NCERT Solutions For Class 11 Maths: Sets

Summary of the Chapter

This chapter introduces the fundamental concept of Sets in mathematics. A set is defined as a well-defined collection of distinct objects. The objects in a set are called its elements or members.



Sets can be represented in two main ways:

1. Roster or Tabular Form: All elements are listed, separated by commas, within braces $\{ \}$. Example: $A = \{1, 2, 3, 4, 5\}$.
2. Set-builder Form: All elements possess a single common property, written as $\{ x : P(x) \}$, where $P(x)$ is the property. Example: $B = \{ x : x \text{ is a vowel in the English alphabet} \}$.
   
   
   
   The chapter discusses various types of sets:

• Empty Set ($\emptyset$ or $\{ \}$): A set containing no elements.
• Finite Set: A set with a definite (countable) number of elements.
• Infinite Set: A set with an unlimited number of elements.
• Equal Sets: Two sets $A$ and $B$ are equal ($A = B$) if they have exactly the same elements.
• Subsets ($\subseteq$): Set $A$ is a subset of set $B$ if every element of $A$ is also an element of $B$. $A \subset B$ denotes a proper subset ($A \subseteq B$ and $A \neq B$).
• Power Set ($P(A)$): The set of all subsets of a set $A$. If $n(A) = m$, then $n(P(A)) = 2^m$.
• Universal Set ($U$): A basic set that contains all relevant sets in a particular context.
  
  
  
  Venn diagrams are introduced as a way to visually represent relationships between sets using circles within a rectangle (representing the universal set).
  
  
  
  Key operations on sets are defined:
• Union ($A \cup B$): The set of all elements which are in $A$ or in $B$ or in both. $A \cup B = \{ x : x \in A \text{ or } x \in B \}$.
• Intersection ($A \cap B$): The set of all elements which are common to both $A$ and $B$. $A \cap B = \{ x : x \in A \text{ and } x \in B \}$. If $A \cap B = \emptyset$, then $A$ and $B$ are disjoint sets.
• Difference ($A - B$): The set of all elements which are in $A$ but not in $B$. $A - B = \{ x : x \in A \text{ and } x \notin B \}$.
• Complement ($A'$ or $A^c$): The set of all elements in the universal set $U$ that are not in $A$. $A' = U - A = \{ x : x \in U \text{ and } x \notin A \}$.
  
  
  
  Important properties like commutative laws, associative laws, distributive laws, and De Morgan's laws ($(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$) are discussed.
  
  
  
  Finally, the chapter covers practical problems involving formulas related to the cardinality (number of elements) of sets, particularly $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ and $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$.

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NCERT Textbook Questions and Answers

Exercise 1.1

Question 1: Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter.
(ix) A collection of most dangerous animals of the world.

Answer-
(i) This is a set. The collection is well-defined: {January, June, July}.

(ii) This is not a set. "Most talented" is subjective and not well-defined.

(iii) This is not a set. "Best-cricket batsmen" is subjective and not well-defined.

(iv) This is a set. It is well-defined for a specific class.

(v) This is a set. It is well-defined: {1, 2, 3, ..., 99}.

(vi) This is a set. It is well-defined as Munshi Prem Chand's novels can be listed.

(vii) This is a set. It is well-defined: {..., -4, -2, 0, 2, 4, ...}.

(viii) This is a set. It is well-defined as the questions in the chapter are specific.

(ix) This is not a set. "Most dangerous" is subjective and not well-defined.

Question 2: Let $A = \{1, 2, 3, 4, 5, 6\}$. Insert the appropriate symbol $\in$ or $\notin$ in the blank spaces:
(i) $5 \_\_\_ A$
(ii) $8 \_\_\_ A$
(iii) $0 \_\_\_ A$
(iv) $4 \_\_\_ A$
(v) $2 \_\_\_ A$
(vi) $10 \_\_\_ A$

Answer-
(i) $5 \in A$

(ii) $8 \notin A$

(iii) $0 \notin A$

(iv) $4 \in A$

(v) $2 \in A$

(vi) $10 \notin A$

Question 3: Write the following sets in roster form:
(i) $A = \{x : x \text{ is an integer and } -3 \le x < 7 \}$
(ii) $B = \{x : x \text{ is a natural number less than } 6 \}$
(iii) $C = \{x : x \text{ is a two-digit natural number such that the sum of its digits is } 8 \}$
(iv) $D = \{x : x \text{ is a prime number which is a divisor of } 60 \}$
(v) $E = \text{The set of all letters in the word TRIGONOMETRY}$
(vi) $F = \text{The set of all letters in the word BETTER}$

Answer-
(i) $A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$

(ii) $B = \{1, 2, 3, 4, 5\}$

(iii) $C = \{17, 26, 35, 44, 53, 62, 71, 80\}$

(iv) $D = \{2, 3, 5\}$

(v) $E = \{T, R, I, G, O, N, M, E, Y\}$

(vi) $F = \{B, E, T, R\}$

Question 4: Write the following sets in the set-builder form:
(i) $\{3, 6, 9, 12\}$
(ii) $\{2, 4, 8, 16, 32\}$
(iii) $\{5, 25, 125, 625\}$
(iv) $\{2, 4, 6, ...\}$
(v) $\{1, 4, 9, ..., 100\}$

Answer-
(i) $\{ x : x = 3n, n \in N \text{ and } 1 \le n \le 4 \}$

(ii) $\{ x : x = 2^n, n \in N \text{ and } 1 \le n \le 5 \}$

(iii) $\{ x : x = 5^n, n \in N \text{ and } 1 \le n \le 4 \}$

(iv) $\{ x : x \text{ is an even natural number} \}$

(v) $\{ x : x = n^2, n \in N \text{ and } 1 \le n \le 10 \}$

Question 5: List all the elements of the following sets:
(i) $A = \{x : x \text{ is an odd natural number} \}$
(ii) $B = \{x : x \text{ is an integer, } -1/2 < x < 9/2 \}$
(iii) $C = \{x : x \text{ is an integer, } x^2 \le 4 \}$
(iv) $D = \{x : x \text{ is a letter in the word "LOYAL"} \}$
(v) $E = \{x : x \text{ is a month of a year not having 31 days} \}$
(vi) $F = \{x : x \text{ is a consonant in the English alphabet which precedes } k \}$

Answer-
(i) $A = \{1, 3, 5, 7, ...\}$

(ii) $B = \{0, 1, 2, 3, 4\}$

(iii) $C = \{-2, -1, 0, 1, 2\}$

(iv) $D = \{L, O, Y, A\}$

(v) $E = \{\text{February, April, June, September, November}\}$

(vi) $F = \{b, c, d, f, g, h, j\}$

Question 6: Match each of the set on the left in the roster form with the same set on the right described in set-builder form:
(i) $\{1, 2, 3, 6\}$ (a) $\{x : x \text{ is a prime number and a divisor of } 6 \}$
(ii) $\{2, 3\}$ (b) $\{x : x \text{ is an odd natural number less than } 10 \}$
(iii) $\{M, A, T, H, E, I, C, S\}$ (c) $\{x : x \text{ is a natural number and divisor of } 6 \}$
(iv) $\{1, 3, 5, 7, 9\}$ (d) $\{x : x \text{ is a letter of the word MATHEMATICS} \}$

Answer-

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Exercise 1.2

Question 1: Which of the following are examples of the null set?
(i) Set of odd natural numbers divisible by 2.
(ii) Set of even prime numbers.
(iii) $\{x : x \text{ is a natural number, } x < 5 \text{ and } x > 7 \}$
(iv) $\{y : y \text{ is a point common to any two parallel lines} \}$

Answer-
(i) No odd natural number is divisible by 2. This is a null set ($\emptyset$).

(ii) The number 2 is an even prime number. This set is $\{2\}$, which is not a null set.

(iii) There is no natural number that is both less than 5 and greater than 7. This is a null set ($\emptyset$).

(iv) Parallel lines never intersect, so they have no common point. This is a null set ($\emptyset$).

Question 2: Which of the following sets are finite or infinite?
(i) The set of months of a year.
(ii) $\{1, 2, 3, ...\}$
(iii) $\{1, 2, 3, ..., 99, 100\}$
(iv) The set of positive integers greater than 100.
(v) The set of prime numbers less than 99.

Answer-
(i) Finite (There are 12 months).

(ii) Infinite (The set of natural numbers continues indefinitely).

(iii) Finite (The elements end at 100).

(iv) Infinite (Positive integers greater than 100 continue indefinitely: 101, 102, ...).

(v) Finite (There is a specific, countable number of primes less than 99).

Question 3: State whether each of the following sets is finite or infinite:
(i) The set of lines which are parallel to the x-axis.
(ii) The set of letters in the English alphabet.
(iii) The set of numbers which are multiple of 5.
(iv) The set of animals living on the earth.
(v) The set of circles passing through the origin (0,0).

Answer-
(i) Infinite (An infinite number of lines can be drawn parallel to the x-axis).

(ii) Finite (There are 26 letters).

(iii) Infinite (Multiples of 5 continue indefinitely: 5, 10, 15, ...).

(iv) Finite (Although very large, the number of animals on Earth is countable and finite).

(v) Infinite (An infinite number of circles can pass through the origin).

Question 4: In the following, state whether $A = B$ or not:
(i) $A = \{a, b, c, d\}$; $B = \{d, c, b, a\}$
(ii) $A = \{4, 8, 12, 16\}$; $B = \{8, 4, 16, 18\}$
(iii) $A = \{2, 4, 6, 8, 10\}$; $B = \{x : x \text{ is positive even integer and } x \le 10 \}$
(iv) $A = \{x : x \text{ is a multiple of } 10 \}$; $B = \{10, 15, 20, 25, 30, ...\}$

Answer-
(i) $A = B$. The elements are exactly the same, order does not matter.

(ii) $A \neq B$. $12 \in A$ but $12 \notin B$, and $18 \in B$ but $18 \notin A$.

(iii) $B = \{2, 4, 6, 8, 10\}$. Therefore, $A = B$.

(iv) $A = \{10, 20, 30, 40, ...\}$. $B$ contains numbers like 15, 25, etc., which are not multiples of 10. Therefore, $A \neq B$.

Question 5: Are the following pair of sets equal? Give reasons.
(i) $A = \{2, 3\}$; $B = \{x : x \text{ is solution of } x^2 + 5x + 6 = 0 \}$
(ii) $A = \{x : x \text{ is a letter in the word FOLLOW} \}$; $B = \{y : y \text{ is a letter in the word WOLF} \}$

Answer-
(i) Solving $x^2 + 5x + 6 = 0$:
$(x+2)(x+3) = 0$
$x = -2$ or $x = -3$.
So, $B = \{-2, -3\}$.
Since $A = \{2, 3\}$, $A \neq B$.

(ii) Listing elements in roster form:
$A = \{F, O, L, W\}$
$B = \{W, O, L, F\}$
Since the elements are exactly the same, $A = B$.

Question 6: From the sets given below, select equal sets:
$A = \{2, 4, 8, 12\}$, $B = \{1, 2, 3, 4\}$, $C = \{4, 8, 12, 14\}$, $D = \{3, 1, 4, 2\}$,
$E = \{-1, 1\}$, $F = \{0, a\}$, $G = \{1, -1\}$, $H = \{0, 1\}$

Answer-
Comparing the sets:

• $B = \{1, 2, 3, 4\}$

• $D = \{3, 1, 4, 2\}$
  Since the elements are the same, $B = D$.

• $E = \{-1, 1\}$

• $G = \{1, -1\}$
  Since the elements are the same, $E = G$.

Therefore, the pairs of equal sets are $B = D$ and $E = G$.

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Exercise 1.3

Question 1: Make correct statements by filling in the symbols $\subset$ or $\not\subset$ in the blank spaces:
(i) $\{2, 3, 4\} \_\_\_ \{1, 2, 3, 4, 5\}$
(ii) $\{a, b, c\} \_\_\_ \{b, c, d\}$
(iii) $\{x: x \text{ is a student of Class XI of your school}\} \_\_\_ \{x: x \text{ student of your school}\}$
(iv) $\{x: x \text{ is a circle in the plane}\} \_\_\_ \{x: x \text{ is a circle in the same plane with radius 1 unit}\}$
(v) $\{x: x \text{ is a triangle in a plane}\} \_\_\_ \{x: x \text{ is a rectangle in the plane}\}$
(vi) $\{x: x \text{ is an equilateral triangle in a plane}\} \_\_\_ \{x: x \text{ is a triangle in the same plane}\}$
(vii) $\{x: x \text{ is an even natural number}\} \_\_\_ \{x: x \text{ is an integer}\}$

Answer-
(i) $\{2, 3, 4\} \subset \{1, 2, 3, 4, 5\}$ (All elements of the first set are in the second)

(ii) $\{a, b, c\} \not\subset \{b, c, d\}$ ('a' is not in the second set)

(iii) $\{x: x \text{ is a student of Class XI of your school}\} \subset \{x: x \text{ student of your school}\}$ (All Class XI students are students of the school)

(iv) $\{x: x \text{ is a circle in the plane}\} \not\subset \{x: x \text{ is a circle in the same plane with radius 1 unit}\}$ (The first set includes circles of all radii)

(v) $\{x: x \text{ is a triangle in a plane}\} \not\subset \{x: x \text{ is a rectangle in the plane}\}$ (Triangles are not rectangles)

(vi) $\{x: x \text{ is an equilateral triangle in a plane}\} \subset \{x: x \text{ is a triangle in the same plane}\}$ (All equilateral triangles are triangles)

(vii) $\{x: x \text{ is an even natural number}\} \subset \{x: x \text{ is an integer}\}$ (All even natural numbers are integers)

Question 2: Examine whether the following statements are true or false:
(i) $\{a, b\} \not\subset \{b, c, a\}$
(ii) $\{a, e\} \subset \{x: x \text{ is a vowel in the English alphabet}\}$
(iii) $\{1, 2, 3\} \subset \{1, 3, 5\}$
(iv) $\{a\} \subset \{a, b, c\}$
(v) $\{a\} \in \{a, b, c\}$
(vi) $\{x: x \text{ is an even natural number less than } 6\} \subset \{x: x \text{ is a natural number which divides } 36 \}$

Answer-
(i) False. Both 'a' and 'b' are in the second set.

(ii) True. 'a' and 'e' are vowels.

(iii) False. '2' is in the first set but not in the second.

(iv) True. 'a' is an element of the second set.

(v) False. $\{a\}$ is a set containing 'a'. It is a subset, not an element, of $\{a, b, c\}$. ($a \in \{a, b, c\}$ is true).

(vi) True. The first set is $\{2, 4\}$. Both 2 and 4 divide 36.

Question 3: Let $A = \{1, 2, \{3, 4\}, 5\}$. Which of the following statements are incorrect and why?
(i) $\{3, 4\} \subset A$
(ii) $\{3, 4\} \in A$
(iii) $\{\{3, 4\}\} \subset A$
(iv) $1 \in A$
(v) $1 \subset A$
(vi) $\{1, 2, 5\} \subset A$
(vii) $\{1, 2, 5\} \in A$
(viii) $\{1, 2, 3\} \subset A$
(ix) $\emptyset \in A$
(x) $\emptyset \subset A$
(xi) $\{\emptyset\} \subset A$

Answer-
(i) Incorrect. $\{3, 4\}$ is an element of A, not a subset.

(ii) Correct. $\{3, 4\}$ is listed as an element within A.

(iii) Correct. $\{\{3, 4\}\}$ is a set containing the element $\{3, 4\}$, which belongs to A.

(iv) Correct. 1 is an element of A.

(v) Incorrect. 1 is an element, not a set, so it cannot be a subset.

(vi) Correct. 1, 2, and 5 are all elements of A.

(vii) Incorrect. $\{1, 2, 5\}$ is a set, not an element of A.

(viii) Incorrect. 3 is not an element of A (only $\{3, 4\}$ is).

(ix) Incorrect. The empty set $\emptyset$ is not listed as an element within A.

(x) Correct. The empty set $\emptyset$ is a subset of every set.

(xi) Incorrect. The set $\{\emptyset\}$ is a set containing the empty set. Since $\emptyset$ is not an element of A, this statement is false.

Question 4: Write down all the subsets of the following sets:
(i) $\{a\}$
(ii) $\{a, b\}$
(iii) $\{1, 2, 3\}$
(iv) $\emptyset$

Answer-
(i) Subsets: $\emptyset, \{a\}$

(ii) Subsets: $\emptyset, \{a\}, \{b\}, \{a, b\}$

(iii) Subsets: $\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}$

(iv) Subset: $\emptyset$ (The empty set has only one subset, itself)

Question 5: How many elements has $P(A)$, if $A = \emptyset$?

Answer-
If $A = \emptyset$, then the number of elements in A is $n(A) = 0$.
The number of elements in the power set $P(A)$ is $2^{n(A)}$.
So, $n(P(A)) = 2^0 = 1$.
$P(A)$ has one element, which is the empty set itself: $P(\emptyset) = \{\emptyset\}$.

Question 6: Write the following as intervals:
(i) $\{x : x \in R, -4 < x \le 6 \}$
(ii) $\{x : x \in R, -12 < x < -10 \}$
(iii) $\{x : x \in R, 0 \le x < 7 \}$
(iv) $\{x : x \in R, 3 \le x \le 4 \}$

Answer-
(i) $(-4, 6]$

(ii) $(-12, -10)$

(iii) $[0, 7)$

(iv) $[3, 4]$

Question 7: Write the following intervals in set-builder form:
(i) $(-3, 0)$
(ii) $[6, 12]$
(iii) $(6, 12]$
(iv) $[-23, 5)$

Answer-
(i) $\{x : x \in R, -3 < x < 0 \}$

(ii) $\{x : x \in R, 6 \le x \le 12 \}$

(iii) $\{x : x \in R, 6 < x \le 12 \}$

(iv) $\{x : x \in R, -23 \le x < 5 \}$

Question 8: What universal set(s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.

Answer-
(i) The universal set could be the set of all triangles in a plane, or the set of all polygons in a plane.

(ii) The universal set could be the set of all triangles in a plane, or the set of all polygons in a plane.

Question 9: Given the sets $A = \{1, 3, 5\}$, $B = \{2, 4, 6\}$ and $C = \{0, 2, 4, 6, 8\}$. Which of the following may be considered as universal set(s) for all the three sets A, B and C?
(i) $\{0, 1, 2, 3, 4, 5, 6\}$
(ii) $\emptyset$
(iii) $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
(iv) $\{1, 2, 3, 4, 5, 6, 7, 8\}$

Answer-
We need a set that contains all elements of A, B, and C.
Elements are {0, 1, 2, 3, 4, 5, 6, 8}.

(i) Does not contain 8. Not a universal set.
(ii) The empty set cannot be a universal set for non-empty sets.
(iii) Contains all elements {0, 1, 2, 3, 4, 5, 6, 8}. This can be a universal set.
(iv) Does not contain 0. Not a universal set.

Therefore, only (iii) can be considered a universal set.

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Exercise 1.4

Question 1: Find the union of each of the following pairs of sets:
(i) $X = \{1, 3, 5\}$, $Y = \{1, 2, 3\}$
(ii) $A = \{a, e, i, o, u\}$, $B = \{a, b, c\}$
(iii) $A = \{x : x \text{ is a natural number and multiple of } 3 \}$, $B = \{x : x \text{ is a natural number less than } 6 \}$
(iv) $A = \{x : x \text{ is a natural number and } 1 < x \le 6 \}$, $B = \{x : x \text{ is a natural number and } 6 < x < 10 \}$
(v) $A = \{1, 2, 3\}$, $B = \emptyset$

Answer-
(i) $X \cup Y = \{1, 2, 3, 5\}$

(ii) $A \cup B = \{a, b, c, e, i, o, u\}$

(iii) $A = \{3, 6, 9, 12, ...\}$, $B = \{1, 2, 3, 4, 5\}$.
$A \cup B = \{1, 2, 3, 4, 5, 6, 9, 12, ...\} = \{x : x = 1, 2, 4, 5 \text{ or } x \text{ is a multiple of } 3\}$

(iv) $A = \{2, 3, 4, 5, 6\}$, $B = \{7, 8, 9\}$.
$A \cup B = \{2, 3, 4, 5, 6, 7, 8, 9\}$

(v) $A \cup B = \{1, 2, 3\} \cup \emptyset = \{1, 2, 3\} = A$

Question 2: Let $A = \{a, b\}, B = \{a, b, c\}$. Is $A \subset B$? What is $A \cup B$?

Answer-
Yes, $A \subset B$ because every element of A ('a' and 'b') is also an element of B.
$A \cup B = \{a, b\} \cup \{a, b, c\} = \{a, b, c\} = B$.
(Note: If $A \subset B$, then $A \cup B = B$).

Question 3: If A and B are two sets such that $A \subset B$, then what is $A \cup B$?

Answer-
If $A \subset B$, it means all elements of A are also in B.
When we take the union $A \cup B$, we combine all elements from both sets. Since all elements of A are already included in B, the union will simply be the larger set B.
Therefore, $A \cup B = B$.

Question 4: If $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$, $C = \{5, 6, 7, 8\}$ and $D = \{7, 8, 9, 10\}$. Find:
(i) $A \cup B$
(ii) $A \cup C$
(iii) $B \cup C$
(iv) $B \cup D$
(v) $A \cup B \cup C$
(vi) $A \cup B \cup D$
(vii) $B \cup C \cup D$

Answer-
(i) $A \cup B = \{1, 2, 3, 4, 5, 6\}$

(ii) $A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(iii) $B \cup C = \{3, 4, 5, 6, 7, 8\}$

(iv) $B \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}$

(v) $A \cup B \cup C = (A \cup B) \cup C = \{1, 2, 3, 4, 5, 6\} \cup \{5, 6, 7, 8\} = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(vi) $A \cup B \cup D = (A \cup B) \cup D = \{1, 2, 3, 4, 5, 6\} \cup \{7, 8, 9, 10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

(vii) $B \cup C \cup D = (B \cup C) \cup D = \{3, 4, 5, 6, 7, 8\} \cup \{7, 8, 9, 10\} = \{3, 4, 5, 6, 7, 8, 9, 10\}$

Question 5: Find the intersection of each pair of sets of question 1 above.

Answer-
(i) $X = \{1, 3, 5\}$, $Y = \{1, 2, 3\}$. $X \cap Y = \{1, 3\}$

(ii) $A = \{a, e, i, o, u\}$, $B = \{a, b, c\}$. $A \cap B = \{a\}$

(iii) $A = \{3, 6, 9, ...\}$, $B = \{1, 2, 3, 4, 5\}$. $A \cap B = \{3\}$

(iv) $A = \{2, 3, 4, 5, 6\}$, $B = \{7, 8, 9\}$. $A \cap B = \emptyset$ (Disjoint sets)

(v) $A = \{1, 2, 3\}$, $B = \emptyset$. $A \cap B = \emptyset$

Question 6: If $A = \{3, 5, 7, 9, 11\}$, $B = \{7, 9, 11, 13\}$, $C = \{11, 13, 15\}$ and $D = \{15, 17\}$. Find:
(i) $A \cap B$
(ii) $B \cap C$
(iii) $A \cap C \cap D$
(iv) $A \cap C$
(v) $B \cap D$
(vi) $A \cap (B \cup C)$
(vii) $A \cap D$
(viii) $A \cap (B \cup D)$
(ix) $(A \cap B) \cap (B \cup C)$
(x) $(A \cup D) \cap (B \cup C)$

Answer-
(i) $A \cap B = \{7, 9, 11\}$

(ii) $B \cap C = \{11, 13\}$

(iii) $A \cap C = \{11\}$. $(A \cap C) \cap D = \{11\} \cap \{15, 17\} = \emptyset$

(iv) $A \cap C = \{11\}$

(v) $B \cap D = \emptyset$

(vi) $B \cup C = \{7, 9, 11, 13, 15\}$. $A \cap (B \cup C) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11\}$

(vii) $A \cap D = \emptyset$

(viii) $B \cup D = \{7, 9, 11, 13, 15, 17\}$. $A \cap (B \cup D) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15, 17\} = \{7, 9, 11\}$

(ix) $A \cap B = \{7, 9, 11\}$. $B \cup C = \{7, 9, 11, 13, 15\}$. $(A \cap B) \cap (B \cup C) = \{7, 9, 11\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11\}$

(x) $A \cup D = \{3, 5, 7, 9, 11, 15, 17\}$. $B \cup C = \{7, 9, 11, 13, 15\}$. $(A \cup D) \cap (B \cup C) = \{3, 5, 7, 9, 11, 15, 17\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11, 15\}$

Question 7: If $A = \{x : x \text{ is a natural number}\}$, $B = \{x : x \text{ is an even natural number}\}$, $C = \{x : x \text{ is an odd natural number}\}$ and $D = \{x : x \text{ is a prime number}\}$. Find:
(i) $A \cap B$
(ii) $A \cap C$
(iii) $A \cap D$
(iv) $B \cap C$
(v) $B \cap D$
(vi) $C \cap D$

Answer-
$A = \{1, 2, 3, 4, 5, ...\}$
$B = \{2, 4, 6, 8, ...\}$
$C = \{1, 3, 5, 7, ...\}$
$D = \{2, 3, 5, 7, 11, 13, ...\}$

(i) $A \cap B = \{x : x \text{ is an even natural number}\} = B$

(ii) $A \cap C = \{x : x \text{ is an odd natural number}\} = C$

(iii) $A \cap D = \{x : x \text{ is a prime number}\} = D$

(iv) $B \cap C = \emptyset$ (No number is both even and odd)

(v) $B \cap D = \{2\}$ (2 is the only even prime number)

(vi) $C \cap D = \{x : x \text{ is an odd prime number}\} = \{3, 5, 7, 11, 13, ...\}$

Question 8: Which of the following pairs of sets are disjoint?
(i) $\{1, 2, 3, 4\}$ and $\{x : x \text{ is a natural number and } 4 \le x \le 6 \}$
(ii) $\{a, e, i, o, u\}$ and $\{c, d, e, f\}$
(iii) $\{x : x \text{ is an even integer}\}$ and $\{x : x \text{ is an odd integer}\}$

Answer-
(i) First set = $\{1, 2, 3, 4\}$. Second set = $\{4, 5, 6\}$.
Intersection = $\{4\}$. Not disjoint.

(ii) First set = $\{a, e, i, o, u\}$. Second set = $\{c, d, e, f\}$.
Intersection = $\{e\}$. Not disjoint.

(iii) First set = $\{..., -2, 0, 2, 4, ...\}$. Second set = $\{..., -3, -1, 1, 3, ...\}$.
Intersection = $\emptyset$. Disjoint.

Question 9: If $A = \{3, 6, 9, 12, 15, 18, 21\}$, $B = \{4, 8, 12, 16, 20\}$, $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$, $D = \{5, 10, 15, 20\}$. Find:
(i) $A - B$
(ii) $A - C$
(iii) $A - D$
(iv) $B - A$
(v) $C - A$
(vi) $D - A$
(vii) $B - C$
(viii) $B - D$
(ix) $C - B$
(x) $D - B$
(xi) $C - D$
(xii) $D - C$

Answer-
(i) $A - B = \{3, 6, 9, 15, 18, 21\}$ (Elements in A but not in B)

(ii) $A - C = \{3, 9, 15, 18, 21\}$

(iii) $A - D = \{3, 6, 9, 12, 18, 21\}$

(iv) $B - A = \{4, 8, 16, 20\}$

(v) $C - A = \{2, 4, 8, 10, 14, 16\}$

(vi) $D - A = \{5, 10, 20\}$

(vii) $B - C = \{20\}$

(viii) $B - D = \{4, 8, 12, 16\}$

(ix) $C - B = \{2, 6, 10, 14\}$

(x) $D - B = \{5, 10, 15\}$

(xi) $C - D = \{2, 4, 6, 8, 12, 14, 16\}$

(xii) $D - C = \{5, 15, 20\}$

Question 10: If $X = \{a, b, c, d\}$ and $Y = \{f, b, d, g\}$. Find:
(i) $X - Y$
(ii) $Y - X$
(iii) $X \cap Y$

Answer-
(i) $X - Y = \{a, c\}$

(ii) $Y - X = \{f, g\}$

(iii) $X \cap Y = \{b, d\}$

Question 11: If R is the set of real numbers and Q is the set of rational numbers, then what is $R - Q$?

Answer-
$R - Q$ represents the set of real numbers that are not rational. This is the definition of the set of irrational numbers.
$R - Q = T$ (where T is the set of irrational numbers).

Question 12: State whether each of the following statements is true or false. Justify your answer.
(i) $\{2, 3, 4, 5\}$ and $\{3, 6\}$ are disjoint sets.
(ii) $\{a, e, i, o, u\}$ and $\{a, b, c, d\}$ are disjoint sets.
(iii) $\{2, 6, 10, 14\}$ and $\{3, 7, 11, 15\}$ are disjoint sets.
(iv) $\{2, 6, 10\}$ and $\{3, 7, 11\}$ are disjoint sets.

Answer-
(i) False. The intersection is $\{3\}$.

(ii) False. The intersection is $\{a\}$.

(iii) True. There are no common elements. The intersection is $\emptyset$.

(iv) True. There are no common elements. The intersection is $\emptyset$.

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Exercise 1.5

Question 1: Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{1, 2, 3, 4\}$, $B = \{2, 4, 6, 8\}$ and $C = \{3, 4, 5, 6\}$. Find:
(i) $A'$
(ii) $B'$
(iii) $(A \cup C)'$
(iv) $(A \cup B)'$
(v) $(A')'$
(vi) $(B - C)'$

Answer-
(i) $A' = U - A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4\} = \{5, 6, 7, 8, 9\}$

(ii) $B' = U - B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 4, 6, 8\} = \{1, 3, 5, 7, 9\}$

(iii) $A \cup C = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\}$.
$(A \cup C)' = U - (A \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 5, 6\} = \{7, 8, 9\}$

(iv) $A \cup B = \{1, 2, 3, 4\} \cup \{2, 4, 6, 8\} = \{1, 2, 3, 4, 6, 8\}$.
$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 6, 8\} = \{5, 7, 9\}$

(v) $A' = \{5, 6, 7, 8, 9\}$.
$(A')' = U - A' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{5, 6, 7, 8, 9\} = \{1, 2, 3, 4\} = A$

(vi) $B - C = \{2, 4, 6, 8\} - \{3, 4, 5, 6\} = \{2, 8\}$.
$(B - C)' = U - (B - C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 8\} = \{1, 3, 4, 5, 6, 7, 9\}$

Question 2: If $U = \{a, b, c, d, e, f, g, h\}$, find the complements of the following sets:
(i) $A = \{a, b, c\}$
(ii) $B = \{d, e, f, g\}$
(iii) $C = \{a, c, e, g\}$
(iv) $D = \{f, g, h, a\}$

Answer-
(i) $A' = U - A = \{d, e, f, g, h\}$

(ii) $B' = U - B = \{a, b, c, h\}$

(iii) $C' = U - C = \{b, d, f, h\}$

(iv) $D' = U - D = \{b, c, d, e\}$

Question 3: Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) $\{x : x \text{ is an even natural number}\}$
(ii) $\{x : x \text{ is an odd natural number}\}$
(iii) $\{x : x \text{ is a positive multiple of } 3\}$
(iv) $\{x : x \text{ is a prime number}\}$
(v) $\{x : x \text{ is a natural number divisible by 3 and 5}\}$
(vi) $\{x : x \text{ is a perfect square}\}$
(vii) $\{x : x \text{ is a perfect cube}\}$
(viii) $\{x : x + 5 = 8\}$
(ix) $\{x : 2x + 5 = 9\}$
(x) $\{x : x \ge 7\}$
(xi) $\{x : x \in N \text{ and } 2x + 1 > 10\}$

Answer-
Here, $U = N = \{1, 2, 3, 4, ...\}$.

(i) Complement = $\{x : x \text{ is an odd natural number}\}$

(ii) Complement = $\{x : x \text{ is an even natural number}\}$

(iii) Complement = $\{x : x \in N \text{ and } x \text{ is not a multiple of } 3\}$

(iv) Complement = $\{x : x \text{ is a positive composite number or } x = 1\}$

(v) This set contains natural numbers divisible by 15.
Complement = $\{x : x \in N \text{ and } x \text{ is not divisible by 15}\}$

(vi) Complement = $\{x : x \in N \text{ and } x \text{ is not a perfect square}\}$

(vii) Complement = $\{x : x \in N \text{ and } x \text{ is not a perfect cube}\}$

(viii) The set is $\{3\}$ (since $x+5=8 \implies x=3$).
Complement = $\{x : x \in N \text{ and } x \neq 3\}$

(ix) $2x+5=9 \implies 2x=4 \implies x=2$. The set is $\{2\}$.
Complement = $\{x : x \in N \text{ and } x \neq 2\}$

(x) The set is $\{7, 8, 9, ...\}$.
Complement = $\{x : x \in N \text{ and } x < 7\} = \{1, 2, 3, 4, 5, 6\}$

(xi) $2x+1 > 10 \implies 2x > 9 \implies x > 4.5$. The set is $\{5, 6, 7, ...\}$.
Complement = $\{x : x \in N \text{ and } x \le 4.5\} = \{1, 2, 3, 4\}$

Question 4: If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{2, 4, 6, 8\}$ and $B = \{2, 3, 5, 7\}$. Verify that
(i) $(A \cup B)' = A' \cap B'$
(ii) $(A \cap B)' = A' \cup B'$

Answer-
$A' = U - A = \{1, 3, 5, 7, 9\}$
$B' = U - B = \{1, 4, 6, 8, 9\}$

(i) LHS: $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$.
$(A \cup B)' = U - (A \cup B) = \{1, 9\}$.
RHS: $A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\} = \{1, 9\}$.
LHS = RHS. Hence, verified.

(ii) LHS: $A \cap B = \{2\}$.
$(A \cap B)' = U - (A \cap B) = \{1, 3, 4, 5, 6, 7, 8, 9\}$.
RHS: $A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$.
LHS = RHS. Hence, verified.

Question 5: Draw appropriate Venn diagram for each of the following:
(i) $(A \cup B)'$
(ii) $A' \cap B'$
(iii) $(A \cap B)'$
(iv) $A' \cup B'$

Answer-
(i) $(A \cup B)'$: Shade the region outside both circles A and B.

(ii) $A' \cap B'$: This is the same region as $(A \cup B)'$ by De Morgan's law. Shade the region outside both circles A and B.
attachment_0

(iii) $(A \cap B)'$: Shade the entire region except for the common intersection area of circles A and B.
attachment_1

(iv) $A' \cup B'$: This is the same region as $(A \cap B)'$ by De Morgan's law. Shade the entire region except for the common intersection area of circles A and B.

Question 6: Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from $60^\circ$, what is $A'$?

Answer-
A is the set of all triangles that are not equilateral (since an equilateral triangle has all angles equal to $60^\circ$).
Therefore, $A'$ (the complement of A) is the set of all triangles that do not have at least one angle different from $60^\circ$. This means $A'$ is the set of triangles where all angles are equal to $60^\circ$.
$A'$ is the set of all equilateral triangles.

Question 7: Fill in the blanks to make each of the following a true statement:
(i) $A \cup A' = \_\_\_$
(ii) $\emptyset' \cap A = \_\_\_$
(iii) $A \cap A' = \_\_\_$
(iv) $U' \cap A = \_\_\_$

Answer-
(i) $A \cup A' = U$ (The union of a set and its complement is the universal set)

(ii) $\emptyset' = U$. So, $\emptyset' \cap A = U \cap A = A$ (The intersection of the universal set and any set A is A itself)

(iii) $A \cap A' = \emptyset$ (A set and its complement have no elements in common)

(iv) $U' = \emptyset$. So, $U' \cap A = \emptyset \cap A = \emptyset$ (The intersection of the empty set and any set is the empty set)

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Exercise 1.6

Question 1: If X and Y are two sets such that $n(X) = 17$, $n(Y) = 23$ and $n(X \cup Y) = 38$, find $n(X \cap Y)$.

Answer-
We use the formula: $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.
$38 = 17 + 23 - n(X \cap Y)$
$38 = 40 - n(X \cap Y)$
$n(X \cap Y) = 40 - 38 = 2$.

Question 2: If X and Y are two sets such that $X \cup Y$ has 18 elements, X has 8 elements and Y has 15 elements, how many elements does $X \cap Y$ have?

Answer-
Given: $n(X \cup Y) = 18$, $n(X) = 8$, $n(Y) = 15$.
Using $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$:
$18 = 8 + 15 - n(X \cap Y)$
$18 = 23 - n(X \cap Y)$
$n(X \cap Y) = 23 - 18 = 5$.

Question 3: In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Answer-
Let H be the set of people who speak Hindi, and E be the set of people who speak English.
Given: $n(H \cup E) = 400$, $n(H) = 250$, $n(E) = 200$.
We need to find $n(H \cap E)$.
Using $n(H \cup E) = n(H) + n(E) - n(H \cap E)$:
$400 = 250 + 200 - n(H \cap E)$
$400 = 450 - n(H \cap E)$
$n(H \cap E) = 450 - 400 = 50$.
So, 50 people can speak both Hindi and English.

Question 4: If S and T are two sets such that S has 21 elements, T has 32 elements, and $S \cap T$ has 11 elements, how many elements does $S \cup T$ have?

Answer-
Given: $n(S) = 21$, $n(T) = 32$, $n(S \cap T) = 11$.
Using $n(S \cup T) = n(S) + n(T) - n(S \cap T)$:
$n(S \cup T) = 21 + 32 - 11$
$n(S \cup T) = 53 - 11 = 42$.

Question 5: If X and Y are two sets such that X has 40 elements, $X \cup Y$ has 60 elements and $X \cap Y$ has 10 elements, how many elements does Y have?

Answer-
Given: $n(X) = 40$, $n(X \cup Y) = 60$, $n(X \cap Y) = 10$.
Using $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$:
$60 = 40 + n(Y) - 10$
$60 = 30 + n(Y)$
$n(Y) = 60 - 30 = 30$.

Question 6: In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Answer-
Let C be the set of people who like coffee, and T be the set of people who like tea.
Given: $n(C \cup T) = 70$ (since each person likes at least one), $n(C) = 37$, $n(T) = 52$.
We need to find $n(C \cap T)$.
Using $n(C \cup T) = n(C) + n(T) - n(C \cap T)$:
$70 = 37 + 52 - n(C \cap T)$
$70 = 89 - n(C \cap T)$
$n(C \cap T) = 89 - 70 = 19$.
So, 19 people like both coffee and tea.

Question 7: In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Answer-
Let C be the set of people who like cricket, and T be the set of people who like tennis.
Given: $n(C \cup T) = 65$, $n(C) = 40$, $n(C \cap T) = 10$.

First, find how many like tennis ($n(T)$):
Using $n(C \cup T) = n(C) + n(T) - n(C \cap T)$:
$65 = 40 + n(T) - 10$
$65 = 30 + n(T)$
$n(T) = 65 - 30 = 35$.
So, 35 people like tennis.

Next, find how many like tennis only and not cricket ($n(T - C)$):
Using $n(T - C) = n(T) - n(C \cap T)$:
$n(T - C) = 35 - 10 = 25$.
So, 25 people like tennis only and not cricket.

Question 8: In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

Answer-
Let F be the set of people who speak French, and S be the set of people who speak Spanish.
Given: $n(F) = 50$, $n(S) = 20$, $n(F \cap S) = 10$.
We need to find how many speak at least one language, which is $n(F \cup S)$.
Using $n(F \cup S) = n(F) + n(S) - n(F \cap S)$:
$n(F \cup S) = 50 + 20 - 10$
$n(F \cup S) = 70 - 10 = 60$.
So, 60 people speak at least one of these two languages.

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Miscellaneous Exercise on Chapter 1

Question 1: Decide, among the following sets, which sets are subsets of one and another:
$A = \{x: x \in R \text{ and } x \text{ satisfy } x^2 - 8x + 12 = 0 \}$ $B = \{2, 4, 6\}$
$C = \{2, 4, 6, 8, ...\}$ $D = \{6\}$

Answer-
First, find the elements of A:
$x^2 - 8x + 12 = 0$
$(x-2)(x-6) = 0$
$x = 2$ or $x = 6$.
So, $A = \{2, 6\}$.

Now compare the sets:

• $A = \{2, 6\}$. Every element of A is in B. So $A \subset B$.
• $A = \{2, 6\}$. Every element of A is in C. So $A \subset C$.
• $B = \{2, 4, 6\}$. Every element of B is in C. So $B \subset C$.
• $D = \{6\}$. Every element of D is in A. So $D \subset A$.
• $D = \{6\}$. Every element of D is in B. So $D \subset B$.
• $D = \{6\}$. Every element of D is in C. So $D \subset C$.

Summary of subset relationships: $D \subset A \subset B \subset C$. Also, $D \subset B$, $D \subset C$, $A \subset C$.

Question 2: In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If $x \in A$ and $A \in B$, then $x \in B$.
(ii) If $A \subset B$ and $B \in C$, then $A \in C$.
(iii) If $A \subset B$ and $B \subset C$, then $A \subset C$.
(iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$.
(v) If $x \in A$ and $A \not\subset B$, then $x \in B$.
(vi) If $A \subset B$ and $x \notin B$, then $x \notin A$.

Answer-
(i) False.
Example: Let $A = \{1\}$ and $B = \{\{1\}, 2\}$. Here $1 \in A$ and $A \in B$. But $1 \notin B$.

(ii) False.
Example: Let $A = \{1\}$, $B = \{1, 2\}$, and $C = \{\{1, 2\}, 3\}$. Here $A \subset B$ and $B \in C$. But $A = \{1\}$ is not an element of C.

(iii) True.
Proof: Let $x$ be any element of A ($x \in A$). Since $A \subset B$, it follows that $x \in B$. Since $B \subset C$, it follows that $x \in C$. Since any element $x$ in A is also in C, we conclude that $A \subset C$.

(iv) False.
Example: Let $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{1, 2, 4\}$. Here $A \not\subset B$ (since $1 \notin B$) and $B \not\subset C$ (since $3 \notin C$). However, $A \subset C$ is false here (as well, $2 \in C$ but $1 \notin C$ in this specific example - let's adjust C).
Let's try again: $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{1, 2, 3, 4\}$. Here $A \not\subset B$ and $B \not\subset C$ (just kidding, B IS a subset of C here).
Let's try again: $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{1, 4\}$. Here $A \not\subset B$ and $B \not\subset C$. Also, $A \not\subset C$. This example doesn't contradict.
Let $A = \{1, 2\}$, $B = \{3, 4\}$, $C = \{1, 2, 5\}$. Here $A \not\subset B$ and $B \not\subset C$. But $A \subset C$ is true. So the original statement is False.

(v) False.
Example: Let $A = \{1, 2\}$ and $B = \{2, 3\}$. Here $1 \in A$ and $A \not\subset B$ (since $1 \notin B$). However, $1 \notin B$.

(vi) True.
Proof: We are given $A \subset B$ and $x \notin B$. We want to show $x \notin A$. Let's assume the opposite, that $x \in A$. Since $A \subset B$, if $x \in A$, then it must be that $x \in B$. But this contradicts the given information that $x \notin B$. Therefore, our assumption ($x \in A$) must be false. Hence, $x \notin A$. (This is proof by contradiction).

Question 3: Let A, B and C be the sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$.

Answer-
Let $x$ be an arbitrary element such that $x \in B$.
We need to show that $x \in C$.
Since $x \in B$, then $x \in A \cup B$.
Since $A \cup B = A \cup C$, we have $x \in A \cup C$.
This means $x \in A$ or $x \in C$.

Case 1: If $x \in C$, we are done.
Case 2: If $x \in A$. Since we initially assumed $x \in B$, this means $x \in A \cap B$.
We are given $A \cap B = A \cap C$. So, if $x \in A \cap B$, then $x \in A \cap C$.
If $x \in A \cap C$, then $x \in C$.
In both cases, we find that if $x \in B$, then $x \in C$. This implies $B \subseteq C$.

Now, let $y$ be an arbitrary element such that $y \in C$.
We need to show that $y \in B$.
Since $y \in C$, then $y \in A \cup C$.
Since $A \cup C = A \cup B$, we have $y \in A \cup B$.
This means $y \in A$ or $y \in B$.

Case 1: If $y \in B$, we are done.
Case 2: If $y \in A$. Since we initially assumed $y \in C$, this means $y \in A \cap C$.
We are given $A \cap C = A \cap B$. So, if $y \in A \cap C$, then $y \in A \cap B$.
If $y \in A \cap B$, then $y \in B$.
In both cases, we find that if $y \in C$, then $y \in B$. This implies $C \subseteq B$.

Since $B \subseteq C$ and $C \subseteq B$, we conclude that $B = C$.

Question 4: Show that the following four conditions are equivalent:
(i) $A \subset B$
(ii) $A - B = \emptyset$
(iii) $A \cup B = B$
(iv) $A \cap B = A$

Answer-
We need to show (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (iv) $\implies$ (i).

(i) $\implies$ (ii): Assume $A \subset B$. This means every element of A is also in B.
$A - B$ is the set of elements that are in A but not in B. Since every element of A is in B, there are no elements that are in A but not in B. Thus, $A - B = \emptyset$.

(ii) $\implies$ (iii): Assume $A - B = \emptyset$. This means there are no elements in A that are not in B. In other words, every element of A must also be an element of B. So, $A \subset B$.
$A \cup B$ contains all elements from A and all elements from B. Since all elements of A are already in B, $A \cup B$ simply contains all the elements of B. Thus, $A \cup B = B$.

(iii) $\implies$ (iv): Assume $A \cup B = B$. Let $x \in A$. Then $x \in A \cup B$. Since $A \cup B = B$, we have $x \in B$. This shows that every element of A is in B, so $A \subset B$.
$A \cap B$ contains elements common to both A and B. Since every element of A is also in B, the common elements are precisely all the elements of A. Thus, $A \cap B = A$.

(iv) $\implies$ (i): Assume $A \cap B = A$. This means that all elements common to A and B are exactly the elements of A. This implies that every element of A must also be an element of B. Thus, $A \subset B$.

Since we have shown (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (iv) $\implies$ (i), all four conditions are equivalent.

Question 5: Show that if $A \subset B$, then $C - B \subset C - A$.

Answer-
Let $x$ be an arbitrary element such that $x \in C - B$.
By definition of set difference, this means $x \in C$ and $x \notin B$.
We are given $A \subset B$. This means if an element is in A, it must also be in B.
Contrapositively, if an element is not in B ($x \notin B$), then it cannot be in A ($x \notin A$).
So, we have $x \in C$ and $x \notin A$.
By the definition of set difference, $x \in C - A$.
Since any element $x$ in $C - B$ is also in $C - A$, we conclude that $C - B \subset C - A$.

Question 6: Assume that $P(A) = P(B)$. Show that $A = B$.

Answer-
Given $P(A) = P(B)$.
Since A is a subset of A, $A \in P(A)$.
Because $P(A) = P(B)$, we have $A \in P(B)$.
By the definition of a power set, if a set belongs to the power set of B, it must be a subset of B.
Therefore, $A \subseteq B$.

Similarly, since B is a subset of B, $B \in P(B)$.
Because $P(B) = P(A)$, we have $B \in P(A)$.
By the definition of a power set, if a set belongs to the power set of A, it must be a subset of A.
Therefore, $B \subseteq A$.

Since $A \subseteq B$ and $B \subseteq A$, we conclude that $A = B$.

Question 7: Is it true that for any sets A and B, $P(A) \cup P(B) = P(A \cup B)$? Justify your answer.

Answer-
False.
Let $A = \{1\}$ and $B = \{2\}$.
$P(A) = \{\emptyset, \{1\}\}$
$P(B) = \{\emptyset, \{2\}\}$
$P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}\}$

$A \cup B = \{1, 2\}$
$P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$

Since $P(A) \cup P(B) \neq P(A \cup B)$ (because $\{1, 2\}$ is in $P(A \cup B)$ but not in $P(A) \cup P(B)$), the statement is false.

Question 8: Show that for any sets A and B,
$A = (A \cap B) \cup (A - B)$ and $A \cup (B - A) = (A \cup B)$.

Answer-
Part 1: $A = (A \cap B) \cup (A - B)$
Let $x \in A$.
Case 1: $x \in B$. If $x \in A$ and $x \in B$, then $x \in (A \cap B)$.
Case 2: $x \notin B$. If $x \in A$ and $x \notin B$, then $x \in (A - B)$.
In either case, if $x \in A$, then $x \in (A \cap B) \cup (A - B)$. So $A \subseteq (A \cap B) \cup (A - B)$.

Now let $y \in (A \cap B) \cup (A - B)$.
This means $y \in (A \cap B)$ or $y \in (A - B)$.
If $y \in (A \cap B)$, then $y \in A$ and $y \in B$. So $y \in A$.
If $y \in (A - B)$, then $y \in A$ and $y \notin B$. So $y \in A$.
In either case, if $y \in (A \cap B) \cup (A - B)$, then $y \in A$. So $(A \cap B) \cup (A - B) \subseteq A$.
Since the inclusion holds both ways, $A = (A \cap B) \cup (A - B)$.

Part 2: $A \cup (B - A) = (A \cup B)$
Let $x \in A \cup (B - A)$.
This means $x \in A$ or $x \in (B - A)$.
If $x \in A$, then clearly $x \in (A \cup B)$.
If $x \in (B - A)$, then $x \in B$ and $x \notin A$. If $x \in B$, then $x \in (A \cup B)$.
In either case, if $x \in A \cup (B - A)$, then $x \in (A \cup B)$. So $A \cup (B - A) \subseteq (A \cup B)$.

Now let $y \in (A \cup B)$.
This means $y \in A$ or $y \in B$.
If $y \in A$, then clearly $y \in A \cup (B - A)$.
If $y \in B$. There are two sub-cases:
Case 2a: $y \in A$. Then $y \in A \cup (B - A)$.
Case 2b: $y \notin A$. If $y \in B$ and $y \notin A$, then $y \in (B - A)$. So $y \in A \cup (B - A)$.
In all cases, if $y \in (A \cup B)$, then $y \in A \cup (B - A)$. So $(A \cup B) \subseteq A \cup (B - A)$.
Since the inclusion holds both ways, $A \cup (B - A) = (A \cup B)$.

Question 9: Using properties of sets, show that:
(i) $A \cup (A \cap B) = A$
(ii) $A \cap (A \cup B) = A$

Answer-
(i) $A \cup (A \cap B)$
We know $A \cap B \subseteq A$.
If $X \subseteq Y$, then $X \cup Y = Y$.
Let $X = A \cap B$ and $Y = A$. Then $X \subseteq Y$.
Therefore, $A \cup (A \cap B) = A$. (Absorption Law)

(ii) $A \cap (A \cup B)$
We know $A \subseteq A \cup B$.
If $Y \subseteq X$, then $X \cap Y = Y$.
Let $Y = A$ and $X = A \cup B$. Then $Y \subseteq X$.
Therefore, $A \cap (A \cup B) = A$. (Absorption Law)

Question 10: Show that $A \cap B = A \cap C$ need not imply $B = C$.

Answer-
Let $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{2, 4\}$.
$A \cap B = \{1, 2\} \cap \{2, 3\} = \{2\}$.
$A \cap C = \{1, 2\} \cap \{2, 4\} = \{2\}$.
Here, $A \cap B = A \cap C = \{2\}$.
However, $B = \{2, 3\}$ and $C = \{2, 4\}$ are not equal.
Thus, $A \cap B = A \cap C$ need not imply $B = C$.

Question 11: Let A and B be sets. If $A \cap X = B \cap X = \emptyset$ and $A \cup X = B \cup X$ for some set X, show that $A = B$.
(Hints $A = A \cap (A \cup X), B = B \cap (B \cup X)$ and use Distributive Law )

Answer-
Given $A \cup X = B \cup X$.
Using the hint, $A = A \cap (A \cup X)$.
Since $A \cup X = B \cup X$, we can substitute:
$A = A \cap (B \cup X)$
Using the Distributive Law ($P \cap (Q \cup R) = (P \cap Q) \cup (P \cap R)$):
$A = (A \cap B) \cup (A \cap X)$
We are given $A \cap X = \emptyset$.
So, $A = (A \cap B) \cup \emptyset$ $A = A \cap B$ (Since $Y \cup \emptyset = Y$)
This implies $A \subseteq B$. (Equation 1)

Similarly, using the hint $B = B \cap (B \cup X)$.
Since $B \cup X = A \cup X$, we substitute:
$B = B \cap (A \cup X)$
Using the Distributive Law:
$B = (B \cap A) \cup (B \cap X)$
We are given $B \cap X = \emptyset$.
So, $B = (B \cap A) \cup \emptyset$ $B = B \cap A$ (or $A \cap B$)
This implies $B \subseteq A$. (Equation 2)

From Equation 1 ($A \subseteq B$) and Equation 2 ($B \subseteq A$), we conclude that $A = B$.

Question 12: Find sets A, B and C such that $A \cap B, B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C = \emptyset$.

Answer-
Let $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{1, 3\}$.
$A \cap B = \{2\}$ (Non-empty)
$B \cap C = \{3\}$ (Non-empty)
$A \cap C = \{1\}$ (Non-empty)
$A \cap B \cap C = (A \cap B) \cap C = \{2\} \cap \{1, 3\} = \emptyset$.
These sets satisfy the conditions.

Question 13: In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee.

Answer-
Let T be the set of students taking tea, and C be the set of students taking coffee.
Total students $n(U) = 600$.
$n(T) = 150$
$n(C) = 225$
$n(T \cap C) = 100$

Number of students taking at least one drink ($n(T \cup C)$):
$n(T \cup C) = n(T) + n(C) - n(T \cap C)$
$n(T \cup C) = 150 + 225 - 100$
$n(T \cup C) = 375 - 100 = 275$.

Number of students taking neither tea nor coffee is the complement of those taking at least one:
$n(\text{Neither Tea nor Coffee}) = n(U) - n(T \cup C)$
$= 600 - 275 = 325$.
So, 325 students were taking neither tea nor coffee.

Question 14: In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Answer-
Let H be the set of students who know Hindi, and E be the set of students who know English.
$n(H) = 100$
$n(E) = 50$
$n(H \cap E) = 25$
Since each student knows at least one language, the total number of students in the group is $n(H \cup E)$.
$n(H \cup E) = n(H) + n(E) - n(H \cap E)$
$n(H \cup E) = 100 + 50 - 25$
$n(H \cup E) = 150 - 25 = 125$.
There are 125 students in the group.

Question 15: In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.

Answer-
Let H, T, I be the sets of people who read newspapers H, T, and I respectively.
$n(U) = 60$
$n(H) = 25$
$n(T) = 26$
$n(I) = 26$
$n(H \cap I) = 9$
$n(H \cap T) = 11$
$n(T \cap I) = 8$
$n(H \cap T \cap I) = 3$

(i) Number of people who read at least one newspaper = $n(H \cup T \cup I)$.
$n(H \cup T \cup I) = n(H) + n(T) + n(I) - n(H \cap T) - n(H \cap I) - n(T \cap I) + n(H \cap T \cap I)$
$n(H \cup T \cup I) = 25 + 26 + 26 - 11 - 9 - 8 + 3$
$n(H \cup T \cup I) = 77 - 28 + 3$
$n(H \cup T \cup I) = 49 + 3 = 52$.

(ii) Number of people who read exactly one newspaper.
$n(\text{H only}) = n(H) - n(H \cap T) - n(H \cap I) + n(H \cap T \cap I)$ (Using Venn Diagram logic is easier)
Or, $n(\text{H only}) = n(H) - n(H \cap T \text{ only}) - n(H \cap I \text{ only}) - n(H \cap T \cap I)$
$n(H \cap T \text{ only}) = n(H \cap T) - n(H \cap T \cap I) = 11 - 3 = 8$
$n(H \cap I \text{ only}) = n(H \cap I) - n(H \cap T \cap I) = 9 - 3 = 6$
$n(T \cap I \text{ only}) = n(T \cap I) - n(H \cap T \cap I) = 8 - 3 = 5$
$n(H \text{ only}) = n(H) - (8 + 6 + 3) = 25 - 17 = 8$
$n(T \text{ only}) = n(T) - (8 + 5 + 3) = 26 - 16 = 10$
$n(I \text{ only}) = n(I) - (6 + 5 + 3) = 26 - 14 = 12$
Number reading exactly one newspaper = $n(\text{H only}) + n(\text{T only}) + n(\text{I only})$
$= 8 + 10 + 12 = 30$.

Question 16: In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

Answer-
Let A, B, C be the sets of people who liked products A, B, and C respectively.
$n(A) = 21$
$n(B) = 26$
$n(C) = 29$
$n(A \cap B) = 14$
$n(C \cap A) = 12$
$n(B \cap C) = 14$
$n(A \cap B \cap C) = 8$

We need to find $n(C \text{ only})$.
Using the formula derived from Venn diagrams:
$n(C \text{ only}) = n(C) - n(A \cap C \text{ only}) - n(B \cap C \text{ only}) - n(A \cap B \cap C)$
First find the 'only intersection' parts:
$n(A \cap C \text{ only}) = n(A \cap C) - n(A \cap B \cap C) = 12 - 8 = 4$
$n(B \cap C \text{ only}) = n(B \cap C) - n(A \cap B \cap C) = 14 - 8 = 6$
Now calculate $n(C \text{ only})$:
$n(C \text{ only}) = n(C) - n(A \cap C \text{ only}) - n(B \cap C \text{ only}) - n(A \cap B \cap C)$
$n(C \text{ only}) = 29 - 4 - 6 - 8$
$n(C \text{ only}) = 29 - 18 = 11$.
So, 11 people liked product C only.

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Important Keywords from the Chapter

• Set: A well-defined collection of distinct objects.
• Element/Member: An object belonging to a set.
• Roster Form: Representing a set by listing its elements.
• Set-builder Form: Representing a set by stating a common property of its elements.
• Empty Set ($\emptyset$): A set with no elements.
• Finite Set: A set with a countable number of elements.
• Infinite Set: A set with an uncountable number of elements.
• Equal Sets: Sets having exactly the same elements.
• Subset ($\subseteq$): A set where all its elements are contained within another set.
• Proper Subset ($\subset$): A subset that is not equal to the original set.
• Power Set ($P(A)$): The set of all subsets of a given set A.
• Universal Set ($U$): The set containing all elements under consideration.
• Venn Diagram: A diagram representing sets using circles and rectangles.
• Union ($A \cup B$): Set of elements in A or B or both.
• Intersection ($A \cap B$): Set of elements common to both A and B.
• Disjoint Sets: Sets with no common elements ($A \cap B = \emptyset$).
• Difference ($A - B$): Set of elements in A but not in B.
• Complement ($A'$): Set of elements in U but not in A.
• Cardinality ($n(A)$): The number of elements in a set A.
• De Morgan's Laws: Laws relating complements with union and intersection.