CBSE Class 11 Mathematics Chapter 2 Notes: Relations and Functions

Chapter Overview

Relations and Functions is Chapter 2 of CBSE Class 11 Mathematics and belongs to the Sets and Functions unit (Unit I). It builds directly on Chapter 1 (Sets) by defining ordered pairs, Cartesian products, relations, and functions. This chapter is foundational for calculus, real analysis, and every branch of higher mathematics.

This guide covers every NCERT topic with definitions, comparison tables, worked examples, common mistakes, and exam tips.

Key Definitions

Cartesian Product of Sets

Given A = {a₁, a₂, …} and B = {b₁, b₂, …}:

• A × B = {(a, b) : a ∈ A, b ∈ B}
• n(A × B) = n(A) × n(B)
• A × B ≠ B × A (unless A = B or one set is empty)
• A × ∅ = ∅ = ∅ × A

Example: A = {1, 2}, B = {a, b}. Then A × B = {(1, a), (1, b), (2, a), (2, b)} and n(A × B) = 2 × 2 = 4.

Types of Relations

Comparison Table

Checking Properties — Worked Example

Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.

• Reflexive? Yes — (1,1), (2,2), (3,3) are all in R.
• Symmetric? Yes — (1,2) ∈ R and (2,1) ∈ R. Every pair's reverse is present.
• Transitive? Yes — (1,2) and (2,1) are in R, and (1,1) is in R. All chains close.
• Equivalence? Yes — all three properties hold.

Common Equivalence Relations

• "Is equal to" on any set
• "Is parallel to" on the set of all lines in a plane
• "Is congruent to" on the set of all triangles
• "Has the same remainder when divided by n" on integers (modulo relation)

Functions — Domain and Range

A function f: A → B assigns exactly one element of B to each element of A.

• Domain: The set A (input set).
• Range (Image set): {f(a) : a ∈ A} ⊆ B.
• Co-domain: The set B (output set, possibly larger than range).

Finding Domain and Range

For real-valued functions f(x):

• Denominator cannot be zero: Exclude values making the denominator 0.
• Square root argument must be non-negative: Require the expression under √ to be ≥ 0.
• Logarithm argument must be positive: Require the argument > 0.

Example: f(x) = 1/(x − 3). Domain: x ≠ 3, i.e., R − {3}. Range: All real numbers except 0, i.e., R − {0}.

Types of Functions

Testing One-One (Horizontal Line Test)

Graphically, a function is one-one if no horizontal line intersects its graph at more than one point.

Testing Onto

Check if the range equals the codomain. For f: R → R, f(x) = x² is not onto because negative numbers have no pre-image.

Composition of Functions

If f: A → B and g: B → C, then the composition g ∘ f : A → C is defined by:

(g ∘ f)(x) = g(f(x))

• f is applied first, then g.
• Domain of g ∘ f = Domain of f (but restricted to values where f(x) is in the domain of g).

Example: f(x) = x + 1, g(x) = 2x. Then (g ∘ f)(x) = g(f(x)) = g(x + 1) = 2(x + 1) = 2x + 2.

Note: In general, f ∘ g ≠ g ∘ f. Composition is not commutative.

Invertible Functions

A function f: A → B is invertible if there exists a function g: B → A such that:

• g ∘ f = identity on A (i.e., g(f(a)) = a for all a ∈ A)
• f ∘ g = identity on B (i.e., f(g(b)) = b for all b ∈ B)

Theorem: A function f is invertible if and only if f is bijective.

Finding the Inverse

Example: f(x) = 3x + 2. Find f⁻¹.

1. Let y = 3x + 2 2. Swap x and y: x = 3y + 2 3. Solve for y: y = (x − 2)/3 4. f⁻¹(x) = (x − 2)/3

Verification: f⁻¹(f(x)) = f⁻¹(3x + 2) = (3x + 2 − 2)/3 = x. Correct.

Binary Operations

A binary operation on a set A is a function : A × A → A that combines two elements of A to produce another element of A.

Properties:

• Commutative: a b = b a for all a, b ∈ A
• Associative: (a b) c = a (b c) for all a, b, c ∈ A
• Identity element e: a e = e a = a for all a ∈ A
• Inverse of a: An element b such that a b = b a = e

Important Numericals

1. Checking Relation Properties

Let A = {1, 2, 3} and R = {(1,2), (2,1)}. Is R reflexive? Symmetric? Transitive?

• Reflexive: No — (1,1), (2,2), (3,3) are missing.
• Symmetric: Yes — (1,2) and (2,1) are both present.
• Transitive: No — (1,2) and (2,1) are in R, but (1,1) is not in R.

2. Finding Domain and Range

f(x) = √(x − 4). Find domain and range.

• Domain: x − 4 ≥ 0 → x ≥ 4 → [4, ∞)
• Range: √(x − 4) ≥ 0 → [0, ∞)

3. Composition

f(x) = x², g(x) = x + 3. Find f ∘ g and g ∘ f.

• (f ∘ g)(x) = f(x + 3) = (x + 3)² = x² + 6x + 9
• (g ∘ f)(x) = g(x²) = x² + 3

4. Inverse Function

f(x) = (x − 1)/(x + 2). Find f⁻¹(x).

1. y = (x − 1)/(x + 2) 2. y(x + 2) = x − 1 → xy + 2y = x − 1 → xy − x = −2y − 1 → x(y − 1) = −(2y + 1) 3. x = −(2y + 1)/(y − 1) = (2y + 1)/(1 − y) 4. f⁻¹(x) = (2x + 1)/(1 − x)

5. One-one and Onto Test

f: R → R defined by f(x) = x³. Is it one-one and onto?

• One-one: If f(a) = f(b), then a³ = b³, so a = b. Yes, one-one.
• Onto: For any y ∈ R, x = ∛y satisfies f(x) = y. Range = R = codomain. Yes, onto.
• Bijective.

Types of Functions — Graphical View

Identity Function

f: R → R defined by f(x) = x. Graph is a straight line through origin at 45°. It is one-one and onto (bijective).

Constant Function

f: R → R defined by f(x) = c. Graph is a horizontal line. It is many-one (all inputs give same output) and into (range = {c} ≠ R).

Modulus Function

Signum Function

f(x) = 1 if x > 0, 0 if x = 0, −1 if x < 0. Range = {−1, 0, 1}. It is many-one and into.

Greatest Integer Function (Floor)

f(x) = [x] = greatest integer ≤ x. Example: [3.7] = 3, [−2.3] = −3. It is many-one and into.

Practice Problems

Problem 1

Let A = {1, 2, 3, 4, 6} and R = {(a, b) : b is divisible by a}. Check if R is reflexive, symmetric, transitive.

• Reflexive: Yes — every number divides itself (a divides a for all a ∈ A).
• Symmetric: No — 2 divides 4 but 4 does not divide 2. (2,4) ∈ R but (4,2) ∉ R.
• Transitive: Yes — if a|b and b|c, then a|c for all integers.

Problem 2

Find the domain and range of f(x) = (2x − 3)/(x + 5).

• Domain: x + 5 ≠ 0, so x ≠ −5. Domain = R − {−5}.
• For range, let y = (2x − 3)/(x + 5). Then y(x + 5) = 2x − 3 → yx + 5y = 2x − 3 → x(y − 2) = −5y − 3 → x = (−5y − 3)/(y − 2). For x to exist, y ≠ 2. Range = R − {2}.

Problem 3

If f(x) = x² + 1 and g(x) = x − 2, find f ∘ g and g ∘ f. Also check if f ∘ g = g ∘ f.

• (f ∘ g)(x) = f(g(x)) = f(x − 2) = (x − 2)² + 1 = x² − 4x + 5
• (g ∘ f)(x) = g(f(x)) = g(x² + 1) = x² + 1 − 2 = x² − 1
• f ∘ g ≠ g ∘ f. Composition is not commutative.

NCERT Exercise Mapping

Common Mistakes

1. Confusing co-domain with range: Co-domain is the entire target set; range is the actual set of images. Range ⊆ Co-domain always. 2. Forgetting to check all elements for reflexivity: (a, a) must be in R for every a ∈ A — missing even one element fails the test. 3. Assuming symmetry from one pair: If (1,2) ∈ R, you must check (2,1) ∈ R. One symmetric pair does not make the entire relation symmetric. 4. Saying f(x) = x² is onto for f: R → R: Negative numbers in the codomain have no pre-image, so range = [0, ∞) ≠ R. It is not onto. 5. Mixing up f ∘ g and g ∘ f: f ∘ g means apply g first, then f. Read right to left. The order matters. 6. Forgetting domain restrictions: f(x) = 1/x is not defined at x = 0. Always exclude values that make the denominator zero or the square root argument negative. 7. Claiming invertibility without checking bijectivity: A function must be both one-one and onto to have an inverse. Check both.

Exam Tips

• 1-mark questions: Define reflexive/symmetric/transitive relations, or identify the type of a given function.
• 2-mark questions: Find domain and range of simple functions, or check if a relation is reflexive/symmetric.
• 4-mark questions: Verify equivalence relation (check all three properties with counter-examples if any fails), or find f ∘ g and g ∘ f.
• 6-mark questions: Find the inverse of a function with full working, or solve a word problem involving function composition.
• Always show each property check separately when verifying equivalence relations.
• For domain/range problems, state the restriction clearly (denominator ≠ 0, square root ≥ 0, etc.).
• When finding the inverse, verify by composing f and f⁻¹ to get the identity.

Study Plan for This Chapter

Quick Revision Notes

• Ordered pair (a, b): order matters; (a, b) ≠ (b, a) unless a = b.
• A × B = {(a, b) : a ∈ A, b ∈ B}; n(A × B) = n(A) × n(B).
• Relation R from A to B is a subset of A × B.
• Reflexive: aRa for all a. Symmetric: aRb ⇒ bRa. Transitive: aRb and bRc ⇒ aRc.
• Equivalence relation = reflexive + symmetric + transitive.
• Function: every element of domain has exactly one image.
• One-one: distinct inputs give distinct outputs. Onto: range = codomain.
• Bijective = one-one + onto. Only bijective functions have inverses.
• Composition: (g ∘ f)(x) = g(f(x)). Not commutative in general.
• Inverse: swap x and y, solve for y. Verify with composition.
• Binary operation: * : A × A → A. May or may not be commutative/associative.