Module 3: Relations and Functions

Dear Class 12 Student! "Relations and Functions" is Chapter 1 of your Class 12 syllabus. You cannot understand advanced concepts like Bijective functions or Invertible matrices without a crystal clear understanding of Class 11 basics. This module is the ultimate survival toolkit for Calculus. Master the graphs, domains, and mapping rules here!

1. Cartesian Product of Sets

Before relations, we must define how sets "interact."

Definition Cartesian Product ($A \times B$): Given two non-empty sets A and B, the Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$. $$A \times B = \{ (a, b) : a \in A, b \in B \}$$ Key Fact: If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = p \times q$.

Practice Problem 1 Question: If $A = \{1, 2\}$ and $B = \{3, 4, 5\}$, list $A \times B$ and calculate the total number of relations from A to B.

Solution:
1. $A \times B = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)\}$.
2. $n(A \times B) = 2 \times 3 = 6$.
3. Total Relations = $2^{n(A \times B)} = 2^6 = \mathbf{64}$.

2. Definition of a Relation and a Function

Relation ($R$): Any subset of $A \times B$. It establishes a "connection" between elements.
Function ($f$): A special type of relation where every element in the Domain (Set A) has exactly one image in the Codomain (Set B).

[AI Image Placeholder: Landscape (16:9) - Manual Insertion Required]
AI Prompt to generate image: "A comparative infographic showing two mapping diagrams. Left diagram (Relation but NOT a Function): Set A has an element '1' pointing to two different elements in Set B, and an element '3' with no arrows. Right diagram (A Valid Function): Every element in Set A has exactly one arrow originating from it, pointing to a single element in Set B. Use clean white/grey circles for sets. Label: 'Relation vs. Function' in bold math style."

Vertical Line Test To check if a graph is a function: Draw a vertical line. If it hits the graph in more than one place, it is NOT a function (one $x$ having multiple $y$ values).

3. Important Functions and their Graphs

You must recognize these graphs instantly to master Class 12 Application of Integrals.

A. Identity & Constant Functions

Identity ($f(x) = x$): A 45° straight line through the origin. Domain: $\mathbb{R}$, Range: $\mathbb{R}$.
Constant ($f(x) = c$): A horizontal line. Domain: $\mathbb{R}$, Range: $\{c\}$.

Function Check A Question: If $f(x)$ is an identity function, find the value of $f(-2) + f(5)$.

Solution: Since $f(x) = x$, then $f(-2) = -2$ and $f(5) = 5$.
Sum = $-2 + 5 = \mathbf{3}$.

B. Polynomial Functions ($x^2, x^3$)

Parabola ($f(x) = x^2$): Upward opening, symmetric about y-axis. Range: $[0, \infty)$.
Cubic ($f(x) = x^3$): Symmetric about origin (Odd function). Range: $\mathbb{R}$.

Function Check B Question: Find the intersection points of the parabola $y = x^2$ and the identity line $y = x$.

Solution: Set them equal: $x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0$.
The $x$-values are 0 and 1. Plugging back in, the points are $\mathbf{(0,0)}$ and $\mathbf{(1,1)}$.

C. Rational Function ($f(x) = 1/x$)

A rectangular hyperbola. The function is never defined at $x=0$.

Function Check C Question: Why is the graph of $f(x) = \frac{1}{x-3}$ disconnected? What is its domain?

Solution: The function creates a rational hyperbola that has an asymptote when the denominator is zero.
$x - 3 \neq 0 \implies x \neq 3$. The domain is $\mathbf{\mathbb{R} - \{3\}}$, causing the break.

D. Modulus Function ($f(x) = |x|$)

The "V-shape" function. Piecewise definition: $x$ if $x \ge 0$ and $-x$ if $x < 0$.

Function Check D Question: Solve the equation for $x$: $|x - 2| = 5$.

Solution: A modulus equation splits into a positive case and a negative case:
Case 1 (Positive): $x - 2 = 5 \implies x = 7$
Case 2 (Negative): $x - 2 = -5 \implies x = -3$
Answer: $\mathbf{x \in \{-3, 7\}}$.

E. Signum Function

Outputs -1 for negatives, 0 for zero, and 1 for positives.

Function Check E Question: What is the value of $sgn(\pi - 4)$?

Solution: Since $\pi \approx 3.14$, the value of $(\pi - 4)$ is intrinsically negative.
The Signum function outputs $-1$ for any negative input. Therefore, $sgn(\pi - 4) = \mathbf{-1}$.

F. Greatest Integer Function ($f(x) = [x]$)

The "Step function". Always rounds DOWN to the nearest integer.

Function Check F Question: Evaluate exactly: $[2.9] + [-2.1]$.

Solution: The Greatest Integer Function always rounds DOWN mapping to the left on the number line.
$[2.9]$ rounds down to $2$.
$[-2.1]$ rounds DOWN to $-3$ (It does not round to $-2$!).
Sum = $2 + (-3) = \mathbf{-1}$.

4. Finding Domain and Range

Mastering these three rules is the only way to survive Class 12 Calculus.

The 3 Golden Rules

Denominator $\neq 0$: If $f(x) = 1/g(x)$, set $g(x) \neq 0$.
Square Root $\ge 0$: If $f(x) = \sqrt{g(x)}$, set $g(x) \ge 0$.
Logarithm $> 0$: If $f(x) = \log(g(x))$, set $g(x) > 0$.

MEGA PRACTICE BANK

Problem 1: Rational Domain Question: Find the domain of $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

Solution: Denominator $\neq 0 \implies x^2 - 8x + 12 \neq 0$.
Factorizing: $(x-6)(x-2) \neq 0 \implies x \neq 2, 6$.
Domain: $\mathbb{R} - \{2, 6\}$.

Problem 2: Square Root Logic Question: Find the domain of $f(x) = \sqrt{x^2 - 9}$.

Solution: $x^2 - 9 \ge 0 \implies (x-3)(x+3) \ge 0$.
Using Wavy Curve: $\mathbf{x \in (-\infty, -3] \cup [3, \infty)}$.

Problem 3: Wavy Curve Challenge Question: Solve for $x$: $\frac{x-1}{(x-2)(x+3)} < 0$.

[AI Image Placeholder: Landscape (16:9)]
Prompt: "Wavy curve diagram on a number line for critical points -3, 1, and 2. The curve starts from top-right (+), crosses down at 2 (-), crosses up at 1 (+), and crosses down at -3 (-). Color negative regions red. Label intervals clearly."

Solution: The $-$ regions are $(-\infty, -3) \cup (1, 2)$.
Answer: $x \in (-\infty, -3) \cup (1, 2)$.

Problem 4: Range Calculation Question: Find the range of $f(x) = \frac{1}{1 + x^2}$.

Solution: Since $x^2 \ge 0$, then $1+x^2 \ge 1$.
Taking reciprocal (flips inequality): $0 < \frac{1}{1+x^2} \le 1$.
Range: $(0, 1]$.

Problem 5: Signum Equation Question: Solve for $x$: $sgn(x^2 - 4) = 1$.

Solution: $sgn(k)=1 \implies k > 0$.
So, $x^2 - 4 > 0 \implies (x-2)(x+2) > 0$.
Answer: $x \in (-\infty, -2) \cup (2, \infty)$.

Problem 6: Combined Domain Question: Find domain of $f(x) = \log_{10}(x - 4) + \sqrt{10 - x}$.

Solution:
1. From log: $x-4 > 0 \implies x > 4$.
2. From root: $10-x \ge 0 \implies x \le 10$.
Intersection of $x > 4$ and $x \le 10$: $\mathbf{x \in (4, 10]}$.

Problem 7: Relation vs Function Status Question: A relation $R$ from set $A = \{1, 2, 3\}$ to set $B = \{4, 5\}$ is defined as $R = \{(1, 4), (2, 5), (1, 5)\}$. Is $R$ a function? Why or why not?

Solution:
For a relation to be a function, every element in the domain must map to exactly one unique element in the codomain.
In the given relation, the element '1' from domain Set A maps to both '4' and '5' in Set B (since we have $(1, 4)$ and $(1, 5)$).
Therefore, $R$ is NOT a function.

Problem 8: Quadratic Range Target Question: Find the range of the quadratic function $f(x) = x^2 - 6x + 13$.

Solution:
To find the range of a quadratic $ax^2 + bx + c$, it's fastest to complete the square.
$f(x) = (x^2 - 6x + 9) + 4$
$f(x) = (x - 3)^2 + 4$

Since the square of any real number is always non-negative, $(x - 3)^2 \ge 0$.
Add 4 to both sides: $(x - 3)^2 + 4 \ge 4$.
Therefore, $f(x) \ge 4$.
Range: $[4, \infty)$.

Problem 9: Piecewise Evaluation Question: Let $f(x) = \begin{cases} 2x + 3, & \text{if } x < -1 \\ x^2, & \text{if } -1 \le x < 2 \\ 5, & \text{if } x \ge 2 \end{cases}$. Find the value of $f(-2) + f(1) + f(5)$.

[AI Image Placeholder: Landscape (16:9)]
Prompt: "A clean math graph showing a 3-part piecewise function. Left part ($x < -1$): A straight line with positive slope ending in a hollow circle at $(-1, 1)$. Middle part ($-1 \le x < 2$): A parabolic curve starting with a solid dot at $(-1, 1)$ curving up to a hollow circle at $(2, 4)$. Right part ($x \ge 2$): A horizontal line starting with a solid dot at $(2, 5)$ and continuing right. Highlight the three distinct colors for each piece."

Solution:
1. For $x = -2$, the condition is $x < -1$. Use the first formula.
$f(-2) = 2(-2) + 3 = -1$.

2. For $x = 1$, the condition is $-1 \le x < 2$. Use the second formula.
$f(1) = 1^2 = 1$.

3. For $x = 5$, the condition is $x \ge 2$. Use the third formula.
$f(5) = 5$.

Summing them up: $-1 + 1 + 5 = \mathbf{5}$.