Official Exam Pattern (2026–27)
The course structure groups Relations and Functions with Inverse Trigonometric Functions in Unit I and lists 8 marks for that unit. The question paper design does not guarantee a fixed chapter or unit allocation, so treat 8 as the broad unit total, not a prediction for this chapter alone.
Question paper design
The 3-hour Mathematics (Code 041) theory paper carries 80 marks. CBSE distributes marks by question skill:
| Question skill | Marks | Share |
|---|---|---|
| Remembering and understanding | 44 | 55% |
| Applying | 20 | 25% |
| Analysing, evaluating and creating | 16 | 20% |
There is no overall choice; internal choices make up 33% of the questions in each section. Read the CBSE Mathematics (Code 041) curriculum for 2026–27.
Revision Checklist
Types of Relations
Basic Definitions
Empty Relation
$R = φ \subset A × A$
No element is related to any element
Universal Relation
$R = A × A$
Every element is related to every element
Equivalence Relation
Reflexive, Symmetric & Transitive
Most important for exams!
Reflexive, Symmetric and Transitive Relations
1. Reflexive (Self-loop)
Every element is related to itself.
$(a, a) \in R$ for every $a \in A$
2. Symmetric (Reverse pair)
Whenever $a$ is related to $b$, $b$ is related to $a$.
$(a, b) \in R \Rightarrow (b, a) \in R$
3. Transitive (Two-step chain)
If $a$ is related to $b$ and $b$ to $c$, then $a$ is related to $c$.
$(a, b), (b, c) \in R \Rightarrow (a, c) \in R$
- Reflexive: $(a, a) \in R$ for every $a \in A$.
- Symmetric: If $(a_1, a_2) \in R$, then $(a_2, a_1) \in R$.
- Transitive: If $(a_1, a_2) \in R$ and $(a_2, a_3) \in R$, then $(a_1, a_3) \in R$.
Equivalence Relation
Reflexive AND Symmetric AND Transitive
- "is equal to" in any set
- "is congruent to" for triangles
- "is similar to" for triangles
- "has same number of pages as" for books
- $R = \{(a, b) : |a - b|$ is even$\}$ in integers
Equivalence Class $[a]$
Denoted by $[a] = \{x \in A : (x, a) \in R\}$.
- All elements of $A$ related to each other belong to the same equivalence class.
- No element of one equivalence class is related to any element of another equivalence class.
- Two equivalence classes are either identical or disjoint (no overlapping elements).
- The union of all equivalence classes gives the original set $A$ (Partition of a set).
- Counting Relations on Set $A$ ($n(A) = n$):
• Total relations $= 2^{n^2}$
• Reflexive relations $= 2^{n(n-1)}$
• Symmetric relations $= 2^{\frac{n(n+1)}{2}}$
• Reflexive and Symmetric relations $= 2^{\frac{n(n-1)}{2}}$ - Smallest & Largest Equivalence Relations:
• Smallest equivalence relation is the Identity relation $I_A = \{(a,a) : a \in A\}$ (contains exactly $n$ elements).
• Largest equivalence relation is the Universal relation $A \times A$ (contains $n^2$ elements). - Equivalence Class Fundamental Properties:
• $a \in [a]$ (every element belongs to its own equivalence class).
• $[a] = [b] \iff (a, b) \in R$.
• If $[a] \neq [b]$, then $[a] \cap [b] = \emptyset$ (any two equivalence classes are either identical or disjoint).
• $\bigcup_{a \in A} [a] = A$ (equivalence classes partition the set $A$). - The Vacuous Transitivity MCQ Trap:
A relation $R$ fails transitivity ONLY when there exist pairs $(a, b) \in R$ and $(b, c) \in R$ such that $(a, c) \notin R$. If no such chaining pair exists (e.g. $R = \{(1, 2)\}$ on $\{1, 2, 3\}$), the hypothesis is false, making the condition vacuously TRUE → Transitive.
Quick Check: If $R = \{(1,1), (2,2), (1,2), (2,1)\}$ on $\{1,2\}$, what is $[1]$?
Injective, Surjective and Bijective Functions
1. One-One (Injective)
Distinct inputs have distinct images: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
Injective, but not onto: $d$ is not reached.
2. Onto (Surjective)
Every element of the codomain is an image: $f(A) = B$.
Onto, but not injective: $q$ has two preimages.
A function is bijective when it is both injective and surjective.
An inverse function $f^{-1}: B \to A$ exists exactly when $f: A \to B$ is bijective. If $f$ is only injective, it has an inverse from its image $f(A)$ to $A$.
| Function | Classification | Reason |
|---|---|---|
| $f: \mathbb{R} \to \mathbb{R}$, $f(x)=2x$ or $x^3$ | Bijective | Every real output has exactly one real preimage. |
| $f: \mathbb{R} \to \mathbb{R}$, $f(x)=x^2$ | Many-one; not onto | $f(-1)=f(1)$, and no negative real number is an output. |
| $f: \mathbb{R} \to [0,\infty)$, $f(x)=x^2$ | Onto; not one-one | Every non-negative output is reached, but $f(-1)=f(1)$. |
| $f: \mathbb{R} \to \mathbb{R}$, $f(x)=|x|$ | Many-one; not onto | Opposite inputs have the same image, and negative outputs are missing. |
| $f: \mathbb{R} \to \mathbb{R}$, $f(x)=\sin x$ | Many-one; not onto | $\sin x$ is periodic and its range $[-1,1]$ is not all of $\mathbb{R}$. |
Codomain check: The rule $x^2$ is not onto when its codomain is $\mathbb{R}$, but it is onto when the codomain is $[0,\infty)$. It is not one-one in either case.
Quick Reference: Function Types
| Function Type | Condition | Does $f^{-1}: B \to A$ exist? |
|---|---|---|
| One-One (Injective) | $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ | Only if it is also onto. $f^{-1}:f(A)\to A$ always exists. |
| Onto (Surjective) | Range = Co-domain | Only if it is also one-one. |
| Bijective | Both One-One and Onto | Yes |
| Many-One | Different inputs can give same output | No; the inverse would assign multiple outputs to one input. |
- Counting Functions ($f: A \to B$ with $n(A) = m$ and $n(B) = n$):
• Total functions $= n^m$.
• One-One (Injective) functions: If $m \le n$, total $= {}^n P_m = \frac{n!}{(n-m)!}$. If $m > n$, total $= 0$ (Pigeonhole Principle).
• Onto (Surjective) functions (when $n=2$ and $m \ge 1$): Total $= 2^m - 2$ (exclude the 2 constant functions).
• Bijective functions (when $m = n$): Total $= n!$. If $m \neq n$, total bijections $= 0$. - Horizontal line test: For a real-valued function with a real domain, it is one-one exactly when every horizontal line meets its graph at most once.
- Strict monotonicity implies injectivity: If $f$ is differentiable on an interval and $f'(x) > 0$ throughout it (or $f'(x) < 0$ throughout it), then $f$ is strictly monotone and one-one.
Equivalence Relations on Product Sets and Bijective Functions
Let $A = \mathbb{N} \times \mathbb{N}$. A relation $R$ on $A$ is defined by: $$(a, b) R (c, d) \iff a + d = b + c$$ Show that $R$ is an equivalence relation, and find the equivalence class $[(2, 5)]$.
- 1. Reflexive: For any $(a, b) \in \mathbb{N} \times \mathbb{N}$, we have $a + b = b + a$ (by commutativity of addition in $\mathbb{N}$).
$\therefore (a, b) R (a, b)$. Hence, $R$ is reflexive. - 2. Symmetric: Let $(a, b) R (c, d) \implies a + d = b + c$.
$\implies b + c = a + d \implies c + b = d + a$
$\implies (c, d) R (a, b)$. Hence, $R$ is symmetric. - 3. Transitive: Let $(a, b) R (c, d)$ and $(c, d) R (e, f)$.
Then $a + d = b + c$ … (i) and $c + f = d + e$ … (ii).
Adding (i) and (ii): $(a + d) + (c + f) = (b + c) + (d + e)$.
Cancelling $c + d$ from both sides: $a + f = b + e \implies (a, b) R (e, f)$.
Hence, $R$ is transitive. - Equivalence Class $[(2, 5)]$:
$[(2, 5)] = \{(x, y) \in \mathbb{N} \times \mathbb{N} : (x, y) R (2, 5)\}$
$(x, y) R (2, 5) \iff x + 5 = y + 2 \iff y - x = 3 \iff y = x + 3$.
Since $x, y \in \mathbb{N} = \{1, 2, 3, \dots\}$: $$[(2, 5)] = \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), \dots\}$$
Bijectivity of a Rational Function
• One-One: Let $f(x_1) = f(x_2) \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$.
Cross-multiplying: $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$
$x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$
$-3x_1 - 2x_2 = -3x_2 - 2x_1 \implies -x_1 = -x_2 \implies x_1 = x_2$.
$\therefore f$ is One-One.
• Onto: Let $y \in B = \mathbb{R} - \{1\}$. We need $f(x) = y$:
$y = \frac{x - 2}{x - 3} \implies y(x - 3) = x - 2 \implies xy - 3y = x - 2$
$xy - x = 3y - 2 \implies x(y - 1) = 3y - 2 \implies x = \frac{3y - 2}{y - 1}$.
Since $y \neq 1$, $x$ is well-defined and real. If $x = 3 \implies 3(y - 1) = 3y - 2 \implies -3 = -2$ (impossible).
Thus $x \in A$ for all $y \in B$.
$\therefore$ Every element in co-domain has a pre-image in domain → $f$ is Onto.
• Since $f$ is both One-One and Onto, $f$ is Bijective.
Proof Templates
Test a relation
- Step 1 (Reflexive?):
Check if $(a,a)$ works for every $a$.
Pass: Yes, for all.
Fail: Find just ONE example where it fails. - Step 2 (Symmetric?):
Assume $(a,b) \in R$. Check if $(b,a) \in R$.
Pass: If equation stays same after swapping. - Step 3 (Transitive?):
Assume $(a,b)$ and $(b,c) \in R$. Check $(a,c)$.
Pass: If you can derive a relation between a & c.
Prove injectivity and surjectivity
- Step 1 (One-One?):
Start with: Let $f(x_1) = f(x_2)$
Solve the equation.
Pass: If you get ONLY $x_1 = x_2$.
Fail: If you get $x_1 = ± x_2$ etc. - Step 2 (Onto?):
Put $f(x) = y$.
Find $x$ in terms of $y$.
Pass: If this $x$ is valid in Domain for ALL $y$.
Solved Examples
Q1. Show that the relation $R$ in $\mathbb{Z}$ given by $R = \{(a, b): 2$ divides $a - b\}$ is an equivalence relation. (3 Marks)
For any $a \in \mathbb{Z}$, we have $a - a = 0 = 2 × 0$.
Therefore, 2 divides $(a - a)$, so $(a, a) \in R$.
Hence, $R$ is reflexive.
Let $(a, b) \in R$. Then 2 divides $(a - b)$.
So, $a - b = 2k$ for some integer $k$.
Therefore, $b - a = -2k = 2(-k)$, where $-k$ is an integer.
Hence, 2 divides $(b - a)$, so $(b, a) \in R$.
Therefore, $R$ is symmetric.
Let $(a, b) \in R$ and $(b, c) \in R$.
Then 2 divides $(a - b)$ and 2 divides $(b - c)$.
So, $a - b = 2m$ and $b - c = 2n$ for integers $m$ and $n$.
Adding: $(a - b) + (b - c) = 2m + 2n \Rightarrow a - c = 2(m + n)$.
Hence, 2 divides $(a - c)$, so $(a, c) \in R$.
Therefore, $R$ is transitive.
Conclusion: $R$ is reflexive, symmetric, and transitive. Hence, $R$ is an equivalence relation.
Q2. Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x$ is one-one and onto. (3 Marks)
Let $f(x_1) = f(x_2)$ where $x_1, x_2 \in \mathbb{R}$.
Then $2x_1 = 2x_2$
Dividing both sides by 2: $x_1 = x_2$
Therefore, $f$ is one-one.
Let $y \in \mathbb{R}$ (co-domain) be arbitrary.
We need to find $x \in \mathbb{R}$ such that $f(x) = y$.
From $f(x) = y$, we have $2x = y \Rightarrow x = \frac{y}{2}$.
Since $y \in \mathbb{R}$, we have $\frac{y}{2} \in \mathbb{R}$ (domain).
Therefore, for every $y$ in co-domain, there exists $x = \frac{y}{2}$ in domain such that $f(x) = y$.
Hence, $f$ is onto.
Q3. Show that the relation $R$ in the set $A = \{x \in \mathbb{Z} : 0 \le x \le 12\}$ given by $R = \{(a, b) : |a - b|\text{ is a multiple of } 4\}$ is an equivalence relation. Find the set of all elements related to $1$. (3 Marks)
For any $a \in A$, $|a - a| = 0 = 0 \times 4$, which is a multiple of 4.
$\therefore (a, a) \in R$ for all $a \in A$. Hence, $R$ is reflexive.
Let $(a, b) \in R \implies |a - b| = 4k$ for some integer $k$.
Then $|b - a| = |-(a - b)| = |a - b| = 4k$, which is also a multiple of 4.
$\therefore (b, a) \in R$. Hence, $R$ is symmetric.
Let $(a, b) \in R$ and $(b, c) \in R \implies a - b = \pm 4m$ and $b - c = \pm 4n$.
Adding: $(a - b) + (b - c) = a - c = 4(\pm m \pm n)$, so $|a - c|$ is a multiple of 4 $\implies (a, c) \in R$.
$\therefore R$ is an equivalence relation.
Elements related to 1: $[1] = \{x \in A : |x - 1|\text{ is a multiple of } 4\} = \mathbf{\{1, 5, 9\}}$.
Previous Year Questions (PYQs)
(A) $\{1, 2, 3, ...\}$ (B) $\{2, 3\}$ (C) $\{1, 2, 3\}$ (D) $\{1, 2, 3, 4\}$
Ans: (C) $\{1, 2, 3\}$. For $x + 2y = 8$: when $y = 1, x = 6$; when $y = 2, x = 4$; when $y = 3, x = 2$. So range = $\{1, 2, 3\}$.
Ans:
• Not Reflexive: For $a = -1$, we have $-1 \not\leq (-1)^2 = 1$ is false. So $(-1, -1) \notin R$.
• Not Symmetric: $(1, 2) \in R$ since $1 \leq 4$, but $(2, 1) \notin R$ since $2 \not\leq 1$.
• Not Transitive: $(3, 2) \in R$ and $(2, 1.5) \in R$, but $(3, 1.5) \notin R$ since $3 \not\leq 2.25$.
Ans:
• Not One-One: $f(1) = 1 = f(2)$, but $1 \neq 2$. Hence not one-one.
• Onto: For any $y \in \mathbb{N}$, if $y = 1$, then $f(1) = 1$. If $y > 1$, choose $x = y + 1$, then $f(y + 1) = (y + 1) - 1 = y$. Hence onto.
Solution Outline:
1. Prove one-one by assuming $f(x_1) = f(x_2)$ and showing $x_1 = x_2$
2. Prove onto by taking arbitrary $y \in B$ and finding $x = \frac{3y - 2}{y - 1} \in A$
3. Since bijective, $f^{-1}$ exists: $f^{-1}(y) = \frac{3y - 2}{y - 1}$
Common Mistakes and Exam Checks
Common Mistakes
Proof checklist
Practice Problems
Level 1: Basic (1-2 Marks Each)
Q1. Check whether the relation $R$ in $\{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ is reflexive, symmetric, or transitive.
We need $(1,1), (2,2), (3,3)$.
Are they present? Yes.
$\therefore$ Reflexive.
We have $(1,2)$. Is $(2,1)$ present? No.
We have $(2,3)$. Is $(3,2)$ present? No.
$\therefore$ Not Symmetric.
We have $(1,2)$ and $(2,3)$.
Transitive requires $(1,3)$ to be present.
Is $(1,3) \in R$? No.
$\therefore$ Not Transitive.
Q2. Show that the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x^2$ is one-one but not onto.
Let $f(x_1) = f(x_2)$
$\Rightarrow x_1^2 = x_2^2$
$\Rightarrow x_1 = x_2$ (Since $x \in \mathbb{N}$, no negative numbers allowed)
$\therefore$ One-One.
Co-domain is $\mathbb{N} = \{1, 2, 3, 4, ...\}$.
Range is $\{1^2, 2^2, 3^2, ...\} = \{1, 4, 9, ...\}$.
Are numbers like 2 or 3 in the range? No.
$\therefore$ Range $\neq$ Co-domain.
$\therefore$ Not Onto.
Q3. Show that the relation $R$ in $\mathbb{R}$ defined by $R = \{(a, b) : a \le b^2\}$ is neither reflexive, nor symmetric, nor transitive.
Take $a = \frac{1}{2} \in \mathbb{R}$.
Is $\frac{1}{2} \le (\frac{1}{2})^2$? $\frac{1}{2} \le \frac{1}{4}$ is FALSE.
$\therefore (\frac{1}{2}, \frac{1}{2}) \notin R$. Hence, $R$ is Not Reflexive.
Take $a = 1, b = 2$. Then $1 \le 2^2 = 4$ is true $\implies (1, 2) \in R$.
Now check $(2, 1)$: Is $2 \le 1^2 = 1$? FALSE.
$\therefore (2, 1) \notin R$. Hence, $R$ is Not Symmetric.
Take $a = 2, b = -2, c = -1$.
• $2 \le (-2)^2 = 4$ is true $\implies (2, -2) \in R$.
• $-2 \le (-1)^2 = 1$ is true $\implies (-2, -1) \in R$.
• Is $2 \le (-1)^2 = 1$? FALSE $\implies (2, -1) \notin R$.
$\therefore R$ is Not Transitive.
Level 2: Intermediate (3 Marks Each)
Q4. Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5, 6\}$ given by $R = \{(a, b): |a - b|$ is even$\}$ is an equivalence relation.
$|a - a| = |0| = 0$, which is even.
$\therefore (a, a) \in R$. Reflexive.
Let $|a - b|$ be even.
$|b - a| = |-(a - b)| = |a - b|$, which is also even.
$\therefore (b, a) \in R$. Symmetric.
Let $|a - b|$ (even) and $|b - c|$ (even).
Sum of two even numbers is even: $(a - b) + (b - c) = a - c$.
So $a - c$ is even $\Rightarrow |a - c|$ is even.
$\therefore (a, c) \in R$. Transitive.
Q5. Show that the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ is bijective.
Let $x_1^3 = x_2^3$. Taking cube root on both sides.
$x_1 = x_2$. (No $\pm $ issue for odd powers)
$\therefore$ One-One.
Let $y \in \mathbb{R}$. We need $x^3 = y \Rightarrow x = y^{1/3}$.
Is cube root of any real number real? Yes (e.g., $\sqrt[3]{-8} = -2$).
$\therefore$ Range = $\mathbb{R}$. Onto.
Since One-One & Onto, it is Bijective.
Level 3: Advanced (4-5 Marks Each)
Q6. Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Prove that the function $f: A \to B$ defined by $f(x) = \frac{x - 2}{x - 3}$ is bijective.
Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$.
$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$
$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$
$x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$
$-x_1 = -x_2 \implies x_1 = x_2$.
$\therefore f$ is One-One.
Let $y \in B = \mathbb{R} - \{1\}$. Set $y = f(x) = \frac{x - 2}{x - 3}$.
$y(x - 3) = x - 2 \implies xy - 3y = x - 2$
$x(y - 1) = 3y - 2 \implies x = \frac{3y - 2}{y - 1}$.
Since $y \neq 1$, $x \in \mathbb{R}$. If $x = 3$, $3(y-1) = 3y - 2 \implies -3 = -2$ (not possible), so $x \neq 3$.
Thus $x \in A$. Every $y \in B$ has a pre-image $x \in A$.
$\therefore f$ is Onto.
Conclusion: Since $f$ is one-one and onto, $f$ is Bijective.
Formula Summary
Relations
1. Empty Relation: $R = φ \subset A × A$2. Universal Relation: $R = A × A$
3. Reflexive: $(a, a) \in R$ for all $a \in A$
4. Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$
5. Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$
6. Equivalence Relation: Reflexive + Symmetric + Transitive
Functions
7. One-One: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$8. Onto: Range = Co-domain
9. Bijective: One-One + Onto
10. Composition: $(gof)(x) = g(f(x))$
11. Invertible: $f$ is invertible $\Leftrightarrow$ $f$ is bijective
12. Inverse Property: $(gof)^{-1} = f^{-1}og^{-1}$
13. Identity: $fof^{-1} = I_Y$ and $f^{-1}of = I_X$
Key Theorems:
• The intersection of two equivalence relations on the same set is also an equivalence relation.
• For finite set $A$, $f: A \to A$ is one-one $\Leftrightarrow$ $f$ is onto.
📝 10-Mark Self-Assessment Test
Time: 25 MinsSolve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation