NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

In our daily lives, we see various connections like student and roll number, person and Aadhaar number, etc., which represent relationships. So, what are relations and functions? A relation shows the connection between two sets of values, such as input and output. When each input has only one output, it becomes a function. For example, assigning marks to students is a function if every student gets only one mark. Functions are special types of relations. They are used in real-life applications like mapping, programming, and data handling.

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1. Relations and Functions Class 12 Questions And Answers PDF Free Download
2. Relations and Functions Class 12 Solutions - Important Formulae
3. Relations and Functions Class 12 NCERT Solutions (Exercise)
4. NCERT Exemplar Class 12th Maths Solutions
5. Importance of solving NCERT questions for class 12 Chapter 1 Relations and Functions
6. NCERT Solutions class-wise
7. Students can check the following links for more in-depth learning.
8. NCERT Books and NCERT Syllabus
9. Students can check the following links for more in-depth learning.

Relations and functions class 12 solutions are essential for students because they comprise quality practice questions. Class 12 Maths Chapter 1 Relations And Functions Notes are very useful for quick revision purposes. After checking the Class 12 Maths Chapter 1 ncert solutions from textbooks, students can also check NCERT Exemplar Solutions For Class 12 Maths Chapter 1 Relations And Functions for a deeper understanding of this chapter.

Relations and Functions Class 12 Questions And Answers PDF Free Download

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>> 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 $A$.
$\mathrm{A}\times \varphi =\varphi$ (where $\varphi$ is the empty set)
A function $f:A\to B$ is denoted as:

$f(x)=y$

Algebra of functions:

- $(f+g)(x)=f(x)+g(x)$
- $(f-g)(x)=f(x)-g(x)$
- $\left(f{}^{\ast }g\right)(x)=f(x{)}^{\ast }g(x)$
- $(kf)(x)=k^{\ast }f(x)$, where $k$ is a real number
- $(f/g)(x)=f(x)/g(x)$, where $g(x)\ne 0$

Relations and Functions Class 12 NCERT Solutions (Exercise)

Question 1 (i) Determine whether each of the following relations are reflexive, symmetric, and transitive:

(i) Relation $R$ in the set $A=\{1,2,3\dots ,13,14\}$ defined as $R=\{(x,y):3x-y=0\}$

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 so 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.

Question1(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

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

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 means 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, 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

(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

Question:2 Show that the relation R in the set R of real numbers is defined as
R={(a,b): a≤b^2} is neither reflexive nor symmetric nor transitive.

Answer:

$R=\{(a,b):a\le b^2\}$
Taking

$(\frac{1}{2},\frac{1}{2})\notin R$

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.

Question:3 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.

Answer:

R 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) Then because 2<4

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question:5 Check whether the relation R in R is defined by R={(a,b): $a\le b^3\}$ is reflexive, symmetric or transitive.

Answer:

$R=\{(a,b):a\le b^3\}$

$(\frac{1}{2},\frac{1}{2})\notin R$ 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.

Question:6 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.

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.

Question:7 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?

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 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 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.

Question:8 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}.

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 number 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 multiple a of 4.

Hence, it is reflexive.

Let, (a,b)∈R i.e. |a−b| be 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 multiple of 4 and (b,c)∈R i.e. |b−c| be 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

|a−c| is multiple a 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 multiple a of 4.

|5−1|=4 is multiple a of 4.

|9−1|=8 is multiple a of 4.

Question:9(ii) Show that each of the relation 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.

Question:11 Show that the relation R in the set A of points in a plane given by R={(P, Q): distant ce of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, shows that the set of All points related to a point P≠(0,0) is the circle passing through P with the origin as the center.

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 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 $(P,Q)\in R$ 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 point $P,Q,S$ 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 $P\ne (0,0)$ 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 $k=OP$ then the set of all points related to $P$ is at a distance $k$ from the origin.

Hence, these set of points form a circle with the center as the origin and this circle passes through point $P$.

Question:12 Show that the relation R defined in the set A of all triangles as R={(T1, T2): T1 is similar to T2}, is an equivalence relation. Consider three right-angle triangles T ₁ with sides 3, 4, 5, T ₂ with sides 5, 12, 13, and T ₃ with sides 6, 8, 10. Which triangles among T ₁ , T _2, and T ₃ are related?

Answer:

$R=\{(T_1,T_2):T_1\text{ is similar to }{T}_2\}$

All triangles are similar to itself, so it is reflexive.

Let,

$(T_1,T_2)\in R$ 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. $(T_2,T_1)\in R$

Hence, it is symmetric.

Let,

$(T_1,T_2)\in R$ and $(T_2,T_3)\in R$ i.e. ${\mathrm{T}}_1$ 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. $(T_1,T_3)\in R$

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 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.

Question:13 Show that the relation R defined in the set A of all polygons as R={(P1, P2): P1 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?

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 ${\mathrm{P}}_2$

${\mathrm{P}}_1$ 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. $(P_2,P_1)\in R$

Hence, it is symmetric.

Let,

$(P_1,P_2)\in R$ and $(P_2,P_3)\in R$ i.e. ${\mathrm{P}}_1$ 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. $(P_1,P_3)\in R$

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 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.

Question 14: 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 $y=2x+4$.

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. ${\mathrm{L}}_1$ is parallel to T 2.

${\mathrm{L}}_1$ is parallel to L 2 is same as L 2 is parallel to ${\mathrm{L}}_1$ 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. $L_1$ is parallel to $L2$ and $L2$ is parallel to $L_3$.

$\Rightarrow {\mathrm{L}}_1$ is parallel to ${\mathrm{L}}_3$ 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 is 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} 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.

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.

(A) (2,4)∈R

(B) (3,8)∈R

(C) (6,8)∈R

(D) (8,7)∈R

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.

Question:1 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 the same as R _∗ ?

Answer:

Given, $f:R_{\ast }⟶R_{\ast }$ 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 $y\in R_{\ast }$, then there exists $x=\frac{1}{y}\in R_{\ast }($ Here $y\ne 0)$ such that

$f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$

$\therefore$ fis onto.
Hence, the function is one-one and onto.

If the domain ${\mathrm{R}}_{\ast }$ is replaced by N with co-domain being same as ${\mathrm{R}}_{\ast }$ i.e. $g:N⟶R_{\ast }$ 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 $1.5\in R_{\ast }$,
$g(x)=\frac{1}{1.5}$ 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:

(i) f:N→N given by f(x)=x^2

Answer:

$f:N\to N$

$f(x)=x^2$
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:

(ii) f:Z→Z given by f(x)=x^2

Answer:

$f:Z\to Z$

$f(x)=x^2$

One-one:

For $-1,1\in Z$ then $f(x)=x^2$

$f(-1)=(-1{)}^2$

$f(-1)=1$ but $-1\ne 1$

$therefore,\mathrm{f}$ 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:

(iii) f:R→R given by f(x)=x^2

Answer:

$f:R\to R$

$f(x)=x^2$
One- one:
For $-1,1\in R$ then $f(x)=x^2$

$f(-1)=(-1{)}^2$

$f(-1)=1\text{ but }-1\ne 1$

$therefore,\mathrm{f}$ 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:

(iv) f:N→N given by f(x)=x^3

Answer:

$f:N\to N$

$f(x)=x^3$

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:

(v) f:Z→Z given by f(x)=x^3

Answer:

$f:Z\to Z$

$f(x)=x^3$

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.

Question:3 Prove that the Greatest Integer Function f: R⟶ , given b f(x)=[x], is neither-one-one nor onto, where [x] denotes the greatest integer less than or equal to.

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.

Question:4 Show that the Modulus Function: R → R, given by f(x)=|x], is neither one-one nor onto, where |x| is if x is positive or 0 and |x| is −x if x is negative.

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:

$f:R\to R$ is given by

$f(x)=\{\begin{array}{cl}1& \text{ if }x>0\\ 0& \text{ if }x=0\\ -1& \text{ if }x<0\end{array}$

As we can see $f(1)=f(2)=1$, but $1\ne 2$
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:

$f:R\to R$

$f(x)=1+x^2$

Let there be $(a,b)\in R$ such that $f(a)=f(b)$

$\begin{array}{c}1+a^2=1+b^2\\ a^2=b^2\\ a=\pm b\end{array}$

For $f(1)=f(-1)=2$ and $1\ne -1$
$\therefore f$ is not one-one.
Let there be $-2\in R(-2$ 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.

Question:9 Let $f:N\to N$ be defined by $f(n)=\{\begin{array}{cl}\frac{n+1}{2}& \text{ if }n\text{ is odd }\\ \frac{n}{2}& \text{ if }n\text{ is even }\end{array}\text{ for all }n\in N.$ State whether the function f is bijective. Justify your answer.

Answer:

$f:N\to N,n\in N$

$f(n)=\{\begin{array}{cc}\frac{n+1}{2}& \text{ if }n\text{ is odd }\\ \frac{n}{2}& \text{ if }n\text{ is even }\end{array}$

Here we can observe,

$f(2)=\frac{2}{2}=1$ and $f(1)=\frac{1+1}{2}=1$
As we can see $f(1)=f(2)=1$ but $1\ne 2$
$\therefore \mathrm{f}$ is not one-one.
Let, $n\in N$ (N=co-domain)
case1 $n$ be even
For $r\in N,n=2r$
then there is $4r\in N$ such that $f(4r)=\frac{4r}{2}=2r$
case2 n be odd
For $r\in N,n=2r+1$
then there is $4r+1\in N$ such that $f(4r+1)=\frac{4r+1+1}{2}=2r+1$
$\therefore \mathrm{f}$ 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 $f:A\to B$ defined by $f(x)=x-2/x-3$. Is $f$ one-one and onto? Justify your answer.

Answer:

$\begin{array}{rl}& A=R-\{3\}\\ & B=R-\{1\}\\ & f:A\to B\\ & f(x)=\left(\frac{x-2}{x-3}\right)\end{array}$

Let $a,b\in A$ such that $f(a)=f(b)$

$\begin{array}{c}\left(\frac{a-2}{a-3}\right)=\left(\frac{b-2}{b-3}\right)\\ (a-2)(b-3)=(b-2)(a-3)\\ ab-3a-2b+6=ab-2a-3b+6\\ -3a-2b=-2a-3b\\ 3a+2b=2a+3b\\ 3a-2a=3b-2b\\ a=b\end{array}$

$\therefore \mathrm{f}$ is one-one.

Let, $b\in B=R-\{1\}$ then $b\ne 1$

$\begin{array}{rl}& a\in A\text{ such that }f(a)=b\\ & \begin{array}{c}\left(\frac{a-2}{a-3}\right)=b\\ (a-2)=(a-3)b\\ a-2=ab-3b\\ a-ab=2-3b\\ a(1-b)=2-3b\\ a=\frac{2-3b}{1-b}\in A\end{array}\end{array}$

For any $b\in B$ there exists $a=\frac{2-3b}{1-b}\in A$ such that

$\begin{array}{rl}f\left(\frac{2-3b}{1-b}\right)& =\frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}\\ & =\frac{2-3b-2+2b}{2-3b-3+3b}\\ & =\frac{-3b+2b}{2-3}\\ & =b\end{array}$

$\therefore \mathrm{f}$ 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.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

$\begin{array}{rl}& f:R\to R\\ & f(x)=x^4\end{array}$

One- one:
For $a,b\in R$ then $f(a)=f(b)$

$\begin{array}{rl}& a^4=b^4\\ & a=\pm b\end{array}$

$\therefore f(a)=f(b)$ does not imply that $a=b$
example : $f(2)=f(-2)=16$ and $2\ne -2$
$\therefore \mathrm{f}$ 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.