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Set - Practice Questions (38)

Question 1: 1. It is known that the set $A = \{ x \mid 1 < x + 2 , , 4 \} , B = \{ x \mid 0 , , x < 6 \}$, then ...

1. It is known that the set $A = \{ x \mid 1 < x + 2 , , 4 \} , B = \{ x \mid 0 , , x < 6 \}$, then $A \cup B =$

  • A. A. $\{ x \mid 0 , , x , , 2 \}$
  • B. B. $\{ x \mid - 1 < x < 6 \}$
  • C. C. $\{ x \mid - 1 < x < 0 \}$
  • D. D. $\{ x \mid 2 < x < 6 \}$

Answer: B

Solution: Because $A = \{ x \mid 1 < x + 2 , , 4 \} = \{ x \mid - 1 < x \leq 2 \} , ~ B = \{ x \mid 0 , , x < 6 \}$. Therefore $A \cup B = \{ x \mid - 1 < x < 6 \}$.

Question 2: 2. Let the set $A = \{ 1 , - 2,3,6,5 \} , B = \{ x \mid 1 \leq x < 5 \}$, then $A \cap B =$

2. Let the set $A = \{ 1 , - 2,3,6,5 \} , B = \{ x \mid 1 \leq x < 5 \}$, then $A \cap B =$

  • A. A. $\{ 1 , - 2 \}$
  • B. B. $\{ 3,1 \}$
  • C. C. $\{ 1,6 \}$
  • D. D. $\oslash$

Answer: B

Solution: According to the question, $A \cap B = \{ 3,1 \}$.

Question 3: 3. It is known that the set $A = \{ x \mid ( x + 1 ) ( 2 x - 1 ) < 0 \}$ and the set $B = \left\{ x ...

3. It is known that the set $A = \{ x \mid ( x + 1 ) ( 2 x - 1 ) < 0 \}$ and the set $B = \left\{ x \left\lvert \, y = \log _ { \frac { 1 } { 2 } } ( 1 - x ) \right. \right\}$ are equal to the set $A \cup B$.

  • A. A. $( - \infty , 1 )$
  • B. B. $( - \infty , - 1 )$
  • C. C. $( - 1,1 )$
  • D. D. $\left( - 1 , \frac { 1 } { 2 } \right)$

Answer: A

Solution: Solution: $\because ( x + 1 ) ( 2 x - 1 ) < 0 , \therefore - 1 < x < \frac { 1 } { 2 } , \therefore A = \left( - 1 , \frac { 1 } { 2 } \right)$, $\because y = \log _ { 1 } ( 1 - x )$ $\because y = \log _ { 1 } ( 1 - x )$ $\therefore A \cup B = ( - \infty , 1 )$

Question 4: 4. If there is only one element in the set $A = \left\{ x \mid a x ^ { 2 } - 2 x + 2 = 0 \right\}$, ...

4. If there is only one element in the set $A = \left\{ x \mid a x ^ { 2 } - 2 x + 2 = 0 \right\}$, then $a =$

  • A. A. 0
  • B. B. 1
  • C. C. 0 or $\frac { 1 } { 2 }$
  • D. D. 0 or 1

Answer: C

Solution: When $a = 0$, the equation $- 2 x + 2 = 0$ has only one solution $x = 1$, and the set $A = \{ 1 \}$ has only one element, hence $a = 0$. When $a \neq 0$ is $a \neq 0$, the set $A$ has only one element, hence $a x ^ { 2 } - 2 x + 2 = 0$ has two equal real roots. $\Delta = 4 - 8 a = 0$, which solves for $a = \frac { 1 } { 2 }$. So ${ } _ { a = 0 }$ or $a = \frac { 1 } { 2 }$.

Question 5: 5. Let the full set $U = \{ 1,2,3,4,5,6,7,8 \}$, the set $A = \{ 1,3,5,7,8 \}$, then $_ { \text {中元素...

5. Let the full set $U = \{ 1,2,3,4,5,6,7,8 \}$, the set $A = \{ 1,3,5,7,8 \}$, then $_ { \text {中元素个数为 } }$

  • A. A. 0
  • B. B. 3
  • C. C. 5
  • D. D. 8

Answer: B

Solution: From the definition of complementary set: ${ } ^ { \circlearrowright } A = \{ 2,4,6 \} , \therefore { } _ { U } A _ { \text {中元素的个数为 } } { } ^ { 3 }$.

Question 6: 6. It is known that $\mathrm { U } = \{ 1,2,3,4,5,6 \} , \mathrm { M } = \{ 1,3,4,5 \} , \mathrm { N...

6. It is known that $\mathrm { U } = \{ 1,2,3,4,5,6 \} , \mathrm { M } = \{ 1,3,4,5 \} , \mathrm { N } = \{ 2,4,5,6 \}$, the

  • A. A. $\mathrm { M } \cap \mathrm { N } = \{ 4,6 \}$
  • B. B. $\mathrm { M } \cup \mathrm { N } = \mathrm { U }$
  • C. C. $\left( \mathrm { C } _ { \mathrm { U } } \mathrm { N } \right) \cup \mathrm { M } = \mathrm { U }$
  • D. D. $\left( \mathrm { C } _ { \mathrm { U } } \mathrm { M } \right) \cap \mathrm { N } = \mathrm { N }$

Answer: C

Solution: Test analysis: According to the basic operations of the set can be concluded. SOLUTION : From the definition of complementary set, we can get $\mathrm { C } _ { \mathrm { U } } \mathrm { M } = \{ 2,6 \}$. Then $\left( \mathrm { C } _ { \mathrm { U } } \mathrm { M } \right) \cup \mathrm { M } = \{ 1,2,3,4,5,6 \} = \mathrm { U }$ , .

Question 7: 7. It is known that the set $A = \{ - 1,0,1 \} , B = \{ 1,2,3 \}$, then $A \cup B =$

7. It is known that the set $A = \{ - 1,0,1 \} , B = \{ 1,2,3 \}$, then $A \cup B =$

  • A. A. $\{ 1 \}$
  • B. B. $\{ - 1,0,1,2,3 \}$
  • C. C. $\{ - 1,0,1,1,2,3 \}$
  • D. D. $[ - 1,3 ]$

Answer: B

Solution: $\because ^ { A = \{ - 1,0,1 \} } , B = \{ 1,2,3 \}$ $\therefore A \cup B = \{ - 1,0,1,2,3 \}$ .

Question 8: 8. It is known that the set $A = \{ - 3 , - 1,2,4,6 \} , B = \{ x \mid - 0.5 < x < 2.5 \}$, then $A ...

8. It is known that the set $A = \{ - 3 , - 1,2,4,6 \} , B = \{ x \mid - 0.5 < x < 2.5 \}$, then $A \cap B =$

  • A. A. $\{ - 1,2,4 \}$
  • B. B. $\{ - 1,2 \}$
  • C. C. $\{ 2 \}$
  • D. D. $\{ 2,4 \}$

Answer: C

Solution: By collection $A = \{ - 3 , - 1,2,4,6 \} , B = \{ x \mid - 0.5 < x < 2.5 \}$ then $A \cap B = \{ 2 \}$.

Question 9: 9. It is known that the set $U = \{ 0,1,2,3,4 \}$ and the set $A = \{ 1,2 \} , B = \{ 2,3 \}$ then $...

9. It is known that the set $U = \{ 0,1,2,3,4 \}$ and the set $A = \{ 1,2 \} , B = \{ 2,3 \}$ then $A \cup$ $B$

  • A. A. $\{ 1 \}$
  • B. B. $\{ 0,2,4 \}$
  • C. C. $\{ 1,2,3 \}$
  • D. D. $\{ 0,1,2,4 \}$

Answer: D

Solution: From the question $\bar { \phi } _ { j } B = \{ 0,1,4 \}$, then $A \cup \bar { \phi } _ { j } B = \{ 0,1,2,4 \}$.

Question 10: 10. It is known that the set $A = \{ - 3 , - 1,0,2,4 \} , B = \{ * - 2 < x < 3 \}$, then $A \cap B =...

10. It is known that the set $A = \{ - 3 , - 1,0,2,4 \} , B = \{ * - 2 < x < 3 \}$, then $A \cap B =$

  • A. A. $\{ - 1,0 \}$
  • B. B. $\{ 0,2 \}$
  • C. C. $\{ - 1,0,2 \}$
  • D. D. $\{ 0,2,4 \}$

Answer: C

Solution: Because $A = \{ - 3 , - 1,0,2,4 \} , B = \{ * - 2 < x < 3 \}$ Therefore $A \cap B = \{ - 3 , - 1,0,2,4 \} \cap \{ * - 2 < x < 3 \} = \{ - 1,0,2 \}$.

Question 11: 11. It is known that the set $A = \left\{ x \mid 3 x ^ { 2 } - 2 x - 1 < 0 \right\} , B = \{ x \mid ...

11. It is known that the set $A = \left\{ x \mid 3 x ^ { 2 } - 2 x - 1 < 0 \right\} , B = \{ x \mid x \geq a \}$ , if $A \cap B = \varnothing$ , then the range of the real number $a$ is

  • A. A. $( 1 , + \infty )$
  • B. B. $\left( - \infty , - \frac { 1 } { 3 } \right)$
  • C. C. $[ 1 , + \infty )$
  • D. D. $\left( - \infty , - \frac { 1 } { 3 } \right]$

Answer: C

Solution:

Question 12: 12. If the full set $U = \{ 1,2,3,4 \} , A = \{ 1,2 \} , B = \{ 2,3 \}$ is known, then $A \cup \left...

12. If the full set $U = \{ 1,2,3,4 \} , A = \{ 1,2 \} , B = \{ 2,3 \}$ is known, then $A \cup \left( \mathrm { D } ^ { \prime } B \right) _ { \text {等于 } }$

  • A. A. $\{ 1,2,3 \}$
  • B. B. $\{ 1,2,4 \}$
  • C. C. $\{ 1 \}$
  • D. D. $\{ 4 \}$

Answer: B

Solution: From the question: $\bar { Q } ^ { B } B = \{ 1,4 \}$ , and $A = \{ 1,2 \}$ . So $A \cup \left( \partial ^ { , } B \right) = \{ 1,2,4 \}$ .

Question 13: 13. In order to meet the requirements of the new college entrance examination, a school has implemen...

13. In order to meet the requirements of the new college entrance examination, a school has implemented a selective classroom teaching mode. A senior class of 40 people, each choose physics, chemistry, biology in one or two subjects, no students at the same time to choose three subjects. Among them, 23 people chose physics, 18 people chose chemistry, 25 people chose biology, the number of students in the class chose two of the subjects is

  • A. A. 24
  • B. B. 25
  • C. C. 26
  • D. D. 27

Answer: C

Solution: Let there be $x$ who choose both Physics and Chemistry, $y$ who choose both Physics and Biology, and $z$ who choose both Biology and Chemistry. Therefore, from the question: 23-x-y + 18-x-z + 25-y-z + x + y + z = 40. So $x + y + z = 26$ Therefore, the number of students who chose two of the subjects in the class is 26 .

Question 14: 14. Let the set $A = \{ 0,2,4,6,10 \} , B = \left\{ x \mid x ^ { 2 } - 8 x + 12 \leq 0 , x \in \math...

14. Let the set $A = \{ 0,2,4,6,10 \} , B = \left\{ x \mid x ^ { 2 } - 8 x + 12 \leq 0 , x \in \mathrm {~N} _ { + } \right\}$, then $A \cap B =$

  • A. A. $\{ 2,3,4,5,6 \}$
  • B. B. $\{ 0,2,6 \}$
  • C. C. $\{ 0,2,4,5,6,10 \}$
  • D. D. $\{ 2,4,6 \}$

Answer: D

Solution: Set $A = \{ 0,2,4,6,10 \} , B = \left\{ x \mid x ^ { 2 } - 8 x + 12 \leq 0 , x \in \mathrm {~N} _ { + } \right\} = \left\{ x \mid 2 \leq x \leq 6 , x \in \mathrm {~N} _ { + } \right\}$ $= \{ 2,3,4,5,6 \}$ So $A \cap B = \{ 2,4,6 \}$.

Question 15: 15. It is known that the set $A = \left\{ x \in Z \mid x ^ { 2 } - 4 \leq 0 \right\} , B = \{ x \| x...

15. It is known that the set $A = \left\{ x \in Z \mid x ^ { 2 } - 4 \leq 0 \right\} , B = \{ x \| x \mid \leq 1 \}$, then $A \cap B =$

  • A. A. $\{ x \mid - 1 \leq x \leq 1 \}$
  • B. B. $\{ x \mid - 2 \leq x \leq 2 \}$
  • C. C. $\{ - 1,0,1 \}$
  • D. D. $\{ - 2 , - 1,0,1,2 \}$

Answer: C

Solution: SOLUTION: The set $\because$, $A = \left\{ x \in Z \mid x ^ { 2 } - 4 , , 0 \right\} = \{ - 2 , - 1,0,1,2 \}$ $B = \{ x | | x \mid , 1 \} = \{ x \mid - 1,1 , x , 1 \}$, $B = \{ x | | x \mid , 1 \} = \{ x \mid - 1,1 , x , 1 \}$, $B = \{ x | | x \mid , 1 \} = \{ x \mid - 1,1 , x , 1 \}$ $\therefore A \cap B = \{ - 1,0,1 \}$

Question 16: 16. "Nanjing Photo Studio", "The Little Monster of Wave Hill" and "Lychee in Chang'an" ranked among ...

16. "Nanjing Photo Studio", "The Little Monster of Wave Hill" and "Lychee in Chang'an" ranked among the top three films at the summer box office in China in 2025. A community survey of the community's part of the public's movie-going situation, the survey results show that: watched the "Nanjing Photo Studio" 63 people, watched the "Wave Hill Monster" 89 people, watched the "Lychee in Chang'an" 47 people, the three movies have watched 24 people, watched two of the movie 46 people, the three movies are not watched by 15 people. The total number of people surveyed is

  • A. A. 100 persons
  • B. B. 120 persons
  • C. C. 144 persons
  • D. D. 178 people

Answer: B

Solution: Represent the set relationship in the question by a Venn diagram, as shown in the figure. ![](/images/questions/set/image-001.jpg) It may be useful to represent the citizens who have watched Nanking Photo Studio, The Little Goblin of Wave Hill, and Lychee in Chang'an by the set $A , B , C$, and Be then $\operatorname { card } ( A ) = 63 , \operatorname { card } ( B ) = 89 , \operatorname { card } ( C ) = 47 , \operatorname { card } ( A \cap B \cap C ) = 24$. Let the total number of people be $n$, and the number of people who have watched Nanking Photo Studio and Wave Hill Gremlins be $x$. The number of people who watched "The Little Demon of Naniwa Mountain" and "Lychee in Chang'an" is $y$. The number of people who watched Nanking Photo Studio and Lychee in Chang'an is $z$, then $\operatorname { card } ( A \cap B ) = 24 + x , \operatorname { card } ( A \cap C ) = 24 + z , \operatorname { card } ( B \cap C ) = 24 + y , x + y + z = 46$. From the formula for the tolerant-exclusive relationship of the three sets $n - 15 = \operatorname { card } ( A ) + \operatorname { card } ( B ) + \operatorname { card } ( C )$ $- \operatorname { card } ( A \cap B ) - \operatorname { card } ( A \cap C ) - \operatorname { card } ( B \cap C ) + \operatorname { card } ( A \cap B \cap C )$ $= 63 + 89 + 47 - ( 24 + x ) - ( 24 + y ) - ( 24 + z ) + 24 = 151 - ( x + y + z ) = 105$ Solve for $n = 120$, so the total number of people surveyed is 120.

Question 17: 17. It is known that the set $A = \{ x \in N | | x - 1 \mid \leq 2 \} , B = \left\{ x \mid x = n ^ {...

17. It is known that the set $A = \{ x \in N | | x - 1 \mid \leq 2 \} , B = \left\{ x \mid x = n ^ { 2 } , n \in A \right\}$, then $A \cap B =$

  • A. A. $\{ 0 \}$
  • B. B. $\{ - 1,0 \}$
  • C. C. $\{ 0,1 \}$
  • D. D. $\{ - 1,0,1 \}$

Answer: C

Solution: Test analysis: Because $$ A = \{ x \in N \| x - 1 \mid \leq 2 \} = \{ x \in N \mid - 1 \leq x \leq 3 \} = \{ 0,1,2,3 \} $$ $$ B = \left\{ x \mid x = n ^ { 2 } , n \in A \right\} = \{ 0,1,4,9 \} $$ , so $A \cap B = \{ 0,1 \} \quad$; Therefore, choose C. Points: 1. set representation; 2. set operations.

Question 18: 18. If the set $M = \{ x \mid 2 x > 3 \} , N = \{ 1,2,3,4 \}$, then $M \cap N =$

18. If the set $M = \{ x \mid 2 x > 3 \} , N = \{ 1,2,3,4 \}$, then $M \cap N =$

  • A. A. $\{ 1,2 \}$
  • B. B. $\{ 3,4 \}$
  • C. C. $\left\{ x \mid 1 < x < 5 , x \in \mathrm {~N} ^ { * } \right\}$
  • D. D. $\left\{ x \mid 1 \leq x \leq 4 , x \in \mathrm {~N} ^ { * } \right\}$

Answer: C

Solution: From the question $M = \{ x \mid 2 x > 3 \} = \left\{ x | x \rangle \frac { 3 } { 2 } \right\} , N = \{ 1,2,3,4 \}$ , the Therefore $M \cap N = \{ 2,3,4 \} = \left\{ x \mid 1 < x < 5 , x \in \mathrm {~N} ^ { * } \right\}$ , .

Question 19: 19. It is known that the set $A = \left\{ a - 2 , a ^ { 2 } + 4 a , 10 \right\}$ , if $- 3 \in A$ , ...

19. It is known that the set $A = \left\{ a - 2 , a ^ { 2 } + 4 a , 10 \right\}$ , if $- 3 \in A$ , then the value of the real number $a$ is

  • A. A. - 3
  • B. B. 1
  • C. C. - 3 or - 1
  • D. D. insoluble (i.e. unable to solve)

Answer: A

Solution: Because $- 3 \in A$. so $a - 2 = - 3$ or $a ^ { 2 } + 4 a = - 3$, the $a ^ { 2 } + 4 a = - 3$ When $a - 2 = - 3$ is $a = - 1$, $A = \{ - 3 , - 3,10 \}$, does not satisfy the reciprocity of the elements of the set. Therefore, $a = - 1$ does not match the meaning of the question and is discarded; When $a ^ { 2 } + 4 a = - 3$ is $a = - 1$ (rounded) or $a = - 3$, $A = \{ - 5 , - 3,10 \}$, which conforms to the meaning. Therefore, the value of $a$ is - 3.

Question 20: 20. It is known that the set $P = \left\{ x \mid x ^ { 2 } - 2 x - 3 < 0 \right\} , Q = \left\{ x \m...

20. It is known that the set $P = \left\{ x \mid x ^ { 2 } - 2 x - 3 < 0 \right\} , Q = \left\{ x \mid 2 ^ { x } > 1 \right\}$, then $P \cap Q =$

  • A. A. $\{ x \mid x > - 1 \}$
  • B. B. $\{ x \mid x < - 1 \}$
  • C. C. $\{ x \mid 0 < x < 3 \}$
  • D. D. $\{ x \mid - 1 < x < 0 \}$

Answer: C

Solution: Because the set $P = \left\{ x \mid x ^ { 2 } - 2 x - 3 < 0 \right\} = \{ x \mid - 1 < x < 3 \} , Q = \left\{ x \mid 2 ^ { x } > 1 \right\} = \{ x \mid x > 0 \}$ Therefore $P \cap Q = \{ x \mid 0 < x < 3 \}$ , .

Question 21: 21. Let the set $A = \{ - 2 , - 1,0,1,2 \} , B = \{ x \mid x \geq 0 \}$ be $A \cap \left( \mathrm { ...

21. Let the set $A = \{ - 2 , - 1,0,1,2 \} , B = \{ x \mid x \geq 0 \}$ be $A \cap \left( \mathrm { C } _ { \mathrm { R } } \mathrm { B } \right) =$.

  • A. A. $\{ 12 \}$
  • B. B. $\{ - 2 , - 1 \}$
  • C. C. $\{ 012 \}$
  • D. D. $\{ - 2,10 \}$

Answer: B

Solution: From the question, the set $A = \{ - 2 , - 1,0,1,2 \} , C _ { U } B = \{ x \mid < 0 \}$, so $A \cap \left( C _ { R } B \right) = \{ - 2 , - 1 \}$ , so choose B .

Question 22: 22. It is known that the set $A = \{ 1,2,3,4,5 \} , B = \{ x \mid - x < - 3 \}$ , then $A \cap B =$

22. It is known that the set $A = \{ 1,2,3,4,5 \} , B = \{ x \mid - x < - 3 \}$ , then $A \cap B =$

  • A. A. $\{ 5 \}$
  • B. B. $\{ 1,2 \}$
  • C. C. $\{ 3,4,5 \}$
  • D. D. $\{ 4,5 \}$

Answer: D

Solution: SOLUTION: Because $\because B = \{ x \mid - x < - 3 \} \therefore B = \{ x \mid x > 3 \} , \because A = \{ 1,2,3,4,5 \}$ $\therefore A \cap B = \{ 4,5 \}$.

Question 23: 23. It is known that the set $M = \{ x \mid x \geq 0 \} , N = \{ x \mid ( x + 1 ) ( x - 3 ) < 0 \}$,...

23. It is known that the set $M = \{ x \mid x \geq 0 \} , N = \{ x \mid ( x + 1 ) ( x - 3 ) < 0 \}$, then $M \cup N =$

  • A. A. $( - 1,3 )$
  • B. B. $( - 1 , + \infty )$
  • C. C. $( 0,3 )$
  • D. D. $[ 0,3 )$

Answer: B

Solution: $N = \{ x \mid ( x + 1 ) ( x + 3 ) < 0 \} = \{ x \mid - 1 < x < 3 \} , M = \{ x \mid x \geq 0 \}$ so $M \cup N = ( - 1 , + \infty )$

Question 24: 24. If $a , b \in R$ is known, and if $\left\{ a , \frac { b } { a } , 1 \right\} = \left\{ a ^ { 2 ...

24. If $a , b \in R$ is known, and if $\left\{ a , \frac { b } { a } , 1 \right\} = \left\{ a ^ { 2 } , a + b , 0 \right\}$ is known, then the value of $a ^ { 2019 } + b ^ { 2019 }$ is ().

  • A. A. 1
  • B. B. 0
  • C. C. - 1
  • D. D. $\pm 1$

Answer: C

Solution: $\because \frac { b } { a } , \therefore a \neq 0$ $\because \left\{ a , \frac { b } { a } , 1 \right\} = \left\{ a ^ { 2 } , a + b , 0 \right\}$ $\therefore \frac { b } { a } = 0$, i.e. $b = 0$. $\therefore \{ a , 0,1 \} = \left\{ a ^ { 2 } , a , 0 \right\}$ $\therefore$ when $\left\{ \begin{array} { l } a ^ { 2 } = 1 \\ a = a \text { 时,} a = - 1 \text { 或 } a = 1 \end{array} \right.$, the When $a = 1$, the set $\{ 1,0,1 \}$ is obtained, which does not satisfy the reciprocity of the elements, so it is rounded off. When $\left\{ \begin{array} { l } a = 1 \\ a ^ { 2 } = a \text { 时,} a = 1 \text { ,即得集合 } \{ 1,0,1 \} \text { ,不符合元素的互异性,故舍去,} \end{array} \right.$ In summary, $a = - 1 , b = 0$ $\therefore a ^ { 2019 } + b ^ { 2019 } = ( - 1 ) ^ { 2019 } + 0 ^ { 2019 } = - 1$.

Question 25: 26. The following groups of objects do not form a set All of the students in the centimeter

26. The following groups of objects do not form a set All of the students in the centimeter

  • A. A. Non-negative real numbers up to 20
  • B. B. Solution of the equation $x ^ { 2 } - 9 = 0$ in the real number range
  • C. C. The whole of the approximation of $\sqrt { 3 }$.
  • D. D. Linchuan Experimental School 2017 School Height Over 170

Answer: C

Solution: A , B , and D are all sets, and the whole of the approximation of $\because \sqrt { 3 }$ does not satisfy determinism and is not a set;

Question 26: 27. If the set $M = \{ - 11 \} , N = \left\{ x \left\lvert \, \frac { 1 } { 2 } < 2 ^ { x + 1 } < 4 ...

27. If the set $M = \{ - 11 \} , N = \left\{ x \left\lvert \, \frac { 1 } { 2 } < 2 ^ { x + 1 } < 4 \right. , x \in Z \right\}$ is known, then $M \cup N =$.

  • A. A. $\{ - 11 \}$
  • B. B. $\{ - 1 \}$
  • C. C. $\{ 0 \}$
  • D. D. $\{ - 10,1 \}$

Answer: D

Solution: $N = \left\{ x \left\lvert \, \frac { 1 } { 2 } < 2 ^ { x + 1 } < 4 \right. , x \in Z \right\}$ is neutralized to: $N = \{ - 1,0 \}$ $M \cup N = \{ - 10,1 \}$

Question 27: 28. Let the set $A = \left\{ x \mid \log _ { 0.5 } ( x - 1 ) > 0 \right\} , B = \left\{ x \mid 2 ^ {...

28. Let the set $A = \left\{ x \mid \log _ { 0.5 } ( x - 1 ) > 0 \right\} , B = \left\{ x \mid 2 ^ { x } < 4 \right\}$, then ( )

  • A. A. $A = B$
  • B. B. $A \cap B = \oslash$
  • C. C. $A \cap B = B$
  • D. D. $A \cup B = B$

Answer: D

Solution: $\log _ { 0.5 } ( x - 1 ) > 0$, which is $\log _ { 0.5 } ( x - 1 ) > \log _ { 0.5 } 1$, then $0 < x - 1 < 1$, which solves for $1 < x < 2$, so $A = \{ x \mid 1 < x < 2 \} , ~ B = \left\{ x \mid 2 ^ { x } < 2 ^ { 2 } \right\} = \{ x \mid x < 2 \}$, and thus $A \subseteq B$, and thus $A \cup B = B$. 5]] and thus $A \cup B = B$.

Question 28: 29. If the set $A = \{ x \| x - 4 < 3 \} , B = \{ x \in N \quad x < 4 \}$ is known, then $A \cap B =...

29. If the set $A = \{ x \| x - 4 < 3 \} , B = \{ x \in N \quad x < 4 \}$ is known, then $A \cap B =$ ( )

  • A. A. $\{ 0,1,2,3 \}$
  • B. B. $\{ 1,2,3 \}$
  • C. C. $\{ \nmid 0 < x < 3 \}$
  • D. D. $\{ \nmid 1 \leq x < 4 \}$

Answer: A

Solution: $\because \mid x - 4 < 3$ is $^ { - 3 < x - 2 < 3 \Rightarrow - 1 < x < 5 }$, so $A = \{ x \mid - 1 < x < 5 \}$, $B = \{ 0,1,2,3 \}$, so $A \cap B = \{ 0,1,2,3 \}$.

Question 29: 31. If the set $A = \{ x \mid x > 0 \} , B = \{ x \mid - 1 < x < 2 \}$ is known, then $A \cap B =$ (...

31. If the set $A = \{ x \mid x > 0 \} , B = \{ x \mid - 1 < x < 2 \}$ is known, then $A \cap B =$ ()

  • A. A. $\{ x \mid x < 2 \}$
  • B. B. $\{ x \mid 0 < x < 2 \}$
  • C. C. $\{ x \mid 1 < x < 2 \}$
  • D. D. $\{ x \mid - 1 < x < 2 \}$

Answer: B

Solution: The set $A = \{ x \mid x > 0 \} , B = \{ x \mid - 1 < x < 2 \}$, by the definition of intersection $A \cap B = \{ x \mid 0 < x < 2 \}$.

Question 30: 32. Let the set $A = \{ x \mid ( x + 3 ) ( x - 2 ) < 0 \} , B = \{ x \mid - 2 < 2 x < 8 \}$ be $A \c...

32. Let the set $A = \{ x \mid ( x + 3 ) ( x - 2 ) < 0 \} , B = \{ x \mid - 2 < 2 x < 8 \}$ be $A \cup B =$, then $A \cup B =$

  • A. A. $\{ x \mid - 1 < x < 2 \}$
  • B. B. $\{ x \mid - 3 < x < 8 \}$
  • C. C. $\{ x \mid - 3 < x < 2 \}$
  • D. D. $\{ x \mid - 3 < x < 4 \}$

Answer: D

Solution: $A = \{ x \mid ( x + 3 ) ( x - 2 ) < 0 \} = \{ x \mid - 3 < x < 2 \}$ $B = \{ x \mid - 2 < 2 x < 8 \} = \{ x \mid - 1 < x < 4 \}$, so $A \cup B = \{ x \mid - 3 < x < 4 \}$.

Question 31: 33. It is known that the set $A = \{ 1,2,3,4,5 \} , B = \{ x \mid ( x - 2 ) ( x - 5 ) < 0 \}$, then ...

33. It is known that the set $A = \{ 1,2,3,4,5 \} , B = \{ x \mid ( x - 2 ) ( x - 5 ) < 0 \}$, then $A \cap B =$

  • A. A. $\{ 1,2,3,4 \}$
  • B. B. $\{ 3,4 \}$
  • C. C. $\{ 2,3,4 \}$
  • D. D. $\{ 4,5 \}$

Answer: B

Solution: $\because \mathrm { A } = \{ 1,2,3,4,5 \} , ~ \mathrm {~B} = \{ \mathrm { x } \mid ( \mathrm { x } - 2 ) \quad ( \mathrm { x } - 5 ) < 0 \} = \{ \mathrm { x } \mid 2 < \mathrm { x } < 5 \}$. and $3,4 \in B$. $\therefore A \cap B = \{ 3,4 \}$.

Question 32: 34. The function $f ( x ) = \cos \pi x , g ( x ) = \mathrm { e } ^ { a x } - a + \frac { 1 } { 2 } (...

34. The function $f ( x ) = \cos \pi x , g ( x ) = \mathrm { e } ^ { a x } - a + \frac { 1 } { 2 } ( a \neq 0 )$ is known. If ${ } ^ { \exists x _ { 1 } , x _ { 2 } \in [ 0,1 ] }$ such that $f \left( x _ { 1 } \right) = g \left( x _ { 2 } \right)$ , then the real number ${ } ^ { a }$ is in the range ( )

  • A. A. $\left[ - \frac { 1 } { 2 } 0 \right)$
  • B. B. $\left[ \frac { 1 } { 2 } , + \infty \right)$
  • C. C. $( - \infty , 0 ) \cup \left[ \frac { 1 } { 2 } , + \infty \right)$
  • D. D. $\left[ - \frac { 1 } { 2 } , 0 \right) \cup \left( 0 , \frac { 1 } { 2 } \right]$

Answer: B

Solution: When $x \in [ 0,1 ]$, $\pi x \in [ 0 , \pi ] , f ( x ) = \cos \pi x \in [ - 1,1 ]$, $x \in [ 0,1 ] , a > 0$, $g ( x ) \in \left[ \frac { 3 } { 2 } - a , \mathrm { e } ^ { a } - a + \frac { 1 } { 2 } \right]$, $g ( x ) \in \left[ \frac { 3 } { 2 } - a , \mathrm { e } ^ { a } - a + \frac { 1 } { 2 } \right]$ When $x \in [ 0,1 ] , a > 0$, $g ( x ) \in \left[ \frac { 3 } { 2 } - a , \mathrm { e } ^ { a } - a + \frac { 1 } { 2 } \right]$, $g ( x ) \in \left[ \frac { 3 } { 2 } - a , \mathrm { e } ^ { a } - a + \frac { 1 } { 2 } \right]$ When $x \in [ 0,1 ] , a < 0$, $g ( x ) \in \left[ \mathrm { e } ^ { a } - a + \frac { 1 } { 2 } , \frac { 3 } { 2 } - a \right]$. such that $h ( a ) = \mathrm { e } ^ { a } - a + \frac { 1 } { 2 }$, then $h ^ { \prime } ( a ) = e ^ { a } - 1 , ~ h ^ { \prime } ( 0 ) = 0$, the When $a > 0$, $h ^ { \prime } ( a ) > 0 , h ( a ) > \mathrm { e } ^ { 0 } - 0 + \frac { 1 } { 2 } = \frac { 3 } { 2 } > 1$; When $a < 0$, $h ^ { \prime } ( a ) < 0 , h ( a ) > \mathrm { e } ^ { 0 } - 0 + \frac { 1 } { 2 } = \frac { 3 } { 2 } > 1$; In summary, $\mathrm { e } ^ { a } - a + \frac { 1 } { 2 } > 1$ ; From the question, we have that the intersection of the value domains of the two functions is non-empty. So $\left\{ \begin{array} { l } a > 0 \\ \frac { 3 } { 2 } - a \leq 1 \end{array} \right.$ , solve $a \geq \frac { 1 } { 2 }$ .

Question 33: 35. Let $M , P$ be two non-empty sets, and define the difference between $M$ and $P$ as $M - P = \{ ...

35. Let $M , P$ be two non-empty sets, and define the difference between $M$ and $P$ as $M - P = \{ x \mid x \in M$ and $x \notin P \}$, and $P - ( M - P )$ and $P - ( M - P )$ as $P - ( M - P )$. FORMULA_5]] is equal to

  • A. A. $P$
  • B. B. $M \cap P$
  • C. C. $M \cup P$
  • D. D. $M$

Answer: A

Solution: When $M \cap P \neq \varnothing$ is shaded in the figure below by the Wayne diagram, $M - P$, then $P - ( M - P ) _ { \text {obviously } }$ is $P$. ![](/images/questions/set/image-002.jpg) $M - P = M$ when $M \cap P = \varnothing$ then $P - ( M - P ) = P - M = \{ x \mid x \in P$ and $x \notin M \} = P$

Question 34: 36. It is known that the set $A = \{ x | | x + 1 \mid < 1 \} , B = \left\{ x \left\lvert \, \left( \...

36. It is known that the set $A = \{ x | | x + 1 \mid < 1 \} , B = \left\{ x \left\lvert \, \left( \frac { 1 } { 2 } \right) x - 2 \geq 0 \right. \right\}$, then $A \cap \left( C _ { R } B \right) =$

  • A. A. $( - 2 , - 1 )$
  • B. B. $( - 2 , - 1 ]$
  • C. C. $( - 1,0 )$
  • D. D. $[ - 1,0 )$

Answer: C

Solution: $| \mathrm { x } + 1 | < 1$ , so $- 1 < \mathrm { x } + 1 < 1$ , so $- 2 < \mathrm { x } < 0$ , so $\mathrm { A } = ( - 2,0 )$ $\left( \frac { 1 } { 2 } \right) ^ { x } - 2 \geq 0 , \therefore 2 ^ { - x } \geq 2 , \therefore - x \geq 1 , \therefore x \leq - 1 , \therefore B = ( - \infty , - 1 ]$. $\therefore C _ { R } B = ( - 1 , + \infty )$ so $A \cap \left( C _ { R } B \right) = ( - 1,0 )$ . Therefore, the answer is C [点睛](1)本题主要考查不等式的解法,考查集的化简和补集交集运算,意意考查学生对这些知识的掌握水平和计算能力. (2) The operation of the set should pay attention to the flexible use of Venn diagrams and axes, in general, the operation of the finite set of Venn diagrams to analyze the operation of the infinite set of axes, which in fact is the combination of the idea of the specific use of the combination of mathematical and physical.

Question 35: A class of students took three math tests, the first time 8 students scored perfect scores, the seco...

A class of students took three math tests, the first time 8 students scored perfect scores, the second time 10 students scored perfect scores, and the third time 12 students scored perfect scores, knowing that 5 students scored perfect scores on the first two tests, and that 15 students scored perfect scores on at least one of the three tests, and that if there are at least $n$ students scoring perfect scores on the last two tests, then $n$ is the value of $n$. [INLINE_FORMULA_1]] is the value of

  • A. A. 7
  • B. B. 8
  • C. C. 9
  • D. D. 10

Answer: A

Solution: As shown, because the distribution of the 15 students who scored perfect on at least one of the three tests: ![](/images/questions/set/image-003.jpg) Since 8 students scored perfect on the first test, 10 students scored perfect on the second test, and Since 8 students scored full marks the first time, 10 students scored full marks the second time, and 5 students scored full marks the first two times, the distribution of students who scored full marks at least once in the first two times is as follows FORMULA_0]], and because 15 students scored perfect on at least one of the three tests, and 12 students scored perfect on the third test, 2 of the 12 students who scored perfect on the third test scored perfect only on the third test, and the remaining 10 students scored perfect on either the first or second test, and $10 - 3 = 7$ was the first time that $10 - 3 = 7$ scored perfect. FORMULA_1]], so at least 7 of the 12 students who scored full marks on the second and third times.

Question 36: 38. For the sets $A$ and $B$, let $A + B = \{ x \mid x = a + b , a \in A , b \in B \}$ be $A + B = \...

38. For the sets $A$ and $B$, let $A + B = \{ x \mid x = a + b , a \in A , b \in B \}$ be $A + B = \{ x \mid x = a + b , a \in A , b \in B \}$, and if $S = \{ x \mid x = 2 k , k \in Z \}$ $T = \{ x \mid x = 2 k + 1 , x \in Z \}$, then $S + T =$ \{ x \mid x = 4 k + 1 , k \in Z \} $$

  • A. A. Set of integers $Z$
  • B. B. S
  • C. C. $T$
  • D. D. $$

Answer: C

Solution: From the question let the elements in the set $S$ be: $2 k , k \in Z$, the elements in the set $T$ be: $2 m + 1 , m \in Z$, and the elements in the set $S + T$ be: $2 k + 2 m + 1 = 2 ( k + m ) + 1 , k , m \in Z$, and the elements in $2 k + 2 m + 1 = 2 ( k + m ) + 1 , k , m \in Z$ be: $2 k + 2 m + 1 = 2 ( k + m ) + 1 , k , m \in Z$, and $S + T = T$ be [INLINE_FORMULA_5]]. Cite the knowable set $S + T = T$.

Question 37: 39. Let the set $M = \left\{ x \mid x ^ { 2 } \leq 4 \right\} , N = \left\{ x \mid \log _ { 2 } x \l...

39. Let the set $M = \left\{ x \mid x ^ { 2 } \leq 4 \right\} , N = \left\{ x \mid \log _ { 2 } x \leq 1 \right\}$ be $M \cap N =$, then $M \cap N =$

  • A. A. $[ - 2,2 ]$
  • B. B. $\{ 2 \}$
  • C. C. $( 0,2 ]$
  • D. D. $( - \infty , 2 ]$

Answer: C

Solution: Test analysis: Solve the inequality $x ^ { 2 } \leq 4$ to get $- 2 \leq x \leq 2$, i.e. the set $\mathrm { A } = \{ x \mid - 2 \leq x \leq 2 \}$, solve the inequality $\log x \leq 1$ to get $0 < x \leq 2$, which is the set $\mathrm { B } =$ $\{ x \mid 0 < x \leq 2 \}$ , which is the set $A \cap B = \{ x \mid 0 < x \leq 2 \} = ( 0,2 ]$ according to the operation of sets, so the choice of this question is C. Point: The operations of sets. $40 . \mathrm { C }$ [Knowledge Points] Solve quadratic inequalities without parameters, the concept and operations of intersection, the concept and operations of complementary sets, and solve inequalities by the monotonicity of logarithmic functions. [Analysis] Solve the inequality to simplify the set, and then use the complementary set, the intersection of the definition of the solution that is obtained. [Detailed Solution] Solve the inequality $| x - 2 | > 1$, get $x < 1$ or $x > 3$, that is, $A = ( - \infty , 1 ) \cup ( 3 , + \infty ) , ~ \Phi _ { \mathrm { k } } A = [ 1,3 ]$, and $A = ( - \infty , 1 ) \cup ( 3 , + \infty ) , ~ \Phi _ { \mathrm { k } } A = [ 1,3 ]$, and $A = ( - \infty , 1 ) \cup ( 3 , + \infty ) , ~ \Phi _ { \mathrm { k } } A = [ 1,3 ]$. Solve the inequality ${ } ^ { \log _ { 2 } x < 1 }$ to get $0 < x < 2$, then $B = ( 0,2 )$, and $B = ( 0,2 )$ to get $B = ( 0,2 )$. So $\left( { \underset { \mathbf { q } } { \mathbf { q } } } _ { \mathbf { R } } A \right) \cap B = [ 1,2 )$.

Question 38: 40. Let the set $A = \{ x | | x - 2 \mid > 1 \} , B = \left\{ x \mid \log _ { 2 } x < 1 \right\}$, t...

40. Let the set $A = \{ x | | x - 2 \mid > 1 \} , B = \left\{ x \mid \log _ { 2 } x < 1 \right\}$, then $\left( \mathrm { c } _ { \mathrm { k } } A \right) \cap B =$ High School Mathematics Assignment, October 29, 2025

  • A. A. $( 0,1 )$
  • B. B. $( 0,2 ) \cup ( 3 , + \infty )$
  • C. C. $[ 1,2 )$
  • D. D. $( 1,2 ) \cup ( 3 , + \infty )$

Answer: C

Solution:
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Set

集合

38 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

A set is a basic concept in mathematics that describes a group of objects and mainly involves operations such as element, subset, union, intersection, and complement. In the CSCA exam, set questions are often combined with inequalities and function definition domains, requiring candidates to accurately understand the meaning of symbols and skillfully perform set operations. Mastering the representation of sets and operations is the key to solving the questions.

Questions:38

Key Points

  • 1Representation of sets (enumeration, description)
  • 2Basic operations on sets (union, intersection, complement)
  • 3Combined applications of sets with inequalities and equations
  • 4Solve set problems using axes or Venn diagrams

Study Tips

Practice by drawing more axes to represent the range of sets, paying special attention to the rounding of endpoint values, and familiarizing yourself with the meaning of symbols such as ∈, ⊆, ∪, and ∩.

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