NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Understanding relations and functions in mathematics is like learning to navigate a map – it helps us move from one point to another with purpose. During the NCERT Solutions of relations and functions class 12, the first question that comes to mind is, what exactly are relations and functions? A relation is like a connection between a student and all the subjects in the syllabus. On the other hand, a function is like a special subject, which is the student's only favourite subject. The main purpose of these relations and functions class 12 NCERT solutions is to make learning easier for students and to explain this topic more easily.

Latest:  JEE Main high scoring chapters | JEE Main 10 year's papers

This Story also Contains

1. NCERT Solution for Class 12 Maths Chapter 1 Solutions: Download PDF
2. Relations and Functions Class 12 Solutions - Important Formulae
3. NCERT Solutions for Class 12 Maths Chapter 1: Exercise Questions
4. Class 12 Maths NCERT Chapter 1: Extra Question
5. Approach to Solve Questions of Relations and Functions Class 12
6. What Extra Should Students Study Beyond the NCERT for JEE?
7. NCERT Solutions for Class 12 Maths: Chapter Wise

LiveCBSE Result 2025 (LIVE): Class 10, 12 marksheet likely today at cbse.gov.in, Digilocker; websites to checkMay 12, 2025 | 10:48 AM IST

Few months back, the Central Board of Secondary Education (CBSE) has made some changes in the curriculum and textbook for applied mathematics and has also plans of allowing the use of calculator in the CBSE Class 12 accounts subject. 

Read More

People use relations and functions in real-life applications, such as mapping, programming, and data handling. These solutions by the NCERT are essential for students because they comprise quality practice questions. After checking the Class 12 Maths NCERT solutions from textbooks, students can also check the NCERT Exemplar Solutions for Class 12 Maths Chapter 1 Relations and Functions to understand this chapter better.

NCERT Solution for Class 12 Maths Chapter 1 Solutions: Download PDF

Students who wish to access the Class 12 Maths Chapter 1 NCERT Solutions can click on the link below to download the complete solution in PDF.

Download PDF

>> Relations:

A relation $R$ is a subset of the Cartesian product of $A\times B$, where $A$ and $B$ are non-empty sets.
$R^{-1}$, the inverse of relation $R$, is defined as:

$R^{-1}=\{(b,a):(a,b)\in R\}$

- Domain of $R=$ Range of $R^{-1}$
- Range of $R=$ Domain of $R^{-1}$

>> Functions:

A relation $f$ from set A to set B is a function if every element in A has one and only one image in B.

$A\times B=\{(a,b):a\in A,b\in B\}$
If $(a,b)=(x,y)$, then $a=x$ and $b=y$
$n(A\times B)=n(A)$ * $n(B)$, where $n(A)$ is the cardinality (number of elements) of set $A$.
$\mathrm{A}\times \varphi =\varphi$ (where $\varphi$ is the empty set)
A function $f:A\to B$ is denoted as:

$f(x)=y$

Algebra of functions:

- $(f+g)(x)=f(x)+g(x)$
- $(f-g)(x)=f(x)-g(x)$
- $\left(f{}^{\ast }g\right)(x)=f(x{)}^{\ast }g(x)$
- $(kf)(x)=k^{\ast }f(x)$, where $k$ is a real number
- $(f/g)(x)=f(x)/g(x)$, where $g(x)\ne 0$

NCERT Solutions for Class 12 Maths Chapter 1: Exercise Questions

Question 1(i): Determine whether each of the following relations are reflexive, symmetric, and transitive:

(i) Relation $R$ in the set $A=\{1,2,3\dots ,13,14\}$ defined as $R=\{(x,y):3x-y=0\}$

Answer:

A={1,2,3...,13,14}

R={(x,y):3x−y=0} ={(1,3),(2,6),(3,9),(4,12)}

Since, (1,1),(2,2),(3,3),(4,4),(5,5)⋅⋅⋅⋅⋅⋅(14,14)∉R so R is not reflexive.

Since (1,3)∈R but (3,1)∉R, R is not symmetric.

Since, (1,3),(3,9)∈R but (1,9)∉R, so R is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive.

Question 1(ii): Determine whether each of the following relations are reflexive, symmetric, and transitive:

(ii) Relation R in the set N of natural numbers defined as

$R=\{(x,y):y=x+5\text{ and }x<4\}$

Answer:

R={(x,y):y=x+5andx<4} ={(1,6),(2,7),(3,8)}

Since (1,1)∉R

So R is not reflexive.

Since, (1,6)∈R but (6,1)∉R

So R is not symmetric.

Since there is no pair in R such that (x,y),(y,x)∈R so this is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive.

Question 1 (iii): Determine whether each of the following relations is reflexive, symmetric, and Transitive:

(iii) Relation R in the set A={1,2,3,4,5,6} as R={(x,y):y is divisible by x}

Answer:

A={1,2,3,4,5,6}

R={(2,4),(3,6),(2,6),(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Any number is divisible by itself, and (x,x)∈R. So it is reflexive.

(2,4)∈R but (4,2)∉R .Hence,it is not symmetric.

(2,4),(4,4)∈R and 4 is divisible by 2 and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

Question 1(iv): Determine whether each of the following relations are reflexive, symmetric, and Transitive:

(iv). Relation R in the set Z of all integers defined as R={(x,y):x−y is an integer}

Answer:

R={(x,y):x−y is an integer}

For x∈Z, (x,x)∈R as x−x=0, which is an integer.

So, it is reflexive.

For x,y∈Z, (x,y)∈R and (y,x)∈R because x−y and y−x are both integers.

So, it is symmetric.

For x,y,z∈Z, (x,y),(y,z)∈R as x−y and y−z are both integers.

Now, x−z=(x−y)+(y−z) is also an integer.

So, (x,z)∈R and hence it is transitive.

Hence, it is reflexive, symmetric, and transitive.

Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and Transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R={(x,y):x and y work at the same place}

Answer:

R={(x,y):x and y work at the same place}

(x,x)∈R ,so it is reflexive

(x,y)∈R means x and y work at the same place.

y and x work at the same place, i.e. (y,x)∈R, so it is symmetric.

(x,y),(y,z)∈R means x and y work at the same place, and also y and z work at the same place. It states that x and z work at the same place, i.e. (x,z)∈R. So, it is transitive.

Hence, it is reflexive, symmetric, and transitive.

Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and Transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(b) R={(x,y):x and y live in the same locality}

Answer:

R={(x,y):x and y live in the same locality}

(x,x)∈R as x and x is same human being.So, it is reflexive.

(x,y)∈R means x and y live in the same locality.

It is the same as y and x live in the same locality, i.e. (y,x)∈R.

So, it is symmetric.

(x,y),(y,z)∈R means x and y live in the same locality, and y and z live in the same locality.

It implies that x and z live in the same locality, i.e. (x,z)∈R.

Hence, it is reflexive, symmetric, and transitive.

Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(c) R={(x,y):x is exactly 7cm taller than y}

Answer:

R={(x,y):x is exactly 7cm taller than y}

(x,y)∈R means x is exactly 7cm taller than y, but x is not taller than x, i.e. (x,x)∉R. So, it is not reflexive.

(x,y)∈R means x is exactly 7cm taller than y, but y is not taller than x, i.e. (y,x)∉R. So, it is not symmetric.

(x,y),(y,z)∈R means x is exactly 7cm taller than y, and y is exactly 7cm taller than z.

x is exactly 7cm taller than z, i.e. (x,z)∉R.

Hence, it is not reflexive, not symmetric, and not transitive.

Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and Transitive:

(v). Relation R in the set A of human beings in a town at a particular time given by

(d) R={(x,y):x is wife of y}

Answer:

R={(x,y):x is wife of y}

(x,y)∈R means x is the wife of y, but x is not the wife of x, i.e. (x,x)∉R.

So, it is not reflexive.

(x,y)∈R means x is wife of y but y is not wife of x i.e. (y,x)∉R .

So, it is not symmetric.

Let (x,y),(y,z)∈R mean x is the wife of y and y is the wife of z.

This case is not possible, so it is not transitive.

Hence, it is not reflexive, symmetric, or transitive.

Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(e) R={(x,y):x is father of y}

Answer:

R={(x,y):x is father of y}

(x,y)∈R means x is the father of y, then x cannot be the father of x, i.e. (x,x)∉R. So, it is not reflexive.

(x,.∈R means x is the father of y, then y cannot be the father of x, i.e. (y,x)∉R. So, it is not symmetric.

Let (x,y),(y,z)∈R mean x is the father of y and y is the father of z, then x cannot be the father of z, i.e. (x,z)∉R.

So, it is not transitive.

Hence, it is neither reflexive nor symmetric nor transitive

Question 2: Show that the relation R in the set R of real numbers is defined as
R={(a,b): a≤b^2} is neither reflexive nor symmetric nor transitive.

Answer:

$R=\{(a,b):a\le b^2\}$
Taking

$(\frac{1}{2},\frac{1}{2})\notin R$

and

$\left(\frac{1}{2}\right)>{\left(\frac{1}{2}\right)}^2$
So,R is not reflexive.
Now,
$(1,2)\in R$ because $1<4$.
But, $4\nless 1$, i.e. 4 is not less than 1
So, $(2,1)\notin R$
Hence, it is not symmetric.
$(3,2)\in R$ and $(2,1.5)\in R$ as $3<4$ and $2<2.25$
Since $(3,1.5)\notin R$ because $3\nless 2.25$
Hence, it is not transitive.
Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.

Question 3: Check whether the relation R defined in the set {1,2,3,4,5,6} as
R={(a,b):b=a+1} is reflexive, symmetric or transitive.

Answer:

R is defined in the set {1,2,3,4,5,6}

R={(a,b):b=a+1}

R={(1,2),(2,3),(3,4),(4,5),(5,6)}

Since, {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}∉R so it is not reflexive.

{(1,2),(2,3),(3,4),(4,5),(5,6)}∈R but {(2,1),(3,2),(4,3),(5,4),(6,5)}∉R

So, it is not symmetric

{(1,2),(2,3),(3,4),(4,5),(5,6)}∈R but {(1,3),(2,4),(3,5),(4,6)}∉R

So, it is not transitive.

Hence, it is neither reflexive, nor symmetric, nor transitive.

Question 4: Show that the relation R in R defined as R={(a,b): a≤b}, is reflexive and transitive but not symmetric.

Answer:

R={(a,b):a≤b}

As (a,a)∈R so I t is reflexive.

Now, we take an example

(2,3)∈R as 2<3

But (3,2)∉R because 2≮3.

So, it is not symmetric.

Now, if we take (2,3)∈Rand(3,4)∈R

Then, (2,4) because 2<4

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question 5: Check whether the relation R in R is defined by R={(a,b): $a\le b^3\}$ is reflexive, symmetric or transitive.

Answer:

$R=\{(a,b):a\le b^3\}$

$(\frac{1}{2},\frac{1}{2})\notin R$ because $\frac{1}{2}\notin {\left(\frac{1}{2}\right)}^3$
So, it is not symmetric.

Now, $(1,2)\in R$ because $1<2^3$
but $(2,1)\notin R$ because $2\nless 1^3$
It is not symmetric
$(3,1.5)\in R$ and $(1.5,1.2)\in R$ as $3<{1.5}^3$ and $1.5<{1.2}^3$.
But, $(3,1.2)\notin R$ because $3\nless {1.2}^3$
So it is not transitive
Thus, it is neither reflexive, nor symmetric, nor transitive.

Question 6: Show that the relation R in the set {1,2,3} given by R={(1,2),(2,1)} is symmetric but neither reflexive nor transitive.

Answer:

Let A= {1,2,3}

R={(1,2),(2,1)}

We can see (1,1),(2,2),(3,3)∉R so it is not reflexive.

As (1,2)∈Rand(2,1)∈R, so it is symmetric.

(1,2)∈Rand(2,1)∈R

But (1,1)∉R, so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question 7: Show that the relation R in the set A of all the books in a library of a college, given by R={(x,y):x and y have the same number of pages}, is an equivalence relation?

Answer:

A = all the books in the library of a college

$R=\{(x,y):x$ and $y$ have same number of pages $\}$

$(x,x)\in R$ because x and x have same number of pages so it is reflexive.

Let $(x,y)\in R$ means x and y have the same number of pages.

Since y and x have the same number of pages, so $(y,x)\in R$.

Hence, it is symmetric.

Let $(x,y)\in R$ means x and y have the same number of pages.

And $(y,z)\in R$ means y and z have the same number of pages.

This states, x and z also have same number of pages i.e. $(x,z)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric, and transitive, i.e. it is an equivalence relation.

Question 8: Show that the relation R in the set A={1,2,3,4,5} given by R={(a,b):|a−b| is even}, is an equivalence relation. Show that all the elements of {1,3,5} are related to each other and all the elements of {2,4} are related to each other. But no element of {1,3,5} is related to any element of {2,4}.

Answer:

A={1,2,3,4,5}

R={(a,b):|a−b| is even}

R={(1,1),(2,2),(3,3),(4,4),(5,5),(1,3),(2,4),(3,5),(3,1),(5,1),(4,2),(5,3)}

Let there be a∈A, then (a, a)∈R as | a−a|=0, which is an even number. Hence, it is reflexive

Let (a,b)∈R where a,b∈A then (b,a)∈R as |a−b|=|b−a|

Hence, it is symmetric

Now, let (a,b)∈Rand(b,c)∈R

|a−b| and |b−c| are even number i.e. (a−b)and(b−c) are even

then, (a−c)=(a−b)+(b−c) is even (sum of even integer is even)

So, (a,c)∈R. Hence, it is transitive.

Thus, it is reflexive, symmetric, and transitive, i.e. it is an equivalence relation.

The elements of {1,3,5} are related to each other because the difference of odd numbers gives even numbers er and in this set, all numbers are odd.

The elements of {2,4} are related to each other because the difference of even numbers is an even number er and in this set, all numbers are even.

The element of {1,3,5} is not related to {2,4} because the difference of odd and even numbers is even.

Question 9(i): Show that each of the relation R in the set A={x∈Z:0≤x≤12}, given by (i) R={(a,b):|a−b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

A={x∈Z:0≤x≤12}

A={0,1,2,3,4,5,6,7,8,9,10,11,12}

R={(a,b):|a−b| is a multiple of 4}

For a∈A, (a, a)∈ as |a−a|=0, which is a multiple of 4.

Hence, it is reflexive.

Let (a,b)∈R, i.e. |a−b| be a multiple of 4.

then |b−a| is also multiple of 4 because |a−b| = |b−a| i.e. (b,a)∈R

Hence, it is symmetric.

Let (a,b)∈R, i.e. |a−b| be a multiple of 4 and (b,c)∈R, i.e. |b−c| be a multiple of 4.

(a−b) is a multiple of 4, and (b−c) is a multiple of 4

(a−c)=(a−b)+(b−c) is multiple of 4

|a−c| is a multiple of 4, i.e. (a,c)∈R

Hence, it is transitive.

Thus, it is reflexive, symmetric, and train, positive, i.e. it is an equivalence relation.

The set of all elements related to 1 is {1,5,9}

|1−1|=0 is a multiple of 4.

|5−1|=4 is a multiple of 4.

|9−1|=8 is a multiple of 4.

Question 9(ii): Show that each of the relations R in the set A={x∈Z:0≤x≤12}, given by (ii) R={(a,b):a=b}, is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

A={x∈Z:0≤x≤12}

A={0,1,2,3,4,5,6,7,8,9,10,11,12}

R={(a,b):a=b}

For a∈A , (a,a)∈R as a=a

Hence, it is reflexive.

Let (a,b)∈R, i.e. a=b

a=b ⇒ b=a i.e. (b,a)∈R

Hence, it is symmetric.

Let, (a,b)∈R i.e. a=b and (b,c)∈R i.e. b=c

∴ a=b=c

a=c i.e. (a,c)∈R

Hence, it is transitive.

Thus, it is reflexive, symmetric, and transitive, i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

Question 10 (i): Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Answer:

Let

A={1,2,3}

R={(1,2),(2,1)}

(1,1),(2,2),(3,3)∉R so it is not reflexive.

(1,2)∈R and (2,1)∈R, so it is symmetric.

(1,2)∈Rand(2,1)∈R but (1,1)∉R so it is not transitive.

Hence, symmetric but neither reflexive nor transitive.

Question 10 (ii): Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Answer:

Let

R={(x,y):x>y}

Now for x∈R , (x,x)∉R so it is not reflexive.

Let (x,y)∈R i.e. x>y

Then y>x is not possible, i.e. (y,x)∉R. So it is not symmetric.

Let (x,y)∈R i.e. x>y and (y,z)∈R i.e. y>z

we can write this as x>y>z

Hence, x>z, i.e. (x,z)∈R. So it's transitive.

Hence, it is transitive but neither reflexive nor symmetric.

Question 10 (iii): Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Answer:

Let

A={1,2,3}

Define a relation R on A as

R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}

If x∈A , (x,x)∈R i.e. {(1,1),(2,2),(3,3)}∈R . So it is reflexive.

If x,y∈A , (x,y)∈R and (y,x)∈R i.e. {(1,2),(2,1),(2,3),(3,2)}∈R . So it is symmetric.

(x,y)∈R and (y,z)∈R i.e. (1,2)∈R . and (2,3)∈R

But (1,3)∉R, so it is not transitive.

Hence, it is Reflexive and symmetric but not transitive.

Question 10 (iv): Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Answer:

Let there be a relation R in R

R={(a,b):a≤b}

(a, a)∈R because a=a

Let (a,b)∈R, i.e. a≤b

But (b, a)∉R, i.e. b⪇a

So it is not symmetric.

Let (a,b)∈R i.e. a≤b and (b,c)∈R i.e. b≤c

This can be written as a≤b≤c i.e. a≤c implies (a,c)∈R

Hence, it is transitive.

Thus, it is Reflexive and transitive but not symmetric.

Question 10 (v): Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Answer:

Let there be a relation A in R

A={1,2}

R={(1,2),(2,1),(2,2)}

(1,1)∉R So R is not reflexive.

We can see (1,2)∈R and (2,1)∈R

So it is symmetric.

Let (1,2)∈R and (2,1)∈R

Also (2,2)∈R

Hence, it is transitive.

Thus, it Symmetric and transitive but not reflexive.

Question 11: Show that the relation R in the set A of points in a plane given by R={(P, Q): distance of the point P from the origin is the same as the distance of the point Q from the origin}, is an equivalence relation. Further, shows that the set of all points related to a point P≠(0,0) is the circle passing through P with the origin as the centre.

Answer:

$R=\{(P,Q)$: distance of the point $P$ from the origin is the same as the distance of the point $Q$ from the origin $\}$

The distance of point P from the origin is always the same as the distance of the same point P from another origin, i.e. $(P,P)\in R$

$\therefore \mathrm{R}$ is reflexive.

Let $(P,Q)\in R$, i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

This is the same as the distance of point Q from the origin, same as the distance of point P from the origin, i.e. $(Q,P)\in R$

$\therefore \mathrm{R}$ is symmetric.

Let $(P,Q)\in R$ and $(Q,S)\in R$

i.e. distance of the point P from the origin is the same as the distance of the point $Q$ from the origin, and also the distance of the point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of points $P,Q,S$ from the origin is the same. This means a distance of point P from the origin is the same as the distance of point S from the origin, i.e. $(P,S)\in R$

$\therefore \mathrm{R}$ is transitive.

Hence, $R$ is an equivalence relation.

The set of all points related to a point $P\ne (0,0)$ are points whose distance from the origin is the same as the distance of point $P$ from the origin.

In other words, we can say there be a point $0(0,0)$ as the origin and the distance between point 0 and point $P$ be $k=OP$; then the set of all points related to $P$ is at a distance $k$ from the origin.

Hence, this set of points forms a circle with the centre as the origin, and this circle passes through point $P$.

Question 12: Show that the relation R defined in the set A of all triangles as R={(T1, T2): T1 is similar to T2} is an equivalence relation. Consider three right-angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13, and T3 with sides 6, 8, 10. Which triangles among T 1 , T 2, and T 3 are related?

Answer:

$R=\{(T_1,T_2):T_1\text{ is similar to }{T}_2\}$

All triangles are similar to themselves, so it is reflexive.

Let,

$(T_1,T_2)\in R$ i.e. ${\mathrm{T}}_1$ is similar to ${\mathrm{T}}_2$

${\mathrm{T}}_1$ is similar to T 2 ; T 2 is similar to T 1 , i.e. $(T_2,T_1)\in R$

Hence, it is symmetric.

Let,

$(T_1,T_2)\in R$ and $(T_2,T_3)\in R$ i.e. ${\mathrm{T}}_1$ is similar to ${\mathrm{T}}_2$ and ${\mathrm{T}}_2$ is similar to ${\mathrm{T}}_3$.

$\Rightarrow {\mathrm{T}}_1$ is similar to ${\mathrm{T}}_3$ i.e. $(T_1,T_3)\in R$

Hence, it is transitive,

Thus, $R=\{(T_1,T_2):T_1$ is similar to $T_2\}$, is equivalence relation.

Now, we see the ratio of the sides of triangles T1 and T3 as shown

$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$

i.e. ratios of sides of T1 and T3 are equal. Hence, ${\mathrm{T}}_1$ and ${\mathrm{T}}_3$ are related.

Question 13: Show that the relation R defined in the set A of all polygons as R={(P1, P2): P1 and P2 have the same number of sides} is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4, and 5?

Answer:

$R=\{(P_1,P_2):P_1$ and $P_2$ have same number of sides $\}$

The same polygon has the same number of sides with itself,i.e. $(P_1,P_2)\in R$, so it is reflexive.

Let,

$(P_1,P_2)\in R$ i.e. $P_1$ have same number of sides as ${\mathrm{P}}_2$

${\mathrm{P}}_1$ have same number of sides as ${\mathrm{P}}_2$ is same as ${\mathrm{P}}_2$ have same number of sides as ${\mathrm{P}}_1$ i.e. $(P_2,P_1)\in R$

Hence, it is symmetric.

Let,

$(P_1,P_2)\in R$ and $(P_2,P_3)\in R$ i.e. ${\mathrm{P}}_1$ have same number of sides as ${\mathrm{P}}_2$ and ${\mathrm{P}}_2$ have same number of sides as ${\mathrm{P}}_3$

$\Rightarrow {\mathrm{P}}_1$ have same number of sides as ${\mathrm{P}}_3$ i.e. $(P_1,P_3)\in R$

Hence, it is transitive,

Thus, $R=\{(P_1,P_2):P_1$ and $P_2$ have same number of sides $\}$, is an equivalence relation.

The elements in A related to the right angle triangle $T$ with sides 3,4 and 5 are those polygons that have 3 sides.

Hence, the set of all elements in A related to the right-angle triangle T is a set of all triangles.

Question 14: Let $L$ be the set of all lines in $XY$ plane and $R$ be the relation in $L$ defined as $R=\{(L1,L2):L1$ is parallel to $L2\}$. Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y=2x+4$.

Answer:

$R=\{(L_1,L_2):L_1$ is parallel to $L_2\}$

All lines are parallel to themselves, so it is reflexive.

Let,

$(L_1,L_2)\in R$ i.e. ${\mathrm{L}}_1$ is parallel to T 2.

${\mathrm{L}}_1$ is parallel to L 2 is the same as L 2 is parallel to ${\mathrm{L}}_1$, i.e. $(L_2,L_1)\in R$

Hence, it is symmetric.

Let,

$(L_1,L_2)\in R$ and $(L_2,L_3)\in R$ i.e. $L_1$ is parallel to $L2$ and $L2$ is parallel to $L_3$.

$\Rightarrow {\mathrm{L}}_1$ is parallel to ${\mathrm{L}}_3$ i.e. $(L_1,L_3)\in R$

Hence, it is transitive,

Thus, $R=\{(L_1,L_2):L_1$ is parallel to $L_2\}$, is equivalence relation.

The set of all lines related to the line $y=2x+4$. Are lines parallel to $y=2x+4$?

Here, Slope $=m=2$ and constant $=c=4$

It is known that the slope of parallel lines is equal.

Lines parallel to this $(y=2x+4$.) line are $y=2x+c,c\in R$

Hence, the set of all parallel lines to $y=2x+4$. are $y=2x+c$.

Question 15: Let R be the relation in the set A= {1,2,3,4} given by R={(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Answer:

A = {1,2,3,4}

R={(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}

For every a∈A, there is (a, a)∈R.

∴ R is reflexive.

Given, (1,2)∈R but (2,1)∉R

∴ R is not symmetric.

For a,b,c∈A there are (a,b)∈Rand(b,c)∈R ⇒ (a,c)∈R

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is option B.

Question 16: Let R be the relation in the set N given by R={(a,b):a=b−2,b>6}. Choose the correct answer.

(A) (2,4)∈R

(B) (3,8)∈R

(C) (6,8)∈R

(D) (8,7)∈R

Answer:

R={(a,b):a=b−2,b>6}

(A) Since b<6, so (2,4)∉R

(B) Since 3≠8−2, so (3,8)∉R

(C) Since, 8>6 and 6=8−2 so (6,8)∈R

(D) Since 8≠7−2, so (8,7)∉R

The correct answer is option C.

Question 1: Show that the function f: R∗⟶R∗ defined by f(x)=1/x is one-one and onto, where R ∗ is the set of all non-zero real numbers. Is the result true if the domain R ∗ is replaced by N, with the co-domain being the same as R ∗ ?

Answer:

Given, $f:R_{\ast }⟶R_{\ast }$ is defined by $f(x)=\frac{1}{x}$.
One - One :

$f(x)=f(y)$

$\frac{1}{x}=\frac{1}{y}$

$x=y$

$\therefore \mathrm{f}$ is one-one.
Onto:
We have $y\in R_{\ast }$, then there exists $x=\frac{1}{y}\in R_{\ast }($ Here $y\ne 0)$ such that

$f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$

$\therefore$ f is onto.
Hence, the function is one-one and onto.

If the domain ${\mathrm{R}}_{\ast }$ is replaced by N with co-domain being same as ${\mathrm{R}}_{\ast }$ i.e. $g:N⟶R_{\ast }$ defined by

$g(x)=\frac{1}{x}$

$g\left(x_1\right)=g\left(x_2\right)$

$\frac{1}{x_1}=\frac{1}{x_2}$

$x_1=x_2$

$\therefore \mathrm{g}$ is one-one.
For $1.5\in R_{\ast }$,
$g(x)=\frac{1}{1.5}$ but there does not exists any x in N.
Hence, function g is one-one but not onto.

Question 2 (i): Check the injectivity and surjectivity of the following functions:

(i) f:N→N given by f(x)=x^2

Answer:

$f:N\to N$

$f(x)=x^2$
One-one:
$x,y\in N$ then $f(x)=f(y)$

$x^2=y^2$

$x=y$

$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3\in N$ there is no x in N such that $f(x)=x^2=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.

Question 2 (ii): Check the injectivity and surjectivity of the following functions:

(ii) f:Z→Z given by f(x)=x^2

Answer:

$f:Z\to Z$

$f(x)=x^2$

One-one:

For $-1,1\in Z$ then $f(x)=x^2$

$f(-1)=(-1{)}^2$

$f(-1)=1$ but $-1\ne 1$

$therefore,\mathrm{f}$ is not one- one i.e. not injective.
For $-3\in Z$ there is no x in Z such that $f(x)=x^2=-3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is neither injective nor surjective

Question 2 (iii): Check the injectivity and surjectivity of the following functions:

(iii) f:R→R given by f(x)=x^2

Answer:

$f:R\to R$

$f(x)=x^2$
One-one:
For $-1,1\in R$ then $f(x)=x^2$

$f(-1)=(-1{)}^2$

$f(-1)=1\text{ but }-1\ne 1$

$therefore,\mathrm{f}$ is not one- one i.e. not injective.
For $-3\in R$ there is no x in R such that $f(x)=x^2=-3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is not injective and not surjective.

Question 2 (iv): Check the injectivity and surjectivity of the following functions:

(iv) f:N→N given by f(x)=x^3

Answer:

$f:N\to N$

$f(x)=x^3$

One-one:
$x,y\in N$ then $f(x)=f(y)$

$x^3=y^3$

$x=y$

$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3\in N$ there is no x in N such that $f(x)=x^3=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.

Question 2 (v): Check the injectivity and surjectivity of the following functions:

(v) f:Z→Z given by f(x)=x^3

Answer:

$f:Z\to Z$

$f(x)=x^3$

One-one:
For $(x,y)\in Z$ then $f(x)=f(y)$

$x^3=y^3$

$x=y$

$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3\in Z$ there is no x in Z such that $f(x)=x^3=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.

Question 3: Prove that the Greatest Integer Function f: R⟶R, given by f(x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Answer:

f:R⟶R

f(x)=[x]

One-one:

For 1.5,1.7∈R then f(1.5)=[1.5]=1 and f(1.7)=[1.7]=1

but 1.5≠1.7

∴ f is not one-one, i.e. not injective.

For 0.6∈R there is no x in R such that f(x)=[0.6]

∴ f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

Question 4: Show that the Modulus Function: R → R, given by f(x)=|x], is neither one-one nor onto, where |x| is if x is positive or 0 and |x| is −x if x is negative.

Answer:

f: R→R

f(x)=|x|

f(x)=|x|=xifx≥0and−xifx<0

One-one:

For −1,1∈R then f(−1)=|−1|=1

f(1)=|1|=1

−1≠1

∴ f is not one-one, i.e. not injective.

For −2∈R,

We know f(x)=|x| is always positive there is no x in R such that f(x)=|x|=−2

∴ f is not onto i.e. not surjective.

Hence, f(x)=|x| is neither one-one nor onto.

Question 5: Show that the Signum Function f: R→R, given by f(x)={1 if x>0 0 if x=0 −1 if x<0, is neither one-one nor onto.

Answer:

$f:R\to R$ is given by

$f(x)=\{\begin{array}{cl}1& \text{ if }x>0\\ 0& \text{ if }x=0\\ -1& \text{ if }x<0\end{array}$

As we can see $f(1)=f(2)=1$, but $1\ne 2$
So it is not one-one.
Now, $\mathrm{f}(\mathrm{x})$ takes only 3 values $(1,0,-1)$ for the element -3 in codomain $R$, there does not exists x in domain $R$ such that $f(x)=-3$.

So it is not onto.
Hence, the signum function is neither one-one nor onto.

Question 6: Let A={1,2,3}, B={4,5,6,7} and let f={(1,4),(2,5),(3,6)} be a function from A to B. Show that f is one-one.

Answer:

A={1,2,3}

B={4,5,6,7}

f={(1,4),(2,5),(3,6)}

f: A→B

∴ f(1)=4,f(2)=5,f(3)=6

Every element of A has a distant value in f.

Hence, it is one-one.

Question 7 (i): In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.

(i) f:R→R defined by f(x)=3−4x

Answer:

f: R→R

f(x)=3−4x

Let there be (a,b)∈R such that f(a)=f(b)

3−4a=3−4b

−4a=−4b

a=b

∴ f is one-one.

Let there be y∈, y=3−4x

x=(3−y)4

f(x)=3−4x

Putting value of x, f(3−y4)=3−4(3−y4)

f(3−y4)=y

∴ f is onto.

f is both one-one and onto; hence, f is bijective.

Question 7 (ii): In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.

(ii) f:R→R defined by f(x)=1+x^2

Answer:

$f:R\to R$

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

Let there be $(a,b)\in R$ such that $f(a)=f(b)$

$\begin{array}{c}1+a^2=1+b^2\\ a^2=b^2\\ a=\pm b\end{array}$

For $f(1)=f(-1)=2$ and $1\ne -1$
$\therefore f$ is not one-one.
Let there be $-2\in R(-2$ in codomain of R$)$

$f(x)=1+x^2=-2$

There does not exists any x in domain R such that $f(x)=-2$
$\therefore \mathrm{f}$ is not onto.
Hence, f is neither one-one nor onto.

Question 8: Let A and B be sets. Show that f: A×B→B×A such that f(a,b)=(b, a) is a bijective function.

Answer:

f: A×B→B×A

f(a,b)=(b,a)

Let (a1,b1),(a2,b2)∈A×B

such that f(a1,b1)=f(a2,b2)

(b1,a1)=(b2,a2)

⇒ b1=b2 and a1=a2

⇒ (a1,b1)=(a2,b2)

∴ f is one- one

Let (b, a)∈B×A

then there exists (a,b)∈A×B such that f(a,b)=(b,a)

∴ f is onto.

Hence, it is bijective.

Question 9: Let $f:N\to N$ be defined by $f(n)=\{\begin{array}{cl}\frac{n+1}{2}& \text{ if }n\text{ is odd }\\ \frac{n}{2}& \text{ if }n\text{ is even }\end{array}\text{ for all }n\in N.$ State whether the function f is bijective. Justify your answer.

Answer:

$f:N\to N,n\in N$

$f(n)=\{\begin{array}{cc}\frac{n+1}{2}& \text{ if }n\text{ is odd }\\ \frac{n}{2}& \text{ if }n\text{ is even }\end{array}$

Here we can observe,

$f(2)=\frac{2}{2}=1$ and $f(1)=\frac{1+1}{2}=1$
As we can see $f(1)=f(2)=1$ but $1\ne 2$
$\therefore \mathrm{f}$ is not one-one.
Let, $n\in N$ (N=co-domain)
case1 $n$ be even
For $r\in N,n=2r$
then there is $4r\in N$ such that $f(4r)=\frac{4r}{2}=2r$
case2 n be odd
For $r\in N,n=2r+1$
then there is $4r+1\in N$ such that $f(4r+1)=\frac{4r+1+1}{2}=2r+1$
$\therefore \mathrm{f}$ is onto.
$f$ is not one-one but onto
Hence, the function $f$ is not bijective

Question 10: Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $f:A\to B$ defined by $f(x)=x-2/x-3$. Is $f$ one-one and onto? Justify your answer.

Answer:

$\begin{array}{rl}& A=R-\{3\}\\ & B=R-\{1\}\\ & f:A\to B\\ & f(x)=\left(\frac{x-2}{x-3}\right)\end{array}$

Let $a,b\in A$ such that $f(a)=f(b)$

$\begin{array}{c}\left(\frac{a-2}{a-3}\right)=\left(\frac{b-2}{b-3}\right)\\ (a-2)(b-3)=(b-2)(a-3)\\ ab-3a-2b+6=ab-2a-3b+6\\ -3a-2b=-2a-3b\\ 3a+2b=2a+3b\\ 3a-2a=3b-2b\\ a=b\end{array}$

$\therefore \mathrm{f}$ is one-one.

Let, $b\in B=R-\{1\}$ then $b\ne 1$

$\begin{array}{rl}& a\in A\text{ such that }f(a)=b\\ & \begin{array}{c}\left(\frac{a-2}{a-3}\right)=b\\ (a-2)=(a-3)b\\ a-2=ab-3b\\ a-ab=2-3b\\ a(1-b)=2-3b\\ a=\frac{2-3b}{1-b}\in A\end{array}\end{array}$

For any $b\in B$ there exists $a=\frac{2-3b}{1-b}\in A$ such that

$\begin{array}{rl}f\left(\frac{2-3b}{1-b}\right)& =\frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}\\ & =\frac{2-3b-2+2b}{2-3b-3+3b}\\ & =\frac{-3b+2b}{2-3}\\ & =b\end{array}$

$\therefore \mathrm{f}$ is onto
Hence, the function is one-one and onto.

Question 11: Let f: R→R be defined as f(x)=x^4. Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

$\begin{array}{rl}& f:R\to R\\ & f(x)=x^4\end{array}$

One-one:
For $a,b\in R$ then $f(a)=f(b)$

$\begin{array}{rl}& a^4=b^4\\ & a=\pm b\end{array}$

$\therefore f(a)=f(b)$ does not imply that $a=b$
example : $f(2)=f(-2)=16$ and $2\ne -2$
$\therefore \mathrm{f}$ is not one- one
For $2\in R$ there is no x in R such that $f(x)=x^4=2$
$\therefore \mathrm{f}$ is not onto.
Hence, $f$ is neither one-one nor onto.
Hence, option D is correct.

Question 12: Let f: R→R be defined as f(x)=3x. Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

f: R→R

f(x)=3x

One - One :

Let (x,y)∈R

f(x)=f(y)

3x=3y

x=y

∴ f is one-one.

Onto:

If we have y∈R, then there exists x=y/3∈R such that

f(y/3)=3×y/3=y

∴f is onto.

Hence, the function is one-one and onto.

The correct answer is A.

Question 1: Show that the function f:R→{x∈R:−1<x<1} defined by f(x)=x/(1+|x|) x∈R is one one and onto function.

Answer:

The function $f:R\to \{x\in R:-1<x<1\}$ defined by

$f(x)=\frac{x}{1+|x|},x\in R$

One-one:

Let $f(x)=f(y),x,y\in R$

$\frac{x}{1+|x|}=\frac{y}{1+|y|}$

It is observed that if x is positive and y is negative.

$\frac{x}{1+x}=\frac{y}{1+y}$

Since x is positive and y is negative.

$x>y\Rightarrow x-y>0\text{ but }2\mathrm{xy}\text{ is negative. }$

$x-y\ne 2xy$

Thus, the case of x being positive and y being negative is removed.

The same happens in the case of y is positive and x is negative, so this case is also removed.

When x and y both are positive:

$\begin{array}{c}f(x)=f(y)\\ \frac{x}{1+x}=\frac{y}{1+y}\\ x(1+y)=y(1+x)\\ x+xy=y+xy\\ x=y\end{array}$

When $x$ and $y$ both are negative:

$\begin{array}{rl}& f(x)=f(y)\\ & \frac{x}{1-x}=\frac{y}{1-y}\\ & x(1-y)=y(1-x)\\ & x-xy=y-xy\\ & x=y\end{array}$

$\therefore \mathrm{f}$ is one-one.

Onto:

Let $y\in R$ such that $-1<y<1$

If y is negative, then $x=\frac{y}{y+1}\in R$

$f(x)=f\left(\frac{y}{y+1}\right)=\frac{\frac{y}{1+y}}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\frac{-y}{1+y}}=\frac{y}{1+y-y}=y$

If y is positive, then $x=\frac{y}{1-y}\in R$

$f(x)=f\left(\frac{y}{1-y}\right)=\frac{\frac{y}{1-y}}{1+\left|\frac{y}{1-y}\right|}=\frac{\frac{y}{1-y}}{1+\frac{-y}{1-y}}=\frac{y}{1-y+y}=y$

Thus, $f$ is onto.

Hence, $f$ is one-one and onto.

Question 2: Show that the function f: R→R given by f(x)=x^3 is injective.

Answer:

f: R→R

f(x)=x^3

One-one:

Let f(x)=f(y)x,y∈R

x^3=y^3

We need to prove x=y. So,

• Let x≠y, then their cubes will not be equal, i.e. x^3≠y^3.

• It will contradict the given condition of cubes being equal.

• Hence, x=y, and it is one-one, which means it is injective

Question 3: Given a nonempty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A⊂B Is R an equivalence relation on P(X)? Justify your answer.

Answer:

Given a nonempty set X, consider $\mathrm{P}(\mathrm{X})$, which is the set of all subsets of X.

Since every set is a subset of itself, ARA for all $A\in P(x)$

$\therefore \mathrm{R}$ is reflexive.

Let $ARB\Rightarrow A\subset B$

This is not the same as $B\subset A$

If $A=\{0,1\}$ and $B=\{0,1,2\}$

then we cannot say that $B$ is related to $A$.

$\therefore \mathrm{R}$ is not symmetric.

If $ARB$ and $BRC$, then $A\subset B$ and $B\subset C$

this implies $A\subset C=ARC$

$\therefore \mathrm{R}$ is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

Question 4: Find the number of all onto functions from the set {1,2,3,...,n} to itself.

Answer:

The number of all onto functions from the set {1,2,3,...,n} to itself is the permutations of n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, the total number of all onto maps from the set {1,2,3,...,n} to itself is the same as permutations on n symbols 1,2,3,4,5...............n, which is n.

Question 5: Let $A=\{-1,0,1,2\},B=\{-4,-2,0,2\}$ and $f,g:A\to B$ be functions defined by $f(x)=x^2-x,x\in A$ and $g(x)=2|x-\frac{1}{2}|-1,x\in A$. Are $f$ and $g$ equal?

Justify your answer. (Hint: One may note that two functions $f:A\to B$ and $g:A\to B$ such that $f(a)=g(a)\mathrm{\forall }a\in A$, are called equal functions).

Answer:

Given :

$A=\{-1,0,1,2\},B=\{-4,-2,0,2\}$

$f,g:A\to B$ are defined by $f(x)=x^2-x,x\in A$ and $g(x)=2|x-\frac{1}{2}|-1,x\in A$.

It can be observed that.

$\begin{array}{rl}& f(-1)=(-1{)}^2-(-1)=1+1=2\\ & g(-1)=2|-1-\frac{1}{2}|-1=2\left|\frac{-3}{2}\right|-1=3-1=2\\ & f(-1)=g(-1)\end{array}$

$\begin{array}{rl}& f(0)=(0{)}^2-(0)=0+0=0\\ & g(0)=2|0-\frac{1}{2}|-1=2\left|\frac{-1}{2}\right|-1=1-1=0\\ & f(0)=g(0)\end{array}$

$\begin{array}{rl}& f(1)=(1{)}^2-(1)=1-1=0\\ & g(1)=2|1-\frac{1}{2}|-1=2\left|\frac{1}{2}\right|-1=1-1=0\\ & f(1)=g(1)\end{array}$

$\begin{array}{rl}& f(2)=(2{)}^2-(2)=4-2=2\\ & g(2)=2|2-\frac{1}{2}|-1=2\left|\frac{3}{2}\right|-1=3-1=2\\ & f(2)=g(2)\end{array}$

$\therefore f(a)=g(a)\mathrm{\forall }a\in A$

Hence, $f$ and $g$ are equal functions.

Question 6: Let A={1,2,3}. The number of relations containing (1,2) and (1,3) which are reflexive and symmetric but not transitive is

(A) 1
(B) 2
(C) 3
(D) 4

Answer:

A={1,2,3}

The smallest relations containing (1,2) and (1,3), which are reflexive and symmetric but not transitive, are given by

R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,1),(3,1)}

(1,1),(2,2),(3,3)∈R , so relation R is reflexive.

(1,2),(2,1)∈R and (1,3),(3,1)∈R , so relation R is symmetric.

(2,1),(1,3)∈R but (2,3)∉R , so realation R is not transitive.

Now, if we add any two pairs (2,3) and (3,2) to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

Question 7: Let A={1,2,3}. The number of equivalence relations containing (1,2) is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

A={1,2,3}

The number of equivalence relations containing (1,2) is given by

R={(1,1),(2,2),(3,3),(1,2),(2,1)}

We are left with four pairs (2,3), (3,2), (1,3),(3,1).

(1,1),(2,2),(3,3)∈R , so relation R is reflexive.

(1,2),(2,1)∈R and (2,3),(3,2)∉R , so relation R is not symmetric.

(1,3),(3,1)∉R , so realation R is not transitive.

Hence, the equivalence relation is bigger than R is the universal relation.

Thus, the total number of equivalence relations containing (1,2) is two.

Thus, option B is correct.

Class 12 Maths NCERT Chapter 1: Extra Question

Question: Let an operation $\ast$ on the set of natural numbers N be defined by $a\ast b=a^b\cdot$
Find (i) whether $\ast$ is a binary or not, and (ii) if it is a binary, then is it commutative or not.

Solution:

(i) As $a^b\in N$ for all $a,b\in N$
ie $a\ast b\in N\mathrm{\forall }a,b\in N$
Hence, $\ast$ is binary.

(ii) As $1^2\ne 2^1$
so $1\ast 2\ne 2\ast 1$

Hence, $\ast$ is not commutative.

Also read,

• Relation And Function Class 12 NCERT Solutions Exercise 1.1

• Relation And Function Class 12 NCERT Solutions Exercise 1.2

• Relation And Function Class 12 NCERT Solutions Miscellaneous Exercise

Approach to Solve Questions of Relations and Functions Class 12

Relations and Functions can feel like a puzzle of definitions, diagrams, and properties—but with the right approach, every question starts to make perfect sense.

• Recognising the type: Build a strong foundation for different kinds of relations, such as reflexive, symmetric, transitive, and equivalence.
  Familiarise yourself with the properties of One-One, onto, and bijective functions.
• Efficient usage of diagrams and graphs: Visual representation of the properties of relations and functions helps to clarify the types.
  Use tables for small sets for binary operations to verify identity, inverse, etc.
• Domain, codomain, range: Grasp these concepts thoroughly to identify whether a relation is a function and determine its behaviour.
  Also, practice extracting the domain and range from relations using ordered pairs or graphs.
• Elimination method: For MCQ-type questions, cancel out the extreme options by checking the basic definitions.
• Shortcut tricks: Smart tricks and shortcuts are always handy during the exam to save time. Write those tricks in a notebook and revise them from time to time. Here are some tricks.
  - If a relation is reflexive and only uses equality or divisibility, it may be equivalence.
  - For transitive relations, check in pairs that share a common middle.
  - Use the formula $2^{n^2}$ to find the total relations over a set of n elements.
  - During the domain checking, avoid the values that cause division by zero or negative square roots.

What Extra Should Students Study Beyond the NCERT for JEE?

NCERT Solutions for Class 12 Maths: Chapter Wise

Also read,

• NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions
• NCERT Notes Class 12 Maths Chapter 1 Relations and Functions

NCERT Solutions class-wise

Students can check the following links for more in-depth learning.

• NCERT solutions for class 11

• NCERT solutions for class 10

• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Students can check the following links for more in-depth learning.

• NCERT Books Class 12 Maths

• NCERT Syllabus Class 12 Maths

• NCERT Books Class 12

• NCERT Syllabus Class 12