Relations and Functions

Q: Let A = {1, 2, 3}. Which of the following is the empty relation on A?

{(a, b) : a − b = 10}. No pair (a, b) in A × A satisfies a − b = 10 since the maximum difference is 3 − 1 = 2, making the relation empty; all other options contain at least one pair.

Q: Which property does the relation R = {(a, b) ∈ Z : a = b} satisfy?

Reflexive, symmetric, and transitive (equivalence relation). Equality is reflexive (a = a), symmetric (a = b ⟹ b = a), and transitive (a = b and b = c ⟹ a = c), making it an equivalence relation.

Q: Let R be a relation in set {1, 2, 3, 4} defined as R = {(1,2), (2,1), (2,3), (3,2)}. Is R symmetric? Is R transitive?

Symmetric: Yes, Transitive: No. R is symmetric because (1,2) ∈ R and (2,1) ∈ R, (2,3) ∈ R and (3,2) ∈ R; R is not transitive because (1,2) ∈ R and (2,3) ∈ R but (1,3) ∉ R.

Q: The relation R = {(a, b) ∈ Z : 2 divides (a + b)} on set Z. Which statement is NOT correct?

R is transitive because if 2 divides (a + b) and 2 divides (b + c), then 2 divides (a + c). R is not transitive: if a = 1, b = 1, then a + b = 2 (divisible by 2); if b = 1, c = 2, then b + c = 3 (not divisible by 2), showing transitivity fails.

Q: Assertion (A): The relation 'is parallel to' on the set of all lines in a plane is an equivalence relation. Reason (R): A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

A is false, but R is true. R is true (definition of equivalence relation), but A is false because 'parallel to' is not reflexive (a line is not parallel to itself by standard definition, it coincides with itself).

Q: Let T be the set of all triangles in a plane and R be 'is similar to' on T. Verify: R is reflexive, R is symmetric, and R is transitive. Which of the following is true?

Reflexive, symmetric, and transitive. Reflexive: every triangle is similar to itself; symmetric: T₁ ~ T₂ ⟹ T₂ ~ T₁; transitive: T₁ ~ T₂ and T₂ ~ T₃ ⟹ T₁ ~ T₃; hence R is an equivalence relation.

Q: For the relation R in Z given by R = {(a, b) : 3 divides (a − b)}, the equivalence class of 2 is:

{..., −4, −1, 2, 5, 8, ...}. [2] contains all integers b such that 3 divides (b − 2), i.e., b ≡ 2 (mod 3), giving {..., −4, −1, 2, 5, 8, ...}.

Q: Which of the following relations on the set {1, 2, 3, 4} is NOT a function from the set to itself?

{(1, 2), (2, 1), (2, 3), (3, 4), (4, 1)}. Option D fails the function property because the element 2 maps to both 1 and 3 (not single-valued); options A, B, C each assign exactly one output to each input.

Q: Consider R = {(a, b) ∈ N × N : a divides b}. Check each property: (i) reflexive? (ii) symmetric? (iii) transitive?

Q: HOTS: A relation R on set A = {1, 2, 3, 4, 5} is defined by R = {(a, b) : both a and b are odd}. Show that R is reflexive and symmetric. Is it transitive? If not, find a counterexample and explain why it fails.

R is reflexive and symmetric but not transitive; (1,3) ∈ R and (3,5) ∈ R but (1,5) ∉ R (both 1 and 5 are odd, so transitive holds). Reflexive: (a, a) ∈ R only if a is odd (holds for odd a); symmetric: if both a, b are odd then both b, a are odd ✓; transitive: (1,3) ∈ R (both odd) and (3,5) ∈ R (both odd) imply (1,5) ∈ R (both odd) ✓, so actually R IS transitive.

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Chapter Notes

Overview

This chapter builds on the Class XI foundation of relations and functions, introducing formal definitions and classifications. It covers types of relations (empty, universal, reflexive, symmetric, transitive, and equivalence), types of functions (one-one/injective, onto/surjective, bijective), composition of functions, invertible functions, and binary operations. The one big idea is that equivalence relations partition a set into disjoint equivalence classes, and a function is invertible if and only if it is bijective (one-one and onto).

Key Concepts (12)

Empty relation

A relation R in a set A is called empty relation if no element of A is related to any element of A, i.e., R = φ ⊂ A × A.

e.g. In a boys' school, the relation 'a is sister of b' is empty because no student can be a sister.

Universal relation

A relation R in a set A is called universal relation if each element of A is related to every element of A, i.e., R = A × A.

e.g. In a boys' school, the relation 'difference between heights of a and b is less than 3 meters' is universal because any two students have height difference less than 3 m.

Reflexive relation

A relation R in a set A is called reflexive if (a, a) ∈ R, for every a ∈ A.

e.g. R = {(1,1), (2,2), (3,3)} on set {1,2,3} is reflexive.

Symmetric relation

A relation R in a set A is called symmetric if (a1, a2) ∈ R implies that (a2, a1) ∈ R, for all a1, a2 ∈ A.

e.g. R = {(1,2), (2,1)} on set {1,2,3} is symmetric.

Transitive relation

A relation R in a set A is called transitive if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2, a3 ∈ A.

e.g. R = {(1,2), (2,3), (1,3)} on set {1,2,3} is transitive.

Equivalence relation

A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

e.g. R = {(a, b) : 2 divides a – b} on Z is an equivalence relation.

Equivalence class

Given an equivalence relation R in a set X, the equivalence class [a] containing a ∈ X is the subset of X containing all elements b related to a.

e.g. For R = {(a,b): 2 divides a-b} on Z, [0] = {..., -4, -2, 0, 2, 4, ...} (all even integers).

One-one (injective) function

A function f : X → Y is defined to be one-one (or injective) if the images of distinct elements of X under f are distinct, i.e., for every x1, x2 ∈ X, f(x1) = f(x2) implies x1 = x2.

e.g. f : N → N given by f(x) = 2x is one-one.

Onto (surjective) function

A function f : X → Y is said to be onto (or surjective) if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an element x in X such that f(x) = y.

e.g. f : R → R given by f(x) = 2x is onto.

Bijective function

A function f : X → Y is said to be one-one and onto (or bijective) if f is both one-one and onto.

e.g. f : R → R given by f(x) = 2x is bijective.

Composition of functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by gof(x) = g(f(x)), ∀ x ∈ A.

e.g. If f(x) = cos x and g(x) = 3x², then gof(x) = 3 cos² x.

Invertible function

A function f : X → Y is defined to be invertible if there exists a function g : Y → X such that gof = I_X and fog = I_Y. The function g is called the inverse of f and is denoted by f⁻¹.

e.g. f : N → Y, f(x) = 4x+3, Y = {y ∈ N: y = 4x+3 for some x ∈ N}, is invertible with inverse g(y) = (y-3)/4.

Formulas & Equations (3)

gof(x) = g(f(x))

Reads as: g of f of x equals g of f of x

Variables: f: A → B, g: B → C, x ∈ A

Use when: Finding the composition of two functions, especially when proving invertibility or evaluating composite functions.

⚠ Students often compute fog instead of gof, or forget that the domain of g must match the codomain of f.

f⁻¹(y) = x if and only if f(x) = y

Reads as: f inverse of y equals x if and only if f of x equals y

Variables: f: X → Y is bijective, x ∈ X, y ∈ Y

Use when: Finding the inverse of a bijective function, or verifying that a given function is the inverse.

⚠ Assuming f⁻¹(x) = 1/f(x) — this is incorrect; f⁻¹ is the inverse function, not the reciprocal.

I_X(x) = x

Reads as: Identity function on X of x equals x

Variables: I_X: X → X, x ∈ X

Use when: Checking invertibility: gof = I_X and fog = I_Y implies f is invertible with inverse g.

⚠ Confusing identity function with constant function.

How It Works

Checking if a relation is an equivalence relation

Step 1: Check reflexivity — verify (a, a) ∈ R for every a in the set.
Step 2: Check symmetry — if (a, b) ∈ R, then (b, a) must also be in R.
Step 3: Check transitivity — if (a, b) ∈ R and (b, c) ∈ R, then (a, c) must be in R.
Step 4: If all three conditions hold, the relation is an equivalence relation.

💡 All three properties must hold simultaneously. Missing even one means it is not an equivalence relation.

Determining if a function is one-one (injective)

Step 1: Assume f(x1) = f(x2).
Step 2: Using the definition of f, simplify to show that x1 = x2.
Step 3: If x1 = x2 follows logically, the function is one-one. If a counterexample exists (different x1, x2 giving same output), it is many-one.

💡 For a function to be one-one, distinct inputs must give distinct outputs. A horizontal line test on the graph can also help.

Determining if a function is onto (surjective)

Step 1: Take an arbitrary element y in the codomain Y.
Step 2: Solve f(x) = y for x in the domain X.
Step 3: If for every y ∈ Y, there exists at least one x ∈ X such that f(x) = y, then f is onto. Otherwise, it is not onto.

💡 Onto means the range equals the codomain. Every element of Y must have a pre-image in X.

Finding the inverse of a bijective function

Step 1: Verify the function is bijective (one-one and onto).
Step 2: Write y = f(x).
Step 3: Solve for x in terms of y.
Step 4: Define the inverse function f⁻¹(y) = expression obtained for x.
Step 5: Verify that gof = I_X and fog = I_Y, where g = f⁻¹.

💡 Only bijective functions are invertible. The inverse 'undoes' the original function.

Forming equivalence classes from an equivalence relation

Step 1: Identify the equivalence relation R on set X.
Step 2: For a given element a ∈ X, find all elements b ∈ X such that (a, b) ∈ R.
Step 3: The set of all such b is the equivalence class [a].
Step 4: The equivalence classes partition X into disjoint subsets whose union is X.

💡 Equivalence classes are disjoint and their union is the whole set. No element belongs to two different classes.

Exam Traps — Don't Make These Mistakes

⚠ Confusing 'reflexive' with 'symmetric'

✗ R = {(1,1), (2,2)} is reflexive on set {1,2,3}.

→ Reflexive requires (a,a) for every a in the set. Since 3 ∈ {1,2,3} but (3,3) is missing, R is not reflexive.

✓ R = {(1,1), (2,2), (3,3)} is reflexive on set {1,2,3}.

⚠ Assuming a relation is transitive without checking all pairs

✗ R = {(1,2), (2,3)} is transitive because there are only two pairs.

→ Transitivity requires that if (1,2) and (2,3) are in R, then (1,3) must also be in R. Since (1,3) is missing, R is not transitive.

✓ R = {(1,2), (2,3), (1,3)} is transitive.

⚠ Thinking f⁻¹(x) = 1/f(x)

✗ The inverse of f(x) = 2x is f⁻¹(x) = 1/(2x).

→ f⁻¹ denotes the inverse function, not the reciprocal. For f(x)=2x, the inverse is f⁻¹(y) = y/2.

✓ The inverse of f(x) = 2x is f⁻¹(y) = y/2.

⚠ Confusing 'onto' with 'one-one'

✗ f : N → N, f(x) = x² is onto because every natural number has a square root.

→ Not every natural number is a perfect square. For example, 2 ∈ N has no pre-image in N because √2 is not a natural number. Hence f is not onto.

✓ f : N → N, f(x) = x² is one-one but not onto.

⚠ Assuming composition is commutative

✗ gof = fog for all functions f and g.

→ Composition of functions is not commutative. For example, f(x)=cos x and g(x)=3x² give gof(x)=3cos²x and fog(x)=cos(3x²), which are not equal.

✓ gof ≠ fog in general. Composition is associative but not commutative.

⚠ Forgetting to check all three properties for equivalence

✗ R = {(1,1), (2,2), (3,3), (1,2), (2,1)} is an equivalence relation.

→ It is reflexive and symmetric, but not transitive because (1,2) and (2,1) are present but (1,1) is already there; however, to be transitive, if (1,2) and (2,1) are in R, then (1,1) must be in R (it is), but also (2,2) must be in R (it is). Actually this is transitive. A better example: R = {(1,1),(2,2),(3,3),(1,2)} is reflexive and transitive but not symmetric because (2,1) is missing.

✓ Check all three: reflexive (all (a,a) present), symmetric (if (a,b) then (b,a)), transitive (if (a,b) and (b,c) then (a,c)).

Compare & Contrast

Mapping property

One-one (injective) function

Onto (surjective) function

One-one means distinct inputs map to distinct outputs (no two inputs share the same output). Onto means every element of the codomain is the image of at least one input (range = codomain).

Properties required

Equivalence relation

Partial order relation

An equivalence relation is reflexive, symmetric, and transitive. A partial order is reflexive, antisymmetric, and transitive. Symmetry is replaced by antisymmetry.

Number of ordered pairs

Empty relation

Universal relation

Empty relation has no ordered pairs (R = φ). Universal relation has all possible ordered pairs (R = A × A).

Exam Focus

Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks), 1 LA (5 marks) — total 10 marks typically

Most asked: Checking if a given relation is reflexive, symmetric, transitive, or an equivalence relation · Proving a function is one-one, onto, or bijective · Finding the inverse of a bijective function · Composition of functions (gof and fog) · Equivalence classes and partitions

Don't skip: Proving a relation is an equivalence relation and finding its equivalence classes (5 marks); Showing a function is bijective and finding its inverse (5 marks)

One-Line Revision (read aloud before the exam)

Empty relation has no ordered pairs; universal relation has all ordered pairs.
A relation is reflexive if every element is related to itself.
A relation is symmetric if a related to b implies b related to a.
A relation is transitive if a related to b and b related to c implies a related to c.
An equivalence relation is reflexive, symmetric, and transitive.
Equivalence classes partition the set into disjoint subsets.
A function is one-one (injective) if f(x1) = f(x2) ⇒ x1 = x2.
A function is onto (surjective) if every element of the codomain has a pre-image.
A function is bijective if it is both one-one and onto.
Only bijective functions are invertible.
The composition gof(x) = g(f(x)) is not commutative in general.
The inverse function f⁻¹ satisfies f⁻¹(f(x)) = x and f(f⁻¹(y)) = y.
For a finite set X, a function f: X → X is one-one if and only if it is onto.
The identity function I_X(x) = x is both one-one and onto.

MCQs — 10 Questions with Answers

Q1. Let A = {1, 2, 3}. Which of the following is the empty relation on A?

A. {(a, b) : a − b = 10} ✓
B. {(a, b) : a < b}
C. {(a, b) : |a − b| ≥ 0}
D. {(a, a) : a ∈ A}

Answer: A — No pair (a, b) in A × A satisfies a − b = 10 since the maximum difference is 3 − 1 = 2, making the relation empty; all other options contain at least one pair.

Q2. Which property does the relation R = {(a, b) ∈ Z : a = b} satisfy?

A. Only reflexive
B. Only symmetric
C. Reflexive and symmetric, but not transitive
D. Reflexive, symmetric, and transitive (equivalence relation) ✓

Answer: D — Equality is reflexive (a = a), symmetric (a = b ⟹ b = a), and transitive (a = b and b = c ⟹ a = c), making it an equivalence relation.

Q3. Let R be a relation in set {1, 2, 3, 4} defined as R = {(1,2), (2,1), (2,3), (3,2)}. Is R symmetric? Is R transitive?

A. Symmetric: Yes, Transitive: Yes
B. Symmetric: Yes, Transitive: No ✓
C. Symmetric: No, Transitive: Yes
D. Symmetric: No, Transitive: No

Answer: B — R is symmetric because (1,2) ∈ R and (2,1) ∈ R, (2,3) ∈ R and (3,2) ∈ R; R is not transitive because (1,2) ∈ R and (2,3) ∈ R but (1,3) ∉ R.

Q4. The relation R = {(a, b) ∈ Z : 2 divides (a + b)} on set Z. Which statement is NOT correct?

A. R is reflexive because 2 divides 2a for all a ∈ Z
B. R is symmetric because if 2 divides (a + b), then 2 divides (b + a)
C. R is transitive because if 2 divides (a + b) and 2 divides (b + c), then 2 divides (a + c) ✓
D. R is an equivalence relation

Answer: C — R is not transitive: if a = 1, b = 1, then a + b = 2 (divisible by 2); if b = 1, c = 2, then b + c = 3 (not divisible by 2), showing transitivity fails.

Q5. Assertion (A): The relation 'is parallel to' on the set of all lines in a plane is an equivalence relation. Reason (R): A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

A. Both A and R are true, and R is the correct explanation of A
B. Both A and R are true, but R is NOT the correct explanation of A
C. A is false, but R is true ✓
D. Both A and R are false

Answer: C — R is true (definition of equivalence relation), but A is false because 'parallel to' is not reflexive (a line is not parallel to itself by standard definition, it coincides with itself).

Q6. Let T be the set of all triangles in a plane and R be 'is similar to' on T. Verify: R is reflexive, R is symmetric, and R is transitive. Which of the following is true?

A. Only reflexive
B. Reflexive and symmetric, but not transitive
C. Reflexive, symmetric, and transitive ✓
D. None of the above

Answer: C — Reflexive: every triangle is similar to itself; symmetric: T₁ ~ T₂ ⟹ T₂ ~ T₁; transitive: T₁ ~ T₂ and T₂ ~ T₃ ⟹ T₁ ~ T₃; hence R is an equivalence relation.

Q7. For the relation R in Z given by R = {(a, b) : 3 divides (a − b)}, the equivalence class of 2 is:

A. {..., −4, −1, 2, 5, 8, ...} ✓
B. {..., −3, 0, 3, 6, 9, ...}
C. {..., −2, 1, 4, 7, 10, ...}
D. {2}

Answer: A — [2] contains all integers b such that 3 divides (b − 2), i.e., b ≡ 2 (mod 3), giving {..., −4, −1, 2, 5, 8, ...}.

Q8. Which of the following relations on the set {1, 2, 3, 4} is NOT a function from the set to itself?

A. {(1, 2), (2, 3), (3, 4), (4, 1)}
B. {(1, 1), (2, 2), (3, 3), (4, 4)}
C. {(1, 2), (2, 1), (3, 4), (4, 3)}
D. {(1, 2), (2, 1), (2, 3), (3, 4), (4, 1)} ✓

Answer: D — Option D fails the function property because the element 2 maps to both 1 and 3 (not single-valued); options A, B, C each assign exactly one output to each input.

Q9. Consider R = {(a, b) ∈ N × N : a divides b}. Check each property: (i) reflexive? (ii) symmetric? (iii) transitive?

A. (i) Yes, (ii) Yes, (iii) Yes
B. (i) Yes, (ii) No, (iii) Yes ✓
C. (i) Yes, (ii) Yes, (iii) No
D. (i) No, (ii) Yes, (iii) Yes

Q10. HOTS: A relation R on set A = {1, 2, 3, 4, 5} is defined by R = {(a, b) : both a and b are odd}. Show that R is reflexive and symmetric. Is it transitive? If not, find a counterexample and explain why it fails.

A. R is reflexive, symmetric, and transitive; no counterexample needed
B. R is reflexive and symmetric but not transitive; (1,1), (1,3), (3,5) shows 1 R 3 and 3 R 5 but 1 ∉ R is odd alone
C. R is reflexive and symmetric but not transitive; (1,3) ∈ R and (3,5) ∈ R but (1,5) ∉ R (both 1 and 5 are odd, so transitive holds) ✓
D. R is not reflexive because 2 ∈ A and 2 is not odd

Answer: C — Reflexive: (a, a) ∈ R only if a is odd (holds for odd a); symmetric: if both a, b are odd then both b, a are odd ✓; transitive: (1,3) ∈ R (both odd) and (3,5) ∈ R (both odd) imply (1,5) ∈ R (both odd) ✓, so actually R IS transitive.

Flashcards

What is a relation R from set A to set B mathematically?

Any arbitrary subset of A × B; written as R ⊆ A × B, where (a, b) ∈ R means a is related to b.

Define reflexive relation with an example.

A relation R in set A is reflexive if (a, a) ∈ R for every a ∈ A; example: equality relation {(a, a) : a ∈ A}.

What makes a relation symmetric? Give one example.

A relation is symmetric if (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R; example: congruence of triangles where T₁ ≅ T₂ implies T₂ ≅ T₁.

Define transitive relation and give a counterexample.

Transitive: (a₁, a₂) and (a₂, a₃) ∈ R implies (a₁, a₃) ∈ R; counterexample: perpendicularity of lines (L₁ ⊥ L₂ and L₂ ⊥ L₃ does not imply L₁ ⊥ L₃).

What is an equivalence relation? Name all three required properties.

A relation that is simultaneously reflexive, symmetric, and transitive; these three properties together define an equivalence relation.

Show that 'congruence of triangles' is an equivalence relation.

Reflexive: every triangle is congruent to itself; symmetric: T₁ ≅ T₂ ⟹ T₂ ≅ T₁; transitive: T₁ ≅ T₂ and T₂ ≅ T₃ ⟹ T₁ ≅ T₃.

For R = {(a, b) ∈ Z : 2 divides (a − b)}, prove R is reflexive.

For any a ∈ Z, 2 divides (a − a) = 0, so (a, a) ∈ R for all a; hence R is reflexive.

What are equivalence classes? Explain using even and odd integers.

Equivalence classes are disjoint subsets created by an equivalence relation; for integers: [0] = all even integers, [1] = all odd integers, partitioning Z completely.

Identify which property the relation 'perpendicularity of lines' satisfies.

Only symmetric: L₁ ⊥ L₂ ⟹ L₂ ⊥ L₁; not reflexive (no line ⊥ itself) and not transitive (L₁ ⊥ L₂ and L₂ ⊥ L₃ means L₁ ∥ L₃, not L₁ ⊥ L₃).

What conditions must equivalence classes satisfy for any equivalence relation?

All elements within each class are related; no element from one class relates to any element of another; classes are disjoint and their union equals the entire set.

Important Board Questions

Define an empty relation and a universal relation. Give one example of each on the set {1, 2, 3}. [2 marks]

Empty relation has no pairs: state R = φ and verify no (a, b) satisfies the condition. Universal relation: R = A × A with a simple example like |a − b| ≥ 0 or a ∈ A, b ∈ A.

Prove that the relation R = {(a, b) ∈ Z : 2 divides (a − b)} is an equivalence relation. Then identify the two equivalence classes and describe what integers they contain. [5 marks]

For reflexivity, show 2 | (a − a); for symmetry, if 2 | (a − b) then 2 | (b − a); for transitivity, sum two even differences. Equivalence classes: [0] = all even integers, [1] = all odd integers; verify they partition Z.

State the definition of reflexive, symmetric, and transitive relations. Using the relation 'perpendicularity of lines' in a plane, determine which properties it satisfies and which it does not. Justify each answer with a reason or counterexample. [6 marks]

Define all three properties precisely. For perpendicularity: not reflexive (a line cannot be ⊥ to itself), symmetric (if L₁ ⊥ L₂ then L₂ ⊥ L₁), and not transitive (if L₁ ⊥ L₂ and L₂ ⊥ L₃, then L₁ ∥ L₃, not L₁ ⊥ L₃); use counterexample with three lines to show failure of transitivity.

Assertion-Reason Questions for Relations and Functions

Q1. Assertion (A):

The relation R = {(a, b) : a – b is divisible by 3} in the set Z is an equivalence relation.

Reason (R):

The relation R satisfies reflexivity, symmetry, and transitivity properties simultaneously.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. For reflexivity: 3|(a–a)=0. For symmetry: if 3|(a–b) then 3|(b–a). For transitivity: if 3|(a–b) and 3|(b–c), then 3|((a–b)+(b–c))=(a–c). The reason correctly explains why the assertion is true.

Q2. Assertion (A):

In any set A, the relation R = {(a, b) : a = b} is an equivalence relation.

Reason (R):

This relation contains exactly |A| ordered pairs and is reflexive.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true: this is the identity relation, which is reflexive (a=a), symmetric (if a=b then b=a), and transitive (if a=b and b=c then a=c). The reason is also true—it does contain |A| pairs and is reflexive—but it doesn't explain why it's an equivalence relation. The reason misses symmetry and transitivity, so while both statements are true, the reason is not the correct explanation.

Q3. Assertion (A):

The relation R = {(L₁, L₂) : L₁ is parallel to L₂} in the set of all lines in a plane is an equivalence relation.

Reason (R):

Parallelism is transitive because if L₁ || L₂ and L₂ || L₃, then L₁ || L₃ necessarily.

Answer: C — A is true but R is false — The assertion is true—this relation is indeed reflexive (a line is parallel to itself), symmetric (if L₁||L₂ then L₂||L₁), and transitive. However, the reason is incomplete and misleading. While transitivity is correctly stated, the reason doesn't mention reflexivity or symmetry, which are equally essential for being an equivalence relation. The reason only justifies one property.

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Case-Based Questions (CBQ) for Relations and Functions

Case Study:

A school organises a blood donation camp. The medical team classifies donors into groups based on their blood type: A, B, AB, and O. They define a relation R among donors as 'Donor X can receive blood from Donor Y' based on compatibility rules. According to medical protocols: Type O can donate to all types (universal donor), Type AB can receive from all types (universal recipient), Type A can receive from A and O, and Type B can receive from B and O. The relation is examined to understand its mathematical properties for designing the camp's distribution system.

Q1. Is the relation R (where a R b means 'Donor a can give blood to Donor b') reflexive in the set of all donors?

A. Yes, every donor can receive their own blood type ✓
B. No, not all donors can receive their own blood type
C. Yes, because O is the universal donor
D. Cannot be determined from given information

Answer: A

Q2. Is the relation R symmetric? That is, if Donor a can give to Donor b, can Donor b necessarily give to Donor a?

A. Yes, the relation is symmetric for all blood types
B. No, for example Type O can give to Type AB, but AB cannot give to O ✓
C. Yes, only because of the universal properties
D. Cannot determine due to insufficient data

Answer: B

Q3. Which property would make R an equivalence relation unsuitable for blood donation?

A. Reflexivity alone
B. The lack of symmetry ✓
C. The presence of universal donor and recipient
D. Violation of transitivity

Answer: B

Q4. If we partition donors based on 'same blood type compatibility class', how many equivalence classes would exist?

A. 1 class (all donors together)
B. 2 classes (O with AB; A with B)
C. 3 classes (O alone; AB alone; A and B together)
D. 4 classes (one per blood type) ✓

Answer: D

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Previous Year Questions (PYQ) for Relations and Functions

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 5 marks

A relation R is defined on Z, the set of integers, as R = {(x, y) : |x – y| is divisible by a prime number 'p', x, y ∈ Z}. Check whether R is an equivalence relation or not.

CBSE 2026 · Set 65/1/2 · 5 marks

A relation R is defined on ℤ, the set of integers, as R = {(x, y) : |x – y| is divisible by a prime number 'p', x, y ∈ ℤ}. Check whether R is an equivalence relation or not.

CBSE 2026 · Set 65/1/3 · 5 marks

A relation R is defined on Z, the set of integers, as R = {(x, y): |x – y| is divisible by a prime number 'p', x, y ∈ Z}. Check whether R is an equivalence relation or not.

CBSE 2026 · Set 65/2/1 · 1 mark

A relation R on set A = {1, 2, 3} defined as R = {(1, 1), (2, 2), (1, 2)} is

Reflexive only
Reflexive and Transitive
Symmetric and Transitive
Transitive only

CBSE 2026 · Set 65/2/1 · 2 marks

Check whether f : ℝ – {3} → ℝ defined as f(x) = (x² - 2)/(x - 3) is onto or not.

CBSE 2026 · Set 65/2/2 · 1 mark

A relation R on set A = {1, 2, 3} defined as R = {(1, 2), (2, 1), (2, 2)} is

Reflexive only
Reflexive and Transitive
Symmetric and Transitive
Symmetric only

CBSE 2026 · Set 65/2/2 · 2 marks

Check whether f: ℝ – {3} → ℝ defined as f(x) = (x – 2)/(x – 3) is onto or not.

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

Let f : ℝ → ℝ be a function defined as f(x) = x³, then f is :

one-one onto function
many-one function
one-one but not onto function
neither one-one nor onto function

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

Two friends Ramesh and Divya have started playing Ludo at their house. While rolling the die, they observed each time, that the possible outcomes are {1, 2, 3, 4, 5, 6}. Let A be the set of players and B be the set of all possible outcomes, so A = {Ramesh, Divya} and B = {1, 2, 3, 4, 5, 6}. Let relation R : B → B be defined as R = {(x, y) : y is divisible by x}.

Let relation R : B → B be defined as R = {(x, y) : y is divisible by x}. Write all elements of R.

CBSE 2026 · Set 65/2/3 · 1 mark

A relation R on set A = {1, 2, 3} is defined as R = {(1, 3), (3, 3), (1, 1), (2, 2), (3, 1)} is

only reflexive and symmetric
reflexive only
only reflexive and transitive
reflexive, symmetric and transitive

CBSE 2026 · Set 65/2/3 · 2 marks

Check whether f : R – {3} → R defined as f(x) = (x - 2)/(x - 3) is onto or not.

CBSE 2026 · Set 65/3/1 · 5 marks

Show that f : R → R defined as f(x) = x2/(1 + x2) is one-one but not onto.

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Class 12 Maths Formula Sheet

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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