NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions:- The performance of students in the 12^{th } board exams is very important to determine their future so that they can take admission into a good college. NCERT solutions for class 12 maths chapter 1 Relations and Functions will help you to understand the concepts and score well in CBSE 12^{th } board exam. This chapter is not only important for mathematics it is also important in real life. We have relations like father, mother, brother, sister, husband, wife. Relation becomes function when there is only one output for every input. In CBSE class 11 maths you have already learnt in brief about relations and functions, range, domain and co-domain with different types of specific real-valued functions and their graphs. In NCERT solutions for class 12 maths chapter 1 Relations and Functions, you are going to learn about different types of relations and functions, invertible functions, the composition of functions, and binary operations. Concepts of this chapter Relations and Functions are very useful in various other topics of calculus and are also very important from the exam point of view. Unit "Relation and Function" of NCERT class 12^{th } includes two chapters i.e. relation and function, and inverse trigonometry which together has 10 % weightage in the CBSE class 12^{th } final examination. This is the reason you should study this chapter carefully, and solve every problem on your own including solved examples. In this article, you will find NCERT solutions for class 12 maths chapter 1 Relations and Functions including miscellaneous exercise which will help you to score more marks in the exam. Here you will find all NCERT solutions at a single place which will be helpful when you are not able to solve the NCERT questions. In this chapter, there are four exercises with 55 questions and one miscellaneous exercise with 19 questions. In this article, you will find the detailed solutions of NCERT for class maths 12 chapter 1 Relations and functions. Here you will get NCERT solutions for class 12 also.
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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.
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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
Question1(i) . Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation in the set defined as
Answer:
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.
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
Answer:
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
Answer:
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
Answer:
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
Answer:
,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
Answer:
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
Answer:
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
Answer:
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
Answer:
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.
Answer:
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.
Answer:
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.
Question:4 Show that the relation R in R defined as , is reflexive and
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
All triangles are similar to itself, so it is reflexive.
Let,
i.e.T _{ 1 } is similar to T2
T _{ 1 } is similar to T2 is the same asT2 is similar to T _{ 1 } i.e.
Hence, it is symmetric.
Let,
and i.e. T _{ 1 } is similar to T2 and T2 is similar toT _{ 3 } .
T _{ 1 } is similar toT _{ 3 } i.e.
Hence, it is transitive,
Thus, , is equivalence relation.
Now, we see the ratio of sides of triangle T _{ 1 } andT _{ 3 } are as shown
i.e. ratios of sides of T _{ 1 } and T _{ 3 } are equal.Hence, T _{ 1 } and T _{ 3 } are related.
Answer:
The same polygon has the same number of sides with itself,i.e. , so it is reflexive.
Let,
i.e.P _{ 1 } have same number of sides as P _{ 2 }
P _{ 1 } have the same number of sides as P _{ 2 } is the same as P _{ 2 } have same number of sides as P _{ 1 } i.e.
Hence,it is symmetric.
Let,
and i.e. P _{ 1 } have the same number of sides as P _{ 2 } and P _{ 2 } have same number of sides as P _{ 3 }
P _{ 1 } have same number of sides as P _{ 3 } 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.
Answer:
All lines are parallel to itself, so it is reflexive.
Let,
i.e.L _{ 1 } is parallel to L _{ 2 } .
_{ L1 } is parallel to L _{ 2 } is same as L _{ 2 } is parallel to L _{ 1 } i.e.
Hence, it is symmetric.
Let,
and i.e. _{ L1 } is parallel to L _{ 2 } and L _{ 2 } is parallel to L _{ 3 } .
L _{ 1 } is parallel to L _{ 3 } 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.
Answer:
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.
Answer:
(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
Answer:
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.
Question:2(i) Check the injectivity and surjectivity of the following functions:
Answer:
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.
Question:2(ii) Check the injectivity and surjectivity of the following functions:
Answer:
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.
Question:2(iii) Check the injectivity and surjectivity of the following functions:
Answer:
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.
Question:2(iv) Check the injectivity and surjectivity of the following functions:
Answer:
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.
Question:2(v) Check the injectivity and surjectivity of the following functions:
Answer:
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.
Answer:
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.
Answer:
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
Answer:
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.
Question:6 Let , and let be a function from A to B. Show that f is one-one.
Answer:
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.
Answer:
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.
Question:7(ii) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
Answer:
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.
Question:8 Let A and B be sets. Show that such that is
bijective function.
Answer:
Let
such that
and
f is one- one
Let,
then there exists such that
f is onto.
Hence, it is bijective.
Question:9 Let be defined by for all . State whether the function f is bijective. Justify your answer.
Answer:
,
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.
Question:10 Let and . Consider the function defined by . Is f one-one and onto? Justify your answer.
Answer:
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.
Question:11 Let be defined as . Choose the correct answer.
(D) f is neither one-one nor onto.
Answer:
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.
Question:12 Let be defined as . Choose the correct answer.
(D) f is neither one-one nor onto.
Answer:
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 .
Answer:
Given : and
and
Hence, =
Question:4 If show that , for all . What is the inverse of ?
Answer:
, for all
Hence,the given function is invertible and the inverse of is itself.
Question:5(i) State with reason whether following functions have inverse
(i)
with
Answer:
(i) with
From the given definition,we have:
f is not one-one.
Hence, f do not have an inverse function.
Question:5(ii) State with reason whether following functions have inverse
Answer:
(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
Answer:
(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.
Question:6 Show that , given by is one-one. Find the inverse of the function
Answer:
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
,
Question:7 Consider given by . Show that f is invertible. Find the inverse of .
Answer:
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 .
Answer:
It is given that
, and
Now, Let f(x) = f(y)
⇒ x ^{ 2 } + 4 = y ^{ 2 } + 4
⇒ x ^{ 2 } = y ^{ 2 }
⇒ x = y
⇒ f is one-one function.
Now, for y [4, ∞), let y = x ^{ 2 } + 4.
⇒ x ^{ 2 } = 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(x ^{ 2 } + 4) =
And, fog(y) = f(g(y)) = =
Therefore, gof = gof = I _{ R } .
Therefore, f is invertible and the inverse of f is given by
f-1(y) = g(y) =
Question:9 Consider given by . Show that is invertible with
Answer:
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
Answer:
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.
Question:11 Consider given by , and . Find and show that .
Answer:
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
Question:12 Let be an invertible function. Show that the inverse of is , i.e.,
Answer:
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:14 Let be a function defined as . The inverse of is the map given by
Answer:
Let f inverse
Let y be the element of range f.
Then there is such that
Now , define as
Hence, g is inverse of f and
The inverse of f is given by .
The correct option is B.
NCERT Solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.4
Answer:
(i) On , define ∗ by
It is not a binary operation as the image of under * is .
Answer:
(ii) On , define ∗ by
We can observe that for ,there is a unique element ab in .
This means * carries each pair to a unique element in .
Therefore,* is a binary operation.
Answer:
(iii) On , define ∗ by
We can observe that for ,there is a unique element in .
This means * carries each pair to a unique element in .
Therefore,* is a binary operation.
Answer:
(iv) On , define ∗ by
We can observe that for ,there is a unique element in .
This means * carries each pair to a unique element in .
Therefore,* is a binary operation.
Answer:
(v) On , define ∗ by
* carries each pair to a unique element in .
Therefore,* is a binary operation.
Question:2(i) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
Answer:
a*b=a-b
b*a=b-a
so * is not commutative
(a*b)*c=(a-b)-c
a*(b*c)=a-(b-c)=a-b+c
(a*b)*c not equal to a*(b*c), so * is not associative
Question:2(ii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
Answer:
(ii) On , define
ab = ba for all
ab+1 = ba + 1 for all
for
where
operation * is not associative.
Question:2(iii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
Answer:
(iii) On , define
ab = ba for all
for all
for
operation * is commutative.
operation * is associative.
Question:2(iv) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
Answer:
(iv) On , define
ab = ba for all
2ab = 2ba for all
for
the operation is commutative.
where
operation * is not associative.
Question:2(v) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
Answer:
(v) On , define
and
for
the operation is not commutative.
where
operation * is not associative.
Question:2(vi) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
Answer:
(iv) On , define
and
for
the operation is not commutative.
where
operation * is not associative.
Question:3 Consider the binary operation on the set defined by . Write the operation table of the operation .
Answer:
for
The operation table of the operation is given by :
| 1 | 2 | 3 | 4 | 5 |
1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 2 | 2 | 2 | 2 |
3 | 1 | 2 | 3 | 3 | 3 |
4 | 1 | 2 | 3 | 4 | 4 |
5 | 1 | 2 | 3 | 4 | 5 |
Question:4(i) Consider a binary operation ∗ on the set given by the following multiplication table (Table 1.2).
(Hint: use the following table)
Answer:
(i)
Question:4(ii) Consider a binary operation ∗ on the set given by the following multiplication table (Table 1.2).
(Hint: use the following table)
Answer:
(ii)
For every , we have . Hence it is commutative.
Question:4(iii) Consider a binary operation ∗ on the set { given by the following multiplication table (Table 1.2).
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
(Hint: use the following table)
Answer:
(iii) (2 ∗ 3) ∗ (4 ∗ 5).
from the above table
Answer:
for
The operation table is as shown below:
| 1 | 2 | 3 | 4 | 5 |
1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 2 | 1 | 2 | 1 |
3 | 1 | 1 | 3 | 1 | 1 |
4 | 1 | 2 | 1 | 4 | 1 |
5 | 1 | 1 | 1 | 1 | 5 |
The operation ∗′ same as the operation ∗ defined in Exercise 4 above.
Question:6(i) let ∗ be the binary operation on N given by . Find
Answer:
a*b=LCM of a and b
(i) 5 ∗ 7, 20 ∗ 16
Question:6(ii) Let ∗ be the binary operation on N given by . Find
Answer:
(ii) for all
Hence, it is commutative.
Question:6(iii) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
Answer:
a b = L.C.M. of a and b
(iii)
Hence, the operation is associative.
Question:6(iv) Let ∗ be the binary operation on N given by . Find
Answer:
(iv) the identity of ∗ in N
We know that
for
Hence, 1 is the identity of ∗ in N.
Question 6(v) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
(v) Which elements of N are invertible for the operation ∗?
Answer:
An element a is invertible in N
if
Here a is inverse of b.
a*b=1=b*a
a*b=L.C.M. od a and b
a=b=1
So 1 is the only invertible element of N
Question:7 Is ∗ defined on the set by a binary operation? Justify your answer.
Answer:
A =
Operation table is as shown below:
| 1 | 2 | 3 | 4 | 5 |
1 | 1 | 2 | 3 | 4 | 5 |
2 | 2 | 2 | 6 | 4 | 10 |
3 | 3 | 6 | 3 | 12 | 15 |
4 | 4 | 4 | 12 | 4 | 20 |
5 | 5 | 10 | 15 | 20 | 5 |
From the table, we can observe that
Hence, the operation is not a binary operation.
Answer:
a ∗ b = H.C.F. of a and b for all
H.C.F. of a and b = H.C.F of b and a for all
Hence, operation ∗ is commutative.
For ,
Hence, ∗ is associative.
An element will be identity for operation * if for .
Hence, the operation * does not have any identity in N.
Question:9(i) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(i) Find which of the binary operations are commutative and which are associative.
Answer:
On the set Q ,the operation * is defined as .It is observed that:
here
Hence, the * operation is not commutative.
It can be observed that
for all
The operation * is not associative.
Question:9(ii) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(ii) Find which of the binary operations are commutative and which are associative.
Answer:
On the set Q ,the operation * is defines as .It is observed that:
For
Hence, the * operation is commutative.
It can be observed that
for all
The operation * is not associative.
Question:9(iii) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(iii) Find which of the binary operations are commutative and which are associative.
Answer:
On the set Q ,the operation * is defines as .It is observed that:
For
for
Hence, the * operation is not commutative.
It can be observed that
for all
The operation * is not associative.
Question:9(iv) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(iv) Find which of the binary operations are commutative and which are associative.
Answer:
On the set Q ,the operation * is defined as .It is observed that:
For
for
Hence, the * operation is commutative.
It can be observed that
for all
The operation * is not associative.
Question:9(v) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(v) Find which of the binary operations are commutative and which are associative.
Answer:
On the set Q ,the operation * is defines as .It is observed that:
For
for
Hence, the * operation is commutative.
It can be observed that
for all
The operation * is associative.
Question:9(vi) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(vi) Find which of the binary operations are commutative and which are associative.
Answer:
On the set Q ,the operation * is defines as .It is observed that:
For
for
Hence, the * operation is not commutative.
It can be observed that
for all
The operation * is not associative.
Question:10 Find which of the operations given above has identity.
Answer:
An element will be identity element for operation *
if for all
when .
Hence, has identity as 4.
However, there is no such element which satisfies above condition for all rest five operations.
Hence, only (v) operations have identity.
Question:11 Let and ∗ be the binary operation on A defined by Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.
Answer:
and ∗ be the binary operation on A defined by
Let
Then,
We have
Thus it is commutative.
Let
Then,
Thus, it is associative.
Let will be a element for operation * if for all .
i.e.
This is not possible for any element in A .
Hence, it does not have any identity.
Question:12(i) State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation ∗ on a set N,
Answer:
(i) For an arbitrary binary operation ∗ on a set N,
An operation * on a set N as
Then , for b=a=2
Hence, statement (i) is false.
Question:12(ii) State whether the following statements are true or false. Justify.
(ii) If ∗ is a commutative binary operation on N, then
Answer:
(ii) If ∗ is a commutative binary operation on N, then
R.H.S
(* is commutative)
( as * is commutative)
= L.H.S
Hence, statement (ii) is true.
Question:13 Consider a binary operation ∗ on N defined as . Choose the correct answer.
Answer:
A binary operation ∗ on N defined as .
For
Thus, it is commutative.
where
Hence, it is not associative.
Hence, B is the correct option.
NCERT solutions for class 12 maths chapter 1 Relations and Functions: Miscellaneous Exercise
Question:1 Let be defined as . Find the function such that .
Answer:
and
For one-one:
Thus, f is one-one.
For onto:
For ,
Thus, for , there exists such that
Thus, f is onto.
Hence, f is one-one and onto i.e. it is invertible.
Let as
and
Hence, defined as
Answer:
if n is odd
if n is even.
For one-one:
Taking x as odd number and y as even number.
Now, Taking y as odd number and x as even number.
This is also impossible.
If both x and y are odd :
If both x and y are even :
f is one-one.
Onto:
Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.
Thus, f is onto.
Hence, f is one-one and onto i.e. it is invertible.
Sice, f is invertible.
Let as if m is even and if m is odd.
When x is odd.
When x is even
Similarly, m is odd
m is even ,
and
Hence, f is invertible and the inverse of f is g i.e. , which is the same as f.
Hence, inverse of f is f itself.
Question:3 If f : R → R is defined by f(x) = x ^{ 2 } – 3x + 2, find f (f (x)).
Answer:
This can be solved as following
f : R → R
Question:4 Show that the function defined by is one one and onto function.
Answer:
The function defined by
,
One- one:
Let ,
It is observed that if x is positive and y is negative.
Since x is positive and y is negative.
but 2xy is negative.
Thus, the case of x is positive and y is negative is removed.
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:
When x and y both are negative :
f is one-one.
Onto:
Let such that
If y is negative, then
If y is positive, then
Thus, f is onto.
Hence, f is one-one and onto.
Question:5 Show that the function given by is injective.
Answer:
One-one:
Let
We need to prove .So,
Let then there cubes will not be equal i.e. .
It will contradict given condition of cubes being equal.
Hence, and it is one -one which means it is injective
Question:6 Give examples of two functions and such that is injective but g is not injective. (Hint : Consider and ).
Answer:
One - one:
Since
As we can see but so is not one-one.
Thus , g(x) is not injective.
Let
Since, so x and y are both positive.
Hence, gof is injective.
Question:7 Give examples of two functions and such that is onto but is not onto.
Answer:
and
and
Onto :
Consider element in codomain N . It is clear that this element is not an image of any of element in domain N .
f is not onto.
Now, it is clear that , there exists such that .
Hence, is onto.
Answer:
Given a non empty set X, consider P(X) which is the set of all subsets of X.
Since, every set is subset of itself , ARA for all
R is reflexive.
Let
This is not same as
If and
then we cannot say that B is related to A.
R is not symmetric.
If
this implies
R is transitive.
Thus, R is not an equivalence relation because it is not symmetric.
Answer:
Given is defined as .
As we know that
Hence, X is the identity element of binary operation *.
Now, an element is invertible if there exists a ,
such that (X is identity element)
i.e.
This is possible only if .
Hence, X is only invertible element in with respect to operation *
Question:10 Find the number of all onto functions from the set to itself.
Answer:
The number of all onto functions from the set to itself is permutations on n symbols 1,2,3,4,5...............n.
Hence, permutations on n symbols 1,2,3,4,5...............n = n
Thus, total number of all onto maps from the set to itself is same as permutations on n symbols 1,2,3,4,5...............n which is n.
Question:11(i) Let and . Find of the following functions F from S to T, if it exists.
Answer:
is defined as
is given by
Question:11(ii) Let and . Find of the following functions F from S to T, if it exists.
Answer:
is defined as
, F is not one-one.
So inverse of F does not exists.
Hence, F is not invertible i.e. does not exists.
Answer:
Given and is defined as
and
For , we have
the operation is commutative.
where
the operation is not associative
Let . Then we have :
Hence,
Now,
for
Hence, operation o does not distribute over operation *.
Question:13 Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).
Answer:
It is given that
be defined as
Now, let .
Then,
And
Therefore,
Therefore, we can say that is the identity element for the given operation *.
Now, an element A P(X) will be invertible if there exists B P(X) such that
Now, We can see that
such that
Therefore, by this we can say that all the element A of P(X) are invertible with
Question:14 Define a binary operation ∗ on the set as Show that zero is the identity for this operation and each element of the set is invertible with being the inverse of .
Answer:
X = as
An element is identity element for operation *, if
For ,
Hence, 0 is identity element of operation *.
An element is invertible if there exists ,
such that i.e.
means or
But since we have X = and . Then .
is inverse of a for .
Hence, inverse of element , is 6-a i.e. ,
Question:15 Let , and be functions defined by and . Are and equal? Justify your answer. (Hint: One may note that two functions and such that , are called equal functions).
Answer:
Given :
,
are defined by and .
It can be observed that
Hence, f and g are equal functions.
Question:16 Let . Then number of relations containing and which are reflexive and symmetric but not transitive is
Answer:
The smallest relations containing and which are
reflexive and symmetric but not transitive is given by
, so relation R is reflexive.
and , so relation R is symmetric.
but , so realation R is not transitive.
Now, if we add any two pairs and 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:17 Let . Then number of equivalence relations containing is
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
The number of equivalence relations containing is given by
We are left with four pairs , , .
, so relation R is reflexive.
and , so relation R is not symmetric.
, so realation R is not transitive.
Hence , equivalence relation is bigger than R is universal relation.
Thus the total number of equivalence relations cotaining is two.
Thus, option B is correct.
Question:18 Let be the Signum Function defined as and be the Greatest Integer Function given by , where is greatest integer less than or equal to . Then, does and coincide in ?
Answer:
is defined as
is defined as
Let
Then we have , if x=1 and
Hence,for , and .
Hence , gof and fog do not coincide with .
Question:19 Number of binary operations on the set are(A) 10(B) 16(C) 20(D ) 8
Answer:
Binary operations on the set are is a function from
i.e. * is a function from
Hence, the total number of binary operations on set is
Hence, option B is correct
NCERT solutions for class 12 maths - Chapter wise
NCERT solutions for class 12 subject wise
NCERT Solutions class wise
- NCERT solutions for class 12
- NCERT solutions for class 11
- NCERT solutions for class 10
- NCERT solutions for class 9
NCERT solutions for class 12 maths chapter 1 Relations and Functions is helpful for the students who wish to perform well in the CBSE 12 board examination.
Some guidelines to follow to make the best use of NCERT solutions:
- Before starting to solve exercise, first solve examples that are given in the NCERT class 12 maths textbook.
- Also, try to solve every exercise including miscellaneous exercise, NCERT chapter examples, miscellaneous examples on your own, if you are not able to do it, then you can take help of NCERT solutions for class 12 maths chapter 1 Relations and functions.
- Reading the solutions is not enough, you have to solve it on your own, even after reading the solutions
- Stick to the syllabus that is provided by NCERT and solve it completely including all examples and all exercises
- If you have solved all NCERT questions, then you can solve previous years papers of CBSE board to get familiar with the pattern of the exam.
Happy learning !!!
F#
Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question: Does CBSE provides the solutions of NCERT class 12 maths ?
Answer:
No, CBSE doesn’t provided NCERT solutions for any class or subject
Question: Where can I find the complete solutions of NCERT class 12 maths ?
Answer:
A Here you will get the detailed NCERT solutions for class 12 maths by clicking on the link.
Question: What are the important topics in chapter relations and functions ?
Answer:
Definitions of relations and functions, types of relations, types of functions, composition of functions, invertible function and binary operations are the important topics in this chapter.
Question: What is the weightage of the chapter relations and functions for CBSE board exam ?
Answer:
Two chapters 'relation and function' and 'inverse trigonometry' combined has 10 % weightage in the CBSE final board exam.
Question: Which is the official website of NCERT ?
Answer:
NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.
Question: How the NCERT solutions are helpful in the board exam ?
Answer:
As CBSE board exam paper is designed entirelly based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. NCERT solutions are not only important when you stuck while solving the problems but students will get how to answer in the board exam in order to get good marks in the board exam.
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