Chapter 2: Relations and Functions

2.1 Cartesian Product of Sets

Let \( A \) and \( B \) be two non-empty sets. The Cartesian product \( A \times B \) is the set of all ordered pairs of elements from \( A \) and \( B \). \[ A \times B = \{ (a, b) : a \in A, b \in B \} \] If either \( A \) or \( B \) is the null set, then \( A \times B \) will also be empty set.

• Number of elements: If \( n(A) = p \) and \( n(B) = q \), then \( n(A \times B) = pq \).
• Three-dimensional product: \( A \times A \times A = \{(a, b, c) : a, b, c \in A\} \).

2.2 Relations

A relation \( R \) from a non-empty set \( A \) to a non-empty set \( B \) is a subset of the Cartesian product \( A \times B \). This subset is derived by describing a relationship between the first element and the second element of the ordered pairs in \( A \times B \).

• Domain: The set of all first elements of the ordered pairs in a relation \( R \) from set \( A \) to \( B \) is called the domain.
• Range: The set of all second elements in a relation \( R \).
• Co-domain: The whole set \( B \) is called the co-domain of the relation \( R \). Note that \( \text{Range} \subseteq \text{Co-domain} \).

Pictorial representation: Relations can be visually represented using arrow diagrams.

2.3 Functions

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 \( B \). If \( (a, b) \in f \), then \( f(a) = b \), where \( b \) is called the image of \( a \) under \( f \), and \( a \) is called the pre-image of \( b \) under \( f \).

Some Functions and their Graphs

• Identity Function: \( f(x) = x \)
• Constant Function: \( f(x) = c \)
• Polynomial Function: \( f(x) = a_n x^n + \dots + a_1 x + a_0 \)
• Rational Function: \( f(x) = \frac{p(x)}{q(x)} \), \( q(x) \neq 0 \)
• Modulus Function: \( f(x) = |x| = \begin{cases} x, & x \ge 0 \ -x, & x < 0 \end{cases} \)
• Signum Function: \( f(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases} \)
• Greatest Integer Function: \( f(x) = [x] \), where \( [x] \) assumes the value of the greatest integer less than or equal to \( x \).

Algebra of Real Functions

For functions \( f : X \to \mathbb{R} \) and \( g : X \to \mathbb{R} \):

• Addition: \( (f + g)(x) = f(x) + g(x) \)
• Subtraction: \( (f - g)(x) = f(x) - g(x) \)
• Multiplication: \( (fg)(x) = f(x) g(x) \)
• Quotient: \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \), provided \( g(x) \neq 0 \)

Competency-Based Questions

Question 1:

An e-commerce company determines that its daily revenue \( R \) (in thousands of rupees) depends on the unit price \( p \) (in rupees) set for a specific electronic item. The relationship is modeled by the function:

\[ R(p) = -2p^2 + 400p - 15000 \]

(a) Find the domain of the price \( p \) for which the company generates a positive revenue.

(b) What is the maximum possible revenue the company can generate per day?

Answer 1:

(a) Domain for positive revenue:

We need to find the values of \( p \) such that \( R(p) > 0 \).

\[ -2p^2 + 400p - 15000 > 0 \]

Divide by -2 (and reverse the inequality):

\[ p^2 - 200p + 7500 < 0 \]

Factor the quadratic equation:

\[ (p - 50)(p - 150) < 0 \]

The roots are \( p = 50 \) and \( p = 150 \). Since this is an upward-opening parabola opening below the x-axis, the solution is the open interval between the roots.

\[ 50 < p < 150 \]

The domain of \( p \) for positive revenue is \( (50, 150) \).

(b) Maximum revenue:

The function \( R(p) = -2p^2 + 400p - 15000 \) is a downward-facing parabola (since \( a = -2 < 0 \)). The maximum value occurs at the vertex.

The p-coordinate of the vertex is:

\[ p = \frac{-b}{2a} = \frac{-400}{2(-2)} = \frac{-400}{-4} = 100 \]

Substitute \( p = 100 \) into the revenue function to find maximum revenue:

\[ R(100) = -2(100)^2 + 400(100) - 15000 \]

\[ R(100) = -20000 + 40000 - 15000 = 5000 \]

Conclusion: The maximum possible revenue is 5000 thousand rupees (or 5,000,000 rupees), achieved at unit price \( p = 100 \).

Question 2:

Determine the domain and range of the real function \( f(x) = \sqrt{9 - x^2} \).

Answer 2:

Domain:

For purely real values of \( f(x) \), the expression inside the square root must be non-negative.

\[ 9 - x^2 \ge 0 \]

\[ x^2 \le 9 \]

Taking the square root on both sides:

\[ |x| \le 3 \]

\[ -3 \le x \le 3 \]

Domain = \( [-3, 3] \).

Range:

Let \( y = \sqrt{9 - x^2} \).

Since a square root yields a principal (non-negative) value, \( y \ge 0 \).

Square both sides:

\[ y^2 = 9 - x^2 \]

\[ x^2 = 9 - y^2 \]

Since \( x^2 \ge 0 \), we must have \( 9 - y^2 \ge 0 \).

\[ y^2 \le 9 \Rightarrow -3 \le y \le 3 \]

Combining \( y \ge 0 \) and \( -3 \le y \le 3 \), we get \( 0 \le y \le 3 \).

Range = \( [0, 3] \).