RD Sharma Solutions for Class 12 Maths Chapter 1 Relations – PDF Download

RD Sharma Solutions for Class 12 Maths Chapter 1 – Relations are designed to help students achieve excellent marks in their board exams. The RD Sharma textbook for Class 12 follows the latest syllabus prescribed by the CBSE (Central Board of Secondary Education). At the end of each chapter, the textbook provides exercises with multiple-choice questions, along with a summary for quick revision of key concepts and formulas.

For success in board exams, it is highly recommended to refer to RD Sharma Solutions for Class 12. These solutions provide clear explanations of all the topics in the chapter.

Students can access and download Chapter 1 – Relations through the links provided. This chapter focuses on the concept of relations and their types, with RD Sharma Solutions covering all related topics. Key topics in this chapter include:

• Types of Relations
• Void Relation
• Universal Relation
• Identity Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Antisymmetric Relation
• Equivalence Relation
• Theorems Based on Relations

Download RD Sharma Solutions for Class 12 Maths Chapter 1 Relations PDF

RD Sharma Class 12 Maths Solutions for Chapter 1 – Relations cover all the questions from the textbook, crafted by expert Mathematics teachers at Infinity Learn. Download our free PDF of Chapter 1 – Relations RD Sharma Solutions for Class 12 to boost your performance in board exams and competitive exams.

Access answers to RD Sharma Solutions Class 12 Maths Chapter 1 – Relations

Q. Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:

R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

R2 = {(a, a)}

R3 = {(b, c)}

R4 = {(a, b), (b, c), (c, a)}.

Find whether or not each of the relations R1, R2, R3, R4 on A is 

(i) reflexive 

(ii) symmetric

(iii) transitive.

Solution:

(i) Consider R₁

Given R₁ = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

Now we have check R₁ is reflexive, symmetric and transitive

Reflexive:

Given (a, a), (b, b) and (c, c) ∈ R₁

So, R₁ is reflexive.

Symmetric:

We see that the ordered pairs obtained by interchanging the components of R₁ are also in R₁.

So, R₁ is symmetric.

Transitive:

Here, (a, b) ∈R₁, (b, c) ∈R₁ and also (a, c) ∈R₁

So, R₁ is transitive.

(ii) Consider R₂

Given R₂ = {(a, a)}

Reflexive:

Clearly (a, a) ∈R₂.

So, R₂ is reflexive.

Symmetric:

Clearly (a, a) ∈R

⇒ (a, a) ∈R.

So, R₂ is symmetric.

Transitive:

R₂ is clearly a transitive relation, since there is only one element in it.

(iii) Consider R₃

Given R₃ = {(b, c)}

Reflexive:

Here,(b, b)∉ R₃ neither (c, c) ∉ R₃

So, R₃ is not reflexive.

Symmetric:

Here, (b, c) ∈R₃, but (c, b) ∉R₃

So, R₃ is not symmetric.

Transitive:

Here, R₃ has only two elements.

Hence, R₃ is transitive.

(iv) Consider R₄

Given R₄ = {(a, b), (b, c), (c, a)}.

Reflexive:

Here, (a, a) ∉ R₄, (b, b) ∉ R₄ (c, c) ∉ R₄

So, R₄ is not reflexive.

Symmetric:

Here, (a, b) ∈ R₄, but (b, a) ∉ R₄.

So, R₄ is not symmetric

Transitive:

Here, (a, b) ∈R₄, (b, c) ∈R₄, but (a, c) ∉R₄

R₄ is not transitive.

Q. Let A be the set of all human beings in a town at a particular time. Determine whether of the following relation is reflexive, symmetric and transitive:

(i) R= {(x, y): x and y work at the same place}

(ii) R= {(x, y): x and y live in the same locality}

(iii) R= {(x, y): x is wife of y}

(iv) R= {(x, y): x is father of y}

Solution:

(i) Given R= {(x, y): x and y work at the same place}

Now we have to check whether the relation is reflexive:

Let x be an arbitrary element of R.

Then, x ∈R

⇒ x and x work at the same place is true since they are the same.

⇒(x, x) ∈R [condition for reflexive relation]

So, R is a reflexive relation.

Now let us check Symmetric relation:

Let (x, y) ∈R

⇒x and y work at the same place [given]

⇒y and x work at the same place

⇒(y, x) ∈R

So, R is a symmetric relation.

Transitive relation:

Let (x, y) ∈R and (y, z) ∈R.

Then, x and y work at the same place. [Given]

y and z also work at the same place. [(y, z) ∈R]

x, y and z all work at the same place.

x and z work at the same place.

(x, z) ∈R

So, R is a transitive relation.

Hence R is reflexive, symmetric and transitive.

(ii) Given R= {(x, y): x and y live in the same locality}

Now we have to check whether the relation R is reflexive, symmetric and transitive.

Let x be an arbitrary element of R.

Then, x ∈R

It is given that x and x live in the same locality is true since they are the same.

So, R is a reflexive relation.

Symmetry:

Let (x, y) ∈ R

⇒ x and y live in the same locality [given]

⇒ y and x live in the same locality

⇒ (y, x) ∈ R

So, R is a symmetric relation.

Transitivity:

Let (x, y) ∈R and (y, z) ∈R.

Then,

x and y live in the same locality and y and z live in the same locality

⇒ x, y and z all live in the same locality

⇒ x and z live in the same locality

⇒ (x, z) ∈ R

So, R is a transitive relation.

Hence R is reflexive, symmetric and transitive.

(iii) Given R= {(x, y): x is wife of y}

Now we have to check whether the relation R is reflexive, symmetric and transitive.

First let us check whether the relation is reflexive:

Let x be an element of R.

Then, x is wife of x cannot be true.

⇒ (x, x) ∉R

So, R is not a reflexive relation.

Symmetric relation:

Let (x, y) ∈R

⇒ x is wife of y

⇒ x is female and y is male

⇒ y cannot be wife of x as y is husband of x

⇒ (y, x) ∉R

So, R is not a symmetric relation.

Transitive relation:

Let (x, y) ∈R, but (y, z) ∉R

Since x is wife of y, but y cannot be the wife of z, y is husband of x.

⇒ x is not the wife of z

⇒(x, z) ∈R

So, R is a transitive relation.

(iv) Given R= {(x, y): x is father of y}

Now we have to check whether the relation R is reflexive, symmetric and transitive.

Reflexivity:

Let x be an arbitrary element of R.

Then, x is father of x cannot be true since no one can be father of himself.

So, R is not a reflexive relation.

Symmetry:

Let (x, y) ∈R

⇒ x is father of y

⇒ y is son/daughter of x

⇒ (y, x) ∉R

So, R is not a symmetric relation.

Transitivity:

Let (x, y) ∈R and (y, z) ∈R.

Then, x is father of y and y is father of z

x is grandfather of z

(x, z) ∉R

So, R is not a transitive relation.

Q. The following relation is defined on the set of real numbers.
 

(i) aRb if a – b > 0

(ii) aRb iff 1 + a b > 0

(iii) aRb if |a| ≤ b.

Find whether relation is reflexive, symmetric or transitive.

Solution:

(i) Consider aRb if a – b > 0

Now for this relation we have to check whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

But a − a = 0 ≯ 0

So, this relation is not reflexive.

Symmetry:

Let (a, b) ∈ R

a − b > 0

− (b − a) >0

b − a < 0

So, the given relation is not symmetric.

Transitivity:

Let (a, b) ∈R and (b, c) ∈R.

Then, a − b > 0 and b − c > 0

Adding the two, we get

a – b + b − c > 0

a – c > 0

(a, c) ∈ R.

So, the given relation is transitive.

(ii) Consider aRb iff 1 + a b > 0

Now for this relation we have to check whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

1 + a × a > 0

i.e. 1 + a² > 0 [Since, square of any number is positive]

So, the given relation is reflexive.

Symmetry:

Let (a, b) ∈ R

1 + a b > 0

1 + b a > 0

(b, a) ∈ R

So, the given relation is symmetric.

Transitivity:

Let (a, b) ∈R and (b, c) ∈R

1 + a b > 0 and 1 + b c >0

But 1+ ac ≯ 0

(a, c) ∉ R

So, the given relation is not transitive.

(iii) Consider aRb if |a| ≤ b.

Now for this relation we have to check whether it is reflexive, transitive and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R [Since, |a|=a]

|a|≮ a

So, R is not reflexive.

Symmetry:

Let (a, b) ∈ R

|a| ≤ b

|b| ≰ a for all a, b ∈ R

(b, a) ∉ R

So, R is not symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R

|a| ≤ b and |b| ≤ c

Multiplying the corresponding sides, we get

|a| × |b| ≤ b c

|a| ≤ c

(a, c) ∈ R

Thus, R is transitive.

Q. Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Solution:

Let A be a set.

Then, Identity relation IA=I_A is reflexive, since (a, a) ∈ A ∀a

The converse of it need not be necessarily true.

Consider the set A = {1, 2, 3}

Here,

Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.

However, R is not an identity relation.