Mathematics · Chapter 2

Relations and Functions

20 min read 6 Topics Class 11 (JEE & NEET)

1. Chapter Overview

Have you ever wondered how a food delivery app matches your address to the nearest rider, or how a UPI app instantly links your phone number to your bank account? Every one of these real-world pairings — one input connected to exactly one output — is a function in disguise. Relations and Functions gives you the formal mathematical language to describe such pairings precisely, and it becomes the backbone for almost every advanced topic you will study later: Calculus, Trigonometry, and Coordinate Geometry all lean on the vocabulary built in this chapter.

In JEE Main, Sets, Relations and Functions together typically contribute 2-3 questions (roughly 6-8% weightage), with domain-range calculations and identifying function types being favourite question styles. JEE Advanced likes to combine this topic with Calculus (composite and inverse functions) in multi-concept problems. Class 11 (JEE & NEET) boards test conceptual clarity on the difference between a relation and a function, and on the standard functions covered in this chapter.

Before starting this chapter, you should be comfortable with Set Theory basics (Chapter 1) — especially set-builder notation, subsets, and union/intersection — since Relations and Functions is built directly on top of the Cartesian product of two sets.

2. Ordered Pairs, Cartesian Product and Relations

Concept Explanation: An ordered pair (a, b) is a pair of elements written in a specific order, where (a, b) = (c, d) if and only if a = c and b = d (unlike a set, where {a,b} = {b,a}). The Cartesian product A × B of two non-empty sets A and B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B. A relation R from set A to set B is simply any subset of A × B. If (a, b) ∈ R, we say "a is related to b", written a R b. Key Formula: $$ n(A \times B) = n(A) \times n(B) $$, $$ \text{Total relations from A to B} = 2^{n(A) \times n(B)} $$
SI Units: None (Relations and Functions deal with pure numbers and sets, not physical quantities).
Common Exam Mistake: Assuming A × B is commutative, i.e. A × B = B × A. Correction: A × B ≠ B × A unless A = B, because (a, b) ≠ (b, a) in general.
JEE/NEET Tip: If n(A × B) is given along with n(A), you can instantly find n(B) by division: n(B) = n(A × B) / n(A).
Solved Example:
If n(A) = 3 and n(B) = 4, find n(A × B) and the total number of relations possible from A to B.
Step-by-Step Solution:
$$ n(A \times B) = n(A) \times n(B) = 3 \times 4 = 12 $$. Since a relation is any subset of A × B, total relations $$ = 2^{12} = 4096 $$.

3. Functions, Domain, Co-domain and Range

Concept Explanation: A function f from set A to set B is a special relation in which every element of A has exactly one image in B (no element of A is left unmapped, and none maps to more than one element). We write f: A → B. Here A is the domain, B is the co-domain, and the actual set of output values {f(x) : x ∈ A} is the range, which is always a subset of the co-domain. For a real-valued function of a real variable, the domain is the largest subset of R for which the formula gives a real, defined output. Key Formula: $$ \text{Range} \subseteq \text{Co-domain} $$, for $$ f(x) = \frac{1}{x-a} $$ domain is $$ \mathbb{R} - \{a\} $$, for $$ f(x) = \sqrt{x-a} $$ domain is $$ x \ge a $$
SI Units: None.
Common Exam Mistake: Confusing co-domain with range and assuming every element of the co-domain must be an output. Correction: Range is only the set of actual outputs, which can be a proper subset of the co-domain.
JEE/NEET Tip: For domain questions with multiple restrictions (square roots, denominators, logarithms together), find the restriction from each piece separately, then take the intersection of all of them.
Solved Example:
Find the domain of $$ f(x) = \frac{1}{\sqrt{x-3}} $$.
Step-by-Step Solution:
Since the term is inside a square root in the denominator, we need $$ x - 3 > 0 $$ (strictly greater than zero, because the denominator cannot be zero). So $$ x > 3 $$, meaning the domain is $$ (3, \infty) $$.

4. Standard Real Functions

Concept Explanation: Certain functions appear repeatedly in JEE/NEET problems and are worth memorizing by name and shape: the Identity function f(x) = x; the Constant function f(x) = c; Polynomial functions like f(x) = a0 + a1x + ... + anxn; Rational functions (the ratio of two polynomials); the Modulus function f(x) = |x|; the Signum function, which returns only -1, 0, or +1 depending on the sign of x; the Greatest Integer function f(x) = [x] (also called the floor function); and Exponential/Logarithmic functions f(x) = ax and f(x) = loga(x), which are inverses of each other. Key Formula: Signum: $$ f(x) = \begin{cases} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{cases} $$, Greatest Integer: $$ [x] \le x < [x]+1 $$
SI Units: None.
Common Exam Mistake: Assuming [x] always rounds to the nearest integer. Correction: The greatest integer function always rounds down (floor), so [2.9] = 2 and [-2.1] = -3, not -2.
JEE/NEET Tip: The range of the Signum function is always exactly {-1, 0, 1} regardless of the domain — a fast way to eliminate wrong options in MCQs.
Solved Example:
Find the value of [3.7] + [-3.7].
Step-by-Step Solution:
$$ [3.7] = 3 $$ (greatest integer ≤ 3.7). $$ [-3.7] = -4 $$ (greatest integer ≤ -3.7, rounding down). Sum $$ = 3 + (-4) = -1 $$.

5. Algebra of Real Functions

Concept Explanation: If f and g are two real functions with a common domain, we can combine them algebraically: their sum (f+g)(x) = f(x) + g(x), difference (f-g)(x) = f(x) - g(x), product (fg)(x) = f(x)·g(x), and quotient (f/g)(x) = f(x)/g(x) provided g(x) ≠ 0. The domain of each combined function is the intersection of the domains of f and g (and, for the quotient, further excluding points where g(x) = 0). Key Formula: $$ (f+g)(x) = f(x)+g(x) $$, $$ (f/g)(x) = \frac{f(x)}{g(x)}, \ g(x) \ne 0 $$
SI Units: None.
Common Exam Mistake: Forgetting to exclude points where g(x) = 0 when finding the domain of f/g. Correction: Always state Domain(f/g) = Domain(f) ∩ Domain(g) − {x : g(x) = 0}.
JEE/NEET Tip: For the domain of combined functions, always find each function's individual domain FIRST, then intersect — never combine the algebraic expressions before checking domain restrictions.
Solved Example:
If f(x) = √x (domain x≥0) and g(x) = 1/(x-2), find the domain of (f/g)(x).
Step-by-Step Solution:
Domain of f = [0,∞). Domain of g = R−{2}. Since g(x) is never 0, domain of f/g $$ = [0, \infty) \cap (\mathbb{R} - \{2\}) = [0, \infty) - \{2\} $$.

Quick Revision Cheat Sheet

**Top 10 Formulas:**
1. Cartesian Product size: $$ n(A \times B) = n(A) \times n(B) $$
2. Total relations A to B: $$ 2^{n(A) \times n(B)} $$
3. Range is a subset of Co-domain: $$ \text{Range} \subseteq \text{Co-domain} $$
4. Sum of functions: $$ (f+g)(x) = f(x) + g(x) $$
5. Quotient of functions: $$ (f/g)(x) = \frac{f(x)}{g(x)}, g(x) \ne 0 $$
6. Signum function range: $$ \{-1, 0, 1\} $$
7. Greatest Integer inequality: $$ [x] \le x < [x] + 1 $$
8. Modulus function: $$ |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} $$
9. Domain of $$ 1/(x-a) $$: $$ \mathbb{R} - \{a\} $$
10. Domain of $$ \sqrt{x-a} $$: $$ x \ge a $$

**Memory Tricks:**
*   *Relation vs Function:* "Every function is a relation, but not every relation is a function — a function is a relation with NO repeats in the domain."
*   *Greatest Integer:* "Floor, not round — [x] always rounds DOWN, even for negative numbers."

**Must-Memorize Facts:**
1. (a, b) = (c, d) if and only if a = c and b = d (order matters).
2. A x B is not equal to B x A unless A = B.
3. Domain of a quotient function excludes where the denominator is zero.
4. The Identity function f(x) = x is both one-one and onto on R.
5. Total number of functions from a set with m elements to a set with n elements is n^m.

**Comparison Table: Relation vs Function**
| Feature | Relation | Function |
| :--- | :--- | :--- |
| Definition | Any subset of A x B | A relation where every element of A has exactly ONE image |
| Repeats in domain | Allowed | NOT allowed (each input maps to exactly one output) |
| Every element of A used | Not necessarily | Yes, always |
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