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WBB • Class XI • Mathematics • Ch 2

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Relations and Functions

Relations and functions constitute the conceptual core of higher mathematics, bridging set-theoretic foundations to calculus, modern algebra, and applied sciences. Starting with ordered pairs and the Cartesian product A × B, a relation R is defined as any subset of A × B, modeling binary connections between elements of sets. The domain encapsulates all first coordinates, while the range comprises all second coordinates actually mapped within the target codomain. A function or mapping f: A → B is a specialized relation satisfying two rigorous criteria: every element in the domain must possess an image, and each element can have exactly one unique image in the codomain. In the Cartesian plane, the Vertical Line Test provides an immediate geometric criterion to identify functions. Real-valued functions form the primary analytical tool of calculus, characterized by specialized mathematical behaviors: the V-shaped modulus function |x|, the staircase-like greatest integer floor function [x], the three-tier signum function sgn(x), polynomials, and rational fractions. The algebra of real functions establishes precise rules for pointwise addition, subtraction, multiplication, and quotient operations under domain intersections. Determining the natural domain and range of algebraic and radical functions requires solving system inequalities and evaluating denominator and square root non-negativity constraints.

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Why This Chapter Matters

A profound mastery of relations and functions is indispensable for success in WBCHSE Board examinations, WBJEE, and JEE Main. In differential and integral calculus, concepts such as limits, continuity, differentiability, maxima-minima, and Riemann integration are meaningless without rigorous definitions of function domains, ranges, and piecewise behaviors. In computer science and software engineering, functions are the fundamental building blocks of functional programming paradigms, lambda calculus, database relational queries, and cryptographic hash algorithms. In artificial intelligence and machine learning, deep neural networks are essentially compositions of multi-dimensional nonlinear activation functions such as the Rectified Linear Unit (ReLU), which is an exact algebraic adaptation of the modulus function f(x) = max(0, x) = (x + |x|)/2. In the WBCHSE Class 11 annual mathematics examination, questions requiring students to determine the domain and range of radical expressions, evaluate greatest integer equations, and verify function criteria represent guaranteed, high-weightage problems carrying 4 to 6 marks.

Chapter Roadmap & Progression

1 1. Ordered Pairs & Cartesian Produc...

2 2. Concept, Representation & Taxono...

3 3. Definition, Criteria & Cardinali...

4 4. Standard Real-Valued Functions &...

5 5. Algebra of Real Functions: Opera...

6 6. Advanced Techniques for Determin...

Complete Concept Guide (100% Curriculum Coverage)

1. Ordered Pairs & Cartesian Product of Sets

The transition from set theory to relations and functions begins with the concept of the ordered pair and the Cartesian product, named in honor of the French philosopher and mathematician René Descartes (1596–1650).

1.1 Ordered Pairs & Equality Criterion

An ordered pair consists of two mathematical objects listed in a specific, immutable order, enclosed within parentheses: $(a, b)$. Here, $a$ is termed the first element (or abscissa) and $b$ is the second element (or ordinate).

Equality of Ordered Pairs: Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if their corresponding first elements are equal and their corresponding second elements are equal: $$(a, b) = (c, d) \iff a = c \quad \text{and} \quad b = d$$

Crucial Contrast with Sets: In set theory, order is irrelevant: $\{1, 2\} = \{2, 1\}$. However, in ordered pairs, order is paramount: $(1, 2) \neq (2, 1)$ unless the elements are identical.

1.2 Cartesian Product of Two Sets ($A \times B$)

Given two non-empty sets $A$ and $B$, the Cartesian Product $A \times B$ (read as "$A$ cross $B$") is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.

Set-Builder Definition: $$A \times B = \{(a, b) : a \in A \land b \in B\}$$

If either $A = \emptyset$ or $B = \emptyset$, then by definition $A \times B = \emptyset$.
Non-Commutative Property: In general, $A \times B \neq B \times A$. They are equal if and only if $A = B$.

Theorem on Cartesian Product Cardinality

If $A$ and $B$ are finite sets with $n(A) = p$ and $n(B) = q$, then:

$$n(A \times B) = n(A) \cdot n(B) = p \cdot q$$

Proof Outline: For each of the $p$ choices of the first element $a \in A$, there are exactly $q$ independent choices for the second element $b \in B$. By the fundamental counting principle of multiplication, the total number of distinct ordered pairs is $p \times q$.

1.3 Distributive Properties of Cartesian Product

The Cartesian product distributes over fundamental set operations:

Distributive over Union: $A \times (B \cup C) = (A \times B) \cup (A \times C)$
Distributive over Intersection: $A \times (B \cap C) = (A \times B) \cap (A \times C)$
Distributive over Difference: $A \times (B - C) = (A \times B) - (A \times C)$
Shared Elements Theorem (High-Yield Exam Identity): If sets $A$ and $B$ have $k$ elements in common (i.e., $n(A \cap B) = k$), then $A \times B$ and $B \times A$ have exactly $k^2$ elements in common: $$(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A) \implies n((A \times B) \cap (B \times A)) = k^2$$

1.4 Cartesian Product in Higher Dimensions

$\mathbb{R} \times \mathbb{R} = \mathbb{R}^2 = \{(x, y) : x, y \in \mathbb{R}\}$ represents the entire continuous two-dimensional Cartesian coordinate plane.
$\mathbb{R} \times \mathbb{R} \times \mathbb{R} = \mathbb{R}^3 = \{(x, y, z) : x, y, z \in \mathbb{R}\}$ represents three-dimensional Euclidean space.

2. Concept, Representation & Taxonomy of Relations

In ordinary language, a relation describes a connection between two objects (e.g., "is father of", "is greater than"). In mathematics, relations are formalized as subsets of Cartesian products.

2.1 Definition of a Relation

A relation $R$ from a non-empty set $A$ to a non-empty set $B$ is any subset of the Cartesian product $A \times B$.

Formal Definition: $R \subseteq A \times B$
If $(a, b) \in R$, we say "$a$ is related to $b$ by relation $R$", written symbolically as $a R b$.
If $(a, b) \notin R$, we write $a \cancel{R} b$.

A relation from a set $A$ to itself ($R \subseteq A \times A$) is called a relation on set $A$.

2.2 Total Number of Possible Relations

Let $A$ and $B$ be finite sets with $n(A) = p$ and $n(B) = q$.

The Cartesian product $A \times B$ contains $p \cdot q$ elements.
Since any subset of $A \times B$ is a relation, the total number of relations from $A$ to $B$ equals the number of subsets of $A \times B$: $$\mathbf{\text{Total Relations} = 2^{n(A \times B)} = 2^{p \cdot q}}$$
The number of non-empty relations is $2^{pq} - 1$.

2.3 Domain, Range & Codomain of a Relation

Term	Mathematical Definition	Key Properties
Domain (ক্ষেত্র / সংজ্ঞার অঞ্চল)	$\text{Dom}(R) = \{a \in A : (a, b) \in R \text{ for some } b \in B\}$	Set of all first coordinates of the ordered pairs in $R$. $\text{Dom}(R) \subseteq A$.
Range (পাল্লা বা প্রসার)	$\text{Range}(R) = \{b \in B : (a, b) \in R \text{ for some } a \in A\}$	Set of all second coordinates of the ordered pairs in $R$. $\text{Range}(R) \subseteq B$.
Codomain (সহ-সংজ্ঞার অঞ্চল)	$\text{Codomain}(R) = B$	The entire target set $B$. Notice that $\text{Range}(R) \subseteq \text{Codomain}(R)$.

2.4 Inverse Relation ($R^{-1}$)

If $R$ is a relation from $A$ to $B$, the inverse relation $R^{-1}$ is a relation from $B$ to $A$ defined by reversing the order of coordinates:

$$R^{-1} = \{(b, a) : (a, b) \in R\} \subseteq B \times A$$ $$\text{Dom}(R^{-1}) = \text{Range}(R) \quad \text{and} \quad \text{Range}(R^{-1}) = \text{Dom}(R)$$

2.5 Special Types of Relations

Empty Relation (Void Relation): The empty set $\emptyset \subseteq A \times A$ is the empty relation on $A$. No element is related to any element.
Universal Relation: The complete Cartesian product $A \times A \subseteq A \times A$ is the universal relation on $A$. Every element is related to every element.
Identity Relation ($I_A$): $I_A = \{(a, a) : a \in A\}$. Every element is related to itself and only to itself.

3. Definition, Criteria & Cardinality of Functions (Mappings)

A function is a special type of relation that provides a deterministic rule matching each input to a single, predictable output.

3.1 Axiomatic Definition of a Function

Let $A$ and $B$ be two non-empty sets. A relation $f$ from $A$ to $B$ is called a function (or mapping), denoted by $f: A \to B$, if it satisfies two essential axioms:

Axiom 1 (Totality / Completeness): Every element $x \in A$ must have an image in $B$. That is, the domain of $f$ must equal the entire set $A$: $\text{Dom}(f) = A$.
Axiom 2 (Uniqueness / Well-Definedness): Each element $x \in A$ must be associated with exactly one unique image in $B$. That is: $$(x, y_1) \in f \quad \text{and} \quad (x, y_2) \in f \implies y_1 = y_2$$

If $(x, y) \in f$, we write $y = f(x)$, where $y$ is called the image of $x$ under $f$, and $x$ is called the pre-image of $y$.

Valid Mappings vs. Invalid Relations

One-to-One (Valid): Distinct inputs map to distinct outputs. (e.g., $1 \to a, 2 \to b, 3 \to c$).
Many-to-One (Valid): Multiple distinct inputs map to the same output. (e.g., $f(x) = x^2$ where $f(-2) = 4$ and $f(2) = 4$). This is a completely valid function!
One-to-Many (STRICTLY FORBIDDEN FOR FUNCTIONS): A single input mapping to two or more distinct outputs. (e.g., $1 \to a$ and $1 \to b$ with $a \neq b$). This violates Axiom 2!
Unmapped Input (STRICTLY FORBIDDEN): An element in domain $A$ having no corresponding arrow/image in $B$. This violates Axiom 1!

3.2 Total Number of Possible Functions

Let $A$ and $B$ be finite sets with $n(A) = p$ and $n(B) = q$.

Total Functions Formula: $$\text{Number of functions from } A \text{ to } B = [n(B)]^{n(A)} = q^p$$

Reasoning: The first element in $A$ has $q$ possible choices of images in $B$. The second element in $A$ also has $q$ choices, and so forth for all $p$ elements. By the multiplication principle: $q \times q \times \dots \times q \text{ ($p$ times)} = q^p$.

3.3 The Vertical Line Test in the Real Plane ($\mathbb{R}^2$)

The Vertical Line Test provides a visual, geometric test to determine whether a given graph represents a real function $y = f(x)$:

Vertical Line Test Theorem: A curve in the Cartesian coordinate plane represents $y$ as a function of $x$ if and only if no vertical line $x = c$ intersects the curve at more than one point.

Parabola $y = x^2$: Every vertical line cuts the curve at exactly one point $\implies$ Valid function!
Circle $x^2 + y^2 = r^2$: Vertical lines between $-r$ and $r$ intersect the circle at two points $(x, \sqrt{r^2-x^2})$ and $(x, -\sqrt{r^2-x^2}) \implies$ Not a function!
Horizontal Parabola $y^2 = 4ax$: For any $x > 0$, vertical line cuts at $y = \pm 2\sqrt{ax} \implies$ Not a function!

4. Standard Real-Valued Functions & Their Graphs

In calculus and higher mathematics, functions whose domain and codomain are subsets of real numbers $\mathbb{R}$ are called real-valued functions. Master these standard archetypes:

Function Name	Mathematical Definition	Domain & Range	Graph Geometry & Key Traits
1. Identity Function	$f(x) = x$	$\text{Dom} = \mathbb{R}$ $\text{Range} = \mathbb{R}$	Straight line passing through the origin at an angle of $45^\circ$ ($\text{slope } m = 1$).
2. Constant Function	$f(x) = c$ ($c \in \mathbb{R}$ constant)	$\text{Dom} = \mathbb{R}$ $\text{Range} = \{c\}$	Horizontal line parallel to the x-axis at height $y = c$. Slope is 0.
3. Modulus (Absolute Value) Function	$f(x) = \|x\| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}$	$\text{Dom} = \mathbb{R}$ $\text{Range} = [0, \infty)$	V-shaped continuous curve with its sharp vertex at the origin $(0, 0)$. Symmetric about the y-axis (even function: $\|-x\| = \|x\|$).
4. Signum Function	$f(x) = \text{sgn}(x) = \begin{cases} \frac{\|x\|}{x} = 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}$	$\text{Dom} = \mathbb{R}$ $\text{Range} = \{-1, 0, 1\}$	Three disconnected horizontal segments: ray $y = 1$ for $x > 0$, isolated point $(0, 0)$, and ray $y = -1$ for $x < 0$. Odd function.
5. Greatest Integer Function (Step / Floor Function)	$f(x) = [x]$ = greatest integer $\le x$	$\text{Dom} = \mathbb{R}$ $\text{Range} = \mathbb{Z}$ (integers)	Staircase/step graph with steps of unit length $[n, n+1)$. Closed dot at left $(n, n)$ and open hollow circle at right $(n+1, n)$. Discontinuous at all integers!
6. Fractional Part Function	$f(x) = \{x\} = x - [x]$	$\text{Dom} = \mathbb{R}$ $\text{Range} = [0, 1)$	Sawtooth wave graph repeating identically on every integer interval $[n, n+1)$. Periodic function with fundamental period $T = 1$.
7. Reciprocal (Rational) Function	$f(x) = \frac{1}{x}$	$\text{Dom} = \mathbb{R} \setminus \{0\}$ $\text{Range} = \mathbb{R} \setminus \{0\}$	Rectangular hyperbola situated in quadrants I and III. Asymptotes are the coordinate axes $x = 0$ and $y = 0$.
8. Square Root Function	$f(x) = \sqrt{x}$	$\text{Dom} = [0, \infty)$ $\text{Range} = [0, \infty)$	Upper half of a horizontal parabola starting at $(0, 0)$ and bending smoothly into quadrant I.

5. Algebra of Real Functions: Operations & Domain Rules

Just as real numbers can be combined via arithmetic operations, functions can be combined to form new composite functions under strict domain constraints.

5.1 Pointwise Function Operations

Let $f: D_1 \to \mathbb{R}$ and $g: D_2 \to \mathbb{R}$ be two real-valued functions with domains $D_1$ and $D_2$ respectively.

Operation	Formula	Resulting Domain
Addition	$(f + g)(x) = f(x) + g(x)$	$\text{Dom}(f + g) = D_1 \cap D_2$
Subtraction	$(f - g)(x) = f(x) - g(x)$	$\text{Dom}(f - g) = D_1 \cap D_2$
Scalar Multiplication	$(c \cdot f)(x) = c \cdot f(x)$ ($c \in \mathbb{R}$)	$\text{Dom}(c \cdot f) = D_1$
Multiplication	$(f \cdot g)(x) = f(x) \cdot g(x)$	$\text{Dom}(f \cdot g) = D_1 \cap D_2$
Quotient (Division)	$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$	$\text{Dom}\left(\frac{f}{g}\right) = (D_1 \cap D_2) \setminus \{x : g(x) = 0\}$

The Denominator Rule: For the quotient function $f/g$, it is NOT sufficient to merely take the intersection $D_1 \cap D_2$. You MUST explicitly exclude every point $x$ where $g(x) = 0$, because division by zero is undefined in real mathematics!

6. Advanced Techniques for Determining Domain and Range

Finding the domain and range of an algebraic function is one of the most heavily tested skills in higher secondary mathematics.

6.1 Systematic Rules for Finding the Domain

The Domain of a real function $f(x)$ is the set of all real numbers $x$ for which the expression $f(x)$ produces a well-defined real number.

Radical Expressions in Numerator: If $f(x) = \sqrt{P(x)}$, enforce: $$P(x) \ge 0$$
Fractions with Non-Radical Denominators: If $f(x) = \frac{P(x)}{Q(x)}$, enforce: $$Q(x) \neq 0$$
Radical Expressions in Denominator: If $f(x) = \frac{P(x)}{\sqrt{Q(x)}}$, enforce: $$Q(x) > 0 \quad (\text{strictly positive})$$
Logarithmic Arguments: If $f(x) = \log_a(P(x))$, enforce: $$P(x) > 0, \quad a > 0, \quad a \neq 1$$

6.2 Standard Quadratic Radical Identities

Identity 1: $f(x) = \sqrt{a^2 - x^2}$
Condition: $a^2 - x^2 \ge 0 \implies x^2 \le a^2 \implies -a \le x \le a$
$$\mathbf{\text{Domain} = [-a, a], \quad \text{Range} = [0, a]}$$ Identity 2: $f(x) = \sqrt{x^2 - a^2}$
Condition: $x^2 - a^2 \ge 0 \implies x^2 \ge a^2 \implies x \le -a \text{ or } x \ge a$
$$\mathbf{\text{Domain} = (-\infty, -a] \cup [a, \infty), \quad \text{Range} = [0, \infty)}$$

6.3 Systematic Inversion Method for Finding Range

To find the Range of $y = f(x)$:

Express $y$ as a function of $x$: $y = f(x)$.
Solve the equation algebraically for $x$ in terms of $y$: $x = \phi(y)$.
Determine the set of all values of $y$ for which $x$ is real and belongs to the domain of $f$.
Intersect with any physical constraints on $y$ (for example, if $y = \sqrt{\dots}$, then $y \ge 0$ must always hold).

Key Formulas, Identities & Theorems

Cartesian Product Cardinality

p · q ordered pairs

For finite sets with p and q elements, the Cartesian product contains p · q ordered pairs.

Total Relations Formula

2^(pq) relations

Every subset of A × B is a relation. The number of non-empty relations is 2^(pq) - 1.

Total Functions Formula

$$q^p functions$$

Each of the p domain elements has q independent choices of images in the codomain.

Sum, Difference & Product Domain Rule

Dom(f) ∩ Dom(g)

Pointwise operations require both functions to be simultaneously well-defined.

Quotient Function Domain Rule

(Dom(f) ∩ Dom(g)) - {x : g(x)=0}

Division by zero is undefined; all roots of the denominator must be excluded.

Semi-Circle Radical Domain & Range

Domain: [-a, a], Range: [0, a]

Classic WBCHSE problem archetype. Range is strictly non-negative.

Conceptual Solved Examples & Case Studies

Example 1

Step-by-Step Solution:

Part (a): Solving for x and y [2 Marks] By the criterion of equality of ordered pairs, equate corresponding elements:

First coordinates:

$$\frac{x}{3} + 1 = \frac{5}{3} \implies \frac{x}{3} = \frac{5}{3} - 1 = \frac{2}{3} \implies x = 2$$

Second coordinates:

$$y - \frac{2}{3} = \frac{1}{3} \implies y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1$$

$$\therefore x = 2, \quad y = 1$$

[Award 1 Mark for x, 1 Mark for y]

Part (b): Tabulating Relation R and Finding Domain, Range & Inverse [3 Marks]

Given equation: $2x + 3y = 20 \implies 3y = 20 - 2x \implies y = \frac{20 - 2x}{3}$. Since $x, y \in \mathbb{N} = \{1, 2, 3, \dots\}$, $(20 - 2x)$ must be a positive multiple of 3:
- For $x = 1: y = 18/3 = 6 \in \mathbb{N} \implies (1, 6) \in R$
- For $x = 4: y = 12/3 = 4 \in \mathbb{N} \implies (4, 4) \in R$
- For $x = 7: y = 6/3 = 2 \in \mathbb{N} \implies (7, 2) \in R$
- For $x = 10: y = 0 \notin \mathbb{N}$ (not natural).

$$\therefore R = \{(1, 6), (4, 4), (7, 2)\}$$

[Award 1 Mark for Roster form] 2. Finding Domain and Range:

$$\text{Dom}(R) = \{1, 4, 7\}$$

$$\text{Range}(R) = \{6, 4, 2\} = \{2, 4, 6\}$$

[Award 1 Mark for Domain & Range] 3. Inverse Relation $R^{-1}$: Reverse the ordered pairs of $R$:

$$R^{-1} = \{(6, 1), (4, 4), (2, 7)\}$$

[Award 1 Mark for Inverse Relation]

Example 2

Step-by-Step Solution:

Analysis of Function Criteria:

(i) Examination of R₁ [1 Mark]

Look at the elements of domain $X = \{1, 2, 3, 4\}$.
The element $4 \in X$ does not appear as a first coordinate in any ordered pair of $R_1$.
Thus, $4 \in X$ has NO image in $Y$.
$\therefore R_1$ violates Axiom 1 (Totality). $R_1$ is NOT a function. [Award 1 Mark]

(ii) Examination of R₂ [1 Mark]

Observe that the element $2 \in X$ appears twice as a first coordinate: $(2, b) \in R_2$ and $(2, c) \in R_2$.
Here, $b \neq c$, so the single input $2$ has two distinct images.
$\therefore R_2$ violates Axiom 2 (Uniqueness of image). $R_2$ is NOT a function. [Award 1 Mark]

(iii) Examination of R₃ [1 Mark]

Every element of $X$ ($1, 2, 3, 4$) appears as a first coordinate.
Each element of $X$ is associated with exactly one image in $Y$ (namely, $b$).
This is a valid constant / many-to-one function.
$\therefore$ $R_3$ IS a function. $\text{Dom}(R_3) = X, \text{Range}(R_3) = \{b\}$. [Award 1 Mark]

(iv) Total Number of Functions from X to Y [1 Mark]

$n(X) = p = 4$ and $n(Y) = q = 3$.
Total number of functions $= [n(Y)]^{[n(X)]} = q^p = 3^4 = 81$.

$$\therefore \text{Total functions} = 81$$

[Award 1 Mark]

Example 3

Step-by-Step Solution:

Part (a): Function f(x) = √(16 - x²) [3 Marks]

Domain Determination: For $f(x)$ to be real, the quantity under the radical must be non-negative:

$$16 - x^2 \ge 0 \implies x^2 \le 16 \implies -4 \le x \le 4$$

$$\therefore \mathbf{\text{Domain of } f = [-4, 4]}$$

[Award 1.5 Marks for Domain] 2. Range Determination: Let $y = \sqrt{16 - x^2}$.

Since the square root symbol represents the principal (non-negative) root, $y \ge 0$.
Squaring both sides: $y^2 = 16 - x^2 \implies x^2 = 16 - y^2$.
For $x$ to be real: $16 - y^2 \ge 0 \implies y^2 \le 16 \implies -4 \le y \le 4$.
Intersecting with the non-negativity constraint $y \ge 0$:

$$0 \le y \le 4$$

$$\therefore \mathbf{\text{Range of } f = [0, 4]}$$

[Award 1.5 Marks for Range]

Part (b): Function g(x) = 1 / √(x² - 25) [2 Marks]

Domain Determination: Here, the radical is in the denominator. Therefore, the expression must be strictly positive (it cannot be zero, nor negative):

$$x^2 - 25 > 0 \implies (x - 5)(x + 5) > 0$$

By the wavy-curve method, the inequality holds when:

$$x < -5 \quad \text{or} \quad x > 5$$

$$\therefore \mathbf{\text{Domain of } g = (-\infty, -5) \cup (5, \infty)}$$

[Award 1 Mark for Domain] 2. Range Determination: As $|x| \to \infty$, $x^2 - 25 \to \infty \implies \sqrt{x^2 - 25} \to \infty \implies y \to 0^+$. As $|x| \to 5^+$, $x^2 - 25 \to 0^+ \implies \sqrt{x^2 - 25} \to 0^+ \implies y \to +\infty$. Since $y$ can take any positive real value strictly greater than 0:

$$\therefore \mathbf{\text{Range of } g = (0, \infty)}$$

[Award 1 Mark for Range]

Example 4

Step-by-Step Solution:

Step 1: Domain Determination [2 Marks]

The given function is a rational function:

$$f(x) = \frac{x^2}{1 + x^2}$$

The denominator is $1 + x^2$.
For any real number $x$, $x^2 \ge 0 \implies 1 + x^2 \ge 1 > 0$.
The denominator is never zero for any $x \in \mathbb{R}$.
Therefore, $f(x)$ is defined for all real numbers.

$$\therefore \mathbf{\text{Domain of } f = \mathbb{R} = (-\infty, \infty)}$$

[Award 2 Marks for Domain]

Step 2: Range Determination via Inversion Method [3 Marks]

Let $y = \frac{x^2}{1 + x^2}$.
Cross-multiply to express $x$ in terms of $y$:

$$y(1 + x^2) = x^2$$

$$y + yx^2 = x^2$$

$$x^2 - yx^2 = y$$

$$x^2(1 - y) = y$$

$$x^2 = \frac{y}{1 - y} \implies x = \pm \sqrt{\frac{y}{1 - y}}$$

For $x$ to be a real number, two mathematical conditions must be satisfied:
- Condition 1: The radicand must be non-negative: $\frac{y}{1 - y} \ge 0$.
- Condition 2: The denominator cannot be zero: $1 - y \neq 0 \implies y \neq 1$.
Multiplying $\frac{y}{1 - y} \ge 0$ by $(1 - y)^2 > 0$:

$$y(1 - y) \ge 0 \implies y(y - 1) \le 0$$

This yields: $0 \le y \le 1$. 5. Excluding $y = 1$ from Condition 2:

$$0 \le y < 1$$

Notice also from the original function: $x^2 \ge 0$ and $x^2 < 1 + x^2$, so $0 \le \frac{x^2}{1 + x^2} < 1$.

$$\therefore \mathbf{\text{Range of } f = [0, 1)}$$

[Award 3 Marks for step-by-step inversion and correct range]

Example 5

Step-by-Step Solution:

Step 1: Determine the Individual Domains

For $f(x) = \sqrt{x + 3}$: Condition: $x + 3 \ge 0 \implies x \ge -3$.

$$\text{Dom}(f) = [-3, \infty)$$

For $g(x) = \sqrt{5 - x}$: Condition: $5 - x \ge 0 \implies x \le 5$.

$$\text{Dom}(g) = (-\infty, 5]$$

Find the intersection of domains:

$$\text{Dom}(f) \cap \text{Dom}(g) = [-3, \infty) \cap (-\infty, 5] = [-3, 5]$$

Part (i): Addition (f + g)(x) [1 Mark]

$$(f + g)(x) = \sqrt{x + 3} + \sqrt{5 - x}$$

$$\text{Domain} = \text{Dom}(f) \cap \text{Dom}(g) = \mathbf{[-3, 5]}$$

[Award 1 Mark]

Part (ii): Subtraction (f - g)(x) [1 Mark]

$$(f - g)(x) = \sqrt{x + 3} - \sqrt{5 - x}$$

$$\text{Domain} = \text{Dom}(f) \cap \text{Dom}(g) = \mathbf{[-3, 5]}$$

[Award 1 Mark]

Part (iii): Multiplication (f · g)(x) [1 Mark]

$$(f \cdot g)(x) = \sqrt{x + 3} \cdot \sqrt{5 - x} = \sqrt{(x + 3)(5 - x)} = \sqrt{15 + 2x - x^2}$$

$$\text{Domain} = \text{Dom}(f) \cap \text{Dom}(g) = \mathbf{[-3, 5]}$$

[Award 1 Mark]

Part (iv): Quotient (f / g)(x) [2 Marks]

$$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x + 3}}{\sqrt{5 - x}} = \sqrt{\frac{x + 3}{5 - x}}$$

For the domain of the quotient:

$$\text{Dom}\left(\frac{f}{g}\right) = [\text{Dom}(f) \cap \text{Dom}(g)] \setminus \{x : g(x) = 0\}$$

Setting $g(x) = 0 \implies \sqrt{5 - x} = 0 \implies x = 5$. Excluding $x = 5$ from $[-3, 5]$ opens the right endpoint:

$$\therefore \mathbf{\text{Domain of } \left(\frac{f}{g}\right) = [-3, 5)}$$

[Award 1 Mark for formula, 1 Mark for excluding x=5 to yield [-3, 5)]

Example 6

Step-by-Step Solution:

Part (a): Solving [x]² - 5[x] + 6 = 0 [2 Marks]

Let $t = [x]$. Note that $t$ must be an integer ($t \in \mathbb{Z}$).
The equation becomes:

$$t^2 - 5t + 6 = 0$$

Factorize:

$$(t - 2)(t - 3) = 0 \implies t = 2 \quad \text{or} \quad t = 3$$

Both $2$ and $3$ are valid integers. Now solve for $x$:
- Case I: $[x] = 2 \implies 2 \le x < 3$
- Case II: $[x] = 3 \implies 3 \le x < 4$
Taking the union of both intervals:

$$[2, 3) \cup [3, 4) = [2, 4)$$

$$\therefore \mathbf{x \in [2, 4)}$$

[Award 2 Marks]

Part (b): Solving 2[x]² - 7[x] + 3 = 0 [2 Marks]

Let $u = [x]$, where $u \in \mathbb{Z}$.
The quadratic equation is:

$$2u^2 - 7u + 3 = 0$$

Factorize by splitting the middle term:

$$2u^2 - 6u - u + 3 = 0$$

$$2u(u - 3) - 1(u - 3) = 0 \implies (2u - 1)(u - 3) = 0$$

$$u = \frac{1}{2} \quad \text{or} \quad u = 3$$

Check the fundamental integer constraint of the step function:
- Since $[x]$ by definition must be an integer, $[x] = 1/2$ is impossible (has no solution).
- Therefore, the only admissible root is:

$$[x] = 3$$

Solving $[x] = 3$ yields:

$$3 \le x < 4$$

$$\therefore \mathbf{x \in [3, 4)}$$

[Award 1 Mark for rejecting 1/2, 1 Mark for interval [3, 4)]

Common Misconceptions & Examiner Traps

Common Misconception

Confusing an ordered pair (a, b) with an unordered set {a, b}, or writing A × B = B × A.

Scientific Reality & Correction

In an ordered pair, order is rigid: (1, 2) represents a completely different point in the Cartesian plane than (2, 1). Consequently, A × B is not equal to B × A unless A = B or one set is empty.

Common Misconception

Computing total number of functions from A to B as p^q instead of q^p.

Scientific Reality & Correction

Each of the p elements in the domain has q choices of images in the codomain. Therefore, the formula is strictly [n(Codomain)]^[n(Domain)] = q^p.

Common Misconception

Assuming that the Range of a function always equals its Codomain.

Scientific Reality & Correction

The codomain is the set B specified in the definition f: A → B. The range is the subset of B consisting only of actual output values f(x). Range equals Codomain only when the function is surjective (onto). In general, Range ⊆ Codomain.

Common Misconception

Forgetting to exclude denominator roots when finding the domain of (f / g)(x).

Scientific Reality & Correction

Division by zero is undefined in real mathematics. Thus, even if a point x lies in both Dom(f) and Dom(g), if g(x) = 0, that point x must be strictly excluded from the domain of f/g.

Common Misconception

Writing the range of f(x) = √(a² - x²) as [-a, a] instead of [0, a].

Scientific Reality & Correction

In mathematics, the radical symbol √ denotes the principal (non-negative) square root. Therefore, f(x) = √(a² - x²) cannot produce negative numbers. Its minimum value is 0 (at x = ±a) and its maximum value is a (at x = 0). Thus, Range = [0, a].

Relations, Functions & Real Graphical Geometry Architectural Roadmap

Chapter Summary & 10 Key Takeaways

Takeaway 1

Two ordered pairs (a, b) and (c, d) are equal if and only if their corresponding coordinates are equal: a = c and b = d.

Takeaway 2

The Cartesian product A × B is the set of all ordered pairs (a, b) with a ∈ A and b ∈ B; its cardinality is n(A × B) = n(A) · n(B).

Takeaway 3

A relation R from set A to set B is any subset of A × B (R ⊆ A × B); if n(A) = p and n(B) = q, the total number of possible relations is 2^(pq).

Takeaway 4

For any relation R, Domain is the set of all first coordinates, Range is the set of all second coordinates, and Codomain is the entire set B. Note that Range(R) ⊆ Codomain(R).

Takeaway 5

A relation f: A → B is a function if and only if every element of A has a unique image in B; the total number of functions from A to B is [n(B)]^[n(A)] = q^p.

Takeaway 6

The Vertical Line Test states that a curve in the Cartesian plane represents y as a function of x if and only if no vertical line intersects the curve at more than one point.

Takeaway 7

The Modulus Function f(x) = |x| has Domain = ℝ and Range = [0, ∞); its graph is a V-shape with vertex at the origin.

Takeaway 8

The Greatest Integer Function f(x) = [x] outputs the largest integer less than or equal to x; its Domain = ℝ and Range = ℤ (set of all integers).

Takeaway 9

For real functions f and g, Dom(f ± g) = Dom(f · g) = Dom(f) ∩ Dom(g), while Dom(f / g) = (Dom(f) ∩ Dom(g)) - {x : g(x) = 0}.

Takeaway 10

To find the domain of radical expressions √(f(x)), enforce the condition f(x) ≥ 0; for 1/√(f(x)), enforce the strict inequality f(x) > 0.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

If A = {1, 2} and B = {3, 4}, write A × B. How many total relations can be defined from A to B?

Reveal Answer & Explanation

Answer: A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. Since n(A) = 2 and n(B) = 2, n(A × B) = 4. The total number of subsets of A × B is 2^4 = 16. Hence, 16 relations can be defined.
List all ordered pairs (a, b) and use the formula 2^(p·q).

Is the relation R = {(2, 1), (3, 1), (4, 2), (5, 2)} a function? Give reasons and find its domain and range.

Reveal Answer & Explanation

Answer: Yes, R is a function because every first coordinate (2, 3, 4, 5) is unique and associated with exactly one second coordinate. Domain = {2, 3, 4, 5} and Range = {1, 2}.
Check if every first coordinate is unique.

Find the domain of the real-valued function f(x) = √(x - 2) / (x - 5).

Reveal Answer & Explanation

Answer: For √(x - 2) to be real, x - 2 ≥ 0 ⇒ x ≥ 2. For the denominator to be non-zero, x - 5 ≠ 0 ⇒ x ≠ 5. Combining both constraints: Domain = [2, 5) ∪ (5, ∞) = [2, ∞) \ {5}.
Enforce x - 2 ≥ 0 for the numerator and x - 5 ≠ 0 for the denominator.

Evaluate [3.7], [-2.3], and [5]. Hence, find the value of [-4.8] + [4.8].

Reveal Answer & Explanation

Answer: [3.7] = 3; [-2.3] = -3 (since -3 ≤ -2.3 < -2); [5] = 5. For [-4.8] + [4.8]: [-4.8] = -5 and [4.8] = 4. Therefore, -5 + 4 = -1. (Note: For non-integer x, [x] + [-x] = -1).
Recall that [x] is the greatest integer less than or equal to x.

Find the range of the function f(x) = 1 / (2 - cos(x)).

Reveal Answer & Explanation

Answer: We know that for all real x: -1 ≤ cos(x) ≤ 1. Multiply by -1: -1 ≤ -cos(x) ≤ 1. Add 2: 2 - 1 ≤ 2 - cos(x) ≤ 2 + 1 ⇒ 1 ≤ 2 - cos(x) ≤ 3. Taking reciprocals reverses the inequalities: 1/3 ≤ 1/(2 - cos(x)) ≤ 1/1 = 1. Therefore, the Range of f is [1/3, 1].
Use the known bounds -1 ≤ cos(x) ≤ 1 and invert the inequality.

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All Class 11 Mathematics Chapters

Ch 1: Sets Ch 2: Relations and Functions Ch 3: Trigonometric Functions Ch 4: Complex Numbers and Quadratic Equations Ch 5: Linear Inequalities Ch 6: Permutations and Combinations Ch 7: Limits and Derivatives Ch 8: Principle of Mathematical Induction Ch 9: Binomial Theorem Ch 10: Sequence and Series Ch 11: Straight Lines Ch 12: Conic Sections Ch 13: Statistics Ch 14: Probability

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Relations and Functions

Why This Chapter Matters

Chapter Roadmap & Progression

Complete Concept Guide (100% Curriculum Coverage)

1. Ordered Pairs & Cartesian Product of Sets

2. Concept, Representation & Taxonomy of Relations

3. Definition, Criteria & Cardinality of Functions (Mappings)

4. Standard Real-Valued Functions & Their Graphs

5. Algebra of Real Functions: Operations & Domain Rules

6. Advanced Techniques for Determining Domain and Range

Key Formulas, Identities & Theorems

Conceptual Solved Examples & Case Studies

Common Misconceptions & Examiner Traps

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Relations, Functions & Real Graphical Geometry Architectural Roadmap

Chapter Summary & 10 Key Takeaways

Check Your Understanding (Diagnostic Practice Questions)

Timed CBT Practice Tests (Exam Simulator)

All Class 11 Mathematics Chapters

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