Pearson | PTE
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
NCERT Solutions for Exercise 1.1 Class 12 Maths Chapter 1 Relations and Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 12 Maths ex 1.1 deals with questions related to various concepts of Relations and Functions which includes types of relations, functions, binary operations etc. Exercise 1.1 Class 12 Maths will help students to grasp the basic concepts of sets and relations. It is highly recommended to students to practise the NCERT Solutions for Class 12 Maths chapter 1 exercise 1.1 to score well in CBSE class 12 board exam. In competitive exams also like JEE main ,some questions can be asked from Class 12 Maths chapter 1 exercise 1.1. Concepts related to functions discussed in Class 12th Maths chapter 1 exercise 1.1 are important for Board examination also.
12th class Maths exercise 1.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
Question1(i) . 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 $R = \{(x,y): 3x - y = 0\}$
Answer:
$A = \{1,2,3 ...,13 ,14\}$
$R = \{(x,y): 3x - y = 0\}$ $= \left \{ \left ( 1,3 \right ),\left ( 2,6 \right ),\left ( 3,9 \right ),\left ( 4,12 \right ) \right \}$
Since, $\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 14,14 \right ) \notin R$ so $R$ is not reflexive.
Since, $\left ( 1,3 \right ) \in R$ but $\left ( 3,1 \right ) \notin R$ so $R$ is not symmetric.
Since, $\left ( 1,3 \right ),\left ( 3,9 \right ) \in R$ but $\left ( 1,9 \right ) \notin R$ so $R$ is not transitive.
Hence, $R$ is neither reflexive nor symmetric and nor transitive.
Question 1(ii) . Determine whether each of the following relations are reflexive, symmetric and
transitive:
(ii) Relation R in the set N of natural numbers defined as
$R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}$
Answer:
$R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}$ $= \left \{ \left ( 1,6 \right ),\left ( 2,7 \right ),\left ( 3,8 \right ) \right \}$
Since, $\left ( 1,1 \right ) \notin R$
so $R$ is not reflexive.
Since, $\left ( 1,6 \right )\in R$ but $\left ( 6,1 \right )\notin R$
so $R$ is not symmetric.
Since there is no pair in $R$ such that $\left ( x,y \right ),\left ( y,x \right )\in R$ so this is not transitive.
Hence, $R$ is neither reflexive nor symmetric and
nor transitive.
Question1(iii) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(iii) Relation R in the set $A = \{1,2,3,4,5,6\}$ as $R = \{(x,y) : y \; \textup{is} \; divisible \; by\; x\}$
Answer:
$A = \{1,2,3,4,5,6\}$
$R = \left \{ \left ( 2,4 \right ),\left ( 3,6 \right ),\left ( 2,6 \right ),\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right )\right \}$
Any number is divisible by itself and $\left ( x,x \right ) \in R$ .So it is reflexive.
$\left ( 2,4 \right ) \in R$ but $\left ( 4,2 \right ) \notin R$ .Hence,it is not symmetric.
$\left ( 2,4 \right ),\left ( 4,4 \right ) \in R$ and 4 is divisible by 2 and 4 is divisible by 4.
Hence, it is transitive.
Hence, it is reflexive and transitive but not symmetric.
Question.1(iv) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(iv). Relation R in the set Z of all integers defined as $R = \{(x,y): x - y \;is\;an\;integer\}$
Answer:
$R = \{(x,y): x - y \;is\;an\;integer\}$
For $x \in Z$ , $\left ( x,x \right ) \in R$ as $x-x = 0$ which is an integer.
So,it is reflexive.
For $x,y \in Z$ , $\left ( x,y \right ) \in R$ and $\left ( y,x \right ) \in R$ because $x-y \, \, and \, \, y-x$ are both integers.
So, it is symmetric.
For $x,y,z \in Z$ , $\left ( x,y \right ),\left ( y,z \right ) \in R$ as $x-y \, \, and \, \, y-z$ are both integers.
Now, $x-z = \left ( x-y \right )+\left ( y-z \right )$ is also an integer.
So, $\left ( x,z \right ) \in R$ and hence it is transitive.
Hence, it is reflexive, symmetric and transitive.
Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) $R = \{(x,y) : x \;and\; y\;work\;at\;the\;same\;place\}$
Answer:
$R = \{(x,y) : x \;and\; y\;work\;at\;the\;same\;place\}$
$\left ( x,x \right )\in R$ ,so it is reflexive
$\left ( x,y \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ .
$y \;and\; x\;work\;at\;the\;same\;place$ i.e. $\left ( y,x \right )\in R$ so it is symmetric.
$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ also $y \;and\; z\;work\;at\;the\;same\;place$ .It states that $x \;and\; z\;work\;at\;the\;same\;place$ i.e. $\left ( x,z \right )\in R$ .So, it is transitive.
Hence, it is reflexive, symmetric and transitive.
Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(v) Relation R in the set A of human beings in a town at a particular time given by
(b) $R = \{(x,y): x\;and\;y\;live\;in\;the\;same\;locality\}$
Answer:
$R = \{(x,y): x\;and\;y\;live\;in\;the\;same\;locality\}$
$\left ( x,x \right )\in R$ as $x$ and $x$ is same human being.So, it is reflexive.
$\left ( x,y \right )\in R$ means $x\;and\;y\;live\;in\;the\;same\;locality$ .
It is same as $y\;and\;x\;live\;in\;the\;same\;locality$ i.e. $\left ( y,x \right )\in R$ .
So,it is symmetric.
$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x\;and\;y\;live\;in\;the\;same\;locality$ and $y\;and\;z\;live\;in\;the\;same\;locality$ .
It implies that $x\;and\;z\;live\;in\;the\;same\;locality$ i.e. $\left ( x,z \right )\in R$ .
Hence, it is reflexive, symmetric and
transitive.
Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(v) Relation R in the set A of human beings in a town at a particular time given by
(c) $R = \{(x, y) : x\;is\;exactly\;7\;cm\;taller\;than\;y\}$
Answer:
$R = \{(x, y) : x\;is\;exactly\;7\;cm\;taller\;than\;y\}$
$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $x\;is\;not\;\;taller\;than\;x$ i.e. $\left ( x,x \right )\notin R$ .So, it is not reflexive.
$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $y\;is\;not\;\;taller\;than\;x$ i.e $\left ( y,x \right )\notin R$ .So, it is not symmetric.
$\left ( x,y\right ),\left ( y,z \right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ and $y\;is\;exactly\;7\;cm\;taller\;than\;z$ .
$x\;is\;exactly\;14\;cm\;taller\;than\;z$ i.e. $\left ( x,z \right )\notin R$ .
Hence, it is not reflexive,not symmetric and
not transitive.
Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(v). Relation R in the set A of human beings in a town at a particular time given by
(d) $R = \{(x, y) : x\;is\;wife\;of\;y\}$
Answer:
$R = \{(x, y) : x\;is\;wife\;of\;y\}$
$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but $x\;is\;not\, wife\;of\;x$ i.e. $\left ( x,x \right ) \notin R$ .
So, it is not reflexive.
$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but $y\;is\;not\, wife\;of\;x$ i.e. $\left ( y,x \right ) \notin R$ .
So, it is not symmetric.
Let, $\left ( x,y \right ),\left ( y,z \right ) \in R$ means $x\;is\;wife\;of\;y$ and $y\;is\;wife\;of\;z$ .
This case is not possible so it is not transitive.
Hence, it is not reflexive, symmetric and
transitive.
Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:
(v) Relation R in the set A of human beings in a town at a particular time given by
(e) $R = \{(x, y) : x \;is \;father \;of \;y \}$
Answer:
$R = \{(x, y) : x \;is \;father \;of \;y \}$
$(x, y) \in R$ means $x \;is \;father \;of \;y$ than $x \;cannot \, be \;father \;of \;x$ i.e. $(x, x) \notin R$ .So, it is not reflexive..
$(x, y) \in R$ means $x \;is \;father \;of \;y$ than $y \;cannot \, be \;father \;of \;x$ i.e. $(y, x) \notin R$ .So, it is not symmetric.
Let, $(x, y),\left ( y,z \right )\in R$ means $x \;is \;father \;of \;y$ and $y \;is \;father \;of \;z$ than $x \;cannot \, be \;father \;of \;z$ i.e. $(x, z) \notin R$ .
So, it is not transitive.
Hence, it is neither reflexive nor symmetric and nor transitive.
Answer:
$R = \{(a, b) : a \leq b^2 \}$
Taking
$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$
and
$\left ( \frac{1}{2} \right )> \left ( \frac{1}{2} \right )^{2}$
So, R is not reflexive.
Now,
$\left ( 1,2 \right )\in R$ because $1< 4$ .
But, $4\nless 1$ i.e. 4 is not less than 1
So, $\left ( 2,1 \right )\notin R$
Hence, it is not symmetric.
$\left ( 3,2 \right )\in R\, \, and \, \, \left ( 2,1.5 \right )\in R$ as $3< 4\, \, and \, \, 2< 2.25$
Since $\left ( 3,1.5 \right )\notin R$ because $3\nless 2.25$
Hence, it is not transitive.
Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.
Answer:
R defined in the set $\{1, 2, 3, 4, 5, 6\}$
$R = \{(a, b) : b = a + 1\}$
$R=\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}$
Since, $\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right ) \right \}\notin R$ so it is not reflexive.
$\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}\in R$ but $\left \{ \left ( 2,1 \right ),\left ( 3,2 \right ),\left ( 4,3 \right ),\left ( 5,4 \right ),\left ( 6,5 \right ) \right \}\notin R$
So, it is not symmetric
$\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}\in R$ but $\left \{ \left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 4,6 \right )\right \}\notin R$
So, it is not transitive.
Hence, it is neither reflexive, nor symmetric, nor transitive.
Question:4 Show that the relation R in R defined as $R = \{(a, b) : a \leq b\}$ , is reflexive and
Answer:
$R = \{(a, b) : a \leq b\}$
As $\left ( a,a \right )\in R$ so it is reflexive.
Now we take an example
$\left ( 2,3 \right )\in R$ as $2< 3$
But $\left ( 3,2 \right )\notin R$ because $2 \nless 3$ .
So,it is not symmetric.
Now if we take, $\left ( 2,3 \right )\in R\, \, and\, \, \left ( 3,4 \right )\in R$
Than, $\left ( 2,4 \right )\in R$ because $2< 4$
So, it is transitive.
Hence, we can say that it is reflexive and transitive but not symmetric.
Answer:
$R = \{(a, b) : a \leq b^3 \}$
$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$ because $\frac{1}{2}\nleqslant (\frac{1}{2}) ^{3}$
So, it is not symmetric
Now, $\left ( 1,2 \right ) \in R$ because $1< 2^{3}$
but $\left ( 2,1 \right )\notin R$ because $2\nleqslant 1^{3}$
It is not symmetric
$\left ( 3,1.5 \right ) \in R\, \, and \, \, \left ( 1.5,1.2 \right ) \in R$ as $3< 1.5^{3} \, \, and \, \, 1.5< 1.2^{3}$ .
But, $\left ( 3,1.2 \right )\notin R$ because $3 \nleqslant 1.2^{3}$
So it is not transitive
Thus, it is neither reflexive, nor symmetric, nor transitive.
Answer:
Let A= $\{1, 2, 3\}$
$R = \{(1, 2), (2, 1)\}$
We can see $\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right )\notin R$ so it is not reflexive.
As $\left ( 1,2 \right )\in R \, and \, \left ( 2,1 \right )\in R$ so it is symmetric.
$(1, 2) \in R \, and\, (2, 1)\in R$
But $(1, 1)\notin R$ so it is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
Answer:
A = all the books in a library of a college
$R = \{(x, y) : x \;and\;y\;have\;same\;number\;of\;pages\}$
$(x,x) \in R$ because x and x have the same number of pages so it is reflexive.
Let $(x,y) \in R$ means x and y have same number of pages.
Since y and x have the same number of pages so $(y,x) \in R$ .
Hence, it is symmetric.
Let $(x,y) \in R$ means x and y have the same number of pages.
and $(y,z) \in R$ means y and z have the same number of pages.
This states,x and z also have the same number of pages i.e. $(x,z) \in R$
Hence, it is transitive.
Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence
relation.?
Answer:
$A = \{1, 2, 3, 4, 5\}$
$R = \{(a, b) : |a - b| \;is\;even\}$
$R=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 4,2 \right ),\left ( 5,3 \right )\right \}$
Let there be $a\in A$ then $(a,a)\in R$ as $\left | a-a \right |=0$ which is even number. Hence, it is reflexive
Let $(a,b)\in R$ where $a,b\in A$ then $(b,a)\in R$ as $\left | a-b \right |=\left | b-a \right |$
Hence, it is symmetric
Now, let $(a,b)\in R \, and\, (b,c)\in R$
$\left | a-b \right | \, and \, \left | b-c \right |$ are even number i.e. $(a-b)\, and\,(b-c)$ are even
then, $(a-c)=(a-b)+(b-c)$ is even (sum of even integer is even)
So, $(a,c)\in R$ . Hence, it is transitive.
Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.
The elements of $\{1, 3, 5\}$ are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.
The elements of $\{2, 4\}$ are related to each other because the difference of even number is even number and in this set, all numbers are even.
The element of $\{1, 3, 5\}$ is not related to $\{2, 4\}$ because a difference of odd and even number is not even.
Question:9(i) Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$ , given by
(i) $R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer:
$A = \{x \in Z : 0 \leq x \leq 12\}$
$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$
$R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$
For $a\in A$ , $(a,a)\in R$ as $\left | a-a \right |=0$ which is multiple of 4.
Henec, it is reflexive.
Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4.
then $\left | b-a \right |$ is also multiple of 4 because $\left | a-b \right |$ = $\left | b-a \right |$ i.e. $(b,a)\in R$
Hence, it is symmetric.
Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4 and $(b,c)\in R$ i.e. $\left | b-c \right |$ is multiple of 4 .
$( a-b )$ is multiple of 4 and $(b-c)$ is multiple of 4
$(a-c)=(a-b)+(b-c)$ is multiple of 4
$\left | a-c \right |$ is multiple of 4 i.e. $(a,c)\in R$
Hence, it is transitive.
Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.
The set of all elements related to 1 is $\left \{1,5,9 \right \}$
$\left | 1-1 \right |=0$ is multiple of 4.
$\left | 5-1 \right |=4$ is multiple of 4.
$\left | 9-1 \right |=8$ is multiple of 4.
Question:9(ii) Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$ , given by
(ii) $R = \{(a, b) : a = b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer:
$A = \{x \in Z : 0 \leq x \leq 12\}$
$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$
$R = \{(a, b) : a = b\}$
For $a\in A$ , $(a,a)\in R$ as $a=a$
Henec, it is reflexive.
Let, $(a,b)\in R$ i.e. $a=b$
$a=b$ $\Rightarrow$ $b=a$ i.e. $(b,a)\in R$
Hence, it is symmetric.
Let, $(a,b)\in R$ i.e. $a=b$ and $(b,c)\in R$ i.e. $b=c$
$\therefore$ $a=b=c$
$a=c$ i.e. $(a,c)\in R$
Hence, it is transitive.
Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.
The set of all elements related to 1 is {1}
Question:10(i) Give an example of a relation.
(i) Which is Symmetric but neither reflexive nor transitive.
Answer:
Let
$A = \left \{ 1,2,3 \right \}$
$R = \left \{ \left ( 1,2 \right ),\left ( 2,1 \right )\right \}$
$\left ( 1,1 \right ),\left ( 2,2 \right ),(3,3) \notin R$ so it is not reflexive.
$(1,2)\in R$ and $(2,1)\in R$ so it is symmetric.
$(1,2)\in R \, and\, (2,1)\in R$ but $(1,1)\notin R$ so it is not transitive.
Hence, symmetric but neither reflexive nor transitive.
Question:10(ii) Give an example of a relation.
(ii) Which is transitive but neither reflexive nor symmetric.
Answer:
Let
$R = \left \{ \left ( x,y \right ): x> y \right \}$
Now for $x\in R$ , $(x,x)\notin R$ so it is not reflexive.
Let $(x,y) \in R$ i.e. $x> y$
Then $y> x$ is not possible i.e. $(y,x) \notin R$ . So it is not symmetric.
Let $(x,y) \in R$ i.e. $x> y$ and $(y,z) \in R$ i.e. $y> z$
we can write this as $x> y> z$
Hence, $x> z$ i.e. $(x,z)\in R$ . So it is transitive.
Hence, it is transitive but neither reflexive nor symmetric.
Question:10(iii) Give an example of a relation.
(iii) Which is Reflexive and symmetric but not transitive.
Answer:
Let
$A = \left \{ 1,2,3 \right \}$
Define a relation R on A as
$R= \left \{ (1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2) \right \}$
If $x\in A$ , $(x,x)\in R$ i.e. $\left \{ (1,1),(2,2),(3,3)\right \} \in R$ . So it is reflexive.
If $x,y\in A$ , $(x,y)\in R$ and $(y,x)\in R$ i.e. $\left \{(1,2),(2,1),(2,3),(3,2) \right \}\in R$ . So it is symmetric.
$(x,y)\in R$ and $(y,z)\in R$ i.e. $(1,2)\in R$ . and $(2,3)\in R$
But $(1,3)\notin R$ So it is not transitive.
Hence, it is Reflexive and symmetric but not transitive.
Question:10(iv) Give an example of a relation.
(iv) Which is Reflexive and transitive but not symmetric.
Answer:
Let there be a relation R in R
$R=\left \{ (a,b):a\leq b \right \}$
$(a,a)\in R$ because $a=a$
Let $(a,b)\in R$ i.e. $a\leq b$
But $(b,a)\notin R$ i.e. $b\nleqslant a$
So it is not symmetric.
Let $(a,b)\in R$ i.e. $a\leq b$ and $(b,c)\in R$ i.e. $b\leq c$
This can be written as $a\leq b\leq c$ i.e. $a\leq c$ implies $(a,c)\in R$
Hence, it is transitive.
Thus, it is Reflexive and transitive but not symmetric.
Question:10(v) Give an example of a relation.
(v) Which is Symmetric and transitive but not reflexive.
Answer:
Let there be a relation A in R
$A= \left \{ 1,2 \right \}$
$R=\left \{ (1,2),(2,1),(2,2)\right \}$
$(1,1)\notin R$ So R is not reflexive.
We can see $(1,2)\in R$ and $(2,1)\in R$
So it is symmetric.
Let $(1,2)\in R$ and $(2,1)\in R$
Also $(2,2)\in R$
Hence, it is transitive.
Thus, it Symmetric and transitive but not reflexive.
Answer:
$R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}$
The distance of point P from the origin is always the same as the distance of same point P from origin i.e. $(P,P)\in R$
$\therefore$ R is reflexive.
Let $(P,Q)\in R$ i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.
this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e. $(Q,P)\in R$
$\therefore$ R is symmetric.
Let $(P,Q)\in R$ and $(Q,S)\in R$
i.e. the distance of point P from the origin is the same as the distance of point Q from the origin, and also the distance of point Q from the origin is the same as the distance of the point S from the origin.
We can say that the distance of point P, Q, S from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e. $(P,S)\in R$
$\therefore$ R is transitive.
Hence, R is an equivalence relation.
The set of all points related to a point $P \neq (0, 0)$ are points whose distance from the origin is the same as the distance of point P from the origin.
In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.
Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.
Answer:
$R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$
All triangles are similar to itself, so it is reflexive.
Let,
$(T_1,T_2) \in R$ i.e.T 1 is similar to T2
T 1 is similar to T2 is the same asT2 is similar to T 1 i.e. $(T_2,T_1) \in R$
Hence, it is symmetric.
Let,
$(T_1,T_2) \in R$ and $(T_2,T_3) \in R$ i.e. T 1 is similar to T2 and T2 is similar toT 3 .
$\Rightarrow$ T 1 is similar toT 3 i.e. $(T_1,T_3) \in R$
Hence, it is transitive,
Thus, $R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$ , is equivalence relation.
Now, we see the ratio of sides of triangle T 1 andT 3 are as shown
$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$
i.e. ratios of sides of T 1 and T 3 are equal.Hence, T 1 and T 3 are related.
Answer:
$R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$
The same polygon has the same number of sides with itself,i.e. $(P_1,P_2) \in R$ , so it is reflexive.
Let,
$(P_1,P_2) \in R$ i.e.P 1 have same number of sides as P 2
P 1 have the same number of sides as P 2 is the same as P 2 have same number of sides as P 1 i.e. $(P_2,P_1) \in R$
Hence,it is symmetric.
Let,
$(P_1,P_2) \in R$ and $(P_2,P_3) \in R$ i.e. P 1 have the same number of sides as P 2 and P 2 have same number of sides as P 3
$\Rightarrow$ P 1 have same number of sides as P 3 i.e. $(P_1,P_3) \in R$
Hence, it is transitive,
Thus, $R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$ , is an equivalence relation.
The elements in A related to the right angle triangle T with sides 3, 4 and 5 are those polygons which have 3 sides.
Hence, the set of all elements in A related to the right angle triangle T is set of all triangles.
Answer:
$R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$
All lines are parallel to itself, so it is reflexive.
Let,
$(L_1,L_2) \in R$ i.e.L 1 is parallel to L 2 .
L1 is parallel to L 2 is same as L 2 is parallel to L 1 i.e. $(L_2,L_1) \in R$
Hence, it is symmetric.
Let,
$(L_1,L_2) \in R$ and $(L_2,L_3) \in R$ i.e. L1 is parallel to L 2 and L 2 is parallel to L 3 .
$\Rightarrow$ L 1 is parallel to L 3 i.e. $(L_1,L_3) \in R$
Hence, it is transitive,
Thus, $R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$ , is equivalence relation.
The set of all lines related to the line $y = 2x + 4.$ are lines parallel to $y = 2x + 4.$
Here, Slope = m = 2 and constant = c = 4
It is known that the slope of parallel lines are equal.
Lines parallel to this ( $y = 2x + 4.$ ) line are $y = 2x + c$ , $c \in R$
Hence, set of all parallel lines to $y = 2x + 4.$ are $y = 2x + c$ .
Question:15 Let R be the relation in the set A= {1,2,3,4}
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Answer:
A = {1,2,3,4}
$R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}$
For every $a \in A$ there is $(a,a) \in R$ .
$\therefore$ R is reflexive.
Given, $(1,2) \in R$ but $(2,1) \notin R$
$\therefore$ R is not symmetric.
For $a,b,c \in A$ there are $(a,b) \in R \, and \, (b,c) \in R$ $\Rightarrow$ $(a,c) \in R$
$\therefore$ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
The correct answer is option B.
Question:16 Let R be the relation in the set N given by $R = \{(a, b) : a = b - 2, b > 6\}$ . Choose the correct answer.
(A) $(2, 4) \in R$
(B) $(3,8) \in R$
(C) $(6,8) \in R$
(D) $(8,7) \in R$
Answer:
$R = \{(a, b) : a = b - 2, b > 6\}$
(A) Since, $b< 6$ so $(2, 4) \notin R$
(B) Since, $3\neq 8-2$ so $(3,8) \notin R$
(C) Since, $8> 6$ and $6=8-2$ so $(6,8) \in R$
(d) Since, $8\neq 7-2$ so $(8,7) \notin R$
The correct answer is option C.
The NCERT Class 12 maths chapter Relations and Functions has a total of 5 exercises including miscellaneous. Exercise 1.1 Class 12 Maths covers solutions to 16 main questions and their sub-questions. The initial 10 questions are based on concepts like symmetric, reflexive and transitive relation and subsequent questions upto 15 are based in equivalence relation etc. NCERT Solutions for Class 12 Maths chapter 1 exercise 1.1 is good source to learn concepts related to symmetric relations, equivalence of a relation etc. Students can get access of NCERT Notes For Class 12 Mathematics Chapter 1 which can be used for quick revision of important concepts of this chapter.
Happy learning!!!
Concepts related to symmetric, reflexive and transitive, equivalence relations etc, are discussed in the Exercise 1.1 Class 12 Maths
Topics like
Two chapters 'relation and function' and 'inverse trigonometry' combined has 10 % weightage in the CBSE final board exam.
From the analysis of previous year questions of Board exams, it is clear that direct questions are asked from the NCERT questions. Also Some of the questions are repeated year after year. Hence it is said that NCERT solutions are low hanging fruits. Every serious student must practice NCERT questions to score well in the exam.
In maths, relation defines the relationship between sets of values of ordered pairs
symmetric, reflexive and transitive, equivalence relations etc, are discussed in the Exercise 1.1 Class 12 Maths
There are 15 questions in Exercise 1.1 Class 12 Maths
In NCERT class 12 maths chapter 1 relations and function, there are a total of 5 exercises which includes a miscellaneous exercise also.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
Re-evaluate Your Study Strategies:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.
I hope this information helps you.
Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
Type C: For candidates scoring between 40% and 50%.
Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE