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Understanding relations and functions in mathematics is like learning to navigate a map – it helps us move from one point to another with purpose. During the NCERT Solutions of relations and functions class 12, the first question that comes to mind is, what exactly are relations and functions? A relation is like a connection between a student and all the subjects in the syllabus. On the other hand, a function is like a special subject, which is the student's only favourite subject. The main purpose of these relations and functions class 12 NCERT solutions is to make learning easier for students and to explain this topic more easily.
More than 42 lakh students appeared in the Class 10 and 12 board exams 2025.
People use relations and functions in real-life applications, such as mapping, programming, and data handling. These solutions by the NCERT are essential for students because they comprise quality practice questions. After checking the Class 12 Maths NCERT solutions from textbooks, students can also check the NCERT Exemplar Solutions for Class 12 Maths Chapter 1 Relations and Functions to understand this chapter better.
Students who wish to access the Class 12 Maths Chapter 1 NCERT Solutions can click on the link below to download the complete solution in PDF.
>> Relations:
A relation $R$ is a subset of the Cartesian product of $A \times B$, where $A$ and $B$ are non-empty sets.
$R^{-1}$, the inverse of relation $R$, is defined as:
$
R^{-1}=\{(b, a):(a, b) \in R\}
$
- Domain of $R=$ Range of $R^{-1}$
- Range of $R=$ Domain of $R^{-1}$
>> Functions:
A relation $f$ from set A to set B is a function if every element in A has one and only one image in B.
$
A \times B=\{(a, b): a \in A, b \in B\}
$
If $(a, b)=(x, y)$, then $a=x$ and $b=y$
$n(A \times B)=n(A)$ * $n(B)$, where $n(A)$ is the cardinality (number of elements) of set $A$.
$\mathrm{A} \times \phi=\phi$ (where $\phi$ is the empty set)
A function $f: A \rightarrow B$ is denoted as:
$ f(x)=y$
Algebra of functions:
- $(f+g)(x)=f(x)+g(x)$
- $(f-g)(x)=f(x)-g(x)$
- $\left(f{ }^* g\right)(x)=f(x)^* g(x)$
- $(k f)(x)=k^* f(x)$, where $k$ is a real number
- $(f / g)(x)=f(x) / g(x)$, where $g(x) \neq 0$
NCERT Relations and Functions Class 11 Solutions: Exercise: 1.1 Page Number: 5-7 Total Questions: 16 |
Question 1(i): Determine whether each of the following relations are reflexive, symmetric, and transitive:
(i) Relation $R$ in the set $A=\{1,2,3 \ldots, 13,14\}$ defined as $R=\{(x, y): 3 x-y=0\}$
Answer:
A={1,2,3...,13,14}
R={(x,y):3x−y=0} ={(1,3),(2,6),(3,9),(4,12)}
Since, (1,1),(2,2),(3,3),(4,4),(5,5)⋅⋅⋅⋅⋅⋅(14,14)∉R so R is not reflexive.
Since (1,3)∈R but (3,1)∉R, R is not symmetric.
Since, (1,3),(3,9)∈R but (1,9)∉R, so R is not transitive.
Hence, R is neither reflexive nor symmetric 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
$R=\{(x, y): y=x+5 \text { and } x<4\}$
Answer:
R={(x,y):y=x+5andx<4} ={(1,6),(2,7),(3,8)}
Since (1,1)∉R
So R is not reflexive.
Since, (1,6)∈R but (6,1)∉R
So R is not symmetric.
Since there is no pair in R such that (x,y),(y,x)∈R so this is not transitive.
Hence, R is neither reflexive nor symmetric nor transitive.
Question 1 (iii): Determine whether each of the following relations is reflexive, symmetric, and Transitive:
(iii) Relation R in the set A={1,2,3,4,5,6} as R={(x,y):y is divisible by x}
Answer:
A={1,2,3,4,5,6}
R={(2,4),(3,6),(2,6),(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
Any number is divisible by itself, and (x,x)∈R. So it is reflexive.
(2,4)∈R but (4,2)∉R .Hence,it is not symmetric.
(2,4),(4,4)∈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∈Z, (x,x)∈R as x−x=0, which is an integer.
So, it is reflexive.
For x,y∈Z, (x,y)∈R and (y,x)∈R because x−y and y−x are both integers.
So, it is symmetric.
For x,y,z∈Z, (x,y),(y,z)∈R as x−y and y−z are both integers.
Now, x−z=(x−y)+(y−z) is also an integer.
So, (x,z)∈R and hence it is transitive.
Hence, it is reflexive, symmetric, and transitive.
Question 1(v): Determine whether each of the following relations is 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}
(x,x)∈R ,so it is reflexive
(x,y)∈R means x and y work at the same place.
y and x work at the same place, i.e. (y,x)∈R, so it is symmetric.
(x,y),(y,z)∈R means x and y work at the same place, and also y and z work at the same place. It states that x and z work at the same place, i.e. (x,z)∈R. So, it is transitive.
Hence, it is reflexive, symmetric, and transitive.
Question 1(v): Determine whether each of the following relations is 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}
(x,x)∈R as x and x is same human being.So, it is reflexive.
(x,y)∈R means x and y live in the same locality.
It is the same as y and x live in the same locality, i.e. (y,x)∈R.
So, it is symmetric.
(x,y),(y,z)∈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. (x,z)∈R.
Hence, it is reflexive, symmetric, and transitive.
Question 1(v): Determine whether each of the following relations is 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 7cm taller than y}
Answer:
R={(x,y):x is exactly 7cm taller than y}
(x,y)∈R means x is exactly 7cm taller than y, but x is not taller than x, i.e. (x,x)∉R. So, it is not reflexive.
(x,y)∈R means x is exactly 7cm taller than y, but y is not taller than x, i.e. (y,x)∉R. So, it is not symmetric.
(x,y),(y,z)∈R means x is exactly 7cm taller than y, and y is exactly 7cm taller than z.
x is exactly 7cm taller than z, i.e. (x,z)∉R.
Hence, it is not reflexive, not symmetric, and not transitive.
Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and Transitive:
(v). Relation R in the set A of human beings in a town at a particular time given by
Answer:
R={(x,y):x is wife of y}
(x,y)∈R means x is the wife of y, but x is not the wife of x, i.e. (x,x)∉R.
So, it is not reflexive.
(x,y)∈R means x is wife of y but y is not wife of x i.e. (y,x)∉R .
So, it is not symmetric.
Let (x,y),(y,z)∈R mean x is the wife of y and y is the wife of z.
This case is not possible, so it is not transitive.
Hence, it is not reflexive, symmetric, or transitive.
Question 1(v): Determine whether each of the following relations is 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)∈R means x is the father of y, then x cannot be the father of x, i.e. (x,x)∉R. So, it is not reflexive.
(x,.∈R means x is the father of y, then y cannot be the father of x, i.e. (y,x)∉R. So, it is not symmetric.
Let (x,y),(y,z)∈R mean x is the father of y and y is the father of z, then x cannot be the father of z, i.e. (x,z)∉R.
So, it is not transitive.
Hence, it is neither reflexive nor symmetric nor transitive
Answer:
$R=\left\{(a, b): a \leq b^2\right\}$
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,
$(1,2) \in R$ because $1<4$.
But, $4 \nless 1$, i.e. 4 is not less than 1
So, $(2,1) \notin R$
Hence, it is not symmetric.
$(3,2) \in R$ and $(2,1.5) \in R$ as $3<4$ and $2<2.25$
Since $(3,1.5) \notin R$ 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 is defined in the set {1,2,3,4,5,6}
R={(a,b):b=a+1}
R={(1,2),(2,3),(3,4),(4,5),(5,6)}
Since, {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}∉R so it is not reflexive.
{(1,2),(2,3),(3,4),(4,5),(5,6)}∈R but {(2,1),(3,2),(4,3),(5,4),(6,5)}∉R
So, it is not symmetric
{(1,2),(2,3),(3,4),(4,5),(5,6)}∈R but {(1,3),(2,4),(3,5),(4,6)}∉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≤b}, is reflexive and transitive but not symmetric.
Answer:
R={(a,b):a≤b}
As (a,a)∈R so I t is reflexive.
Now, we take an example
(2,3)∈R as 2<3
But (3,2)∉R because 2≮3.
So, it is not symmetric.
Now, if we take (2,3)∈Rand(3,4)∈R
Then, (2,4) because 2<4
So, it is transitive.
Hence, we can say that it is reflexive and transitive but not symmetric.
Answer:
$R=\left\{(a, b): a \leq b^3\right\}$
$\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$ because $\frac{1}{2} \notin\left(\frac{1}{2}\right)^3$
So, it is not symmetric.
Now, $(1,2) \in R$ because $1<2^3$
but $(2,1) \notin R$ because $2 \nless 1^3$
It is not symmetric
$(3,1.5) \in R$ and $(1.5,1.2) \in R$ as $3<1.5^3$ and $1.5<1.2^3$.
But, $(3,1.2) \notin R$ because $3 \nless 1.2^3$
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 (1,1),(2,2),(3,3)∉R so it is not reflexive.
As (1,2)∈Rand(2,1)∈R, so it is symmetric.
(1,2)∈Rand(2,1)∈R
But (1,1)∉R, so it is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
Answer:
A = all the books in the library of a college
$R=\{(x, y): x$ and $y$ have same number of pages $\}$
$(x, x) \in R$ because x and x have same number of pages so it is reflexive.
Let $(x, y) \in R$ means x and y have the 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 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={(1,1),(2,2),(3,3),(4,4),(5,5),(1,3),(2,4),(3,5),(3,1),(5,1),(4,2),(5,3)}
Let there be a∈A, then (a, a)∈R as | a−a|=0, which is an even number. Hence, it is reflexive
Let (a,b)∈R where a,b∈A then (b,a)∈R as |a−b|=|b−a|
Hence, it is symmetric
Now, let (a,b)∈Rand(b,c)∈R
|a−b| and |b−c| 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)∈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 numbers er and in this set, all numbers are odd.
The elements of {2,4} are related to each other because the difference of even numbers is an even number er and in this set, all numbers are even.
The element of {1,3,5} is not related to {2,4} because the difference of odd and even numbers is even.
Question 9(i): 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} is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer:
A={x∈Z:0≤x≤12}
A={0,1,2,3,4,5,6,7,8,9,10,11,12}
R={(a,b):|a−b| is a multiple of 4}
For a∈A, (a, a)∈ as |a−a|=0, which is a multiple of 4.
Hence, it is reflexive.
Let (a,b)∈R, i.e. |a−b| be a multiple of 4.
then |b−a| is also multiple of 4 because |a−b| = |b−a| i.e. (b,a)∈R
Hence, it is symmetric.
Let (a,b)∈R, i.e. |a−b| be a multiple of 4 and (b,c)∈R, i.e. |b−c| be 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 multiple of 4
|a−c| is a multiple of 4, i.e. (a,c)∈R
Hence, it is transitive.
Thus, it is reflexive, symmetric, and train, positive, i.e. it is an equivalence relation.
The set of all elements related to 1 is {1,5,9}
|1−1|=0 is a multiple of 4.
|5−1|=4 is a multiple of 4.
|9−1|=8 is a multiple of 4.
Question 9(ii): Show that each of the relations R in the set A={x∈Z:0≤x≤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∈Z:0≤x≤12}
A={0,1,2,3,4,5,6,7,8,9,10,11,12}
R={(a,b):a=b}
For a∈A , (a,a)∈R as a=a
Hence, it is reflexive.
Let (a,b)∈R, i.e. a=b
a=b ⇒ b=a i.e. (b,a)∈R
Hence, it is symmetric.
Let, (a,b)∈R i.e. a=b and (b,c)∈R i.e. b=c
∴ a=b=c
a=c i.e. (a,c)∈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={1,2,3}
R={(1,2),(2,1)}
(1,1),(2,2),(3,3)∉R so it is not reflexive.
(1,2)∈R and (2,1)∈R, so it is symmetric.
(1,2)∈Rand(2,1)∈R but (1,1)∉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={(x,y):x>y}
Now for x∈R , (x,x)∉R so it is not reflexive.
Let (x,y)∈R i.e. x>y
Then y>x is not possible, i.e. (y,x)∉R. So it is not symmetric.
Let (x,y)∈R i.e. x>y and (y,z)∈R i.e. y>z
we can write this as x>y>z
Hence, x>z, i.e. (x,z)∈R. So it's 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={1,2,3}
Define a relation R on A as
R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}
If x∈A , (x,x)∈R i.e. {(1,1),(2,2),(3,3)}∈R . So it is reflexive.
If x,y∈A , (x,y)∈R and (y,x)∈R i.e. {(1,2),(2,1),(2,3),(3,2)}∈R . So it is symmetric.
(x,y)∈R and (y,z)∈R i.e. (1,2)∈R . and (2,3)∈R
But (1,3)∉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={(a,b):a≤b}
(a, a)∈R because a=a
Let (a,b)∈R, i.e. a≤b
But (b, a)∉R, i.e. b⪇a
So it is not symmetric.
Let (a,b)∈R i.e. a≤b and (b,c)∈R i.e. b≤c
This can be written as a≤b≤c i.e. a≤c implies (a,c)∈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={1,2}
R={(1,2),(2,1),(2,2)}
(1,1)∉R So R is not reflexive.
We can see (1,2)∈R and (2,1)∈R
So it is symmetric.
Let (1,2)∈R and (2,1)∈R
Also (2,2)∈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 the 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 the same point P from another origin, i.e. $(P, P) \in R$
$\therefore \mathrm{R}$ 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 the distance of point Q from the origin, same as the distance of point P from the origin, i.e. $(Q, P) \in R$
$\therefore \mathrm{R}$ is symmetric.
Let $(P, Q) \in R \quad$ and $(Q, S) \in R$
i.e. distance of the point P from the origin is the same as the distance of the point $Q$ from the origin, and also the distance of the 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 points $P, Q, S$ from the origin is the same. This means a distance of point P from the origin is the same as the distance of point S from the origin, i.e. $(P, S) \in R$
$\therefore \mathrm{R}$ 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 $0(0,0)$ as the origin and the distance between point 0 and point $P$ be $k=O P$; then the set of all points related to $P$ is at a distance $k$ from the origin.
Hence, this set of points forms a circle with the centre as the origin, and this circle passes through point $P$.
Answer:
$R=\left\{\left(T_1, T_2\right): T_1 \text { is similar to } T_2\right\}$
All triangles are similar to themselves, so it is reflexive.
Let,
$\left(T_1, T_2\right) \in R$ i.e. $\mathrm{T}_1$ is similar to $\mathrm{T}_2$
$\mathrm{T}_1$ is similar to T 2 ; T 2 is similar to T 1 , i.e. $\left(T_2, T_1\right) \in R$
Hence, it is symmetric.
Let,
$\left(T_1, T_2\right) \in R$ and $\left(T_2, T_3\right) \in R$ i.e. $\mathrm{T}_1$ is similar to $\mathrm{T}_2$ and $\mathrm{T}_2$ is similar to $\mathrm{T}_3$.
$\Rightarrow \mathrm{T}_1$ is similar to $\mathrm{T}_3$ i.e. $\left(T_1, T_3\right) \in R$
Hence, it is transitive,
Thus, $R=\left\{\left(T_1, T_2\right): T_1\right.$ is similar to $\left.T_2\right\}$, is equivalence relation.
Now, we see the ratio of the sides of triangles T1 and T3 as shown
$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$
i.e. ratios of sides of T1 and T3 are equal. Hence, $\mathrm{T}_1$ and $\mathrm{T}_3$ are related.
Answer:
$R=\left\{\left(P_1, P_2\right): P_1\right.$ and $P_2$ have same number of sides $\}$
The same polygon has the same number of sides with itself,i.e. $\left(P_1, P_2\right) \in R$, so it is reflexive.
Let,
$\left(P_1, P_2\right) \in R$ i.e. $P_1$ have same number of sides as $\mathrm{P}_2$
$\mathrm{P}_1$ have same number of sides as $\mathrm{P}_2$ is same as $\mathrm{P}_2$ have same number of sides as $\mathrm{P}_1$ i.e. $\left(P_2, P_1\right) \in R$
Hence, it is symmetric.
Let,
$\left(P_1, P_2\right) \in R$ and $\left(P_2, P_3\right) \in R$ i.e. $\mathrm{P}_1$ have same number of sides as $\mathrm{P}_2$ and $\mathrm{P}_2$ have same number of sides as $\mathrm{P}_3$
$\Rightarrow \mathrm{P}_1$ have same number of sides as $\mathrm{P}_3$ i.e. $\left(P_1, P_3\right) \in R$
Hence, it is transitive,
Thus, $R=\left\{\left(P_1, P_2\right): P_1\right.$ 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 that have 3 sides.
Hence, the set of all elements in A related to the right-angle triangle T is a set of all triangles.
Answer:
$R=\left\{\left(L_1, L_2\right): L_1\right.$ is parallel to $\left.L_2\right\}$
All lines are parallel to themselves, so it is reflexive.
Let,
$\left(L_1, L_2\right) \in R$ i.e. $\mathrm{L}_1$ is parallel to T 2.
$\mathrm{L}_1$ is parallel to L 2 is the same as L 2 is parallel to $\mathrm{L}_1$, i.e. $\left(L_2, L_1\right) \in R$
Hence, it is symmetric.
Let,
$\left(L_1, L_2\right) \in R$ and $\left(L_2, L_3\right) \in R$ i.e. $L_1$ is parallel to $L 2$ and $L 2$ is parallel to $L_3$.
$\Rightarrow \mathrm{L}_1$ is parallel to $\mathrm{L}_3$ i.e. $\left(L_1, L_3\right) \in R$
Hence, it is transitive,
Thus, $R=\left\{\left(L_1, L_2\right): L_1\right.$ is parallel to $\left.L_2\right\}$, 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 is equal.
Lines parallel to this $(y=2 x+4$.) line are $y=2 x+c, c \in R$
Hence, the set of all parallel lines to $y=2x+4$. are $y=2 x+c$.
(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∈A, there is (a, a)∈R.
∴ R is reflexive.
Given, (1,2)∈R but (2,1)∉R
∴ R is not symmetric.
For a,b,c∈A there are (a,b)∈Rand(b,c)∈R ⇒ (a,c)∈R
∴ 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.
Answer:
R={(a,b):a=b−2,b>6}
(A) Since b<6, so (2,4)∉R
(B) Since 3≠8−2, so (3,8)∉R
(C) Since, 8>6 and 6=8−2 so (6,8)∈R
(D) Since 8≠7−2, so (8,7)∉R
The correct answer is option C.
NCERT Relations and Functions Class 11 Solutions: Exercise: 1.2 Page Number: 10-11 Total Questions: 12 |
Answer:
Given, $f: R_* \longrightarrow R_*$ is defined by $f(x)=\frac{1}{x}$.
One - One :
$f(x)=f(y)$
$\frac{1}{x}=\frac{1}{y}$
$x=y$
$\therefore \mathrm{f}$ is one-one.
Onto:
We have $y \in R_*$, then there exists $x=\frac{1}{y} \in R_* \quad($ Here $y \neq 0)$ such that
$f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$
$\therefore$ f is onto.
Hence, the function is one-one and onto.
If the domain $\mathrm{R}_*$ is replaced by N with co-domain being same as $\mathrm{R}_*$ i.e. $g: N \longrightarrow R_*$ defined by
$g(x)=\frac{1}{x}$
$g\left(x_1\right)=g\left(x_2\right)$
$\frac{1}{x_1}=\frac{1}{x_2}$
$x_1=x_2$
$\therefore \mathrm{g}$ is one-one.
For $1.5 \in R_*$,
$g(x)=\frac{1}{1.5}$ but there does not exists any x in N.
Hence, function g is one-one but not onto.
Question 2 (i): Check the injectivity and surjectivity of the following functions:
Answer:
$f: N \rightarrow N$
$f(x)=x^2$
One-one:
$x, y \in N$ then $f(x)=f(y)$
$x^2=y^2$
$x=y$
$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3 \in N$ there is no x in N such that $f(x)=x^2=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.
Question 2 (ii): Check the injectivity and surjectivity of the following functions:
Answer:
$f: Z \rightarrow Z$
$f(x)=x^2$
One-one:
For $-1,1 \in Z$ then $f(x)=x^2$
$f(-1)=(-1)^2$
$f(-1)=1$ but $-1 \neq 1$
$\, therefore, \mathrm{f}$ is not one- one i.e. not injective.
For $-3 \in Z$ there is no x in Z such that $f(x)=x^2=-3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is neither injective nor surjective
Question 2 (iii): Check the injectivity and surjectivity of the following functions:
Answer:
$f: R \rightarrow R$
$f(x)=x^2$
One-one:
For $-1,1 \in R$ then $f(x)=x^2$
$f(-1)=(-1)^2$
$f(-1)=1 \text { but }-1 \neq 1$
$\, therefore, \mathrm{f}$ is not one- one i.e. not injective.
For $-3 \in R$ there is no x in R such that $f(x)=x^2=-3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is not injective and not surjective.
Question 2 (iv): Check the injectivity and surjectivity of the following functions:
Answer:
$f: N \rightarrow N$
$f(x)=x^3$
One-one:
$x, y \in N$ then $f(x)=f(y)$
$x^3=y^3$
$x=y$
$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3 \in N$ there is no x in N such that $f(x)=x^3=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.
Question 2 (v): Check the injectivity and surjectivity of the following functions:
Answer:
$f: Z \rightarrow Z$
$f(x)=x^3$
One-one:
For $(x, y) \in Z$ then $f(x)=f(y)$
$x^3=y^3$
$x=y$
$\therefore \mathrm{f}$ is one- one i.e. injective.
For $3 \in Z$ there is no x in Z such that $f(x)=x^3=3$
$\therefore \mathrm{f}$ is not onto i.e. not surjective.
Hence, $f$ is injective but not surjective.
Answer:
f:R⟶R
f(x)=[x]
One-one:
For 1.5,1.7∈R then f(1.5)=[1.5]=1 and f(1.7)=[1.7]=1
but 1.5≠1.7
∴ f is not one-one, i.e. not injective.
For 0.6∈R there is no x in R such that f(x)=[0.6]
∴ f is not onto i.e. not surjective.
Hence, f is not injective but not surjective.
Answer:
f: R→R
f(x)=|x|
f(x)=|x|=xifx≥0and−xifx<0
One-one:
For −1,1∈R then f(−1)=|−1|=1
f(1)=|1|=1
−1≠1
∴ f is not one-one, i.e. not injective.
For −2∈R,
We know f(x)=|x| is always positive there is no x in R such that f(x)=|x|=−2
∴ f is not onto i.e. not surjective.
Hence, f(x)=|x| is neither one-one nor onto.
Question 5: Show that the Signum Function f: R→R, given by f(x)={1 if x>0 0 if x=0 −1 if x<0, is neither one-one nor onto.
Answer:
$f: R \rightarrow R$ is given by
$f(x)=\left\{\begin{array}{cl}1 & \text { if } x>0 \\ 0 & \text { if } x=0 \\ -1 & \text { if } x<0\end{array}\right.$
As we can see $f(1)=f(2)=1$, but $1 \neq 2$
So it is not one-one.
Now, $\mathrm{f}(\mathrm{x})$ takes only 3 values $(1,0,-1)$ for the element -3 in codomain $R$, there does not exists x in domain $R$ such that $f(x)=-3$.
So it is not onto.
Hence, the signum function is neither one-one nor onto.
Question 6: 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.
Answer:
A={1,2,3}
B={4,5,6,7}
f={(1,4),(2,5),(3,6)}
f: A→B
∴ f(1)=4,f(2)=5,f(3)=6
Every element of A has a distant value in f.
Hence, it is one-one.
Question 7 (i): 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
Answer:
f: R→R
f(x)=3−4x
Let there be (a,b)∈R such that f(a)=f(b)
3−4a=3−4b
−4a=−4b
a=b
∴ f is one-one.
Let there be y∈, y=3−4x
x=(3−y)4
f(x)=3−4x
Putting value of x, f(3−y4)=3−4(3−y4)
f(3−y4)=y
∴ f is onto.
f is both one-one and onto; hence, f is bijective.
Question 7 (ii): In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.
(ii) f:R→R defined by f(x)=1+x^2
Answer:
$f: R \rightarrow R$
$f(x)=1+x^2$
Let there be $(a, b) \in R$ such that $f(a)=f(b)$
$\begin{gathered}1+a^2=1+b^2 \\ a^2=b^2 \\ a= \pm b\end{gathered}$
For $f(1)=f(-1)=2$ and $1 \neq-1$
$\therefore f$ is not one-one.
Let there be $-2 \in R \quad(-2$ in codomain of R$)$
$f(x)=1+x^2=-2$
There does not exists any x in domain R such that $f(x)=-2$
$\therefore \mathrm{f}$ is not onto.
Hence, f is neither one-one nor onto.
Question 8: Let A and B be sets. Show that f: A×B→B×A such that f(a,b)=(b, a) is a bijective function.
Answer:
f: A×B→B×A
f(a,b)=(b,a)
Let (a1,b1),(a2,b2)∈A×B
such that f(a1,b1)=f(a2,b2)
(b1,a1)=(b2,a2)
⇒ b1=b2 and a1=a2
⇒ (a1,b1)=(a2,b2)
∴ f is one- one
Let (b, a)∈B×A
then there exists (a,b)∈A×B such that f(a,b)=(b,a)
∴ f is onto.
Hence, it is bijective.
Answer:
$f: N \rightarrow N, n \in N$
$f(n)=\left\{\begin{array}{cc}\frac{n+1}{2} & \text { if } n \text { is odd } \\ \frac{n}{2} & \text { if } n \text { is even }\end{array}\right.$
Here we can observe,
$f(2)=\frac{2}{2}=1 \quad$ and $\quad f(1)=\frac{1+1}{2}=1$
As we can see $f(1)=f(2)=1$ but $1 \neq 2$
$\therefore \mathrm{f}$ is not one-one.
Let, $n \in N \quad$ (N=co-domain)
case1 $n$ be even
For $r \in N, \quad n=2 r$
then there is $4 r \in N$ such that $f(4 r)=\frac{4 r}{2}=2 r$
case2 n be odd
For $r \in N, n=2 r+1$
then there is $4 r+1 \in N$ such that $f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1$
$\therefore \mathrm{f}$ is onto.
$f$ is not one-one but onto
Hence, the function $f$ is not bijective
Question 10: Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $f: A \rightarrow B$ defined by $f(x)=x-2 / x-3$. Is $f$ one-one and onto? Justify your answer.
Answer:
$\begin{aligned} & A=R-\{3\} \\ & B=R-\{1\} \\ & f: A \rightarrow B \\ & f(x)=\left(\frac{x-2}{x-3}\right)\end{aligned}$
Let $a, b \in A$ such that $f(a)=f(b)$
$\begin{gathered}\left(\frac{a-2}{a-3}\right)=\left(\frac{b-2}{b-3}\right) \\ (a-2)(b-3)=(b-2)(a-3) \\ a b-3 a-2 b+6=a b-2 a-3 b+6 \\ -3 a-2 b=-2 a-3 b \\ 3 a+2 b=2 a+3 b \\ 3 a-2 a=3 b-2 b \\ a=b\end{gathered}$
$\therefore \mathrm{f}$ is one-one.
Let, $b \in B=R-\{1\} \quad$ then $b \neq 1$
$\begin{aligned} & a \in A \text { such that } f(a)=b \\ & \begin{array}{c}\left(\frac{a-2}{a-3}\right)=b \\ (a-2)=(a-3) b \\ a-2=a b-3 b \\ a-a b=2-3 b \\ a(1-b)=2-3 b \\ a=\frac{2-3 b}{1-b} \quad \in A\end{array}\end{aligned}$
For any $b \in B$ there exists $a=\frac{2-3 b}{1-b} \quad \in A$ such that
$\begin{aligned} f\left(\frac{2-3 b}{1-b}\right) & =\frac{\frac{2-3 b}{1-b}-2}{\frac{2-3 b}{1-b}-3} \\ & =\frac{2-3 b-2+2 b}{2-3 b-3+3 b} \\ & =\frac{-3 b+2 b}{2-3} \\ & =b\end{aligned}$
$\therefore \mathrm{f}$ is onto
Hence, the function is one-one and onto.
Question 11: Let f: R→R be defined as f(x)=x^4. Choose the correct answer.
(D) f is neither one-one nor onto.
Answer:
$\begin{aligned} & f: R \rightarrow R \\ & f(x)=x^4\end{aligned}$
One-one:
For $a, b \in R$ then $f(a)=f(b)$
$\begin{aligned} & a^4=b^4 \\ & a= \pm b\end{aligned}$
$\therefore f(a)=f(b)$ does not imply that $a=b$
example : $f(2)=f(-2)=16$ and $2 \neq-2$
$\therefore \mathrm{f}$ is not one- one
For $2 \in R$ there is no x in R such that $f(x)=x^4=2$
$\therefore \mathrm{f}$ is not onto.
Hence, $f$ is neither one-one nor onto.
Hence, option D is correct.
Question 12: Let f: R→R be defined as f(x)=3x. Choose the correct answer.
(D) f is neither one-one nor onto.
Answer:
f: R→R
f(x)=3x
One - One :
Let (x,y)∈R
f(x)=f(y)
3x=3y
x=y
∴ f is one-one.
Onto:
If we have y∈R, then there exists x=y/3∈R such that
f(y/3)=3×y/3=y
∴f is onto.
Hence, the function is one-one and onto.
The correct answer is A.
NCERT Relation and Functions Class 12 Solutions: Exercise: Miscellaneous Exercise Page Number: 15-16 Total Questions: 7 |
Question 1: Show that the function f:R→{x∈R:−1<x<1} defined by f(x)=x/(1+|x|) x∈R is one one and onto function.
Answer:
The function $f: R \rightarrow\{x \in R:-1<x<1\}$ defined by
$f(x)=\frac{x}{1+|x|}, x \in R$
One-one:
Let $f(x)=f(y) \quad, x, y \in R$
$\frac{x}{1+|x|}=\frac{y}{1+|y|}$
It is observed that if x is positive and y is negative.
$\frac{x}{1+x}=\frac{y}{1+y}$
Since x is positive and y is negative.
$x>y \Rightarrow x-y>0 \quad \text { but } 2 \mathrm{xy} \text { is negative. }$
$x-y \neq 2 x y$
Thus, the case of x being positive and y being negative is removed.
The same happens in the case of y is positive and x is negative, so this case is also removed.
When x and y both are positive:
$\begin{gathered}f(x)=f(y) \\ \frac{x}{1+x}=\frac{y}{1+y} \\ x(1+y)=y(1+x) \\ x+x y=y+x y \\ x=y\end{gathered}$
When $x$ and $y$ both are negative:
$\begin{aligned} & f(x)=f(y) \\ & \frac{x}{1-x}=\frac{y}{1-y} \\ & x(1-y)=y(1-x) \\ & x-x y=y-x y \\ & \quad x=y\end{aligned}$
$\therefore \mathrm{f}$ is one-one.
Onto:
Let $y \in R$ such that $-1<y<1$
If y is negative, then $x=\frac{y}{y+1} \in R$
$f(x)=f\left(\frac{y}{y+1}\right)=\frac{\frac{y}{1+y}}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\frac{-y}{1+y}}=\frac{y}{1+y-y}=y$
If y is positive, then $x=\frac{y}{1-y} \in R$
$f(x)=f\left(\frac{y}{1-y}\right)=\frac{\frac{y}{1-y}}{1+\left|\frac{y}{1-y}\right|}=\frac{\frac{y}{1-y}}{1+\frac{-y}{1-y}}=\frac{y}{1-y+y}=y$
Thus, $f$ is onto.
Hence, $f$ is one-one and onto.
Question 2: Show that the function f: R→R given by f(x)=x^3 is injective.
Answer:
f: R→R
f(x)=x^3
One-one:
Let f(x)=f(y)x,y∈R
x^3=y^3
We need to prove x=y. So,
Let x≠y, then their cubes will not be equal, i.e. x^3≠y^3.
It will contradict the given condition of cubes being equal.
Hence, x=y, and it is one-one, which means it is injective
Answer:
Given a nonempty set X, consider $\mathrm{P}(\mathrm{X})$, which is the set of all subsets of X.
Since every set is a subset of itself, ARA for all $A \in P(x)$
$\therefore \mathrm{R}$ is reflexive.
Let $A R B \Rightarrow A \subset B$
This is not the same as $B \subset A$
If $A=\{0,1\}$ and $B=\{0,1,2\}$
then we cannot say that $B$ is related to $A$.
$\therefore \mathrm{R}$ is not symmetric.
If $A R B$ and $B R C$, then $A \subset B$ and $B \subset C$
this implies $A \subset C=A R C$
$\therefore \mathrm{R}$ is transitive.
Thus, R is not an equivalence relation because it is not symmetric.
Question 4: Find the number of all onto functions from the set {1,2,3,...,n} to itself.
Answer:
The number of all onto functions from the set {1,2,3,...,n} to itself is the permutations of n symbols 1,2,3,4,5...............n.
Hence, permutations on n symbols 1,2,3,4,5...............n = n
Thus, the total number of all onto maps from the set {1,2,3,...,n} to itself is the same as permutations on n symbols 1,2,3,4,5...............n, which is n.
Answer:
Given :
$A=\{-1,0,1,2\}, B=\{-4,-2,0,2\}$
$f, g: A \rightarrow B$ are defined by $f(x)=x^2-x, x \in A$ and $g(x)=2\left|x-\frac{1}{2}\right|-1, x \in A$.
It can be observed that.
$\begin{aligned} & f(-1)=(-1)^2-(-1)=1+1=2 \\ & g(-1)=2\left|-1-\frac{1}{2}\right|-1=2\left|\frac{-3}{2}\right|-1=3-1=2 \\ & \quad f(-1)=g(-1)\end{aligned}$
$\begin{aligned} & f(0)=(0)^2-(0)=0+0=0 \\ & g(0)=2\left|0-\frac{1}{2}\right|-1=2\left|\frac{-1}{2}\right|-1=1-1=0 \\ & \quad f(0)=g(0)\end{aligned}$
$\begin{aligned} & f(1)=(1)^2-(1)=1-1=0 \\ & g(1)=2\left|1-\frac{1}{2}\right|-1=2\left|\frac{1}{2}\right|-1=1-1=0 \\ & \quad f(1)=g(1)\end{aligned}$
$\begin{aligned} & f(2)=(2)^2-(2)=4-2=2 \\ & g(2)=2\left|2-\frac{1}{2}\right|-1=2\left|\frac{3}{2}\right|-1=3-1=2 \\ & \quad f(2)=g(2)\end{aligned}$
$\therefore f(a)=g(a) \forall a \in A$
Hence, $f$ and $g$ are equal functions.
Answer:
A={1,2,3}
The smallest relations containing (1,2) and (1,3), which are reflexive and symmetric but not transitive, are given by
R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,1),(3,1)}
(1,1),(2,2),(3,3)∈R , so relation R is reflexive.
(1,2),(2,1)∈R and (1,3),(3,1)∈R , so relation R is symmetric.
(2,1),(1,3)∈R but (2,3)∉R , so realation R is not transitive.
Now, if we add any two pairs (2,3) and (3,2) to relation R, then relation R will become transitive.
Hence, the total number of the desired relation is one.
Thus, option A is correct.
Question 7: Let A={1,2,3}. The number of equivalence relations containing (1,2) is
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
A={1,2,3}
The number of equivalence relations containing (1,2) is given by
R={(1,1),(2,2),(3,3),(1,2),(2,1)}
We are left with four pairs (2,3), (3,2), (1,3),(3,1).
(1,1),(2,2),(3,3)∈R , so relation R is reflexive.
(1,2),(2,1)∈R and (2,3),(3,2)∉R , so relation R is not symmetric.
(1,3),(3,1)∉R , so realation R is not transitive.
Hence, the equivalence relation is bigger than R is the universal relation.
Thus, the total number of equivalence relations containing (1,2) is two.
Thus, option B is correct.
Question: Let an operation $\ast$ on the set of natural numbers N be defined by $a\ast b= a^{b}\cdot$
Find (i) whether $\ast$ is a binary or not, and (ii) if it is a binary, then is it commutative or not.
Solution:
(i) As $a^b \in N$ for all $a, b \in N$
ie $a * b \in N \forall a, b \in N$
Hence, $*$ is binary.
(ii) As $1^2 \neq 2^1$
so $1 * 2 \neq 2 * 1$
Hence, $\ast$ is not commutative.
Also read,
Relations and Functions can feel like a puzzle of definitions, diagrams, and properties—but with the right approach, every question starts to make perfect sense.
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Also read,
Students can check the following links for more in-depth learning.
Students can check the following links for more in-depth learning.
NCERT Class 12 Maths Chapter 1 "Relations and Functions" covers essential standards of set principles and mappings. Key subjects encompass:
Types of Relations – Reflexive, symmetric, transitive, and equivalent members of the family.
Types of Functions – One-one (injective), onto (surjective), and bijective functions.
Composition of Functions – Understanding how functions integrate.
The inverse of a Function – Conditions for the lifestyles of an inverse.
Binary Operations – Definition consisting of commutativity and associativity.
In NCERT Class 12 Maths, one-one (injective) and onto (surjective) functions are two important types of functions:
- One-One Function (Injective): A function is one-one if different inputs give different outputs. This means no two elements in the domain map to the same element in the codomain. Example: f(x)=2x is one-one.
- Onto Function (Surjective): A function is onto if every element in the codomain has at least one pre-image in the domain. This means the function covers the entire codomain. Example: f(x)=x3 is onto for real numbers.
A function can be both one-one and onto (bijective).
In NCERT Class 12 Maths Chapter 1, relations and functions are different concepts:
- Relation: A relation connects elements of one set to another. It is simply a pairing of elements but may not follow specific rules. Example: If Set A={1,2,3} and Set B={4,5}, a relation can be {(1,4), (2,5)}.
- Function: A function is a special type of relation where each input has exactly one output. Example: f(x)=x+2 maps every x to a unique value.
Thus, every function is a relation, but not every relation is a function.
In NCERT Class 12 Maths Chapter 1, the composition of functions means applying one function to the result of another. If we have two functions f(x) and g(x), their composition is written as (f o g)(x), which means f(g(x)).
Steps to find composition:
1. First, find g(x) (solve for x in g ).
2. Substitute g(x) into f(x).
3. Simplify the expression.
For example, if f(x)=x2 and g(x)=x+1, then:
(f o g)(x)=f(g(x))=f(x+1)=(x+1)2
In NCERT Class 12 Maths Chapter 1, the domain and range of a function describe its input and output values:
- Domain: The set of all possible input values (x) for which the function is defined. Example: For f(x)=1/x, the domain is all real numbers except x=0.
- Range: The set of all possible output values f(x). Example: For f(x)=x2, the range is [0, infinity) because squares are always non-negative.
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.
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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!
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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.
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