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Exercise 1.1 NCERT Solutions Relations & Functions Class 12 Math Chapter 1 free PDF Download

Relation and Function are two ways of establishing links between two sets in mathematics. Relation and Function in maths are analogous to the relation that we see in our daily lives i.e., two persons are related by the relation of father-son, mother-daughter, brother-sister, and many more. On a similar pattern, two numbers can be related to each other as one number is the square of another number and many more. A function is a special kind of relation that is defined as a unique relation between two mathematical entities.

Class 12 NCERT Solutions- Mathematics Chapter 1 Relations And Functions – Exercise 1.1

Table of Contents

Question 1. Determine whether each of the following relations is reflexive, symmetric or transitive:

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

Solution:

A={1 ,2 ,3,…,13, 14}

R={(x,y): 3x-y=0}

Therefore R={(1,3),(2,6),(3,9),(4,12)}

R is not reflexive since (1,1),(2,2),(3,3),…,(14,14)∉R

Also, R is not reflexive since (1,3)∈R, but (3,1)∉R.[since 3(3)-1≠0]

Also, R is not transitive as (1,3), (3,9)∈R, but (1,9)∉R.[since 3(1)-9≠0]

Hence, R is not reflexive, nor symmetric nor transitive.

(ii) Relation R in the set N of natural numbers defined as R={ ( x, y) : y=x+5 and x<4}

Solution:

R={(x, y): y=x+5 and x<4}={(1,6), (2, 7), (3,8)}

It is seen that (1, 1)∉R. Therefore, R is not reflexive.

(1, 6)∈R. But, (6, 1)∉R so, R is not symmetric.

Now, since their is no pair in R such that (x, y) and (y, z)∈R, so (x, z) can not belong to R. Therefore, R is not transitive.

So, we can conclude that R is neither reflexive, nor symmetric, nor transitive.

(iii) Relation R in the set A= {1, 2, 3, 4, 5, 6} as R={(x, y): y is divisible by x}

Solution:

A={1, 2, 3, 4, 5, 6}

R={(x, y): y is divisible by x}

We know that any number is always divisible by itself.⇒ (x, x)∈ R. Therefore, R is reflexive.

We see, (2, 4)∈R [4 is divisible by 2]. But, (4, 2)∉R [2 is not divisible by 4]. Therefore, R is not symmetric.

Let’s assume (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. Therefore, z is divisible by x.⇒ (x, z) ∈ R. Therefore, R is transitive.

(iv) Relation R in the set Z of all integers defined as R={(x, y): x-y is an integer}

Solution:

R={(x, y): x-y is an integer}

For every x ∈ Z, (x, x) ∈ R [x-x=0 which is an integer]. Therefore, R is reflexive.

For every x, y ∈ Z if (x, y) ∈ R, then x-y is an integer. ⇒-(x-y) is also an integer. ⇒ (y-x) is also an integer. Therefore, R is symmetric.

Let’s assume, (x, y) and (y, z) ∈ R, where x, y and z ∈ Z. ⇒ (x-y) and (y-z) are integers. ⇒ x-z=(x-y)+(y-z) is an integer. ⇒ (x, z) ∈ R. Therefore, R is transitive.

(v) Relation R in a 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. 

Solution: 

We can see (x,x) ∈ R .Therefore, R is reflexive.

If (x,y) ∈ R, then x and y work at the same place. So, (y,x) ∈ R. Therefore, R is symmetric.

Let, (x,y), (y,z) ∈ R. 

⇒ x and y work mat the same place and y and z work at the same place.

⇒ x and z work at the same place.

Therefore, R is transitive.

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

Solution: 

We can see (x,x) ∈ R. Therefore, R is reflexive.

If (x,y) ∈ R, then x and y live in same locality. So, (y,x) ∈ R. Therefore, R is symmetric.

Let (x,y) ∈ R and (y,z) ∈ R. So, x, y and z live in the same locality. So, (x,z) ∈ R. Therefore, R is transitive.

(c) R={(x,y): x is exactly 7 cm taller than y}. 

Solution: 

(x,x)∉R since, human being can not be taller than himself. So, R is not reflexive.

Let (x,y) ∈ R ,then x is exactly 7 cm taller than y. Then, y is not taller than x. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is exactly 7 cm taller than y and y is exactly 7 cm taller than z which means x is 14 cm taller than z. So, (x,z)∉R . Therefore, R is not transitive.

(d) R={(x,y): x is wife of y}. 

Solution: 

(x,x) ∉ R. Since, x can not be the wife of herself. Therefore, R is not reflexive.

Let (x,y) ∈ R, then x is the wife of y. So, y is not the wife of x ,i.e., (y,x) ∉ R. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is the wife of y and y is the wife of z which is not possible. So, R can not be transitive.

(e) R={(x,y): x is father of y}. 

Solution:

(x,x) ∉ R. Since, x can not be the father of himself. Therefore, R is not reflexive.

Let (x,y) ∈ R, then x is the father of y. So, y can not be the father of x. So, (y,x) ∉ R. Therefore, R is not symmetric.

Let (x,y), (y,z) ∈ R, then x is the father of y and y is the father of z which means x is the grandfather of z. So, So, (x, z) ∉ R. Therefore, R is not transitive.

Question 2. Show that the relation R in the set R of real numbers, defined as R= {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.

Solution:

It can be observed that (½, ½) ∉ R, since ½>(½)2 =¼. Therefore, R is not reflexive.

(1,4) ∈ R as 1<42 .But, (4,1) ∉  R. Therefore, R is not symmetric.

(3,2), (2,1.5) ∈ R. But, 3> (1.5)2 =2.25 . So, (3,1.5) ∉  R. Therefore, R is not transitive.

Hence, R 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}. 

Solution:

Let the set {1, 2, 3, 4, 5, 6} be named A.

R={(1,2), (2, 3), (3, 4), (4, 5
), (5, 6)}

We can see (x, x) ∉  R. Since, x ≠ x+1. Therefore, R is not reflexive.

It is observed that (1,2) ∈ R but, (2,1) ∉  R. Therefore, R is not symmetric.

We can see, (1,2), (2, 3) ∈ R, but (1,3) ∉  R. Therefore, R is not transitive.

Hence, R 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.

Solution:

Clearly, (a,a) ∈ R as a=a. Therefore, R is reflexive.

(2,4) ∈ R (as 2<4) but (4,2) ∉  R as 4 is greater than 2. Therefore, R is not symmetric.

Let (a,b), (b,c) ∈ R. Then, a≤ b and b≤ c.

⇒a ≤ c. 

(a, c) ∈ R. Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

Question 5. Check whether the relation R in R  defined as R ={ (a, b): a ≤ b3 } is reflexive, symmetric or transitive. 

Solution:

It is observed that (½, ½) ∉  R as ½ > (½)3 =(1/8). Therefore, R is not reflexive.

(1,2) ∈ R(as 1<8) but, (2,1) ∉  R. Therefore, R is not symmetric.

We have, (3, 3/2), (3/2, 6/5) ∈ R but, (3, 6/5) ∉  R. Therefore, R is not transitive.

Hence, R 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. 

Solution:

Let the set {1, 2, 3} be named A.

It is seen that, (1, 1), (2,2), (3,3)∉  R. Therefore, R is not reflexive.

As (1, 2) ∈ R and (2, 1) ∈ R. Therefore, R is symmetric.

However, (1, 1)∉  R. Therefore, R 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 the same number of pages} is a equivalence relation.

Solution:

Set A is the set of all books in the library of a college.

R={(x,y):x and y have the same number of pages} 

R is reflexive since (x,x) ∈ R as x and x have the same number of pages.

Let (x,y) ∈ R 

⇒x and  y have the same number of pages

⇒y and x have the same number of pages.

⇒(y,x)∈ R

Therefore , R is symmetric.

Let (x,y) ∈ R and (y,z)∈ R.

⇒x and y have the same number of pages and y and z have the same number of pages.

⇒x and z have the same number of pages.

⇒(x,z) ∈ R

Therefore, R is transitive.

Hence, R is an equivalence relation.

Solution:

A={ 1,2,3,4,5}

R={(a,b):|a-b| is even }

It is clear that for any element a∈ A, we have |a-a|=0 (which is even).

Therefore, R is reflexive.

Let (a,b) ∈ R.

⇒|a-b| is even.

⇒|-(a-b)|=|b-a| is also even.

⇒(b,a)∈ R

Therefore, R is symmetric.

Now , let (a,b)∈ R and (b,c)∈ R.

⇒|a-b| is even and |b-c| is even.

⇒(a-b) is even and (b-c) is even.

⇒(a-c)=(a-b)+(b-c) is even.   [Sum of two even integers is even]

⇒|a-c| is even.

Therefore, R is transitive.

Hence, R is an equivalence relation.

All elements of the set {1,2,3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2,4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1,3,5} can be related to any element of {2,4} as all elements of {1,3,5} are odd and all elements of {2,4} are even . Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

Question 9: Show that each of the relation R in the set A={x∈ Z:0<=x<=12} , given by

(i) R={(a,b):|a-b| is a multiple of 4}

(ii) R={(a,b):a=b}

is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution:

A={x∈ Z:0<=x<=12}={0,1,2,3,4,5,6,7,8,9,10,11,12}

 (i) R={(a,b):|a-b| is a multiple of 4}

For any element a ∈A , we have (a,a)∈R as |a-a|=0 is a multiple of 4.

Therefore, R is reflexive.

Now , let (a,b)∈R ⇒|a-b| is a multiple of 4.

⇒|-(a-b)|=|b-a| is a multiple of 4.

⇒(b,a)∈R

Therefore, R is symmetric.

Let (a,b) ,(b,c) ∈ R.

⇒|a-b| is a multiple of 4 and |b-c| is 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 a multiple of 4.

⇒|a-c| is a multiple of 4.

⇒(a,c)∈R

Therefore, R is transitive.

Hence, R is an equivalence relation .

The set of elements related to 1 is {1,5,9} since |1-1|=0 is a multiple of 4,

|5-1|=4 is a multiple of 4,and

|9-1|=8 is a multiple of 4.

(ii) R={(a,b):a=b}

For any element a∈A, we have (a,a) ∈ R , since a=a.

Therefore, R is reflexive.

Now , let (a,b)∈R.

⇒a=b

⇒b=a

⇒(b,a)∈R

Therefore, R is symmetric.

Now, let (a,b)∈R and (b,c)∈R.

⇒a=b and b=c

⇒a=c

⇒(a,c)∈R

Therefore, R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1].

Question 10: Give an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii)Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Solution:

(i) Let A ={5,6,7}.

Define a relation R on A as R ={(5,6),(6,5)}.

Relation R is not reflexive as (5,5) , (6,6),(7,7)∉R.

Now, as (5,6)∈R and also (6,5)∈R, R is symmetric.

⇒(5,6),(6,5)∈R , but (5,5)∉R

Therefore , R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

(ii) Consider a relation R in R defined as:

R= {(a,b): a

For any a∈R, we have (a,a)∉R since a cannot be strictly less than itself . In fact a=a.

Therefore, R is not reflexive. Now,

(1,2)∈R (as 1<2)

But, 2 is not less than 1.

Therefore, (2,1)∉R

Therefore, R is not symmetric.

Now, let (a,b),(b,c)∈R.

⇒a

⇒a

⇒ (a,c)∈R

Therefore, R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

(iii) Let A={4,6,8}.

Define a relation R on A as:

A={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)}

Relation R is reflexive since for every a∈A , (a,a)∈R i.e.,(4,4),(6,6),(8,8)∈R.

Relation R is symmetric since (a,b)∈R⇒(b,a)∈R for all a,b∈R.

Relation R is not transitive since (4,6),(6,8)∈R , but (4,8)∉R.

Hence, relation R is reflexive and symmetric but not transitive.

(iv) Define a relation R in R as:

R={ (a,b): a3 ≥ b3 }

Clearly (a,a)∈R as a^3=a^3

Therefore, R is reflexive.

Now, (2,1)∈R(as 23>=13)

But, (1,2)∉ R (as 13< 23)

Therefore, R is not symmetric

Let (a,b), (b,c) ∈ R.

⇒a3 >= b3 and b3 >=c3

⇒a3>=c3

⇒(a, c)∈R

Therefore, R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

(v) Let A={-5,-6},

Define a relation R on A as:

R={(-5,-6), (-6,-5), (-5,-5)}

Relation R  is not reflexive as (-6,-6)∉ R.

Relation R is symmetric as (-5,-6)∈ R and (-6,-5)∈ R.

It is seen that (-5,-6),(-6,-5)∈R . Also , (-5,-5)∈R.

Therefore, the relation R is transitive .

Hence, relation R is symmetric and transitive but not reflexive.

Solution:

We can see (P, P) ∈ R since, the distance of point P from the origin is always the same as the distance of the same point P from the origin. Therefore, R is reflexive.

Let (P,Q)∈ R.

⇒The distance of point P from the origin is the same as the distance of point Q from the origin.

⇒The distance of point Q from the origin is the same as the distance of point P from the origin.

So, (Q,P) ∈ R. Therefore, R is symmetric.

Let (P,Q), (Q,S) ∈ R.

⇒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 P from the origin is the same.

⇒ The distance of points P and S from the origin is the same.

⇒(P, S) ∈ R. Therefore, R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

In other words, if O(0 0) is the origin and OP = k, then the set of all points related to P is at the same as k from the origin. Hence, this set of points forms a circle with the centre as the origin and the circle passes through point P.

Solution:

R is reflexive since, every triangle is similar to itself.

If (T1, T2) ∈ R, then T1 is similar to T2. So, (T2, T1) ∈ R. Therefore, R is symmetric.

Let (T1, T2), (T2, T3) ∈ R, then T1 is similar to T2 and T2 is similar to T3. So, T1 is also similar to T3. Therefore, (T1, T3) ∈ R so, R is transitive.

Thus, R is an equivalence relation.

We observe,

(3/6)=(4/8)=(5/10)=1/2

Therefore, the corresponding sides of triangles T1 and T3 are in the same ratio. Then, triangle T1 is similar to triangle T3.

Hence, T1 is related to T3.

Solution:

R is reflexive since(P1, P2) ∈ R as the same polygon has the same number of sides with itself.

Let (P1, P2) ∈ R, then P1 and P2 have the same number of sides. So, (P2, P1) ∈ R. Therefore, R is symmetric.

Let (P1, P2), (P2, P3) ∈ R, then P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides. So, P1 and P3 have the same number of sides ,i.e., (P1, P3) ∈ R. Therefore, R is transitive.

Hence, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides(since, T is a polygon with 3 sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.

Solution:

R is reflexive as any line L1 is parallel to itself i.e., (L1, L2) ∈ R.

Let (L1, L2) ∈ R, then L1 is parallel to L2. So, (L2, L1) ∈ R. Therefore, R is symmetric.

Let (L1, L2), (L2, L3) ∈ R, then L1 is parallel to L2 and L2 is parallel to L3. So, L1 is parallel to L3. Therefore, R is transitive.

Hence, R is an equivalence relation.

The set of all lines related to the line y=2x +4 is the set of all lines that are parallel to the line y=2x+4.Slope of the line is m=2.

It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y=2x +c, where c ∈ R.

Hence, the set of all lines related to the given line is given by y=2x +c, where c ∈ R.

Question 15. Let R be the relation in the set {1, 2, 3, 4} given by R={(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is equivalence relation.

Solution:

It is seen that (a,a) ∈ R, for every a ∈ {1, 2, 3, 4}. Therefore, R is reflexive.

It is seen that (1,2) ∈ R but (2,1) ∉ R. Therefore, R is not symmetric.

Also, it is observed that (a,b), (b,c) ∈ R⇒ (a,c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}. Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric. The correct answer is 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)

(D) (8,7) ∈ R

Solution:

Since b>a, (2,4) ∉ R also, as 3≠8-2, (3,8)∉R and as 8≠7-2. Therefore, (8,7)∉R

Consider (6,8). We have 8>6 and also, 6=8-2. Therefore, (6,8) ∈ R.

The correct answer is C.

Relations And Functions- FAQs

What is Relation?

Relation in mathematics is defined as the well-defined relationship between two sets. The relation connects the value of the first set with the value of the second set. We represent relation in mathematics using the ordered pair. If we are given two sets Set X and Set Y then the relation between them is represented using the ordered pair (x, y) such that, x ϵ X and y ϵ Y.

What is Reflexive Relation?

The Reflexive Relation means if element “a” is present in set A, then a relation “a” to “a” (aRa) should be present in relation R. If any such aRa is not present in R then R is not a reflexive relation.

What is Transitive Relation?

A relation R on a set A is called transitive relation if and only if

∀ a, b, c ∈ A, if (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R,

where R is a subset of (A x A), i.e. the cartesian product of set A with itself.

What is Symmetric Relation?

A relation R on a set A is called symmetric relation if and only if

∀ a, b ∈ A, if (a, b) ∈ R then (b, a) ∈ R and vice versa i.e.,

∀ a, b ∈ A, (a, b) ∈ R\ <=> (b, a) ∈ R,

where R is a subset of (A x A), i.e. the cartesian product of set A with itself.

What is Equivalence Relation?

A relation R on a set A is called an equivalence relation if it is

Reflexive Relation: (a, a) ∈ R ∀ a ∈ A, i.e. aRa for all a ∈ A.

Symmetric Relation: ∀ a, b ∈ A, (a, b) ∈ R ↔ (b, a) ∈ R.

Transitive Relation: ∀ a, b, c ∈ A, if (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R.

where R is a subset of (A x A), i.e. the cartesian product of set A with itself.

What is Function?

A function is a relation between two sets set A and set B. Such that every element of set A has an image in set B and no element in set A has more than one image in set B.

What are types of Functions?

There are several types of functions in maths. Some of the important types are:

One to One (Injective) function

Many to One function

Onto (Surjective) Function

Into Function

Weightage of chapter Relation and Function in Board Exam

The chapter Relation and Function consists of 8 Marks Weightage in Board Exams.

It is predicted according to previous year’s papers the chapter Relation and Function will be asked in 2 Marks which will be seen in MCQ.

And rest the marks can be distributed in any section of the Paper.

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Transpose of a Matrix-Addition & Multiplication Property of Transpose, Solved Examples, Class 12 Matrices Chapter Notes Study Material Download pdf

Transpose of a matrix is one of the most commonly used methods in matrix transformation. For a given matrix, the transpose of a matrix is obtained by interchanging rows into columns or columns to rows. In this article, we are going to learn the definition of the transpose of a matrix, steps to find the transpose of a matrix, properties and examples with a complete explanation. Before learning how to find the transpose of a matrix, first let us learn, what a matrix is? Table of Contents What is a Matrix? Transpose of a Matrix Definition How to Find the Transpose of a Matrix? Properties of Transpose of a Matrix (i) Transpose of the Transpose Matrix (ii) Addition Property of Transpose (iii) Multiplication by Constant (iv) Multiplication Property of Transpose Transpose of a Matrix Examples Practice Problems Frequently Asked Questions What is the transpose of a matrix? How to calculate the transpose of a Matrix? What is the Addition Property of Transpose? What is the Multiplication Property...
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