CBSE Class 12 Mathematics Relations and Functions Notes Set 04

This page gives the important notes and formulas of Chapter 1 Relations and Functions for Class 12 CBSE Mathematics, covering relations, equivalence relations and classes, and types of functions. It also has solved examples with step-wise marks, Do Yourself practice questions and their answers.

Important Notes and Formulas

1. Cartesian Product of Two Sets

Given two non-empty sets A and B. The cartesian product A × B is the set of all ordered pairs of the form (a, b) where the first entry comes from set A & second comes from set B.
A × B = {(a, b) ∀ a ∈ A, b ∈ B}
e.g. A = {1, 2, 3} and B = {p, q}
A × B = {(1, p), (1, q), (2, p), (2, q), (3, p), (3, q)}

Note:
(i) If either A or B is a null set, then A × B will also be empty set, i.e. A × B = ϕ
(ii) If n(A) = p, n(B) = q, then n(A × B) = pq, where n(X) denotes the number of elements in set X.

2. Relations

Let A and B be two sets. Then a relation R from A to B is a subset of A × B.
Thus, R is a relation from A to B ⇔ R ⊆ A × B.

Note:
(i) If (a, b) ∈ R then b is the image of a under R and a is the preimage of b under R.
(ii) If n(A) = m, n(B) = n, then the number of relations defined from set A to B are 2^mn.

3. Domain and Range of a Relation

Let R be a relation from set A to set B. Then the set of all first components of the ordered pairs belonging to R is called the domain of R, while the set of all second components of the ordered pairs in R is called the range of R.
Thus, Domain (R) = {a : (a, b) ∈ R} ⊆ A
and, Range (R) = {b : (a, b) ∈ R} ⊆ B

4. Types of Relations

Reflexive Relation: A relation R on set A is said to be reflexive if (a, a) ∈ R, ∀ a ∈ A.
Thus, R on a set A is not reflexive if there exists at least one element a ∈ A such that (a, a) ∉ R.

Symmetric Relation: A relation R on set A is said to be symmetric relation if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A
i.e. a R b ⇒ b R a for all a, b ∈ A.

Transitive Relation: A relation R on set A is said to be transitive relation if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A
i.e. a R b and b R c ⇒ a R c for all a, b, c ∈ A

Equivalence Relation:
A relation R on a set A is said to be an equivalence relation on A if
(i) it is reflexive i.e. (a, a) ∈ R ∀ a ∈ A
(ii) it is symmetric i.e. (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A
(iii) it is transitive i.e. (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R ∀ a, b, c ∈ A.

5. Equivalence Class

Let X be a nonempty set and R be an equivalence relation on X. Let a be an element of X. The set of all elements of X which are in relation R related to a is denoted by [a] and is called an equivalence class.
Thus [a] = {x : x ∈ X, x R a i.e., (x, a) ∈ R}
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:
(i) all elements of A_i are related to each other, for all i.
(ii) no element of A_i is related to any element of A_j, i ≠ j.
(iii) \( \bigcup_{i=1}^{n} A_i = X \) and A_i ∩ A_j = ϕ, i ≠ j.
The subsets A_i are called equivalence classes.

6. Function

A relation f from set A to set B is called a function if each element of A is related to unique element of B. It is denoted by the symbol:
f : A → B or \( A \xrightarrow{\;f\;} B \)
which reads ‘f’ is a function from A to B ‘or’ f maps A to B.
If an element a ∈ A is associated with an element b ∈ B, then b is called ‘the f image of a’ or ‘image of a under f’ or ‘the value of the function f at a’; it is written as b = f(a). Also a is called the ‘pre-image of b’; it is written as a = f^–1(b).
Thus a function ‘f’ from set A to set B is a subset of A × B in which each element of A as first component of ordered pair belonging to f appears one and only one time.

Representation of Function:
Ordered pair: Every function from A → B satisfies the following conditions:
(i) f ⊆ A × B
(ii) ∀ a ∈ A there exists b ∈ B such that f(a) = b
(iii) (a, b) ∈ f & (a, c) ∈ f ⇒ b = c

7. Domain, Co-domain and Range of a Function

Let f : A → B, then the set A is known as the domain of f & the set B is known as co-domain of f. The set of images of all the elements of A is known as the range of f.
Domain of f = {a | a ∈ A, (a, f(a)) ∈ f}
Range of f = {f(a) | a ∈ A, f(a) ∈ B}

8. Classification of Functions

One-One Function (Injective mapping):
A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different f images in B.
Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ⇔ x₁ = x₂ or x₁ ≠ x₂ ⇔ f(x₁) ≠ f(x₂).
Diagrammatically an injective mapping can be shown as

Many-one function (non injective):
A function f : A → B is said to be a many one function if two or more elements of A have the same f image in B.
Thus f : A → B is many one if there exists x₁, x₂ ∈ A such that f(x₁) = f(x₂) but x₁ ≠ x₂.
Diagrammatically a many one mapping can be shown as

Onto function (Surjective mapping):
A function f : A → B is such that each element in B (co-domain) is the f image of at least one element of A, then we say that f is a function of A ‘onto’ B. Thus f : A → B is surjective if ∀ b ∈ B, ∃ a ∈ A such that f(a) = b.
Diagrammatically surjective mapping can be shown as

Note:
If range is same as co-domain, then f(x) is onto.

Into function:
A function f : A → B is such that there exists at least one element in B (co-domain) which is not the image of any element in A (domain), then f(x) is into, i.e. ∃ b ∈ B such that f(a) ≠ b ∀ a ∈ A.
Diagrammatically into function can be shown as

Note:
One-one onto function is also known as Bijective function.

Solved Examples

Illustration 1. Let A = {1, 2, 3, 4} and B = {x, y, z}. Consider the subset R = {(1, x), (1, y), (2, z), (3, x)} of A × B. Is R, a relation from A to B? If yes, find domain and range of R. Draw arrow diagram of R. [1 Mark]
Answer:
Since every subset of A × B is a relation from A to B, therefore, R is also a relation from A to B.
Domain of R = {1, 2, 3}
Range of R = {x, y, z}

Illustration 2. If A = {1, 2, 3, 4} and B = {a, b, c}, then find the domain and range of the relation R = {(1, a), (2, b), (3, a)} from A to B. [1 Mark]
Answer:
Domain = {1, 2, 3}
Range = {a, b}

Illustration 3. If A = {1, 2, 3}, then determine the Reflexive, symmetric and transitive relations on A. [4 Marks]
(a) R₁ = {(1, 2), (2, 1)}
(b) R₂ = {(1, 1), (2, 2), (3, 3), (1, 2)}
(c) R₃ = {(1, 2), (2, 1), (1, 1), (2, 2)}
(d) R₄ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Answer:
(a) R₁ = {(1, 2), (2, 1)} is a relation on A, which is not reflexive as (1, 1) ∉ R. Also R is not transitive as (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R. However, R is symmetric on A as a R b ⇒ b R a. [1]
(b) R₂ = {(1, 1), (2, 2), (3, 3), (1, 2)} is a relation on A which is reflexive and transitive but not symmetric as (1, 2) ∈ R but (2, 1) ∉ R. [1]
(c) R₃ = {(1, 2), (2, 1), (1, 1), (2, 2)} is a relation on A which is symmetric and transitive but not reflexive as (3, 3) ∉ R. [1]
(d) R₄ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} is reflexive, symmetric and transitive also. [1]

Illustration 4. Let L be the set of all lines in XY-plane and R be the relation in L defined as R = {(L₁, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4. [4 Marks]
Answer:
Since every line ℓ ∈ L is parallel to itself, therefore (ℓ, ℓ) ∈ R ∀ ℓ ∈ L
∴ R is reflexive. [1]
If (L₁, L₂) ∈ R ⇒ L₁ ∥ L₂ ⇒ L₂ ∥ L₁ ⇒ (L₂, L₁) ∈ R.
∴ R is symmetric. [1]
If (L₁, L₂) ∈ R and (L₂, L₃) ∈ R.
⇒ L₁ ∥ L₂ ⇒ L₂ ∥ L₃
⇒ L₁ ∥ L₃ ⇒ (L₁, L₃) ∈ R
∴ R is transitive. [1]
Since R is reflexive, symmetric and transitive on set L.
Hence, R is an equivalence relation.
Req. set = {ℓ : ℓ is a line related to the line y = 2x + 4}
= {ℓ : ℓ is a line parallel to y = 2x + 4}
= {ℓ : ℓ is a line whose equation is y = 2x + k, k being any real}. [1]

Illustration 5. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric. [4 Marks]
Answer:
If (a, a) ∈ R ⇒ a ≤ a, which is true ∀ a ∈ R.
⇒ R is reflexive. [1]
Also (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c ⇒ a ≤ c ⇒ (a, c) ∈ R
⇒ R is transitive. [1½]
However, R is not symmetric as (a, b) ∈ R ⇏ (b, a) ∈ R; for example (3, 4) ∈ R but (4, 3) ∉ R.
(∵ 3 ≤ 4 but 4 > 3) [1½]

Illustration 6. Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d) ⇔ ad(b + c) = bc(a + d). Check whether R is an equivalence relation on N × N. [6 Marks]
Answer:
Reflexive:
Let (a, b) ∈ N × N i.e. a, b ∈ N such that
if (a, b) R (a, b)
⇒ ab(b + a) = ba(a + b) [by commu. of add. and mult. on N]
Which is true ∀ (a, b) ∈ N × N
So R is reflexive on N × N. [1½]
Symmetric:
Let (a, b), (c, d) ∈ N × N be such that
if (a, b) R (c, d)
⇒ ad(b + c) = bc(a + d)
⇒ cb(d + a) = da(c + b)
⇒ (c, d) R (a, b)
Thus, (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N
So, R is symmetric on N × N. [1½]
Transitive:
Let (a, b), (c, d), (e, f) ∈ N × N such that
if (a, b) R (c, d) and (c, d) R (e, f).
(a, b) R (c, d) ⇒ ad(b + c) = bc(a + d) ⇒ \( \frac{b+c}{bc} = \frac{a+d}{ad} \Rightarrow \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \) ....(i)
and, (c, d) R (e, f) ⇒ cf(d + e) = de(c + f) ⇒ \( \frac{d+e}{de} = \frac{c+f}{cf} \Rightarrow \frac{1}{e} + \frac{1}{d} = \frac{1}{f} + \frac{1}{c} \) ....(ii)
Adding (i) and (ii), we get
\( \left(\frac{1}{c} + \frac{1}{b}\right) + \left(\frac{1}{e} + \frac{1}{d}\right) = \left(\frac{1}{d} + \frac{1}{a}\right) + \left(\frac{1}{f} + \frac{1}{c}\right) \)
⇒ \( \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \Rightarrow \frac{b+e}{be} = \frac{a+f}{af} \)
⇒ af(b + e) = be(a + f) ⇒ (a, b) R (e, f)
So, R is transitive on N × N. [2]
Since R is reflexive, symmetric and transitive on set N × N.
Hence R is an equivalence relation on N × N. [1]

Illustration 7. If the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 15} given by R = {(a, b) : a, b ∈ Z, |a – b| is multiple of 5} is an equivalence relation, then find the equivalence class [2].
Answer:
Let x ∈ A which is related to 2 by given relation.
∵ x R 2 ⇒ |x – 2| is a multiple of 5
⇒ |x – 2| = 0, 5, 10, 15, .......
⇒ x = 2, 7, 12
Hence equivalence class [2] = {2, 7, 12}.

Illustration 8. If A = {1, 2, 3, ......, 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for a, b, c, d ∈ A is an equivalence relation, then find the equivalence class [(2, 5)].
Answer:
Given set A = {1, 2, 3, ......., 9}
Let (x, y) ∈ A × A which is related to (2, 5)
∵ (x, y) R (2, 5) ⇒ x + 5 = y + 2 ⇒ y = x + 3
⇒ (1, 4) R (2, 5), (2, 5) R (2, 5), (3, 6) R (2, 5), (4, 7) R (2, 5), (5, 8) R (2, 5), (6, 9) R (2, 5)
Hence equivalence class [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

Illustration 9. Let f : R → R be defined by f(x) = x² + 1 then find the pre-images of 17 and –3. [(Exemplar) 1 Mark]
Answer:
We have f^–1(17) = x
⇒ f(x) = 17
⇒ x² + 1 = 17 ⇒ x = ±4
Hence f^–1(17) = {4, –4}
and f^–1(–3) = x
⇒ f(x) = –3
⇒ x² + 1 = –3 ⇒ x² = –4
⇒ \( x = \sqrt{-4} \notin R \)
Hence f^–1(–3) = ϕ

Illustration 10. The domain of the function f : R → R defined by f(x) = \( \sqrt{x^2 - 3x + 2} \) [(Exemplar) 1 Mark]
Answer:
Here x² – 3x + 2 ≥ 0 ⇒ (x – 1)(x – 2) ≥ 0 ⇒ x ≤ 1 or x ≥ 2
Hence the domain of f = (–∞, 1] ∪ [2, ∞)

Illustration 11. Let f : R → R be the function defined by f(x) = \( \frac{1}{2 - \cos x} \); x ∈ R then find the range of f. [(Exemplar) 1 Mark]
Answer:
We know that range of cos x is –1 ≤ cos x ≤ 1
⇒ 1 ≥ –cos x ≥ –1
⇒ 3 ≥ 2 – cos x ≥ 1
⇒ \( \frac{1}{3} \le \frac{1}{2 - \cos x} \le 1 \)
⇒ \( \frac{1}{3} \le f(x) \le 1 \)
Hence range of f(x) = [1/3, 1].

Illustration 12. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = \( \left(\frac{x-2}{x-3}\right) \). Is f one-one and onto? Justify your answer. [2013 Comp., 4 Marks]
Answer:
Let x₁, x₂ ∈ A be such that
f(x₁) = f(x₂)
⇒ \( \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \)
⇒ (x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 2x₂ – 3x₁ + 6 = x₁x₂ – 2x₁ – 3x₂ + 6
⇒ x₂ = x₁
∴ f is one-one. [2]
Let f(x) = y
⇒ \( \frac{x-2}{x-3} = y \)
⇒ x – 2 = yx – 3y
⇒ x – xy = 2 – 3y
⇒ x(1 – y) = 2 – 3y
⇒ \( x = \frac{2 - 3y}{1 - y} \) ∈ A ∀ y ∈ B
Hence, f is onto. [2]

Do Yourself

Do Yourself 1:
(i) Check whether the relation R₁ in R defined by R₁ = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.
(ii) Let L be the set of all lines in a plane and R be the relation on L defined as R = {(ℓ, m) : ℓ is perpendicular to m}. Check whether R is reflexive, symmetric or transitive.

Answers for Do Yourself

Do Yourself 1:
(i) Not reflexive, not symmetric & not transitive.
(ii) Not reflexive, not transitive but symmetric.