NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Komal MiglaniUpdated on 20 Aug 2025, 04:37 PM IST

Understanding relations and functions in mathematics is like learning to navigate a map – it helps students move from one point to another with purpose. In the Relations and Functions Class 12 NCERT Solutions, the first question that comes to students' minds is, "What are relations and functions?". The answer is simple: 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 favourite subject. The main purpose of these NCERT Solutions for class 12, Relations and Functions, is to make learning easier for students and to explain this topic more easily.

This Story also Contains

  1. NCERT Solution for Class 12 Maths Chapter 1 Solutions: Download PDF
  2. NCERT Solutions for Class 12 Maths Chapter 1: Exercise Questions
  3. Class 12 Maths NCERT Chapter 1: Extra Question
  4. Relations and Functions Class 12 NCERT Solutions: Topics
  5. Class 12 Maths NCERT Chapter 1, Relations and Functions: Important Formulae
  6. Approach to Solve Questions of Relations and Functions Class 12
  7. What Extra Should Students Study Beyond the NCERT for JEE?
  8. NCERT Solutions for Class 12 Maths: Chapter Wise
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Relations and functions are not just theory-based, but they also come in handy in real life. People can apply it during mapping, programming, and data handling. These NCERT solutions, created by Careers360 experts with years of experience in the field, provide students with quality practice questions. After checking these NCERT solutions for Class 12 Maths, students can also check the NCERT Exemplar Solutions and notes for Class 12 Maths NCERT Chapter 1, Relations and Functions, to understand this chapter better. For a structured syllabus, concise notes, and PDFs, click this NCERT link.

NCERT Solution for Class 12 Maths Chapter 1 Solutions: Download PDF

Students who wish to access the Class 12 Maths Chapter 1 Solutions PDF can click on the link below to download the complete solution in PDF format.

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NCERT Solutions for Class 12 Maths Chapter 1: Exercise Questions

NCERT Relations and Functions Class 12 Solutions: Exercise: 1.1

Page Number: 5-7

Total Questions: 16

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

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

Answer:

A={1,2,3...,13,14}

R={(x,y):3x−y=0} ={(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 is reflexive, symmetric, and transitive:

(ii) Relation R in the set N of natural numbers defined as

R={(x,y):y=x+5 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 is 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 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

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

Answer:

R={(a,b):a≤b2}
Taking

(12,12)∉R

and

(12)>(12)2
So,R is not reflexive.
Now,
(1,2)∈R because 1<4.
But, 4≮1, i.e. 4 is not less than 1
So, (2,1)∉R
Hence, it is not symmetric.
(3,2)∈R and (2,1.5)∈R as 3<4 and 2<2.25
Since (3,1.5)∉R because 3≮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 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.

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

Answer:

R={(a,b):a≤b3}

(12,12)∉R because 12∉(12)3
So, it is not symmetric.

Now, (1,2)∈R because 1<23but(2,1) \notin Rbecause2 \nless 1^3Itisnotsymmetric(3,1.5) \in Rand(1.5,1.2) \in Ras3<1.5^3and1.5<1.2^3.But,(3,1.2) \notin Rbecause3 \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 the 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)∈R because x and x have the same number of pages, so it is reflexive.

Let (x,y)∈R mean x and y have the same number of pages.

Since y and x have the same number of pages, so (y,x)∈R.

Hence, it is symmetric.

Let (x,y)∈R mean x and y have the same number of pages.

And (y,z)∈R means y and z have the same number of pages.

This states, x and z also have the same number of pages, i.e. (x,z)∈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 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 relations 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 is 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): distance of the point P from the origin is the 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 centre.

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)∈R

∴R is reflexive.

Let (P,Q)∈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)∈R

∴R is symmetric.

Let (P,Q)∈R and (Q,S)∈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)∈R

∴R is transitive.

Hence, R is an equivalence relation.

The set of all points related to a point P≠(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 is 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, this set of points forms a circle with the centre 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 T1 with sides 3, 4, 5, T2 with sides 5, 12, 13, and T3 with sides 6, 8, 10. Which triangles among T 1 , T 2, and T 3 are related?

Answer:

R={(T1,T2):T1 is similar to T2}

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

Let,

(T1,T2)∈R i.e. T1 is similar to T2

T1 is similar to T 2 ; T 2 is similar to T 1 , i.e. (T2,T1)∈R

Hence, it is symmetric.

Let,

(T1,T2)∈R and (T2,T3)∈R i.e. T1 is similar to T2 and T2 is similar to T3.

⇒T1 is similar to T3 i.e. (T1,T3)∈R

Hence, it is transitive,

Thus, R={(T1,T2):T1 is similar to T2}, is equivalence relation.

Now, we see the ratio of the sides of triangles T1 and T3 as shown

36=48=510=12

i.e. ratios of sides of T1 and T3 are equal. Hence, T1 and T3 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 the 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={(P1,P2):P1 and P2 have same number of sides }

The same polygon has the same number of sides as itself,i.e. (P1,P2)∈R, so it is reflexive.

Let,

(P1,P2)∈R, i.e. P1 have the same number of sides as P2

P1 have same number of sides as P2 is same as P2 have same number of sides as P1 i.e. (P2,P1)∈R

Hence, it is symmetric.

Let,

(P1,P2)∈R and (P2,P3)∈R i.e. P1 have same number of sides as P2 and P2 have same number of sides as P3

⇒P1 have same number of sides as P3 i.e. (P1,P3)∈R

Hence, it is transitive,

Thus, R={(P1,P2):P1 and P2 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 the 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={(L1,L2):L1 is parallel to L2}

All lines are parallel to themselves, so it is reflexive.

Let,

(L1,L2)∈R, i.e. L1 is parallel to T 2.

L1 is parallel to L 2 is the same as L 2 being parallel to L1, i.e. (L2,L1)∈R

Hence, it is symmetric.

Let,

(L1,L2)∈R and (L2,L3)∈R i.e. L1 is parallel to L2 and L2 is parallel to L3.

⇒L1 is parallel to L3 i.e. (L1,L3)∈R

Hence, it is transitive,

Thus, R={(L1,L2):L1 is parallel to L2}, 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∈R

Hence, the 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.

NCERT Relations and Functions Class 12 Solutions: Exercise: 1.2

Page Number: 10-11

Total Questions: 12

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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∗⟶R∗ is defined by f(x)=1x.
One - One :

f(x)=f(y)

1x=1y

x=y

∴f is one-one.
Onto:
We have y∈R∗, then there exists x=1y∈R∗( Here y≠0) such that

f(x)=1(1y)=y

∴ f is onto.
Hence, the function is one-one and onto.

If the domain R∗ is replaced by N with the co-domain being the same as R∗ i.e. g:N⟶R∗ defined by

g(x)=1x

g(x1)=g(x2)

1x1=1x2

x1=x2

∴g is one-one.
For 1.5∈R∗,
g(x)=11.5, but there does not exist 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→N

f(x)=x2
One-one:
x,y∈N then f(x)=f(y)

x2=y2

x=y

∴f is one- one i.e. injective.
For 3∈N there is no x in N such that f(x)=x2=3
∴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→Z

f(x)=x2

One-one:

For −1,1∈Z then f(x)=x2

f(−1)=(−1)2

f(−1)=1 but −1≠1

Therefore,f is not one- one i.e. not injective.
For −3∈Z there is no x in Z such that f(x)=x2=−3
∴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→R

f(x)=x2
One-one:
For −1,1∈R then f(x)=x2

f(−1)=(−1)2

f(−1)=1 but −1≠1

Therefore,f is not one- one i.e. not injective.
For −3∈R there is no x in R such that f(x)=x2=−3
∴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→N

f(x)=x3

One-one:
x,y∈N then f(x)=f(y)

x3=y3

x=y

∴f is one- one i.e. injective.
For 3∈N there is no x in N such that f(x)=x3=3
∴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→Z

f(x)=x3

One-one:
For (x,y)∈Z then f(x)=f(y)

x3=y3

x=y

∴f is one- one i.e. injective.
For 3∈Z there is no x in Z such that f(x)=x3=3
∴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⟶R, given by f(x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

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→R is given by

f(x)={1 if x>00 if x=0−1 if x<0

As we can see f(1)=f(2)=1, but 1≠2
So it is not one-one.
Now, f(x) takes only 3 values (1,0,−1) for the element -3 in codomain R, there does not exist 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→R

f(x)=1+x2

Let there be (a,b)∈R such that f(a)=f(b)

1+a2=1+b2a2=b2a=±b

For f(1)=f(−1)=2 and 1≠−1
∴f is not one-one.
Let there be −2∈R(−2 in codomain of R)

f(x)=1+x2=−2

There does not exist any x in dthe omain R such that f(x)=−2
∴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→N be defined by f(n)={n+12 if n is odd n2 if n is even for all n∈N. State whether the function f is bijective. Justify your answer.

Answer:

f:N→N,n∈N

f(n)={n+12 if n is odd n2 if n is even

Here we can observe,

f(2)=22=1 and f(1)=1+12=1
As we can see f(1)=f(2)=1 but 1≠2
∴f is not one-one.
Let, n∈N (N=co-domain)
case1 n be even
For r∈N,n=2r
then there is 4r∈N such that f(4r)=4r2=2r
case2 n be odd
For r∈N,n=2r+1
then there is 4r+1∈N such that f(4r+1)=4r+1+12=2r+1
∴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→B defined by f(x)=x−2/x−3. Is f one-one and onto? Justify your answer.

Answer:

A=R−{3}B=R−{1}f:A→Bf(x)=(x−2x−3)

Let a,b∈A such that f(a)=f(b)

(a−2a−3)=(b−2b−3)(a−2)(b−3)=(b−2)(a−3)ab−3a−2b+6=ab−2a−3b+6−3a−2b=−2a−3b3a+2b=2a+3b3a−2a=3b−2ba=b

∴f is one-one.

Let, b∈B=R−{1} then b≠1

a∈A such that f(a)=b(a−2a−3)=b(a−2)=(a−3)ba−2=ab−3ba−ab=2−3ba(1−b)=2−3ba=2−3b1−b∈A

For any b∈B there exists a=2−3b1−b∈A such that

f(2−3b1−b)=2−3b1−b−22−3b1−b−3=2−3b−2+2b2−3b−3+3b=−3b+2b2−3=b

∴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:

f:R→Rf(x)=x4

One-one:
For a,b∈R then f(a)=f(b)

a4=b4a=±b

∴f(a)=f(b) does not imply that a=b
example : f(2)=f(−2)=16 and 2≠−2
∴f is not one- one
For 2∈R there is no x in R such that f(x)=x4=2
∴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.

(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:

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→{x∈R:−1<x<1} defined by

f(x)=x1+|x|,x∈R

One-one:

Let f(x)=f(y),x,y∈R

x1+|x|=y1+|y|

It is observed that if x is positive and y is negative.

x1+x=y1+y

Since x is positive and y is negative.

x>y⇒x−y>0 but 2xy is negative.

x−y≠2xy

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:

f(x)=f(y)x1+x=y1+yx(1+y)=y(1+x)x+xy=y+xyx=y

When x and y both are negative:

f(x)=f(y)x1−x=y1−yx(1−y)=y(1−x)x−xy=y−xyx=y

∴f is one-one.

Onto:

Let y∈R such that −1<y<1

If y is negative, then x=yy+1∈R

f(x)=f(yy+1)=y1+y1+|y1+y|=y1+y1+−y1+y=y1+y−y=y

If y is positive, then x=y1−y∈R

f(x)=f(y1−y)=y1−y1+|y1−y|=y1−y1+−y1−y=y1−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

Question 3: Given a nonempty set X, consider P(X), which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A⊂B Is R an equivalence relation on P(X)? Justify your answer.

Answer:

Given a nonempty set X, consider P(X), which is the set of all subsets of X.

Since every set is a subset of itself, ARA for all A∈P(x)

∴R is reflexive.

Let ARB⇒A⊂B

This is not the same as B⊂A

If A={0,1} and B={0,1,2}

Then we cannot say that B is related to A.

∴R is not symmetric.

If ARB and BRC, then A⊂B and B⊂C

This implies A⊂C=ARC

∴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 the permutations on n symbols 1,2,3,4,5...............n, which is n.

Question 5: Let A={−1,0,1,2},B={−4,−2,0,2} and f,g:A→B be functions defined by f(x)=x2−x,x∈A and g(x)=2|x−12|−1,x∈A. Are f and g equal?

Justify your answer. (Hint: One may note that two functions f:A→B and g:A→B such that f(a)=g(a)∀a∈A, are called equal functions.)

Answer:

Given :

A={−1,0,1,2},B={−4,−2,0,2}

f,g:A→B are defined by f(x)=x2−x,x∈A and g(x)=2|x−12|−1,x∈A.

It can be observed that.

f(−1)=(−1)2−(−1)=1+1=2g(−1)=2|−1−12|−1=2|−32|−1=3−1=2f(−1)=g(−1)

f(0)=(0)2−(0)=0+0=0g(0)=2|0−12|−1=2|−12|−1=1−1=0f(0)=g(0)

f(1)=(1)2−(1)=1−1=0g(1)=2|1−12|−1=2|12|−1=1−1=0f(1)=g(1)

f(2)=(2)2−(2)=4−2=2g(2)=2|2−12|−1=2|32|−1=3−1=2f(2)=g(2)

∴f(a)=g(a)∀a∈A

Hence, f and g are equal functions.

Question 6: Let A={1,2,3}. The number of relations containing (1,2) and (1,3), which are reflexive and symmetric but not transitive, is

(A) 1
(B) 2
(C) 3
(D) 4

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.

Also read,

Class 12 Maths NCERT Chapter 1: Extra Question

Question: Let an operation ∗ on the set of natural numbers N be defined by a∗b=ab⋅
Find (i) whether ∗ is a binary or not, and (ii) if it is a binary, then is it commutative or not.

Solution:

(i) As ab∈N for all a,b∈N
ie a∗b∈N∀a,b∈N
Hence, ∗ is binary.

(ii) As 12≠21
so 1∗2≠2∗1

Hence, ∗ is not commutative.

Relations and Functions Class 12 NCERT Solutions: Topics

Relations and functions are an integral part of mathematics, and the NCERT Class 12 textbook discusses the following Maths topics.

Class 12 Maths NCERT Chapter 1, Relations and Functions: Important Formulae

Relations:

A relation R is a subset of the Cartesian product of A×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)∈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×B={(a,b):a∈A,b∈B}
If (a,b)=(x,y), then a=x and b=y
n(A×B)=n(A) * n(B), where n(A) is the cardinality (number of elements) of set A.
A×ϕ=ϕ (where ϕ is the empty set)
A function f:A→B is denoted as:
f(x)=y
This means (x,y)∈f.

Algebra of functions:

  • (f+g)(x)=f(x)+g(x)
  • (f−g)(x)=f(x)−g(x)
  • (f⋅g)(x)=f(x)⋅g(x)
  • (k⋅f)(x)=k⋅f(x), where k∈R
  • (f/g)(x)=f(x)/g(x), where g(x)≠0

Approach to Solve Questions of Relations and Functions Class 12

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.

  • Recognising the type: Build a strong foundation for different kinds of relations, such as reflexive, symmetric, transitive, and equivalence.
    Familiarise yourself with the properties of One-One, onto, and bijective functions.
  • Efficient usage of diagrams and graphs: Visual representation of the properties of relations and functions helps to clarify the types.
    Use tables for small sets for binary operations to verify identity, inverse, etc.
  • Domain, codomain, range: Grasp these concepts thoroughly to identify whether a relation is a function and determine its behaviour.
    Also, practice extracting the domain and range from relations using ordered pairs or graphs.
  • Elimination method: For MCQ-type questions, cancel out the extreme options by checking the basic definitions.
  • Shortcut tricks: Smart tricks and shortcuts are always handy during the exam to save time. Write those tricks in a notebook and revise them from time to time. Here are some tricks.
    - If a relation is reflexive and only uses equality or divisibility, it may be an equivalence.
    - For transitive relations, check in pairs that share a common middle.
    - Use the formula 2n2 to find the total relations over a set of n elements.
    - During the domain checking, avoid the values that cause division by zero or negative square roots.

NCERT Solutions for Class 12 Maths: Chapter Wise

Students can access all NCERT Class 12 Maths solutions from the links below all-in-one place.


Also read,

NCERT Solutions class-wise

Students can check the following links for more in-depth learning.

NCERT Books and NCERT Syllabus

Students can check the following links for more in-depth learning.

Frequently Asked Questions (FAQs)

Q: How do you find the composition of functions in Chapter 1?
A:

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 

Q: What is the range and domain of functions in NCERT Chapter 1?
A:

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.
Q: What is the difference between one-one and onto functions in NCERT Class 12 Maths?
A:

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).

Q: What are the key topics covered in NCERT Class 12 Maths Chapter 1?
A:

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.
Q: What is the difference between relation and function?
A:

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 it 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.

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Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

Hello Aspirant,

Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.

From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .

If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.

The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.