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Class 12 Mathematics • Comprehensive Chapter Notes
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Chapter 1: Relations and Functions
Syllabus Reference: NCERT Class 12 | RD Sharma Vol. 1 | RS Aggarwal | KC Sinha |
Board Level + JEE Mains Level
Dear Student 💡
Welcome to the most foundational chapter of Class 12 Mathematics. This
chapter is a direct extension of sets and relations from Class 11 — but now we go much deeper. It lays
the groundwork for Calculus, Matrices, and all of higher mathematics. In Board exams, you will be asked
to prove equivalence relations, verify types of functions, find inverses, and solve binary
operations. In JEE Mains, you will additionally deal with counting functions,
piecewise composition, and anti-symmetry. Pay attention to every definition — precision
matters here!
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📚 Detailed Table of Contents — Relations and Functions
1. Types of Relations on a Set
When we talk about "types of relations" in Class 12, we almost always mean a relation defined on a
single set $A$, i.e., $R \subseteq A \times A$.
Number of Possible Relations
- If $n(A) = p$ and $n(B) = q$, the total number of relations from $A$ to $B$ is $2^{pq}$.
- If a relation is defined on a single set $A$ with $n$ elements, the total number of relations is $2^{n^2}$.
- The number of Reflexive Relations on a set with $n$ elements is $2^{n(n-1)}$.
- The number of Symmetric Relations on a set with $n$ elements is $2^{\frac{n(n+1)}{2}}$.
1.1 Special Relations: Empty, Universal, Identity
| Relation |
Definition |
Example (A = {1,2,3}) |
| Empty / Void |
$R = \phi$. No element is related to any element. |
$R = \{\}$ |
| Universal |
$R = A \times A$. Every element is related to every element. |
$R = \{(1,1),(1,2),\ldots,(3,3)\}$ — 9 pairs |
| Identity ($I_A$) |
$R = \{(a,a) : a \in A\}$. Each element is related only to itself. |
$R = \{(1,1),(2,2),(3,3)\}$ |
⚠️ Common Confusion: Identity vs Reflexive
- An Identity Relation allows ONLY the pairs $(a,a)$ — nothing else.
- A Reflexive Relation MUST contain all $(a,a)$ pairs, but can also have additional
pairs like $(a,b)$ where $a \neq b$.
- Therefore: Every Identity relation is Reflexive, but not every Reflexive relation is an
Identity relation.
- Also note: The Empty Relation and the Universal Relation on a set
are BOTH reflexive and transitive. The empty relation is symmetric; the universal relation
is also symmetric. So both the Empty and Universal relations are equivalence relations.
1.2 The Big Three Properties
Reflexive, Symmetric, Transitive — Formal Definitions
Let $R$ be a relation on a non-empty set $A$.
① Reflexive: $R$ is reflexive if $(a, a) \in R$ for every $a \in A$.
Symbolically: $\forall\, a \in A,\; aRa$.
② Symmetric: $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in
A$.
Symbolically: $aRb \implies bRa$.
③ Transitive: $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$
for all $a, b, c \in A$.
Symbolically: $aRb$ and $bRc \implies aRc$.
2.4 Checking Reflexive, Symmetric, Transitive - A Step-by-Step Method
- Reflexive Check: Write down every element of $A$. Check if $(a, a)$ is in $R$ for ALL
of them. If even one is missing, not reflexive.
- Symmetric Check: For every pair $(a, b) \in R$ where $a \neq b$, check if $(b, a)$ is
also in $R$. If any such pair is missing, not symmetric.
- Transitive Check: For every pair of pairs $(a,b)$ and $(b,c)$ in $R$, check that
$(a,c)$ is also in $R$. If not, not transitive.
🔑 Vacuous Truth — Critical for Transitive & Symmetric (RD Sharma Tip)
- For Transitivity: If $(a,b) \in R$ but there is NO pair $(b, c)$ in $R$ (for any
$c$), then the transitive condition is vacuously satisfied for that pair. The relation is
still transitive.
- For Symmetry: If $R = \phi$ (empty relation), it is vacuously symmetric (no pair
exists to violate symmetry) and also vacuously transitive.
- Conclusion: The empty relation on any set is reflexive only if $A = \phi$ itself.
On a non-empty set, the empty relation is not reflexive (since $(a,a)$ is missing), but it
IS symmetric and transitive.
Practice Problem 2 — NCERT / Board Level
Q: Let $A = \{1, 2, 3, 4, 5, 6\}$, $R = \{(a, b) : b = a + 1\}$. Check if $R$ is reflexive,
symmetric, or transitive.
Roster form: $R = \{(1,2),(2,3),(3,4),(4,5),(5,6)\}$
① Reflexive: $(1,1) \notin R$ → Not Reflexive.
② Symmetric: $(1,2) \in R$ but $(2,1) \notin R$ → Not Symmetric.
③ Transitive: $(1,2) \in R$ and $(2,3) \in R$, but $(1,3) \notin R$ → Not
Transitive.
Conclusion: $R$ is none of the three.
Practice Problem 2B — JEE Main 2005 Focus
Q: Let $R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\}$ be a relation on the set $A = \{3, 6, 9, 12\}$. Show that $R$ is reflexive and transitive but not symmetric.
Solution:
① Reflexive:
For a relation to be reflexive, $(a, a) \in R$ for all $a \in A$.
Since $(3,3), (6,6), (9,9), (12,12)$ are all elements of $R$, the relation is Reflexive. ✓
② Symmetric:
A relation is symmetric if $(a, b) \in R \implies (b, a) \in R$.
Here, we observe that $(6, 12) \in R$ but its reverse $(12, 6) \notin R$.
Therefore, the relation is Not Symmetric. ✗
③ Transitive:
A relation is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Let's check the non-diagonal pairs:
- $(3,6) \in R$ and $(6,12) \in R \implies (3,12) \in R$ (which is in $R$) ✓
- $(3,3) \in R$ and $(3,9) \in R \implies (3,9) \in R$ ✓
- $(3,3) \in R$ and $(3,12) \in R \implies (3,12) \in R$ ✓
All combinations of $(a,b)$ and $(b,c)$ are verified and their transitives are present in $R$.
Therefore, the relation is Transitive. ✓
Conclusion: $R$ is reflexive and transitive but not symmetric.
Practice Problem 3 — RD Sharma Type
Q: Let $A = \{1, 2, 3\}$ and $R = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}$. Is $R$ reflexive,
symmetric, transitive? Is it an equivalence relation?
① Reflexive: $(1,1),(2,2),(3,3)$ all present → Yes, Reflexive.
② Symmetric: $(1,2) \in R$ and $(2,1) \in R$. No other asymmetric pairs. → Yes,
Symmetric.
③ Transitive: Check: $(1,2)\in R$ and $(2,1)\in R \Rightarrow$ need $(1,1)\in R$ ✓.
$(2,1)\in R$ and $(1,2)\in R \Rightarrow$ need $(2,2)\in R$ ✓. All other checks pass. → Yes,
Transitive.
Since all three hold → $R$ is an Equivalence Relation.
Practice Problem 4 — KC Sinha / RS Aggarwal Type
Q: Let $R$ be the relation on the set of real numbers defined by $R = \{(a, b) : a - b
\text{ is an integer}\}$. Show that $R$ is an equivalence relation.
① Reflexive: For any $a \in \mathbb{R}$, $a - a = 0 \in \mathbb{Z}$. So $(a,a) \in R$.
✓ Reflexive.
② Symmetric: If $(a,b) \in R \implies a-b \in \mathbb{Z} \implies -(a-b) = b-a \in
\mathbb{Z} \implies (b,a) \in R$. ✓ Symmetric.
③ Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $a-b = m \in \mathbb{Z}$ and
$b-c = n \in \mathbb{Z}$. So $a-c = (a-b)+(b-c) = m+n \in \mathbb{Z} \implies (a,c) \in R$. ✓
Transitive.
Hence $R$ is an equivalence relation.
Practice Problem — NCERT Favorite
Q: Show that the relation $R$ in the set of real numbers $\mathbb{R}$ defined as $R = \{(a, b) : a \le b^2\}$ is neither reflexive, nor symmetric, nor transitive.
① Not Reflexive: We need to find an $a \in \mathbb{R}$ such that $a \not\le a^2$. Let $a = \frac{1}{2}$. Since $\frac{1}{2} \not\le \left(\frac{1}{2}\right)^2 = \frac{1}{4}$, the pair $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$. So, it is not reflexive. ✗
② Not Symmetric: Let $a = 1, b = 4$. We see $1 \le 4^2$ (i.e., $1 \le 16$), so $(1, 4) \in R$. But $4 \not\le 1^2$ (i.e., $4 \not\le 1$), so $(4, 1) \notin R$. So, it is not symmetric. ✗
③ Not Transitive: Let $a = 3, b = -2, c = -1$. We see $3 \le (-2)^2$ (i.e., $3 \le 4$), so $(3, -2) \in R$. And $-2 \le (-1)^2$ (i.e., $-2 \le 1$), so $(-2, -1) \in R$. But $3 \not\le (-1)^2$ (i.e., $3 \not\le 1$), so $(3, -1) \notin R$. So, it is not transitive. ✗
Practice Problem — Constructing Examples (NCERT)
Q: Give an example of a relation which is symmetric but neither reflexive nor transitive.
Let set $A = \{1, 2, 3\}$.
Let relation $R = \{(1, 2), (2, 1)\}$.
Check Reflexive: $(1, 1), (2, 2), (3, 3) \notin R$. So, not reflexive. ✗
Check Symmetric: $(1, 2) \in R$ and $(2, 1) \in R$. No other pairs exist. So, symmetric. ✓
Check Transitive: $(1, 2) \in R$ and $(2, 1) \in R$. For transitivity, we must have $(1, 1) \in R$, which is not true. So, not transitive. ✗
2. Equivalence Relations and Equivalence Classes
Equivalence Relation — Definition
A relation $R$ on a set $A$ is called an
Equivalence Relation if and only if it is
simultaneously:
- Reflexive AND
- Symmetric AND
- Transitive
Fig 3.1 — Equivalence Relation
partitions Set A into disjoint equivalence classes [a], [b], [c]
2.1 Equivalence Classes
If $R$ is an equivalence relation on set $A$, the equivalence class of an element $a \in A$,
denoted $[a]$ or $\bar{a}$, is defined as:
$$ [a] = \{ x \in A : (x, a) \in R \} = \{ x \in A : x \text{ is related to } a \text{ under } R \} $$
Key Properties of Equivalence Classes:
- Every element of $A$ belongs to exactly one equivalence class.
- $[a] = [b]$ if and only if $(a, b) \in R$ (i.e., $a$ and $b$ are in the same equivalence class).
- Two distinct equivalence classes are always disjoint: $[a] \cap [b] = \phi$ if $[a]
\neq [b]$.
- The union of all equivalence classes equals $A$ (this is called a Partition of $A$).
2.2 Congruence Modulo $n$ - Most Important Equivalence Relation
Congruence Modulo n (RD Sharma + JEE Focus)
Definition: $a \equiv b \pmod{n}$ if and only if $n$ divides $(a - b)$, i.e., $a - b = kn$
for some integer $k$.
Theorem: The relation $R = \{(a,b) : n \mid (a-b)\}$ is always an equivalence relation on
$\mathbb{Z}$.
Equivalence Classes for $n = 3$:
$[0] = \{\ldots, -6, -3, 0, 3, 6, \ldots\}$ (multiples of 3)
$[1] = \{\ldots, -5, -2, 1, 4, 7, \ldots\}$ (leave remainder 1 when divided by 3)
$[2] = \{\ldots, -4, -1, 2, 5, 8, \ldots\}$ (leave remainder 2 when divided by 3)
These 3 classes form a partition of $\mathbb{Z}$.
Practice Problem 5 — NCERT Exemplar / Board
Q: Show that $R = \{(a, b) : 2 \mid (a - b)\}$ on $\mathbb{Z}$ is an equivalence relation.
Find $[0]$ and $[1]$.
① Reflexive: $a - a = 0 = 2(0)$, so $2 \mid 0$ → $(a,a) \in R$. ✓
② Symmetric: $(a,b) \in R \Rightarrow a-b=2k \Rightarrow b-a=-2k=2(-k)$, so $(b,a) \in
R$. ✓
③ Transitive: $(a,b),(b,c) \in R \Rightarrow a-b=2m,\; b-c=2n \Rightarrow a-c=2(m+n)
\Rightarrow (a,c) \in R$. ✓
$\therefore$ $R$ is an equivalence relation.
$[0] = \{x \in \mathbb{Z} : 2 \mid x\} = \{\ldots, -4,-2,0,2,4, \ldots\}$ (All even integers)
$[1] = \{x \in \mathbb{Z} : 2 \mid (x-1)\} = \{\ldots, -3,-1,1,3,5, \ldots\}$ (All odd integers)
Practice Problem 6 — RD Sharma Board Level
Q: Let $T$ be the set of all triangles in a plane and $R = \{(T_1, T_2) : T_1 \text{ is
similar to } T_2\}$. Show that $R$ is an equivalence relation.
① Reflexive: Every triangle is similar to itself → $(T_1, T_1) \in R$ ✓
② Symmetric: If $T_1 \sim T_2$ then $T_2 \sim T_1$ (similarity is symmetric) →
$(T_1,T_2)\in R \Rightarrow (T_2,T_1)\in R$ ✓
③ Transitive: $T_1 \sim T_2$ and $T_2 \sim T_3 \Rightarrow T_1 \sim T_3$ (transitivity
of similarity) → $(T_1,T_3)\in R$ ✓
$\dots$ $R$ is an equivalence relation.
Practice Problem 5B — JEE Main 2021 Focus
Q: If $R = \{(x, y) \mid x, y \in \mathbb{Z}, x^2 + y^2 \le 4\}$ is a relation defined on the set of integers $\mathbb{Z}$, then find the domain of $R$.
Solution:
A relation $R$ consists of ordered pairs $(x, y)$ of integers such that $x^2 + y^2 \le 4$. The domain is the set of all possible first elements $x$.
Since $x$ and $y$ are integers, their squares $x^2$ and $y^2$ must be non-negative perfect squares (i.e., $0, 1, 4, 9, \dots$).
For $x^2 + y^2 \le 4$, the possible values of $x^2$ are $0$, $1$, or $4$ (since $x^2 \le 4$). Let's analyze each case:
1. If $x^2 = 0 \implies x = 0$:
$$0^2 + y^2 \le 4 \implies y^2 \le 4 \implies y = 0, \pm 1, \pm 2$$
This gives valid integer pairs: $(0, 0), (0, \pm 1), (0, \pm 2)$.
2. If $x^2 = 1 \implies x = \pm 1$:
$$1^2 + y^2 \le 4 \implies y^2 \le 3 \implies y = 0, \pm 1$$
This gives valid integer pairs: $(\pm 1, 0), $(\pm 1, \pm 1)$.
3. If $x^2 = 4 \implies x = \pm 2$:
$$2^2 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$$
This gives valid integer pairs: $(\pm 2, 0)$.
All these cases yield valid integer pairs $(x, y)$ that belong to $R$.
Therefore, the possible integer values of $x$ are:
$$\text{Domain}(R) = \{-2, -1, 0, 1, 2\}$$
Practice Problem 5C — JEE Main 2021 Focus
Q: If $R = \{(x, y) \mid x, y \in \mathbb{Z}, x^2 + 3y^2 \le 8\}$ is a relation on the set of integers $\mathbb{Z}$, then find the domain of its inverse relation $R^{-1}$.
Solution:
We know that the domain of the inverse relation $R^{-1}$ is exactly equal to the range of the original relation $R$:
$$\text{Domain}(R^{-1}) = \text{Range}(R) = \{ y \in \mathbb{Z} : (x, y) \in R \text{ for some } x \in \mathbb{Z} \}$$
We need to find all possible integer values of $y$ for which there exists at least one integer $x$ satisfying:
$$x^2 + 3y^2 \le 8$$
Since $y$ must be an integer, let's test integer values of $y$ systematically:
1. If $y = 0$:
$$x^2 + 3(0)^2 \le 8 \implies x^2 \le 8 \implies x = 0, \pm 1, \pm 2$$
(Valid pairs exist: e.g., $(0, 0)$ is in $R$, so $y=0$ is in the range).
2. If $y = \pm 1$:
$$x^2 + 3(\pm 1)^2 \le 8 \implies x^2 + 3 \le 8 \implies x^2 \le 5 \implies x = 0, \pm 1, \pm 2$$
(Valid pairs exist: e.g., $(0, 1)$ is in $R$, so $y=\pm 1$ are in the range).
3. If $y = \pm 2$:
$$x^2 + 3(\pm 2)^2 \le 8 \implies x^2 + 12 \le 8 \implies x^2 \le -4$$
Since $x^2 \ge 0$ for all real numbers $x$, there are no integer values of $x$ satisfying this inequality. Thus, $y = \pm 2$ are not in the range.
4. If $|y| \ge 3$:
$3y^2 \ge 27$, so $x^2 \le 8 - 3y^2 < 0$, which yields no real solutions.
Thus, the only possible integer values of $y$ are $0, 1, -1$.
Therefore, the domain of the inverse relation $R^{-1}$ is:
$$\text{Domain}(R^{-1}) = \{-1, 0, 1\}$$
3. Functions - Definition and Types
Number of Possible Functions
Let set $A$ have $m$ elements ($n(A) = m$) and set $B$ have $n$ elements ($n(B) = n$).
- Total number of functions from $A$ to $B$: $n^m$.
- Total number of One-One (Injective) functions:
If $n \ge m$: $^nP_m = \dfrac{n!}{(n-m)!}$
If $n < m$: $0$
- Total number of Onto (Surjective) functions:
If $n = 2$: $2^m - 2$ (most common case in boards)
If $n > m$: $0$
- Total number of Bijective functions:
If $m = n$: $n!$
If $m \neq n$: $0$
Definition: Function (Mapping)
A relation $f$ from set $A$ to set $B$ is called a function if:
(i) Every element of $A$ has an image in $B$ (no element of $A$ is left out).
(ii) Each element of $A$ has exactly one image in $B$ (no element maps to two
different elements).
We write: $f: A \rightarrow B$ and say "f maps A to B."
For each $a \in A$, $f(a)$ denotes the unique image of $a$ in $B$.
Fig 4.1 — The four types of
functions: One-One (Injective), Many-One, Onto (Surjective), Into
3.1 One-One Function (Injective)
$f: A \rightarrow B$ is One-One (Injective) if distinct elements of $A$ have distinct images
in $B$.
$f(x_1) = f(x_2) \implies x_1 = x_2$ OR equivalently: $x_1 \neq x_2 \implies f(x_1) \neq
f(x_2)$
Methods to Prove One-One
Method 1 — Algebraic (Standard Board Method):
Step 1: Let $x_1, x_2 \in A$ (domain).
Step 2: Assume $f(x_1) = f(x_2)$.
Step 3: Solve algebraically. If this necessarily gives $x_1 = x_2$, function is One-One.
Method 2 — Derivative Test (JEE / RD Sharma):
If $f$ is continuous and differentiable:
$f'(x) > 0$ for all $x$ in domain → $f$ is strictly increasing → One-One.
$f'(x) < 0$ for all $x$ in domain → $f$ is strictly decreasing → One-One.
Method 3 — Horizontal Line Test (Graphical):
Draw any horizontal line. If it intersects the graph at most once, the function is One-One. If
it intersects more than once at any point, it is Many-One.
Fig 4.2 — Horizontal Line Test: A
horizontal line cuts y = x² at 2 points (Many-One) but cuts y = x³/3 at only 1 point (One-One)
3.2 Onto Function (Surjective)
$f: A \rightarrow B$ is Onto (Surjective) if every element of the co-domain $B$ has at least
one pre-image in $A$.
$\text{Range of } f = \text{Co-domain of } f$ (i.e., $f(A) = B$)
Algorithm to Prove Onto (Board Standard)
Step 1: Let $y$ be an arbitrary element of the co-domain $B$.
Step 2: Set $f(x) = y$.
Step 3: Solve for $x$ in terms of $y$ to get $x = g(y)$.
Step 4: Verify that this $x$ actually lies in the domain $A$. If yes for all $y \in B$, the function is
Onto.
3.3 Into, Many-One, and Bijective Functions
| Type |
Condition |
Key Property |
| One-One (Injective) |
$f(x_1)=f(x_2) \Rightarrow x_1=x_2$ |
No two inputs share an output |
| Many-One |
$\exists\; x_1 \neq x_2$ with $f(x_1)=f(x_2)$ |
At least two inputs share an output (Not injective) |
| Onto (Surjective) |
Range $=$ Co-domain |
Every output has a pre-image |
| Into |
Range $\subsetneq$ Co-domain |
At least one co-domain element has no pre-image (Not surjective) |
| Bijective |
One-One AND Onto |
Perfect pairing — every input has a unique output and every output is covered |
| Constant Function |
$f(x) = c$ for all $x$ |
Many-One and Into (unless $|B|=1$) |
🧠 Memory Trick: INTO vs ONTO
Think of it as archery:
• ONTO: Every target (element of B) is HIT by at least one arrow.
• INTO: Some targets are left UN-HIT (range is a strict subset of
co-domain).
For One-One vs Many-One:
• ONE-ONE: Each target hit by exactly ONE arrow.
• MANY-ONE: At least one target hit by MANY arrows.
Practice Problem 7 — Board Level (5 Marks)
Q: Show that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 3x + 4$ is
bijective.
① One-One: Let $x_1, x_2 \in \mathbb{R}$. Assume $f(x_1) = f(x_2)$.
$\Rightarrow 3x_1+4 = 3x_2+4 \Rightarrow 3x_1 = 3x_2 \Rightarrow x_1 = x_2$.
$\therefore f$ is One-One. ✓
② Onto: Let $y \in \mathbb{R}$ (co-domain). Set $f(x) = y$.
$3x+4 = y \Rightarrow x = \dfrac{y-4}{3}$.
Since $y \in \mathbb{R}$, $x = \dfrac{y-4}{3} \in \mathbb{R}$ = domain. So for every $y$, a
pre-image $x$ exists.
$\therefore f$ is Onto. ✓
Since $f$ is both One-One and Onto, $\mathbf{f}$ is Bijective.
Practice Problem 8 — RD Sharma / KC Sinha Type
Q: Show that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^2$ is neither
one-one nor onto.
Not One-One: $f(2) = 4 = f(-2)$, but $2 \neq -2$. So two different inputs give the
same output. $\therefore$ Many-One (not one-one).
Not Onto: $f(x) = x^2 \geq 0$ for all $x \in \mathbb{R}$. So the range is $[0,
\infty)$, which is a proper subset of $\mathbb{R}$ (the co-domain). For example, $-1 \in \mathbb{R}$
but $-1$ has no pre-image. $\therefore$ Into (not onto).
Practice Problem 9 — NCERT Exemplar / JEE Mains Type
Q: Let $A = \mathbb{R} \setminus \{3\}$ and $B = \mathbb{R} \setminus \{1\}$. Show that
$f: A \rightarrow B$ defined by $f(x) = \dfrac{x-2}{x-3}$ is bijective.
① One-One: Assume $f(x_1) = f(x_2)$.
$\dfrac{x_1-2}{x_1-3} = \dfrac{x_2-2}{x_2-3}$
$(x_1-2)(x_2-3) = (x_2-2)(x_1-3)$
$x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$
$-3x_1 - 2x_2 = -3x_2 - 2x_1$
$-3x_1 + 2x_1 = -3x_2 + 2x_2 \Rightarrow -x_1 = -x_2 \Rightarrow x_1 = x_2$. ✓ One-One.
② Onto: Let $y \in B$ (so $y \neq 1$). Set $f(x) = y$:
$\dfrac{x-2}{x-3} = y \Rightarrow x-2 = y(x-3) \Rightarrow x-2 = yx-3y$
$x - yx = 2 - 3y \Rightarrow x(1-y) = 2-3y \Rightarrow x = \dfrac{2-3y}{1-y} =
\dfrac{3y-2}{y-1}$.
Since $y \neq 1$, $x$ is defined. Also $x \neq 3$: if $x = 3$, then $3y-2 = 3(y-1) = 3y-3
\Rightarrow -2=-3$, contradiction. So $x \in A$. ✓ Onto.
$\therefore$ $f$ is bijective.
Practice Problem — NCERT Classic (Modulus Function)
Q: Show that the Modulus Function $f : \mathbb{R} \rightarrow \mathbb{R}$, given by $f(x) = |x|$, is neither one-one nor onto.
Not One-One: $f(1) = |1| = 1$ and $f(-1) = |-1| = 1$. Since $f(1) = f(-1)$ but $1 \neq -1$, two different inputs have the same output. Therefore, it is Many-One (not one-one). ✗
Not Onto: The co-domain is $\mathbb{R}$. However, the output of the modulus function is always non-negative ($f(x) \ge 0$). This means negative real numbers in the co-domain (like $-2$) have no pre-image in the domain. Since the range $[0, \infty)$ is a proper subset of the co-domain $\mathbb{R}$, it is Into (not onto). ✗
Practice Problem — NCERT Classic (Greatest Integer Function)
Q: Show that the Greatest Integer Function $f : \mathbb{R} \rightarrow \mathbb{R}$, given by $f(x) = [x]$, is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.
Not One-One: Consider $x_1 = 1.2$ and $x_2 = 1.5$. Both are in $\mathbb{R}$.
$f(1.2) = [1.2] = 1$ and $f(1.5) = [1.5] = 1$. Since $f(1.2) = f(1.5)$ but $1.2 \neq 1.5$, the function is Many-One. ✗
Not Onto: The co-domain is $\mathbb{R}$. The range of $f(x) = [x]$ is the set of integers $\mathbb{Z}$. Thus, non-integers in the co-domain (e.g., $0.5, 0.7$) do not have any pre-image in the domain. ✗
