UG Mathematics Booster Test 3 - Types of Relations -

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UG Mathematics Booster Test 3 - Types of Relations

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 3 - Types of Relations

1 / 20

QUESTION 1 OF 20

Match the set and condition to its resulting relation status:

List I	List II
1. A = {1,2,3}, R = {(a,b) : a - b = 10}	a. Universal relation
2. A = Boys school, R = {(a,b) : a is sister of b}	b. Equivalence relation
3. A = {1,2}, R = {(1,2), (2,1), (1,1), (2,2)}	c. Empty relation (Algebraic)
4. A = {1,2}, R = {(a,b) : a + b > 0}	d. Empty relation (Contextual)

Answer Review

1 Match the set and condition to its resulting relation status:

List I	List II
1. A = {1,2,3}, R = {(a,b) : a - b = 10}	a. Universal relation
2. A = Boys school, R = {(a,b) : a is sister of b}	b. Equivalence relation
3. A = {1,2}, R = {(1,2), (2,1), (1,1), (2,2)}	c. Empty relation (Algebraic)
4. A = {1,2}, R = {(a,b) : a + b > 0}	d. Empty relation (Contextual)

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

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

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

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

💡 Short Explanation

a−b=10 impossible in given set "Sister" impossible in boys school Remaining relations satisfy all ordered pairs

✅ Correct Answer Explanation

Relation 1 is algebraically impossible, so it is an empty relation → c. Relation 2 is contextually impossible in a boys school → d. Relation 3 contains all pairs on set {1,2}, making it universal and equivalence → b fits best. Relation 4 satisfies every pair because sums are always positive → universal relation → a.

❓ Why Other Options Are Incorrect

Option A → Swaps contextual and algebraic empty relations incorrectly.
Option B → Universal and equivalence relations are mismatched.
Option D → Relation 3 is not contextual empty relation.

🎯 Strategy

Used: Option Grouping

Application:

Classify each relation carefully before matching.

Final Logic:

Impossible conditions give empty relations; all-valid conditions give universal relation.

🧠 Memory Trick

"Impossible → Empty, All valid → Universal"

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

2 Arrange the deductive sequence proving R = {(a,b) : |a - b| ≥ 0} on set A={1,2,3,4} is a universal relation:
1. Thus, all pairs (a,b) in A × A satisfy the condition.
2. Given A = {1,2,3,4}, select any two elements a, b.
3. Therefore, R = A × A, making it the universal relation.
4. The absolute value of any real number difference |a - b| is always ≥ 0.

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 4, 1, 3

D 3, 1, 4, 2

💡 Short Explanation

Start with arbitrary elements Apply absolute value property Conclude every pair satisfies relation

✅ Correct Answer Explanation

First choose arbitrary elements a,b from A. Then use the property: \(∣a-b∣\geq 0\) Thus every ordered pair satisfies the condition, implying R=A×A. Hence the relation is universal. Therefore sequence 2→4→1→3 is correct.

❓ Why Other Options Are Incorrect

Option A → Starts with conclusion before assumptions.
Option B → Logical proof order is reversed.
Option D → Final conclusion appears too early.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange proof steps from assumption to conclusion.

Final Logic:

Choose elements → apply property → conclude universal relation.

🧠 Memory Trick

"Pick → Prove → Conclude"

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

3 Assertion (A): The relation R = {(1, 2), (2, 3)} on set {1, 2, 3} is reflexive because the elements form an ordered chain.
Reason (R): For a relation to be reflexive on set A, it must contain (x, x) for every element x in A.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Reflexive needs self-pairs Relation lacks (1,1),(2,2),(3,3) Ordered chains do not imply reflexivity

✅ Correct Answer Explanation

A relation is reflexive only if every element relates to itself: \((x,x)\in R ∀x\in A\) The given relation lacks all self-pairs. Hence Assertion is false. Reason correctly states the definition of reflexive relation. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is mathematically correct.
Option B → Assertion is not true because self-pairs are missing.
Option C → Assertion itself is false.

🎯 Strategy

Used: Elimination

Application:

Check the formal definition of reflexivity directly.

Final Logic:

No self-pairs means not reflexive.

🧠 Memory Trick

"Reflexive means self-related"

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

4 Let R be a relation defined on planar shapes where (X, Y) ∈ R if Area(X) / Area(Y) = 1. Analytically, why does this enforce symmetry?

A Because if Area(X)/Area(Y) = 1, then taking the reciprocal gives Area(Y)/Area(X) = 1.

B Because areas cannot be negative.

C Because X and Y must be the exact same shape.

D It is not symmetric, it is only reflexive.

💡 Short Explanation

Equality works both ways Reciprocal of 1 remains 1 Hence relation reverses symmetrically

✅ Correct Answer Explanation

If: \(\frac{Area(X)}{Area(Y)}=1\) then reciprocally: \(\frac{Area(Y)}{Area(X)}=1\) Thus if (X,Y) ∈ R, then (Y,X) ∈ R. Hence the relation is symmetric. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Non-negativity does not establish symmetry.
Option C → Equal areas do not require identical shapes.
Option D → The relation satisfies symmetry.

🎯 Strategy

Used: Substitution

Application:

Reverse the ratio directly to test symmetry.

Final Logic:

Reciprocal equality preserves the relation.

🧠 Memory Trick

"1 reversed stays 1"

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

5 Let R be a relation on independent events where (A, B) ∈ R if P(A ∩ B) = P(A)P(B). If event A relates to B, and B relates to C, does A necessarily relate to C transitively?

A Not necessarily; pairwise independence does not guarantee mutual independence or transitive pairwise independence.

B Always, because probability is multiplicative.

C Yes, because P(C) is positive.

D Yes, if P(A) = 1.

💡 Short Explanation

Independence is not transitive Pairwise conditions are insufficient Mutual independence is stronger concept

✅ Correct Answer Explanation

Even if A is independent of B and B is independent of C, A need not be independent of C. Probability independence does not behave transitively. Therefore pairwise independence alone cannot guarantee transitivity of the relation. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Multiplication property alone does not ensure transitivity.
Option C → Positive probability is irrelevant here.
Option D → P(A)=1 does not establish general transitivity.

🎯 Strategy

Used: Extreme Word Filter

Application:

Reject "always" type claims lacking universal proof.

Final Logic:

Independence relations are not generally transitive.

🧠 Memory Trick

"Independent ≠ Transitive"

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

6 Let relation R be defined on functions such that (f,g) ∈ R if
∫f(x)dx > ∫g(x)dx
Logically, the transitivity of this relation relies on:

A The symmetric property of definite integrals.

B The strict inequality property (a>b and b>c ⇒ a>c) applied to real numbers resulting from the integrals.

C The property that all integrals evaluate to zero.

D The fact that f(x) must be a constant function.

💡 Short Explanation

Integrals produce real numbers Greater-than relation is transitive Hence relation inherits transitivity

✅ Correct Answer Explanation

If: \(\int f(x)dx>\int g(x)dx\) and \(\int g(x)dx>\int h(x)dx\) then by transitivity of inequalities: \(\int f(x)dx>\int h(x)dx\) Thus relation R is transitive. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Symmetry is unrelated to inequalities.
Option C → Integrals need not evaluate to zero.
Option D → Functions need not be constant.

🎯 Strategy

Used: Substitution

Application:

Translate the relation into real-number inequalities.

Final Logic:

Transitivity comes from numerical ordering.

🧠 Memory Trick

"Greater-than stays transitive"

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

7 Relation R evaluates financial time series data. (S1, S2) ∈ R if the maximum moving average of S1 equals the maximum moving average of S2. Does this definition inherently satisfy reflexivity?

A No, moving averages fluctuate.

B No, S1 cannot be compared to itself.

C Yes, any dataset has a maximum moving average identical to its own maximum moving average.

D Yes, but only if the average is positive.

💡 Short Explanation

Every quantity equals itself Same dataset gives same maximum average Hence reflexive property holds

✅ Correct Answer Explanation

For any dataset S1, its maximum moving average equals itself automatically. Thus: (S1,S1) ∈ R for every dataset. Therefore the relation satisfies reflexivity. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Fluctuation does not affect self-equality.
Option B → Reflexivity specifically compares elements to themselves.
Option D → Positivity is unnecessary.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the self-equality idea of reflexivity.

Final Logic:

Every dataset matches its own maximum value.

🧠 Memory Trick

"Anything equals itself"

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

8 Consider the set of all points in the first quadrant. R = {(x, y) : y = x}. This relation graphed forms a line bisecting the quadrant. Which relational property is exclusively represented by the points strictly ON this line compared to the whole quadrant set?

A Transitivity

B Symmetry

C Universal relation

D Reflexivity (identity elements)

💡 Short Explanation

Points on line satisfy x=y Self-pairs represent identity relation Reflexive property uses identical coordinates

✅ Correct Answer Explanation

The graph y=x consists of all ordered pairs of the form: \(\left(x,\ x\right)\) These are identity/self-related pairs, which characterize reflexivity. Hence the line y=x represents reflexive or identity elements. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Transitivity cannot be visualized by this line alone.
Option B → Symmetry involves reversed ordered pairs.
Option C → Universal relation covers entire plane region.

🎯 Strategy

Used: Odd One Out

Application:

Identify the unique property represented by identical coordinates.

Final Logic:

Line y=x corresponds to self-related elements.

🧠 Memory Trick

"y=x means self-pairs"

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

9 If R is a symmetric relation on a set A, consider its representation as an adjacency matrix. Which of the following properties correctly model R?
I. The matrix is equal to its transpose.
II. The matrix has only 1s on the main diagonal.
III. If entry (i,j)=1, then entry (j,i)=1.
IV. The determinant is always 0.

A I and II

B II and IV

C III and IV

D I and III

💡 Short Explanation

Symmetric matrices equal transpose Reverse entries must match Diagonal entries are not compulsory

✅ Correct Answer Explanation

A symmetric relation satisfies: if (i,j) exists, then (j,i) also exists. Hence adjacency matrix entries mirror across diagonal: \(A=A^{T}\) Thus statements I and III are correct. Reflexivity is needed for all diagonal 1s, not symmetry alone. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Symmetry does not require all diagonal entries as 1.
Option B → Determinant need not be zero.
Option C → Statement IV is mathematically false.

🎯 Strategy

Used: Elimination

Application:

Separate symmetry conditions from reflexivity conditions.

Final Logic:

Symmetric relation means mirrored matrix entries.

🧠 Memory Trick

"Symmetric matrix mirrors itself"

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

10 Identify the INCORRECT analytical statement about symmetric relations:

A If R is an empty relation on a non-empty set A, it cannot be symmetric because it lacks elements.

B The universal relation A × A is always symmetric.

C A relation can be symmetric without being reflexive.

D R = {(1,2), (2,1)} on set {1,2,3} is symmetric but not transitive.

💡 Short Explanation

Empty relation vacuously satisfies symmetry No violating pairs exist Hence statement A is incorrect

✅ Correct Answer Explanation

A relation is symmetric if whenever (a,b) ∈ R, then (b,a) ∈ R. In an empty relation, no ordered pairs exist to violate symmetry. Therefore empty relations are symmetric vacuously. Hence Option A is the incorrect statement.

❓ Why Other Options Are Incorrect

Option B → Universal relation always contains reversed pairs.
Option C → Symmetry does not guarantee reflexivity.
Option D → Relation lacks (1,1),(2,2), so transitivity fails.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether "cannot" creates an absolute false claim.

Final Logic:

No counterexample means empty relation is symmetric.

🧠 Memory Trick

"Empty relations violate nothing"

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

11 Let relation R on vectors be defined as (A,B) ∈ R if
A = kB for some scalar k > 0
If (A,B) ∈ R and (B,C) ∈ R, what is the analytic proof of transitivity?

A A = B + C

B A = k₁B and B = k₂C, so A = (k₁·k₂)C, where (k₁·k₂) > 0

C A·C = 0

D A × C = 1

💡 Short Explanation

Scalar multiples chain together Product of positive scalars remains positive Hence transitivity holds

✅ Correct Answer Explanation

If A=k₁B and B=k₂C with k₁,k₂>0, then substituting gives: \(A=(k_{1}k_{2})C\) Since product of positive scalars remains positive, A is also a positive scalar multiple of C. Hence (A,C) ∈ R and the relation is transitive. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Vector addition does not define the relation.
Option C → Dot product condition is unrelated to scalar multiples.
Option D → Cross product value does not establish transitivity here.

🎯 Strategy

Used: Substitution

Application:

Replace B using the second relation directly into the first.

Final Logic:

Positive scalar multiplication remains preserved through chaining.

🧠 Memory Trick

"Positive multiples stay positive"

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

12 In a mixture problem, relation R specifies (M1, M2) ∈ R if the ratio of milk to water in M1 is strictly greater than in M2. If M1 is 3:1 (75%), M2 is 1:1 (50%), and M3 is 1:3 (25%). Which analytic observation proves transitivity?

A M2 and M3 have the same volume.

B 75% > 50% and 50% > 25%, implying strictly that 75% > 25%, confirming (M1, M3) ∈ R.

C Mixtures cannot be transitive because liquids dilute.

D M1 + M2 = M3.

💡 Short Explanation

Greater-than relation is transitive Ratios preserve numerical ordering Final comparison confirms M1 relates to M3

✅ Correct Answer Explanation

The relation compares milk concentration percentages. Since: 75% > 50% and 50% > 25%, it follows mathematically that: 75% > 25%. Therefore (M1,M3) ∈ R. This directly proves transitivity of the relation. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Volumes are irrelevant to transitivity here.
Option C → Numerical ordering still satisfies transitivity.
Option D → Adding mixtures does not prove relation properties.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply transitivity of strict inequalities to percentages.

Final Logic:

If a>b and b>c, then a>c.

🧠 Memory Trick

"Greater-than flows forward"

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

📄 Passage

In the study of relations, the equivalence class represents the subset of all elements related to zero under a specific equivalence relation. For the relation defined by "2 divides a - b" on the set of integers, [0] equals the set of all even integers, while [1] equals the set of all odd integers. These subsets E and O are disjoint and their union forms the entire set of integers Z.

Analytically, what ensures that the subsets E and O discussed in the passage do not overlap?

A The symmetric property alone

B The fact that Z is infinite

C The combined properties of an equivalence relation guarantee that elements of distinct equivalence classes are mutually disjoint.

D The empty relation property

💡 Short Explanation

Equivalence classes form partitions Distinct classes are always disjoint Even and odd integers never overlap

✅ Correct Answer Explanation

An equivalence relation partitions a set into mutually disjoint equivalence classes. Therefore no integer can belong simultaneously to both E (even integers) and O (odd integers). This disjointness follows from the combined reflexive, symmetric, and transitive properties of equivalence relations. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Symmetry alone cannot guarantee disjoint classes.
Option B → Infiniteness of integers is unrelated to overlap.
Option D → Empty relation does not generate equivalence classes here.

🎯 Strategy

Used: Elimination

Application:

Identify which property specifically creates partitions.

Final Logic:

Equivalence relations always produce disjoint equivalence classes.

🧠 Memory Trick

"Different classes never overlap"

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

📄 Passage

If an arbitrary integer 'x' belongs to [0] as defined in the passage, which defining property of equivalence relations explicitly guarantees that (x, 0) ∈ R?

A Symmetry (since 0 relates to x, x relates to 0)

B Transitivity

C Universal relation

D The empty set definition

💡 Short Explanation

Membership in [0] means relation with 0 Symmetry reverses related pairs Hence (x,0) belongs to R

✅ Correct Answer Explanation

If x belongs to [0], then 0Rx by definition of equivalence class. Since equivalence relations are symmetric, 0Rx implies xR0. Therefore (x,0) ∈ R is guaranteed by symmetry. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Transitivity requires three linked elements.
Option C → Universal relation is unrelated here.
Option D → Empty relation contradicts existence of equivalence classes.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the reverse-pair property of symmetry directly.

Final Logic:

0Rx implies xR0 by symmetry.

🧠 Memory Trick

"Symmetry flips the pair"

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

15 Let A = {1,2,3,4,5,6,7}. R = {(a,b): both are odd or both are even}. Why can no element of the equivalence class {1,3,5,7} relate to {2,4,6}?

A Because 7 + 6 = 13

B Because 1 is a prime number

C Because the set is finite

D Because an odd number and an even number fail the relation condition "both odd or both even".

💡 Short Explanation

Relation groups numbers by parity Odd-even pairs violate condition Hence classes remain disjoint

✅ Correct Answer Explanation

The relation requires both numbers to have identical parity: both odd or both even. Any pair consisting of one odd and one even number fails this condition. Therefore no element from {1,3,5,7} can relate to {2,4,6}. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Sum value is irrelevant to parity relation.
Option B → Primality does not affect the relation condition.
Option C → Finiteness of the set is unrelated.

🎯 Strategy

Used: Elimination

Application:

Check the exact condition defining the relation.

Final Logic:

Odd-even combinations violate same-parity requirement.

🧠 Memory Trick

"Parity groups stay separate"

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

16 Match the equivalence class representation to its algebraic set-builder form under modulo 3 arithmetic:

List I	List II
1. A₁=[0]	a. Set of all integers Z
2. A₂=[1]	b. {x∈Z:x=3r+2, r∈Z}
3. A₃=[2]	c. {x∈Z:x=3r, r∈Z}
4. A₁∪A₂∪A₃	d. {x∈Z:x=3r+1, r∈Z}

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

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

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

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

💡 Short Explanation

Modulo 3 creates three classes Multiples of 3 form [0] Their union gives all integers

✅ Correct Answer Explanation

[0] contains integers of form 3r → c. [1] contains integers of form 3r+1 → d. [2] contains integers of form 3r+2 → b. The union of all three classes equals Z → a. Thus Option B is correct.

❓ Why Other Options Are Incorrect

Option A → [0] is incorrectly matched with 3r+1 form.
Option C → Union cannot equal only one class.
Option D → [0] cannot represent numbers of form 3r+2.

🎯 Strategy

Used: Option Grouping

Application:

Match modulo classes with their remainder forms carefully.

Final Logic:

Modulo 3 partitions integers by remainders 0,1,2.

🧠 Memory Trick

"Modulo = remainder groups"

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

17 If sets A₁,A₂,A₃ form a valid partition of a set X under an equivalence relation, which of the following strictly hold true analytically?
I. A₁∩A₂=∅
II. A₁∪A₂∪A₃=X
III. (a,b)∈R for a∈A₁ and b∈A₂
IV. A₁,A₂,A₃ are mutually disjoint

A I and II only

B III and IV only

C I, II, and IV

D II, III, and IV

💡 Short Explanation

Partitions are mutually disjoint Their union equals original set Different classes contain unrelated elements

✅ Correct Answer Explanation

Partitions generated by equivalence relations satisfy: A₁∩A₂=∅, A₁∪A₂∪A₃=X, and all subsets are mutually disjoint. Statement III is false because elements from distinct equivalence classes are not related. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Omits the important mutual disjointness property.
Option B → Statement III contradicts partition definition.
Option D → Distinct classes cannot contain related elements.

🎯 Strategy

Used: Elimination

Application:

Test each statement against partition properties individually.

Final Logic:

Partitions are disjoint and collectively exhaustive.

🧠 Memory Trick

"Partition = disjoint + complete"

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

18 If the union of partition subsets equals the original set X, what does this analytically imply about the equivalence relation R mapping the elements?

A R maps every element to zero.

B R is a universal relation.

C R maps elements to infinity.

D Every element in X belongs to exactly one equivalence class defined by R.

💡 Short Explanation

Partitions cover the entire set Every element appears in one class Classes remain mutually disjoint

✅ Correct Answer Explanation

If the union of equivalence classes equals X, then every element of X belongs to at least one equivalence class. Since partitions are mutually disjoint, each element belongs to exactly one equivalence class. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Equivalence relations do not map everything to zero.
Option B → Partition property does not imply universal relation.
Option C → Infinity is irrelevant to equivalence classes.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret partition union property directly.

Final Logic:

Complete union plus disjointness gives unique class membership.

🧠 Memory Trick

"One element, one class"

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

19 Assertion (A): The relation "is congruent to" on triangles is an equivalence relation, but "is similar to" is not, because similar triangles can be different sizes.
Reason (R): Similarity fails the transitive property because scaling is not additive.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Similarity is also an equivalence relation Similarity satisfies transitivity Different sizes do not break equivalence

✅ Correct Answer Explanation

Congruence and similarity are both equivalence relations on triangles. Similarity satisfies reflexive, symmetric, and transitive properties even though triangle sizes may differ. Therefore Assertion is false. The Reason is also false because similarity is transitive under scaling transformations. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion incorrectly rejects similarity as equivalence relation.
Option C → Similarity does satisfy transitivity.
Option D → Reason is mathematically incorrect.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether the claim incorrectly excludes similarity relation.

Final Logic:

Similarity preserves equivalence despite scaling differences.

🧠 Memory Trick

"Similarity is also equivalence"

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

20 Let R = {(a,b): 4 divides |a - b|} defined on Z. Which of the following elements are in the exact same equivalence class as 3?

A 4, 8, 12

B 7, 11, -1

C 2, 4, 6

D 0, 1, 2

💡 Short Explanation

Numbers in same class differ by multiples of 4 Each listed number differs from 3 by 4 or 8 Hence same equivalence class

✅ Correct Answer Explanation

For numbers equivalent to 3 under the relation: \(4∣∣a-3∣\) Check Option B: |7−3|=4, |11−3|=8, |-1−3|=4. All are divisible by 4, so these numbers belong to the same equivalence class as 3. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → 4 differs from 3 by 1, not divisible by 4.
Option C → 2 differs from 3 by 1.
Option D → 1 differs from 3 by 2, not divisible by 4.

🎯 Strategy

Used: Substitution

Application:

Compute differences with 3 and test divisibility by 4.

Final Logic:

Equivalent numbers differ by multiples of 4.

🧠 Memory Trick

"Modulo 4 keeps same remainder"

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

UG Mathematics Booster Test 3 - Types of Relations

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