Number line representation [ a,b ] 2．Set of real numbers R It can be expressed as an interval (-∞, +∞), and "∞" is read as "infinity", "-∞" is read as "negative infinity" and "+∞" is read as "positive infinity". 3．Representation of special intervals -4Discipline Network (Beijing) Co ., Ltd definition symbol xa ( a , + ) xa  a, + ) xb ( −, b ) xb ( −, b  Number line representation Module 2 Draw inferences from the explanations of the test sites Test 1: Operations between sets Typical case 3 【Example 1】Known set A = −4, 0,1, 2,8 , B =  x | x = x then A B = ( ). A． 0,1, 2 B． 1, 2, 8 C． 3,8 D． 0,1 【Answer】D 【Analysis】After solving the equation to find the set B , it can be solved according to the intersection operation of the set 2 [Detailed explanation] by x 3 = x , get x 3 − x = 0 , that is, x ( x − 1) = x ( x − 1)( x + 1) = 0 solve or x = 0 x = 1 or x = −1 , 3 Then B =  x | x = x =  − 1, 0,1 Again A = −4, 0,1, 2,8 , then A  B = −4, 0,1, 2,8  −1, 0,1 = 0,1 Selection: D. 【Example 2】Known set A =  x − 1  x  2 , B =  x 1  x  3 then A B = ( ). A．  x 1  x  2 B．  x 1  x  2 C．  x − 1  x  3 D．  x − 1  x  3 【Answer】D 【Analysis】The answer can be found from the definition of union. -5Discipline Network (Beijing) Co ., Ltd Because A =  x − 1  x  2 , B =  x 1  x  3 of the collection, so A  B =  x − 1  x  3 ... Selection: D. 【Example 3】A known set A = { x∣− 2  x  5} , B = −5, −4, 3, 6 then ( ðR A ) I B = ( ). B． −5, −4, 6 A． −5, −4 C． 3, 6 D． − 4, 3 【Answer】B 【Analysis】It can be calculated directly by using the algorithm of complement and intersection of sets. [Detailed explanation] A = { x∣− 2  x  5} ,  ðR A = { x∣ x  − 2 or x ≥ 5 }, Therefore ( ðR A ) B = −5, −4, 6 , Therefore, B. draw inferences 4 2 [Variant 1] known set A =  − 4, − 2, − 1, 0,1, 2, 4 , B =  x∣ x − x − 12  0 , then A B = ( ). A． −1, 0,1 B． 0,1, 2 C． 0,1, 2, 4 D． −2, −1, 0,1, 2 【Answer】A 【Analysis】Find the set first B , and then solve it according to the intersection operation 4 2 2 2 2 [Detailed explanation] by the meaning x − x − 12  0  ( x − 4 )( x + 3 )  0  0  x  4  − 2  x  2 of the title, So B =  x − 2  x  2 , so A B = −1, 0,1 , Therefore, A. 【Variant 2】A known set A =  x | −1  x  1 , B =  x | 0  x  2 then A B = ( ). A． { x | −1  x  2} B． { x | −1  x  2} C． { x | 0  x  1} D． { x | 0  x  2} 【Answer】B 【Analysis】With the help of union definition calculation. Because of A = {x | −1  x  1} the set, B = { x | 0  x  2} so A B = { x | −1  x  2} Therefore, B. 【Variant 3】Known set A = {−1, 0,1, 2} , B = {−1, 0} then ðA B = ( ). -6Discipline Network (Beijing) Co ., Ltd A． {−1, 0} B． {0,1} D． {− 1,1} C． {1, 2} 【Answer】C 【Analysis】It can be solved by the concept of complements. Since A there is only 1 element in the 2  B set, therefore... ðA B = 1, 2 Selection: C. 【Variant 4】Known set A = −1, 0,1 , B = −2, 0, 2 ,, C = 1, 2, 3 ,then ( A A． 1, 2 B． 1, 3 C．  2, 3 B) C = ( ). D． 1, 2, 3 【Answer】A 【Analysis】It is found by the definition of conditional union and A B conclusion, and then according to the definition of intersection 【Explanation】Because A = −1, 0,1 , B = −2, 0, 2 , B =  − 2, − 1, 0, 1, 2 , So A So ( A B) again C = 1, 2, 3 , C = 1, 2 , Therefore, A. 【Variant 5】Let the set U = 0,1, 2, 3, 4 , P = 0,1, 2, 3 , Q = 1, 2, 4 ,then P  ( ðUQ ) = ( ). A． 0 B． 0, 3 C． 3 D． 1, 3 【Answer】B 【Analysis】Find out ðU Q what you can find P [Detailed explanation], ðU Q = 0, 3 therefore, P (ð Q ) U ( ðU Q ) = 0, 3 Therefore, B. Test point 2: Operations between point sets Typical case 【Example 4】Knowing the set M={(x,y)|y= x − 1 2,xy≤0},N={(x,y)|y=x2 −4 }, then M  N The number of elements in is ( ). A．0 B．1 C．2 D.1 or 2 -7- Discipline Network (Beijing) Co ., Ltd 【Answer】A 【解析】∵集合 M={(x,y)|y=2x-1,xy≤0},N={(x,y)|y=x2-4},  y = 2 x − 1， xy„ 0 The  number of elements in }=.∴M∩N is 0 ∴M∩N＝{（x，y）|  2 y = x − 4 Test   3 [Variant 1] If the set is known M = ( x , y ) x − y = 0 , N = ( x , y ) y = x , the M  N number of elements in is ( ). A．0 B．1 C．2 D.3 【Answer】D   3 [Analysis] because of the set M = ( x , y ) x − y = 0 , N = ( x , y ) y = x ,  y = x    = (0, 0), (1,1), ( −1, −1) , the A Therefore M  N =  ( x , y )  3 y = x     B number of elements in the medium is 3, so choose D ) 3 x − y = 0} set, N = {( x , y ) | x + y = 0} then ( ). 【Variant 2】If M = {( x , y∣ 2 A． M  N = M B． M  N = M 2 C． M  N = N D． M  N =  【Answer】B 【Analysis】∵ set M = {( x, y ) | 3 x − y = 0} , N = ( x , y ) | x 2 + y 2 = 0 = ( 0, 0 ) ,  3x − y = 0 x = 0  Because  2 of the ∴ M  N = ( 0, 0 ) = N , therefore M  N = M , I choose: B 2 x + y = 0 y = 0 2 2 [Variant 3] Known set M ＝( x , y ) |x + y  1, x  Z , y  Z  , N＝{( x , y ) | x + y  2 }, the M  N number of elements in the set is ( ). A．6 B．7 C．8 D.9 【Answer】C [Analysis] is obtained from x 2 + y 2  2 the available, x 2  2, y 2  2 i.e x  2, y  2 ., , The satisfying whole point x  Z , y  Z in N is: ( 0, 0 ) , ( 0,1) , ( 0, −1) , ( −1, 0 ) , (1, 0 ) , (1,1) , (1, −1) , ( −1,1) , ( −1, −1) , a total of 9 points, Among them, only (1,1) is not satisfied x + y  1 , so M  N the number of elements in it is 8, so C -8Discipline Network (Beijing) Co ., Ltd Test 3: The use of Wayne diagrams in sets Typical case 【Example 5】If you set A = 3, 4, 5 a set, B = 2, 3, 4,8 then the set represented by the shaded part in the figure is ( ). A．  2, 8 B． 3, 4 C． 2, 5,8 D． 3, 4, 5 【Answer】C 【Analysis】First find A, B the intersection of the A  B set, then the A, B union of the set A  B , and finally find the element composition of the shadow part 【Explanation】Because A = 3, 4, 5 , B = 2, 3, 4,8 , So A  B = 3, 4 , A  B = 2, 3, 4, 5,8 , So the shaded part represents the set as 2, 5,8 Selection: C. Test 【Variant 1】If you know the complete set U = R , set A = 0,1, 2, 3 , B = −2, −1, 0,1 then the set represented by the shaded part in the figure is ( ). A． −2, −1, 0,1, 2, 3 B． 0,1 C．  2, 3 D．  − 2, −1 【Answer】D 【Analysis】From the shaded part, we can know that the corresponding set is B  ( ðU A ) , and the conclusion can be obtained -9Discipline Network (Beijing) Co ., Ltd The corresponding set of shaded parts is B  ( ðU A ) , ∵ Complete Works U = R , A = 0,1, 2, 3 Collections, B = −2, −1, 0,1 , ∴ B  ( ðU A ) = −2, −1 . Selection: D.  【Variant 2】If the whole set U = R is known A = 0,1, 2, 3, 4, 5 , B =  x  N 3  x  8 the set, then the set represented by the shaded part in the figure is ( ). A． 0,1, 2 B． 1, 2 C． 1, 2, 3 D． 0,1, 2, 3 【Answer】D 【Analysis】The shaded part can be represented by the operation of the set, and then substituted into the calculation, and the result can be obtained.  [Detailed explanation], B =  x  N 3  x  8 =  4, 5, 6, 7, 8 and A = 0,1, 2, 3, 4, 5 , Then A B = 4, 5 , The shaded part represents the collection that is A the element that has A  B been removed from the collection The shaded part represents the set of 0,1, 2, 3 Therefore, D [Variant 3] If the set is known A = 1, 2, 4 , B = 2, 4, 5 the set represented by the shaded part in the figure is ( ). A． 1 B． 2, 4 C． 5 D． 1, 2, 4, 5 【Answer】B 【Analysis】The results are obtained according to the meaning of the set represented by the shaded part. - 10 Discipline Network (Beijing) Co ., Ltd 【Explanation】The set represented by the shaded part of the title is A  B = 2, 4 , Therefore, B. Test point 4: Calculate the parameters according to the operation of the set Typical case [Example 6], A =  x − 2  x  5 B =  x m + 1  x  2 m − 1 if A B = A , then m the range of values of real numbers is ( ). A． ( − , 3 B． ( − , 2 ) C．  −3, 3 D． ( −, 3 ) 【Answer】A 【Analysis】The analysis shows that in the B  A B =  B   first case, the inequality about the real number can be obtained m ; In the second case, a group of inequalities about real numbers can be obtained m , and the range of values of real m numbers can be obtained by synthesis 【Explanation】Because A =  x − 2  x  5 , B =  x m + 1  x  2 m − 1 ,, A B = A ,, B  A , If B =  , then m + 1  2m − 1 , the solution m  2 ; m + 1  2m − 1  If so B   B  A , then  m + 1  − 2 , the solution 2  m  3 2m − 1  5  To sum up, the m range of values for ( − , 3 real numbers is Therefore, A. [Example 7] set the set A =  x | x + m  0 , B = { x | −2  x  4} the complete works U = R (1) ( ðU A ) B =  If, find m the range of values of real numbers; (2) If ( ðU A )  B   , find m the range of values of real numbers; (3) If ( ðU B ) A = R so, find m the range of values of real numbers 【Answer】(1) {m | m  2} (2) {m | m  2} (3) {m | m  2} ． - 11 Discipline Network (Beijing) Co ., Ltd Solution: (1) Solution 1 is easy to know A = { x | x  − m} , so ðU A = { x | x  − m} and B = { x | −2  x  4} ( ðU A ) B =  , and, so − m  −2 , the solution is obtained m ≥ 2 , so m the range of values of real numbers is {m | m  2} Solution 2 From ( ðU A ) B =  , know B  A , and B = { x | −2  x  4} , A = { x | x  − m} so − m  −2 , solve m ≥ 2 , so the m range of values of real numbers is {m | m  2} (2) Because ðU A = { x | x  − m} , B = { x | −2  x  4} ( ðU A )  B   ,, −m  −2 so, the solution is obtained m  2 , so m the range of values of real numbers is {m | m  2} (3) Because A = { x | x  − m} , ðU B = { x | x  − 2 or x  4} , ( ðU B ) A = R , so −m  −2 , the solution is obtained m ≥ 2 , so m the range of values of real numbers is {m | m  2} Test 【Variant 1】Given A =  x | x  −1 or x  3 , B =  x | m − 2  x  m + 2 if (ðR A )  B   , then the range of m is 【Answer】 m | −3  m  5 【Analysis】Find ðR A , by (ðR A )  B   establishing an inequality, the solution can be obtained [Details] by A =  x | x  −1 or x  3 , can be obtained ðR A =  −1, 3 , Because (ðR A )  B   , B =  x | m − 2  x  m + 2 , So 3  m − 2 and −1  m + 2 , Solution −3  m  5 , So the answer is: m | −3  m  5 [Variant 2] known set A = { x∣− 5  x  2}, B = { x∣2 a − 3  x  a + 1} 2 (1) If C = 3, 4, a + 2 a − 3 , 0  ( B C ) , find the value of the real number a; (2) Select one of the conditions (1), (2), and (3) as the known condition, and find the value range of the real number a. Conditions: (1); A B = B ② B  ( ðR A ) =  ； ③ A (ð B ) = R ． R 【Answer】(1) a = 1 (2) See analysis for the answer - 12 Discipline Network (Beijing) Co ., Ltd [Detailed explanation] solution: (1) Because 0  ( B (2) If you choose (1), you A  a 2 + 2 a − 3 = 0, C ) of, so  the solution a = 1  2 a − 3  0  a + 1, B = B will get B  A it B =  When the time comes, then 2a − 3  a + 1 the solution is satisfied a  4 ;  2 a − 3  a + 1,  When B   it is, it is  2 a − 3  − 5, solved −1  a  1  a + 1  2,  In summary, the range of values of the real number a is [ −1,1] [4, + ) If you choose (2), B  ( ðR A ) =  B =  When, 2a − 3  a + 1 the solution is solved a  4 , and the conditions are satisfied:  2 a − 3  a + 1,  ð A = { x ∣ x  2 x  − 5} When, or  2 a − 3  − 5, , then −1  a  1 the solution B R  a + 1  2,  In summary, the range of values of the real number a is [ −1,1] [4, + ) If you choose (3), A (ð B ) = R R When B =  the time comes, 2a − 3  a + 1 the solution is satisfied a  4 , and the conditions are met;  2 a − 3  a + 1,  B   ðR B = { x∣ x  a + 1 When, x  2 a − 3} or  2 a − 3  − 5, , then −1  a  1 the solution  a + 1  2,  In summary, the range of values of the real number a is [ −1,1] [4, + ) 2 [Variant 3] known complete works U = {1, 2, 3, 4, 5}, A =  x∣ x − 5qx + 4 = 0, q  R  , A  U (1) If ðU A there are four elements, sum ðU A the value of q; (2) If ðU A = U so, find the range of the value of the real number q 【Answer】(1), ðU A = {1, 3, 4, 5}  4  5 (2)  q − q q= 4 5 4  5 【Analysis】(1) According to the complete set and conditions, it can be judged that the equation can be solved if it has equal real roots; (2) It is transformed into an equation without real roots, which can be solved by using discriminant formula. - 13 Discipline Network (Beijing) Co ., Ltd [Explanation] (1) Because ðU A there are four elements, A is a single element set Then the equation x 2 − 5qx + 4 = 0 has two equal solutions to real numbers From the relationship between the root and the coefficient, the product of these two equal solutions is 4, So only A = {2} , thus, 2 + 2 = 5q so q = 4 5 . So ðU A = {1, 3, 4, 5} ... (2) By ðU A = U knowing A =  , that is, x 2 − 5qx + 4 = 0 there is no solution to the equation So  = (5 q ) 2 − 16  0 , the solution − 4 5 q 4 5 ,  4  5 Therefore, the range of values of the real number q is  q − q 4  5 Test 5: The application of set operations in real life 【Example 8】In order to enrich students' extracurricular life, a school has set up a basketball club, an AI club, and a Go club, and a total of 30 students in a class in the first year of high school have participated in the school club, of which 15 people have participated in the basketball club, 8 have participated in the AI club, 14 have participated in the Go club, 3 people have participated in the basketball club and the AI club at the same time, 3 people have participated in the basketball club and the Go club at the same time, no one has participated in the three clubs at the same time, and the number of people who only participate in the Go club is ( ). A．10 B．9 C．7 D.4 【Answer】A 【Analysis】It can be solved by combining the operation of the set according to the principle of repulsion according to the meaning of the problem. There are 15 people who participate in the basketball club, 3 people who participate in the basketball club and the AI club at the same time, and participate in the basketball club and Go at the same time There are 3 people in the club, and no one participates in three clubs at the same time, so only 9 people participate in the basketball club; There are people who participate in both the AI club and the Go club x , because there are 8 people who participate in the AI club There are 3 people who participate in the basketball club and the AI club at the same time, so there are only people - 14 Discipline Network (Beijing) Co ., Ltd who participate in the AI club 8 − 3 − x = 5 − x ; And because there are 14 people who participate in the Go club, and 3 people who participate in the basketball club and the Go club at the same time, So there are people who only participate in the Go club 14 − 3 − x = 11 − x . To sum up, a total of 30 people participated in the school club So 9 + 3 + 3 + ( 5 − x ) + x + (11 − x ) = 30 , the solution x = 1 , Therefore, the number of people who only participate in the Go club is people 10 Therefore, A. Test 【Variant 1】A school in our city has a total of 1500 students in the school lunch, each lunch can only choose to eat in an upstairs or downstairs canteen, according to statistics, among the students who have lunch in the upstairs canteen that day, some 10 0 0 students will go to the downstairs canteen for lunch the next day: and among the students who have lunch in the downstairs canteen on the same day, some 15 0 0 students will go to the upstairs canteen for lunch the next day, then one semester later, The number of students having lunch in the upstairs cafeteria is about ( ). A．700 B．800 C．900 D.1000 【Answer】C 【Analysis】According to the meaning of the question, list the equations and substitute them for the calculation, and the result can be obtained. After one semester, the number of students having lunch in the upstairs cafeteria is about x , The number of students having lunch in the downstairs cafeteria is about 1500 − x , Students who were originally in the upstairs cafeteria and stayed: percentage1 − 10 0 0 = 90 0 0 , i.e 0.9x ., Students transferred from the downstairs cafeteria: the number of people in the downstairs cafeteria 15 0 0 , i.e 0.15 (1500 − x ) So x = 0.9 x + 0.15 (1500 − x ) , the solution x = 900 So after a semester, the number of students who have lunch in the upstairs cafeteria is about 900 Therefore, C - 15 Discipline Network (Beijing) Co ., Ltd 【Variant 2】If each student in a class takes at least one course, one 33 student takes the "Football" course, one 26 student takes the "Basketball" course, and 10 one student takes both courses, the number of students in the class is ( ). A． 39 B． 49 C． 59 D． 69 【Answer】B 【Analysis】The number of students in the class can be obtained by A assuming that the composition of students taking "Football" is and B the set of students who are taking "Basketball" is made to make a Venn diagram 【Explanation】The set of students who take the "Football" course is , A and the set of students who take the "Basketball" course is B , As shown in the figure below: As can be seen from the figure, the number of students in the class is 23 + 10 + 16 = 49 Therefore, B. 【Variant 3】The school counted 30 students in a class participating in three interest groups of music, science and physical education, and it was known that each person participated in at least one interest group, of which the number of participants in the music, science and physical education groups was 19, 19, and 18 respectively, and the number of people who only participated in the music and science groups at the same time was 4, the number of people who only participated in the music and physical education groups at the same time was 2, and the number of people who participated in the science and physical education groups at the same time was 4, and the number of people who participated in the three groups at the same time was ( A．5 B．6 C．7 ） D.8 【Answer】D 【Analysis】Let the number of participants in 3 groups at the same time be, x and then use the Venn diagram to solve the problem according to the meaning. - 16 Discipline Network (Beijing) Co ., Ltd As shown in the figure, if the number of people who participated in 3 groups at the same time is x, then 13 − x + 11 − x + 12 − x + 2 + 4 + 4 + x = 30 , The x = 8 number of people who participated in 3 groups at the same time was 8 Selection: D. Module 3 Knowledge Testing Test 1: Operations between sets 1. (25-26 high school one, national, holiday homework) known set A =  x | −2  x  0 , B =  x | x  1 then B = ( ). A A．  x | −1  x  1 B．  x | −2  x  0 C．  x | −1  x  0 D．  x | −2  x  1 【Answer】C 【Analysis】First solve the absolute value inequality, and then find the intersection. 【Detailed explanation】by B =  x | x  1 =  x | − 1  x  1 , Then A B =  x | −2  x  0  x | −1  x  1 =  x | −1  x  0 , Selection: C. 2. (25-26 high school one, national, holiday homework) known set A =  x − 1  x  2 , B =  x x − 1  A then B = ( ). A A．  x 0  x  3 B．  x − 1  x  3 C．  x 0  x  2 D．  x − 1  x  2 【Answer】C - 17 Discipline Network (Beijing) Co ., Ltd 【Analysis】Find the set B first, and then solve it according to the operation of the intersection. 【Explanation】Because A =  x − 1  x  2 , B =  x x − 1  A =  x − 1  x − 1  2 =  x 0  x  3 , Therefore A  B =  x 0  x  2 Selection: C. 3. (25-26 high school one, national holiday homework) known set A =  x | −1  x  1 , B =  x | 0  x  2 , then A B = ( ). A．  x | −1  x  2 B．  x | 0  x  2 C．  x | 0  x  1 D．  x | −1  x  2 【Answer】D 【Analysis】According to the meaning of the problem, combined with the concept and operation of set union, it can be solved. 【Detailed explanation】By the collection A =  x | −1  x  1 , B =  x | 0  x  2 , According to the operation of the union of sets, we can get A B =  x | −1  x  2 , Selection: D. 4. Known complete works U = { x∣− 3  x  3} , sets A = { x∣0  x  1} , then ðU A = ( ). A． { x∣− 3  x  0 or 1  x  3} B． { x∣− 3  x  0 or 1  x  3} C． { x∣0  x  1} D． { x∣0  x  1} 【Answer】A 【Analysis】The solution is solved by the operation of the complement. 【Explanation】Because U = { x∣− 3  x  3} , set A = { x∣0  x  1} , then the collection ðU A = | −3  x  0 or 1  x  3 Therefore, A. 5. The known complete set is R , set A =  x 1  x  2 , B =  x x  4 , then ( ). A． A B=A C． B ( ðR A ) = B B． A B = R D． A  (ðR B ) = A 【Answer】C - 18 Discipline Network (Beijing) Co ., Ltd 【Analysis】Judge A according to the operation of the intersection, judge B according to the counterexample of the operation of the union, judge C according to the operation of the complement and intersection, judge D according to the operation of the complement and union. For option A, because A =  x 1  x  2 , B =  x x  4 , so A B =   A , so A is incorrect; For option B, because 3 R , but 3  A B , yes, A  B  R so B is incorrect; For option C, by B =  x x  4 , A =  x 1  x  2 , , or ðR A =  x x  1 x  2 , ( ) Therefore B  ðR A = B , C is correct; For option D, by B =  x x  4 , get ðR B =  x x  4 , Again A =  x 1  x  2 , therefore A  (ðR B ) =  x x  4  A , D is incorrect Selection: C. 6. (24-25 high school sophomore lower Zhejiang Wenzhou stage exercises) known complete works U = R , A =  x 0  x  1 , then ðU A = ( ). A．  x x  0 B．  x x  1 C． { x x  0 or x  1} D． { x x  0 or x  1} 【Answer】C 【Analysis】Solve according to the definition of the complement. Because of the complete works U = R , A =  x 0  x  1 , So ðU A = { x x  0 or x  1} Therefore, C 7. Known complete works U = { x | x  4} , collections, A = { x | −2  x  3} B = { x | −3  x  2} finds, A  B and (ðU A )  B Answer, A  B = { x | −2  x  2} (ðU A )  B = { x | x  2 or 3  x  4} 【Analysis】It can be solved directly by using the operation of set intersection, set complement and union. Because of the set U = { x | x  4} , the set A = { x | −2  x  3} , B = { x | −3  x  2} , So A  B = { x | −2  x  2} ðU A = { x | x ≤ − 2 or 3  x  4}  (ðU A )  B = { x | x  2 or 3  x  4} - 19 Discipline Network (Beijing) Co ., Ltd 8. Known set A =  x 2  x  5 , set B =  x 3  x  7 , set C =  x x  4 (1) Seeking A  B ; A B (2) Seeking A  ( B  C ) , C  ( A  B ) . . 【Answer】(1) A  B =  x 3  x  5 , A  B =  x 2  x  7 ; (2) A  ( B  C ) =  x 4  x  5 ， C  ( A  B ) =  x x  3 . 【Analysis】(1) Solve according to the definition of the intersection and union of sets; (2) Find B  C , re-find, A I ( B I C ) and re-combine according to the intersection definition C  ( A  B ) 【Explanation】(1) Because A =  x 2  x  5 , B =  x 3  x  7 , So A  B =  x 3  x  5 , A  B =  x 2  x  7 , (2) Because B =  x 3  x  7 , C =  x x  4 , So B  C =  x 4  x  7 , again A =  x 2  x  5 , So A  ( B  C ) =  x 4  x  5 , by (1), A  B =  x 3  x  5 C =  x x  4 , So C  ( A  B ) =  x x  3 ... Test 2: The use of Venn diagrams in sets 9. A Venn diagram that is known to represent sets M = {- 1, 0,1} and P = {0,1, 2,3} relationships is shown in the figure, then the set represented by the shaded part is ( ). A． {0,1} B． {0} C． {−1, 2, 3} D． {− 1, 0,1, 2, 3} 【Answer】A 【Analysis】First of all, by identifying the Venn diagram, we know that the shaded part represents the set M  P , and then judge it according to the connotation of the intersection 【Explanation】From the Venn diagram in the question, the set represented by the shaded part is M - 20 Discipline Network (Beijing) Co ., Ltd P , Because M = {−1, 0,1}, P = {0,1, 2, 3} , So M I P = {−1, 0,1} I {0,1, 2, 3} = {0,1} ... Therefore, A. 10. The set represented by the shaded part of the Venn diagram as shown in the figure is ( ) A． ðU ( A  B ) B． ðU ( A B) C． ( ðU A ) B D． A (ð B ) U 【Answer】B 【Analysis】The answer can be obtained according to the meaning of the intersection and complementation of the set. 【Detailed explanation】by the meaning of the title. The shaded part of the figure represents the set of ðU ( A B) Therefore, B. 11. If the collection is known A = {0,1, 2, 3, 4, 5}, B = {0, 2, 4, 6,8} , the number of elements of the collection shown in the shaded part of the figure is ( ). A．2 B．3 C．4 D.6 【Answer】B 【Analysis】According to the meaning of the question, the set represented by the shaded part is obtained ðAU B B , and combined with the set algorithm, it can be solved 【Explanation】From the title, the set represented by the shaded part in the figure is ðA B ( B ) , Because of the collection A = {0,1, 2, 3, 4, 5}, B = {0, 2, 4, 6,8} , it is available ðA B ( B ) = 1, 3, 5 So the number of elements of the collection shown in the shaded section is 3 one Therefore, B. 12．已知集合 U =  x 0  x  m ， M =  x 1  x  3 ， N =  x 2  x  6 ， - 21 Discipline Network (Beijing) Co ., Ltd (1) If ( M N )  U , find the range of values of m; (2) When m = 10 the time comes, find the set corresponding to the shaded part of the diagram 【Answer】(1)  m m  6 (2)  x 3  x  6 【Analysis】(1) First, find the union set, and then find the parameters according to the inclusion relationship of the set; (2) First, the corresponding set relationship is determined according to the shaded part, and then the intersection complement definition is applied to solve the problem. [Details] (1), M  N =  x 1  x  6 By ( M N )  U , there is: m  6 . Therefore, the value range of m is  m m  6 (2) The shaded part is: N ðU M , ðU M =  x 0  x  1或 3  x  10 , Therefore: N  ðU M =  x 3  x  6 Test point 3: Calculate the parameters according to the operation of the set 13. Known A = −1, 0,1, 2 , B = { x∣ a  x  2} Collections. If so A  B = 1 , the a value range is . 【Answer】 0  a  1 【Analysis】Using the result of the given intersection, the relationship between the elements and the set is combined to solve the problem. 【Detailed explanation】According to the meaning of the topic, 1  A, 0  A ,, 0  a  1 , Therefore a , the value range is 0  a  1 So the answer is: 0  a  1 14. Known set A =  − 1, 2 , B =  x mx + 1 = 0 , if A B = A , then the set of possible values of m is - 22 Discipline Network (Beijing) Co ., Ltd  1  2 【Answer】  0,1, −  【Analysis】It can be obtained from the meaning B  A of the question, and it can be solved by using the meaning of the sub-meaning [Detailed explanation], Q A U B = A ∴ B  A ∴ when B =  , m = 0 ; When − 1  B , m = 1 ; When 2  B , m = − The value of ∴m is 0,1, − 1 2  1  2  1  2 1 2 , , and the value of ∴m is  0,1, −  . So the answer is:  0,1, −  . 2 15. Set m  R , set A = ( x , y ) y = x + 1, x  R  , B = ( x , y ) | y = − x + 2 x + m , x  R  3 (1) If m = , ask A  B ; 4 (2) If A B =  so, find m the range of values of real numbers  1 3   【Answer】(1)   ,    2 2    3  4 (2)  − ,  2 【Analysis】(1) Transform the y = x + 1 intersection problem into summation y = − x + 2 x + 3 4 and solve it jointly (2) Transform into no intersection y = x + 1 with non-existent y = − x 2 + 2 x + m , synonymous, discriminant   0 , and you can get the answer   y = x +1  x = 1  2 x − x + = 0 [Detailed explanation] (1) by  , get , solve 3  2 4  y = −x + 2x + y = 4   1 2 3 2  1 3   B =   ,   ...  2 2    y = x + 1, (2) By  , de, x 2 − x + 1 − m = 0 2 y = − x + 2 x + m  So A x 2 − x + 1 − m = 0 Discriminant formula of known equations Δ = ( − 1) − 4 (1 − m ) = 4 m − 3  0 , 2 From so m  3 4 ...  3  4 Therefore, the range of values for m  − ,  real numbers is 16．已知集合 A = { x | x 2 − px + 15 = 0} ， B = { x | x 2 − ax − b = 0} ， A - 23 Discipline Network (Beijing) Co ., Ltd B = {2, 3, 5} ， A  B = {3} . (1) Find the values of p, a, b; (2) If C = { x | mx + 2 = 0} , and C  B , find the value of m 【Answer】(1) p = 8 , a = 5, b = −6 ; (2) m = 0 or m = −1 or m = − 2 3 【Analysis】(1) According to the intersection result 3 2 − 3 p + 15 = 0 p , the B = {2, 3} parameter value is calculated by combining the relationship between the root and the coefficient; (2) From the inclusion relationship and discussion C =  , the C   corresponding parameter value can be obtained [Explanation] (1) by A  B = {3} , therefore, 3 2 − 3 p + 15 = 0 can be obtained p = 8 , then A = { x | x 2 − 8 x + 15 = 0} = {3, 5} , Again A B = {2, 3, 5} , then B = { x | x 2 − ax − b = 0} = {2, 3} , therefore; a = 5, b = − 6 So p = 8 , a = 5, b = −6 ; (2) By C  B , If C =  , i.e m = 0 ., satisfies the question, 2 if C   , i.e m  0 ., then 2m + 2 = 0  m = −1 , or 3m + 2 = 0  m = − , 3 In summary, m = 0 or m = −1 or m = − 2 3 Test 4: The application of set operations in real life 17. There are 12 students in Class 1 of the third year of high school who have read "The Peony Pavilion", 8 students who have read "Xingshi Hengyan", and 4 students who have read both of them, then at least one of the students in the class has read "The Peony Pavilion" and "Xingshi Hengyan" ( ) A.16 persons B.18 people C. 20 persons D.24 【Answer】A 【Analysis】It can be solved according to the repulsion principle of the set. 【Detailed explanation】Set up a collection A = of "students who have read "The Peony Pavilion" in Class 1 of the third year of high school", and the number of elements is recorded as card ( A ) ; The number B = of elements of the collection of "students who have read "Xingshi Hengyan" in Class 1 of the third year of high school" is recorded as card ( B ) ; - 24 Discipline Network (Beijing) Co ., Ltd Then card( A) = 12, card( B ) = 8, card( A B) = 4 , B ) = card( A) + card( B ) − card( A Then card( A B ) = 12 + 8 − 4 = 16 Therefore, there are 16 students in the class who have read at least one of the "Peony Pavilion" and "Xingshi Hengyan". Therefore, A. 18. In order to enrich the students' extracurricular life, Qiujing Middle School has carried out a variety of student club activities, among which the Psychology Club, the Animation Club and the Geography Club are the most popular, and 35 students in a certain class of the first year of high school have participated in these three clubs, of which 19 have participated in the Psychology Club, 16 have participated in the Geography Club, 15 have participated in the Animation Club, 6 have participated in the Psychology Club and the Geography Club, 5 have participated in the Geography Club and the Animation Club, and it is known that everyone has participated in at least one club, and no one has participated in three clubs at the same time, There are only ( ) students who have participated in only one club A．16 B．18 C．20 D.24 【Answer】C 【Analysis】It can be solved by combining the operation of the set according to the principle of repulsion according to the meaning of the problem. 【Details】Set the psychology club as A, the geography club as B, and the animation club as C. Then card( A B C ) = 35, card( A B C) = 0 , card( A) = 19, card( B ) = 16, card(C ) = 15, card( A get card( A B B ) = 6, card( B C ) = card( A) + card( B ) + card(C ) − card( A That is 35 = 19 + 16 + 15 − 6 − 5 − card( A C) , yes, card( A C) = 5 ， B ) − card( B C ) − card( A C ) − card( A B C) C) = 4 Therefore, the number of people who only participate in one club is shared 19 − (4 + 6) + 16 − (6 + 5) + 15 − (4 + 5) = 20 Therefore, C 19. In order to carry forward the red culture and inherit the cultural spirit, a school arranged a social practice activity assignment for the study of red culture on the occasion of the holiday, and randomly checked the completion of 100 students after the start of school (each student participated in at least one activity), of which 52 people watched the red movie, 43 people visited the martyr's cemetery, 49 people visited the red education base, 24 people watched both the red movie and the martyr's cemetery, and 20 people watched both the red movie and the - 25 Discipline Network (Beijing) Co ., Ltd red education base. There were 17 people who visited both the martyrs' cemetery and the red education base, and the number of participants in all three activities was . 【Answer】17 【Analysis】The results are calculated according to the method of finding the number of elements in the set and the principle of repulsion. 【Details】Set𝐴 = {𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑤𝑎𝑡𝑐ℎ𝑖𝑛𝑔 𝑟𝑒𝑑 𝑚𝑜𝑣𝑖𝑒𝑠} set,Set𝐵 = {𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑣𝑖𝑠𝑖𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑀𝑎𝑟𝑡𝑦𝑟𝑠 𝐶𝑒𝑚𝑒𝑡𝑒𝑟𝑦}, Collection𝐶 = {𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑤𝑎𝑡𝑐ℎ𝑖𝑛𝑔 𝑟𝑒𝑑 𝑚𝑜𝑣𝑖𝑒𝑠}, The number of participants in all three activities is x , Then card ( A B C) = x, It can be obtained from the title card ( A  B  C ) = card ( A ) + card ( B ) + card ( C ) − card ( A B ) − card ( B C ) − card ( A C ) + card ( A B C) Namely 100 = 52 + 43 + 49 − 24 − 17 − 20 + x , Solution x = 17 So the answer is: 17 20. In the "track and field games" of a certain school, a total of 12 students participated in the three events of 100 meters, 400 meters and 1500 meters, of which 8 people participated in the "100 meters race", 7 people participated in the "400 meters race", 5 people participated in the "1500 meters race", "100 meters and 400 meters" both participated in 4 people, "100 meters and 1500 meters" both participated in 3 people, "400 meters and 1500 meters" participated in 3 people, then there were people who participated in all three competitions. 【Answer】2 【Analysis】According to the principle of repulsion, the number of participants in all three items can be analyzed. According to the title, it is set to A =  x | x be a student who participates in 100 meters participates in 400 meters  , and a C =  x | x student who participates in 1500 meters - 26 Discipline Network (Beijing) Co ., Ltd , a  B =  x | x student who ， Then card ( A ) = 8, card ( B ) = 7, card ( C ) = 5 , And card ( A  B ) = 4, card ( A  C ) = 3, card ( B  C ) = 3 , Then card ( A  B  C ) = 12 −  ( 8 + 7 + 5 ) − ( 4 + 3 + 3 )  = 2 , So there are 2 people who participate in all three competitions, So the answer is:2. Test 1. Let the whole set U = −3, −2, −1, 0,1, 2, 3 , set S = −3, 0,1 , T = −1, 0, 2 is ðU ( S  T ) equal to ( ). A．  B． −3, −1, 0,1, 2 C． −2, −1, 2, 3 D．  − 2, 3 【Answer】D 【Analysis】The solution can be obtained from the concepts of union and complement. 【Detailed explanation】Collection S = −3, 0,1 , T = −1, 0, 2 , So S U T = −3, −1, 0,1, 2 , ðU ( S  T ) = −2, 3 , Selection: D. 2. Known complete set U = −1, 0,1, 2, 3, 4, 5 , set A = 1, 2 , B = −1, 0, 2, 3 then set ( ðU A )  B = ( ). A． 3 B．  − 1, 0, 3 C． −1, 0, 2,3, 4,5 D． 2, 3, 4, 5 【Answer】C 【Analysis】Apply the union operation of the set to find the set. [Detailed explanation] is set by the question ðU A = {− 1, 0, 3, 4, 5} , again B = −1, 0, 2, 3 , then ( ðU A )  B = - 27 Discipline Network (Beijing) Co ., Ltd −1, 0, 2,3, 4,5 Therefore, C 3. Set the whole set U = 1, 2, 3, ,10 , and the non-empty set A, B meets the following conditions: B=； ① A B =U ， A (2) if x  A , y  B then and x + y  A xy  B When 5  A ( 1 【Answer】 B  fill  or  ), the B number of elements in this time is 8 【Analysis】First assume 1  A , the introduction contradicts the condition, and obtains 1  B , according to the meaning x + y  U of the question, if, xy  U then x + y  B , xy  A , exclude each element, so as to obtain the set A and the set B B =  ,So 1  A and 1  B there is only one established, 【Explanation】Because A  B = U , A If 1  A , for any y  B one, 1  y = y  B , and if x  A , y  B then xy  B contradictory, So 1  A it doesn't hold, only;1  B Because 5  A , 1  B ,so 5 + 1 = 6  B , If 4  A then, then 4 + 1 = 5  B 5  A contradictory, so, 4  B From 5  A , 4  B ,Available 5 + 4 = 9  B :, If 2  A , because 6  B , so 2 + 6 = 8  B , Again 4  B , so 2  4 = 8  A , with A B =  contradictions, so 2  B , Then 5  A , 2  B , 5 + 2 = 7  B If 3  A , because 2  B , so, 3 + 2 = 5  B , and 5  A contradictory, so 3  B , From 5  A , 3  B ,Available 5 + 3 = 8  B :, If 10  B , because 5  A , 2  B ,so, 5  2 = 10  A ,and A B =  contradiction, So 10  A , So A = 5,10 B = 1, 2, 3, 4, 6, 7,8, 9 , so B the number of elements in is 8 one So the answer is: , 8 4. There are 54 students in a class in the first year of high school, and each student should choose 3 courses from the six courses of physics, chemistry, biology, history, geography and politics. It is known that 32 people chose physics, 24 people chose chemistry, and 22 people chose biology, of which 18 people chose physics and chemistry, - 28 Discipline Network (Beijing) Co ., Ltd 10 people chose chemistry and biology, and 16 people chose physics and biology. Then the most students in the class choose Physics or Chemistry or Biology. 【Answer】44 【Analysis】According to the question, if 54 students are regarded as a set U , people who choose physics form a set A , people who choose chemistry form a set B , and people who choose biology form a set C , combined with the Venn diagram and the principle of repulsion, it can be seen that when card ( A  B  C ) the maximum value is taken card ( A  B  C ) , the verification can be obtained 【Explanation】Consider the 54 students as a set U , and choose the people who choose physics to form the A set, the people who choose chemistry to form the set B , and the people who choose biology to form the set C It is known from the meaning card (U ) = 54, card ( A ) = 32, card ( B ) = 24, card ( C ) = 22 of the title And card ( A  B ) = 18,card ( B  C ) = 10, card ( C  A ) = 16 , Then card ( A  B  C )  10 , composed card ( A  B  C ) = card ( A ) + card ( B ) + card ( C ) − card ( A  B ) − card ( B  C ) − card ( C  A ) + card ( A  B  C ) ， Available card ( A  B  C ) = 32 + 24 + 22 − 18 − 10 − 16 + card ( A  B  C )  34 + 10 = 44 , If and only if card ( A  B  C ) = 10 , i.e card ( A  B  C ) = 44 . Verification: At this time, the number of people in each area is as shown in the figure, and all the conditions of the question are met. - 29 Discipline Network (Beijing) Co ., Ltd Therefore, there are the most students in the class who choose Physics, Chemistry or Biology 44 So the answer is: 44 . 5. Known complete works U = R , set A =  x − 1  x  2 , B =  x 0  x  3 .find: (1) A  B and A B; (2) ðU ( A  B ) and A  ( ðU B ) 【Answer】(1), A  B =  x 0  x  2 A (2) ðU ( A  B ) =  x x  − 1 or x  3 , A B =  x − 1  x  3 ( ð B ) =  x −1  x  0 U 【Analysis】(1) It can be solved by the intersection and complement of the set; (2) It can be solved by intersection, union and complement set operations; [Explanation] (1) because A =  x − 1  x  2 , B =  x 0  x  3 , So A  B =  x 0  x  2 , B =  x − 1  x  3 A (2) From (1) obtainable: ðU ( A  B ) =  x x  − 1 or x  3 , From B =  x 0  x  3 , may obtain: ðU B =  x x  0 or x  3 , So A  ( ðU B ) =  x − 1  x  0 6. Set the whole set U = R , the set, A =  x 1  x  5 the set B =  x − 1 − 2 a  x  a − 2 (1) If a = 4 , ask A B ; A  ( ðU B ) (2) If B  A so, find a the range of values of real numbers 【Answer】(1) A  B =  x − 9  x  5 , A  ( ðU B ) =  x 2  x  5 ;  1  3 (2)  a a   . 【Analysis】(1) According to the concepts of union, intersection and complement, it is directly calculated. (2) According to the inclusion relationship between sets, column inequalities can be solved by solving inequalities. [Explanation] (1) because a = 4 , so B =  x − 9  x  2 Because A =  x 1  x  5 , so A  B =  x − 9  x  5 - 30 Discipline Network (Beijing) Co ., Ltd U = R Because, so ðU B =  x x  − 9 or x  2 , so A  ( ðU B ) =  x 2  x  5 ... (2) Because B  A (1) B =  When, satisfied B  A , at this time − 1 − 2 a  a − 2 , solved a  1 3 ;  − 1 − 2 a  1,  (2) When B   the time comes, if you want to be satisfied B  A , you  a − 2  5, will be solved a    − 1 − 2 a  a − 2,   1  3 To sum up, the a range of values for  a a   real numbers is 7. It is known that the complete set U is the set of real numbers, and the set A = { x∣− 2  x  3},B = { x∣2 m − 1  x  m + 2} (1) If m = −1 so, find the set C represented by the shaded part in the figure; (2) If A B = B so, find m the range of values of real numbers 【Answer】(1) { x∣− 3  x  − 2} (2) { m | − 1 2  m  1 or m  3 【Analysis】(1) According to the Venn diagram, the shadow part is a set ( ðU A )  B , and it is solved according to the complement set and intersection operation; (2) Transform to B  A , classify and discuss, list inequalities, and solve them 【Explanation】(1) The shaded part in the figure indicates that the set is ( ðU A )  B m = −1 When B = { x∣− 3  x  1} , ðU A = { x∣ x  − 2 ,or x  3} , So ( ðU A )  B = { x∣− 3  x  −2} ; (2) Because A B = B , so B  A , When B =  , 2m − 1  m + 2 ,Solution m  3 . 2m − 1  m + 2  When B   , if B  A , then there is  2 m − 1  − 2 m + 2  3  - 31 Discipline Network (Beijing) Co ., Ltd 1 Solution − ≤ m ≤ 1 , 2 To sum up, the m range of values for { m | − 1 2  m  1 real numbers is or m  3 8. Known A =  x − 2  x  5 , B =  x a + 1  x  2 a − 1 (1) If a = 3 the time comes, ask A  B ; A  ( ðR B ) (2) If B  ( ðR A ) =  the a value range is obtained 【Answer】(1), A  B =  4, 5  A (ð B ) = R R (2) ( − , 3 【Analysis】(1) When a = 3 the set is found, B the set can be obtained by using the definition of intersection A  B , and the set can be obtained by using the definition of complement set and union set A  ( ðR B ) ; (2) From the meaning of the topic, it can be seen that B  A B =  in the B   first case, the inequality about the real number can be obtained a ; In the second case, according to the inclusion relationship of the set, a group of inequalities about the real numbers can be obtained a , and the range of values of the a real numbers can be obtained by synthesis 【Explanation】(1) When a = 3 , B =  x a + 1  x  2 a − 1 =  4, 5  , A =  x − 2  x  5 ,,, A  B =  4, 5  , So ðR B = ( − , 4 ) ( 5, + ) , then A ( ð B ) = R ... R (2) Because B  ( ðR A ) =  , then B  A , At B =  that time, a + 1  2a − 1 the solution is in a  2 line with the topic;  a + 1  −2 When B   , that is a  2 , when, there is  2a − 1  5 , solve −3  a  3 , i.e 2  a  3 . In summary, a  3 that is, the a range of values of real numbers is ( − , 3 9. Known A =  x | m  x  2 m collections, B = { x | x  −5 or x  4} (1) When m = 3 the time comes, ask; A  ( ðR B ) (2) in (1); A  ðR B ② A B =  ； (3) A ( ð B ) = A Choose one of these three conditions, supplement it on the R horizontal line below, and solve it. If ___________, find m the range of values of the real numbers - 32 Discipline Network (Beijing) Co ., Ltd 【Answer】(1) A ( ð B ) =  x | −5  x  6 R (2)  m | m  2 . 【Analysis】(1) Find ðR B , the result can be obtained by using the concept of union (2) If (1) is selected, the A =  result can be obtained by analyzing the sum, and using the concept of subsets. If you choose (2), analyze A   and A =  , use A   it to get the result. If (3) is selected, A ( ) A  ð B = A results can be obtained from the analysis of (1). A  ðR B R 【Explanation】(1) When m = 3 , A =  x | 3  x  6 , Because B = { x | x  −5 or x  4} , so ðR B =  x | − 5  x  4 , ( ) Therefore A  ðR B =  x | −5  x  6 (2) If (1) is chosen A =  : when, m  2 m ,, m  0 A  ðR B is established  m  −5 the available −5  m  2 , the solution 0  m  2 , so A   m  0 When, A  ðR B by  2m  4 In summary, the m value range is m | m  2 If (2): when A =  , m  2 m , m  0 ,True A B= When A   , m  0 , From A  m  −5 B =  the obtainable  , the solution −5  m  2 , so 0  m  2 2m  4 In summary, the m value range is m | m  2 ( ) If you choose (3): You A  ðR B = A can get A  ðR B it A =  When, m  2 m , m  0 ,was A  ðR B established  m  −5 A   When, m  0 by A  ðR B the can be solved  −5  m  2 , so 0  m  2 2m  4 In summary, the m value range is m | m  2 - 33 Discipline Network (Beijing) Co ., Ltd B =  the