Sets: Exercise 1.3 – Guided Solutions
Work through every question with crystal‑clear steps, stress–strain visuals and exam‑oriented hints – all in a sleek deep‑space layout.
Q1 Make correct statements by filling in the symbols \( \subset \) or \( \not\subset \).
(i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 }
(ii) { a, b, c } . . . { b, c, d }
(iii) {x : x is a student of Class XI of your school} . . . {x : x is a student of your school}
(iv) {x : x is a circle in the plane} . . . {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} . . . {x : x is an integer}
📘 Concept Theory
A set \(A\) is called a subset of another set \(B\) if every element of \(A\) is also an element of \(B\).
Mathematically:
\[ A \subset B \quad \text{if} \quad x \in A \Rightarrow x \in B \]
If even one element of \(A\) does not belong to \(B\), then
\[ A \not\subset B \]
Thus, to determine subset relations we simply check whether all elements of the first set appear in the second set.
🧠 Solution Roadmap
To solve each statement, follow these steps:
- Step 1: Identify the first set (candidate subset).
- Step 2: Identify the second set (reference set).
- Step 3: Check whether every element of the first set belongs to the second set.
- Step 4: If yes → use \( \subset \); otherwise → use \( \not\subset \).
✔ Solution
\( \begin{aligned} (i)\;& \{2,3,4\} \subset \{1,2,3,4,5\} \end{aligned} \)
Every element of the first set appears in the second set.
\( \begin{aligned} (ii)\;& \{a,b,c\} \not\subset \{b,c,d\} \end{aligned} \)
The element \(a\) is not present in the second set.
\( \begin{aligned} (iii)\;& \{x : x \text{ is a Class XI student of your school}\} \subset \{x : x \text{ is a student of your school}\} \end{aligned} \)
All Class XI students are obviously students of the same school.
\( \begin{aligned} (iv)\;& \{x : x \text{ is a circle in the plane}\} \not\subset \{x : x \text{ is a circle with radius 1 unit}\} \end{aligned} \)
Circles may have many radii (2, 3, 5, etc.), so not all circles have radius 1.
\( \begin{aligned} (v)\;& \{x : x \text{ is a triangle}\} \not\subset \{x : x \text{ is a rectangle}\} \end{aligned} \)
Triangles and rectangles are completely different geometric figures.
\( \begin{aligned} (vi)\;& \{x : x \text{ is an equilateral triangle}\} \subset \{x : x \text{ is a triangle}\} \end{aligned} \)
Every equilateral triangle satisfies the definition of a triangle.
\( \begin{aligned} (vii)\;& \{x : x \text{ is an even natural number}\} \subset \{x : x \text{ is an integer}\} \end{aligned} \)
Even natural numbers like \(2,4,6\) are also integers.
🎯 Significance for Board & Competitive Exams
- Understanding subset relations is fundamental for set algebra.
- Used extensively in Venn diagram reasoning questions.
- Important for probability, relations, and functions in later chapters.
- Frequently tested in JEE Main, NDA, Olympiads, and other entrance exams.
Mastering subset identification helps students quickly solve questions involving union, intersection, complement, and power sets.
Q2 Examine whether the following statements are true or false.
(i) { a, b } ⊄ { b, c, a }
(ii) { a, e } ⊂ { x : x is a vowel in the English alphabet }
(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }
(iv) { a } ⊂ { a, b, c }
(v) { a } ∈ { a, b, c }
(vi) { x : x is an even natural number less than 6 } ⊂ { x : x is a natural number which divides 36 }
📘 Concept Theory
In set theory we use the following symbols:
- \(A \subset B\) → every element of set \(A\) belongs to set \(B\)
- \(A \not\subset B\) → at least one element of \(A\) does not belong to \(B\)
- \(x \in A\) → element \(x\) belongs to set \(A\)
- \(x \notin A\) → element \(x\) does not belong to set \(A\)
Students often confuse the difference between
subset ( ⊂ ) and element ( ∈ ).
Example:
\[ a \in \{a,b,c\} \]
but
\[ \{a\} \subset \{a,b,c\} \]
because the set \( \{a\} \) contains the element \(a\).🧠 Solution Roadmap
To verify each statement:
- Step 1: Identify whether the symbol refers to subset or element.
- Step 2: Compare elements of the first set with the second set.
- Step 3: Check whether the statement agrees with the definition.
- Step 4: Declare the statement true or false.
✔ Solution
\( \begin{aligned} (i)\;& \{a,b\} \not\subset \{b,c,a\} \end{aligned} \)
Both elements \(a\) and \(b\) are present in the second set.
Therefore,
\[ \{a,b\} \subset \{b,c,a\} \]
Hence the given statement is FALSE.
\( \begin{aligned} (ii)\;& \{a,e\} \subset \{x : x \text{ is a vowel}\} \end{aligned} \)
The vowel set in English is
\[ \{a,e,i,o,u\} \]
Both \(a\) and \(e\) belong to this set.
Hence the statement is TRUE.
\( \begin{aligned} (iii)\;& \{1,2,3\} \subset \{1,3,5\} \end{aligned} \)
Element \(2\) does not belong to the second set.
Therefore the statement is FALSE.
\( \begin{aligned} (iv)\;& \{a\} \subset \{a,b,c\} \end{aligned} \)
The only element \(a\) belongs to the second set.
Hence the statement is TRUE.
\( \begin{aligned} (v)\;& \{a\} \in \{a,b,c\} \end{aligned} \)
The elements of the set \( \{a,b,c\} \) are
\[ a,\; b,\; c \]
The set \( \{a\} \) itself is not an element of the set.
Hence the statement is FALSE.
\( \begin{aligned} (vi)\;& \{x : x \text{ is an even natural number less than }6\} \subset \{x : x \text{ divides }36\} \end{aligned} \)
First set:
\[ \{2,4\} \]
Since
\[ 36 \div 2 = 18, \quad 36 \div 4 = 9 \]
both numbers divide 36.
Hence the statement is TRUE.
🎯 Significance for Board & Competitive Exams
- This question tests the difference between subset (⊂) and element (∈).
- Such conceptual questions are frequently asked in JEE Main, NDA, Olympiads and foundation tests.
- Understanding these relations is essential for solving problems involving:
- Venn diagrams
- Set algebra
- Relations and functions
- Probability
Mastering these basic concepts helps students avoid common mistakes in advanced set theory problems.
Q3 Let \( A = \{ 1, 2, \{3,4\}, 5 \} \). Which of the following statements are incorrect? Give reasons.
(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) \( \varnothing \in A \)
(x) \( \varnothing \subset A \)
(xi) \( \{\varnothing\} \subset A \)
📘 Concept Theory
The given set is
\[ A = \{1, 2, \{3,4\}, 5\} \]
The elements of \(A\) are:
\[ 1,\; 2,\; \{3,4\},\; 5 \]
Important Observation
The set \( \{3,4\} \) appears as a single element inside \(A\). However, the numbers \(3\) and \(4\) themselves are not elements of \(A\).
This question tests three important ideas:
- Difference between element (∈) and subset (⊂)
- Sets that contain other sets as elements
- Properties of the empty set \( \varnothing \)
🧠 Solution Roadmap
To determine correctness:
- Step 1: Identify whether the statement uses \( \in \) or \( \subset \).
- Step 2: Check the elements of the given set \(A\).
- Step 3: Verify whether the required elements belong to \(A\).
- Step 4: Conclude whether the statement is true or false.
✔ Solution
\( \begin{aligned} (i)\;& \{3,4\} \subset A \end{aligned} \)
For subset relation, elements \(3\) and \(4\) must belong to \(A\). But \(A\) contains the set \( \{3,4\} \) as a single element, not the numbers 3 and 4 individually.
Incorrect statement
\( \begin{aligned} (ii)\;& \{3,4\} \in A \end{aligned} \)
The set \( \{3,4\} \) itself appears as an element of \(A\).
Correct statement
\( \begin{aligned} (iii)\;& \{\{3,4\}\} \subset A \end{aligned} \)
The set \( \{\{3,4\}\} \) has one element: \( \{3,4\} \). Since this element belongs to \(A\), it is a subset of \(A\).
Correct statement
\( \begin{aligned} (iv)\;& 1 \in A \end{aligned} \)
1 is explicitly listed as an element of \(A\).
Correct statement
\( \begin{aligned} (v)\;& 1 \subset A \end{aligned} \)
The symbol \( \subset \) is defined only between sets. Since \(1\) is a number and not a set, this relation is meaningless.
Incorrect statement
\( \begin{aligned} (vi)\;& \{1,2,5\} \subset A \end{aligned} \)
All elements \(1,2,5\) belong to \(A\).
Correct statement
\( \begin{aligned} (vii)\;& \{1,2,5\} \in A \end{aligned} \)
The set \( \{1,2,5\} \) is not listed as an element of \(A\).
Incorrect statement
\( \begin{aligned} (viii)\;& \{1,2,3\} \subset A \end{aligned} \)
Although 1 and 2 belong to \(A\), the element 3 does not belong to \(A\).
Incorrect statement
\( \begin{aligned} (ix)\;& \varnothing \in A \end{aligned} \)
The empty set does not appear as an element of \(A\).
Incorrect statement
\( \begin{aligned} (x)\;& \varnothing \subset A \end{aligned} \)
The empty set is a subset of every set.
Correct statement
\( \begin{aligned} (xi)\;& \{\varnothing\} \subset A \end{aligned} \)
This would require \( \varnothing \in A \), which is not true.
Incorrect statement
Thus the incorrect statements are:
(i), (v), (vii), (viii), (ix), (xi)
🎯 Significance for Board & Competitive Exams
- This problem tests the difference between element and subset.
- It also examines understanding of sets containing other sets.
- Concepts of the empty set and nested sets are frequently tested in JEE and Olympiad questions.
- Mastering such questions helps avoid common mistakes in:
- Venn diagram problems
- Power set questions
- Set identities
- Relations and functions
Such conceptual clarity is essential for solving advanced set theory problems in JEE Main, NDA, Olympiads, and other competitive exams.
Q4 Write down all the subsets of the following sets.
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) \( \varnothing \)
📘 Concept Theory – Subsets & Power Set
A set \(A\) is called a subset of a set \(B\) if every element of \(A\) belongs to \(B\).
The collection of all subsets of a set is called the power set.
If a set contains \(n\) elements, then the total number of subsets is
\[ 2^n \]
because each element may either be included or excluded from a subset.
🧠 Solution Roadmap
To list all subsets of a set:
- Step 1: Always include the empty set \( \varnothing \).
- Step 2: Write subsets containing single elements.
- Step 3: Write subsets containing two elements.
- Step 4: Continue until the complete set itself appears.
- Step 5: Verify the total number using \(2^n\).
✔ Solution
\( \begin{aligned} (i)\;& \{a\} \end{aligned} \)
Number of elements = 1 Total subsets = \(2^1 = 2\)
Subsets:
\[ \{\varnothing, \{a\}\} \]
\( \begin{aligned} (ii)\;& \{a,b\} \end{aligned} \)
Number of elements = 2 Total subsets = \(2^2 = 4\)
Subsets:
\[ \{\varnothing,\{a\},\{b\},\{a,b\}\} \]
\( \begin{aligned} (iii)\;& \{1,2,3\} \end{aligned} \)
Number of elements = 3 Total subsets = \(2^3 = 8\)
All subsets are
\[ \{\varnothing, \{1\},\{2\},\{3\}, \{1,2\},\{1,3\},\{2,3\}, \{1,2,3\}\} \]
\( \begin{aligned} (iv)\;& \varnothing \end{aligned} \)
The empty set has zero elements.
Total subsets:
\[ 2^0 = 1 \]
Therefore its only subset is
\[ \{\varnothing\} \]
🎯 Significance for Board & Competitive Exams
- The formula \(2^n\) for number of subsets is frequently asked in JEE Main and Olympiad questions.
- Understanding subset generation is essential for:
- Power set problems
- Set identities
- Combinatorics questions
- Probability problems involving sample spaces
- Board examinations often ask students to list all subsets of small sets.
A strong understanding of subsets forms the foundation for advanced topics such as relations, functions, and combinatorial counting.
Q5 Write the following sets in interval notation.
(i) {x : x ∈ R, –4 < x ≤ 6}
(ii) {x : x ∈ R, –12 < x < –10}
(iii) {x : x ∈ R, 0 ≤ x < 7}
(iv) {x : x ∈ R, 3 ≤ x ≤ 4}
📘 Concept Theory – Interval Notation
A continuous set of real numbers between two endpoints can be written using interval notation.
The type of bracket used depends on whether the endpoint is included.
| Condition | Interval Notation | Meaning |
|---|---|---|
| \(a < x < b\) | \((a,b)\) | Both endpoints excluded |
| \(a \le x \le b\) | \([a,b]\) | Both endpoints included |
| \(a < x \le b\) | \((a,b]\) | Left excluded, right included |
| \(a \le x < b\) | \([a,b)\) | Left included, right excluded |
🧠 Solution Roadmap
- Step 1: Identify the lower and upper limits.
- Step 2: Check whether the inequality is strict (< or>).
- Step 3: Use round brackets ( ) for excluded endpoints.
- Step 4: Use square brackets [ ] for included endpoints.
✔ Solution
\( \begin{aligned} (i)\;& \{x : x \in \mathbb{R}, -4 < x \le 6\} \end{aligned} \)
Since \(-4\) is excluded and \(6\) is included:
\[ (-4,6] \]
\( \begin{aligned} (ii)\;& \{x : x \in \mathbb{R}, -12 < x < -10\} \end{aligned} \)
Both endpoints are excluded.
\[ (-12,-10) \]
\( \begin{aligned} (iii)\;& \{x : x \in \mathbb{R}, 0 \le x < 7\} \end{aligned} \)
0 is included but 7 is excluded.
\[ [0,7) \]
\( \begin{aligned} (iv)\;& \{x : x \in \mathbb{R}, 3 \le x \le 4\} \end{aligned} \)
Both endpoints are included.
\[ [3,4] \]
🎯 Significance for Board & Competitive Exams
- Interval notation is fundamental for understanding real number sets.
- Used extensively in:
- Inequalities
- Graphs of functions
- Domain and range problems
- Calculus intervals
- Such conversions are frequently asked in JEE Main, NDA, Olympiads, and board exams.
Mastery of interval notation helps students correctly interpret solution sets of inequalities and function domains.
Q6 Write the following intervals in set-builder form.
(i) (–3, 0)
(ii) [6 , 12]
(iii) (6, 12]
(iv) [–23, 5)
📘 Concept Theory – Interval Form vs Set-Builder Form
Real number intervals can be expressed in two common forms:
- Interval notation → compact form using brackets
- Set-builder form → describes numbers using inequalities
The type of bracket indicates whether endpoints are included.
| Interval | Meaning | Set-Builder Form |
|---|---|---|
| (a,b) | endpoints excluded | \(a |
| [a,b] | endpoints included | \(a\le x\le b\) |
| (a,b] | left excluded | \(a |
| [a,b) | right excluded | \(a\le x |
🧠 Solution Roadmap
- Step 1: Identify the lower and upper endpoints.
- Step 2: Determine whether each endpoint is included or excluded.
- Step 3: Convert brackets to inequality symbols.
- Step 4: Express the result using set-builder notation.
✔ Solution
\( \begin{aligned} (i)\;& (-3,0) \end{aligned} \)
Both endpoints are excluded.
\[ \{x : x \in \mathbb{R},\; -3 < x < 0\} \]
\( \begin{aligned} (ii)\;& [6,12] \end{aligned} \)
Both endpoints are included.
\[ \{x : x \in \mathbb{R},\; 6 \le x \le 12\} \]
\( \begin{aligned} (iii)\;& (6,12] \end{aligned} \)
6 is excluded while 12 is included.
\[ \{x : x \in \mathbb{R},\; 6 < x \le 12\} \]
\( \begin{aligned} (iv)\;& [-23,5) \end{aligned} \)
−23 is included while 5 is excluded.
\[ \{x : x \in \mathbb{R},\; -23 \le x < 5\} \]
🎯 Significance for Board & Competitive Exams
- Converting between interval notation and set-builder form is a basic skill in real number analysis.
- It is frequently required in:
- inequality problems
- domain and range of functions
- graphing functions
- solution sets of equations
- Such questions are common in JEE Main, NDA, Olympiads, and board examinations.
Understanding this conversion helps students interpret continuous sets of real numbers in algebra and calculus.
Q7 What universal set(s) would you propose for each of the following?
(i) The set of right triangles
(ii) The set of isosceles triangles
📘 Concept Theory – Universal Set
A universal set is the set that contains all objects under consideration in a particular discussion.
It is usually denoted by
\[ U \]
Every set considered in the problem must be a subset of the universal set.
Thus, if \(A\) is a given set, the universal set must satisfy
\[ A \subset U \]
🧠 Solution Roadmap
- Step 1: Identify the given set.
- Step 2: Determine a larger set containing all its elements.
- Step 3: Choose a natural mathematical category as the universal set.
✔ Solution
\( \begin{aligned} (i)\;& \text{Set of right triangles} \end{aligned} \)
Every right triangle belongs to the class of triangles.
Therefore a suitable universal set is
\[ U = \{\text{all triangles in a plane}\} \]
Hence,
\[ \text{Right triangles} \subset U \]
\( \begin{aligned} (ii)\;& \text{Set of isosceles triangles} \end{aligned} \)
Every isosceles triangle is also a triangle.
Thus the universal set can again be chosen as
\[ U = \{\text{all triangles in a plane}\} \]
Hence,
\[ \text{Isosceles triangles} \subset U \]
🔎 Visual Illustration
🎯 Significance for Board & Competitive Exams
- The concept of universal sets is essential for understanding complements of sets.
- It is used extensively in:
- Venn diagrams
- Set operations
- Probability theory
- Questions involving identification of universal sets appear in JEE foundation, Olympiad, and board examinations.
Recognizing an appropriate universal set helps students correctly perform set operations and logical classifications.
Q8
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 a universal set for the sets \(A, B,\) and \(C\)?
(i) {0,1,2,3,4,5,6}
(ii) \( \varnothing \)
(iii) {0,1,2,3,4,5,6,7,8,9,10}
(iv) {1,2,3,4,5,6,7,8}
📘 Concept Theory – Universal Set
A universal set is a set that contains all elements under discussion.
If \(U\) is the universal set and \(A,B,C\) are given sets, then
\[ A \subset U, \quad B \subset U, \quad C \subset U \]
Therefore, the universal set must contain every element present in the sets \(A\), \(B\), and \(C\).
🧠 Solution Roadmap
- Step 1: List all elements present in sets \(A\), \(B\), and \(C\).
- Step 2: Check whether each proposed set contains these elements.
- Step 3: A valid universal set must contain all of them.
✔ Solution
First list the elements:
\[ A = \{1,3,5\} \] \[ B = \{2,4,6\} \] \[ C = \{0,2,4,6,8\} \]
Combined elements involved are:
\[ \{0,1,2,3,4,5,6,8\} \]
\( \begin{aligned} (i)\;& \{0,1,2,3,4,5,6\} \end{aligned} \)
Element \(8\) (present in set \(C\)) is missing.
Not a universal set
\( \begin{aligned} (ii)\;& \varnothing \end{aligned} \)
The empty set contains no elements.
It cannot contain the elements of \(A\), \(B\), and \(C\).
\( \begin{aligned} (iii)\;& \{0,1,2,3,4,5,6,7,8,9,10\} \end{aligned} \)
This set contains all elements of \(A\), \(B\), and \(C\).
Valid universal set
\( \begin{aligned} (iv)\;& \{1,2,3,4,5,6,7,8\} \end{aligned} \)
Element \(0\) from set \(C\) is missing.
Not a universal set
Hence only option (iii) can serve as the universal set.
🔎 Visual Illustration
🎯 Significance for Board & Competitive Exams
- Identifying universal sets is important for performing set operations.
- The concept is essential for:
- complements of sets
- Venn diagram problems
- probability sample spaces
- Such classification questions appear in JEE foundation tests, Olympiads, and board examinations.
Understanding universal sets helps students correctly apply union, intersection, and complement operations.
🧠 Interactive Universal Set Finder
Enter the sets A, B, and C (comma separated). The system will compute the minimum universal set, check candidate sets, and display a visual Venn diagram.