CBSE Class 11 Mathematics Relations and Functions Notes Set 01

Important Notes and Formulas

Cartesian Product of Sets

Definition: Given two non-empty sets P and Q, the cartesian product P × Q is the set of all ordered pairs of elements from P and Q, i.e.
P × Q = {(p, q) : p ∈ P, q ∈ Q}

Relations

Definition: Let A and B be two non-empty sets. Then any subset R of A × B is a relation from A to B.
If (a, b) ∈ R, then we write it as a R b which is read as 'a is related to b by the relation R'; b is also called the image of a under R.

Domain and range of a relation: If R is a relation from A to B, then the set of first elements in R is called the domain and the set of second elements in R is called the range of R. Symbolically,
Domain of R = {x : (x, y) ∈ R}
Range of R = {y : (x, y) ∈ R}
The set B is called the co-domain of the relation R. Note that range ⊂ co-domain.
Note: Total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. If n(A) = p and n(B) = q, then n(A × B) = pq and the total number of relations is 2^pq.

Functions

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

We write f : A → B (read as: f is a function from set A to set B). Here A is the domain and B is the co-domain of f.

Relations which cannot be categorised as a function

Here not all elements of set A are associated with some element of set B (every element of A must have an image), so the relation is not a function.

Here an element of set A is associated with more than one element of set B (each element of A must have only one image), so the relation is not a function.

Domain, Co-domain and Range of a Function

Domain: When we define y = f(x) with a formula and the domain is not stated explicitly, the domain is assumed to be the largest set of x-values for which the formula gives real y-values.
The domain of y = f(x) is the set of all real x for which f(x) is defined (real).
Rules for finding Domain:
(i) Expression under even root (i.e. square root, fourth root etc.) should be non-negative.
(ii) Denominator ≠ 0.
(iii) If the domains of y = f(x) and y = g(x) are D₁ and D₂ respectively, then the domain of f(x) ± g(x) or f(x) . g(x) is D₁ ∩ D₂, while the domain of \( \frac{f(x)}{g(x)} \) is D₁ ∩ D₂ − {x : g(x) = 0}.

Range: The set of all f-images of elements of A is known as the range of f and is denoted by f(A).
Range = f(A) = {f(x) : x ∈ A};
f(A) ⊆ B {Range ⊆ Co-domain}.
Rule for finding range: First of all find the domain of y = f(x).
(i) If the domain consists of a finite number of points ⇒ range is the set of corresponding f(x) values.
(ii) If the domain is R or R − {some finite points}: put y = f(x), then express x in terms of y. From this find the values of y for which x is defined (i.e., find the values of y for which x exists).

Note: Two functions f and g are said to be equal iff
1. Domain of f = Domain of g
2. Co-domain of f = Co-domain of g
3. f(x) = g(x) ∀ x ∈ Domain.

Some Standard Real Functions and their Graphs

(1) Identity Function: The function f : R → R defined by y = f(x) = x ∀ x ∈ R is called the identity function.

(2) Constant Function: The function f : R → R defined by y = f(x) = c, ∀ x ∈ R, where c is a constant, is called a constant function. The graph below is for f(x) = 3.

(3) Modulus Function: The function f : R → R defined by \( f(x) = \begin{cases} x, & x \ge 0 \\ -x, & x \lt 0 \end{cases} \) is called the modulus function. It is denoted by y = f(x) = |x|.
It is also known as the 'Absolute value function'.

Properties of Modulus Function: The modulus function has the following properties:
1. For any real number x, we have \( \sqrt{x^2} = |x| \)
2. |xy| = |x| |y|
3. |x + y| ≤ |x| + |y| (triangle inequality)
4. |x − y| ≥ ||x| − |y|| (triangle inequality)

(4) Signum Function: The function f : R → R defined by \( f(x) = \begin{cases} 1, & x \gt 0 \\ 0, & x = 0 \\ -1, & x \lt 0 \end{cases} \) is called the signum function. It is usually denoted by y = f(x) = sgn(x).

Note: \( \text{sgn}(x) = \begin{cases} \frac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} \)

(5) Greatest Integer Function: The function f : R → R defined as the greatest integer less than or equal to x is called the greatest integer function. It is usually denoted as y = f(x) = [x].

Algebra of Real Functions

Here we learn how to add two real functions, subtract a real function from another, multiply a real function by a scalar (here by a scalar we mean a real number), multiply two real functions and divide one real function by another.

(1) Addition of two real functions: Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define (f + g) : X → R by
(f + g)(x) = f(x) + g(x), for all x ∈ X.

(2) Subtraction of a real function from another: Let f : X → R and g : X → R be any two real functions, where X ⊂ R. Then, we define (f − g) : X → R by
(f − g)(x) = f(x) − g(x), for all x ∈ X.

(3) Multiplication by a scalar: Let f : X → R be a real valued function and α be a scalar. Here by scalar, we mean a real number. Then the product αf is a function from X to R defined by (αf)(x) = αf(x), x ∈ X.

(4) Multiplication of two real functions: The product (or multiplication) of two real functions f : X → R and g : X → R is a function fg : X → R defined by
(fg)(x) = f(x) g(x), for all x ∈ X.
This is also called pointwise multiplication.

(5) Quotient of two real functions: Let f and g be two real functions defined from X → R where X ⊂ R. The quotient of f by g, denoted by \( \frac{f}{g} \), is a function defined by
\( \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} \), provided g(x) ≠ 0, x ∈ X.

Solved Examples

Example 1. If \( \left( \frac{x}{3} + 1,\ y - 1 \right) = (2, 1) \), find the values of x and y.
Answer:
By equality of ordered pairs, \( \frac{x}{3} + 1 = 2 \) and y − 1 = 1
⇒ x = 3 and y = 2.

Example 2. If A = {1, 2}, find A × A × A.
Answer:
A × A × A = {(x, y, z) : x ∈ A, y ∈ A, z ∈ A}
So, A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (2, 2, 2), (2, 2, 1), (2, 1, 2), (1, 2, 2)}.

Example 3. The following figure shows a relation between sets P and Q. Write this relation in (i) set builder form, (ii) roster form.

Answer:
It is clear that the relation R is "y is the square of x".
(i) In set builder form, R = {(x, y) : y = x², x ∈ P, y ∈ Q}
(ii) In roster form, R = {(1, 1), (−1, 1), (2, 4), (−2, 4), (−4, 16)}.

Example 4. Let R be the relation on Z defined by R = {(a, b) : a, b ∈ Z, a − b is an integer}. Find the domain and range of R.
Answer:
For any two integers a and b, a − b is an integer.
Hence every pair of integers belongs to R, so the domain and range of R are both Z (all integers).

Example 5. Determine the domain and range of: \( R = \left\{ \left( x - 4,\ \frac{2 + x}{2 - x} \right) : 4 \le x \le 6,\ x \in N \right\} \)
Answer:
Here x = 4, 5, 6.
x = 4: \( \left( 0, \frac{6}{-2} \right) = (0, -3) \); x = 5: \( \left( 1, \frac{7}{-3} \right) = \left( 1, -\frac{7}{3} \right) \); x = 6: \( \left( 2, \frac{8}{-4} \right) = (2, -2) \)
So, \( R = \left\{ (0, -3),\ \left( 1, -\frac{7}{3} \right),\ (2, -2) \right\} \)
Domain = {0, 1, 2}
Range = \( \left\{ -3, -\frac{7}{3}, -2 \right\} \)

Example 6. Let A = {1, 2}. List all the relations on A.
Answer:
Given A = {1, 2}
A × A = {(1, 1), (1, 2), (2, 1), (2, 2)}
Since a relation R from set A to set A is a subset of A × A,
∴ All the relations on A are:
φ, {(1, 1)}, {(1, 2)}, {(2, 1)}, {(2, 2)}, {(1, 1), (1, 2)}, {(1, 1), (2, 1)}, {(1, 1), (2, 2)}, {(1, 2), (2, 1)}, {(1, 2), (2, 2)}, {(2, 1), (2, 2)}, {(1, 1), (1, 2), (2, 1)}, {(1, 1), (1, 2), (2, 2)}, {(1, 1), (2, 1), (2, 2)}, {(1, 2), (2, 1), (2, 2)}, {(1, 1), (1, 2), (2, 1), (2, 2)}.
Since n(A × A) = 4, the number of all relations on the set A = 2⁴, i.e., 16.

Example 7. Find the domain and range of the following functions
(i) \( \left\{ \left( x, \frac{x^2 - 1}{x - 1} \right) : x \in R, x \ne 1 \right\} \)
(ii) \( \left\{ \left( x, \frac{1}{1 - x^2} \right) : x \in R, x \ne \pm 1 \right\} \)
Answer:
(i) Let \( f(x) = \frac{x^2 - 1}{x - 1} \). Clearly, f is not defined when x = 1
∴ f is defined for all real values of x except x = 1
∴ Domain = R − {1}
Let \( y = \frac{x^2 - 1}{x - 1} = x + 1 \) (as x ≠ 1)
∴ x = y − 1. Clearly, x is not defined when y = 2 as x ≠ 1
∴ Range = R − {2}.
(ii) Let \( f(x) = \frac{1}{1 - x^2} \). Clearly, f(x) is not defined when 1 − x² = 0, i.e., when x = ±1
∴ Domain = R − {1, −1}
Further, \( y = \frac{1}{1 - x^2} \Rightarrow 1 - x^2 = \frac{1}{y} \Rightarrow x = \pm\sqrt{1 - \frac{1}{y}} = \pm\sqrt{\frac{y - 1}{y}} \)
∴ x is defined when \( \frac{y - 1}{y} \ge 0 \), i.e., y ∈ (−∞, 0) ∪ [1, ∞).
Range = (−∞, 0) ∪ [1, ∞).

Example 8. Let \( f = \left\{ \left( x, \frac{x^2}{1 + x^2} \right) : x \in R \right\} \) be a function from R to R. Determine the range of f.
Answer:
Clearly f : R → R is a function such that \( f(x) = \frac{x^2}{1 + x^2} \)
Let \( y = \frac{x^2}{1 + x^2} \). So x² = y(1 + x²). Therefore, x²(1 − y) = y, which implies \( x = \pm\sqrt{\frac{y}{1 - y}} \).
Since x ∈ R, \( \frac{y}{1 - y} \ge 0 \), i.e. y ∈ [0, 1).
Thus range is [0, 1).

Example 9. Let f, g : R → R be defined respectively by f(x) = x + 1, g(x) = 2x − 3. Find f + g, f − g and f/g.
Answer:
Let f(x) = x + 1, g(x) = 2x − 3
∴ (f + g)(x) = f(x) + g(x) = (x + 1) + (2x − 3) = 3x − 2
(f − g)(x) = f(x) − g(x) = (x + 1) − (2x − 3) = x + 1 − 2x + 3 = −x + 4
\( \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3},\ x \ne \frac{3}{2} \)

Example 10. A function f is defined on the set {1, 2, 3, 4, 5} as follows: \( f(x) = \begin{cases} 1 + x, & \text{if } 1 \le x \lt 2 \\ 2x - 1, & \text{if } 2 \le x \lt 4 \\ 3x - 10, & \text{if } 4 \le x \lt 6 \end{cases} \)
(i) Find the domain of the function.
(ii) Find the range of the function.
(iii) Find the values of f(2), f(3), f(4), f(5).
Answer:
(i) Domain: {1, 2, 3, 4, 5}
(ii) Range:
f(1) = 1 + 1 = 2,   f(2) = 2(2) − 1 = 3,   f(3) = 2(3) − 1 = 5,   f(4) = 3(4) − 10 = 2,   f(5) = 3(5) − 10 = 5
So, range is {2, 3, 5}.
(iii) f(2) = 3, f(3) = 5, f(4) = 2, f(5) = 5.