UG Mathematics Booster Test 2 - Types of Relations -

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UG Mathematics Booster Test 2 - Types of Relations

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Mathematics Booster Test 2 - Types of Relations

1 / 20

QUESTION 1 OF 20

Let S be the set of all squares. A relation R on S is defined by
(S₁,S₂) ∈ R if Area(S₁) − Area(S₂) = −50.
Given that every square in S has area 10 square units, determine the nature of R.

Answer Review

1 Let S be the set of all squares. A relation R on S is defined by
(S₁,S₂) ∈ R if Area(S₁) − Area(S₂) = −50.
Given that every square in S has area 10 square units, determine the nature of R.

A Universal relation

B Empty relation

C Reflexive relation

D Transitive relation

💡 Short Explanation

Every square has equal area Difference of areas becomes 0 Condition −50 is impossible

✅ Correct Answer Explanation

Since every square has area 10 square units, for any pair S₁ and S₂: Area(S₁) − Area(S₂) = 10 − 10 = 0. The required condition −50 can never occur. Hence no ordered pair satisfies the relation, making R an empty relation. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Universal relation requires all possible ordered pairs to satisfy the condition.
Option C → Reflexivity requires every element related to itself, which fails here.
Option D → Empty relation trivially satisfies transitivity, but its primary classification here is empty relation.

🎯 Strategy

Used: Substitution

Application:

Substitute given area values directly into the relation condition.

Final Logic:

10 − 10 ≠ −50, so no ordered pair exists.

🧠 Memory Trick

"Impossible condition ⇒ Empty relation"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Empty Relation

2 Let E be the set of all impossible events in a sample space. A relation R on E is defined by
(E₁,E₂) ∈ R if P(E₁)+P(E₂)=0.
What best describes the relation R?

A Empty relation

B Universal relation on E

C It lacks symmetry

D It is only transitive

💡 Short Explanation

Impossible events have probability 0 Sum of two impossible events remains 0 Every pair satisfies relation

✅ Correct Answer Explanation

For impossible events, P(E₁)=0 and P(E₂)=0. Therefore: \(0+0=0\) Every ordered pair satisfies the relation condition. Hence all pairs belong to R, making it the universal relation on E. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Relation is not empty because every pair satisfies the condition.
Option C → Relation is symmetric since addition is commutative.
Option D → Universal relation has more properties than only transitivity.

🎯 Strategy

Used: Substitution

Application:

Use probability value of impossible events directly in relation equation.

Final Logic:

All ordered pairs satisfy the condition, giving universal relation.

🧠 Memory Trick

"Impossible event → Probability zero"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Universal Relation

3 Let R be a relation on the set of continuous functions defined by
(f,g) ∈ R if
∫ₐᵇ(f(x)−g(x)) dx = 0.
For the relation R to be reflexive, the value of the integral must be:

A 1

B f(x)

C 0

D ∞

💡 Short Explanation

Reflexive means f relates to itself f(x) − f(x) becomes 0 Integral of zero equals zero

✅ Correct Answer Explanation

For reflexivity, every function f must satisfy (f,f) ∈ R. Then: \(\int_{a}^{b}\,(f(x)-f(x)) dx=\int_{a}^{b}\,0 dx=0\) Hence the integral must equal 0. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Reflexive condition does not require integral value 1.
Option B → f(x) is a function, not a constant integral value.
Option D → Integral of zero cannot become infinite.

🎯 Strategy

Used: Substitution

Application:

Replace g(x) with f(x) to test reflexivity directly.

Final Logic:

f(x)−f(x)=0, so integral equals 0.

🧠 Memory Trick

"Self minus self = zero"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Definite Integrals in Relations

4 Let R be a relation defined on the set of all lines in the Cartesian plane such that
(L₁,L₂) ∈ R if L₁ ⟂ L₂.
Why is this relation symmetric but not reflexive?

A A line is perpendicular to itself but not to other lines.

B Two lines cannot intersect at 90°.

C A line is parallel to itself and hence perpendicular.

D If L₁ ⟂ L₂, then L₂ ⟂ L₁, but no line is perpendicular to itself.

💡 Short Explanation

Perpendicularity works both ways No line is perpendicular to itself Hence symmetric but not reflexive

✅ Correct Answer Explanation

If L₁ is perpendicular to L₂, then L₂ is also perpendicular to L₁, satisfying symmetry. However, a line cannot be perpendicular to itself, so reflexivity fails. Therefore Option D correctly explains both properties simultaneously.

❓ Why Other Options Are Incorrect

Option A → A line is never perpendicular to itself.
Option B → Perpendicular lines intersect exactly at 90°.
Option C → Parallelism and perpendicularity are different concepts.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret geometric meaning of perpendicular lines carefully.

Final Logic:

Perpendicularity reverses symmetrically but fails self-relation.

🧠 Memory Trick

"Perpendicular both ways, never self"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Symmetric Relation

5 Consider a relation R on stock data arrays. (A, B) ∈ R if array A's 5-day moving average is greater than array B's. If (A, B) ∈ R and (B, C) ∈ R, what is the required condition for R to be transitive?

A A's average < C's average

B A's average = C's average

C B's average is 0

D A's average > C's average

💡 Short Explanation

Greater-than relation is transitive A > B and B > C imply A > C Hence transitivity holds

✅ Correct Answer Explanation

If A's moving average is greater than B's, and B's is greater than C's, then by transitivity of inequalities: A > C. Thus the relation remains transitive only when A's average exceeds C's average. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Opposite inequality contradicts transitivity.
Option B → Equality cannot follow from strict greater-than conditions.
Option C → Value of B's average is irrelevant.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply transitivity property of inequalities directly.

Final Logic:

Greater-than relation naturally chains forward.

🧠 Memory Trick

"Greater-than keeps flowing"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Transitive Relation

6 For relation R = {(a, b): a is a subset of b} applied to set operations, which of the following logically implies transitivity?
I. If Set A ⊂ Set B
II. And Set B ⊂ Set C
III. Then Set A ⊂ Set C
IV. Set A ∩ Set B = φ

A I, II, and III

B I and IV

C II and IV

D III and IV

💡 Short Explanation

Subset relation is transitive Chain of subsets implies larger inclusion Intersection condition unnecessary

✅ Correct Answer Explanation

If A ⊂ B and B ⊂ C, then logically A ⊂ C. This is the exact transitive property of subset relation. Statement IV about intersection being empty is unrelated to transitivity. Hence Statements I, II, and III together correctly define transitivity, making Option A correct.

❓ Why Other Options Are Incorrect

Option B → Intersection condition is unrelated to transitivity.
Option C → Missing initial subset condition.
Option D → Statement III alone cannot establish implication chain.

🎯 Strategy

Used: Option Grouping

Application:

Identify statements forming logical implication chain.

Final Logic:

Subset relation transfers through intermediate sets.

🧠 Memory Trick

"Subset chain passes forward"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Transitive Relation

7 Match the specific relation on Set A = {1, 2, 3} to its reflexive status:

List I	List II
1. {(1,1), (2,2), (3,3)}	a. Not reflexive, missing (1,1), (2,2), (3,3)
2. {(1,1), (2,2), (3,3), (1,2)}	b. Reflexive, exact identity relation
3. {(1,1), (2,2)}	c. Reflexive, contains extra elements
4. {(1,2), (2,3)}	d. Not reflexive, missing (3,3)

A 1-b, 2-d, 3-c, 4-a

B 1-b, 2-c, 3-d, 4-a

C 1-a, 2-b, 3-c, 4-d

D 1-c, 2-a, 3-b, 4-d

💡 Short Explanation

Reflexive relation contains all self-pairs Missing any self-pair breaks reflexivity Extra pairs do not matter

✅ Correct Answer Explanation

Relation 1 is exact identity relation → b. Relation 2 contains all self-pairs plus extra element → c. Relation 3 misses (3,3) → d. Relation 4 misses all self-pairs → a. Thus the correct matching is Option B.

❓ Why Other Options Are Incorrect

Option A → Relation 2 is still reflexive despite extra pair.
Option C → Relation 1 is reflexive, not non-reflexive.
Option D → Relation 4 lacks all required self-pairs.

🎯 Strategy

Used: Option Grouping

Application:

Check presence or absence of all self-pairs systematically.

Final Logic:

Reflexive requires every (a,a) pair.

🧠 Memory Trick

"All self-pairs must appear"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Reflexive Relation

8 Arrange the steps to check if relation R = {(x,y) : x - y is an integer} is reflexive for all real numbers x:
1. Thus, R is reflexive.
2. Substitute y with x to check (x, x).
3. 0 is an integer.
4. Calculate x - x = 0.

A 2, 3, 4, 1

B 4, 3, 2, 1

C 2, 4, 3, 1

D 1, 2, 4, 3

💡 Short Explanation

Start by checking self-pair Difference becomes zero Since zero is integer, reflexive holds

✅ Correct Answer Explanation

To test reflexivity, first substitute y=x. Then calculate x−x=0. Since 0 is an integer, the relation condition holds for every x. Therefore the relation is reflexive. Hence correct order is 2 → 4 → 3 → 1, making Option C correct.

❓ Why Other Options Are Incorrect

Option A → Integer conclusion appears before calculation.
Option B → Begins with conclusion instead of substitution.
Option D → Final conclusion cannot appear first.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange proof statements in logical mathematical order.

Final Logic:

Substitute → calculate → verify → conclude.

🧠 Memory Trick

"Self-check gives zero integer"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Reflexive Relation

9 Assertion (A): The relation R = {(1, 2), (2, 1)} on set {1, 2, 3} is symmetric.
Reason (R): For every (a, b) in R, the pair (b, a) is also present in R.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Symmetry requires reverse pair presence Reverse ordered pairs exist here Reason correctly explains assertion

✅ Correct Answer Explanation

The relation contains both (1,2) and (2,1). Hence whenever (a,b) belongs to R, the reverse pair (b,a) also belongs to R. This exactly satisfies the definition of symmetry. Therefore both Assertion and Reason are true, and Reason correctly explains Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements match symmetric relation definition.
Option B → Reason is mathematically correct.
Option D → Assertion correctly identifies symmetric relation.

🎯 Strategy

Used: Elimination

Application:

Check whether every ordered pair has its reverse pair.

Final Logic:

Presence of reverse pairs guarantees symmetry.

🧠 Memory Trick

"Symmetric means pair reversal"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Symmetric Relation

10 Relation R on a set of vectors defines (A, B) ∈ R if the cross product A × B = 0. If (A, B) ∈ R, does it imply (B, A) ∈ R?

A No, because B × A = 1

B , and -0 is still 0.

C Yes, because A × B = B × A exactly

D Yes, because B × A = -(A ×

💡 Short Explanation

Cross product reverses sign Zero vector remains unchanged under sign change Hence relation is symmetric

✅ Correct Answer Explanation

For vectors: A × B = −(B × A). If A × B = 0, then: B × A = −0 = 0. Therefore the reverse ordered pair also satisfies the condition, making the relation symmetric. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Cross product reversal does not always equal 1.
Option B → Parallel vectors can have zero cross product.
Option C → Cross product is anti-commutative, not commutative.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use anti-commutative property of vector cross product.

Final Logic:

Negative of zero is still zero.

🧠 Memory Trick

"−0 is still 0"

📚 Class XI – NCERT Mathematics – Chapter: Vector Algebra – Topic: Cross Product and Symmetry

11 An EMI relation R on loan accounts asserts (L1, L2) ∈ R if EMI of L1 is exactly Rs. 500 less than EMI of L2. Is this relation transitive?

A No, if L1 is 500 less than L2, and L2 is 500 less than L3, L1 is 1000 less than L3.

B Yes, the difference remains 500.

C Yes, because loan amounts are fixed.

D No, because EMIs cannot be compared.

💡 Short Explanation

Transitivity needs same relation preserved Differences accumulate numerically Final difference becomes Rs.1000

✅ Correct Answer Explanation

Suppose EMI(L1)=EMI(L2)−500 and EMI(L2)=EMI(L3)−500. Then: \(EMI(L1)=EMI(L3)-1000\) Thus L1 is not Rs.500 less than L3. Therefore the relation fails transitivity. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Difference changes to Rs.1000, not Rs.500.
Option C → Fixed loan amounts do not affect logical transitivity.
Option D → EMIs are numerical quantities and are comparable.

🎯 Strategy

Used: Substitution

Application:

Apply the relation step-by-step using actual numerical differences.

Final Logic:

500 + 500 = 1000, so transitivity fails.

🧠 Memory Trick

"Differences add up"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Transitive Relation

12 Regarding the relation R = {(L1, L2) : L1 is perpendicular to L2} on a set of lines, identify the INCORRECT statement:

A R is a transitive relation because L1 ⊥ L2 and L2 ⊥ L3 means L1 ⊥ L3.

B L1 ⊥ L2 and L2 ⊥ L3 implies L1 is parallel to L3.

C R is symmetric.

D R is not reflexive.

💡 Short Explanation

Perpendicularity is not transitive Two lines perpendicular to same line become parallel Hence statement A is incorrect

✅ Correct Answer Explanation

If L1 ⊥ L2 and L2 ⊥ L3, then L1 and L3 are parallel, not perpendicular. Therefore perpendicularity is not transitive. Hence Option A is incorrect. Options B, C, and D correctly describe perpendicular relation properties.

❓ Why Other Options Are Incorrect

Option B → Correct because perpendicular lines to same line become parallel.
Option C → Perpendicularity works both ways, so symmetry holds.
Option D → No line is perpendicular to itself, so reflexivity fails.

🎯 Strategy

Used: Elimination

Application:

Test geometric properties of perpendicular lines carefully.

Final Logic:

Perpendicular-to-same-line implies parallelism, not perpendicularity.

🧠 Memory Trick

"⊥ then ⊥ gives ∥"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Transitive Relation

📄 Passage

An equivalence relation R on a set X partitions X into mutually disjoint subsets Aᵢ. Elements within the same subset are related, elements from different subsets are not, and the union of all subsets equals X.

The existence of mutually disjoint subsets Aᵢ that together cover the entire set X is a consequence of:

A Only transitive

B An equivalence relation

C Only reflexive

D An empty relation

💡 Short Explanation

Equivalence relations partition sets Partitions create disjoint subsets Their union covers the entire set

✅ Correct Answer Explanation

The passage states that equivalence relations partition a set into mutually disjoint subsets whose union equals the original set. This property belongs specifically to equivalence relations, which combine reflexive, symmetric, and transitive properties. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Transitivity alone cannot create complete partitions.
Option C → Reflexivity alone is insufficient for equivalence classes.
Option D → Empty relation does not partition meaningfully.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the exact statement given in the passage.

Final Logic:

Partitions arise directly from equivalence relations.

🧠 Memory Trick

"Equivalence ⇒ Partition"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Equivalence Relation

📄 Passage

If x ∈ A₁ and y ∈ A₂ where A₁ ≠ A₂, then:

A x is symmetrically related to y

B x = y

C x is reflexively related to y

D x is not related to y under R

💡 Short Explanation

Different equivalence classes are disjoint Related elements stay in same class Hence x and y are unrelated

✅ Correct Answer Explanation

In equivalence relations, elements from different equivalence classes are not related to each other. Since x belongs to A₁ and y belongs to distinct class A₂, they cannot satisfy the relation R. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Symmetry applies only when relation already exists.
Option B → Different class membership does not imply equality.
Option C → Reflexivity concerns self-relations only.

🎯 Strategy

Used: Elimination

Application:

Use the disjoint property of equivalence classes.

Final Logic:

Different equivalence classes contain unrelated elements.

🧠 Memory Trick

"Different classes, no relation"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Equivalence Classes

15 For R = {(a, b) : 2 divides a - b} defined on integers, which elements form the equivalence class [0]?

A All positive integers

B All odd integers

C All even integers

D All prime numbers

💡 Short Explanation

[0] contains numbers related to 0 Difference with 0 must be divisible by 2 Hence all even integers belong

✅ Correct Answer Explanation

For any integer a to belong to [0], we need: \(2∣(a-0)\) This means a must be divisible by 2. Therefore [0] consists of all even integers. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Positive odd numbers are not divisible by 2.
Option B → Odd integers differ from 0 by odd numbers.
Option D → Prime numbers are not necessarily even.

🎯 Strategy

Used: Substitution

Application:

Replace b by 0 in divisibility condition directly.

Final Logic:

Numbers related to 0 must be even.

🧠 Memory Trick

"Class [0] = even numbers"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Equivalence Classes

16 For the equivalence relation R = {(a, b): 3 divides a - b} on set Z, the equivalence classes can be represented correctly by which combinations?
I. = {..., -6, -3, 0, 3, 6, ...}
II. = {..., -5, -2, 1, 4, 7, ...}
III. = {..., -4, -1, 2, 5, 8, ...}
IV. = {1, 2, 3, 4}

A I, II, and III

B I and IV

C II and IV

D III and IV

💡 Short Explanation

Modulo 3 gives three equivalence classes Numbers differ by multiples of 3 Finite set IV is incorrect

✅ Correct Answer Explanation

Integers modulo 3 form exactly three equivalence classes: [0], [1], and [2]. Sets I, II, and III correctly represent these classes since consecutive elements differ by 3. Set IV is not closed under modulo-3 equivalence. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Set IV is incomplete and not an equivalence class.
Option C → Omits one valid modulo-3 class.
Option D → Includes invalid finite representation.

🎯 Strategy

Used: Option Grouping

Application:

Identify sets formed by numbers differing by multiples of 3.

Final Logic:

Modulo 3 partitions integers into three infinite classes.

🧠 Memory Trick

"Modulo 3 gives three groups"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Equivalence Classes

17 If subsets E and O form a partition of the set of Integers (Z) under a specific equivalence relation, what is the intersection of E and O?

A Z

B φ (Empty set)

C {0}

D {1}

💡 Short Explanation

Partition subsets are disjoint Even and odd integers never overlap Their intersection is empty

✅ Correct Answer Explanation

A partition divides a set into mutually disjoint subsets. Since E and O represent separate equivalence classes (even and odd integers), no integer can belong to both simultaneously. Therefore: \(E\cap O=\emptyset\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Entire set cannot equal the intersection.
Option C → 0 is only even, not odd.
Option D → 1 is only odd, not even.

🎯 Strategy

Used: Odd One Out

Application:

Use the mutually disjoint property of partitions.

Final Logic:

Distinct equivalence classes never overlap.

🧠 Memory Trick

"Partition means no overlap"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Partition of Sets

18 Assertion (A): For equivalence classes A₁,A₂,A₃ forming a partition of set Z,
A₁ ∪ A₂ ∪ A₃ = Z
Reason (R): Partitions of a set generated by an equivalence relation must completely cover the original set.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Partitions cover the whole set Equivalence classes form partitions Reason directly explains assertion

✅ Correct Answer Explanation

Equivalence classes generated by an equivalence relation partition the original set completely. Therefore the union of all equivalence classes equals the full set Z. Both Assertion and Reason are true, and the Reason correctly explains the Assertion. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct.
Option B → Reason is true and foundational.
Option D → Assertion follows directly from partition property.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Connect partition definition with union property directly.

Final Logic:

Partitions always cover the entire original set.

🧠 Memory Trick

"All classes together give whole set"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Partition of Sets

19 Match the geometry relation to its properties:

List I	List II
1. T1 is congruent to T2	a. Reflexive, Symmetric, Transitive (Equivalence on lines)
2. T1 is similar to T2	b. Symmetric only
3. L1 is parallel to L2	c. Equivalence on triangles (size and shape identical)
4. L1 is perpendicular to L2	d. Equivalence on triangles (shape identical, size varies)

A 1-b, 2-c, 3-a, 4-d

B 1-d, 2-a, 3-b, 4-c

C 1-a, 2-b, 3-c, 4-d

D 1-c, 2-d, 3-a, 4-b

💡 Short Explanation

Congruence preserves size and shape Similarity preserves shape only Perpendicularity is symmetric only

✅ Correct Answer Explanation

Congruence is an equivalence relation on triangles → c. Similarity preserves shape but not size → d. Parallelism on lines is reflexive, symmetric, and transitive → a. Perpendicularity is symmetric but neither reflexive nor transitive → b. Thus Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Congruence is not merely symmetric.
Option B → Parallelism is not represented correctly.
Option C → Perpendicularity is not equivalence relation.

🎯 Strategy

Used: Option Grouping

Application:

Match geometric relations with standard relation properties.

Final Logic:

Only parallelism and congruence form equivalence relations.

🧠 Memory Trick

"Congruent = same size and shape"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Special Relations

20 Arrange the steps proving the symmetric property FAILS for the relation "a divides b" on Natural numbers:
1. Therefore, 4 does not divide 2, so (4, 2) ∉ R.
2. Let a = 2 and b = 4.
3. This proves the relation is not symmetric.
4. 2 divides 4, so (2, 4) ∈ R.

A 2, 4, 1, 3

B 1, 2, 3, 4

C 4, 3, 2, 1

D 2, 1, 4, 3

💡 Short Explanation

Begin with example values Check forward divisibility relation Reverse relation fails, disproving symmetry

✅ Correct Answer Explanation

Choose a=2 and b=4. Since 2 divides 4, (2,4) ∈ R. But 4 does not divide 2, so (4,2) ∉ R. Therefore the reverse pair fails, proving the relation is not symmetric. Thus the correct sequence is 2 → 4 → 1 → 3.

❓ Why Other Options Are Incorrect

Option B → Begins with conclusion before choosing example.
Option C → Logical order is completely reversed.
Option D → Reverse divisibility checked before original relation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange statements according to mathematical proof sequence.

Final Logic:

Example → verify → reverse fails → conclude.

🧠 Memory Trick

"2 divides 4, not reverse"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Symmetric Relation Failure

UG Mathematics Booster Test 2 - Types of Relations

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