Relations and Functions Class 12: Notes, Methods and Solved Questions

What Relations and Functions is about

Weightage: This chapter is part of Unit I, Relations and Functions, which carries 8 of the 80 theory marks together with Inverse Trigonometric Functions.Source: CBSE Mathematics syllabus 2026-27, Unit I Relations and Functions

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5.1 marks per paper. See the questions

Key concepts

Relations: empty, universal and the three basic properties

NCERT section 1.2

A relation R from A to B is any subset of $A\times B$; a relation on A is a subset of $A\times A$. We write $(a,b)\in R$ or $aRb$.

• Empty relation: no element is related to any element, $R=\varphi$.
• Universal relation: every element is related to every element, $R=A\times A$.

A relation R on A is

• reflexive if $(a,a)\in R$ for every $a\in A$;
• symmetric if $(a,b)\in R\Rightarrow (b,a)\in R$ for all $a,b$;
• transitive if $(a,b)\in R$ and $(b,c)\in R\Rightarrow (a,c)\in R$ for all $a,b,c$.

To prove a property, argue for general elements. To disprove it, one counterexample is enough. If no pair $(a,b),(b,c)$ exists, the transitive condition is satisfied vacuously.

Worked example

Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Is R reflexive, symmetric, transitive? [2 marks]

Try it first, then show the solutionSolution

1. Reflexive: (1, 1), (2, 2), (3, 3) are all in R, so yes.
2. Symmetric: (1, 2) ∈ R but (2, 1) ∉ R, so no.
3. Transitive: (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R, so no.

Answer: Reflexive, but neither symmetric nor transitive.

Checking properties of relations on infinite sets

NCERT section 1.2

For relations defined by a rule on sets such as N, Z or R, test each property with general elements:

• Reflexive: put b = a in the rule; check it holds for every a.
• Symmetric: assume the rule for (a, b); rewrite it to get the rule for (b, a).
• Transitive: assume the rule for (a, b) and (b, c); combine them to get the rule for (a, c).

When a property fails, give specific numbers. Watch for rules like $a\le b^2$ or $a-b\le 2$ that fail for fractions or negative numbers even though they seem to hold for small integers.

Worked example

Check whether the relation $R=\{(a,b):a-b\le 2\}$ on the set of real numbers is reflexive, symmetric or transitive. [3 marks]

Try it first, then show the solutionSolution

1. Reflexive: $a-a=0\le 2$ for every real a, so $(a,a)\in R$: reflexive.
2. Symmetric: $(-5,0)\in R$ since $-5\le 2$, but $(0,-5)\notin R$ since $0-(-5)=5>2$: not symmetric.
3. Transitive: $(4,2)\in R$ ($2\le 2$) and $(2,0)\in R$ ($2\le 2$), but $(4,0)\notin R$ ($4>2$): not transitive.

Answer: Reflexive only; not symmetric, not transitive.

Equivalence relations and equivalence classes

NCERT section 1.2

A relation that is reflexive, symmetric and transitive is an equivalence relation. It divides the set A into disjoint subsets called equivalence classes: the class of a is $[a]=\{x\in A:(x,a)\in R\}$.

• Every element lies in exactly one class.
• Two elements are related if and only if they lie in the same class.
• Different classes have no common element, and together they make up A (a partition).

The standard example: on Z, $R=\{(a,b):a-b\text{ is divisible by }m\}$ is an equivalence relation with m classes, $[0],[1],\dots ,[m-1]$ (numbers with the same remainder on division by m).

Worked example

Show that $R=\{(a,b):a-b\text{ is divisible by }4\}$ is an equivalence relation on Z. Write its equivalence classes. [5 marks]

Try it first, then show the solutionSolution

1. Reflexive: $a-a=0=4\times 0$, divisible by 4, so $(a,a)\in R$ for all $a\in Z$.
2. Symmetric: if $a-b=4k$, then $b-a=4(-k)$, divisible by 4, so $(b,a)\in R$.
3. Transitive: if $a-b=4k$ and $b-c=4l$, then $a-c=4(k+l)$, divisible by 4, so $(a,c)\in R$.
4. Hence R is an equivalence relation.
5. Classes: $[0]=\{\dots ,-8,-4,0,4,8,\dots \}$, $[1]=\{\dots ,-7,-3,1,5,\dots \}$, $[2]=\{\dots ,-6,-2,2,6,\dots \}$, $[3]=\{\dots ,-5,-1,3,7,\dots \}$.

Answer: R is an equivalence relation; its classes are the four remainder classes [0], [1], [2], [3].

One-one (injective) functions

NCERT section 1.3

A function $f:X\to Y$ is one-one (injective) if different elements of X have different images: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$. Otherwise it is many-one.

Method: take $f(x_1)=f(x_2)$, simplify, and show $x_1=x_2$. To disprove, find two different inputs with the same image.

Quick checks: a strictly increasing or strictly decreasing function is one-one. Functions such as $x^2$, $|x|$ and $\cos x$ on all of R are many-one, because $f(-a)=f(a)$ or values repeat. On a graph, a function is one-one if every horizontal line meets the graph at most once. The domain matters: $f(x)=x^2$ is many-one on R but one-one on N.

Worked example

Show that $f:R\to R$, $f(x)=3x-7$ is one-one, and that $g:R\to R$, $g(x)=x^2+1$ is not. [2 marks]

Try it first, then show the solutionSolution

1. $f(x_1)=f(x_2)\Rightarrow 3x_1-7=3x_2-7\Rightarrow x_1=x_2$: f is one-one.
2. $g(2)=5=g(-2)$ with $2\ne -2$: g is not one-one.

Answer: f is one-one; g is many-one.

Onto (surjective) and bijective functions

NCERT section 1.3

A function $f:X\to Y$ is onto (surjective) if every element of Y is the image of some element of X, i.e. the range equals the codomain.

Method: take any y in the codomain, solve $y=f(x)$ for x, and check that this x belongs to the domain (and is defined). If some y has no pre-image, f is not onto (give that y).

A function that is both one-one and onto is bijective (a one-one correspondence).

For a function between finite sets with the same number of elements, one-one and onto are equivalent: a one-one function is automatically onto, and vice versa.

Worked example

Show that $f:R\to R$, $f(x)=3x-7$ is onto, and that $g:R\to R$, $g(x)=x^2+1$ is not onto. [2 marks]

Try it first, then show the solutionSolution

1. For any real y, $y=3x-7$ gives $x=\frac{y+7}{3}$, a real number, and $f\left(\frac{y+7}{3}\right)=y$: f is onto (so f is bijective).
2. $x^2+1\ge 1$ for all real x, so y = 0 (or any y < 1) has no pre-image: g is not onto.

Answer: f is onto (bijective); g is not onto.

Bijectivity of rational functions

NCERT section 1.3

Board questions often give a function such as $f(x)=\frac{ax+b}{cx+d}$ from $R-\{-\frac{d}{c}\}$ to $R-\{\frac{a}{c}\}$ and ask you to prove it is bijective.

One-one: set $f(x_1)=f(x_2)$, cross-multiply, cancel the common terms and arrive at $x_1=x_2$ (needs $ad-bc\ne 0$).

Onto: solve $y=\frac{ax+b}{cx+d}$ for $x=\frac{dy-b}{a-cy}$. This is defined because $y\ne \frac{a}{c}$. Also check that x is not the excluded value $-\frac{d}{c}$ (setting it equal leads to a contradiction).

Worked example

Show that $f:R-\{3\}\to R-\{2\}$, $f(x)=\frac{2x+1}{x-3}$ is one-one and onto. [5 marks]

Try it first, then show the solutionSolution

1. One-one: $\frac{2x_1+1}{x_1-3}=\frac{2x_2+1}{x_2-3}\Rightarrow (2x_1+1)(x_2-3)=(2x_2+1)(x_1-3)$
2. $2x_1{x}_2-6x_1+x_2-3=2x_1{x}_2-6x_2+x_1-3\Rightarrow 7x_2=7x_1\Rightarrow x_1=x_2$: one-one.
3. Onto: let $y\in R-\{2\}$ and $y=\frac{2x+1}{x-3}$. Then $xy-3y=2x+1\Rightarrow x(y-2)=3y+1\Rightarrow x=\frac{3y+1}{y-2}$, defined since $y\ne 2$.
4. Is $x\ne 3$? If $\frac{3y+1}{y-2}=3$ then $3y+1=3y-6$, impossible. So $x\in R-\{3\}$ and $f(x)=y$.
5. Hence f is onto, and therefore bijective.

Answer: f is one-one and onto (bijective).

Counting relations and functions on finite sets

NCERT section 1.2 and 1.3

If A has m elements and B has n elements:

• Number of relations from A to B: $2^{mn}$; relations on A: $2^{m^2}$.
• Number of reflexive relations on A: $2^{m^2-m}$ (the m diagonal pairs are forced in).
• Number of functions from A to B: $n^m$.
• Number of one-one functions from A to B (m ≤ n): $\frac{n!}{(n-m)!}$; none if m > n.
• Number of bijections from A to itself: $m!$.

Onto functions from A to B can exist only if m ≥ n.

Worked example

Let A = {1, 2, 3} and B = {a, b, c, d}. Find (a) the number of functions from A to B, (b) the number of one-one functions from A to B, (c) the number of reflexive relations on A. [2 marks]

Try it first, then show the solutionSolution

1. (a) $4^3=64$
2. (b) $\frac{4!}{1!}=4\times 3\times 2=24$
3. (c) $2^{9-3}=2^6=64$

Answer: (a) 64 (b) 24 (c) 64

Maths in real life: Relations and Functions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Reflexive, Symmetric and Transitive Relations
• Equivalence Relations and Equivalence Classes
• One-One and Onto Functions

Relations and Functions formula sheet

Reflexive

$(a,a)\in R\forall a\in A$

Symmetric

$(a,b)\in R\Rightarrow (b,a)\in R$

Transitive

$(a,b)\in R,(b,c)\in R\Rightarrow (a,c)\in R$

Equivalence class

$[a]=\{x\in A:(x,a)\in R\}$

One-one

$f(x_1)=f(x_2)\Rightarrow x_1=x_2$

Onto

$\forall y\in Y\exists x\in X:f(x)=y,\text{range}=\text{codomain}$

Number of relations

$\text{from }A\text{ to }B:2^{mn},\text{on }A:2^{m^2}$

Number of reflexive relations on A

$2^{m^2-m}$

Number of functions

$n^m$

Number of one-one functions

$\frac{n!}{(n-m)!}(m\le n)$

Bijections from A to A

$m!$

How the board asks Relations and Functions

The question patterns to expect, and the method that works for each.

Tricks to remember Relations and Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

• RST = Equivalence

  Reflexive, Symmetric, Transitive: all three together make an equivalence relation.

  Use it when: Recalling the definition.

• Prove generally, disprove with one example

  To prove a property use letters; to disprove it, one counterexample with numbers is enough.

  Use it when: Every relation and function check.

• Onto = solve for x and check

  Solve $y=f(x)$ for x; then check two things: x exists (no division by zero) and x is in the domain.

  Use it when: Onto proofs.

• Squares and modulus are two-faced

  $x^2$, $|x|$ and even functions give the same output for x and −x, so they are not one-one on R.

  Use it when: Fast counterexamples.

• 2 to the power of the pairs

  Relations on A: $2^{m^2}$; reflexive ones: drop the m diagonal pairs from the exponent, $2^{m^2-m}$.

  Use it when: Counting MCQs.

Relations and Functions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Relations and Functions

The slips examiners see most often, and what to write instead.

• ✗Mistake: Checking reflexivity on only a few elements

  Testing (1, 1) and (2, 2) and stopping.

  ✓Do this: Reflexive needs $(a,a)\in R$ for every element of A; write it for a general a (or check every element of a finite set).

  Usually costs: 1 mark.

• ✗Mistake: Giving a general argument where a property fails

  Not looking for a counterexample.

  ✓Do this: If a property is false, show one specific pair (with numbers) that breaks it.

  Usually costs: 1-2 marks.

• ✗Mistake: Proving onto without checking that x is in the domain

  Stopping after solving for x.

  Wrong working

  $f:N\to N$, $f(x)=2x$: $x=\frac{y}{2}$, so onto

  Right working

  For y = 3, $x=1.5\notin N$, so f is not onto

  ✓Do this: Always check the solved x belongs to the domain.

  Usually costs: Full marks for onto.

• ✗Mistake: Ignoring the excluded value in a rational function

  Forgetting the codomain restriction.

  ✓Do this: For $f(x)=\frac{ax+b}{cx+d}$, show the solved x is defined (y ≠ a/c) and is not the excluded point.

  Usually costs: 1 mark.

• ✗Mistake: Writing about composition or inverse functions

  Old syllabi included them.

  ✓Do this: Composition of functions and invertible functions are not in the 2026-27 syllabus.

  Usually costs: Time.

NCERT exercise guide: Relations and Functions

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Relations and Functions: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5.1 marks per paper; 19 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Relations and Functions

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Relations and Functions question, by year

Practice questions on Relations and Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The number of relations on a set with 2 elements is:
   (a) 4
   (b) 8
   (c) 16
   (d) 2

   Hint

   $2^{n^2}$.

   Answer

   (c) 16

2. Q2 · 1 mark · MCQBoard level

   Which of these functions from R to R is one-one?
   (a) $f(x)=x^2$
   (b) $f(x)=|x|$
   (c) $f(x)=x^3$
   (d) $f(x)=\cos x$

   Hint

   Look for a strictly increasing function.

   Answer

   (c) $x^3$

3. Q3 · 1 mark · Very short answerBoard level

   Let A = {1, 2, 3} and R = {(1, 2), (2, 1)}. Is R symmetric? Is it transitive?

   Hint

   For transitivity check (1, 2), (2, 1).

   Answer

   Symmetric: yes. Transitive: no, since (1, 2), (2, 1) ∈ R but (1, 1) ∉ R.

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): The relation R = {(1, 1), (2, 2), (3, 3)} on A = {1, 2, 3} is an equivalence relation.
   Reason (R): The identity relation on any set is reflexive, symmetric and transitive.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   R is the identity relation on A.

   Answer

   (a)

5. Q5 · 2 marks · Very short answerHarder board level

   Show that $f:N\to N$, $f(x)=2x$ is one-one but not onto.

   Hint

   Odd numbers.

   Answer

   $2x_1=2x_2\Rightarrow x_1=x_2$: one-one. 3 ∈ N has no pre-image (x would be 1.5): not onto.

6. Q6 · 2 marks · Very short answerHarder board level

   Check whether $f:R\to R$, $f(x)=x^2-4x+5$ is one-one and onto.

   Hint

   Complete the square.

   Answer

   $f(x)=(x-2{)}^2+1$. $f(1)=2=f(3)$: not one-one. $f(x)\ge 1$, so 0 has no pre-image: not onto.

7. Q7 · 3 marks · Short answerHarder board level

   Show that the relation R on Z given by $R=\{(a,b):5\text{ divides }a-b\}$ is an equivalence relation. Write the class [2].

   Hint

   Reflexive, symmetric, transitive.

   Answer

   Reflexive: 5 | 0. Symmetric: 5 | (a − b) ⇒ 5 | (b − a). Transitive: 5 | (a − b) and 5 | (b − c) ⇒ 5 | (a − c). [2] = {…, −8, −3, 2, 7, 12, …}.

8. Q8 · 3 marks · Short answerHarder board level

   Show that $f:R\to R$ defined by $f(x)=4x^3+7$ is a bijection.

   Hint

   Cube root.

   Answer

   One-one: $4x_1^3+7=4x_2^3+7\Rightarrow x_1^3=x_2^3\Rightarrow x_1=x_2$. Onto: $x=\sqrt[3]{\frac{y-7}{4}}$ is real for every real y and f(x) = y.

9. Q9 · 5 marks · Long answerHarder board level

   Let R be the relation on the set of real numbers defined by $R=\{(a,b):1+ab>0\}$. Show that R is reflexive and symmetric but not transitive.

   Hint

   Try a = 1, b = −1/2, c = −4.

   Answer

   Reflexive: $1+a^2>0$. Symmetric: $1+ab=1+ba$. Not transitive: $(1,-\frac{1}{2})\in R$ since $1-\frac{1}{2}>0$; $(-\frac{1}{2},-4)\in R$ since $1+2>0$; but $(1,-4)\notin R$ since $1-4<0$.

10. Q10 · 2 marks · Very short answerHOTS / competency

    Find the number of one-one functions and the number of onto functions from A = {1, 2, 3} to B = {x, y}. Explain.

    Hint

    Compare the sizes.

    Answer

    One-one: 0, since 3 elements cannot map to 2 distinct images. Onto: total functions $2^3=8$, minus the 2 constant functions = 6.

11. Q11 · 3 marks · Short answerHOTS / competency

    Show that $f:[-1,1]\to R$, $f(x)=\frac{x}{x+2}$ is one-one. Is it onto R? If not, find its range.

    Hint

    f is increasing on its domain.

    Answer

    One-one: $\frac{x_1}{x_1+2}=\frac{x_2}{x_2+2}\Rightarrow 2x_1=2x_2$. Not onto R: for example y = 1 would need $x=x+2$. Since f is increasing ($f(x)=1-\frac{2}{x+2}$), its range is $[f(-1),f(1)]=[-1,\frac{1}{3}]$.

12. Q12 · 2 marks · Very short answerHarder board level

    Give an example of a relation on A = {1, 2, 3} that is (a) symmetric and transitive but not reflexive, (b) reflexive and symmetric but not transitive.

    Hint

    Leave out one diagonal pair for (a).

    Answer

    (a) {(1, 1), (1, 2), (2, 1), (2, 2)} (misses (3, 3)). (b) {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} ((1, 2), (2, 3) present but (1, 3) missing).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define an equivalence relation.Show answer

   A relation that is reflexive, symmetric and transitive.

2. 2.When is a function one-one?Show answer

   When $f(x_1)=f(x_2)$ implies $x_1=x_2$.

3. 3.When is a function onto?Show answer

   When every element of the codomain has a pre-image (range = codomain).

4. 4.Number of relations on a set with 3 elements?Show answer

   $2^9=512$.

5. 5.Number of reflexive relations on a set with 2 elements?Show answer

   $2^{4-2}=4$.

6. 6.Is $f:N\to N$, $f(x)=x^2$ one-one? Onto?Show answer

   One-one (x > 0); not onto (2 has no pre-image).

7. 7.How many bijections are there from {1, 2, 3, 4} to itself?Show answer

   4! = 24.

8. 8.Write the equivalence class [1] for 'a − b divisible by 3' on Z.Show answer

   {…, −5, −2, 1, 4, 7, …}.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state reflexive, symmetric and transitive in symbols?
• ☐Can you check all three properties for a relation given as a list of pairs?
• ☐Can you disprove a property with a counterexample on R?
• ☐Can you prove that 'a − b divisible by m' is an equivalence relation and list its classes?
• ☐Can you prove that a relation on pairs such as (a, b)R(c, d) ⇔ a + d = b + c is an equivalence relation?
• ☐Can you show a linear or cubic function on R is one-one and onto?
• ☐Can you show a quadratic or modulus function on R is not one-one and not onto?
• ☐Can you prove a rational function between restricted sets is bijective?
• ☐Can you count relations, reflexive relations, functions and one-one functions?
• ☐Can you explain why one-one and onto are equivalent for functions between finite sets of the same size?

Relations and Functions: frequently asked questions