Class 12 Mathematics Notes Chapter 1 (Relations and Functions) – Examplar Problems (English) Book

Alright class, let's begin our focused revision of Chapter 1: Relations and Functions from the NCERT Exemplar perspective, keeping government exam requirements in mind. This chapter lays the foundation for understanding mappings and structures, which are crucial in many areas of mathematics.

Chapter 1: Relations and Functions - Detailed Notes for Exam Preparation

1. Relations

• Definition: A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. If B = A, the relation R is on set A (i.e., R ⊆ A × A).

• Representation: An element (a, b) ∈ R means 'a' is related to 'b' under the relation R, often written as aRb.

• Domain & Range:

  • Domain of R = {a ∈ A | (a, b) ∈ R for some b ∈ B}
  • Range of R = {b ∈ B | (a, b) ∈ R for some a ∈ A}
  • Codomain of R (from A to B) is the entire set B.

• Types of Relations (on a set A):

  • Empty Relation: R = ∅ ⊂ A × A. No element is related to any element.
  • Universal Relation: R = A × A. Every element is related to every element.
  • (Key Focus) Reflexive Relation: A relation R on A is reflexive if (a, a) ∈ R for every a ∈ A. (Every element must be related to itself).
    • Exam Tip: To prove not reflexive, find just one element 'a' such that (a, a) ∉ R.
  • (Key Focus) Symmetric Relation: A relation R on A is symmetric if (a, b) ∈ R implies (b, a) ∈ R for all a, b ∈ A. (If a is related to b, then b must be related to a).
    • Exam Tip: To prove not symmetric, find one pair (a, b) ∈ R such that (b, a) ∉ R.
  • (Key Focus) Transitive Relation: A relation R on A is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R for all a, b, c ∈ A. (If a is related to b, and b is related to c, then a must be related to c).
    • Exam Tip: Transitivity check often involves considering different cases. If the "if" condition ((a,b)∈R and (b,c)∈R) is never met for certain elements, transitivity is not violated for those cases. To prove not transitive, find specific a, b, c such that (a, b) ∈ R and (b, c) ∈ R, but (a, c) ∉ R.

• Equivalence Relation: A relation R on A is an equivalence relation if it is Reflexive, Symmetric, AND Transitive.

  • Importance: Equivalence relations partition the set A into disjoint subsets called Equivalence Classes.

• Equivalence Class: For an equivalence relation R on A, the equivalence class of an element a ∈ A, denoted by [a], is the set of all elements in A which are related to 'a'.

  • [a] = {x ∈ A | (x, a) ∈ R}
  • Properties:
    • For any a ∈ A, a ∈ [a] (due to reflexivity).
    • Either [a] = [b] or [a] ∩ [b] = ∅. (Classes are identical or disjoint).
    • The union of all distinct equivalence classes equals the original set A.

2. Functions (Mappings)

• Definition: A relation f from a set A to a set B is called a function if every element of set A has one and only one image in set B.

  • Notation: f: A → B
  • A is the Domain, B is the Codomain.
  • The set of all images f(a) for a ∈ A is the Range of f. Range(f) ⊆ Codomain(f).

• Types of Functions (Crucial for Exams):

  • One-one (Injective) Function: A function f: A → B is one-one if distinct elements in the domain A have distinct images in the codomain B.
    • Mathematically: f(x₁) = f(x₂) ⇒ x₁ = x₂ for all x₁, x₂ ∈ A.
    • Alternatively: x₁ ≠ x₂ ⇒ f(x₁) ≠ f(x₂) for all x₁, x₂ ∈ A.
    • Test: Assume f(x₁) = f(x₂), and algebraically derive x₁ = x₂. If you can find two different x values giving the same f(x), it's not one-one.
  • Onto (Surjective) Function: A function f: A → B is onto if every element in the codomain B is the image of at least one element in the domain A.
    • Mathematically: For every y ∈ B, there exists an x ∈ A such that f(x) = y.
    • Condition: Range(f) = Codomain(f).
    • Test: Take an arbitrary element y from the codomain B. Try to find an x in the domain A (in terms of y) such that f(x) = y. Ensure this x actually belongs to the domain A for all y in B.
  • Bijective Function: A function f: A → B is bijective if it is both One-one (Injective) AND Onto (Surjective).
    • Importance: Bijective functions are invertible.

3. Composition of Functions

• Definition: Let f: A → B and g: B → C be two functions. The composition of f and g, denoted by gof (read as 'g of f'), is a function gof: A → C defined by:
  • gof(x) = g(f(x)), for all x ∈ A.
• Domain of gof: Domain of f (which is A).
• Codomain of gof: Codomain of g (which is C).
• Existence: gof exists only if Range(f) ⊆ Domain(g). (In the standard definition f: A→B, g: B→C, this is implicitly satisfied as Domain(g)=B and Range(f)⊆B).
• Properties:
  • Composition is generally not commutative (i.e., gof ≠ fog).
  • Composition is associative: If f: A → B, g: B → C, h: C → D, then ho(gof) = (hog)of.
  • Composition with Identity Function (I): If f: A → B, then foI<0xE2><0x82><0x90> = f and I<0xE2><0x82><0x8B>of = f, where I<0xE2><0x82><0x90> and I<0xE2><0x82><0x8B> are identity functions on A and B respectively (I<0xE2><0x82><0x90>(x)=x, I<0xE2><0x82><0x8B>(y)=y).

4. Invertible Functions

• Definition: A function f: A → B is invertible if there exists a function g: B → A such that:
  • gof = I<0xE2><0x82><0x90> (Identity function on A)
  • fog = I<0xE2><0x82><0x8B> (Identity function on B)
  • The function g is called the inverse of f, denoted by f⁻¹.
• Condition for Invertibility: A function f: A → B is invertible if and only if f is bijective (one-one and onto).
• Finding the Inverse:
  1. Ensure f is bijective.
  2. Write y = f(x).
  3. Solve the equation y = f(x) for x in terms of y. This gives x = g(y).
  4. Replace y with x to get the standard notation g(x) or f⁻¹(x).
  5. Verify that gof = I<0xE2><0x82><0x90> and fog = I<0xE2><0x82><0x8B>.
• Property: If f: A → B and g: B → C are both invertible (hence bijective), then gof: A → C is also invertible, and (gof)⁻¹ = f⁻¹og⁻¹. (Note the reversal of order).

5. Binary Operations

• Definition: A binary operation * on a non-empty set A is a function *: A × A → A. It takes two elements from A and produces a unique element in A.
  • Notation: a * b represents the result of the operation on (a, b).
• Properties: Let * be a binary operation on A.
  • 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: There exists an element e ∈ A such that a * e = e * a = a for all a ∈ A. 'e' is the identity element. It must be unique if it exists.
  • Inverse Element: For an element a ∈ A, an element b ∈ A is called the inverse of 'a' (often denoted a⁻¹) if a * b = b * a = e, where 'e' is the identity element. An element may or may not have an inverse, but if it does and the operation is associative, the inverse is unique.

Key Exam Focus Areas:

1. Determining if a given relation is Reflexive, Symmetric, Transitive, or an Equivalence Relation.
2. Finding Equivalence Classes.
3. Determining if a given function is One-one, Onto, or Bijective.
4. Calculating Composite Functions (gof, fog).
5. Finding the Inverse of a Bijective function.
6. Checking properties (Commutativity, Associativity, Identity, Inverse) of Binary Operations.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts discussed, similar to what you might encounter:

1. Let R be a relation on the set N of natural numbers defined by R = {(x, y) : x, y ∈ N, 2x + y = 10}. Which of the following is true?
   (A) R is reflexive
   (B) R is symmetric
   (C) R is transitive
   (D) R is not an equivalence relation

2. Let A = {1, 2, 3}. The number of equivalence relations containing (1, 2) is:
   (A) 1
   (B) 2
   (C) 3
   (D) 4

3. Let f: R → R be defined by f(x) = x² + 1. Then f is:
   (A) One-one and Onto
   (B) One-one but not Onto
   (C) Onto but not One-one
   (D) Neither One-one nor Onto

4. The function f: N → N defined by f(n) = n + 1 if n is odd, and f(n) = n - 1 if n is even, is:
   (A) One-one but not Onto
   (B) Onto but not One-one
   (C) One-one and Onto (Bijective)
   (D) Neither One-one nor Onto

5. Let f: R → R be defined by f(x) = 3x - 4. Then f⁻¹(x) is given by:
   (A) (x + 4) / 3
   (B) (x / 3) - 4
   (C) 3x + 4
   (D) (x - 4) / 3

6. Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Then gof is:
   (A) {(1, 3), (3, 1), (4, 3)}
   (B) {(1, 1), (3, 3), (4, 1)}
   (C) {(1, 3), (3, 3), (4, 1)}
   (D) {(1, 1), (3, 1), (4, 3)}

7. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
   (A) R is reflexive and symmetric but not transitive
   (B) R is reflexive and transitive but not symmetric
   (C) R is symmetric and transitive but not reflexive
   (D) R is an equivalence relation

8. Let A = R - {3} and B = R - {1}. Consider the function f: A → B defined by f(x) = (x - 2) / (x - 3). Then f is:
   (A) One-one but not onto
   (B) Onto but not one-one
   (C) Bijective
   (D) Neither one-one nor onto

9. Let * be a binary operation on the set Q of rational numbers defined by a * b = ab / 4. The identity element for * is:
   (A) 1
   (B) 0
   (C) 4
   (D) 1/4

10. If f: R → R is given by f(x) = (3 - x³)¹ᐟ³, then fof(x) is:
    (A) x¹ᐟ³
    (B) x³
    (C) x
    (D) 3 - x³

Answers to MCQs:

1. (D) R = {(1, 8), (2, 6), (3, 4), (4, 2)}. Not reflexive (1,1)∉R. Not symmetric (1,8)∈R but (8,1)∉R. Not transitive (3,4)∈R, (4,2)∈R but (3,2)∉R.
2. (B) A = {1, 2, 3}. Smallest equivalence relation containing (1,2) must have (1,1), (2,2), (3,3) [Reflexive], (2,1) [Symmetric], and potentially more for transitivity.
   • R₁ = {(1,1), (2,2), (3,3), (1,2), (2,1)} - This is an equivalence relation.
   • R₂ = A × A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)} - This is also an equivalence relation containing (1,2).
     There are only 2 such relations.
3. (D) Not one-one because f(1) = 1²+1 = 2 and f(-1) = (-1)²+1 = 2. Not onto because the range is [1, ∞), which is not equal to the codomain R (e.g., 0 has no pre-image).
4. (C) Check one-one: Case 1: n₁, n₂ odd. f(n₁)=f(n₂) ⇒ n₁+1=n₂+1 ⇒ n₁=n₂. Case 2: n₁, n₂ even. f(n₁)=f(n₂) ⇒ n₁-1=n₂-1 ⇒ n₁=n₂. Case 3: n₁ odd, n₂ even. f(n₁)=n₁+1 (even), f(n₂)=n₂-1 (odd). So f(n₁)≠f(n₂). Hence, one-one. Check onto: For any odd y in codomain N, let n=y+1 (even) in domain N. f(n)=f(y+1)=(y+1)-1=y. For any even y in codomain N, let n=y-1 (odd) in domain N. f(n)=f(y-1)=(y-1)+1=y. Hence, onto. So, bijective.
5. (A) Let y = 3x - 4. Then y + 4 = 3x, so x = (y + 4) / 3. Thus f⁻¹(y) = (y + 4) / 3. Replacing y with x, f⁻¹(x) = (x + 4) / 3.
6. (A) gof(1) = g(f(1)) = g(2) = 3. gof(3) = g(f(3)) = g(5) = 1. gof(4) = g(f(4)) = g(1) = 3. So gof = {(1, 3), (3, 1), (4, 3)}.
7. (B) Reflexive: (1,1), (2,2), (3,3), (4,4) are all in R. Symmetric: (1,2)∈R but (2,1)∉R. Not symmetric. Transitive: Check pairs. (1,2)∈R, (2,2)∈R ⇒ (1,2)∈R. (1,3)∈R, (3,3)∈R ⇒ (1,3)∈R. (1,3)∈R, (3,2)∈R ⇒ (1,2)∈R. All required transitive pairs are present. So, Reflexive and Transitive but not Symmetric.
8. (C) Check one-one: f(x₁) = f(x₂) ⇒ (x₁-2)/(x₁-3) = (x₂-2)/(x₂-3) ⇒ (x₁-2)(x₂-3) = (x₂-2)(x₁-3) ⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 3x₂ - 2x₁ + 6 ⇒ -3x₁ - 2x₂ = -3x₂ - 2x₁ ⇒ -x₁ = -x₂ ⇒ x₁ = x₂. So, one-one. Check onto: Let y ∈ B (y ≠ 1). Set 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, x is well-defined. We need to ensure x ∈ A (i.e., x ≠ 3). If x=3, then 3 = (3y-2)/(y-1) ⇒ 3y-3 = 3y-2 ⇒ -3 = -2, which is impossible. So x ≠ 3. Thus, for every y ∈ B, there exists x ∈ A such that f(x)=y. So, onto. Hence, bijective.
9. (C) We need e such that a * e = a for all a ∈ Q. So, ae / 4 = a. Since this must hold for non-zero 'a', we can divide by 'a': e / 4 = 1 ⇒ e = 4. Check: e * a = 4a / 4 = a. So, identity element is 4.
10. (C) fof(x) = f(f(x)) = f((3 - x³)¹ᐟ³) = [3 - {(3 - x³)¹ᐟ³}³]¹ᐟ³ = [3 - (3 - x³)]¹ᐟ³ = [3 - 3 + x³]¹ᐟ³ = (x³)¹ᐟ³ = x.

Make sure you understand the reasoning behind each answer, especially for the relation and function type identification questions. Practice more problems from the Exemplar book itself. Good luck!