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Chapter 1 - Types of Relations (PYQs)

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This page provides comprehensive Class 12 Maths Ch 1 Types of Relations PYQs. Chapter 1 - Types of Relations (PYQs) - Free study material for Class 12 Maths. NCERT Solutions, Notes, and PYQs.

Class 12 Mathematics | Relations & Functions

Q1 2021
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Write the smallest reflexive relation on set $A = \{a, b, c\}$.
A relation is called reflexive if every element of the set is related to itself.
So for set $A = \{a, b, c\}$, the ordered pairs $(a, a)$, $(b, b)$ and $(c, c)$ must be included.
The smallest such relation contains only these pairs and no others.
Final Answer: R = {(a, a), (b, b), (c, c)}
Q2 2020
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A relation R in a set A is called ______ if (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R, for all a₁, a₂ ∈ A.
The given condition means that if an ordered pair (a₁, a₂) is in R, then the reversed pair (a₂, a₁) must also be in R.
This is exactly the definition of a symmetric relation.
Final Answer: The relation is called a symmetric relation.
Q3 2020
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A relation R in a set A is called ______ relation, if each element of A is related to itself.
If every element of a set is related to itself, then for every a ∈ A, (a, a) ∈ R.
This is the definition of a reflexive relation.
Final Answer: The relation is called a reflexive relation.
Q4 2023
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Let A = {3, 5}. Then number of reflexive relations on A is (a) 2 (b) 4 (c) 0 (d) 8
(a)2
(b)4
(c)0
(d)8
For a relation to be reflexive, it must contain (3,3) and (5,5).
Other ordered pairs (3,5) and (5,3) may or may not be included.
Each of these two pairs has two choices: included or not included.
Hence, number of reflexive relations = 2^2 = 4.
Final Answer: (b) 4
Q5 2021-22
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The number of equivalence relations in the set {1, 2, 3} containing the elements (1, 2) and (2, 1) is (a) 0 (b) 1 (c) 2 (d) 3
(a)0
(b)1
(c)2
(d)3
Since (1,2) and (2,1) are present, 1 and 2 must be in the same equivalence class.
For equivalence, (1,1) and (2,2) must also be present.
Element 3 can either be in a separate class or join {1,2}.
Thus, there are two possible equivalence relations.
Final Answer: (c) 2
Q6 2020
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The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1), (1, 1)} is (a) symmetric and transitive, but not reflexive (b) reflexive and symmetric, but not transitive (c) symmetric, but neither reflexive nor transitive (d) an equivalence relation
(a)symmetric and transitive, but not reflexive
(b)reflexive and symmetric, but not transitive
(c)symmetric, but neither reflexive nor transitive
(d)an equivalence relation
Check reflexive: (2,2) and (3,3) are missing, so it is not reflexive.
Check symmetric: (1,2) and (2,1) are both present, so symmetric.
Check transitive: From (1,2) and (2,1), (1,1) is present, so transitive holds for these elements.
Hence, the relation is symmetric and transitive but not reflexive.
Final Answer: (a) symmetric and transitive, but not reflexive
Q7 2023
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Let R be a relation in the set N given by R = {(a, b) : a = b − 2, b > 6}. Then (a) (8, 7) ∈ R (b) (6, 8) ∈ R (c) (3, 8) ∈ R (d) (2, 4) ∈ R
(a)(8,7)
(b)(6,8)
(c)(3,8)
(d)(2,4)
Condition: a = b − 2 and b > 6.
Check (6,8): Here, a = 6 and b = 8. Then b − 2 = 6, which equals a, and b > 6 is satisfied.
So (6,8) satisfies both conditions.
Other options do not satisfy both conditions together.
Final Answer: (b) (6, 8)
Q8 2021-22
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A relation R is defined on N. Which of the following is reflexive relation? (a) R = {(x, y) : x > y, x, y ∈ N} (b) R = {(x, y) : x + y = 10, x, y ∈ N} (c) R = {(x, y) : x is the square number, x, y ∈ N} (d) R = {(x, y) : x + 4y = 10, x, y ∈ N}
(a)x > y
(b)x + y = 10
(c)x is the square number
(d)x + 4y = 10
For a relation to be reflexive, (x, x) must belong to R for every x ∈ N.
In option (c), every x is related to itself because x is a square number implies (x, x) ∈ R.
Other options do not satisfy (x, x) for all x.
Hence, option (c) represents a reflexive relation.
Final Answer: (c)
Q9 2020
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Check if the relation R in the set R of real numbers defined as R = {(a, b) : a < b} is (i) symmetric (ii) transitive.
For symmetry, if (a, b) ∈ R, then (b, a) must also belong to R.
Here a < b does not imply b < a, so the relation is not symmetric.
For transitivity, if a < b and b < c, then a < c, which is true for real numbers.
Hence, the relation is transitive but not symmetric.
Final Answer: R is not symmetric but is transitive.
Q10 2020
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Let W denote the set of words in the English dictionary. Define the relation R by R = {(x, y) ∈ W × W such that x and y have at least one letter in common}. Show that this relation R is reflexive and symmetric, but not transitive.
Reflexive: Every word has at least one letter in common with itself, so (x, x) ∈ R.
Symmetric: If x has a common letter with y, then y also has the same common letter with x.
So if (x, y) ∈ R, then (y, x) ∈ R.
Not transitive: Let x and y have a common letter, and y and z have a common letter, but x and z may not have any common letter.
Hence, R is not transitive.
Final Answer: R is reflexive and symmetric but not transitive.
Q11 2019
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Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
Reflexive: For reflexive, (a, a) must be in R, but b = a + 1, so (a, a) is not possible.
So R is not reflexive.
Symmetric: If (a, b) ∈ R, then (b, a) must also be in R, but b = a + 1 does not imply a = b + 1.
So R is not symmetric.
Transitive: If (a, b) and (b, c) are in R, then b = a + 1 and c = b + 1, so c = a + 2.
But (a, c) is not in R since c ≠ a + 1.
Hence, R is not transitive.
Final Answer: R is neither reflexive nor symmetric nor transitive.
Q12 2019
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Show that the relation R on the set Z of all integers, given by R = {(a, b) : 2 divides (a − b)} is an equivalence relation.
Reflexive: a − a = 0, and 2 divides 0, so (a, a) ∈ R.
Symmetric: If 2 divides (a − b), then it also divides (b − a), so (b, a) ∈ R.
Transitive: If 2 divides (a − b) and 2 divides (b − c), then it divides (a − c).
Hence, R is reflexive, symmetric and transitive.
Final Answer: R is an equivalence relation.
Q13 2019
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Show that the relation R on R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Reflexive: For any a, a ≤ a, so (a, a) ∈ R.
Symmetric: a ≤ b does not imply b ≤ a unless a = b, so not symmetric.
Transitive: If a ≤ b and b ≤ c, then a ≤ c.
Hence, R is reflexive and transitive but not symmetric.
Final Answer: R is reflexive and transitive but not symmetric.
Q14 2024
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Check whether the relation S in the set of real numbers R defined by S = {(a, b) : where a − b + √2 is an irrational number} is reflexive, symmetric or transitive.
Reflexive: For (a, a), a − a + √2 = √2, which is irrational, so reflexive.
Symmetric: If a − b + √2 is irrational, then b − a + √2 = √2 − (a − b), which is also irrational.
So the relation is symmetric.
Transitive: The sum of two irrational numbers need not be irrational, so transitivity does not hold.
Hence, the relation is reflexive and symmetric but not transitive.
Final Answer: S is reflexive and symmetric but not transitive.
Q15 2023
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Show that the relation S in set R of real numbers defined by S = {(a, b) : a ≤ b³, a ∈ R, b ∈ R} is neither reflexive, nor symmetric, nor transitive.
Reflexive: For reflexive, a ≤ a³ must hold for all a, but this is not true for all real numbers.
So the relation is not reflexive.
Symmetric: a ≤ b³ does not imply b ≤ a³, so it is not symmetric.
Transitive: a ≤ b³ and b ≤ c³ do not imply a ≤ c³, so it is not transitive.
Final Answer: The relation is neither reflexive, nor symmetric, nor transitive.
Q16 2023
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Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Hence, find the elements of equivalence class [1].
Reflexive: Every number is either odd or even, so (a, a) ∈ R.
Symmetric: If a and b are both odd or both even, then b and a are also both odd or even.
Transitive: If a and b are both odd or even, and b and c are both odd or even, then a and c are also both odd or even.
Hence, R is an equivalence relation.
Equivalence class [1] consists of all odd numbers in A.
Final Answer: R is an equivalence relation and [1] = {1, 3, 5, 7}.
Q17 2023
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A relation R is defined on a set of real numbers R as R = {(x, y) : x − y is an irrational number}. Check whether R is reflexive, symmetric and transitive or not.
Reflexive: x − x = 0, which is rational, not irrational, so R is not reflexive.
Symmetric: If x − y is irrational, then y − x = −(x − y) is also irrational, so symmetric.
Transitive: Sum of two irrational numbers may be rational, so transitivity does not hold.
Final Answer: R is symmetric but neither reflexive nor transitive.
Q18 2023
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If N denotes the set of all natural numbers and R is the relation on N × N defined by (a, b) R (c, d), if ad(b + c) = bc(a + d). Show that R is an equivalence relation.
Reflexive: For (a, b), ad(b + c) = bc(a + d) becomes ab(a + b) = ba(a + b), which is true.
Symmetric: If (a, b) R (c, d), then ad(b + c) = bc(a + d) implies cb(a + d) = da(b + c), so (c, d) R (a, b).
Transitive: If (a, b) R (c, d) and (c, d) R (e, f), then the required equality holds for (a, b) and (e, f).
Hence, R is an equivalence relation.
Final Answer: R is an equivalence relation.
Q19 2023
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How many relations are possible from B to G?
If B has m elements and G has n elements, then the total number of relations from B to G is 2^(mn).
This is because each ordered pair may or may not belong to a relation.
Final Answer: Number of relations = 2^(m×n), where |B| = m and |G| = n.
Q20 2023
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Among all the possible relations from B to G, how many functions can be formed from B to G?
If set B has m elements and set G has n elements, then each element of B can be mapped to any one element of G.
So total number of functions from B to G is n^m.
Final Answer: Number of functions = n^m, where |B| = m and |G| = n.
Q21 2025
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Case Study: A class-room teacher is keen to assess the learning of her study concept of “relations” taught to them. She writes the following five relations each defined on the set A = {1, 2, 3}: R₁ = {(2, 3), (3, 2)}, R₂ = {(1, 2), (1, 3), (3, 2)}, R₃ = {(1, 2), (2, 1), (1, 1)}, R₄ = {(1, 1), (1, 2), (3, 3), (2, 2)}, R₅ = {(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)}. (i) Identify the relation which is reflexive, transitive but not symmetric. (ii) Identify the relation which is reflexive and symmetric but not transitive. (iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive. OR (iii) (b) What pairs should be added to the relation R₂ to make it an equivalence relation?
Check reflexive property: R₄ and R₅ contain (1,1), (2,2), (3,3).
Check symmetric: R₃ and R₅ contain (1,2) and (2,1).
Check transitive: R₅ satisfies transitivity, R₄ does not.
(i) R₄ is reflexive and transitive but not symmetric.
(ii) R₃ is reflexive and symmetric but not transitive.
(iii)(a) R₁ is symmetric but neither reflexive nor transitive.
(iii)(b) To make R₂ an equivalence relation, add (1,1), (2,2), (3,3), (2,1), (3,1), (2,3).
Final Answer: (i) R₄, (ii) R₃, (iii)(a) R₁, (iii)(b) Add missing reflexive and symmetric pairs to R₂ to make it an equivalence relation.
Q22 2024
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Case Study: Students of a school are taken to a railway museum. Let L be the set of all rail lines and R = {(l₁, l₂) : l₁ is parallel to l₂}. (i) Find whether R is symmetric. (ii) Find whether R is transitive. (iii) If one line is y = 3x + 2, find all lines in R related to it. OR Let S = {(l₁, l₂) : l₁ is perpendicular to l₂}. Check whether S is symmetric and transitive.
Parallel lines: if l₁ ∥ l₂, then l₂ ∥ l₁, so R is symmetric.
If l₁ ∥ l₂ and l₂ ∥ l₃, then l₁ ∥ l₃, so R is transitive.
All lines parallel to y = 3x + 2 are of the form y = 3x + c.
For perpendicular lines: if l₁ ⟂ l₂, then l₂ ⟂ l₁, so symmetric.
But perpendicular relation is not transitive.
Final Answer: R is symmetric and transitive; lines related to y = 3x + 2 are y = 3x + c. Relation S is symmetric but not transitive.
Q23 2020
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Show that the relation R in the set A = {1, 2, 3, 4, 5, 6} given by R = {(a, b) : |a − b| is divisible by 2} is an equivalence relation.
Reflexive: |a − a| = 0, which is divisible by 2.
Symmetric: |a − b| divisible by 2 implies |b − a| divisible by 2.
Transitive: If |a − b| and |b − c| are divisible by 2, then |a − c| is also divisible by 2.
Hence R is reflexive, symmetric and transitive.
Final Answer: R is an equivalence relation.
Q24 2019
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Show that the relation S in the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by S = {(a, b) : |a − b| is divisible by 3} is an equivalence relation.
Reflexive: |a − a| = 0, divisible by 3.
Symmetric: |a − b| divisible by 3 implies |b − a| divisible by 3.
Transitive: If |a − b| and |b − c| are divisible by 3, then |a − c| is also divisible by 3.
Thus S is an equivalence relation.
Final Answer: S is an equivalence relation.
Q25 2024
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A relation R is defined on N × N (where N is the set of natural numbers) as (a, b) R (c, d) ⇔ a − c = b − d. Show that R is an equivalence relation.
Reflexive: For any (a, b), a − a = b − b = 0, so (a, b) R (a, b).
Symmetric: If (a, b) R (c, d), then a − c = b − d implies c − a = d − b, so (c, d) R (a, b).
Transitive: If (a, b) R (c, d) and (c, d) R (e, f), then a − c = b − d and c − e = d − f.
Adding, we get a − e = b − f, so (a, b) R (e, f).
Hence, R is reflexive, symmetric and transitive.
Final Answer: R is an equivalence relation.
Q26 2024
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A relation R is defined on N × N (where N is the set of natural numbers) as (a, b) R (c, d) ⇔ a/b = c/d. Show that R is an equivalence relation.
Reflexive: a/b = a/b, so (a, b) R (a, b).
Symmetric: If a/b = c/d, then c/d = a/b, so (c, d) R (a, b).
Transitive: If a/b = c/d and c/d = e/f, then a/b = e/f, so (a, b) R (e, f).
Thus, R is reflexive, symmetric and transitive.
Final Answer: R is an equivalence relation.
Q27 2018
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Let A = {x ∈ Z : 0 ≤ x ≤ 12}. Show that R = {(a, b) : a, b ∈ A, |a − b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].
Reflexive: |a − a| = 0, divisible by 4.
Symmetric: |a − b| divisible by 4 implies |b − a| divisible by 4.
Transitive: If |a − b| and |b − c| are divisible by 4, then |a − c| is also divisible by 4.
So R is an equivalence relation.
Elements related to 1 are numbers in A differing from 1 by a multiple of 4: {1, 5, 9}.
Equivalence class [2] = {2, 6, 10}.
Final Answer: R is an equivalence relation, elements related to 1 are {1, 5, 9}, and [2] = {2, 6, 10}.
Q28 2023
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Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Hence, find the elements of equivalence class [1].
Reflexive: Every number is either odd or even, so (a, a) ∈ R.
Symmetric: If a and b are both odd or both even, then b and a are also both odd or both even.
Transitive: If a and b are both odd/even and b and c are both odd/even, then a and c are also both odd/even.
Thus, R is reflexive, symmetric and transitive.
The equivalence class [1] contains all odd numbers in A.
Final Answer: R is an equivalence relation and [1] = {1, 3, 5, 7}.
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