NCERT Solutions Class 12 Maths Chapter 1 Relation and Function: Exercise PDFs, Weightage & Important Formulae

The Relation and Function chapter carries up to 8 marks weightage in the Board exams. This chapter is also important to understand complex JEE Mains topics such as calculus and inverse trigonometry. Several key topics are discussed in the exercises;  

• Types of relations: Reflexive, Symmetric, Transitive, and Equivalence relations 

• Types of functions: one-one, many-one, onto and into functions 
• Other Important Functions: Modulus function, identity function, inverse function, and invertibility

• Binary Operations: Commutativity, Associativity, Existence of identity, and inverse of Function

Read more: Most scoring topics in Mathematics for JEE Main

Check out Dropped topics of Relation and Function- 1.4 Composition of Functions and Invertible Function, Page 13-14 Full Pages, Page 15 Examples 24 and 25, Page 16-25 Full Pages, Page 26 Question 12 and 13, Page 27-28 Examples 45 and 49, Page 29-31 Ques. 1–3, 6–7, 9, 11–14, 18–19, Page 31-32 Summary points 11-13 and 15-19.

Relation and Function Important Formulae for CBSE and Competitive Exams

Relations

• Domain of a Relation: The set of all first elements (x) in the ordered pairs of a relation $R$:

  $$\text{Domain of }R=\{x\mathrm{\mid }(x,y)\in R\}\mathrm{.}$$

• Range of a Relation: The set of all second elements (y) in the ordered pairs of a relation $R$:

  $$\text{Range of }R=\{y\mathrm{\mid }(x,y)\in R\}\mathrm{.}$$

• The inverse of a Relation: If $R$ is a relation from set $A$ to set $B$, then the inverse $R^{-1}$ is a relation from set $B$ to set $A$:

  $$R^{-1}=\{(y,x)\mathrm{\mid }(x,y)\in R\}\mathrm{.}$$

• Types of Relations

  • Reflexive Relation: $R$ is reflexive if $(a, a) \in R$ $a \in A$
    
  • Symmetric Relation: $R$ is symmetric if $(a, b) \in R \Rightarrow (b, a) \in R$
    
  • Transitive Relation: $R$ is transitive if $(a, b) \in R \text{ and } (b, c) \in R \Rightarrow (a, c) \in R$
    
  • Equivalence Relation: $R$ is an equivalence relation if it is reflexive, symmetric, and transitive.

Functions

• Function Definition: A function $f: A \to B$maps every element of $A$ to exactly one element of $B$:

  $$f(a)=b,\mathrm{\forall }a\in A\mathrm{.}$$

• Domain and Range of a Function

  • Domain: Set of all permissible inputs $x$of $f(x)$
    
  • Range: Set of all possible outputs ($f(x)$).

• Types of Functions

  • One-One (Injective): $f(a_1) = f(a_2) \Rightarrow a_1 = a_2$
    
  • Onto (Surjective): Every $y \in B$has a pre-image $x \in A$
    
  • Bijective: $f$is both one-one and onto.

• Composition of Functions: If $f: A \to B$and $g: B \to C$
  

  $$(g\circ f)(x)=g(f(x))\mathrm{.}$$

• Inverse of a Function: A function $f$is invertible if it is bijective:

  $$f^{-1}(y)=x\text{ such that }f(x)=y\mathrm{.}$$

• Binary Operation: A binary operation $*$ on $A$

  $$\ast :A\times A\to A\mathrm{.}$$

• Modulo Function

  $$f(x)=x\mathrm{m}\mathrm{o}\mathrm{d}n\text{(remainder when }x\text{ is divided by }n\text{)}\mathrm{.}$$

• Greatest Integer Function

  $$f(x)=\lfloor x\rfloor \text{(greatest integer less than or equal to }x\text{)}\mathrm{.}$$

Students can find the solutions of all the exercises of relation and function in one place. This Relation and Function NCERT solution PDF will be really beneficial for all categories of students whether boards or competitive exams. access the PDF below;

Class 12 Math Chapter 1 Relation and Function Solution: Free PDF Download

The class 12 Relation and Function chapter deals with many important topics, such as the type of relation, symmetric and asymmetric relations, functions, bijective functions, and surjective functions. Students will get problems related to different concepts and properties of relations and functions in the exercises.  Candidates can check the exercise-wise chapter 1 math solutions below;

Chapter 1 Relation and Function Exercise 1.1 focuses on problems based on identifying reflexive, symmetric, and transitive relations & checking whether the relation is an equivalence relation or not. Class 12 Math Relation and Function Exercise 1.1 consists of 16 Questions (14 Short Answers, 2 MCQ). Students can check the complete the exercise 1.1 solutions below;

Relation and Function exercise 1.1 solutions

Q1. Determine whether each of the following relations are reflexive, 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}

A.1. (i) We have,  $R=\{\left(x,y\right):3x-y=0\}$ a relation in set A=  $\{1,2,3..........14\}$

For  $x\in A,y=3x$ or  $y\ne x$ i.e.,

$\left(x,x\right)$ does not exist in R

$\therefore$ R is not reflexive.

For  $\left(x,y\right)\in R,y=3x$

Then  $\left(y,x\right)x\ne 3y$

So  $\left(y,x\right)\notin R$

$\therefore$ R is not symmetric

For  $\left(x,y\right)\in R$ and  $\left(y,z\right)\in R$ . We have

$y=3x$ and  $z=3y$

Then  $z=3\left(3x\right)=9x$

i.e.,  $\left(x,z\right)\notin R$

$\therefore$ R is not Transitive

(ii) We have,

R=  $\{\left(x,y\right):y=x+5\&x<4\}$ is a relation in N

=  $\{\left(1,1+5\right),\left(2,2+5\right),\left(3,3+5\right)\}$

=  $\{\left(1,6\right),\left(2,7\right),\left(3,8\right)\}$

Clearly, R is not reflexive as  $\left(x,x\right)\notin R$ and  $x<4\&x\in N$

Also, R is not symmetric as  $\left(1,6\right)\in R$ but  $\left(6,1\right)\notin R$

And for  $\left(x,y\right)\in R\left(y,z\right)\notin R$ . Hence, R is not Transitive.

(iii) R=  $\{\left(x,y\right);y$ is divisible by x  $\}$ is a relation in set

A=  $\{1,2,3,4,5,6\}$

So, R=  $\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,2\right),\left(2,4\right),\left(2,6\right),\left(3,3\right),\left(3,6\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\}$

Hence, R is reflexive because  $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\in R$ i.e.,  $\left(x,x\right)\in R$

R is not symmetric as  $\left(1,2\right)\in R$ but  $\left(2,1\right)\notin R$

And for  $x,y,z\in A$ and  $\left(x,y\right)\in R\&\left(y,z\right)\in R$

$\frac{y}{z}=n$ and  $\frac{z}{y}=m$ where  $n\&m\in N$

Then,  $z=my=m(nx)=mn.x$

$\frac{z}{x}=m.n,m,mn\in N$

Hence,  $\left(x,z\right)\in R$

$\therefore$ R is transitive

(iv) R=  $\{\left(x,y\right):x-y$ is an integer  $\}$ is a relation in set Z

For  $x\in Z$

$x-x=0$ is an integer

So,  $\left(x,x\right)\in R$ i.e., R is reflexive

For  $x,y\in Z$ and  $\left(x,y\right)\in R$

$x-y$ is an integer

$-\left(y-x\right)$ is an integer

$\left(y-x\right)$ is an integer

So,  $\left(y,x\right)\in R$ i.e., R is symmetric

For  $x,y,z\in R$ and  $\left(x,y\right)\in R\&\left(y,z\right)\in R$ We have,

$x-y$ is an integer

$y-z$ is an integer

So,  $\left(x-y\right)+\left(y-z\right)$ is also an integer

$x-z$ is an integer

So,  $\left(x,z\right)\in R$ i.e., R is transitive.

(v) (a) R=  $\{\left(x,y\right):x$ and  $y$ work at same place  $\}$ in set A of human being.

For  $x\in A$ we get

$x\&x$ work at same place

$\left(x,x\right)\in R$ So, R is reflexive.

For  $x,y\in A$ and  $\left(x,y\right)\in R$ We get,

$x\&y$ work at same place

$y\&x$ work at same place

$\left(y,x\right)\in R$ . So, R is symmetric.

For  $x,y,z\in A$ and  $\left(x,y\right)$ and  $\left(y,z\right)\in R$ . We have,

$x\&y$ work at same place

$y\&z$ work at same place

$x\&z$ work at same place

i.e.,  $\left(x,z\right)\in R$ . So, R is Transitive.

(b) R=  $\{\left(x,y\right):x\&y$ live in same locality  $\}$

For  $x\in A$ , we get,

$x\&x$ live in same locality

$\left(x,x\right)\in R$ So, R is reflexive.

For  $x,y\in A$ and  $\left(x,y\right)\in R$ we get,

$x\&y$ live in same locality

$y\&x$ live in same locality

i.e.,  $\left(y,x\right)\in R$ . So, R is symmetric.

For  $x,y,z\in A$ and  $\left(x,y\right)$ &  $\left(y,z\right)\in R$ . Then

$x,y$ live in same locality

$y,z$ live in same locality

$x,z$ live in same locality

So,  $\left(x,z\right)\in R$ i.e., R is transitive.

(c) R=  $\{\left(x,y\right):x$ is exactly 7 cm taller than y  $\}$

For  $x\in A$ ,

Height  $\left(x\right)\ne 7+$ height of  $\left(x\right)$

So,  $\left(x,x\right)\notin R$ . i.e., R is not reflexive.

For,  $x,y\in A$ and  $\left(x,y\right)\in R$ we have,

Height  $\left(x\right)$  $=7+height\left(y\right)$

But  $height\left(y\right)\ne 7+height\left(x\right)$

So,  $\left(y,x\right)\in R$ i.e., R is not symmetric.

For  $x,y,z\in A$ and  $\left(x,y\right)\in R$ and  $\left(y,z\right)\in R$ we have,

Height  $\left(x\right)=7+height\left(y\right)$

And  $height\left(y\right)=7+height\left(z\right)$

So,  $height\left(x\right)=7+7+height\left(z\right)=14+height\left(z\right)$

i.e.,  $\left(x,z\right)\notin R$

So, R is not transitive.

(d) R=  $\{\left(x,y\right):x$ is wife of y  $\}$

For  $x\in A,$

X is not wife of x.

So,  $\left(x,x\right)\notin R$ i.e., R is not reflexive.

For  $x,y\in A$ and  $\left(x,y\right)\in R$ ,

X is wife of y but y is not wife of x

So,  $\left(y,x\right)\notin R$ . i.e., R is not symmetric.

For  $x,y,z\in A$ and  $\left(x,y\right)$ and  $\left(y,z\right)\in R$

X is wife of y

Y is wife of z

But y can never be husband & wife simultaneously

So, R is not transitive.

(e) R=  $\{\left(x,y\right):x$ is father of y  $\}$

For  $x\in A$ ,

X is not a father of x

So,  $\left(x,x\right)\notin R$ i.e., R is not reflexive.

For  $x,y\in A$ and  $\left(x,y\right)\in R$

X is father of y

Y is father of z

But x is not father of z

So,  $\left(x,z\right)\notin R$ i.e., R is not transitive.

Q2. Show that the relation R in the set R of real numbers, defined as 

R = {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.

A.2. We have,

R=  $\{\left(a,b\right):a\le b^2\}$ is a relation in R.

For  $a\in R$ then is  $b=a,a\le a^2$ is not true for all real number less than 1.

Hence, R is not reflexive.

Let  $\left(a,b\right)\in R$ and a=1 and b=2

Then,  $a\le b^2$ =  $1\le 2^2$ =  $1\le 4$ so,  $\left(1,2\right)\in R$

But  $\left(b,a\right)=\left(2,1\right)$

i.e.,  $2\le 1^2$ =  $2\le 1$ is not true

so,  $\left(2,1\right)\notin R$

hence, R is not symmetric.

For,  $\left(a,b\right)=\left(10,4\right)\&\left(b,c\right)=\left(4,2\right)\in R$

We have,  $a=10\le 4^2=b^2$ =>  $10\le 16$ is true

So,  $\left(10,4\right)\in R$

And  $4\le 2^2$ =>  $4\le 4$ So,  $\left(4,2\right)\in R$

But  $10\le 2^2$ =>  $10\le 4$ is not true.

So,  $\left(10,2\right)\notin R$

Hence, R is not transitive.

Q3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

A.3. We have,

R=  $\{\left(a,b\right):b=a+1\}$ is a relation in set  $\{1,2,3,4,5,6\}$

So, R=  $\{\left(1,2\right),\left(2,3\right),\left(3,4\right),\left(4,5\right),\left(5,6\right),\left(6,7\right)\}$

As,  $\left(1,1\right)\notin R$ , R is not reflexive

As,  $\left(1,2\right)\in R$ but  $\left(2,1\right)\notin R$ , R is not symmetric

And as  $\left(1,2\right)$ &  $\left(2,3\right)\in R$ but  $\left(1,3\right)\notin R$

Hence, R is not transitive.

Q4. Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

A.4. We have, R=  $\{\left(a,b\right):a\le b\}$ is a relation in R.

For,  $a\in R$ ,

$a\le b$ but  $b\le a$ is not possible i.e.,  $\left(b,a\right)\notin R$

Hence, R is not symmetric.

For  $\left(a,b\right)\in R\&\left(b,c\right)\in R$ and  $a,b,c\in R$

$a\le b$ and  $b\le c$

So,  $a\le c$

i.e.,  $\left(a,c\right)\in R$

$\therefore$ R is transitive.

Q5. Check whether the relation R in R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

A.5. We have,

R=  $\{\left(a,b\right):a\le b^3\}$ is a relation in R.

For,  $\left(a,b\right)\in R$ and  $a=\frac{1}{2}$ we can write

$a\le a^3$ =>  $\frac{1}{2}\le {\left(\frac{1}{2}\right)}^3$ =>  $\frac{1}{2}\le \frac{1}{8}$ which is not true.

So, R is not reflexive.

For  $\left(a,b\right)=\left(1,2\right)\in R$ we have,

$a\le b^3$ =>  $1\le 2^3$ =>  $1\le 8$ is true.

So,  $\left(1,2\right)\in R$

But  $2\le 1^3$ =>  $2\le 1$ is not true

So,  $\left(2,1\right)\notin R$ and  $\left(b,a\right)\notin R$

Hence, R is not symmetric.

For,  $\left(a,b\right)=\left(10,4\right)$ and  $\left(b,c\right)=\left(4,2\right)\in R$

$10\le 4^3$ =>  $10\le 64$ is true=>  $\left(10,4\right)\in R$

$4\le 2^3$ =>  $4\le 8$ is true=>  $\left(4,2\right)\in R$

But  $10\le 2^3$ =>  $10\le 8$ is not true=>  $\left(10,2\right)\notin R$

Hence, for  $\left(a,b\right),\left(b,c\right)\in R,\left(a,c\right)\notin R$

So, R is not transitive.

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

A.6. We have,

R=  $\{\left(1,2\right),\left(2,1\right)\}$ is a relation in set  $\{1,2,3\}$

Then, as  $\left(1,1\right)\notin R$ and  $\left(2,2\right)\notin R$

So, R is not reflective

As  $\left(1,2\right)\in R$ and  $\left(2,1\right)\in R$

So, R is symmetric

And as  $\left(1,2\right)\in R,\left(2,1\right)\in R$ but  $\left(1,1\right)\notin R$

So, R is not transitive.

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.

A.7. We have,

R=  $\{\left(x,y\right):x\&y$ have same number of pages  $\}$ is a relation in set of A of all books in

For  $\left(x,y\right)\in R\&x,y\in A$

As x=y=same no. of pages

Then,  $\left(x,x\right)\in R$

Hence, R is reflexive.

For  $\left(x,y\right)\in R$ and  $x,y\in A$

Also,  $\left(y,x\right)\in R$ ,  $\therefore x=y$

Hence, R is symmetric.

For  $x,y,z\in A$ and  $\left(x,y\right)\in R$ and  $\left(y,z\right)\in R$

x=y and y=z

x=z

i.e.,  $\left(x,z\right)\in R$

hence, R is also transitive

$\therefore$ R is an equivalence relation.

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.8. We have,

R=  $\{\left(a,b\right)\therefore \left|a-b\right|$ is even  $\}$ is a relation in set A=  $\{1,2,3,4,5\}$

For all  $a\in A$ ,  $\left|a-a\right|=0$ is even.

So,  $\left(a,a\right)\in R$ . Hence R is reflexive

For  $a,b\in A$ and  $\left(a,b\right)\in R$

$\left|a-b\right|$ is even

$\left|-b+a\right|$ is even  $|$

$\left|-\left(b-a\right)\right|$ is even

$\left|b-a\right|$ is even

i.e.,  $\left(b,a\right)\in R$

Hence, R is symmetric.

For  $a,b,c\in A$ and  $\left(a,b\right)\in R$ and  $\left(b,c\right)\in R$

We have  $\left|a-b\right|$ is even

and  $\left|b-c\right|$ is even

then,  $\left|a-b\right|+\left|b-c\right|$ is even as even + even=even

$\left|a-b+b-c\right|$ is even

$\left|a-c\right|$ is even

$\therefore$  $\left(a,c\right)\in R$

So, R is transitive.

$\therefore$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly, all elements of [2,4] are even positive numbers and its subset are even and their difference gives an even number. Hence, they are related to each other.

However, elements of [1,3,5] are related to elements of [2,4] since the difference of the two subsets is never even.

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.9. We have,

A=  $\{x\in 2,0\le x\le 12\}$

The relation in set A is defined by

R= {  $\left(a,b\right):\left|a-b\right|$ is a multiple of 4}

For all  $a\in A$ ,

$\left|a-a\right|=0$ is a multiple of 4

So,  $\left(a,a\right)\in R$ i.e., R is reflexive

For  $a,b\in A\&\left(a,b\right)\in R$ we have,

$\left|a-b\right|$ is multiple of 4

$\left|-\left(b-a\right)\right|$ is multiple of 4

$\left|b-a\right|$ is multiple of 4

So,  $\left(b,a\right)\in R$

i.e., R is symmetric

for  $a,b,c\in A$  $\&\left(a,b\right)\in R\&\left(b,c\right)\in R$

$\left|a-b\right|$ &  $\left|b-c\right|$ is a multiple of 4

So  $\left|a-b\right|+\left|b-c\right|$ is also a multiple of 4

$\left|a-b+b-c\right|$ is a multiple of 4

$\left|a-c\right|$ is a multiple of 4

So,  $\left(a,c\right)\in R$

i.e., R is transitive

Hence, R is an equivalence relation.

Finding all set of elements related to 1

For  $a\in A$

Then,  $\left(a,1\right)\in R$ i.e.,  $\left|a-1\right|$ is a multiple of 4

So, a can be 0 ≤ a ≤ 12

Only,

$\left|1-1\right|=0$

$\left|5-1\right|=4$ is a multiple of 4

$\left|9-1\right|=8$

Hence, sets of elements related to A=  $\{1,5,9\}$

The relation in set A is defined as R=  $\{\left(a,b\right):a=b\}$

For all  $a\in A$ ,

a=a is true

so,  $\left(a,a\right)\in R$

i.e., R is reflexive

for  $a,b\in A\&\left(a,b\right)\in R$ we have,

a=b

b=a i.e.,  $\left(b,a\right)\in R$

so, R is symmetric

for  $a,b,c\in A$ ,  $\left(a,b\right)\in R$ and  $\left(b,c\right)\in R$ . We have,

a=b and b=c

a=c i.e.,  $\left(a,c\right)\in R$

$\therefore$ R is transitive

Hence, R is an equivalence relation

Find all sets of elements related to 1

For  $a\in A$ , we need  $\left(a,1\right)\in R$

i.e., a=1

so, set of elements related to 1 in A is {1}.

Q10. Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

A.10. Let A=  $\left\{a,b,c\right\}$

(i) R= $\left\{\left(a,b\right),\left(b,a\right)\right\}$ is a relation in set A

So,  $\left(a,b\right)\in R$ and  $\left(b,a\right)\in R\to$ Symmetric

$\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R,\left(b,a\right)\in R$ but  $\left(a,a\right)\notin R$  $\to$ not transitive

(ii) R= $\left\{\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is a relation in set A

So,  $\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ but  $\left(b,a\right)\notin R\to$ not symmetric

$\left(a,b\right)\in R\&\left(b,c\right)\in R$ and also  $\left(a,c\right)\in R\to$ transitive

(iii) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,a\right),\left(a,c\right),\left(c,a\right)\right\}$

So,  $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$ Reflexive

$\left(a,b\right)\in R\to \left(b,a\right)\in R\to$ Symmetric

$\left(a,c\right)\in R\to \left(c,a\right)\in R$

$\left(b,a\right)\in R$ and  $\left(a,c\right)\in R$

But  $\left(b,c\right)\notin R\to$ not transitive

(iv) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is s relation in set A

So,  $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$ reflexive

$\left(a,b\right)\&\left(b,c\right)\in R$ so,  $\left(a,c\right)\in R\to$ transitive

$\left(a,b\right)\in R$ but  $\left(b,a\right)\notin R\to$ not symmetric

(v) R= $\left\{\left(a,a\right),\left(a,b\right),\left(b,a\right)\right\}$

So,  $\left(b,b\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ and  $\left(b,a\right)\in R\to$ symmetric

And  $\left(a,b\right)\in R\&\left(b,a\right)\in R$

and also  $\left(a,a\right)\in R\to$ transitive

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.

A.11. The given relation in set A of points in a plane is

R=  $\{\left(P,Q\right):$ distance of point P from origin=distance of point Q from  $origin\}$

If O is the point of origin

R=  $\left\{\left(P,Q\right):PO=QO\right\}$

Then, for  $P\in A$ we have PO=PO

So,  $\left(P,P\right)\in R$

i.e., P is reflexive

for,  $P,Q\in A$ and  $\left(P,Q\right)\in R$ we have

PO=QO

QO=PO i.e.,  $\left(Q,P\right)\in R$

i.e., R is symmetric

for  $P,Q,S\in A$ and  $\left(P,Q\right)\&\left(Q,S\right)\in R$

PO=QO and QO=SO

PO=SO

i.e.,  $\left(P,S\right)\in R$

so, R is transitive

Hence, R is an equivalence relation

For a point  $P\ne \left(o,o\right)$ the set of all points related to P i.e., distance from origin to the points are equal is a circle with center at origin (o,o) by the definition of circle

Q12. Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂): T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

A.12. The given relation to set A of all triangles is defined as

R=  $\{\left(T_1,T_2\right):T_1$ is similar to  $T_2\}$

For  $T_1\in A$ ,

$T_1$ is always similar to  $T_1$

So,  $\left(T_1,T_1\right)\in R$ . Hence R is reflexive.

For  $T_{1,}{T}_2\in A$ and  $\left(T_1,T_2\right)\in R$ we have

$T_1\sim T_2\left(similar\right)$

$T_2\sim T_1$ i.e.,  $\left(T_2,T_1\right)\in R$

so, R is symmetric.

for,  $T_1,T_2,T_3\in A$ and  $\left(T_1,T_2\right)\in R$ and  $\left(T_2,T_3\right)\in R$

$T_1\sim T_2$ and  $T_2\sim T_3$

i.e.,  $T_1\sim T_3$  $\to \left(T_1,T_3\right)\in R$

so, R is transitive

$\therefore$ R is an equivalence relation.

Given, sides of  $T_1$ are 3,4,5

Sides of  $T_2$ are 5,12,13

Sides of  $T_3$ are 6,8,10

As  $\frac{3}{5}\ne \frac{4}{12}\ne \frac{5}{13}$ we conclude that  $T_1$ is not similar to  $T_2$

As  $\frac{5}{6}\ne \frac{12}{8}\ne \frac{13}{10}$ we conclude that  $T_2$ is not similar to  $T_3$

But as  $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$ we conclude that  $T_1\sim T_3$

Hence,  $T_1$ is related to  $T_3$

Q13. Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P_{1 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?

A.13. The given relation in set A of all polygons is defined as

R=  $\{\left(P_1,P_2\right):P_1$ and  $P_2$ have same number of sides  $\}$

Let  $P_1\in A$ ,

As number of sides  $\left(P_1\right)$ = number of sides  $\left(P_1\right)$

$\left(P_1,P_1\right)\in R$

So, R is reflexive.

Let  $P_1,P_2\in A$ and  $\left(P_1,P_2\right)\in R$

Then, number of sides of  $P_1$ = number of sides of  $P_2$

Number of sides of  $P_2$ = number of sides of  $P_1$

i.e.,  $\left(P_2,P_1\right)\in R$

so, R is symmetric.

Let  $P_1,P_2,P_3\in A$ and  $\left(P_1,P_2\right)$ and  $\left(P_2,P_3\right)\in R$

Then, number of sides  $\left(P_1\right)$ = number of sides  $\left(P_2\right)$

Number of sides  $\left(P_2\right)$ = number of sides  $\left(P_3\right)$

So, number of sides  $\left(P_1\right)$ = number of sides  $\left(P_3\right)$

I.e.,  $\left(P_1,P_3\right)\in R$

So, R is transitive.

Hence, R is an equivalence relation.

The set of all triangles will be the elements in A related to the right-angle triangle T with sides 3,4 and 5 as they all have three number of sides.

Q14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂): L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

A.14. The given relation in the set  $L=$ all lines in  $XY-$ plane is defined as

$R=\{\left(L_1,L_2\right):L_1$ is parallel to  $L_2\}$

Let  $L_1\in A$ then as  $L_1$ is parallel to  $L_1$ ,

$\left(L_1,L_1\right)\in R$

So,  $R$ is reflexive

Let  $L_1,L_2\in A$ and  $\left(L_1,L_1\right)\in R$

Then,  $L_1$ is parallel to  $L_2$

$L_2$ is parallel to  $L_1$

So,  $\left(L_2,L_1\right)\in R$

i.e.,  $R$ is symmetric

Let  $L_1,L_2,L_3\in A$ and  $\left(L_1,L_2\right)$ and  $\left(L_2,L_3\right)\in R$

Then,  $L_1\parallel L_2$ and  $L_2\parallel L_3$

So,  $L_1\parallel L_3$

i.e.,  $(L_1,L_2)\in R$

So,  $R$ is transitive

Hence,  $R$ is an equivalence relation

The set of lines related to  $y=2x+4$ is given by the equation  $y=2x+C$ where  $C$ is some constant.

Q15.  Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 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.

A.15. The set in  $A=\left\{1,2,3,4\right\}$

The relation in this set  $A$ is given by

$R=\left\{\left(1,2\right),\left(2,2\right),\left(1,1\right),\left(4,4\right),\left(1,3\right),\left(3,3\right),\left(3,2\right)\right\}$

$R$ is reflexive as  $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\in R$

As,  $\left(1,2\right)\in R$ but  $\left(2,1\right)\notin R$

$R$ is not symmetric

For  $\left(1,2\right)\in R$ and  $\left(2,2\right)\in R;\left(1,2\right)\in R$

And for  $\left(1,3\right)\in R$ and  $\left(3,2\right)\in R;\left(1,3\right)\in R$

∴  $R$ is transitive

Hence, option (B) is correct

Q16. Let R be the relation in the set N given by R = {(a, b): a = b − 2, b > 6}. Choose the correct answer.

(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R

A.16. The given relation in set N defined by

$R=\left\{\left(a,b\right):a=b-2,b>6\right\}$

For (2,4),       4>6 is not true

For (3,8),    8>6  but  3= 8-2 ⇒3=6 is not true

For (6,8),     8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

Chapter 1 Relation and Function Exercise 1.2 mainly focuses on functions and their types, such as many-one, onto, and into, and one-one functions. It also deals with problems of finding the range, domain, co-domain, and other elements of functions. Class 12 Math Relation and Function Exercise 1.2 consists of 12 Questions (10 Short Answers, 2 MCQ).  students can check the complete solutions for exercise 1.2 below;

Relation and Function Exercise 1.2 Solutions

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

A.1. The  $fx$ n is  $f(x)=\frac{1}{x}$ , which is a  $f:R_{*}$  $\to$  $R_{*}$ and  $R_{*}$ is set of all non-zero real numbers

For,  $x_1,x_2\in R_{*},f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So,  $f$ is one-one

For,  $y\in R_{*},$  $x=\frac{1}{f(x)}=\frac{1}{y}$ such that

So,  $f(x)=y$

So, every element in the co-domain has a pre-image in  $f$

So,  $f$ is onto

If  $f:N\to R_{*}$ such that  $f(x)=\frac{1}{x}$

For,  $x_1,x_2\in N,$  $f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So,  $f$ is one-one

For,  $y\in R_{*}$ and  $f(x)=y$ we have  $x=\frac{1}{y}\notin N$

Eg.,  $3\in R_{*}$ so  $x=\frac{1}{3}\notin N$

So,  $f$ is not onto

Q2. Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: R → R given by f(x) = x²

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

A.2. (i) $f:N\to N$ given by  $f(x)=x_{}^2$

For,  $x_1,x_2\in N$ ,  $f(x_1)=f(x_2)$

$\Rightarrow x_1^2=x_2^2$

$\Rightarrow x_1^{}=x_2^{}\notin N$

So,  $f$ is one-one/ injective

For  $x\in N$ , i.e.,  $x=1,2,3....$

Range of  $f(x)=\{1_{}^2,2_{}^2,3_{}^2...\}=\{1,4,9...\}\ne N$

i.e., co-domain of  $N$

So,  $f$ is not onto/ subjective

(ii) $f:Z\to Z$ given by  $f(x)=x_{}^2$

For,  $x_1,x_2\in Z$ ,  $f(x_1)=f(x_2)$

$\Rightarrow x_1^2=x_2^2$

$\Rightarrow x_1^{}=\pm x_2^{}\notin Z$

i.e.,  $x_1=x_2$ and  $x_1=-x_2$

So,  $f$ is not one-one/ injective

For  $x\in Z$ ,  $x=0,\pm 1,\pm 2,\pm 3....$

Range of  $f(x)=\{0_{}^2,{(\pm 1)}_{}^2,{(\pm 2)}_{}^2,{(\pm 3)}_{}^2...\}$

$\{0,1,4,9....\}\ne$ co-domain  $Z$

So,  $f$ is not onto/ subjective

(iii) $f:R\to R$ given by  $f(x)=x_{}^2$

For,  $x_1,x_2\in R$ ,  $f(x_1)=f(x_2)$

$\Rightarrow x_1^2=x_2^2$

$\Rightarrow x_1^{}=\pm x_2^{}$

So,  $f$ is not injective

For  $x\in R$

Range of  $f(x)=\{x_{}^2,x\in R\}$ gives a set of all positive real numbers

Hence, range of  $f(x\ne )$ co-domain of  $R$

So,  $f$ is not subjective

(iv) $f:N\to N$ given by  $f(x)=x_{}^3$