SETS-Exercise 1.5

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.

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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...

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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...

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January 14, 2026  |  By Academia Aeternum

SETS-Exercise 1.5

Maths - Exercise

Q1. 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 ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′

Solution

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

Since complements are taken with respect to the universal set \(U\), we determine each result accordingly.

\( \begin{aligned} \text{(i)}\quad A' &= U - A = \{5,6,7,8,9\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\quad B' &= U - B = \{1,3,5,7,9\} \end{aligned} \)

First, \(A \cup C = \{1,2,3,4,5,6\}\).

\( \begin{aligned} \text{(iii)}\quad (A \cup C)' &= U - (A \cup C) = \{7,8,9\} \end{aligned} \)

Next, \(A \cup B = \{1,2,3,4,6,8\}\).

\( \begin{aligned} \text{(iv)}\quad (A \cup B)' &= U - (A \cup B) = \{5,7,9\} \end{aligned} \)

\( \begin{aligned} \text{(v)}\quad (A')' &= A = \{1,2,3,4\} \end{aligned} \)

Now, \(B - C = \{2,8\}\).

\( \begin{aligned} \text{(vi)}\quad (B - C)' &= U - (B - C) = \{1,3,4,5,6,7,9\} \end{aligned} \)


Q2. 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}

Solution

\( \begin{aligned} U &= \{a,b,c,d,e,f,g,h\} \end{aligned} \)

The complement of a set is obtained by removing all its elements from the universal set \(U\).

\( \begin{aligned} \text{(i)}\quad A &= \{a,b,c\} \\ A' &= \{d,e,f,g,h\} \end{aligned} \)


\( \begin{aligned} \text{(ii)}\quad B &= \{d,e,f,g\} \\ B' &= \{a,b,c,h\} \end{aligned} \)


\( \begin{aligned} \text{(iii)}\quad C &= \{a,c,e,g\} \\ C' &= \{b,d,f,h\} \end{aligned} \)


\( \begin{aligned} \text{(iv)}\quad D &= \{f,g,h,a\} \\ D' &= \{b,c,d,e\} \end{aligned} \)


Q3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x : x is an even natural number}
(ii) { x : x is an odd natural number }
(iii) {x : x is a positive multiple of 3}
(iv) { x : x is a prime number }
(v) {x : x is a natural number divisible by 3 and 5}
(vi) { x : x is a perfect square }
(vii) { x : x is a perfect cube}
(viii) { x : x + 5 = 8 }
(ix) { x : 2x + 5 = 9}
(x) { x : x ≥ 7 }
(xi) { x : x ∈ N and 2x + 1 > 10 }

Solution

\( U=\{1,2,3,4,5,\ldots\} \)

Here, the universal set is the set of all natural numbers. The complement of each set consists of all natural numbers that do not satisfy the given condition.

\( \begin{aligned} \text{(i)}\quad A &= \{x : x \text{ is an even natural number}\} \\ A' &= \{x : x \text{ is an odd natural number}\} \end{aligned} \)

\( \begin{aligned} \text{(ii)}\quad B &= \{x : x \text{ is an odd natural number}\} \\ B' &= \{x : x \text{ is an even natural number}\} \end{aligned} \)

\( \begin{aligned} \text{(iii)}\quad C &= \{x : x \text{ is a positive multiple of } 3\} \\ C' &= \{x : x \text{ is not a positive multiple of } 3\} \end{aligned} \)

\( \begin{aligned} \text{(iv)}\quad D &= \{x : x \text{ is a prime number}\} \\ D' &= \{x : x \text{ is a non-prime natural number}\} \end{aligned} \)

\( \begin{aligned} \text{(v)}\quad E &= \{x : x \text{ is divisible by both } 3 \text{ and } 5\} \\ E' &= \{x : x \text{ is not divisible by both } 3 \text{ and } 5\} \end{aligned} \)

\( \begin{aligned} \text{(vi)}\quad F &= \{x : x \text{ is a perfect square}\} \\ F' &= \{x : x \text{ is not a perfect square}\} \end{aligned} \)

\( \begin{aligned} \text{(vii)}\quad G &= \{x : x \text{ is a perfect cube}\} \\ G' &= \{x : x \text{ is not a perfect cube}\} \end{aligned} \)

\( \begin{aligned} \text{(viii)}\quad H &= \{x : x+5=8\} \\ H' &= \{x : x+5 \neq 8\} \end{aligned} \)

\( \begin{aligned} \text{(ix)}\quad I &= \{x : 2x+5=9\} \\ I' &= \{x : 2x+5 \neq 9\} \end{aligned} \)

\( \begin{aligned} \text{(x)}\quad J &= \{x : x \ge 7\} \\ J' &= \{x : 1 \le x < 7,\; x \in \mathbb{N}\} \end{aligned} \)

\( \begin{aligned} \text{(xi)}\quad K &= \{x : x \in \mathbb{N} \text{ and } 2x+1>10\} \\ K' &= \{x : x \in \mathbb{N} \text{ and } 2x+1 \le 10\} \end{aligned} \)


Q4. 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 ∪ B)′ = A′ ∩ B′
(ii) (A ∩ B)′ = A′ ∪ B′

Solution

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

We verify each identity by evaluating both sides using the given universal set.

\( \begin{aligned} \text{(i)}\quad (A \cup B)' &= A' \cap B' \end{aligned} \)

\( \begin{aligned} A \cup B &= \{2,3,4,5,6,7,8\} \\ (A \cup B)' &= U - (A \cup B) = \{1,9\} \end{aligned} \)

\( \begin{aligned} A' &= U - A = \{1,3,5,7,9\} \\ B' &= U - B = \{1,4,6,8,9\} \end{aligned} \)

\( \begin{aligned} A' \cap B' &= \{1,9\} \end{aligned} \)

Thus, \( (A \cup B)' = A' \cap B' \) and the identity is verified.

\( \begin{aligned} \text{(ii)}\quad (A \cap B)' &= A' \cup B' \end{aligned} \)

\( \begin{aligned} A \cap B &= \{2\} \\ (A \cap B)' &= U - (A \cap B) = \{1,3,4,5,6,7,8,9\} \end{aligned} \)

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

\( \begin{aligned} A' \cup B' &= \{1,3,4,5,6,7,8,9\} \end{aligned} \)

Hence, \( (A \cap B)' = A' \cup B' \) and the second identity is also verified.


Q5. Draw appropriate Venn diagram for each of the following :
(i) (A ∪ B)′,
(ii) A′ ∩ B′,
(iii) (A ∩ B)′,
(iv) A′ ∪ B′

Solution

(A ∪ B)′
\((A\cup B)'\)
A' Ո B'
\(A'\cap B'\)
(A Ո B)'
\((A\cap B)'\)
A' U B'
\(A'\cup B'\)

Q6. 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°, what is A′?

Solution

\( U=\{x : x \text{ is a triangle in a plane}\} \)

Here, the universal set \(U\) consists of all possible triangles in a plane.

\( A=\{x : x \text{ is a triangle with at least one angle different from } 60^\circ\} \)

The complement of \(A\) consists of all triangles that do not satisfy the given condition.

\( A'=\{x : x \text{ is a triangle in which all angles are } 60^\circ\} \)

Thus, \(A'\) is the set of all equilateral triangles.


Q7. Fill in the blanks to make each of the following a true statement :
(i) A ∪ A′ = . . .
(ii) φ′ ∩ A = . . .
(iii) A ∩ A′ = . . .
(iv) U′ ∩ A = . . .

Solution

Let \(U\) denote the universal set and \(A\) be any subset of \(U\). We evaluate each expression using basic properties of sets and their complements.

\( \begin{aligned} \text{(i)}\quad A \cup A' &= U \end{aligned} \)

The union of a set with its complement contains every element of the universal set.

\( \begin{aligned} \text{(ii)}\quad \varnothing' \cap A &= U \cap A = A \end{aligned} \)

Since the complement of the empty set is the universal set, its intersection with \(A\) gives back the set \(A\).

\( \begin{aligned} \text{(iii)}\quad A \cap A' &= \varnothing \end{aligned} \)

A set and its complement have no common elements, so their intersection is the empty set.

\( \begin{aligned} \text{(iv)}\quad U' \cap A &= \varnothing \cap A = \varnothing \end{aligned} \)

The complement of the universal set is the empty set, and its intersection with any set is again the empty set.


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.

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