SETS-Exercise 1.4

The chapter Sets forms the very foundation of modern mathematics and logical thinking in Class XI. Almost every advanced topic in mathematics—relations, functions, probability, calculus, and even statistics—rests on the basic language of sets. This chapter introduces students to the idea of well-defined collections, helping them distinguish between precise mathematical reasoning and vague descriptions. Through concepts such as roster form, set-builder form, types of sets, subsets, power sets, and operations on sets, learners develop clarity, accuracy, and logical discipline. The solutions presented here are written strictly in accordance with the NCERT syllabus and examination pattern. Each solution emphasizes conceptual understanding, step-by-step reasoning, and correct mathematical notation. Special care has been taken to justify answers clearly, avoid ambiguity, and use language that mirrors the NCERT textbook style. These solutions are designed not only to help students score well in school and competitive exams, but also to build a strong conceptual base for higher mathematics.

Continue Reading →
Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.2

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

Continue Reading →
Exercise
Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

Continue Reading →
Exercise
January 13, 2026  |  By Academia Aeternum

SETS-Exercise 1.4

Maths - Exercise

Q1. 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 is a natural number and multiple of 3}
B = {x : x is a natural number less than 6} (iv) A = {x : x is a natural number and 1 < x ≤6 } B={x : x is a natural number and 6 < x < 10 }
(v) A={1, 2, 3}, B=φ

Solution

Solution: The union of two sets is the set containing all elements which belong to either of the two sets or to both of them, without repetition.

\( \begin{aligned} \text{(i)}\;& X=\{1,3,5\} \\ & Y=\{1,2,3\} \\ & X\cup Y=\{1,2,3,5\}. \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& A=\{a,e,i,o,u\} \\ & B=\{a,b,c\} \\ & A\cup B=\{a,b,c,e,i,o,u\}. \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& A=\{3,6,9,\ldots\} \\ & B=\{1,2,3,4,5\} \\ & A\cup B=\{1,2,3,4,5,6,9,\ldots\}. \end{aligned} \)

\( \begin{aligned} \text{(iv)}\;& A=\{2,3,4,5,6\} \\ & B=\{7,8,9\} \\ & A\cup B=\{2,3,4,5,6,7,8,9\}. \end{aligned} \)

\( \begin{aligned} \text{(v)}\;& A=\{1,2,3\} \\ & B=\varnothing \\ & A\cup B=\{1,2,3\}. \end{aligned} \)


Q2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

Solution

Solution: If one set is a subset of another, then every element of the smaller set already belongs to the larger set. Hence, their union will be the larger set itself.

\( \begin{aligned} A &= \{a, b\} \\ B &= \{a, b, c\} \\ A &\subset B \\ \Rightarrow A \cup B &= \{a, b, c\}. \end{aligned} \)

Thus, when \(A \subset B\), the union of \(A\) and \(B\) is equal to \(B\).


Q3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

Solution

Solution: If a set \(A\) is a subset of a set \(B\), then every element of \(A\) already belongs to \(B\). Therefore, the union of \(A\) and \(B\) does not add any new elements to \(B\).

\( \begin{aligned} \text{If } A \subset B,\\ A \cup B = B. \end{aligned} \)


Q4. 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 ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vi) B ∪ C ∪ D

Solution

Solution: The union of two or more sets is the set containing all distinct elements that belong to at least one of the given sets.

\( \begin{aligned} A &= \{1,2,3,4\} \\ B &= \{3,4,5,6\} \\ C &= \{5,6,7,8\} \\ D &= \{7,8,9,10\} \end{aligned} \)

\( \begin{aligned} \text{(i)}\;& A \cup B = \{1,2,3,4,5,6\} \\ \text{(ii)}\;& A \cup C = \{1,2,3,4,5,6,7,8\} \\ \text{(iii)}\;& B \cup C = \{3,4,5,6,7,8\} \\ \text{(iv)}\;& B \cup D = \{3,4,5,6,7,8,9,10\} \end{aligned} \)

\( \begin{aligned} \text{(v)}\;& A \cup B \cup C = (A \cup B) \cup C \\ &= \{1,2,3,4,5,6\} \cup \{5,6,7,8\} \\ &= \{1,2,3,4,5,6,7,8\} \end{aligned} \)

\( \begin{aligned} \text{(vi)}\;& A \cup B \cup D = (A \cup B) \cup D \\ &= \{1,2,3,4,5,6\} \cup \{7,8,9,10\} \\ &= \{1,2,3,4,5,6,7,8,9,10\} \end{aligned} \)

\( \begin{aligned} \text{(vii)}\;& B \cup C \cup D = (B \cup C) \cup D \\ &= \{3,4,5,6,7,8\} \cup \{7,8,9,10\} \\ &= \{3,4,5,6,7,8,9,10\} \end{aligned} \)


Q5. Find the intersection of each pair of sets of question 1 above.

Solution

Solution: The intersection of two sets consists of all elements that are common to both sets.

\( \begin{aligned} \text{(i)}\;& X=\{1,3,5\} \\ & Y=\{1,2,3\} \\ & X \cap Y=\{1,3\}. \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& A=\{a,e,i,o,u\} \\ & B=\{a,b,c\} \\ & A \cap B=\{a\}. \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& A=\{3,6,9,12,\ldots\} \\ & B=\{1,2,3,4,5\} \\ & A \cap B=\{3\}. \end{aligned} \)

\( \begin{aligned} \text{(iv)}\;& A=\{2,3,4,5,6\} \\ & B=\{7,8,9\} \\ & A \cap B=\varnothing. \end{aligned} \)

\( \begin{aligned} \text{(v)}\;& A=\{1,2,3\} \\ & B=\varnothing \\ & A \cap B=\varnothing. \end{aligned} \)


Q6. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find
(i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) ( A ∩ B ) ∩ ( B ∪ C )
(x) ( A ∪ D) ∩ ( B ∪ C)

Solution

Solution: The intersection of sets contains only those elements which are common to all the sets involved in the operation.

\( \begin{aligned} A &= \{3,5,7,9,11\} \\ B &= \{7,9,11,13\} \\ C &= \{11,13,15\} \\ D &= \{15,17\} \end{aligned} \)

\( \begin{aligned} \text{(i)}\;& A \cap B = \{7,9,11\} \\ \text{(ii)}\;& B \cap C = \{11,13\} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& A \cap C \cap D = (A \cap C) \cap D \\ &= \{11\} \cap \{15,17\} \\ &= \varnothing \end{aligned} \)

\( \begin{aligned} \text{(iv)}\;& A \cap C = \{11\} \\ \text{(v)}\;& B \cap D = \varnothing \end{aligned} \)

\( \begin{aligned} \text{(vi)}\;& A \cap (B \cup C) \\ &= \{3,5,7,9,11\} \cap \{7,9,11,13,15\} \\ &= \{7,9,11\} \end{aligned} \)

\( \begin{aligned} \text{(vii)}\;& A \cap D = \varnothing \end{aligned} \)

\( \begin{aligned} \text{(viii)}\;& A \cap (B \cup D) \\ &= \{3,5,7,9,11\} \cap \{7,9,11,13,15,17\} \\ &= \{7,9,11\} \end{aligned} \)

\( \begin{aligned} \text{(ix)}\;& (A \cap B) \cap (B \cup C) \\ &= \{7,9,11\} \cap \{7,9,11,13,15\} \\ &= \{7,9,11\} \end{aligned} \)

\( \begin{aligned} \text{(x)}\;& (A \cup D) \cap (B \cup C) \\ &= \{3,5,7,9,11,15,17\} \cap \{7,9,11,13,15\} \\ &= \{7,9,11,15\} \end{aligned} \)


Q7. If A = {x : x is a natural number }, B = {x : x is an even natural number} C = {x : x is an odd natural number}andD = {x : x is a prime number }, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D

Solution

Solution: The intersection of two sets consists of all elements which are common to both sets.

\( \begin{aligned} A &= \{1,2,3,4,\ldots\} \\ B &= \{2,4,6,8,\ldots\} \\ C &= \{1,3,5,7,\ldots\} \\ D &= \{2,3,5,7,11,\ldots\} \end{aligned} \)

\( \begin{aligned} \text{(i)}\;& A \cap B = \{2,4,6,\ldots\} = B \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& A \cap C = \{1,3,5,7,\ldots\} = C \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& A \cap D = \{2,3,5,7,11,\ldots\} = D \end{aligned} \)

\( \begin{aligned} \text{(iv)}\;& B \cap C = \varnothing \end{aligned} \)

\( \begin{aligned} \text{(v)}\;& B \cap D = \{2\} \end{aligned} \)

\( \begin{aligned} \text{(vi)}\;& C \cap D = \{3,5,7,11,\ldots\} \end{aligned} \)


Q8.Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }
(ii) { a, e, i, o, u } and { c, d, e, f }
(iii) {x : x is an even integer } and {x : x is an odd integer}

Solution

Solution: Two sets are said to be disjoint if they have no element in common, that is, if their intersection is the empty set.

\( \begin{aligned} \text{(i)}\;& \{1,2,3,4\} \text{ and } \{x : x \text{ is a natural number and } 4 \le x \le 6\}. \\ &\text{The second set is } \{4,5,6\}. \\ &\text{Since } 4 \text{ is common to both sets,} \\ &\Rightarrow \text{these sets are not disjoint.} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& \{a,e,i,o,u\} \text{ and } \{c,d,e,f\}. \\ &\text{The element } e \text{ is common to both sets.} \\ &\Rightarrow \text{these sets are not disjoint.} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& \{x : x \text{ is an even integer}\} \text{ and } \{x : x \text{ is an odd integer}\}. \\ &\text{No integer can be both even and odd at the same time.} \\ &\Rightarrow \text{the intersection is } \varnothing, \text{ so these sets are disjoint.} \end{aligned} \)

Hence, only the pair given in (iii) consists of disjoint sets.


Q9. 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 –
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii) D – C

Solution

Solution: The difference of two sets \(A - B\) consists of all those elements of \(A\) which do not belong to \(B\).

\( \begin{aligned} A &= \{3,6,9,12,15,18,21\} \\ B &= \{4,8,12,16,20\} \\ C &= \{2,4,6,8,10,12,14,16\} \\ D &= \{5,10,15,20\} \end{aligned} \)

\( \begin{aligned} \text{(i)}\;& A - B = \{3,6,9,15,18,21\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& A - C = \{3,9,15,18,21\} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& A - D = \{3,6,9,12,18,21\} \end{aligned} \)

\( \begin{aligned} \text{(iv)}\;& B - A = \{4,8,16,20\} \end{aligned} \)

\( \begin{aligned} \text{(v)}\;& C - A = \{2,4,8,10,14,16\} \end{aligned} \)

\( \begin{aligned} \text{(vi)}\;& D - A = \{5,10,20\} \end{aligned} \)

\( \begin{aligned} \text{(vii)}\;& B - C = \{20\} \end{aligned} \)

\( \begin{aligned} \text{(viii)}\;& B - D = \{4,8,12,16\} \end{aligned} \)

\( \begin{aligned} \text{(ix)}\;& C - B = \{2,6,10,14\} \end{aligned} \)

\( \begin{aligned} \text{(x)}\;& D - B = \{5,10,15\} \end{aligned} \)

\( \begin{aligned} \text{(xi)}\;& C - D = \{2,4,6,8,12,14,16\} \end{aligned} \)

\( \begin{aligned} \text{(xii)}\;& D - C = \{5,15,20\} \end{aligned} \)


Q10. If X= { a, b, c, d } and Y = { f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y

Solution

Solution: The difference of two sets consists of elements belonging to the first set but not to the second, while the intersection consists of elements common to both sets.

\( \begin{aligned} X &= \{a, b, c, d\} \\ Y &= \{f, b, d, g\} \end{aligned} \)

\( \begin{aligned} \text{(i)}\;& X - Y = \{a, c\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& Y - X = \{f, g\} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& X \cap Y = \{b, d\} \end{aligned} \)


Q11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Solution

Solution: Let \(R\) denote the set of all real numbers and \(Q\) denote the set of all rational numbers. Every real number is either rational or irrational.

\( \begin{aligned} R &= \{\text{all real numbers}\} \\ Q &= \{\text{all rational numbers}\} \\ R - Q &= \{\text{real numbers which are not rational}\}. \end{aligned} \)

Hence, \(R - Q\) represents the set of all irrational numbers.


Q12. State whether each of the following statement 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.

Solution

Solution: Two sets are said to be disjoint if they have no element in common, that is, if their intersection is the empty set.

\( \begin{aligned} \text{(i)}\;& \{2,3,4,5\} \text{ and } \{3,6\}. \\ &\text{The element } 3 \text{ is common to both sets.} \\ &\Rightarrow \text{The statement is false.} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\;& \{a,e,i,o,u\} \text{ and } \{a,b,c,d\}. \\ &\text{The element } a \text{ is common to both sets.} \\ &\Rightarrow \text{The statement is false.} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\;& \{2,6,10,14\} \text{ and } \{3,7,11,15\}. \\ &\text{No element of the first set appears in the second set.} \\ &\Rightarrow \text{The statement is true.} \end{aligned} \)

\( \begin{aligned} \text{(iv)}\;& \{2,6,10\} \text{ and } \{3,7,11\}. \\ &\text{These two sets also have no element in common.} \\ &\Rightarrow \text{The statement is true.} \end{aligned} \)


Frequently Asked Questions

A set is a well-defined collection of distinct objects called elements.

So that it is possible to clearly decide whether a given object belongs to the set or not.

The individual objects or members contained in a set are called its elements.

Sets are generally denoted by capital letters such as \(A,\, B,\, C\).

Elements are represented by small letters such as \(a, \,b,\, x\).

It means “belongs to” or “is an element of”.

It means “does not belong to” a given set.

A method of listing all elements of a set within curly braces.

A representation describing a set by a common property satisfied by its elements.

(\A = {2,4,6,8}\).

\(A = {x : x \text{ is an even natural number}}\).

A set containing no elements, denoted by \(\varnothing\).

Yes, there is only one empty set.

A set containing exactly one element.

A set with a definite number of elements.

A set with an unlimited number of elements.

A set containing all objects under consideration for a particular discussion.

It is usually denoted by \(U\).

A set \(A\) is a subset of \(B\) if every element of \(A\) is also an element of \(B\).

A subset that is not equal to the original set.

\(\subseteq\) represents subset, and \(\subset\) represents proper subset.

A set with \(n\) elements has \(2^n\) subsets.

The set of all subsets of a given set.

It is denoted by \(P(A)\).

The union of two sets contains all elements belonging to either or both sets.

The symbol for union is \(\cup\).

The intersection contains only those elements common to both sets.

The symbol for intersection is \(\cap\).

Sets having no common elements.

The set of all elements in the universal set not belonging to the given set.

The complement of \(A\) is written as \(A^'\) or \(A^c\).

\(\varnothing\) has no elements, while \({\varnothing}\) has one element.

\(A \cup B = B \cup A\).

\(A \cap B = B \cap A\).

\((A \cup B) \cup C = A \cup (B \cup C)\).

\((A \cap B) \cap C = A \cap (B \cap C)\).

\(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\).

\((A \cup B)' = A' \cap B'\).

\((A \cap B)' = A' \cup B'\).

Diagrams using closed curves to represent sets and their relationships visually.

They simplify understanding of set operations and relationships.

The number of elements in a set.

It is denoted by \(n(A)\).

\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\).

A method to calculate the number of elements in unions of sets accurately.

No, as per standard definition, a set does not contain itself.

No, all elements of a set must be distinct.

Conceptual definitions, Venn diagram problems, formulas, and numerical applications.

It provides the basic language and structure for all other topics in mathematics.

Sets are used in data classification, logic, probability, and computer science.

Recent posts


    Important Links

    Leave Your Message & Comments