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Exercise 1.3 NCERT Solutions Relations & Functions Class 12 Math Chapter 1 free PDF Download

Class 12 NCERT Solutions – Mathematics Chapter 1 Relations and Functions – Exercise 1.3

Table of Contents

Question 1: Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(i) On Z+, define ∗ by a ∗ b = a – b

Solution: 

If a, b belongs to Z+

a * b = a – b which may not belong to Z+

For eg:  1 – 3 = -2 which doesn’t belongs to Z+ 

Therefore, * is not a Binary Operation on Z+

(ii) On Z+, define * by a * b = ab

Solution: 

If a, b belongs to Z+ 

a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

(iii) On R, define * by a * b = ab²

Solution:

If a, b belongs to R

a * b = ab2  which belongs to R

Therefore, * is Binary Operation on R

(iv) On Z+, define * by a * b = |a – b|

Solution:

If a, b belongs to Z+

a * b = |a – b| which belongs to Z+

 Therefore, * is Binary Operation on Z+

(v) On Z+, define * by a * b = a

Solution:

If a, b belongs to Z+

a * b = a which belongs to Z+

Therefore, * is Binary Operation on Z+

Question 2: For each binary operation * defined below, determine whether * is binary, commutative or associative.

(i) On Z, define a * b = a – b 

Solution:

a) Binary: 

If a, b belongs to Z

a * b = a – b which belongs to Z

Therefore, * is Binary Operation on Z

b) Commutative: 

If a, b belongs to Z, a * b = b * a 

LHS = a * b = a – b

RHS = b * a = b – a

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a – b + c

RHS = (a – b) * c = a – b- c

Since, LHS is not equal to RHS

Therefore, * is not Associative

(ii) On Q, define a * b = ab + 1

Solution:

a) Binary:

If a, b belongs to Q, a * b = ab + 1 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative: 

If a, b belongs to Q, a * b = b * a 

LHS = a * b = ab + 1

RHS = b * a = ba + 1 = ab + 1

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc + 1) = abc + a + 1

RHS = (a * b) * c = abc + c + 1

Since, LHS is not equal to RHS

Therefore, * is not Associative

(iii) On Q, define a ∗ b = ab/2

Solution :

a) Binary:

If a, b belongs to Q, a * b = ab/2 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative:

If a, b belongs to Q, a * b = b * a

LHS = a * b = ab/2

RHS = b * a = ba/2

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc/2) = (abc)/2

RHS = (a * b) * c = (ab/2) * c = (abc)/2

Since, LHS is equal to RHS

Therefore, * is Associative

(iv) On Z+, define a * b = 2ab

Solution:

a) Binary:

If a, b belongs to Z+, a * b = 2ab which belongs to Z+

Therefore, * is Binary Operation on Z+

b) Commutative:

If a, b belongs to Z+, a * b = b * a

LHS = a * b = 2ab

RHS = b * a = 2ba = 2ab

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * 2bc = 2a * 2^(bc)

RHS = (a * b) * c = 2ab * c = 22abc

Since, LHS is not equal to RHS

Therefore, * is not Associative

(v) On Z+, define a * b = ab

Solution:

a) Binary:

If a, b belongs to Z+, a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

b) Commutative:

If a, b belongs to Z+, a * b = b * a

LHS = a * b = ab

RHS = b * a = ba

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * bc = ab^c

RHS = (a * b) * c = ab * c = abc

Since, LHS is not equal to RHS

Therefore, * is not Associative

(vi) On R – {– 1}, define a ∗ b = a / (b + 1)

Solution: 

a) Binary:

If a, b belongs to R, a * b = a / (b+1) which belongs to R

Therefore, * is Binary Operation on R

b) Commutative:

If a, b belongs to R, a * b = b * a

LHS = a * b = a / (b + 1)

RHS = b * a = b / (a + 1)

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to A, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1

RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)

Since, LHS is not equal to RHS

Therefore, * is not Associative

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

Solution: 

^12345
111111
212222
312333
412344
512345

Question 4: Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table.

(Hint: use the following table) 

*12345
111111
212121
311311
412141
511115

(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)

Solution:

Here, (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) Is ∗ commutative?

Solution:

The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

Solution:

(2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5: Let ∗′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a ∗′ b = H.C.F. of a and b. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer.

Solution:

Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.

*’12345
111111
212121
311311
412141
511115

We see that the operation *’ is the same as the operation * in Exercise 4 above.

Question 6: Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(i) 5 ∗ 7, 20 ∗ 16

Solution:

If a, b belongs to N

a * b = LCM of a and b

5 * 7 = 35

20 * 16 = 80

(ii) Is ∗ commutative?

Solution:

If a, b belongs to N

LCM of a * b = ab

LCM of b * a = ba = ab

a*b = b*a

Thus, * binary operation is commutative.

(iii) Is ∗ associative?

Solution:

a * (b * c) = LCM of a, b, c

(a * b) * c = LCM of a, b, c

Since, a * (b * c) = (a * b) * c

Thus, * binary operation is associative.

(iv) Find the identity of ∗ in N

Solution:

Let ‘e’ is an identity 

a * e = e * a, for a belonging to N

LCM of a * e = a, for a belonging to N

LCM of e * a = a, for a belonging to N

e divides a 

e divides 1

Thus, e = 1

Hence, 1 is an identity element

(v) Which elements of N are invertible for the operation ∗? 

Solution:

a * b = b * a = identity element

LCM of a and b = 1

a = b = 1

only ‘1’ is invertible element in N. 

Question 7: 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. 

Solution:

The operation * on the set {1, 2, 3, 4, 5} is defined as

a * b = L.C.M. of a and b

Let a=3, b=5

3 * 5 = 5 * 3 = L.C.M. of 3 and 5 = 15 which does not belong to the given set

Thus, * is not a Binary Operation.

Question 8: 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?

Solution:

If a, b belongs to N

LHS = a * b = HCF of a and b

RHS = b * a = HCF of b and a

Since LHS = RHS

Therefore, * is Commutative

Now, If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = HCF of a, b and c

RHS = (a – b) * c = HCF of a, b and c

Since, LHS = RHS

Therefore, * is Associative

Now, 1 * a = a * 1 ≠ a

Thus, there doesn’t exist any identity element.

Question 9: Let ∗ be a binary operation on the set Q of rational numbers as follows:

(i) a ∗ b = a – b 

(ii) a ∗ b = a2 + b2

(iii) a ∗ b = a + ab 

(iv) a ∗ b = (a – b)2

(v) a ∗ b = ab / 4

(vi) a ∗ b = ab2

Find which of the binary operations are commutative and which are associative. 

Solution:

(i) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a – b

RHS = b * a = b – a

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a – (b – c) = a – b + c

RHS = (a – b) * c = a – b – c

Since, LHS is not equal to RHS

Therefore, * is not Associative

(ii) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a2 + b2

RHS = b * a = b2 + a2

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b2 + c2) = a2 + (b2 + c2)2

RHS = (a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2 

Since, LHS is not equal to RHS

Therefore, * is not Associative

(iii) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a + ab

RHS = b * a = b + ba

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b + bc) = a + a(b + bc)

RHS = (a * b) * c = (a + ab) * c = a + ab + (a + ab)c

Since, LHS is not equal to RHS

Therefore, * is not Associative

(iv) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = (a – b)2

RHS = b * a = (b – a)2

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b – c)2 = [a – (b – c)2]2 

RHS = (a * b) * c = (a – b)2 * c = [(a – b)2  – c]2

Since, LHS is not equal to RHS

Therefore, * is not Associative

(v) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = ab / 4

RHS = b * a = ba / 4

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * bc/4 = abc/16

RHS = (a * b) * c = ab/4 * c = abc/16

Since, LHS is equal to RHS

Therefore, * is Associative

(vi) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = ab2

RHS = b * a = ba2

Since, LHS is not equal to RHS

Therefore, * is not Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc)2 = a(bc2)2

RHS = (a * b) * c = (ab2) * c = ab2c2

Since, LHS is not equal to RHS

Therefore, * is not Associative

Question 10: Find which of the operations given above has identity

Solution:

An element e ∈ Q will be the identity element for the operation * if

a * e = a = e * a, for a ∈ Q

for (v) a * b = ab/4

Let e be an identity element 

a * e = a = e * a

LHS : ae/4 = a

   => e = 4

RHS : ea/4 = a

  => e = 4

LHS = RHS

Thus, Identity element exists

Other operations doesn’t satisfy the required conditions. 

Hence, other operations doesn’t have identity.

Question 11: Let A = N × N and ∗ be the binary operation on A defined by :

(a, b) ∗ (c, d) = (a + c, b + d)

Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any. 

Solution:

Given (a, b) * (c, d) = (a+c, b+d) on A

Let (a, b), (c, d), (e,f) be 3 pairs ∈ A

Commutative :

LHS = (a, b) * (c, d) = (a+c, b+d)

RHS = (c, d) * (a, b) = (c+a, d+b) = (a+c, b+d)

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f)

RHS = [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f)  

Since, LHS is equal to RHS

Therefore, * is Associative

Existence of Identity element:

For a, e ∈ A, a * e = a

(a, b) * (e, e) = (a, b)

(a+e, b+e) = (a, b)

a + e = a    

=> e = 0

b + e = b

=> e = 0

As 0 is not a part of set of natural numbers. So, identity function does not exist.

Question 12: 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

Solution:

(i) Let * be an operation on N, defined as:

a * b =  a + b ∀ a, b ∈ N

Let us consider b = a = 6, we have:

6 * 6 = 6 + 6 = 12 ≠ 6

Therefore, this statement is false. 

(ii) Since, * is commutative

LHS = a ∗ (b ∗ c) = a * (c * b) = (c * b) * a = RHS

Therefore, this statement is true.

Question 13: Consider a binary operation ∗ on N defined as a ∗ b = a3+ b3. 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? 

Solution:

On N, * is defined as a * b = a3 + b3

Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a3 + b3

RHS = b * a = b3 + a3

Since, LHS is equal to RHS

Therefore, * is Commutative

Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3

RHS = (a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3

Since, LHS is not equal to RHS

Therefore, * is not Associative

Thus, Option (B) is correct

Summary

Exercise 1.4 serves as a critical juncture in the study of functions, bridging basic concepts with more advanced applications. Through exploring function composition, students learn to create complex functions from simpler ones, a skill essential in modeling real-world phenomena. The focus on invertible functions introduces the idea of reversibility in mathematical operations, a concept with far-reaching implications in various fields of study. By working through problems involving finding inverses and composing functions, students develop analytical skills and a deeper intuition about functional relationships. This exercise not only reinforces previous learning but also prepares students for more advanced topics in calculus and beyond, such as differential equations and functional analysis. Mastery of these concepts provides a solid foundation for understanding transformation of functions, a key element in many areas of higher mathematics and its applications in science and engineering.

FAQs on Relations and Functions

What is function composition?

Function composition is the operation of combining two functions to create a new function. If f and g are functions, their composition (f ∘ g)(x) is defined as f(g(x)).

When is a function invertible?

A function is invertible if it is both injective (one-to-one) and surjective (onto). In other words, each element in the codomain is paired with exactly one element in the domain.

How do you find the inverse of a function?

To find the inverse of a function f(x), replace f(x) with y, swap x and y, then solve for y. The resulting expression in terms of x is f^(-1)(x).

What is the relationship between (f ∘ g) and (g ∘ f)?

In general, (f ∘ g) is not equal to (g ∘ f). Function composition is not commutative.

Why is function composition important in mathematics?

Function composition is crucial for creating complex functions from simpler ones, modeling intricate relationships, and solving equations involving multiple transformations. It’s a fundamental concept in advanced mathematics and its applications.

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