Chapter 1: Relations and Functions

Syllabus Reference: NCERT Class 12 | RD Sharma Vol. 1 | RS Aggarwal | KC Sinha  |  Board Level + JEE Mains Level

0. Class 11 Quick Revision (Basics of Sets)

Before diving into Relations and Functions, it is crucial to recall the basic terminology of Sets from Class 11.

0.1 Sets and Subsets

• Set: A well-defined collection of distinct objects. Often denoted by capital letters $A, B, C$.
• Subset ($\subseteq$): Set $A$ is a subset of $B$ if every element of $A$ is also in $B$. The empty set $\phi$ is a subset of every set.
• Number of Subsets: If a set has $n$ elements, it has exactly $2^n$ subsets.

0.2 Operations on Sets

1. Cartesian Product & Basics of Relations

1.1 Cartesian Product (Quick Revision)

For two non-empty sets $A$ and $B$, the Cartesian Product is defined as:

• If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = p \times q$.
• $A \times B \neq B \times A$ (in general), but $n(A \times B) = n(B \times A)$.
• $A \times \phi = \phi$ (Cartesian product with an empty set is empty).
• If $A = B$, then $A \times A = A^2$ is called the Cartesian Square of $A$.

1.2 What is a Relation?

Fig 1.1 — Arrow diagram of a Relation: showing Domain, Range, and Co-domain of R ⊆ A × B

2. 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., $R \subseteq A \times A$.

2.1 Special Relations: Empty, Universal, Identity

2.2 The Big Three Properties

2.3 Anti-Symmetric Relation (JEE Focus)

2.4 Checking Reflexive, Symmetric, Transitive - A Step-by-Step Method

1. 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.
2. 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.
3. 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.

3. Equivalence Relations and Equivalence Classes

Fig 3.1 — Equivalence Relation partitions Set A into disjoint equivalence classes [a], [b], [c]

3.1 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:

Key Properties of Equivalence Classes:

1. Every element of $A$ belongs to exactly one equivalence class.
2. $[a] = [b]$ if and only if $(a, b) \in R$ (i.e., $a$ and $b$ are in the same equivalence class).
3. Two distinct equivalence classes are always disjoint: $[a] \cap [b] = \phi$ if $[a] \neq [b]$.
4. The union of all equivalence classes equals $A$ (this is called a Partition of $A$).

3.2 Congruence Modulo $n$ - Most Important Equivalence Relation

4. Functions - Definition and Types

Fig 4.1 — The four types of functions: One-One (Injective), Many-One, Onto (Surjective), Into

4.1 One-One Function (Injective)

$f: A \rightarrow B$ is One-One (Injective) if distinct elements of $A$ have distinct images in $B$.

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)

4.2 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 $A$.

4.3 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

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.

\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

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.

.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.,

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.

.5, 0.7$) do not have any pre-image in the domain. ✗

5. Special Functions (JEE & Boards Focus)

Mastering modulus, greatest integer, fractional part, signum, exponential, and logarithmic functions is critical for Board MCQs, but absolutely vital for JEE Mains. Let's arrange their properties and graphs systematically.

5.1 Modulus Function (Absolute Value Function)

Definition: Modulus Function The Modulus Function $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined as: $$ f(x) = |x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases} $$

• Domain: $\mathbb{R}$ (all real numbers)
• Range: $[0, \infty)$ (all non-negative real numbers)
• Geometric Meaning: Represents the absolute distance of point $x$ from the origin on the number line. Distance is never negative, hence $|x| \ge 0$.

Opening Modulus (Critical Points Method)

To free an expression from modulus, locate its critical point (CP) by setting the expression inside the mod to zero. On the right of the critical point, the modulus opens with a plus sign; on the left, it opens with a minus sign:

$$ |g(x)| = \begin{cases} g(x), & \text{if } g(x) \ge 0 \\ -g(x), & \text{if } g(x) < 0 \end{cases} $$

Practice Problem 7B — slide 80 Q: Define the function $f(x) = |2x - 3|$ free of modulus and draw its graph.

Solution:
Step 1: Locate the critical point (CP)
Set the expression inside the mod to zero:
$$2x - 3 = 0 \implies x = \dfrac{3}{2}$$ So $x = 1.5$ is the critical point.

Step 2: Open the modulus piecewise
- If $x \ge \frac{3}{2} \implies 2x - 3 \ge 0 \implies f(x) = \mathbf{2x - 3}$
- If $x < \frac{3}{2} \implies 2x - 3 < 0 \implies f(x) = -(2x - 3) = \mathbf{-2x + 3}$
Therefore: $$ f(x) = \begin{cases} 2x - 3, & \text{if } x \ge 1.5 \\ -2x + 3, & \text{if } x < 1.5 \end{cases} $$ Step 3: Graph the function
The graph is a symmetrical V-shape with its vertex at the critical point $(1.5, 0)$. - For $x \ge 1.5$, it follows the straight line $y = 2x - 3$. - For $x < 1.5$, it follows the straight line $y = -2x + 3$.

Double Modulus Graphs (The Boat/Trough Shortcut)

When drawing graphs of the form $f(x) = |x - a| + |x - b|$, locate the critical points at $x = a$ and $x = b$. Between these two points, the graph forms a flat bottom (like a boat or trough), rising symmetrically to infinity on both the left and right sides.

Practice Problem 7C — slide 85 Q: Draw the graph of the double modulus function $f(x) = |x - 1| + |x - 2|$ using the shortcut method.

Solution:
Step 1: Find Critical Points
Set each modulus expression to zero:
$$x - 1 = 0 \implies x = 1 \quad \text{and} \quad x - 2 = 0 \implies x = 2$$ So the critical points are $x = 1$ and $x = 2$.

Step 2: Plot Critical Points and Boundary Values
Let's calculate $y = f(x)$ at critical points and some nearby integers:

• At $x = 1 \implies f(1) = |1 - 1| + |1 - 2| = 0 + 1 = \mathbf{1}$
• At $x = 2 \implies f(2) = |2 - 1| + |2 - 2| = 1 + 0 = \mathbf{1}$
• At $x = 0 \implies f(0) = |0 - 1| + |0 - 2| = 1 + 2 = \mathbf{3}$
• At $x = 3 \implies f(3) = |3 - 1| + |3 - 2| = 2 + 1 = \mathbf{3}$

Step 3: Connect the points
- Connect the vertices $(1, 1)$ and $(2, 1)$ with a flat horizontal line $y = 1$.
- For $x \ge 2$, draw the ray starting at $(2, 1)$ passing through $(3, 3)$ with slope $+2$ ($y = 2x - 3$).
- For $x \le 1$, draw the ray starting at $(1, 1)$ passing through $(0, 3)$ with slope $-2$ ($y = -2x + 3$).

Fig 5.1 — Trough-shaped double modulus curve of f(x) = |x - 1| + |x - 2| showing the flat bottom between vertices

Solving Modulus Equations

Modulus equations can be solved using either algebraic piecewise definition or graphical intersection methods.

Practice Problem 7D — slide 113 Q: Solve for real $x$: $||x + 1| - 4| = 5$.

Solution:
Using the property $|f(x)| = a \implies f(x) = \pm a$:
$$|x + 1| - 4 = \pm 5 \implies |x + 1| = 4 \pm 5$$ This splits into two cases:

1. Case 1: $|x + 1| = 4 + 5 = 9$
   $$x + 1 = \pm 9 \implies x = -1 \pm 9 \implies \mathbf{x = 8} \quad \text{or} \quad \mathbf{x = -10}$$
2. Case 2: $|x + 1| = 4 - 5 = -1$
   Since the absolute value of a real number is always non-negative, $|x + 1| = -1$ has no real solutions.

Therefore, the only real solutions are: $$\boxed{x = 8, -10}$$

Practice Problem 7E — slide 114 Q: Solve for real $x$: $|x - 1| - |x| = 1$ using algebraic and graphical methods.

1. Algebraic Method:
Identify critical points: $x = 0$ and $x = 1$. This divides the real line into three intervals:

1. Interval 1 ($x < 0$): Both terms inside the mod are negative:
   $$-(x - 1) - (-x) = 1 \implies -x + 1 + x = 1 \implies 1 = 1$$ This identity is always true! Thus, the entire interval is a solution:
   $$\mathbf{x \in (-\infty, 0)}$$
2. Interval 2 ($0 \le x \le 1$): $(x-1)$ is non-positive, but $x$ is non-negative:
   $$-(x - 1) - x = 1 \implies -2x + 1 = 1 \implies 2x = 0 \implies \mathbf{x = 0}$$
3. Interval 3 ($x > 1$): Both terms inside the mod are positive:
   $$(x - 1) - x = 1 \implies -1 = 1 \quad \text{(False)}$$ No solution exists in this interval.

Combining the results, the final solution set is: $$\boxed{x \in (-\infty, 0]}$$
2. Graphical Method:
We plot the curve $y = |x - 1| - |x|$ and locate its intersection with the horizontal line $y = 1$: - For $x > 1$, $y = -1$. - For $0 \le x \le 1$, $y = 1 - 2x$, which is a straight line segment joining $(0, 1)$ and $(1, -1)$. - For $x < 0$, $y = 1$. The curve intersects and merges with the line $y = 1$ exactly for all $x \le 0$. Hence, the solution is $x \in (-\infty, 0]$.

5.2 Greatest Integer Function (G.I.F. / Step Function)

Definition: Greatest Integer Function The Greatest Integer Function $f(x) = [x]$ (also known as the floor function) maps a real number $x$ to the greatest integer that is less than or equal to $x$. $$ f(x) = [x] = n \iff n \le x < n+1, \quad n \in \mathbb{Z} $$

• Domain: $\mathbb{R}$
• Range: $\mathbb{Z}$ (set of all integers)
• Examples: $[3.2] = 3$, $[2.1] = 2$, $[0.7] = 0$, $[-3.14] = -4$

Fig 5.2 — Staircase-style step graph of the Greatest Integer Function y = [x] showing closed left endpoints and open right endpoints

Core Properties of G.I.F. (Must Memorize)

Properties of Floor Function

1. $x - 1 < [x] \le x$
2. $[x + n] = [x] + n$ if and only if $n$ is an integer ($n \in \mathbb{Z}$).
3. $[x + [x]] = 2[x]$ (Since $[x]$ is always an integer, it pulls out).
4. $[x] + [-x] = \begin{cases} 0, & \text{if } x \in \mathbb{Z} \\ -1, & \text{if } x \notin \mathbb{Z} \end{cases}$

5.3 Fractional Part Function (F.P.F.)

Definition: Fractional Part Function The Fractional Part Function is denoted as $f(x) = \{x\}$ and is defined as the difference between a real number $x$ and its greatest integer part: $$ f(x) = \{x\} = x - [x] $$

• Domain: $\mathbb{R}$
• Range: $[0, 1)$ (always non-negative and strictly less than 1)
• Examples: $\{3.2\} = 3.2 - 3 = 0.2$, $\{-2.1\} = -2.1 - [-2.1] = -2.1 - (-3) = 0.9$

Fig 5.3 — Sawtooth periodic curve of the Fractional Part Function y = {x} with period 1

Core Properties of F.P.F.

Properties of {x}

1. $0 \le \{x\} < 1$
2. $x = [x] + \{x\}$ (Fundamental Decomposition).
3. $\{x + n\} = \{x\}$ if and only if $n \in \mathbb{Z}$. (FPF is a periodic function with period 1).
4. $\{x\} + \{-x\} = \begin{cases} 0, & \text{if } x \in \mathbb{Z} \\ 1, & \text{if } x \notin \mathbb{Z} \end{cases}$

Practice Problem 7F — slide 174 Q: Solve the equation: $4[x] = x + \{x\}$.

Solution:
Step 1: Decompose x into integer and fractional parts
Substitute $x = [x] + \{x\}$ into the equation:
$$4[x] = ([x] + \{x\}) + \{x\}$$ $$4[x] = [x] + 2\{x\}$$ $$3[x] = 2\{x\} \implies \{x\} = \dfrac{3[x]}{2}$$ Step 2: Apply the range constraint $0 \le \{x\} < 1$
$$0 \le \dfrac{3[x]}{2} < 1$$ Multiply by $\frac{2}{3}$:
$$0 \le [x] < \dfrac{2}{3}$$ Step 3: Solve for the integer [x]
Since $[x]$ must be an integer, the only integer in the interval $[0, \frac{2}{3})$ is:
$$[x] = 0$$ Step 4: Find the final value of x
Substitute $[x] = 0$ back into the formula for $\{x\}$:
$$\{x\} = \dfrac{3(0)}{2} = 0$$ Thus, the solution is:
$$x = [x] + \{x\} = 0 + 0 = \mathbf{0}$$

Practice Problem 7G — slide 176 Q: Find the number of real solutions of the equation $[2x] - 3\{2x\} = 1$.

Solution:
Step 1: Simplify by substitution
Let $t = 2x$. The equation becomes:
$$[t] - 3\{t\} = 1 \implies 3\{t\} = [t] - 1 \implies \{t\} = \dfrac{[t] - 1}{3}$$ Step 2: Apply FPF range constraint $0 \le \{t\} < 1$
$$0 \le \dfrac{[t] - 1}{3} < 1$$ $$0 \le [t] - 1 < 3 \implies 1 \le [t] < 4$$ Step 3: Solve for integers [t]
Since $[t]$ must be an integer, the possible values are:
$$[t] = 1, 2, 3$$ Step 4: Find values of t and x
- If $[t] = 1 \implies \{t\} = \frac{1 - 1}{3} = 0 \implies t = [t] + \{t\} = 1 \implies x = \dfrac{1}{2}$
- If $[t] = 2 \implies \{t\} = \frac{2 - 1}{3} = \frac{1}{3} \implies t = 2 + \frac{1}{3} = \frac{7}{3} \implies x = \dfrac{7}{6}$
- If $[t] = 3 \implies \{t\} = \frac{3 - 1}{3} = \frac{2}{3} \implies t = 3 + \frac{2}{3} = \frac{11}{3} \implies x = \dfrac{11}{6}$

All these three solutions are distinct and valid.
Therefore, the number of real solutions is 3, and the solution set is: $$\boxed{x \in \left\{ \frac{1}{2}, \frac{7}{6}, \frac{11}{6} \right\}}$$

Practice Problem 7H — slide 178 Q: Find the number of real solutions of the equation $[x] + \{-x\} = 2x$.

Solution:
We know that the fractional part of a negative number is:
$$\{-x\} = \begin{cases} 0, & \text{if } x \in \mathbb{Z} \\ 1 - \{x\}, & \text{if } x \notin \mathbb{Z} \end{cases}$$ This splits the problem into two cases:

1. Case 1: $x \in \mathbb{Z}$ (Integers)
   For $x \in \mathbb{Z}$, we have $[x] = x$ and $\{-x\} = 0$.
   The equation becomes:
   $$x + 0 = 2x \implies x = 0$$ Since $0$ is an integer, this is a valid solution. ✓
2. Case 2: $x \notin \mathbb{Z}$ (Non-Integers)
   Here, $[x] = x - \{x\}$ and $\{-x\} = 1 - \{x\}$.
   The equation becomes:
   $$(x - \{x\}) + (1 - \{x\}) = 2x$$ $$x - 2\{x\} + 1 = 2x \implies 1 - 2\{x\} = x$$ Substitute $x = [x] + \{x\}$:
   $$1 - 2\{x\} = [x] + \{x\}$$ $$\implies 3\{x\} = 1 - [x] \implies \{x\} = \dfrac{1 - [x]}{3}$$ Apply the constraint for non-integers $0 < \{x\} < 1$:
   $$0 < \dfrac{1 - [x]}{3} < 1 \implies 0 < 1 - [x] < 3$$ $$\implies -2 < -[x] < 1 \implies -1 < [x] < 2$$ Since $[x]$ is an integer, the possible values are: - Subcase 2.1: $[x] = 0$
   $$\{x\} = \dfrac{1 - 0}{3} = \dfrac{1}{3} \implies x = 0 + \dfrac{1}{3} = \dfrac{1}{3}$$ *(This is valid because $1/3 \notin \mathbb{Z}$ and $[1/3] = 0$).* ✓ - Subcase 2.2: $[x] = 1$
   $$\{x\} = \dfrac{1 - 1}{3} = 0 \implies x = 1 + 0 = 1$$ *(This is rejected because $1 \in \mathbb{Z}$, which violates the non-integer assumption).* ✗

Therefore, the equation has exactly 2 solutions: $$\boxed{x = 0, \frac{1}{3}}$$

5.4 Signum Function

Definition: Signum Function The Signum Function $f: \mathbb{R} \rightarrow \mathbb{R}$ extracts the sign of a real number: $$ f(x) = \text{sgn}(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases} $$ Or equivalently: $$ f(x) = \text{sgn}(x) = \begin{cases} \dfrac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} $$

• Domain: $\mathbb{R}$
• Range: $\{-1, 0, 1\}$ (Contains exactly three values)

Fig 5.4 — Graph of the Signum Function y = sgn(x) showing open intervals at (0,1) and (0,-1), and a solid dot at (0,0)

5.5 Exponential and Logarithmic Functions

1. Exponential Function: Defined as $f(x) = a^x$ where base $a > 0$ and $a \neq 1$. - Domain: $\mathbb{R}$, Range: $(0, \infty)$ - If $a > 1$, it is strictly increasing (e.g. $y = e^x$). - If $0 < a < 1$, it is strictly decreasing (e.g. $y = (1/2)^x$).

2. Logarithmic Function: Defined as $f(x) = \log_a x$, which is the inverse of the exponential function. It is defined ONLY if: - Base Constraint: $a > 0$ and $a \neq 1$. - Input Constraint: $x > 0$. - Domain: $(0, \infty)$, Range: $\mathbb{R}$.

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6. Counting Functions - Combinatorics (JEE Focus)

Let $n(A) = m$ and $n(B) = n$ (both finite sets).

Important Counting Formulas — Must Memorize ① Total number of functions $f: A \rightarrow B$: $\;n^m$ (each of m elements can go to any of n elements)

② Total number of One-One (Injective) functions:
  If $n \geq m$: $\;^nP_m = \dfrac{n!}{(n-m)!}$
  If $n < m$: $\;0$ (Pigeonhole Principle — cannot inject m elements into fewer slots)

③ Total number of Onto (Surjective) functions:
  If $n > m$: $\;0$ (cannot cover more co-domain elements than domain elements)
  If $n \leq m$: Inclusion-Exclusion Formula: $$ \sum_{r=0}^{n} (-1)^r \,{}^nC_r\, (n-r)^m $$ Special case $n = 2$: $\;2^m - 2$   (simplest and most asked in exams!)

④ Total number of Bijective functions ($m = n$): $\;n!$
  If $m \neq n$: $\;0$.

Practice Problem 10 — JEE Mains Style Q: Let $A = \{1,2,3,4,5\}$ and $B = \{x,y\}$. Find: (a) Total functions, (b) One-One functions, (c) Onto functions.

$m = n(A) = 5$, $n = n(B) = 2$.
(a) Total functions $= n^m = 2^5 = \mathbf{32}$
(b) One-One functions: Since $n < m$ (i.e., $2 < 5$), $=\mathbf{0}$ (Pigeonhole).
(c) Onto functions: $n=2 \leq m=5$, use $2^m - 2 = 2^5 - 2 = 32 - 2 = \mathbf{30}$.

Practice Problem 11 — RD Sharma Counting Q: Let $A = \{1,2,3\}$ and $B = \{4,5,6,7\}$. Find: (a) Total functions, (b) One-One functions, (c) Onto functions.

$m = 3$, $n = 4$.
(a) Total functions $= 4^3 = \mathbf{64}$
(b) One-One functions: $n \geq m$, so $^4P_3 = \dfrac{4!}{(4-3)!} = \dfrac{24}{1} = \mathbf{24}$
(c) Onto functions: $n > m$ ($4 > 3$), so $= \mathbf{0}$.

7. Composition of Functions

Definition: Composition Let $f: A \rightarrow B$ and $g: B \rightarrow C$. The composite function $g \circ f$ is defined as: $$ (g \circ f)(x) = g(f(x)) \quad \forall\; x \in A $$ The composite function $g \circ f : A \rightarrow C$.

Fig 6.1 — Composition g∘f: the red curved arrow shows the direct mapping from A to C, combining f and g

Key Properties of Composition

• Condition for $g \circ f$ to exist: Range of $f \subseteq$ Domain of $g$.
• Non-commutative: In general, $f \circ g \neq g \circ f$.
• Associative: $h \circ (g \circ f) = (h \circ g) \circ f$ (when defined).
• Identity Composition: $f \circ I_A = f = I_B \circ f$.
• Composition preserves bijectivity: If $f$ and $g$ are both bijections, then $g \circ f$ is also a bijection.
• Important: If $g \circ f$ is one-one, then $f$ must be one-one (but $g$ need not be).
• Important: If $g \circ f$ is onto, then $g$ must be onto (but $f$ need not be).

Practice Problem 12 — NCERT Board Level Q: If $f(x) = x^2 - 3x + 2$, find $f(f(x))$.

$f(f(x)) = (f(x))^2 - 3(f(x)) + 2$
$= (x^2-3x+2)^2 - 3(x^2-3x+2) + 2$
Expand $(x^2-3x+2)^2 = x^4 + 9x^2 + 4 - 6x^3 + 4x^2 - 12x = x^4 - 6x^3 + 13x^2 - 12x + 4$
$\therefore\; f(f(x)) = x^4 - 6x^3 + 13x^2 - 12x + 4 - 3x^2 + 9x - 6 + 2$
$= \mathbf{x^4 - 6x^3 + 10x^2 - 3x}$

Practice Problem 13 — RD Sharma / JEE Type Q: If $f: \mathbb{R} \rightarrow \mathbb{R}$ is $f(x) = 2x+1$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ is $g(x) = x^2-2$, find $f \circ g$ and $g \circ f$ and show they are not equal.

$(f \circ g)(x) = f(g(x)) = f(x^2-2) = 2(x^2-2)+1 = 2x^2-4+1 = \mathbf{2x^2-3}$
$(g \circ f)(x) = g(f(x)) = g(2x+1) = (2x+1)^2-2 = 4x^2+4x+1-2 = \mathbf{4x^2+4x-1}$
Since $2x^2-3 \neq 4x^2+4x-1$ in general (e.g., at $x=1$: $-1 \neq 7$), we have $f \circ g \neq g \circ f$.

Practice Problem 14 — RS Aggarwal Type Q: If $f(x) = \dfrac{4x+3}{6x-4}$, $x \neq \dfrac{2}{3}$, show that $f \circ f = I$ (the identity function).

$(f \circ f)(x) = f(f(x)) = f\!\left(\dfrac{4x+3}{6x-4}\right)$

$= \dfrac{4\cdot\dfrac{4x+3}{6x-4}+3}{6\cdot\dfrac{4x+3}{6x-4}-4} = \dfrac{\dfrac{16x+12}{6x-4}+3}{\dfrac{24x+18}{6x-4}-4}$

$= \dfrac{\dfrac{16x+12+18x-12}{6x-4}}{\dfrac{24x+18-24x+16}{6x-4}} = \dfrac{34x+0}{34} = \dfrac{34x}{34} = \mathbf{x}$

Since $(f \circ f)(x) = x = I(x)$, $f \circ f = I$. Hence $f$ is its own inverse.

8. Invertible Functions and Inverse of a Function

Definition: Invertible Function A function $f: A \rightarrow B$ is invertible if there exists a function $g: B \rightarrow A$ such that:
$$ g \circ f = I_A \quad \text{and} \quad f \circ g = I_B $$ The function $g$ is called the inverse of $f$, denoted by $f^{-1}$.

Fundamental Theorem: $f$ is invertible $\iff$ $f$ is bijective (both one-one and onto).

Algorithm to Find $f^{-1}(x)$ — Board Standard Method Step 1: First prove $f$ is bijective.
Step 2: Let $y = f(x)$ (write the function).
Step 3: Solve the equation to get $x$ expressed entirely in terms of $y$. You get $x = h(y)$.
Step 4: Write $f^{-1}(y) = h(y)$.
Step 5: Replace $y$ by $x$ to get the final answer: $f^{-1}(x) = h(x)$.

Fig 7.1 — The graph of f(x) = eˣ and its inverse f⁻¹(x) = ln(x) are mirror images across the line y = x

Properties of Inverse Functions

• $(f^{-1})^{-1} = f$
• Reversal Rule: $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$   (Note the reversal of order!)
• The inverse of a bijection is also a bijection.
• The graph of $f$ and $f^{-1}$ are symmetric about the line $y = x$.
• $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Practice Problem 15 — NCERT Board (5 Marks) Q: Consider $f: \mathbb{R}_+ \rightarrow [4, \infty)$ defined by $f(x) = x^2 + 4$. Show that $f$ is invertible and find $f^{-1}$.

① One-One: Let $f(x_1) = f(x_2) \Rightarrow x_1^2+4 = x_2^2+4 \Rightarrow x_1^2 = x_2^2$. Since $x_1, x_2 \in \mathbb{R}_+$ (positive), $x_1 = x_2$. ✓
② Onto: Let $y \in [4,\infty)$. Set $x^2+4 = y \Rightarrow x = \sqrt{y-4}$ (positive root, since domain = $\mathbb{R}_+$). Since $y \geq 4$, $y-4 \geq 0$, so $x$ is real and positive, i.e., $x \in \mathbb{R}_+$. ✓
$\therefore$ $f$ is bijective, hence invertible.

Finding $f^{-1}$: From above, $x = \sqrt{y-4}$.
$f^{-1}(y) = \sqrt{y-4}$. Replacing $y$ by $x$:
$\boxed{f^{-1}(x) = \sqrt{x-4}, \quad x \in [4,\infty)}$

Practice Problem 16 — RD Sharma Type Q: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x) = \dfrac{2x-7}{4}$. Find $f^{-1}$.

Step 1: (One-One) $f(x_1)=f(x_2) \Rightarrow 2x_1-7=2x_2-7 \Rightarrow x_1=x_2$. ✓
Step 2: (Onto) Let $y = \dfrac{2x-7}{4} \Rightarrow 4y=2x-7 \Rightarrow x = \dfrac{4y+7}{2} \in \mathbb{R}$ for all $y \in \mathbb{R}$. ✓
Step 3: $f^{-1}(y) = \dfrac{4y+7}{2}$. Replace $y$ by $x$:
$\boxed{f^{-1}(x) = \dfrac{4x+7}{2}}$

9. Binary Operations

Note: Binary Operations are an important part of the NCERT syllabus for Class 12 and are regularly asked in both Board exams and RS Aggarwal exercises. RD Sharma also has a dedicated chapter.

Definition: Binary Operation A Binary Operation $*$ on a non-empty set $A$ is a function: $$ * : A \times A \rightarrow A $$ It combines any two elements $a, b \in A$ to produce a unique element $a * b \in A$ (still in the same set $A$). This is called the Closure Property.

9.1 Properties of Binary Operations

Property	Definition	Example
Closure	$a * b \in A$ for all $a,b \in A$	$+$ on $\mathbb{N}$: $3+5=8 \in \mathbb{N}$ ✓
Commutative	$a * b = b * a$ for all $a,b \in A$	$+$ and $\times$ on $\mathbb{R}$ are commutative; $-$ is not
Associative	$(a*b)*c = a*(b*c)$ for all $a,b,c \in A$	$+$ and $\times$ on $\mathbb{R}$ are associative; $-$ is not
Identity Element	$a*e = e*a = a$ for all $a \in A$	$0$ for $+$; $1$ for $\times$
Inverse Element	$a*b = b*a = e$ (identity)	$-a$ for $+$; $1/a$ for $\times$ (if $a\neq 0$)

Important Notes on Identity and Inverse

• The identity element is unique for a given binary operation on a set.
• The inverse of each element is unique (when it exists) if $*$ is associative and has an identity.
• An element may not always have an inverse (e.g., $0$ has no multiplicative inverse in $\mathbb{R}$).
• If an identity exists, then the inverse of the identity element is the identity itself: $e^{-1} = e$.

Practice Problem 17 — NCERT Board Level (5 Marks) Q: Let $*$ be a binary operation on $\mathbb{Q}$ defined by $a * b = \dfrac{ab}{2}$.
(i) Show that $*$ is commutative and associative.
(ii) Find the identity element.
(iii) Find the inverse of a general element $a \in \mathbb{Q}$ ($a \neq 0$).

(i) Commutative: $a*b = \dfrac{ab}{2} = \dfrac{ba}{2} = b*a$. ✓
Associative: $(a*b)*c = \dfrac{ab}{2}*c = \dfrac{\frac{ab}{2} \cdot c}{2} = \dfrac{abc}{4}$
$a*(b*c) = a*\dfrac{bc}{2} = \dfrac{a \cdot \frac{bc}{2}}{2} = \dfrac{abc}{4}$ ✓ Associative.

(ii) Identity element $e$: $a*e = a \Rightarrow \dfrac{ae}{2}=a \Rightarrow e=2$. Check: $2 \in \mathbb{Q}$ ✓. Identity = $\mathbf{2}$.

(iii) Inverse: $a*b = e = 2 \Rightarrow \dfrac{ab}{2} = 2 \Rightarrow ab = 4 \Rightarrow b = \dfrac{4}{a}$.
Inverse of $a$ is $\boxed{\dfrac{4}{a}}$ (defined for $a \neq 0$).

Practice Problem 18 — RD Sharma / RS Aggarwal Type Q: Define $*$ on $\mathbb{Z}$ by $a * b = a + b - ab$. Is $*$ a binary operation? Is it commutative? Find identity. Find inverse (if exists) of $a$.

Binary Operation: $a+b-ab \in \mathbb{Z}$ for all $a,b \in \mathbb{Z}$. ✓ (closure)
Commutative: $a*b = a+b-ab = b+a-ba = b*a$. ✓
Identity $e$: $a*e = a \Rightarrow a+e-ae = a \Rightarrow e-ae=0 \Rightarrow e(1-a)=0$. This gives $e=0$ (for all $a \neq 1$). Check: $a*0 = a+0-0 = a$ ✓. Identity = 0.
Inverse $b$ of $a$: $a*b = 0 \Rightarrow a+b-ab=0 \Rightarrow b(1-a)=-a \Rightarrow b = \dfrac{-a}{1-a} = \dfrac{a}{a-1}$ (provided $a \neq 1$).
The element $1$ has no inverse since the identity condition fails at $a=1$.

Practice Problem 19 — KC Sinha Type (Multiplication Table) Q: Let $A = \{1, \omega, \omega^2\}$ where $\omega$ is a complex cube root of unity. Define $*$ as ordinary multiplication. Verify it is a binary operation and find the identity and inverse of each element.

Multiplication Table (Cayley Table):

$*$	$1$	$\omega$	$\omega^2$
$1$	$1$	$\omega$	$\omega^2$
$\omega$	$\omega$	$\omega^2$	$1$
$\omega^2$	$\omega^2$	$1$	$\omega$

All entries are in $A$ → Closed. The table is symmetric → Commutative.
Identity = 1 (first row = first column pattern).
Inverse of 1 = 1 (since $1*1=1=e$).
Inverse of $\omega$ = $\omega^2$ (since $\omega * \omega^2 = \omega^3 = 1 = e$).
Inverse of $\omega^2$ = $\omega$ (since $\omega^2 * \omega = 1 = e$).

10. Important Results, Theorems & JEE Mains Tricks

10.1 Useful Theorems (NCERT + RD Sharma)

Theorems — Must Know for Boards and JEE

1. If $f: A \rightarrow B$ and $g: B \rightarrow C$ are both one-one, then $g \circ f: A \rightarrow C$ is also one-one.
2. If $f: A \rightarrow B$ and $g: B \rightarrow C$ are both onto, then $g \circ f: A \rightarrow C$ is also onto.
3. If $g \circ f$ is one-one, then $f$ is one-one (but $g$ need not be).
4. If $g \circ f$ is onto, then $g$ is onto (but $f$ need not be).
5. The composition of two bijections is a bijection, and $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.
6. A function $f: A \rightarrow B$ has an inverse $f^{-1}: B \rightarrow A$ if and only if $f$ is bijective.

10.2 Domain Restriction Trick (JEE Mains)

A function that is NOT one-one or NOT onto over a large domain can often be made bijective (and hence invertible) by restricting its domain and/or co-domain.

Practice Problem 20 — Domain Restriction (JEE Type) Q: $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = x^2$ is neither one-one nor onto. Restrict domain and co-domain to make it bijective.

Define $f: [0, \infty) \rightarrow [0, \infty)$, $f(x) = x^2$.
One-One: On $[0,\infty)$, $f$ is strictly increasing, so $x_1 = x_2$ whenever $f(x_1) = f(x_2)$. ✓
Onto: Range $= [0,\infty) =$ Co-domain. ✓
Now $f$ is bijective and $f^{-1}(x) = \sqrt{x}$, $x \geq 0$.

10.3 Common JEE MCQ Trap Questions

⚡ JEE Traps — Be Careful!

1. Is the empty relation on $\{1,2,3\}$ an equivalence relation? — NO. It is symmetric and transitive (vacuously), but NOT reflexive ($(1,1) \notin R$). So NOT an equivalence relation.
2. Is the universal relation an equivalence relation? — YES. It is reflexive, symmetric, and transitive.
3. $f(x) = |x|$: is it one-one? — NO. $f(2) = f(-2) = 2$, but $2 \neq -2$.
4. If $f \circ g$ is bijective, are both $f$ and $g$ bijective? — NOT necessarily. Only $g$ must be one-one and $f$ must be onto.
5. $f(x) = \sin x$, $f: \mathbb{R} \rightarrow \mathbb{R}$: is it onto? — NO. Range $= [-1,1] \neq \mathbb{R}$. But $f: [-\pi/2, \pi/2] \rightarrow [-1,1]$ IS bijective.
6. Identity element of $a*b = a+b+ab$ on $\mathbb{R}$: $a+e+ae = a \Rightarrow e(1+a)=0$. This must hold for all $a$, which is impossible unless $e=0$. But then: $a*0 = a+0+0 = a$ ✓. So identity $= 0$.

11. Chapter Summary at a Glance

Fig 10.1 — Complete Chapter Mind Map: Relations (left branch) and Functions (right branch)

Concept	Key Condition / Formula	Boards Weightage
Total Relations from A to B	$2^{mn}$ (where $m = |A|$, $n = |B|$)	1 Mark MCQ
Equivalence Relation	Reflexive + Symmetric + Transitive	5 Marks
Equivalence Class $[a]$	$\{x \in A : (x,a) \in R\}$	Part of 5 Mark Q
Bijective Function	One-One AND Onto; Range = Co-domain AND unique pre-images	5 Marks
Total Functions (finite sets)	$n^m$	1–2 Marks MCQ
Total One-One Functions	$^nP_m$ (if $n \geq m$) else $0$	1–2 Marks MCQ
Total Onto Functions ($n=2$)	$2^m - 2$	1–2 Marks MCQ
Invertible Function	Bijective only; $f^{-1}(x)$ found by solving $y=f(x)$ for $x$	5 Marks
Composition $g \circ f$	$(g \circ f)(x) = g(f(x))$; Range of $f \subseteq$ Domain of $g$	3–5 Marks
Binary Operation	$*: A \times A \to A$; check closure, identity, inverse, commutativity, associativity	5 Marks

12. Previous Year Board Exam Questions (CBSE)

PYQ 1 — CBSE 2023 (5 Marks) Q: Check whether the relation $R$ defined on the set $A = \{1, 2, 3, 4, 5, 6\}$ as $R = \{(a,b) : b = a + 1\}$ is reflexive, symmetric, or transitive.

Refer to Practice Problem 2 above. $R$ is none of the three.

PYQ 2 — CBSE 2022 (5 Marks) Q: Show that the relation $R$ in $\mathbb{R}$ defined as $R = \{(a,b) : a \leq b\}$ is reflexive and transitive but not symmetric.

Reflexive: $a \leq a$ ✓ for all $a \in \mathbb{R}$. → Reflexive.
Not Symmetric: $1 \leq 2$ but $2 \not\leq 1$. So $(1,2) \in R$ but $(2,1) \notin R$. → Not Symmetric.
Transitive: If $a \leq b$ and $b \leq c$, then $a \leq c$. ✓ → Transitive.

PYQ 3 — CBSE 2020 (5 Marks) Q: Let $A = \{-1, 0, 1, 2\}$, $B = \{-4, -2, 0, 2\}$ and $f, g: A \rightarrow B$ be defined as $f(x) = x^2 - x$ and $g(x) = 2|x-\frac{1}{2}|-1$. Are $f$ and $g$ equal? Justify.

Compute $f$ and $g$ at each element of $A$:
$f(-1) = 1+1=2$,   $g(-1)=2|\!-3/2\!|-1=2(3/2)-1=3-1=2$. ✓
$f(0) = 0-0=0$,   $g(0)=2|{-1/2}|-1=1-1=0$. ✓
$f(1) = 1-1=0$,   $g(1)=2|1/2|-1=1-1=0$. ✓
$f(2) = 4-2=2$,   $g(2)=2|3/2|-1=3-1=2$. ✓
Since $f(x) = g(x)$ for all $x \in A$, $f$ and $g$ are equal functions.

PYQ 4 — CBSE 2019 (3 Marks) Q: If $f: \mathbb{R} \rightarrow \mathbb{R}$ is given by $f(x) = (3-x^3)^{1/3}$, find $f \circ f(x)$.

$f(f(x)) = f((3-x^3)^{1/3})$
$= \Big(3 - \big((3-x^3)^{1/3}\big)^3\Big)^{1/3}$
$= \big(3 - (3-x^3)\big)^{1/3}$
$= (3 - 3 + x^3)^{1/3}$
$= (x^3)^{1/3} = \mathbf{x}$
So $f \circ f = I$ (identity function) → $f$ is its own inverse.

PYQ 5 — CBSE 2018 (5 Marks) Q: Let $*$ be defined on $\mathbb{Q}^+$ (positive rationals) by $a * b = \dfrac{a+b}{2}$. Is $*$ a binary operation? Is it commutative? Associative? Find identity element if it exists.

Binary Operation: For $a,b \in \mathbb{Q}^+$, $\dfrac{a+b}{2} > 0$ and rational → $a*b \in \mathbb{Q}^+$. ✓
Commutative: $a*b = \dfrac{a+b}{2} = \dfrac{b+a}{2} = b*a$. ✓
Not Associative: $(a*b)*c = \dfrac{a+b}{2}*c = \dfrac{\frac{a+b}{2}+c}{2} = \dfrac{a+b+2c}{4}$.
$a*(b*c) = a*\dfrac{b+c}{2} = \dfrac{a+\frac{b+c}{2}}{2} = \dfrac{2a+b+c}{4}$. These are unequal in general. → Not Associative.
Identity: $a*e=a \Rightarrow \dfrac{a+e}{2}=a \Rightarrow a+e=2a \Rightarrow e=a$. But identity must be a fixed element independent of $a$. So no identity element exists.

13. Quick Revision - Formulae & Facts Sheet

🎯 Board Exam Checklist — Before You Write For Proving Equivalence Relation (5 Marks):

1. State all three properties to check.
2. Prove each one formally, not just "it is obvious."
3. Write "Hence $R$ is an equivalence relation" at the end.

For Proving Bijective Function (5 Marks):

1. Prove One-One: "Assume $f(x_1) = f(x_2)$, then show $x_1 = x_2$."
2. Prove Onto: "Let $y \in B$, find $x \in A$ such that $f(x) = y$, verify $x \in$ domain."
3. State: "Since $f$ is one-one and onto, $f$ is bijective."

For Finding Inverse (3–5 Marks):

1. Prove bijectivity first (or state "clearly $f$ is bijective").
2. Set $y = f(x)$, solve for $x$, write $f^{-1}(y)$, replace $y$ with $x$.

All Key Formulas — One Place

Quantity	Formula
Total Relations from A to B	$2^{mn}$
Total Functions from A to B	$n^m$
Total One-One Functions	$^nP_m = \frac{n!}{(n-m)!}$ (if $n \geq m$), else 0
Total Onto Functions ($|B|=2$)	$2^m - 2$
Total Bijective Functions	$n!$ (if $m=n$), else 0
Composition	$(g \circ f)(x) = g(f(x))$
Reversal Rule for Inverse	$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$
Congruence Modulo $n$	$a \equiv b \pmod n \iff n \mid (a-b)$
Equivalence Class	$[a] = \{x \in A : xRa\}$

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