NCERT Solutions Class 12 Mathematics Chapter 1 Relations and Functions

These NCERT Solutions Class 12 Mathematics Chapter 1 Relations and Functions provide step-by-step guidance designed to help students master abstract algebraic concepts and secure maximum marks in their CBSE board examinations. Relations and Functions form the absolute foundation of higher mathematics, advanced calculus, and algebra in Class 12. In this definitive guide, you will find complete, curriculum-aligned NCERT Mathematics Class 12 solutions covering Exercise 1.1, Exercise 1.2, and key Miscellaneous exercise questions with clear whiteboard-style proofs, counterexamples, and key exam strategies.

Chapter Overview: Relations and Functions

Chapter 1 of Class 12 Mathematics bridges foundational set theory learned in Class 11 with the advanced language of mappings and equivalence structures. Under the latest CBSE syllabus, Chapter 1 is structured around two major exercises:

• Exercise 1.1 (Types of Relations): Empty relation, universal relation, reflexive, symmetric, transitive relations, equivalence relations, and equivalence classes.
• Exercise 1.2 (Types of Functions): Injective (one-to-one), surjective (onto), and bijective (one-to-one and onto) functions.

In the CBSE Class 12 board examination, Unit I: Relations and Functions (which includes Chapter 1 and Chapter 2: Inverse Trigonometric Functions) carries a total weightage of 8 marks. Mastering this chapter ensures confidence in 1-mark objective questions, 2-mark concept verifications, and 3-mark or 5-mark long algebraic proofs.

Key Concepts, Definitions & Theorems

Before diving into the detailed exercise solutions, let us review the core mathematical definitions and theoretical frameworks tested in CBSE board exams.

1. Types of Relations

Let A be a non-empty set. A relation R on A is a subset of the Cartesian product A × A (i.e., R ⊆ A × A).

• Empty Relation: A relation R in a set A is called an empty relation if no element of A is related to any element of A, denoted by R = φ ⊆ A × A.
• Universal Relation: A relation R in a set A is called a universal relation if each element of A is related to every element of A, denoted by R = A × A.
• Reflexive Relation: A relation R on A is reflexive if (a, a) ∈ R for every a ∈ A.
• Symmetric Relation: A relation R on A is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A.
• Transitive Relation: A relation R on A is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A.
• Equivalence Relation: A relation R on a set A is an equivalence relation if it is simultaneously reflexive, symmetric, and transitive.
• Equivalence Class: Given an equivalence relation R on set A and an element a ∈ A, the equivalence class of a, denoted by [a], is the set of all elements in A related to a:
  [a] = {x ∈ A : (x, a) ∈ R}

2. Types of Functions (Mappings)

Let f: X → Y be a function from domain X to codomain Y.

• One-to-One (Injective) Function: A function f: X → Y is injective if distinct elements in X have distinct images in Y. Formally:
  f(x₁) = f(x₂) ⇒ x₁ = x₂ for all x₁, x₂ ∈ X.
• Onto (Surjective) Function: A function f: X → Y is surjective if every element in the codomain Y is the image of at least one element in the domain X. Formally:
  For every y ∈ Y, there exists x ∈ X such that f(x) = y (i.e., Range of f = Codomain Y).
• Bijective Function: A function f: X → Y is bijective if it is both injective (one-to-one) and surjective (onto).

NCERT Solutions Class 12 Mathematics Chapter 1: Exercise 1.1

Question 1 (i): Determine whether the relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x - y = 0} is reflexive, symmetric, and transitive.

Step-by-Step Solution:

1. Given: Set A = {1, 2, 3, ..., 14} and relation R = {(x, y) : 3x - y = 0}, which means y = 3x where x, y ∈ A.
2. Roster Form of R:
   • For x = 1: y = 3(1) = 3 ∈ A ⇒ (1, 3) ∈ R
   • For x = 2: y = 3(2) = 6 ∈ A ⇒ (2, 6) ∈ R
   • For x = 3: y = 3(3) = 9 ∈ A ⇒ (3, 9) ∈ R
   • For x = 4: y = 3(4) = 12 ∈ A ⇒ (4, 12) ∈ R
   • For x = 5: y = 3(5) = 15 ∉ A
   Thus, R = {(1, 3), (2, 6), (3, 9), (4, 12)}.
3. Test for Reflexivity: A relation is reflexive if (a, a) ∈ R for all a ∈ A. Here, (1, 1) ∉ R because 3(1) - 1 = 2 ≠ 0. Hence, R is not reflexive.
4. Test for Symmetry: A relation is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R. Here, (1, 3) ∈ R (since 3(1) - 3 = 0), but (3, 1) ∉ R (since 3(3) - 1 = 8 ≠ 0). Hence, R is not symmetric.
5. Test for Transitivity: A relation is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. Here, (1, 3) ∈ R and (3, 9) ∈ R, but (1, 9) ∉ R (since 3(1) - 9 = -6 ≠ 0). Hence, R is not transitive.

Final Answer: The relation R is neither reflexive, nor symmetric, nor transitive.

Question 2: Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.

Step-by-Step Solution:

1. Given: Relation R = {(a, b) : a ≤ b², a, b ∈ ℝ}.
2. Reflexivity Test (Counterexample Method):

   Let a = 1/2 ∈ ℝ.

   Then, a² = (1/2)² = 1/4.

   Since 1/2 ≤ 1/4 is false (1/2 > 1/4), (1/2, 1/2) ∉ R.

   Therefore, R is not reflexive.

3. Symmetry Test (Counterexample Method):

   Let a = 1 and b = 2.

   1 ≤ 2² (1 ≤ 4) is true, so (1, 2) ∈ R.

   However, 2 ≤ 1² (2 ≤ 1) is false, so (2, 1) ∉ R.

   Therefore, R is not symmetric.

4. Transitivity Test (Counterexample Method):

   Let a = 2, b = -2, and c = -1 (or choose a = 3, b = -2, c = -1).

   Take a = 2, b = -2, c = -1:

   • Is (a, b) ∈ R? 2 ≤ (-2)² = 4 (True), so (2, -2) ∈ R.
   • Is (b, c) ∈ R? -2 ≤ (-1)² = 1 (True), so (-2, -1) ∈ R.
   • Is (a, c) ∈ R? 2 ≤ (-1)² = 1 is False (2 > 1), so (2, -1) ∉ R.

   Therefore, R is not transitive.

Final Answer: Hence proved, the relation R = {(a, b) : a ≤ b²} on ℝ is neither reflexive, nor symmetric, nor transitive.

Question 5: Show that the relation R in the set Z of integers given by R = {(a, b) : 2 divides (a - b)} is an equivalence relation.

Step-by-Step Solution:

1. Given: R = {(a, b) : a, b ∈ ℤ, (a - b) is divisible by 2}.
2. Reflexivity:

   For any a ∈ ℤ:

   a - a = 0 = 2 × 0, which is divisible by 2.

   Thus, (a, a) ∈ R for all a ∈ ℤ. Hence, R is reflexive.

3. Symmetry:

   Let (a, b) ∈ R.

   ⇒ (a - b) is divisible by 2.

   ⇒ a - b = 2k for some integer k.

   ⇒ -(a - b) = -2k ⇒ b - a = 2(-k).

   Since -k is also an integer, (b - a) is divisible by 2.

   ⇒ (b, a) ∈ R. Hence, R is symmetric.

4. Transitivity:

   Let (a, b) ∈ R and (b, c) ∈ R.

   ⇒ a - b = 2k and b - c = 2m for some integers k, m ∈ ℤ.

   Adding the two equations:

   (a - b) + (b - c) = 2k + 2m

   ⇒ a - c = 2(k + m).

   Since (k + m) is an integer, (a - c) is divisible by 2.

   ⇒ (a, c) ∈ R. Hence, R is transitive.

Final Answer: Since R is reflexive, symmetric, and transitive, R is an equivalence relation on ℤ.

Question 9: Show that the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by R = {(a, b) : |a - b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1.

Step-by-Step Solution:

1. Given: Set A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} and R = {(a, b) : |a - b| = 4k, k ∈ ℤ_≥0}.
2. Reflexive: For any a ∈ A, |a - a| = 0 = 4(0), which is a multiple of 4. Thus (a, a) ∈ R ∀ a ∈ A. Hence, R is reflexive.
3. Symmetric: Let (a, b) ∈ R ⇒ |a - b| is a multiple of 4 ⇒ |-(b - a)| = |b - a| is a multiple of 4 ⇒ (b, a) ∈ R. Hence, R is symmetric.
4. Transitive: Let (a, b) ∈ R and (b, c) ∈ R.
   ⇒ a - b = ±4k₁ and b - c = ±4k₂ for some integers k₁, k₂.
   Adding them: (a - b) + (b - c) = a - c = 4(±k₁ ± k₂) = 4m.
   ⇒ |a - c| is a multiple of 4 ⇒ (a, c) ∈ R. Hence, R is transitive.
5. Finding Elements Related to 1:

   The set of elements related to 1 is the equivalence class [1]:

   [1] = {x ∈ A : (x, 1) ∈ R} = {x ∈ A : |x - 1| is a multiple of 4}.

   Checking elements of A:

   • |1 - 1| = 0 (multiple of 4) ⇒ x = 1
   • |5 - 1| = 4 (multiple of 4) ⇒ x = 5
   • |9 - 1| = 8 (multiple of 4) ⇒ x = 9

Final Answer: R is an equivalence relation. The set of all elements related to 1 is {1, 5, 9}.

NCERT Solutions Class 12 Mathematics Chapter 1: Exercise 1.2

Question 1: Show that the function f: R_* → R_* defined by f(x) = 1/x is one-one and onto, where R_* is the set of all non-zero real numbers. Is the result true if the domain R_* is replaced by N with codomain being same as R_*?

Step-by-Step Solution:

1. Given: Function f: ℝ_* → ℝ_* where f(x) = 1/x and ℝ_* = ℝ - {0}.
2. Injective (One-One) Test:

   Let x₁, x₂ ∈ ℝ_* such that f(x₁) = f(x₂).

   ⇒ 1 / x₁ = 1 / x₂

   ⇒ x₁ = x₂.

   Since f(x₁) = f(x₂) implies x₁ = x₂, f is one-one (injective).

3. Surjective (Onto) Test:

   Let y ∈ ℝ_* (codomain). We must show there exists x ∈ ℝ_* such that f(x) = y.

   Set y = 1/x ⇒ x = 1/y.

   Since y ∈ ℝ_*, y ≠ 0, which means x = 1/y is a well-defined non-zero real number (x ∈ ℝ_*).

   Now, evaluate f(x): f(1/y) = 1 / (1/y) = y.

   Since every element in the codomain has a pre-image in the domain, f is onto (surjective).

4. Case when Domain is replaced by N (f: N → R_*):
   • Injectivity: For x₁, x₂ ∈ ℕ, 1/x₁ = 1/x₂ ⇒ x₁ = x₂. So f is still one-one.
   • Surjectivity: Take y = 2 ∈ ℝ_*. For f(x) = 2, we need 1/x = 2 ⇒ x = 1/2 ∉ ℕ. Hence, y = 2 has no pre-image in ℕ. Therefore, f is not onto.

Final Answer: f: ℝ_* → ℝ_* is both one-one and onto. When domain is ℕ, f remains one-one but is no longer onto.

Question 2: Check the injectivity and surjectivity of the function f(x) = x² across different domains and codomains.

Understanding how the domain and codomain alter the properties of a function is a classic CBSE board examination testing point:

Question 10: Let A = R - {3} and B = R - {1}. Consider the function f: A → B defined by f(x) = (x - 2) / (x - 3). Is f one-one and onto? Justify your answer.

Step-by-Step Solution:

1. Given: Domain A = ℝ - {3}, Codomain B = ℝ - {1}, and f(x) = (x - 2)/(x - 3).
2. Injectivity (One-to-One):

   Let x₁, x₂ ∈ A such that f(x₁) = f(x₂).

   ⇒ (x₁ - 2) / (x₁ - 3) = (x₂ - 2) / (x₂ - 3)

   Cross-multiplying both sides:

   (x₁ - 2)(x₂ - 3) = (x₂ - 2)(x₁ - 3)

   ⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 3x₂ - 2x₁ + 6

   Subtracting (x₁x₂ + 6) from both sides:

   -3x₁ - 2x₂ = -3x₂ - 2x₁

   ⇒ -3x₁ + 2x₁ = -3x₂ + 2x₂

   ⇒ -x₁ = -x₂ ⇒ x₁ = x₂.

   Thus, f is injective (one-one).

3. Surjectivity (Onto):

   Let y ∈ B = ℝ - {1} (so y ≠ 1). We need to express x in terms of y:

   y = (x - 2) / (x - 3)

   ⇒ y(x - 3) = x - 2

   ⇒ yx - 3y = x - 2

   ⇒ yx - x = 3y - 2

   ⇒ x(y - 1) = 3y - 2

   ⇒ x = (3y - 2) / (y - 1).

   Since y ≠ 1, the denominator (y - 1) ≠ 0, so x is a valid real number.

   Let us check if x could ever equal 3:

   If (3y - 2) / (y - 1) = 3 ⇒ 3y - 2 = 3(y - 1) = 3y - 3 ⇒ -2 = -3, which is impossible!

   Therefore, x ≠ 3, which confirms x ∈ A.

   Now evaluate f(x):

   f(x) = f((3y - 2)/(y - 1)) = [((3y - 2)/(y - 1)) - 2] / [((3y - 2)/(y - 1)) - 3]

   = [(3y - 2 - 2y + 2)/(y - 1)] / [(3y - 2 - 3y + 3)/(y - 1)] = y / 1 = y.

   Hence, every y ∈ B has a pre-image in A. Therefore, f is surjective (onto).

Final Answer: f is both one-one and onto; hence, f is a bijective function.

Miscellaneous High-Yield Question: Equivalence Relation on N × N

Question: Let N be the set of natural numbers and R be the relation on N × N defined by (a, b) R (c, d) if and only if a + d = b + c. Show that R is an equivalence relation. Also, find the equivalence class [(2, 5)].

Step-by-Step Solution:

1. Reflexivity:

   Let (a, b) ∈ ℕ × ℕ.

   Since addition is commutative on natural numbers: a + b = b + a.

   By definition of R, this implies (a, b) R (a, b).

   Thus, R is reflexive on ℕ × ℕ.

2. Symmetry:

   Let (a, b) R (c, d) for (a, b), (c, d) ∈ ℕ × ℕ.

   ⇒ a + d = b + c

   ⇒ b + c = a + d

   ⇒ c + b = d + a (using commutativity)

   ⇒ (c, d) R (a, b).

   Thus, R is symmetric.

3. Transitivity:

   Let (a, b) R (c, d) and (c, d) R (e, f).

   ⇒ a + d = b + c   --- (Equation 1)

   ⇒ c + f = d + e   --- (Equation 2)

   Adding Equation 1 and Equation 2:

   (a + d) + (c + f) = (b + c) + (d + e)

   ⇒ a + d + c + f = b + c + d + e

   Canceling common terms (c + d) from both sides:

   ⇒ a + f = b + e

   ⇒ (a, b) R (e, f).

   Thus, R is transitive.

4. Equivalence Class [(2, 5)]:

   [(2, 5)] = {(x, y) ∈ ℕ × ℕ : (x, y) R (2, 5)}

   ⇒ x + 5 = y + 2 ⇒ y - x = 3 (where x, y ∈ ℕ).

   Listing ordered pairs:

   • For x = 1: y = 1 + 3 = 4 ⇒ (1, 4)
   • For x = 2: y = 2 + 3 = 5 ⇒ (2, 5)
   • For x = 3: y = 3 + 3 = 6 ⇒ (3, 6)
   • For x = 4: y = 4 + 3 = 7 ⇒ (4, 7), and so on.

   [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), ..., (n, n + 3) : n ∈ ℕ}.

Final Answer: R is an equivalence relation. The equivalence class [(2, 5)] = {(n, n + 3) : n ∈ ℕ}.

Important Formulas and Theorems Summary

Use this quick-reference table for revision before tests and board exams:

Common Mistakes and Tips for Relations and Functions

• Disproving vs. Proving: To prove a property (like reflexivity or transitivity), you must provide a general algebraic proof for arbitrary elements. To disprove a property, providing a single concrete counterexample is sufficient and required.
• Transitivity Pitfall: Remember the logical implication: If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R. If there is no pair where the second element matches the first element of another ordered pair, transitivity is vacuously true.
• Domain & Codomain Awareness: Always check the specified set (ℕ, ℤ, ℝ, or ℝ_*). As shown in Exercise 1.2, f(x) = x² is injective on ℕ but non-injective on ℤ and ℝ.
• Writing the Pre-Image Step: In subjective / board questions, always explicitly prove that your calculated x belongs to the domain (e.g., verifying x ≠ 3 when domain is ℝ - {3}).

CBSE Board Exam Relevance & Marks Distribution

In the CBSE Class 12 Mathematics paper, questions from Relations and Functions are formatted across multiple sections:

• Section A (1-Mark MCQs): Questions on finding the number of equivalence relations, determining if a given relation is symmetric/transitive, or calculating total bijections.
• Section B & C (2-Mark & 3-Mark Short Answers): Checking injectivity and surjectivity of rational algebraic functions, modulus functions, or greatest integer functions.
• Section D / E (Case Study / Long Answer): Proving equivalence relations on Cartesian sets like ℕ × ℕ or ℤ × ℤ, followed by finding specific equivalence classes.

More NCERT Solutions and Practice

Scoring a perfect 100/100 in CBSE Class 12 Mathematics requires consistent practice with step-by-step problem-solving. For chapter-wise test papers, customized mock exams, previous years' solved CBSE board papers, and instant question paper generators, visit qptool.theorify.in.