UG Mathematics Booster Test 3 - Advanced Concepts and Exercises -

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UG Mathematics Booster Test 3 - Advanced Concepts and Exercises

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 3 - Advanced Concepts and Exercises

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QUESTION 1 OF 20

Assertion (A): The relation \(R=\{(x,y):x-y is an integer\}\)on \(Z\) is strictly reflexive.
Reason (R): For any integer \(x\), \(x-x=0\), and 0 is an integer. (Assertion-Reason)

Answer Review

1 Assertion (A): The relation \(R=\{(x,y):x-y is an integer\}\)on \(Z\) is strictly reflexive.
Reason (R): For any integer \(x\), \(x-x=0\), and 0 is an integer. (Assertion-Reason)

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Reflexive means \((x,x)\in R\) for all \(x\) \(x-x=0\) for every integer Since 0 is integer, relation is reflexive

✅ Correct Answer Explanation

A relation is reflexive if every element is related to itself. Here: \(x-x=0\) and \(0\in Z\). Therefore \((x,x)\in R\) for every integer \(x\). Thus Assertion is true. Reason correctly explains why the relation is reflexive. Hence both are true and Reason explains Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct.
Option B → Reason is not false because 0 is indeed an integer.
Option D → Assertion is true since reflexive condition is satisfied.

🎯 Strategy

Used: Substitution

Application:

Substitute \(y=x\) into relation definition.

Final Logic:

Since \(x-x=0\in Z\), every element relates to itself.

🧠 Memory Trick

"Reflexive ⇒ self-related"

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

2 Which of the following relations are symmetrically valid? (Multiple Correct)
I. \(R=\{(x,y):3x-y=0\}\)on \(\left\{1,2,…,14\right\}\)
II. \(R=\{(L_{1},L_{2}):L_{1}∥L_{2}\}\)
III. \(R=\{(x,y):x is father of y\}\)

A I only

B III only

C I and II

D II only

💡 Short Explanation

Parallelism works both ways Father relation is directional \(3x-y=0\) is not reversible generally

✅ Correct Answer Explanation

A relation is symmetric if: \((x,y)\in R\Rightarrow (y,x)\in R\) Parallel lines satisfy symmetry because if \(L_{1}∥L_{2}\), then \(L_{2}∥L_{1}\). Relation I fails because \(3x=y\) does not imply \(3y=x\). Father relation is directional and not symmetric.

❓ Why Other Options Are Incorrect

Option A → Relation I is not symmetric mathematically.
Option B → Father relation cannot reverse direction.
Option C → Relation I again violates symmetry property.

🎯 Strategy

Used: Elimination

Application:

Check whether relation reverses logically.

Final Logic:

Only parallelism remains unchanged after swapping.

🧠 Memory Trick

"Symmetric = reversible relation"

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

3 Let \(T\) be the set of all triangles in a plane. \(R=\{(T_{1},T_{2}):T_{1} is congruent to T_{2}\}\). Is \(R\) transitive? (Graph/Region-based)

A Yes, if \(T_{1}≅T_{2}\)and \(T_{2}≅T_{3}\), then \(T_{1}≅T_{3}\).

B No, congruence is only symmetric.

C No, triangles can have different areas.

D Yes, but only for right-angled triangles.

💡 Short Explanation

Congruence preserves dimensions Transitivity follows logical chaining Same shape and size pass through relation

✅ Correct Answer Explanation

A relation is transitive if: \(aRb and bRc\Rightarrow aRc\) If triangle \(T_{1}\)is congruent to \(T_{2}\), and \(T_{2}\)to \(T_{3}\), then all corresponding sides and angles are equal. Hence \(T_{1}≅T_{3}\). Therefore congruence is transitive.

❓ Why Other Options Are Incorrect

Option B → Congruence is not only symmetric; it is also transitive.
Option C → Area difference is irrelevant here.
Option D → Property holds for all triangles, not only right triangles.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply definition of transitivity directly.

Final Logic:

Congruence chaining guarantees transitivity.

🧠 Memory Trick

"Congruent chain stays congruent"

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

4 \(R=\{(a,b):a\leq b^{2}\}\)on the set of real numbers \(R\). Which pair breaks the rule of transitivity? (MCQ)

A (1, 2) and (2, 3)

B (2, -2) and (-2, 1)

C (0, 0) and (0, 1)

D (1, 1) and (1, 1)

💡 Short Explanation

Need \(aRb\) and \(bRc\) but not \(aRc\) Counterexample disproves transitivity Option B violates final condition

✅ Correct Answer Explanation

For Option B: \(2\leq (-2)^{2}=4\) and \(-2\leq 1^{2}=1\) Both relations hold. But: \(2\leq 1^{2}=1\) is false. Hence transitivity fails. Therefore Option B gives a valid counterexample.

❓ Why Other Options Are Incorrect

Option A → All conditions satisfy transitivity.
Option C → No contradiction occurs.
Option D → Reflexive pair naturally satisfies transitivity.

🎯 Strategy

Used: Substitution

Application:

Test ordered pairs numerically.

Final Logic:

Option B satisfies first two relations but breaks third.

🧠 Memory Trick

"Counterexample breaks final link"

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

5 Let \(A\) be the set of books in a library. \(R=\{(x,y):x and y have the same number of pages\}\). What type of relation is \(R\)? (MCQ)

A Only Reflexive

B Only Symmetric

C Equivalence relation

D Not Transitive

💡 Short Explanation

Same-page relation is reflexive Relation is symmetric and transitive All three properties imply equivalence

✅ Correct Answer Explanation

Books always have same pages as themselves, so reflexive. If book \(x\) has same pages as \(y\), then \(y\) has same pages as \(x\), making it symmetric. If \(x\) and \(y\) have same pages, and \(y\) and \(z\) also, then \(x\) and \(z\) match too. Hence equivalence relation.

❓ Why Other Options Are Incorrect

Option A → Relation is more than reflexive.
Option B → It also satisfies reflexive and transitive properties.
Option D → Transitivity clearly holds.

🎯 Strategy

Used: Option Grouping

Application:

Check all three equivalence properties.

Final Logic:

Reflexive + symmetric + transitive ⇒ equivalence relation.

🧠 Memory Trick

"RST makes equivalence"

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

6 Let \(R\) be defined on the set of all rectangles such that \(ARB\) if Area of \(A=\)Area of \(B\). What is the equivalence class of a rectangle with sides 3 and 4? (Area)

A All squares of side 12

B Only rectangles with sides 3 and 4

C All triangles with area 12

D All rectangles whose product of adjacent sides is 12

💡 Short Explanation

Rectangle area equals length × breadth Equivalence class contains equal-area rectangles Area here equals 12

✅ Correct Answer Explanation

A rectangle with sides 3 and 4 has area: \(3\times 4=12\) Its equivalence class includes every rectangle whose area is also 12. Thus all rectangles satisfying: \(l\times b=12\) belong to the same equivalence class.

❓ Why Other Options Are Incorrect

Option A → Square side 12 has area 144.
Option B → Class contains many rectangles, not one.
Option C → Relation applies only to rectangles.

🎯 Strategy

Used: Substitution

Application:

Compute area and identify matching objects.

Final Logic:

Equal area defines equivalence class.

🧠 Memory Trick

"Same area ⇒ same class"

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

7 Let \(f:R\rightarrow R\) be defined by \(f(x)=3-4x\). Is \(f\) injective? (MCQ)

A Yes, because \(f(x_{1})=f(x_{2})\Rightarrow -4x_{1}=-4x_{2}\Rightarrow x_{1}=x_{2}\).

B No, because it is a linear function.

C No, because \(f(0)=3\).

D Yes, because \(f(1)\neq f(-1)\).

💡 Short Explanation

Injective means equal outputs imply equal inputs Linear function slope is non-zero Algebra proves uniqueness

✅ Correct Answer Explanation

Assume: \(f(x_{1})=f(x_{2})\) Then: \(3-4x_{1}=3-4x_{2}-4x_{1}=-4x_{2}x_{1}=x_{2}\) Hence distinct inputs cannot give same output. Therefore function is injective.

❓ Why Other Options Are Incorrect

Option B → Non-zero slope linear functions are injective.
Option C → Single output value does not affect injectivity.
Option D → One example alone cannot prove injectivity fully.

🎯 Strategy

Used: Substitution

Application:

Apply injective definition algebraically.

Final Logic:

Equal outputs force equal inputs.

🧠 Memory Trick

"Non-zero slope ⇒ one-one"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: One-One Function

8 Let the signum function \(f:R\rightarrow R\) be defined as \(f(x)=1\) if \(x>0\), \(0\) if \(x=0\), \(-1\) if \(x<0\). What mathematically makes it NOT surjective? (MCQ)

A It maps multiple values to 1.

B Elements like 2 or -5 in the co-domain have no pre-image.

C It is discontinuous at \(x=0\).

D It is an odd function.

💡 Short Explanation

Range contains only \(-1,0,1\) Codomain is all real numbers Many real numbers lack pre-images

✅ Correct Answer Explanation

The signum function outputs only: \(-1, 0, 1\) But codomain is \(R\). Numbers like 2 or -5 are never obtained as outputs. Therefore not every codomain element has a pre-image, so function is not surjective.

❓ Why Other Options Are Incorrect

Option A → Many-one behavior does not determine surjectivity.
Option C → Continuity is unrelated to onto property.
Option D → Oddness does not affect surjectivity.

🎯 Strategy

Used: Elimination

Application:

Compare range with codomain.

Final Logic:

Missing codomain elements imply not onto.

🧠 Memory Trick

"Onto covers entire codomain"

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

9 Match the given function to its mapped nature (Domain \(R\rightarrow R\)):

List I	List II
1. f(x)=x^3	a. One-one but not onto
2. f(x)=x^2	b. Bijective (One-one and Onto)
3. f(x)=x^2+1	c. Neither one-one nor onto
4. f(x)=e^x	d. Many-one but not onto

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

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

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

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

💡 Short Explanation

x^3is one-one and onto on R.

x^2is many-one and not onto.

x^2+1is neither one-one nor onto, while e^xis one-one but not onto.

✅ Correct Answer Explanation

Consider each function with domain and codomain R:

f(x)=x^3: It is strictly increasing and takes every real value. Hence, it is bijective (one-one and onto).

1 → b

f(x)=x^2: Since f(2)=f(-2)=4, it is many-one. Also, it never takes negative values, so it is not onto.

2 → d

f(x)=x^2+1: It is many-one and its range is [1,∞), so it is not onto R. Therefore, it is neither one-one nor onto.

3 → c

f(x)=e^x: It is strictly increasing, so it is one-one, but its range is (0ⓜ,∞), so it is not onto R.

4 → a

Thus,

1 → b

2 → d

3 → c

4 → a

Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly identifies x^3as not onto and e^xas bijective.
Option C → Incorrectly classifies x^2and x^2+1.
Option D → Incorrectly matches all four functions.

🎯 Strategy

Concept Matching

Application:

Determine whether each function is one-one and/or onto by examining its monotonicity and range.

Final Logic:

x^3→ Bijective

x^2→ Many-one, Not Onto

x^2+1→ Neither One-one nor Onto

e^x→ One-one, Not Onto

🧠 Memory Trick

Odd Power → Bijective • Square → Many-one • Square + 1 → Neither • Exponential → One-one, Positive Only

📚 Class XII – Mathematics – Chapter: Relations and Functions – Topic: One-One, Onto and Bijective Functions

10 Which of the following is a strict example of a bijective function? (MCQ)

A \(f(x)=[x]\)on \(R\)

B \(f(x)=x^{2}\)on \(Z\)

C \(f(x)=∣x∣\)on \(R\)

D \(f(x)=x\) on \(N\)

💡 Short Explanation

Identity mapping preserves uniqueness Every natural number maps to itself Function is one-one and onto

✅ Correct Answer Explanation

The identity function: \(f(x)=x\) maps each natural number uniquely onto itself. Hence it is injective and surjective simultaneously. Therefore it is bijective.

❓ Why Other Options Are Incorrect

Option A → Greatest integer function is not onto \(R\).
Option B → \(x^{2}\)fails injectivity since \(1^{2}=(-1)^{2}\).
Option C → Modulus function is neither one-one nor onto.

🎯 Strategy

Used: Elimination

Application:

Reject functions failing one-one or onto tests.

Final Logic:

Identity mapping satisfies both conditions perfectly.

🧠 Memory Trick

"Identity ⇒ always bijective"

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

11 A function is randomly generated from set \(A=\{1,2\}\)to \(B=\{a,b,c\}\). What is the exact probability that it is a constant function? (Probability)

A 1/3

B 1/9

C 2/9

D 6/9

💡 Short Explanation

Total functions from \(A\rightarrow B\) are \(3^{2}=9\) Constant functions map all inputs to one output Exactly 3 constant functions exist

✅ Correct Answer Explanation

For \(A=\{1,2\}\)and \(B=\{a,b,c\}\): Total functions: \(3^{2}=9\) Constant functions are: \((1,2)\rightarrow a,(1,2)\rightarrow b,(1,2)\rightarrow c\) So number of constant functions \(=3\). Hence probability: \(\frac{3}{9}=\frac{1}{3}\)

❓ Why Other Options Are Incorrect

Option B → Counts only one constant mapping.
Option C → Incorrect simplification of favorable outcomes.
Option D → Exceeds actual probability calculation.

🎯 Strategy

Used: Substitution

Application:

Use probability = favorable outcomes ÷ total outcomes.

Final Logic:

3 constant mappings among 9 total functions give \(1/3\).

🧠 Memory Trick

"Constant → one output only"

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

12 How many strictly onto functions are there from a set of 3 elements to a set of 2 elements? (Case/Numerical)

A 8

B 6

C 4

D 2

💡 Short Explanation

Total functions \(=2^{3}=8\) Exclude non-onto constant mappings Remaining functions are onto

✅ Correct Answer Explanation

From a 3-element set to a 2-element set: \(2^{3}=8\) total functions exist. Non-onto functions are the two constant functions where all elements map to only one codomain element. Thus onto functions: \(8-2=6\) Hence answer is 6.

❓ Why Other Options Are Incorrect

Option A → Counts all functions, not onto only.
Option C → Under-counts valid onto mappings.
Option D → Counts only constant cases.

🎯 Strategy

Used: Elimination

Application:

Subtract non-surjective mappings from total functions.

Final Logic:

Total minus constant functions gives onto count.

🧠 Memory Trick

"Onto covers every output"

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

13 The total mathematical number of relations on a set with \(n\) elements is: (Case/Numerical)

A \(2^{n}\)

B \(n^{2}\)

C \(2^{\left(n^{2}\right)}\)

D \(2^{2n}\)

💡 Short Explanation

Relations are subsets of \(A\times A\) \(A\times A\) has \(n^{2}\)elements Number of subsets is \(2^{n^{2}}\)

✅ Correct Answer Explanation

If set \(A\) has \(n\) elements, then: \(∣A\times A∣=n^{2}\) A relation is any subset of \(A\times A\). Number of subsets of a set containing \(n^{2}\)elements is: \(2^{n^{2}}\) Hence total relations equal \(2^{n^{2}}\).

❓ Why Other Options Are Incorrect

Option A → Counts subsets of \(A\), not relations.
Option B → Gives ordered-pair count only.
Option D → Incorrect exponent formula.

🎯 Strategy

Used: Substitution

Application:

Use Cartesian product size and subset counting.

Final Logic:

Relations are subsets of \(A\times A\).

🧠 Memory Trick

"Relations = subsets of square set"

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

14 Which statement is INCORRECT regarding the relations formed on set \(A=\{1,2\}\)? (Incorrect Statement)

A Total relations = 16

B Empty relation is a valid relation

C Universal relation has 4 elements

D There are 8 equivalence relations

💡 Short Explanation

Total relations on 2-element set = 16 Empty and universal relations are valid Equivalence relations are fewer than 8

✅ Correct Answer Explanation

For \(A=\{1,2\}\): \(∣A\times A∣=4\) So total relations: \(2^{4}=16\) Universal relation contains all 4 ordered pairs. Empty relation is valid. However, equivalence relations on a 2-element set are only: 1. \(\left\{\{1\},\{2\}\right\}\) 2. \(\left\{\{1,2\}\right\}\) Hence only 2 equivalence relations exist.

❓ Why Other Options Are Incorrect

Option A → Correct because \(2^{4}=16\).
Option B → Empty set is always a subset relation.
Option C → Universal relation equals entire Cartesian product.

🎯 Strategy

Used: Elimination

Application:

Verify each statement using relation formulas.

Final Logic:

Equivalence relations correspond to partitions, not 8.

🧠 Memory Trick

"Equivalence ↔ partitions"

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

15 A mixture expands over time. \(f(t)=2t\) represents volume. \(g(v)=5v\) represents cost based on volume. Evaluate the composite function for cost based on time \(g(f(t))\). (Case/Numerical)

A \(10t\)

B \(7t\)

C \(2.5t\)

D \(10t^{2}\)

💡 Short Explanation

First compute volume from time Then compute cost from volume Composition multiplies factors

✅ Correct Answer Explanation

Given: \(f(t)=2t,g(v)=5v\) Then: \(g(f(t))=g(2t)=5(2t)=10t\) Hence the composite function representing cost based on time is \(10t\).

❓ Why Other Options Are Incorrect

Option B → Incorrect arithmetic.
Option C → Division instead of multiplication used.
Option D → Squaring is not involved in composition.

🎯 Strategy

Used: Substitution

Application:

Replace variable in outer function with inner output.

Final Logic:

Outer function acts on inner result.

🧠 Memory Trick

"Inside first, outside next"

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

16 Let \(f:R^{*}\rightarrow R^{*}\)be given by \(f(x)=1/x\). What is the value of \(f(f(x))\)? (MCQ)

A \(1/x^{2}\)

B \(x\)

C \(1\)

D \(-x\)

💡 Short Explanation

Reciprocal applied twice restores original First inversion gives \(1/x\) Second inversion returns \(x\)

✅ Correct Answer Explanation

Given: \(f(x)=\frac{1}{x}\) Then: \(f(f(x))=f\left(\frac{1}{x}\right)=\frac{1}{1/x}=x\) Thus the function is self-invertible.

❓ Why Other Options Are Incorrect

Option A → Multiplies reciprocals incorrectly.
Option C → Simplifies only for \(x=1\).
Option D → Sign never changes here.

🎯 Strategy

Used: Substitution

Application:

Substitute output back into same function.

Final Logic:

Reciprocal of reciprocal equals original value.

🧠 Memory Trick

"Double reciprocal returns original"

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

17 Find the inverse formula of \(f(x)=\frac{4x}{3x+4}\)where domain is \(R-\{-4/3\}\)and Range is \(f(x)\). (MCQ)

A \(\frac{3y}{4-4y}\)

B \(\frac{4y}{3y-4}\)

C \(\frac{4y}{4-3y}\)

D \(\frac{4-3y}{4y}\)

💡 Short Explanation

Replace \(f(x)\)by \(y\) Solve equation for \(x\) Express \(x\) in terms of \(y\)

✅ Correct Answer Explanation

Start with: \(y=\frac{4x}{3x+4}\) Cross multiply: \(y(3x+4)=4x3xy+4y=4x4y=x(4-3y)x=\frac{4y}{4-3y}\) Hence inverse is: \(f^{-1}(y)=\frac{4y}{4-3y}\)

❓ Why Other Options Are Incorrect

Option A → Numerator incorrect.
Option B → Denominator sign error.
Option D → Reciprocal arrangement incorrect.

🎯 Strategy

Used: Substitution

Application:

Interchange variables and isolate \(x\).

Final Logic:

Algebraic solving yields inverse directly.

🧠 Memory Trick

"Swap, solve, simplify"

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

18 Arrange the steps required to verify \(g(y)=y^{1/3}\)is the strict inverse of \(f(x)=x^{3}\): (Arrange in Order)
I. Calculate \(g(f(x))=g(x^{3})=(x^{3})^{1/3}=x\)
II. Conclude \(gof=I\) and \(fog=I\)
III. Calculate \(f(g(y))=f(y^{1/3})=(y^{1/3})^{3}=y\)
IV. State that \(f\) and \(g\) are inverses

A I, II, III, IV

B III, I, IV, II

C I, IV, III, II

D I, III, II, IV

💡 Short Explanation

Verify both compositions separately Show each gives identity mapping Then conclude inverse relation

✅ Correct Answer Explanation

First compute: \(g(f(x))=x\) Then compute: \(f(g(y))=y\) After verifying both identities, conclude: \(gof=I,fog=I\) Hence \(f\) and \(g\) are inverses. Therefore correct order is: \(I, III, II, IV\)

❓ Why Other Options Are Incorrect

Option A → Conclusion appears before second verification.
Option B → Final statement order incorrect.
Option C → Inverse claim stated prematurely.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow logical proof sequence carefully.

Final Logic:

Both compositions must equal identity before conclusion.

🧠 Memory Trick

"Check both ways first"

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

📄 Passage

Given an arbitrary equivalence relation \(R\) in an arbitrary set \(X\), \(R\) divides \(X\) into mutually disjoint subsets \(A_{i}\)called partitions or subdivisions of \(X\) satisfying: all elements of \(A_{i}\)are related to each other, no element of \(A_{i}\)is related to any element of \(A_{j}\), and their union is \(X\).

If \(X=\{1,2,3\}\)and \(R\) divides it into \(A_{1}=\{1,2\}\)and \(A_{2}=\{3\}\). Which of the following is true?

A \(1R2\)

B \(1R3\)

C \(2R3\)

D \(A_{1}\)and \(A_{2}\)have common elements

💡 Short Explanation

Same partition elements are related Different partition elements are unrelated Partitions are disjoint

✅ Correct Answer Explanation

Since \(1\) and \(2\) belong to the same equivalence class \(A_{1}\), they are related under \(R\): \(1R2\) Element \(3\) belongs to a different class, so it is not related to \(1\) or \(2\).

❓ Why Other Options Are Incorrect

Option B → Different equivalence classes are unrelated.
Option C → Same reason; classes differ.
Option D → Partitions are mutually disjoint.

🎯 Strategy

Used: Option Grouping

Application:

Use equivalence-class definition directly.

Final Logic:

Elements in same partition are related.

🧠 Memory Trick

"Same class ⇒ related"

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

📄 Passage

What is the logical union of all the generated partitions \(A_{i}\)?

A Empty set \(\emptyset\)

B The original set \(X\)

C \(X\times X\)

D The relation \(R\)

💡 Short Explanation

Equivalence classes partition entire set Every element belongs to one class Union recreates original set

✅ Correct Answer Explanation

Partitions formed by an equivalence relation satisfy: \(⋃A_{i}=X\) Thus the union of all equivalence classes gives the complete original set \(X\).

❓ Why Other Options Are Incorrect

Option A → Partition union cannot be empty.
Option C → Cartesian product differs from partition union.
Option D → Relation is a set of ordered pairs, not union of classes.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply direct property of partitions.

Final Logic:

Partitions together cover the entire set.

🧠 Memory Trick

"Partitions rebuild the set"

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

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