NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions :- This is the first chapter of Class 12 math. NCERT solutions class 12 maths chapter 1 Relations and Functions contains the answer and step-by-step solution to each question asked in the exercise of NCERT Class 12 maths book. NCERT Class 12 maths solutions Chapter 1 will help you to understand the concepts and score well in CBSE 12th board exam.
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This chapter 1 12th class is not only important for mathematics, but also it is important in real life too. In this article, you will find NCERT class 12 maths chapter 1 solutions including miscellaneous exercise which will help you to score more marks in the exam. Here you will find all NCERT solutions of chapter 1 maths class 12 at a single place which will be helpful when you are not able to solve the NCERT questions.
In NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions, there are four exercises with 55 questions and one miscellaneous exercise with 19 questions. relations and functions class 12 solutions are very important for students because they comprise quality practice questions. In this article, you will find the detailed NCERT solutions for class 12 maths chapter 1. Here you will get NCERT solutions for class 12 also.
Also Read:
- Class 12 Maths Chapter 1 Relations And Functions Notes
- NCERT Exemplar Solutions For Class 12 Maths Chapter 1 Relations And Functions
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 A= {1,2,3,4}
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 = {1,2,3,4}
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 Λ on the set {1, 2, 3, 4, 5} defined by
a Λ b = min {a, b}. Write the operation table of the operation Λ .
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. ,
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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
.