NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Maths
Chapter : 1
Chapter Name : Relation And Functions

Q1 Determine whether each of the following relations are re exive, symmetric and transitive:
(i)Relation R in the set A = {1, 2, 3…13, 14} defined as
R = {(x, y) : 3x − y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y) : x − y is as integer }
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y) : x and y work at the same place }
(b) R = {(x, y) : x and y live in the same locality }
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}

( i) A = {1, 2, 3…13, 14}
R = {(x, y) : 3x − y = 0}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2)…(14, 14) ∉ R.
Also,R is not symmetric as (1, 3) ∈ R,but (3, 1) ∉ R. [3(3) − 1 ≠ 0]
Also,R is not transitive as (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R
[3(1) − 9 ≠ 0]
Hence, R is neither re exive, nor symmetric, nor transitive.

(ii) R = {(x, y) : y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}
It is seen that (1, 1) ∉ R .
∴ R is not reflexive.
(1, 6) ∈ R
But,

(1, 6) ∉ R
∴ R is not symmetric.
Now, since there is no pair in R such that (x, y) and (y, z) ∈ R, then (x, z) cannot belong
to R.
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

(iii) A = {1, 2, 3, 4, 5, 6}
R = {(x, y) : y is divisible by x}
We know that any number (x) is divisible by itself.
⇒ (x, x) ∈ R
∴ R is reflexive.
Now,
(2, 4) ∈ R[ as 4 is divisible by 2]
But,
(4, 2) ∉ R. [as 2 is not divisible by 4
∴ R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y .
∴ z is divisible by x .
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is re exive and transitive but not symmetric.

( iv )R = {(x, y) : x − y is an integer }
Now, for every x ∈ Z, (x, x) ∈ R as x − x = 0 is an integer.
∴ R is reflexive.
Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer.
⇒ − (x − y) is also an integer.
⇒ (y − x) is an integer.
∴ (y, x) ∈ R
∴ R is symmetric.
Now,
Let (x, y) and (y, z) ∈ R, where x, y, z ∈ Z .
⇒ (x − y) and (y − z) are integers.
⇒ (x − z = (x − y) + (y − z) is an integer.
∴ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.

(v)(a)R = {(x, y) : x and y work at the same place }
⇒ (x, x) ∈ R
∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
⇒ y and x work at the same place.
⇒ (y, x) ∈ R.
∴ R is symmetric.
Now, let (x, y), (y, z) ∈ R
⇒ x and y work at the same place and y and z work at the same place.
⇒ x and z work at the same place.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.

(b) R = {(x, y) : x and y live in the same locality }
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈ R, then x and y live in the same locality.
⇒ y and x live in the same locality.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R .
⇒ x and y live in the same locality and y and z live in the same locality.
⇒ x and z live in the same locality.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.

(c) R = {(x, y) : x is exactly 7cm taller than y}
Now,
(x, x) ∉ R
since human being x cannot be taller than himself.
∴ R is not reflexive.
Now, let (x, y) ∈ R .
⇒ x is exactly 7cm taller than y
Then, y is not taller than x .
∴ (y, x) ∉ R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
.•.R is not symmetric.

Let (x, y), (y, z) ∈ R .
⇒ x is exactly 7cm taller than y and y is exactly 7cm taller than z.
⇒ x is exactly 14cm taller than z .
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

(d) R = {(x, y) : x is the wife of y}
Now,
(x, x) ∉ R
since x cannot be the wife of herself.
∴ R is not reflexive.
Now, let (x, y) ∈ R
⇒ x is the wife of y
Clearly y is not the wife of x.
∴ (y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x.
∴ R is not transitive.
Let (x, y), (y, z) ∈ R
⇒ x is the wife of y and y is the wife of z
This case is not possible. Also, this does not imply that x is the wife of z .
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

(e) R = {(x, y) : x is the father of y}
(x, x) ∉ R
As x cannot be the father of himself.
∴ R is not reflexive.
Now, let (x, y) ∈ R .
⇒ x is the father of y
⇒ y cannot be the father of y
Indeed, y is the son or the daughter of y.
∴ (y, x) ∉ R
∴ R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R .
⇒ x is the father of y and y is the father of z .
⇒ x is not the father of z
Indeed x is the grandfather of z .
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Page : 5 , Block Name : Exercise 1.1

Q2 Show that the relation R in the set R of real numbers,de ned as R = {(a, b) : a ≤ b } is neither
2

re exive, nor symmetric, nor transitive.

R= {(a, b) : a ≤ b }
2

It can be observed that
( )1 1
,
2 2
∉ R, since 2 >
1
()
1
2
2 1
= 4

∴ R is not reflexive.
Now, (1, 4) ∈ R as 1 < 4 2
But, 4 is not less than 1 2 .
∴ (4, 1) ∉ R
∴ R is not symmetric.
Now,
(3, 2), (2, 1.5) ∈ R
(as 3 < 2 2 = 4 and 2 < (1.5) 2 = 2.25 )
But, 3 > (1.5) 2 = 2.25
∴ (3, 1.5) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Page : 5 , Block Name : Exercise 1.1

Q3 Check whether the relation R de ned in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is
re exive, symmetric or transitive.

Let A = {1, 2, 3, 4, 5, 6}
A relation R is defined on set A as:
R = {(a, b) : b = a + 1}
∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We can find (a, a) ∉ R, where a ∈ A
For instance,
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
∴ R is not reflexive.
It can be observed that (1, 2) ∈ R, but (2, 1) ∈ R .
∴ R is not symmetric.

Now, (1, 2), (2, 3) ∈ R
But,
(1, 3) ∉ R
∴ R is not transitive
Hence, R is neither reflexive, nor symmetric, nor transitive.

Page : 5 , Block Name : Exercise 1.1

Q4 Show that the relation R in R de ned as R = {(a, b) : a ≤ b}, is re exive andtransitive but not
symmetric.

R = {(a, b); a ≤ b}
Clearly (a, a) ∈ R as a = a
∴ R is reflexive.
Now,
(2, 4) ∈ R( as 2 < 4)
But, (4, 2) ∉ R as 4 is greater than 2 .
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R
Then,
a ≤ b and b ≤ c
⇒a≤c
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.

Page : 5 , Block Name : Exercise 1.1

Q5 Check whether the relation R in R de ned by R = {(a, b) : a ≤ b } is re exive,symmetric or
3

transitive.

R= {(a, b) : a ≤ b }3

It is observed that ( ) 1 1
,
2 2
1
∉ R as 2 > ()
1
2
3 1
= 8

∴ R is not reflexive.
Now,

(
(1, 2) ∈ R as 1 < 2 3 = 8 )
But,

(
(2, 1) ∉ R as 2 3 > 1 )
∴ R is not symmetric.

( )( ) 3
We have 3, 2 , 2 , 5
3 6
∈ R as 3 <
() 3
2
3 3
and 2 <
()
6
5
3

( )
But 3, 5
6
()∉ R as 3 >
6
5
3

∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Page : 5 , Block Name : Exercise 1.1

Q6 Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither
re exive nor transitive.

Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen that (1, 1), (2, 2), (3, 3) ∉ R .
∴ R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R
However,
(1, 1) ∉ R
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.

Page : 6 , Block Name : Exercise 1.1

Q7 Show that the relation R in the set A of all the books in a library of a college, given by R = { (x,
y): x and y have same number of pages} is an equivalence relation.

Answer. Set A is the set of all books in the library of a college.
R = {x, y): x and y have the same number of pages}
Now, R is re exive since (x, x) R as x and x has the same number of pages.

Let (x, y) ∈ R ⇒ x and y have the same number of pages.
⇒ y and x have the same number of pages.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R
⇒ x and y and have the same number of pages and y and z have the same number of
pages.
⇒ x and z have the same number of pages.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.

Page : 6 , Block Name : Exercise 1.1

Q8 Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an
equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the
elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of
{2, 4}.

A = {1, 2, 3, 4, 5}
R = {(a, b) : | a − b | is even }
It is clear that for any element a ∈ A, we have | a − a | = 0 (which is even).
∴ R is reflexive.
Let (a, b) ∈ R
⇒ | a − b | is even.
⇒ | − (a − b) | = | b − a | is also even.
⇒ (b, a) ∈ R
∴ R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R .
⇒ | a − b | is even and | b − c | is even.
⇒ (a − b) is even and (b − c) is even.
⇒ (a − c) = (a − b) + (b − c) is even.
⇒ | a − c | is even.
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 2, 3} are related to each other as all the elements of this subset are
odd. Thus, the modulus of the difference between any two elements will be even
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset
are even. Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all
elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the

difference between the two elements (from each of these two subsets) will not be even.

Page : 6 , Block Name : Exercise 1.1

Q9
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : | a − b | is a multiple of 4}
(ii) R = {(a, b); a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.

A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {(a, b) : | a − b | is a multiple of 4}
For any element a ∈ A, we have (a, a) ∈ R as | a − a | = 0 is a multiple of 4.
∴ R is reflexive.
Now, let (a, b) ∈ R ⇒ | a − b | is a multiple of 4.
⇒ | − (a − b) | = | b − a | is a multiple of 4
⇒ (b, a) ∈ R
∴ R is symmetric.
Now, let (a, b), (b, c) ∈ R .
⇒ | a − b | is a multiple of 4 and | b − c | is a multiple of 4 .
⇒ (a − b) is a multiple of 4 and (b − c) is a multiple of 4.
⇒ (a − c) = (a − b) + (b − c) is a multiple of 4.
⇒ | a − c | is a multiple of 4.
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9} since
| 1 − 1 | = 0 is a multiple of 4
| 5 − 1 | = 4 is a multiple of 4, and
| 9 − 1 | = 8 is a multiple of 4.
(ii) R = {(a, b) : a = b}
For any element a ∈ A, we have (a, a) ∈ R, since a = a .
∴ R is reflexive.
Now, let (a, b) ∈ R
⇒a=b
⇒b=a
⇒ (b, a) ∈ R

∴ R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R .
⇒ a = b and b = c
⇒a=c
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A which are equal
to 1 .
Hence, the set of elements related to 1 is {1}

Page : 6 , Block Name : Exercise 1.1

Q10 Given an example Of a relation. Which is
(i) Symmetric but neither re exive nor transitive.
(ii) Transitive but neither re exive nor symmetric.
(iii) Re exive and symmetric but not transitive.
(iv) Re exive and transitive but not symmetric.
(v) Symmetric and transitive but not re exive.

(i) Let A = {5, 6, 7} .
Define a relation R on A as R = {(5, 6), (6, 5)} .
Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R .
Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.
⇒ (5, 6), (6, 5) ∈ R, but (5, 5) ∈ R
∴ R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.

(ii) Consider a relation R in R defined as:
R = {(a, b) : a < b}
For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself. In fact, a =
a.
∴ R is not reflexive.
Now,
(1, 2) ∈ R( as 1 < 2)
But, 2 is not less than 1
∴ (2, 1) ∉ R
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R
∴ R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.

(iii) Let A = {4, 6, 8}.
Define a relation R on A as:
A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}
Relation R is reflexive since for every a ∈ A, (a, a) ∈ R i.e., (4, 4), (6, 6), (8, 8)} ∈ R .
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R .
Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R .
Hence, relation R is reflexive and symmetric but not transitive.

(iv) Define a relation R in R as:
R = {a, b) : a 3 ≥ b 3
Clearly (a, a) ∈ R as a 3 = a 3
∴ R is reflexive.
Now,

(
(2, 1) ∈ R as 2 3 ≥ 1 3 )
But,

(
(1, 2) ∉ R as 1 3 < 2 3 )
∴ R is not symmetric.
Now,
Let (a, b), (b, c) ∈ R
⇒ a 3 ≥ b 3 and b 3 ≥ c 3
⇒ a3 ≥ c3
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, relation R is reflexive and transitive but not symmetric.

(v) Let A = { − 5, − 6}.
Define a relation R on A as:
R = {( − 5, − 6), ( − 6, − 5), ( − 5, − 5)}
Relation R is not reflexive as ( − 6, − 6) ∉ R
Relation R is symmetric as ( − 5, − 6) ∈ R and ( − 6, − 5} ∈ R .
It is seen that ( − 5, − 6), ( − 6, − 5) ∈ R. Also, ( − 5, − 5) ∈ R
∴ The relation R is transitive.
Hence, relation R is symmetric and transitive but not re exive.

Page : 6 , Block Name : Exercise 1.1

Q11 Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the
point P from the origin is same as the distance of the point Q from the origin}, is an equivalence
relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing

through p with origin as centre.

Answer. R = { (P,Q): distance of point p from the origin is the same as the distance of point Q from
the origin}
Clearly, (P, P) ϵ R since the distance Of point P from the origin is always the same as the distance
of the same point P from the origin.
∴ R is reflexive.
Now,
Let (P, Q) ∈ R
The distance of point P from the origin is the same as the distance of point Q from the
origin.
The distance of point Q from the origin is the same as the distance of point P from the origin.
⇒ (Q, P) ∈ R
∴ R is symmetric.
Now,
Let (P, Q), (Q, S) ∈ R
The distance of points p and Q from the origin is the same and also, the distance of points Q and S
from the origin is the same.
The distance of points p and S from the origin is the Same.
⇒ (P, S) ∈ R
∴ R is transitive.
The set of all points related to P not ≠ (O, O) will be those points whose distance from the origin
is the same as the distance of point P from the origin.
In other words, if O (O, O) is the origin and OP = k, then the set of all points related to P is at a
distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through
point P.

Page : 6 , Block Name : Exercise 1.1

Q12
Show that the relation R defined in the set A of all triangles as R = {(T1, T2 ) : T1 is similar
to T 2}, is equivalence relation. Consider three right angle triangles T 1 with sides 3, 4, 5,
T 2 with sides 5, 12, 13 and T 3 with sides 6, 8, 10. Which triangles among T 1, T 2 and T 3
are related?

R= {(T1, T2 ) : T1 is similar to T2}
R is reflexive since every triangle is similar to itself.

( )
Further, if T 1, T 2 ∈ R, then T 1 is similar to T 2

⇒ T 2 is similar to T 1 .

(
⇒ T 2, T 1 ∈ R )
Now,

( )(
Let T 1, T 2 , T 2, T 3 ∈ R . )
⇒ T 1 is similar to T 2 and T 2 is similar to T 3
⇒ T 1 is similar to T 3

(
⇒ T 1, T 3 ∈ R )
∴ R is transitive.
Thus, R is an equivalence relation.
Now, we can observe that:

3
6
4
= 8 = 10
5
( ) 1
= 2

∴ R is transitive.
Thus, R is an equivalence relation.
Now, we can observe that:

3
6
4
= 8 = 10
5
( ) 1
= 2

∴ The corresponding sides of triangles T 1 and T 3 are in the same ratio.
Then, triangle T 1 is similar to triangle T 3 .
Hence, T 1 is related to T 3 .

Page : 6 , Block Name : Exercise 1.1

Q13
Show that the relation R defined in the set A of all polygons as R = {(P1, P2 ) : P1 and P2
have same number of sides }, is an equivalence relation. What is the set of all elements
in A related to the right angle triangle T with sides 3, 4 and 5 ?

R= {(P1, P2 ) : P1 and P2 have same the number of sides }
R is reflexive since (P 1, P 1 ) ∈ R as the same polygon has the same number of sides with

itself.

( )
Let P 1, P 2 ∈ R .

⇒ P 1 and P 2 have the same number of sides.
⇒ P 2 and P 1 have the same number of sides.

( )
⇒ P 2, P 1 ∈ R

∴ R is symmetric.
Now,

( )( )
Let P 1, P 2 , P 2, P 3 ∈ R .

⇒ P 1 and P 2 have the same number of sides. Also, P 2 and P 3 have the same number of
sides.
⇒ P 1 and P 3 have the same number of sides.

( )
⇒ P 1, P 3 ∈ R

∴ R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons
which have 3 sides (since Tis a polygon with 3 sides).
Hence, the set of all elements in A related to triangle Tis the set of all triangles.

Page : 6 , Block Name : Exercise 1.1

Q14
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1,
L 2) : L 1 is parallel to L 2}. Show that R is an equivalence relation. Find the set ofl lines
related to the line y = 2x + 4 .

R= {(L1, L2 ) : L1 is parallel to L2 }
( )
R is reflexive as any line L 1 is parallel to itself i.e., L 1, L 1 ∈ R

Now,

( )
Let L 1, L 2 ∈ R

⇒ L 1 is parallel to L 2
⇒ L 2 is parallel to L 1

( )
⇒ L 2, L 1 ∈ R

∴ R is symmetric.
Now,

( )(
Let L 1, L 2 , L 2, L 3 ∈ R)
⇒ L 1 is parallel to L 2 . Also, L 2 is parallel to L 3
⇒ L 1 is parallel to L 3
∴ R is transitive.
The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to
the line y = 2x + 4 .
Slope of line y = 2x + 4 is m = 2
It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form y = 2x + c, where c ∈ R .
Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R .

Page : 6 , Block Name : Exercise 1.1

Q15
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (2, 2), (1, 1), (4, 4),
(1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.

R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈ {1, 2, 3, 4}
∴ R is reflexive.
It is seen that (1, 2) ∈ R, but (2, 1) ∉ R .
∴ R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4} .
∴ R is transitive.
Hence, R is re exive and transitive but not symmetric. The correct answer is B.

Page : 7 , Block Name : Exercise 1.1

Q16

Let R be the relation in the set N given by R = {(a, b) : a = b − 2, b > 6}. Choose th
correct answer.
(A) (2, 4) ∈ R(B)(3, 8) ∈ R(C)(6, 8) ∈ R(D)(8, 7) ∈ R

R = {(a, b) : a = b − 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 − 2, (3, 8) ∉ R
And, as 8 ≠ 7 − 2
∴ (8, 7) ∉ R
∴ (8, 7) ∉ R
Now, consider (6, 8)
We have 8 > 6 and also, 6 = 8 − 2
∴ (6, 8) ∈ R
The correct answer is C.

Page : 7 , Block Name : Exercise 1.1

1
Q1 Show that the function f : R ∗ → R ∗ de ned by f(x) = x is one-one and onto, where is the set
of all non-zero real numbers. Is the result true, if the domain R. is replaced by N with co-domain
same as R. ?

1
f : R ∗ → R ∗ is defined by f(x) = x one-one :
f(x) = f(y)
1 1
⇒ x = y

⇒x=y
∴ f is one-one.
Onto:
1
It is clear that for y ∈ R ∗ , there existx = y ∈ R ∗ ( Exists as y ≠ 0) such that
1
f(x) = =y
()
1
y

∴ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g : N → R ∗ defined by

1
g(x) = x
We have,
1 1
( ) ( )
g x1 = g x2 ⇒ x = x ⇒ x1 = x2
1 2

a is one-one.
Further, it is clear that g is not onto as for 1.2 ∈ R ∗ there does not exist any x in N such
1
that g(x) = 1.2

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

Page : 10 , Block Name : Exercise 1.2

Q2 Check the injective and surjective of the following activity of the following functions :
(i) f : N → N given by f(x) = x 2
(ii) f : Z → Z given by f(x) = x 2
(iii) f : R → R given by f(x) = x 2
( iv )f : N → N given by f(x) = x 3
(v)f : Z → Z given by f(x) = x 3

(i) f : N → N is given by,
f(x) = x 2
It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x 2 = y 2 ⇒ x = y .
if is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x 2 = 2 .
∴ f is not surjective.
Hence, function f is injective but not surjective.

(ii) f : z → z is given by,
f(x) = x 2
It is seen that f( − 1) = f(1) = 1, but − 1 ≠ 1
∴ f is not injective.
Now, − 2 ∈ Z . But, there does not exist any element x ∈ Z such that f(x) = x 2 = − 2 .
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iii) f : R → R is given by,
f(x) = x 2
It is seen that f( − 1) = f(1) = 1, but − 1 ≠ 1
∴ f is not injective.

Now, − 2 ∈ R . But, there does not exist any element x ∈ R such that f(x) = x 2 = − 2 .
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iv) f : N → N given by,
f(x) = x 3
It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x 3 = y 3 ⇒ x = y
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x 3 =
2.
∴ f is not surjective
Hence, function f is injective but not surjective.

(v) f : Z → Z is given by,
f(x) = x 3
It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x 3 = y 3 ⇒ x = y .
∴ f is injective.
Now, 2 ∈ Z . But, there does not exist any element x in domain z such that f(x) = x 3 = 2 .
∴ f is not surjective.
Hence, function f is injective but not surjective.

Page : 10 , Block Name : Exercise 1.2

Q3 Prove that the greatest integer function f : R → R given by f(x) = [x] in neither one-once nor
onto, where [x] denotes the greatest integer less than or equal to x .

f : R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9
∴ f is not one-one.
Now, consider 0.7 ∈ R .
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x
∈ R such that f(x) = 0.7
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.

Page : 10 , Block Name : Exercise 1.2

Q4 Show that the modulus function f : R → R given by f(x) = | x | , is either one on one nor onto,
where | x | is x, if x is positive or 0 and | x | is − x, if x is negative.

f : R → R is given by,

f(x) = | x | =
{ x, if x ≥ 0
− x, if x < 0

It is seen that f( − 1) = | − 1 | = 1, f(1) = | 1 | = 1
∴ f( − 1) = f(1), but − 1 ≠ 1
∴ f is not one-one.
Now, consider − 1 ∈ R .
It is known that f(x) = | x | is always non-negative. Thus, there does not exist any
element x in domain R such that f(x) = | x |
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.

Page : 11 , Block Name : Exercise 1.2

{
1, if x > 0
Q5 Show that the signum Function f : R → R, given by f(x) = 0, if x = 0 is neither one-one nor
− 1, if x < 0
onto.

f : R → R is given by,
1, if x > 0
0, if x = 0
− 1, if x < 0
It is seen that f(1) = f(2) = 1, but 1 ≠ 2 .
∴ f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or − 1) for the element − 2 in co-domain R , there
does not exist any x in domain R such that f(x) = − 2 .
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.

Page : 11 , Block Name : Exercise 1.2

Q6
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from
A to B . Show that f is one-one.

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7} .
f : A → B is defined as f = {(1, 4), (2, 5), (3, 6)}
∴ f(1) = 4, f(2) = 5, f(3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.

Page : 11 , Block Name : Exercise 1.2

Q7
In each of the following cases, state whether the function is one-one, onto or bijective.
Justify your answer.
(i) f : R → R defined by f(x) = 3 − 4x
(ii) f : R → R defined by f(x) = 1 + x 2

(i) f : R → R is defined as f(x) = 3 − 4x

( ) ( )
Let x 1, x 2 ∈ R such that f x 1 = f x 2

⇒ 3 − 4x 1 = 3 − 4x 2
⇒ − 4x 1 = − 4x 2
∴ x1 = x2
∴ f 1 is one-one.
3−y
For any real number (y) in R, there exists 4 in R such that

f
( )
3−y
4 =3−4
( ) 3−y
4 =y

∴ f is onto.
Hence, f is bijective.

(ii) f : R → R is defined as
f(x) = 1 + x 2

( ) ( )
Let x 1, x 2 ∈ R such that f x 1 = f x 2
2 2
⇒ 1 + x1 = 1 + x2

⇒ x 21 = x 22
⇒ x1 = ± x2

( ) ( )
∴ f x 1 = f x 2 does not imply that x 1 = x 2

For instance,

f(1) = f( − 1) = 2
∴ f is not one-one.
Consider an element − 2 in co-domain R .
It is seen that f(x) = 1 + x 2 is positive for all x ∈ R .
Thus, there does not exist any x in domain R such that f(x) = − 2 .
∴ f is not onto.
Hence, f is neither one-one nor onto.

Page : 11 , Block Name : Exercise 1.2

Q8
Let A and B be sets. Show that f : A × B → B × A such that (a, b) = (b, a) is bijective
function.

f : A × B → B × A is defined as f(a, b) = (b, a)

( )( ) ( ) (
Let a 1, b 1 , a 2, b 2 ∈ A × B such that f a 1, b 1 = f a 2, b 2 )
⇒ ( b 1, a 1 ) = ( b 2, a 2 )

⇒ b 1 = b 2 and a 1 = a 2

( ) (
⇒ a 1, b 1 = a 2, b 2 )
∴ f is one-one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). [By definition of f
∴ f is onto.
Hence, f is bijective.

Page : 11 , Block Name : Exercise 1.2

Q9 Let

{
n+1
2
, if n is odd
f : N → N be de ned by f(n) = n
for all n ∈ N.
2
, if n is even

State whether the function f is bijective. Justify your answer.

{
n+1
2
, if n is odd
f : N → N be de ned is f(n) = n
for all n ∈ N.
2
, if n is even

It can be observe that
1+1 2
f(1) = 2
= 1 and f(2) = 2 = 1 [ By definition of f]

∴ f(1) = f(2), where 1 ≠ 2
∴ f(1) not one-one.
Consider a natural number (n) in co-domain N .
Case I: n is odd
∴ n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1 ∈ N such that
4r + 1 + 1
f(4r + 1) = 2
= 2r + 1
Case II: n is even
4r
∴ n = 2r for some r ∈ N. Then, there exists 4r ∈ N such that f(4r) = 2 = 2r
∴ f is onto.
Hence, f is not a bijective function.

Page : 11 , Block Name : Exercise 1.2

Q10
Let A = R − {3} and B = R − {1}. Consider the function f : A → B defined by

f(x) = ( )
x−2
x−3
. Is f one-one and onto? Justify your answer.

A = R − {3}, B = R − {1}
f : A → B is defined as
f(x) = ( ) x−2
x−3

Let x, y ∈ A such that f(x) = f(y)
x−2 y−2
⇒ x−3 = y−3
⇒ (x − 2)(y − 3) = (y − 2)(x − 3)
⇒ xy − 3x − 2y + 6 = xy − 3y − 2x + 6
⇒ − 3x − 2y = − 3y − 2x
⇒x=y
∴ f is one-one.
Let y ∈ B = R − {1}. Then , y ≠ 1.
The function f is onto if there exists x ∈ A such that f(x) = y .
Now,

2 − 3y
Thus, for any y ∈ B , there exists 1 − y ∈ A

( )
2 − 3y
−2

( )= ( ) =
1−y
2 − 3y 2 − 3y − 2 + 2y −y
f 1−y 2 − 3y − 3 + 3y
= −1 = y
2 − 3y
1−y
−3

∴ f is onto.
Hence, function f is one-one and onto.

Page : 11 , Block Name : Exercise 1.2

Q11
Let f : R → R be defined as f(x) = x 4. Choose the correct answer.
(A) f is one-one onto (B)f is many-one onto
(C) f is one-one but not onto (D)f is neither one-one nor onto

f : R → R is defined as f(x) = x 4
Let x, y ∈ R such that f(x) = f(y)
⇒ x4 = y4
⇒x= ±y

( ) ( )
∴ f x 1 = f x 2 does not imply that x 1 = x 2

For instance,
f(1) = f( − 1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain R . It is clear that there does not exist any x in
domain R such that f(x) = 2 .
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.

Page : 11 , Block Name : Exercise 1.2

Q12
Let f : R → R be defined as f(x) = 3x . Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto

f : R → R is defined as f(x) = 3x
Let x, y ∈ R such that f(x) = f(y)
⇒ 3x = 3y
⇒x=y
∴ f is one-one.
y
Also, for any real number (y) in co-domain R , there exists 3 in R such that

() ()
f 3
y
=3 3
y
=y

∴ f is onto.
Hence , function f is one-one and onto. The correct answer is A

Page : 11 , Block Name : Exercise 1.2

Q1
Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5),
(4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

The functions f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} are defined as
f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}
gof(1) = g(f(1)) = g(2) = 3 [f(1) = 2 and g(2) = 3]
gof(3) = g(f(3)) = g(5) = 1 [f(3) = 5 and g(5) = 1]
gof(4) = g(f(4)) = g(1) = 3 [f(4) = 1 and g(1) = 3]
∴ gof = {(1, 3), (3, 1), (4, 3)}

Page : 18 , Block Name : Exercise 1.3

Q2 Let f,g and h be function from R to R, show that
(f + g)oh = foh + goh
(f ⋅ g)oh = (foh) ⋅ (goh)

To prove:
(f + g)oh = foh + goh
Consider:
((f + g) oh )(x)
= (f + g)(h(x))
= f(h(x)) + g(h(x))

( )
= f Oh (x) + (goh)(x)
= {(foh) + (goh)}(x)
∴ ((f + g) oh )(x) = {(foh) + (goh)}(x) ∀x ∈ R
Hence(f + g) oh = f ∘ h + goh

To prove:
(f ⋅ g)oh = (foh) ⋅ (goh)
Consider:
((f ⋅ g) oh )(x)
= (f ⋅ g)(h(x))
= f(h(x)) ⋅ g(h(x))
= (f oh )(x) ⋅ (g ∘ h)(x)
= {(foh) ⋅ (goh)}(x)
∴ ((f ⋅ g) ∘ h)(x) = {(f ∘ h) ⋅ (goh)}(x)∀x ∈ R
Hence(f ⋅ g) oh = (f ∘ h) ⋅ (g ∘ h)

Page : 18 , Block Name : Exercise 1.3

Q3
Find g of and f ∘ g, if
(i) f(x) = | x | and g(x) = | 5x − 2 |
1
(ii) f(x) = 8x 3 and g(x) = x 3

f(x) = | x | and g(x) = | 5x − 2 |
(i) ∴ (g ∘ f)(x) = g(f(x)) = g( | x | ) = | 5 | x | − 2 |
(f ∘ g)(x) = f(g(x)) = f( | 5x − 2 | ) = | | 5x − 2 | | = | 5x − 2 |
1
f(x) = 8x 3 and g(x) = x 3
1

(ii)
∴ (g of )(x) = g(f(x)) = g 8x 3 ( ) = ( ) = 2x
8x 3 3

() ()
(f ∘ g)(x) = f(g(x)) = f x 3
1 1
= 8 x3
3
= 8x

Page : 18 , Block Name : Exercise 1.3

( 4x + 3 ) 2 2
Q4 If f(x) = ( 6x − 4 ) , x ≠ 3 ,show that f o f(x) = x for all x ≠ 3 . What is the inverse of f ?

(4x + 3) 2
f(x) = ,x ≠
(6x − 4) 3
It is given that (6x − 4)

f(x) = f(f(x)) = f ( ) 4x + 3
6x − 4

( )
4
4x + 3
6x − 4 +3
16x + 12 + 18x − 12 34x
= = 24x + 18 − 24x + 16 = 34 = x

( )
6
4x + 3
6x − 4
−4

2
Therefore, fof (x) = x, for all x ≠ 3

⇒ f of = 1
Hence, the given function f is invertible and the inverse of s f itself.

Page : 18 , Block Name : Exercise 1.3

Q5 State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}

(i) f : {1, 2, 3, 4} → {10} defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) =
f(3) = f(4) = 10
f is not one-one.
Hence, function f does not have an inverse.

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4
∴ g is not one-one,
Hence, function g does not have an inverse.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h .
∴ Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set
{2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.

Page : 18 , Block Name : Exercise 1.3

x
Q6 Show that f : [ − 1, 1] → R, . Given by f(x) = ( x + 2 ) is one-one. Find the inverse of the function
f : [ − 1, 1] → Range f
x 2y
(Hint y ∈ Range f, y = f ( x ) = x + 2 for some x in[-1 , 1] i.e., x = ( 1 − y ) ).

x
f(x) = ( x + 2 )

f : [ − 1, 1] → R is given as
Let f(x) = f(y)
x y
⇒ x+2 = y+2
⇒ xy + 2x = xy + 2y
⇒ 2x = 2y
⇒x=y
∴ f is a one-one function.
It is clear that f : [ − 1, 1] → Range f is onto.
∴ f : [ − 1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f : [ − 1, 1] → Range f exists.
Let g : Range f → [ − 1, 1] be the inverse of f .
Let y be an arbitrary element of range f .
since f : [ − 1, 1] → Range f is onto, we have:
y = f(x) for same x ∈ [ − 1, 1]
x
⇒ y = x+2
⇒ xy + 2y = x
⇒ x(1 − y) = 2y
2y
⇒ x = 1−y, y ≠ 1
Now, let us define g : Range f → [ − 1, 1] as
2y
g(y) = 1 − y , y ≠ 1

2 ( )
x

( )
x+2
x 2x 2x
Now, (gof)(x) = g(f(x)) = g x + 2 = x = x+2−x = 2 = x
1− x+2
2y

(fog)(y) = f(g(y)) = f 1 − y
( )
2y
=
1−y
1−y
2y
+2
2y
= 2y + 2 − 2y = 2 = y
2y

2y
∴ f − 1(y) = 1 − y , y ≠ 1

Page : 18 , Block Name : Exercise 1.3

Q7

f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f

f : R → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒x=y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3
y−3
⇒x= 4 ∈R
y−3
Therefore, for any y ∈ R, there exists x = 4 ∈ R

f(x) = f
( ) ( )
y−3
4
=4
y−3
4
+3=y

∴ f is onto.
Thus, f is one-one and onto and therefore, f − 1 exists.
y−3
Let us define g : R → R by g ( x ) = 4
( 4x + 3 ) − 3
Now , (g of )(x) = g(f(x)) = g(4x + 3) = 4
=x

(f ∘ g)(y) = f(g(y)) = f ( ) ( )
y−3
4
=4
y−3
4
+3=y−3+3=y

∴ g ∘ f = f ∘ g = IR
Hence, f is invertible and the inverse of f is given by is re exive,
y−3
f − 1(y) = g(y) = 4
symmetric or transitive.

Page : 18 , Block Name : Exercise 1.3

Q8
Consider f : R + → [4, ∞) given by f(x) = x 2 + 4. Show that f is invertible with the inverse

f − 1 of given f by f − 1(y) = √y − 4, where R + is the set of all non-negative real numbers.

f : R + → [4, ∞) is given as f(x) = x 2 + 4
One-one:
Let f(x) = f(y)
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒x=y as x = y ∈ R +
∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x 2 + 4
⇒ x 2 = y − 4 ≥ 0 [ as y ≥ 4]
⇒ x = √y − 4 ≥ 0
Therefore, for any y ∈ R, there exists x = √y − 4 ∈ R such that

f(x) = f(√y − 4) = (√y − 4) 2 + 4 = y − 4 + 4 = y
∴ f is onto.
Thus, f is one-one and onto and therefore, f − 1 exists.
Let us define g : [4, ∞) → R + by,

g(y) = √y − 4

Now, gof (x) = g(f(x)) = g x 2 + 4 = ( ) √ ( x + 4 ) − 4 = √x = x
2 2

And , f ∘ g(y) = f(g(y)) = f(√y − 4) = (√y − 4) 2 + 4 = (y − 4) + 4 = y
∴ g ∘ f = f ∘ g = In +
Hence, f is invertible and the inverse of f is given by
f − 1(y) = g(y) = √y − 4

Page : 18 , Block Name : Exercise 1.3

Q9
Consider f : R + → [ − 5, ∞) given by f(x) = 9x 2 + 6x − 5. Show that f is invertible with

( )
( √y + 6 ) − 1
f − 1(y) = 3

f : R + → [ − 5, ∞) is given as f(x) = 9x 2 + 6x − 5 .
Let y be an arbitrary element of [ − 5, ∞)
Let y = 9x 2 + 6x − 5 .
⇒ y = (3x + 1) 2 − 1 − 5 = (3x + 1) 2 − 6
⇒ (3x + 1) 2 = y + 6
⇒ 3x + 1 = √y + 6

√y + 6 − 1
⇒x= 3
∴ f is onto, thereby range f = [ − 5, ∞)
√y + 6 − 1
Let us de ne g : [ − 5, ∞) → R + as g(y) = 3
We now have:

(
(g ∘ f)(x) = g(f(x)) = g 9x 2 + 6x − 5 )
= g ((3x + 1) − 6 )
2

√(3x + 1) 2 − 6 + 6 − 1
=
3
3x + 1 − 1
= =x
3

And, (fog)(y) = f(g(y)) = f
( √y + 6 − 1
3 )
[( √y + 6 − 1
) ]
2
= 3 +1 −6
3

= (√y + 6) 2 − 6 = y + 6 − 6 = y
gof = I R , and fo g = I [ − 5 , ∞ )
Hence, f is invertible and the inverse of f is given by
√y + 6 − 1
f − 1(y) = g(y) = 3

Page : 19 , Block Name : Exercise 1.3

Q10 Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint : Suppose g 1 and g 2 are two inverse of f. Then for all y ∈ Y f 0g 1(y) = I γ(y) = fog 2(y). Use one-
one ness of f).

Let f : x → Y be an invertible function.
Also, suppose f has two inverses (say g 1 and g 2
Then, for all y ∈ Y, we have:
fog 1(y) = I Y(y) = fog 2(y)

( ) (
⇒ f g 1(y) = f g 2(y) )
⇒ g 1(y) = g 2(y)
⇒ g1 = g2
Hence,f has a unique inverse.

Page : 19 , Block Name : Exercise 1.3

Q11
Consider f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f − 1 and

( ) =f
show that f − 1
−1

Function f : {1, 2, 3} → {a, b, c} is given by,
f(1) = a, f(2) = b, and f(3) = c
If we define g : {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:
(f ∘ g)(a) = f(g(a)) = f(1) = a
(f ∘ g)(b) = f(g(b)) = f(2) = b
(f ∘ g)(c) = f(g(c)) = f(3) = c
And
(gof)(1) = g(f(1)) = g(a) = 1
(gof)(2) = g(f(2)) = g(b) = 2
(gof)(3) = g(f(3)) = g(c) = 3
∴ gof = I x and fog = I γWhereX = {1, 2, 3} and Y = {a, b, c}
Thus, the inverse of f exists and f − 1 = g .
∴ f − 1 : {a, b, c} → {1, 2, 3} is given by,
f − 1(a) = 1, f − 1(b) = 2, f 1(c) = 3
Let us now find the inverse of f − 1 i.e., find the inverse of g.
If we define h : {1, 2, 3} → {a, b, c} as
h(1) = a, h(2) = b, h(3) = c, then we have:
(g ∘ h)(1) = g(h(1)) = g(a) = 1
(g ∘ h)(2) = g(h(2)) = g(b) = 2
(g ∘ h)(3) = g(h(3)) = g(c) = 3
And
(h ∘ g)(a) = h(g(a)) = h(1) = a
(h ∘ g)(b) = h(g(b)) = h(2) = b
(h ∘ g)(c) = h(g(c)) = h(3) = c

goh = I x and hog = I γ, where x = {1, 2, 3} and Y = {a, b, c}

Thus the inverse of g exists and g − 1 = h ⇒ f − 1 ( ) =h−1

It can be noted that h = f.

Hence f − 1 ( ) =f −1

Page : 19 , Block Name : Exercise 1.3

Q12
Let f : X → Y be an invertible function. Show that the inverse of f − 1 is f, i.e.,

(f ) = f
−1 −1

Let f : X → Y be an invertible fuction
Then , there exists a function g : Y → X such that g ∘ f = I x and f ∘ g = I Y
Here , f − 1 = g.
Now g ∘ f = I x and f ∘ g = I Y
⇒ f − 1 of = I X and fof − 1 = I Y
Hence, f − 1 : Y → X is invertible and f is the inverse of f − 1

( ) =f
i.e., f − 1
−1

Page : 19 , Block Name : Exercise 1.3

Q13
1

If f : R → R be given by f(x) = ( 3 − x3 ) , then f ∘ f(x) is
3

( )
1
(A) (B)x 3 (C)x 3 − x 3
x3

1

f : R → R is given as f(x) = ( 3 − x3 ) 3

1

f(x) = (3 − x3 ) 3

(( ) ) [ (( ) ) ]
1
1 1 3 3

∴ f 0f(x) = f(f(x)) = f 3 − x3 3 = 3− 3 − x3 3

1 1

= 3−[ (3 − x3 )] = ( ) = x
3 x3 5

∴ fof (x) = x
The correct answer is C

Page : 19 , Block Name : Exercise 1.3

Q14 SLet f : R −
{ } − 3
4 4x
→ R be a function de ned as f(x) = 3x + 4 . The inverse of f is map g:

Range f : R − { }
− 3
4
→R

3y
(A) g(y) = 3 − 4y
4y
(B) g(y) = 4 − 3y
4y
(C) g(y) = 3 − 4y
3y
(D) g(y) = 4 − 3y

Answer. It is given that

f: R −
{ }− 3
4 4x
→ R is defined as f(x) = 3x + 4

Let y be an arbitrary element of range f.Then, there exists

x ∈ R− { } y = f(x)
such4x
−3
that
4

⇒ y = 3x + 4
⇒ 3xy + 4y = 4x

⇒ x(4 − 3y) = 4y Let us de ne g: Range f → R −
4y
{ } 4
− 3
4y
as g(y) = 4 − 3y

⇒ x = 4 − 3y

Now, (g ∘ f)(x) = g(f(x)) = g 3x + 4
( )
4x

( )4
4x
3x + 4
16x 16x
= = 12x + 16 − 12x
= 16 = x

( )
4−3
4x
3x + 4

And, (f ∘ g)(y) = f(g(y)) = f
( ) 4y
4 − 3y

=
( ) 4y
4 4 − 3y
=
16y
=
16y
=y
12y + 16 − 12y 16
( )4y
3 4 − 3y +4

g is the inverse of fi.e., f 1 = g
Hence, the inverse of f is the map g : Range

f→R− { }
4y
4
− 3 , which is given by

g(y) = 4 − 3y
The correct answer is B

Page : 19 , Block Name : Exercise 1.3

Q1 Determine whether or not each Of the de nition Of given below gives a binary operation. In
the event that * is not a binary operation, give justi cation for this.
(i) On z + , define ∗ by a ∗ b = a − b
(ii) On z + , define ∗ by a ∗ b = ab
(iii) On R, define ∗ by a ∗ b = ab 2
(iv) On z + , define ∗ by a ∗ b = | a − b |
(v) On z + , define ∗ by a ∗ b = a

(i) On z + , ∗ is defined by a ∗ b = a − b
It is not a binary operation as the image of (1, 2) under ∗ is 1 ∗ 2 = 1 − 2
= − 1 ∉ Z+

(ii) on Z + , ∗ is defined by a ∗ b = ab.
It is seen that for each a, b ∈ Z + ,there is a unique element ab in Z +
This means that * carries each pair (a, b) to a unique element a * b = ab in Z + .
Therefore, * is a binary operation.

(iii) On R, ∗ is defined by a ∗ b = ab 2
It is seen that for each a, b ∈ R, there is a unique element ab 2 in R .
This means that * carries each pair (a, b) to a unique element a ∗ b = ab 2 in R .
( iv ) On Z + , ∗ is defined by a ∗ b = | a − b |
It is seen that for each a, b ∈ Z + , there is a unique element | a − b | in z + .
This means that * carries each pair (a, b) to a unique element a ∗ b =
| a − b | in z +

Therefore * is a binary operation.
(v) On z + , ∗ is defined by a ∗ b = a .
∗ carries each pair (a, b) to a unique element a ∗ b = a in z +
Therefore, ∗ is a binary operation.

Page : 24 , Block Name : Exercise 1.4

Q2
For each binary operation * defined below, determine whether* is commutative or
associative.
(i) On z, define a ∗ b = a − b
(ii) On Q, define a ∗ b = ab + 1
ab
(iii) On Q, define a ∗ b = 2
(iv) On z + , define a ∗ b = 2 ab
(v) On z + , define a ∗ b = a b
a
(vi) On R − { − 1}, definea ∗ b = b + 1

(i) On z, ∗ is defined by a ∗ b = a − b
It can be observed that 1 ∗ 2 = 1 − 2 = 1 and 2 ∗ 1 = 2 − 1 = 1
∴ 1 ∗ 2 ≠ 2 ∗ 1; where 1, 2 ∈ Z
Hence, the operation ∗ is not commutative.
Also we have:
(1 ∗ 2) ∗ 3 = (1 − 2) ∗ 3 = − 1 ∗ 3 = − 1 − 3 = − 4
1 ∗ (2 ∗ 3) = 1 ∗ (2 − 3) = 1 ∗ − 1 = 1 − ( − 1) = 2
∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3); where 1, 2, 3 ∈ Z
Hence, the operation * is not associative.

(ii) Q, ∗ is defined by a ∗ b = ab + 1
It is known that: ab = ba for all a, b ∈ Q
⇒ ab + 1 = ba + 1 for all a, b ∈ Q
⇒ a ∗ b = a ∗ b for all a, b ∈ Q
Therefore, the operation * is commutative.
It can be observed that
(1 ∗ 2) ∗ 3 = (1 × 2 + 1) ∗ 3 = 3 ∗ 3 = 3 × 3 + 1 = 10
1 ∗ (2 ∗ 3) = 1 ∗ (2 × 3 + 1) = 1 ∗ 7 = 1 × 7 + 1 = 8
∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3), where 1, 2, 3 ∈ Q Therefore, the operation ∗

(iii)
ab
On Q, ∗ is defined by a ∗ b = 2

It is known that: ab = ba for all a, b ∈ Q

ab ba
⇒ 2 = 2 for all a, b ∈ Q
⇒ a ∗ b = b ∗ a for all a, b ∈ Q
Therefore, the operation ∗ is commutative.
For all a, b, c ∈ Q, we have

( ) ab
c

()
2
ab abc
(a ∗ b) ∗ c = 2
∗c= 2
= 4

a ( )
bc

()
2
bc abc
a ∗ (b ∗ c) = a ∗ 2
= 2
= 4

( ) c = a (b ), where a, b, c ∈ Q
∴ a ∗b
∗ ∗ ∗

Therefore, the operation ∗ is associative.

(iv)
On Z + , ∗ is defined by a ∗ b = 2 ab
It is known that: ab = ba for all a, b ∈ Z +
⇒ 2 ab = 2 ba for all a, b ∈ Z +
⇒ a ∗ b = b ∗ a for all a, b ∈ Z +
Therefore, the operation ∗ is commutative.
It can be observed that
(1 ∗ 2) ∗ 3 = 2 1 × 2 ∗ 3 = 4 ∗ 3 = 2 4 × 3 = 2 12 and
1 ∗ (2 ∗ 3) = 1 ∗ 2 2 × 3 = 1 ∗ 2 6 = 1 ∗ 64 = 2 1 × 64 = 2 64
Therefore, the operation * is not associative.

(v) Z + , ∗ is defined by a ∗ b = a b
It can be observed that
1 ∗ 2 = 1 2 = 1 and 2 ∗ 1 = 2 1 = 2
∴ 1 ∗ 2 ≠ 2 ∗ 1, where 1, 2 ∈ Z +
Therefore, the operation * is not commutative.
It can also be observed that
(2 ∗ 3) ∗ 4 = 2 3 ∗ 4 = 8 ∗ 4 = 8 12 and 2 ∗ (3 ∗ 4) = 2 ∗ 31 = 2 81
∴ (2 ∗ 3) ∗ 4 ≠ 2 ∗ (3 ∗ 4), where 2, 3, 4 ∈ Z +
Therefore, the operation ∗ is not associative.

a
(vi) R, ∗ − { − 1} is defined by a ∗ b = b + 1

It can be observed that
1 1 2 2
1 ∗ 2 = 2 + 1 = 3 and 2 ∗ 1 = 1 + 1 = 2 = 1
∴ 1 ∗ 2 ≠ 2 ∗ 1, where 1, 2 ∈ R − { − 1}
Therefore, the operation * is not commutative.
It can also be observed that
1
1 1 3 1
(1 ∗ 2) ∗ 3 = 2 + 1 ∗ 3 = 3 ∗ 3 = 3 + 1 = 12
2 2 1 1 1 2
1 ∗ (2 ∗ 3) = 1 ∗ 3 + 1 = 1 ∗ 4 = 1 ∗ 2 = 1 = 3 = 3
2 +1 2
∗
∴ (1 ∗ 2) ∗ 3 ≠ 1 (2 ∗ 3), where 1, 2, 3 ∈ R − { − 1} Therefore, the operation ∗

Page : 24 , Block Name : Exercise 1.4

Q3 Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} de ned by a ∧ b = min {a, b}. Write the
operation table of the operation ∧ .

The binary operation von the set {1, 2, 3, 4, 5} is defined as a ∨ b = min {a, b}
for all a, b ∈ {1, 2, 3, 4, 5}
Thus, the operation table for the given operation v can be given as:

Page : 24 , Block Name : Exercise 1.4

Q4

Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following
multiplication table.
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
(ii) Is * commutative?
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5)
(Hint: use the following table)

(i) (2 ∗ 3) ∗ 4 = 1 ∗ 4 = 1
2 ∗ (3 ∗ 4) = 2 ∗ 1 = 1
(ii) For every a, b ∈ {1, 2, 3, 4, 5}, we have a ∗ b = b ∗ a. Therefore, the operation ∗ is
commutative.
(iii) (2 ∗ 3) = 1 and (4 ∗ 5) = 1
∴ (2 ∗ 3) ∗ (4 ∗ 5) = 1 ∗ 1 = 1

Page : 25 , Block Name : Exercise 1.4

Q5 Let * be the binary operation on the set {1, 2, 3, 4, 5} de ned by a b = H.C.F. of a and b. Is the
operation * same as the operation * de ned in Exercise 4 above? Justify your answer.

Answer. The binary operation * on the set {1, 2, 3 4, 5} is de ned as a * b = H.C.F of a and b. The
operation table for the operation can be given as:

We observe that the operation tables for the operations * and are the same. Thus, the operation is
same as the operation *.

Page : 25 , Block Name : Exercise 1.4

Q6
Let ∗ be the binary operation on N given by a ∗ b = L ⋅ C ⋅ M. of a and b. Find
(i) 5 ∗ 7, 20 ∗ 16
(ii) Is ∗ commutative?
(iii) Is ∗ associative?
(iv) Find the identity of ∗ in N
(v) Which elements of N are invertible for the operation ∗?

The binary operation ∗ on N is defined as a ∗ b = L. C. M. of a and b
(i) 5 ∗ 7 = L.C.M. of 5 and 7 = 35
20 ∗ 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:
L.C.M of a and b = L.C.M of b and afor alla, b ∈ N .
∴a∗b=b∗a
Thus, the operation ∗ is commutative.

(iii) For a, b, c ∈ N, we have:
(a ∗ b) ∗ c = ( L. C.M of a and b) ∗ c = LCM of a, b, and c
a ∗ (b ∗ c) = a ∗ (LCM of b and c) = L.C.M of a, b, and c
∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
Thus, the operation ∗ is associative.

(iv) It is known that:
L.C.M. of a and 1 = a = L.C.M. 1 and a for alla ∈ N

⇒ a ∗ 1 = a = 1 ∗ a◻a ∈ N
Thus, 1 is the identity of ∗ in N .

(v) An element a in N is invertible with respect to the operation ∗ if there exists an
element b in N, such that a ∗ b = e = b ∗ a .
Here, e = 1
This means that:
L.C.M of a and b = 1 = L.C.M of b and a
This case is possible only when a and b are equal to 1 .
Thus, 1 is the only invertible element of N with respect to the operation *

Page : 25 , Block Name : Exercise 1.4

Q7
Is * defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b a binary operation?
Justify your answer.

The operation ∗ on the set A = {1, 2, 3, 4, 5} is defined as
a ∗ b = L.C.M. of a and b.
Then, the operation table for the given operation * can be given as:

It can be observed from the obtained table that:
3 ∗ 2 = 2 ∗ 3 = 6 ∉ A, 5 ∗ 2 = 2 ∗ 5 = 10 ∉ A, 3 ∗ 4 = 4 ∗ 3 = 12 ∉ A
3 ∗ 5 = 5 ∗ 3 = 15 ∉ A, 4 ∗ 5 = 5 ∗ 4 = 20 ∉ A
Hence, the given operation * is not a binary operation.

Page : 25 , Block Name : Exercise 1.4

Q8
Let ∗ be the binary operation on N defined by a ∗ b = H. C. F. of a and b. Is *
commutative? Is * associative? Does there exist identity for this binary operation on N?

The binary operation * on N is defined as:
a ∗ b = H. C. F. of a and b
It is known that:
H.C.F of a and b = H. C. F. of b and a ◻a, b ∈ N
∴a∗b=b∗a
Thus, the operation * is commutative.
For a, b, c ∈ N, we have:
(a ∗ b) ∗ c = (H. C. F. of a and b) ∗ c = H.C.F. of a, b, and c
a ∗ (b ∗ c) = a ∗ (H. C. F. of b and c) = H. C. F. of a, b, and c
∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
Thus, the operation ∗ is associative.
Now, an element e ∈ N will be the identity for the operation ∗ if a ∗ e = a = e ∗ a∀a ∈
N
But this relation is not true for any a ∈ N .
Thus, the operation * does not have any identity in N.

Page : 25 , Block Name : Exercise 1.4

Q9 Let ∗ be a binary operation on the set Q of rational number as follows :
(i) a ∗ b = a − b (ii)a ∗ b = a 2 + b 2
(iii) a ∗ b = a + ab (iv)a ∗ b = (a − b) 2
ab
(v)a ∗ b = 4 (vi)a ∗ b = ab 2

(i) On Q, the operation * is defined as a ∗ b = a − b .
It can be observed that:
1 1 1 1 3−2 1 1 1 1 1 2−3 −1
2
∗ 3 = 2 − 3 = 6
= 6 6
∗ 3 = 3 − 2 = 6
= 6
1 1 1 1 1 1
2
∗ 3 ≠ 3 ∗ 2 ; where 2 , 3 ∈ Q
Thus, the operation ∗ is not commutative.
It can also be observed that:

( ) ( )
1
2 ∗ 3
1
∗ 4 =
1 1
2 − 3
1 1 1 1 1
∗ 4 = 6 ∗ 4 = 6 − 4 = 12 = 12
1 2−3 −1

1
( ) ( )
2 ∗
1
3 ∗ 4
1 1
= 2 ∗
1
3 − 4
1 1 1 1 1
= 2 ∗ 12 = 2 − 12 = 12 = 12
6−1 5

( )
∴
( )
1
2 ∗ 3
1 1
∗ 4 ≠ 2 ∗
1 1
3 ∗ 4
1 1 1 1
; where 2 , 3 , 4 ∈ Q

Thus, the operation ∗ is not associative.
Thus, the operation * is not commutative.

(ii) On Q, the operation ∗ is defined as a ∗ b = a 2 + b 2
For a, b ∈ Q, we have:
a ∗ b = a2 + b2 = b2 + a2 = b ∗ a
∴a∗b=b∗a

( )
(1 ∗ 2) ∗ 3 = 1 2 + 2 2 ∗ 3 = (1 + 4) ∗ 4 = 5 ∗ 4 = 5 2 + 4 2 = 41

( )
1 ∗ (2 ∗ 3) = 1 ∗ 2 2 + 3 2 = 1 ∗ (4 + 9) = 1 ∗ 13 = 1 2 + 13 2 = 169
∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3); where 1, 2, 3 ∈ Q
Thus, , the operation ∗ is not associative.

(iii) Q, the operation ∗ is defined as a ∗ b = a + ab
It can be observed that:
1∗2=1+1×2=1+2=3
2∗1=2+2×1=2+2=4
∴ 1 ∗ 2 ≠ 2 ∗ 1; where 1, 2 ∈ Q
Thus, the operation ∗ is not commutative.
It can also be observed that:
(1 ∗ 2) ∗ 3 = (1 + 1 × 2) ∗ 3 = 3 ∗ 3 = 3 + 3 × 3 = 3 + 9 = 12
1 ∗ (2 ∗ 3) = 1 ∗ (2 + 2 × 3) = 1 ∗ 8 = 1 + 1 × 8 = 9
∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3); where 1, 2, 3 ∈ Q
Thus, the operation ∗ is not associative.

(iv) On Q, the operation ∗ is defined by a ∗ b = (a − b) 2
For a, b ∈ Q, we have:
a ∗ b = (a − b) 2
b ∗ a = (b − a) 2 = [ − (a − b)] 2 = (a − b) 2
∴a∗b=b∗a
Thus, the operation * is commutative.
It can be observed that:
(1 ∗ 2) ∗ 3 = (1 − 2) 2 ∗ 3 = ( − 1) 2 ∗ 3 = 1 ∗ 3 = (1 − 3) 2 = ( − 2) 2 = 4
1 ∗ (2 ∗ 3) = 1 ∗ (2 − 3) 2 = 1 ∗ ( − 1) 2 = 1 ∗ 1 = (1 − 1) 2 = 0
∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3); where 1, 2, 3 ∈ Q
thus, the operation is not associative.

(v) On Q the operation is de ned as

ab
a∗b= 4
For a, b ∈ Q, we have:
ab ba
a∗b= 4 = 4 =b∗a
∴a∗b=b∗a
Thus, the operation ∗ is commutative.
For a, b, c ∈ Q, we have:
ab
ab 4 ⋅c abc
(a ∗ b) ∗ c = 4 ∗ c = 4
= 16
bc
bc a⋅ 4 abc
a ∗ (b ∗ c) = a ∗ 4 = 4
= 16

∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
Thus, the operation * is associative.

(vi) On Q , the operation * is defined as a ∗ b = ab 2
It can be observed that:

1
2
∗ 3 = 2 ⋅
1 1
() 1
3
2 1
= 2 ⋅ 9 = 18
1 1

1
3
∗ 2 = 3 ⋅
1
1

1
1
()
1
1
2
2

1
1
= 3 ⋅ 4 = 12
1

1 1
1

∴ 2 ∗ 3 ≠ 3 ∗ 2 ; where 2 , 3 ∈ Q
Thus, the operation ∗ is not commutative.
It can also be observed that:

( ) 1
2
∗ 3
[ ( )]
1
∗ 4 =
1 1
2
⋅
1
3
2 1 1
∗ 4 = 18 ∗ 4 = 18 ⋅
1 1
()
1
4
2 1
= 18 × 16

1
2
∗
( ) [ ( )]
1
3
∗ 4
1
= 2 ∗
1 1
3
⋅
1
4
2 1 1 1
= 2 ∗ 48 = 2 ⋅
()
1
48
2
=
1
2 × ( 48 ) 2

∴
( ) 1
2
∗ 3
1 1
∗ 4 ≠ 2 ∗
1
( ) 1
3
1
∗ 4 ; where 2 , 3 , 4 ϵQ
1 1 1

Thus, the operation ∗ is not associative.
Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined
in (v) is associative.

Page : 25 , Block Name : Exercise 1.4

Q10 Find which of the operations given above has identity.

An element e ∈ Q will be the identity element for the operation ∗ if
a ∗ e = a = e ∗ a, ∀a ∈ Q
satisfying the above condition.
Thus, none of the six operations has identity.

Page : 25 , Block Name : Exercise 1.4

Q11
Let A = N × N and ∗ be the binary operation on A defined by
(a, b) ∗ (c, d) = (a + c, b + d)
∗ is commutative and associative. Find the identity element for ∗ on A,

A=N×N
∗ is a binary operation on A and is defined by:
(a, b) ∗ (c, d) = (a + c, b + d)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have:
(a, b) ∗ (c, d) = (a + c, b + d)
(c, d) ∗ (a, b) = (c + a, d + b) = (a + c, b + d)
Therefore, the operation ∗ is commutative.
Now, let (a, b), (c, d), (e, f) ∈ A
Then, a, b, c, d, e, f ∈ N
We have:
((a, b) ∗ (c, d)) ∗ (e, f) = (a + c, b + d) ∗ (e, f) = (a + c + e, b + d + f)
(a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (c + e, d + f) = (a + c + e, b + d + f)
∴ ((a, b) ∗ (c, d)) ∗ (e, f) = (a, b) ∗ ((c, d) ∗ (e, f))
Therefore, the operation ∗ is associative.

( )
An element e = e 1, e 2 ∈ A will be an identity element for the operation ∗ if

a ∗ e = a = e ∗ a∀a = (a 1, a 2 ) ∈ A

not true for any element in A .
Therefore, the operation * does not have any identity element.

Page : 25 , Block Name : Exercise 1.4

Q12
State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N, a ∗ a = a∀a ∗ N .
(ii) If * is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a

(i) Define an operation ∗ on N as:
a∗b=a+b ∀a, b ∈ N
Then, in particular, for b = a = 3, we have:
3×3=3+3=6≠3
Therefore, statement (i) is false.
(ii) R.H.S. = (c ∗ b) ∗ a
= (b ∗ c) ∗ a[ ∗ is commutative ]
= a ∗ (b ∗ c)[ Again, as ∗ is commutative]
= L. H.S.
∴ a ∗ (b ∗ c) = (c ∗ b) ∗ a
Therefore, statement (ii) is true.

Page : 26 , Block Name : Exercise 1.4

Q13
Consider a binary operation ∗ on N defined as a ∗ b = a 3 + b 3 . Choose the correct
answer.
(A) Is * both associative and commutative?
(B) Is * commutative but not associative?
(C) Is * associative but not commutative?
(D) Is * neither commutative nor associative?

On N, the operation ∗ is defined as a ∗ b = a 3 + b 3 .
For, a, b, ∈ N, we have:
a ∗ b = a 3 + b 3 = b 3 + a 3 = b ∗ a [Addition is commutative in N]
Therefore, the operation * is commutative.
It can be observed that:

( )
(1 ∗ 2) ∗ 3 = 1 3 + 2 3 ∗ 3 = 9 ∗ 3 = 9 3 + 3 3 = 729 + 27 = 756

( )
1 ∗ (2 ∗ 3) = 1 ∗ 2 3 + 3 3 = 1 ∗ (8 + 27) = 1 × 35 = 1 3 + 35 3 = 1 + (35) 3 = 1 + 42875 = 42876

∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3); where 1, 2, 3 ∈ N
Therefore, the operation * is not associative.
Hence, the operation * is commutative, but not associative. Thus, the correct answer is
B.

Page : 26 , Block Name : Exercise 1.4

Q1

Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g ∘ f = f
o g = 1R

It is given that f : R → R is defined as f(x) = 10x + 7.
One-one:
Let f(x) = f(y), where x, y ∈ R .
⇒ 10x + 7 = 10y + 7
⇒x=y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 10x + 7
y−7
⇒ x = 10 ∈ R
y−7
Therefore, for any y ∈ R, there exists x = 10 ∈ R such that

f(x) = f ( ) ( )
y−7
10
= 10
y−7
10
+7=y−7+7=y

∴ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
y−7
Let us define g : R → R as g ( y ) = 10
Now, we have:
( 10x + 7 ) − 7 10x
g ∘ f(x) = g(f(x)) = g(10x + 7) = 10
= 10 = 10

And

f 0g(y) = f(g(y)) = f ( ) ( )
y−7
10
= 10
y−7
10
+7=y−7+7=y

∴ g ∘ f = I R and f ∘ g = I R
y−7
R → R is defined as g ( y ) = 10

Page : 29 , Block Name : Miscellaneous Exercise

Q2
Let f : W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show
that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer. It is given that

f : W → W is de ned as f(n) =
{ n − 1, if n is odd
n + 1, if n is even

It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1 .
⇒n−m=2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also be ignored under a
similar argument.
∴ Both n and m must be either odd or even.
Now, if both n and m are odd, then we have:
f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m
Again, if both n and m are even, then we have:
f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m
is one-one.
It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N
and any even number 2r in co-domain N is the image of 2r + 1 in domain N .
∴ f is onto.
Hence, f is an invertible function.
Let us define g : W → W as:

g(m) = { m + 1, if m is even
m − 1, if m is odd
Now, when n is odd:
g of (n) = g(f(n)) = g(n − 1) = n − 1 + 1 = n
And, when n is even:
g of (n) = g(f(n)) = g(n + 1) = n + 1 − 1 = n
Similarly, when m is odd:
f0g(m) = f(g(m)) = f(m − 1) = m − 1 + 1 = m
fog(m) = f(g(m)) = f(m + 1) = m + 1 − 1 = m
∴ g ∘ f = I w and fog = I w
Thus, f is invertible and the inverse of f is given by f − 1 = g, which is the same as f .
Hence, the inverse of f is f itself.

Page : 29 , Block Name : Miscellaneous Exercise

Q3
f : R → R is defined by f(x) = x 2 − 3x + 2, find f(f(x))

It is given that f : R → R is defined as f(x) = x 2 − 3x + 2 .

( )
f(f(x)) = f x 2 − 3x + 2

= (x − 3x + 2 ) − 3 (x − 3x + 2 ) + 2
2 2 2

= x 4 + 9x 2 + 4 − 6x 3 − 12x + 4x 2 − 3x 2 + 9x − 6 + 2
= x 4 − 6x 3 + 10x 2 − 3x

Page : 29 , Block Name : Miscellaneous Exercise

x
Q4 Show that the function f : R → {x ∈ R : − 1 < x < 1} defined by f(x) = 1 + | x |
x ∈ R is one one and onto function.

Answer. It is given that
x
f : R → {x ∈ R : − 1 < x < 1} defined by f(x) = 1 + | x | x ∈R.
Suppose f(x) = f(y), where x, y ∈ R.
x y
⇒ 1+ |x| = 1+ |y|
It can be observed that if x is positive and y is negative, then we have:
x y
1 + x = 1 − y ⇒ 2xy = x − y
Since x is positive and y is negative:
x>y⇒x−y>0
But, 2xy is negative.
then 2xy ≠ x − y
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out
x and y have to be either positive or negative.
When x and y are both positive, we have:
x y
f(x) = f(y) ⇒ 1 + x = 1 + y ⇒ x + xy = y + xy ⇒ x = y

When x and y are both negative, we have:
x y
f(x) = f(y) ⇒ 1 − x = 1 − y ⇒ x − xy = y − yx ⇒ x = y
∴ f is one-one.
Now, let y ∈ R such that −1 < y < 1.
y
If x is negative, then there exists x = 1 + y ∈ R such that,

( ) y
y

( )= | | = ( ) =
1+y
y 1+y y
f(x) = f 1 + y 1+y−y
=y
y −y
1+ 1+y
1+ 1+y

y
If x is positive, then there exists x = 1 − y ∈ R such that

( ) y
y

( )= ( ) =
1−y
y 1−y y
f(x) = f 1 − y y = 1−y+y = y
y 1+ 1−y
1+ 1−y
|

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

Page : 29 , Block Name : Miscellaneous Exercise

Q5
Show that the function f : R → R given by f(x) = x 3is injective

f : R → R is given as f(x) = x 3 .
Suppose f(x) = f(y), where x, y ∈ R .
⇒ x 3 = y 3…(1)
Now, we need to show that x = y .
Suppose x ≠ y, their cubes will also not be equal.
⇒ x3 ≠ y3
However, this will be a contradiction to (1)
∴x=y
Hence, f is injective.

Page : 29 , Block Name : Miscellaneous Exercise

Q6
Give examples of two functions f : N → Z and g : Z → Z such that g ∘ f is injective but g is
not injective.

Define f : N → z as f(x) = x and g : z → z as g(x) = | x |
We first show that g is not injective.
It can be observed that:
g( − 1) = | − 1 | = 1
∴ g is not injective.
Now, gof: N → Z is defined as g of (x) = g(f(x)) = g(x) = | x |
Let x, y ∈ N such that gof (x) = gof(y)
Let x, y ∈ N such that gof(x) = g ∘ f(y)
⇒ |x| = |y|
since x and y ∈ N, both are positive.
∴ |x| = |y| ⇒ x = y
Hence, gof is injective

Page : 29 , Block Name : Miscellaneous Exercise

Q7
Given examples of two functions f : N → N and g : N → N such that g ∘ f is onto but f is
not onto.

Define f : N → Nby
f(x) = x + 1

g(x) = { x−1
1
if x > 1
if x = 1
We first show that g is not onto.
For this, consider element 1 in co-domain N . It is clear that this element is not an image
of any of the elements in domain N .
∴ ∴ f is not onto.
gof(x) = g(f(x)) = g(x + 1) = (x + 1) − 1 [x ∈ N ⇒ (x + 1) > 1]
=x
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that g ∘ f(x) = y .
Hence, gof is onto.

Page : 29 , Block Name : Miscellaneous Exercise

Q8
Given a non empty set X , consider P(X) which is the set of all subsets of x.
Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation
Justify you answer:

since every set is a subset of itself, ARA for all A ∈ P(X) .
∴ R is reflexive.
Let ARB ⇒ A ⊂ B .
This cannot be implied to B ⊂ A .
∴ R is not symmetric.
Further, if ARB and BRC, then A ∈ B and B ⊂ C .
⇒A∈C
⇒ ARC
∴ R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.

Page : 29 , Block Name : Miscellaneous Exercise

Q9
Given a non-empty set X , consider the binary operation ∗ : P(X) × P(X) → P(X) given by
A ∗ B = A ∩ B◻A, B in P(X) is the power set of X . Show that x is the identity element
for this operation and X is the only invertible element in P(X) with respect to the
operation*.

it is given that ∗ : P(X) × P(X) → P(X) is defined as A ∗ B = A ∩ B∀A, B ∈ P(X)
we know that A ∩ X = A = X ∩ A∀A ∈ P(X)
⇒ A ∗ X = A = X ∗ A∀A ∈ P(X)

⇒ A ∗ X = A = X ∗ A∀A ∈ P(X)
Thus, X is the identity element for the given binary operation *
A ∈ P(X) is invertible if there exists B ∈ P(X) such that
A ∗ B = X = B ∗ A.
i.e.,
A∩B=X=B∩A
This case is possible only when A = X = B .
Thus, X is the only invertible element in P(X) with respect to the given operation*.
Hence, the given result is proved.

Page : 30 , Block Name : Miscellaneous Exercise

Q10 Find the number of all onto functions from the set {1,2,3,.....,n} to itself.

Onto functions from the set {1, 2, 3, …, n} to itself is simply a permutation on n symbols
1, 2, …, n.
Thus, the total number of onto maps from {1, 2, …, n} to itself is the same as the total
number of permutations on n symbols 1, 2, …, n, which is the same as the total

Page : 30 , Block Name : Miscellaneous Exercise

Q11
Let S = {a, b, c} and T = {1, 2, 3}. Find F − 1 of the following functions F from S to T, if it
exists.
(i) F = {(a, 3), (b, 2), (c, 1)}( (ii) F = {(a, 2), (b, 1), (c, 1)}

S = {a, b, c}, T = {1, 2, 3}
(i)F : S → T is defined as:
F = {(a, 3), (b, 2), (c, 1)}
⇒ F(a) = 3, F(b) = 2, F(c) = 1
Therefore, F − 1 : T → S is given by
F − 1 = {(3, a), (2, b), (1, c)}
since F(b) = F(c) = 1, F is not one-one.
Hence, F is not invertible i.e., F − 1 does not exist.

Page : 30 , Block Name : Miscellaneous Exercise

Q12 Consider the binary operations ∗ : R × R → R and o : R × R → R de ned as a ∗b = |a – b| and a
o b = a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not
commutative. Further, show that ∀ a, b, c ∈ R, a ∗ (b o c) = (a ∗ b) o (a ∗ c). [If it is so, we say that

the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer.

It is given that ∗ : R × R → and 0 : R × R → R
a ∗ b = | a − b | and a ∘ b = a, ◻a, b ∈ R
For a, b ∈ R, we have:
a ∗ b = |a − b|
b ∗ a = | b − a | = | − (a − b) | = | a − b |
∴a∗b=b∗a
∴ The operation ∗ is commutative.
It can be observed that,
(1 ∗ 2) ∗ 3 = ( | 1 − 2 | ) ∗ 3 = 1 ∗ 3 = | 1 − 3 | = 2
1 ∗ (2 ∗ 3) = 1 ∗ ( | 2 − 3 | ) = 1 ∗ 1 = | 1 − 1 | = 0
∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3) (where 1, 2, 3 ∈ R)
Now, consider the operation o:
It can be observed that 1 o 2 = 1 and 2 o 1 = 2 .
∴ 1 ∘ 2 ≠ 2 o 1 (where 1, 2 ∈ R)
. The operation o is not commutative.
Let a, b, c ∈ R. Then, we have:
(a ∘ b) ∘ c = a ∘ c = a
a ∘ (b ∘ c) = a ∘ b = a
⇒ a ∘ b) ∘ c = a ∘ (b ∘ c)
∴ The operation o is associative.
Now, let a, b, c ∈ R, then we have:
Now, let a, b, c ∈ R, then we have:
a ∗ (b ∘ c) = a ∗ b = | a − b |
(a ∗ b) ∘ (a ∗ c) = ( | a − b | ) ∘ ( | a − c | ) = | a − b |
Hence, a ∗ (b ∘ c) = (a ∗ b) ∘ (a ∗ c)
Now,
10(2 ∗ 3) = 10( | 2 − 3 | ) = 1 o 1 = 1
(1 o 2) ∗ (1 ∘ 3) = 1 ∗ 1 = | 1 − 1 | = 0
∴ 1 ∘ (2 ∗ 3) ≠ (1 ∘ 2) ∗ (1 ∘ 3)( where 1, 2, 3 ∈ R)
∴ The operation o does not distribute over*.

Page : 30 , Block Name : Miscellaneous Exercise

Q13
Given a non empty set X, let ∗ : P(X) × P(X) → P(X) → be defined as A ∗ B = (A − B) ∪ (B
− A), for allA, B ∈ P(X). Show that the empty set Φ is the identity for the operation ∗
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)

It is given that ∗ : P(x) × P(X) → P(X) is defined as
A ∗ B = (A − B) ∪ (B − A)◻A, B ∈ P(X)
Let A ∈ P(X). Then, we have:
A ∗ Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A
Φ ∗ A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A
∴ A ∗ Φ = A = ∅ ∗ A. ◻A ∈ P(X)
Thus, Φ is the identity element for the given operation ∗
A ∗ B = Φ = B ∗ A. ( As Φ is the identity element)
Now, we observed that A ∗ A = (A − A) ∪ (A − A) = ϕ ∪ ϕ = ϕ∀A ∈ P(X)
A of P(X) are invertible with A − 1 = A

Page : 30 , Block Name : Miscellaneous Exercise

Q14 De ne a binary operation *on the set {0, 1, 2, 3, 4, 5} as

a∗b=
{ a + b,
a+b−6
if a + b < 6
if a + b ≥ 6
Show that zero is the identity for this operation and each element a ≠ 0 of the set is
invertible with 6 − a being the inverse of a.

Let X = {0, 1, 2, 3, 4, 5}
The operation ∗ on X is defined as:

a∗b= { a+b
a+b−6
if a + b < 6
if a + b ≥ 6
An element e ∈ X is the identity element for the operation ∗ , if a ∗ e = a = e ∗ a∀a ∈ X
For a ∈ X, we observed that:
a∗0=a+0=a [a ∈ X ⇒ a + 0 < 6]
0∗a=0+a=a [a ∈ X ⇒ 0 + a < 6]
∴ a ∗ 0 = a = 0 ∗ a∀a ∈ X
Thus, 0 is the identity element for the given operation *
An element a ∈ X is invertible if there exists b ∈ X such that a ∗ b = 0 = b ∗ a.

{ a + b = 0 = b + a, if a + b < 6
a + b − 6 = 0 = b + a − 6, if a + b ≥ 6
a = − b or b = 6 − a
But, x = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ − b
∴ b = 6 − a is the inverse of a ◻a ∈ X.
∈ X, a ≠ 0 is 6 − a i.e., a − 1 = 6 − a

Page : 30 , Block Name : Miscellaneous Exercise

Q15
A = { − 1, 0, 1, 2}, B = { − 4, − 2, 0, 2} and f, g : A → B be functions defined by f(x) =

| |
1
x 2 − x 1x ∈ A and g(x) = 2 x − 2 − 1, x ∈ A . Are f and g equal ?

It is given that A = { − 1, 0, 1, 2}, B = { − 4, − 2, 0, 2}
Also, it is given that f, g : A → B are defined by f(x) = x 2 − x, x ∈ A and

| | 1
g(x) = 2 x − 2 − 1, x ∈ A

It is observed that:
f( − 1) = ( − 1) 2 − ( − 1) = 1 + 1 = 2

| 1
|
g( − 1) = 2 ( − 1) − 2 − 1 = 2 2 ()
3
−1=3−1=2

⇒ f( − 1) = g( − 1)
f(0) = (0) 2 − 0 = 0

| | 1
g(0) = 2 0 − 2 − 1 = 2 2
()
1
−1=1−1=0

⇒ f(0) = g(0)
f(1) = (1) 2 − 1 = 1 − 1 = 0

| | 1
g(1) = 2 1 − 2 − 1 = 2 2 ()
1
−1=1−1=0

⇒ f(1) = g(1)
f(2) = (2) 2 − 2 = 4 − 2 = 2

| | 1
g(2) = 2 2 − 2 − 1 = 2 2 ()
3
−1=3−1=2

⇒ f(2) = g(2)
∴ f(a) = g(a)∀a ∈ A
Hence, the functions f and g are equal.

Page : 30 , Block Name : Miscellaneous Exercise

Q16 Let A {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are
re exive and symmetric but not transitive is
(A) 1
(B) 2
(C) 3
(D) 4

The given set is A = {1, 2, 3} .
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but
not transitive is given by:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R .
Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R .
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R .
Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R , then relation R will
become transitive.
Hence, the total number of desired relations is one.
The correct answer is A.

Page : 30 , Block Name : Miscellaneous Exercise

Q17 Let A — {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C) 3
(D) 4

It is given that A = {1, 2, 3}.
The smallest equivalence relation containing (1, 2) is given by,
R 1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1) .
If we odd any one pair [say(2, 3)] to R 1, then for symmetry we must add (3, 2). Also,
for transitivity we are required to add (1, 3) and (3, 1) .
Hence, the only equivalence relation (bigger than R 1 ) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two.
The correct answer is B .

Page : 30 , Block Name : Miscellaneous Exercise

Q18 Let f : R → R be the sigmum function de ned as

{
1, x > 0
f(x) = 0, x = 0 and g: R -9 R be the Greatest Integer Function given by g(x) = [x] where [x] is
− 1, x < 0
greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1) ?

{
1, x > 0
f : R → R is de ned as f(x) = 0, x=0
− 1, x < 0
Also, g : R → R is defined as g(x) = [x], where [x] is the greatest integer less than or
equal to x.

Thus, when x ∈ (0, 1), we have f ∘ g(x) = 0 and g ∘ f(x) = 1
Hence, f ∘ g and g do not coincide in (0, 1] .

Page : 31 , Block Name : Miscellaneous Exercise

Q19
Number of binary operations on the set {a, b} are
(A) 10(B)16 (C) 20 (D) 8

A binary operation * on {a, b}, is a function from {a, b} × {a, b} → {a, b}
i.e., ∗ is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}
Hence, the total number of binary operations on the set {a, b} is 2 4 i.e., 16 .
The correct answer is B .

Page : 31 , Block Name : Miscellaneous Exercise