# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Exercise:1.1

Exercise:1.2

Exercise:1.3

Exercise:1.4

Miscellaneous Exercise

These CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions are explained in a step-by-step method, so it will be very easy to understand the concepts. Still if you are in a doubt anywhere, you can contact our subject matter experts who are available to help you out and make learning easier for you.

## What is the Relation?

The meaning of the term ‘relation’ in mathematics is the same as the meaning of 'relation' in the English language. Relation means two quantities or objects are related if there is a link between them. In other words, we can say that it is a connection between or among things.

Let's understand with an example - let A is the set of students of class XII of a school and B is the set of students of class XI of the same school. Then some of the examples of relations from A to B are-

(i) {(a, b) ∈A × B: a is a brother of b},

(ii) {(a, b) ∈A × B: a is a sister of b},

(iii) {(a, b) ∈A × B: age of a is less than the age of b}.

If (a, b) ∈ R, we can say that ‘a’ is related to ‘b’ under the relation ‘R’ and we write as  ‘a R b’. To understand the topic in-depth, after every concept, some topic wise questions are given in the textbook of CBSE class 12. In this article, you will find solutions of NCERT for class 12 maths chapter 1 Relations and Functions for such type of questions also.

## Topics of NCERT Grade 12 Maths Chapter-1  Relations and Functions

1.1 Introduction

1.2 Types of Relations

1.3 Types of Functions

1.4 Composition of Functions and Invertible Function

1.5 Binary Operations

## NCERT Solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.1

Since,   so  is not reflexive.

Since,  but   so  is not symmetric.

Since,  but  so  is not transitive.

Hence,  is neither reflexive nor symmetric and nor transitive.

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

Since,

so  is not  reflexive.

Since,  but

so  is not symmetric.

Since there is no pair in   such that  so this is not transitive.

Hence,  is neither reflexive nor symmetric and
nor transitive.

### Question1(iii) Determine whether each of the following relations are reflexive, symmetric and transitive:

(iii) Relation R in the set  as

Any number is divisible by itself  and  .So it is  reflexive.

but  .Hence,it is not symmetric.

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

For  as  which is an integer.

So,it is reflexive.

For  , and  because   are both integers.

So, it is symmetric.

For  , as  are both integers.

Now,  is also an integer.

So, and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

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

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

(a)

,so it is reflexive

means  .

i.e.  so it is symmetric.

means  also .It states that  i.e. .So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

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

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

(b)

as  and  is same human being.So, it is reflexive.

means  .

It is same as   i.e. .

So,it is symmetric.

means   and .

It implies that  i.e. .

Hence, it is reflexive, symmetric and
transitive.

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

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

(c)

means  but  i.e. .So, it is not reflexive.

means   but  i.e  .So, it is not symmetric.

means  and .

i.e.  .

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

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

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

(d)

means  but   i.e..

So, it is not reflexive.

means  but   i.e..

So, it is not symmetric.

Let,  means  and .

This case is not possible so it is not transitive.

Hence, it is not  reflexive, symmetric and
transitive.

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

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

(e)

means  than   i.e. .So, it is not reflexive..

means  than   i.e. .So, it is not symmetric.

Let,  means  and  than  i.e. .

So, it is not transitive.

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

## Question:2 Show that the relation R in the set R of real numbers defined as is neither reflexive nor symmetric nor transitive.

Taking

and

So, R is not reflexive.

Now,

because    .

But,   i.e. 4 is not less than 1

So,

Hence, it is not symmetric.

as

Since   because

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 as is reflexive, symmetric or transitive.

R defined in the set

Since,  so it is not reflexive.

but

So, it is not symmetric

but

So, it is not transitive.

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

transitive but not symmetric.

As  so it is reflexive.

Now we take an  example

as

But   because .

So,it is not symmetric.

Now if we take,

Than,  because

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 defined by is reflexive, symmetric or transitive.

because

So, it is not symmetric

Now,     because

but   because

It is not symmetric

as   .

But,    because

So it is not transitive

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

### Question:6 Show that the relation R in the set given by is symmetric but neither reflexive nor transitive.

Let A=

We can see   so it is not reflexive.

As  so it is symmetric.

But   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 is an equivalence relation.?

A = all the books in a library of a college

because x and x have the same number of pages so it is reflexive.

Let    means x and y have same number of pages.

Since y and x have the same number of pages so    .

Hence, it is symmetric.

Let    means x and y have the same number of pages.

and    means y and z have the same number of pages.

This states,x and z also have the same number of pages i.e.

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 given by , is an equivalence relation. Show that all the elements of are related to each other and all the elements of are related to each other. But no element of  is related to any element of .

Let there be  then  as   which is even number. Hence, it is reflexive

Let   where  then  as

Hence, it is symmetric

Now, let

are even number i.e.  are even

then,  is even                  (sum of even integer is even)

So, . Hence, it is transitive.

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

The elements of are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of  is not related to   because a difference of odd and even number is not even.

## Question:9(i) Show that each of the relation R in the set , given by

(i)  is an equivalence relation. Find the set of all elements related to 1 in each case.

For  ,   as  which is multiple of 4.

Henec, it is reflexive.

Let,  i.e.  is multiple of 4.

then   is also multiple of 4 because   =    i.e.

Hence, it is symmetric.

Let,  i.e.  is multiple of 4   and      i.e.  is multiple of 4 .

is multiple of 4  and    is multiple of 4

is multiple of 4

is multiple of 4 i.e.

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

is multiple of 4.

is multiple of 4.

is multiple of 4.

## Question:9(ii) Show that each of the relation R in the set , given by

(ii)   is an equivalence relation. Find the set of all elements related to 1 in each case.

For  ,   as

Henec, it is reflexive.

Let,  i.e.

i.e.

Hence, it is symmetric.

Let,  i.e.    and      i.e.

i.e.

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.

Let

so it is not reflexive.

and    so it is symmetric.

but   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.

Let

Now for  , so it is not reflexive.

Let   i.e.

Then  is not possible i.e.  . So it is not symmetric.

Let   i.e.     and   i.e.

we can write this as

Hence,  i.e. . So it is 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.

Let

Define a relation R on A as

If    , i.e.. So it is reflexive.

If    ,     and   i.e.. So it is symmetric.

and   i.e. .  and

But  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.

Let there be a relation R in R

because

Let   i.e.

But  i.e.

So it is not symmetric.

Let   i.e.   and   i.e.

This can be written as   i.e.   implies

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.

Let there be a relation A in R

So R is not reflexive.

We can see    and

So it is symmetric.

Let      and

Also

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 , is an equivalence relation. Further, show that the set of all points related to a point is the circle passing through P with origin as centre.

The distance of point P from the origin is always the same as the distance of same point P from origin i.e.

R is reflexive.

Let  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 distance of the point Q from the origin is the same as the distance of the point P from the origin i.e.

R is symmetric.

Let         and

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

R is transitive.

Hence, R  is an equivalence relation.

The set of all points related to a point  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 O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.

### Question:12 Show that the relation R defined in the set A of all triangles as , is 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 T1 , T2 and T3 are related?

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

Let,

i.e.T1 is similar to T2

T1 is similar to T2 is the same asT2 is similar to T1 i.e.

Hence, it is symmetric.

Let,

and    i.e. T1 is similar to T2  and T2 is similar toT3 .

T1 is similar toT3   i.e.

Hence, it is transitive,

Thus,  , is equivalence relation.

Now, we see the ratio of sides of triangle T1 andT3 are as shown

i.e. ratios of sides of T1 and T3 are equal.Hence, T1 and T3 are related.

## Question:13 Show that the relation R defined in the set A of all polygons as , 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?

The same polygon has the same number of sides with itself,i.e. , so it is reflexive.

Let,

i.e.P1 have same number of sides as  P2

P1 have the same number of sides as Pis the same as P2 have same number of sides as P1 i.e.

Hence,it is symmetric.

Let,

and    i.e. P1 have the same number of sides as P2  and Phave same number of sides as P3

Phave same number of sides as P3   i.e.

Hence, it is transitive,

Thus,  , is an equivalence relation.

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

Hence, the set of all elements in A related to the right angle triangle T is 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 . Show that R is an equivalence relation. Find the set of all lines related to the line

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

Let,

i.e.L1 is parallel to L2.

L1 is parallel to L2 is same as L2 is parallel to L1 i.e.

Hence, it is symmetric.

Let,

and    i.e. L1 is parallel to L2  and L2 is parallel  to L3 .

L1 is parallel to L3   i.e.

Hence, it is transitive,

Thus,   , is equivalence relation.

The set of all lines related to the line  are lines parallel to

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

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

Lines parallel to this (  )  line  are   ,

Hence, set of all parallel lines to  are .

### Question:15 Let R be the relation in the set given by . 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.

A =

For every    there is

R is reflexive.

Given,   but

R is not symmetric.

For   there are

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 . Choose the correct answer.

(A)
(B)
(C)
(D)

(A) Since,  so

(B) Since,   so

(C) Since,   and    so

(d) Since, so

The correct answer is option C.

## NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.2

Given,  is defined by  .

One - One :

f is one-one.

Onto:

We have  , then there exists      ( Here ) such that

.

Hence, the function is one-one and onto.

If the domain R is replaced by N with co-domain being same as R∗   i.e.    defined by

g is one-one.

For    ,

but there does not exists any x in N.

Hence, function g is one-one but not onto.

(i)  given by

One- one:

then

f is one- one i.e. injective.

For  there is no x in N such that

f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

(ii)  given by

One- one:

For     then

but

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

For    there is no x in Z such that

f is not onto i.e. not surjective.

Hence, f is neither injective nor surjective.

(iii)  given by

One- one:

For     then

but

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

For    there is no x in R such that

f is not onto i.e. not surjective.

Hence, f is not injective and not surjective.

(iv)  given by

One- one:

then

f is one- one i.e. injective.

For  there is no x in N such that

f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

(v)   given by

One- one:

For     then

f is one- one i.e. injective.

For    there is no x in Z such that

f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

One- one:

For      then      and

but

f is not one- one i.e. not injective.

For    there is no x in R such that

f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

One- one:

For      then

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

For  ,

We know    is always positive there is no x in R such that

f is not onto i.e. not surjective.

Hence,  , is neither one-one nor onto.

Question:5 Show that the Signum Function , given by

is given by

As we can see       ,  but

So it is not one-one.

Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain  ,there does not exists x in domain  such that .

So it is not onto.

Hence, signum function is neither one-one nor onto.

Every element of A has a distant value in f.

Hence, it is one-one.

(i)  defined by

Let  there  be    such that

f is one-one.

Let there be ,

Puting value of x,

f is onto.

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

(ii)  defined by

Let  there  be    such that

For      and

f is not one-one.

Let there be    (-2 in codomain of R)

There does not exists any x in domain R  such that

f is not onto.

Hence, f is neither one-one nor onto.

Let

such that

and

f is one- one

Let,

then there exists   such that

f is onto.

Hence, it is bijective.

,

Here we can observe,

and

As we can see   but

f is not one-one.

Let,    (N=co-domain)

case1   n be even

For ,

then there is  such that

case2   n be odd

For  ,

then there is  such that

f is onto.

f is not one-one but onto

hence, the function f is not bijective.

Let  such that

f is one-one.

Let,        then

such that

For any  there exists     such that

f is onto

Hence, the function is one-one and onto.

(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.

One- one:

For     then

does not imply that

example: and

f is not  one- one

For    there is no x in R such that

f is not onto.

Hence, f is neither one-one nor onto.

Option D is correct.

(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.

One - One :

Let

f is one-one.

Onto:

We have  , then there exists     such that

.

Hence, the function is one-one and onto.

The correct answer is A .

## NCERT Solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.3

Question:1 Let and be given by and . Write down .

Given :                    and

and

Hence,    =

To prove :

Hence,

To prove:

Hence,

Question:3(i) Find  and , if

(i)   and

and

Question:3(ii) Find gof and fog, if

(ii) and

The solution is as follows

(ii)  and

,  for all

Hence,the given function is invertible and the inverse of  is  itself.

(i)

with

(i)  with

From the given definition,we have:

f is not one-one.

Hence, f do not have an inverse function.

(ii)   with

(ii)   with

From the definition, we can conclude :

g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse

(iii)    with

(iii)    with

From the definition, we can see the set  have distant values under h.

h is one-one.

For every element y of set ,there exists an element x  in  such that

h is onto

Thus, h is one-one and onto so h has an inverse function.

One -one:

f is one-one.

It is clear that   is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in

let y be an arbitrary element of range f

Since,  is onto, so

for

,

is given by

One-one :

Let

f is one-one function.

Onto:

So, for  there is     ,such that

f  is onto.

Thus, f is one-one and onto so  exists.

Let,  by

Now,

and

Hence, function f is invertible and inverse of f is .

It is given that
,    and

Now, Let f(x) = f(y)

⇒ x2 + 4 = y2 + 4

⇒ x2 = y2

⇒ x = y

⇒ f is one-one function.

Now, for y [4, ∞), let y = x2 + 4.

⇒ x2 = y -4 ≥ 0 ⇒ for any y  R, there exists x = R such that = y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) = Now, gof(x) = g(f(x)) = g(x2 + 4) = And, fog(y) = f(g(y)) = = Therefore, gof = gof = IR.

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) = One- one:

Let

Since, x and y are positive.

f is one-one.

Onto:

Let for    ,

f is onto  and range is .

Since f is one-one and onto so it is invertible.

Let      by

Hence,   is invertible with the inverse of  given by

Let be an invertible function

Also, suppose f has two inverse

For , we have

[f is invertible implies f is one - one]

[g is one-one]

Thus,f has a unique inverse.

It is given that

Now,, lets define a function g :
such that

Now,

Similarly,

And

Hence,  and, where and

Therefore, the inverse of f exists and

Now,
is given by

Now, we need to  find the inverse of  ,

Therefore, lets define  such that

Now,

Similarly,

Hence,   and , where and

Therefore,  inverse of    exists and

Therefore,

Hence proved

To prove:

Let    be a invertible function.

Then there is      such that     and

Also,

and

and

Hence,     is invertible function and f is inverse of .

i.e.

Question:13 If  be given by , then  is

(A)

(B)

(C)

(D)

Thus, is x.

Hence, option c is correct answer.

(A)

(B)

(C)

(D)

Let f inverse

Let y be the element of range f.

Then there is     such that

Now , define     as