Aakash Repeater Courses
Take Aakash iACST and get instant scholarship on coaching programs.
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
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.
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.
NCERT Relations and Functions Class 12 Solutions: Exercise: 1.1 Page Number: 5-7 Total Questions: 16 |
Take Aakash iACST and get instant scholarship on coaching programs.
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
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
Answer:
R={(x,y):x is wife of y}
(x,y)∈R means x is the wife of y, but x is not the wife of x, i.e. (x,x)∉R.
So, it is not reflexive.
(x,y)∈R means x is wife of y but y is not wife of x i.e. (y,x)∉R .
So, it is not symmetric.
Let (x,y),(y,z)∈R mean x is the wife of y and y is the wife of z.
This case is not possible, so it is not transitive.
Hence, it is not reflexive, symmetric, or transitive.
Question 1(v): Determine whether each of the following relations is reflexive, symmetric, and transitive:
(v) Relation R in the set A of human beings in a town at a particular time, given by
(e) R={(x,y):x is father of y}
Answer:
R={(x,y):x is father of y}
(x,y)∈R means x is the father of y, then x cannot be the father of x, i.e. (x,x)∉R. So, it is not reflexive.
(x,.∈R means x is the father of y, then y cannot be the father of x, i.e. (y,x)∉R. So, it is not symmetric.
Let (x,y),(y,z)∈R mean x is the father of y and y is the father of z, then x cannot be the father of z, i.e. (x,z)∉R.
So, it is not transitive.
Hence, it is neither reflexive nor symmetric nor transitive
Answer:
R={(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.
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.
Answer:
Let A= {1,2,3}
R={(1,2),(2,1)}
We can see (1,1),(2,2),(3,3)∉R so it is not reflexive.
As (1,2)∈Rand(2,1)∈R, so it is symmetric.
(1,2)∈Rand(2,1)∈R
But (1,1)∉R, so it is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
Answer:
A = all the books in the library of a college
R={(x,y):x and y have same number of pages }
(x,x)∈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.
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.
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.
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.
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.
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.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Answer:
A = {1,2,3,4}
R={(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}
For every a∈A, there is (a, a)∈R.
∴ R is reflexive.
Given, (1,2)∈R but (2,1)∉R
∴ R is not symmetric.
For a,b,c∈A there are (a,b)∈Rand(b,c)∈R ⇒ (a,c)∈R
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
The correct answer is option B.
Question 16: Let R be the relation in the set N given by R={(a,b):a=b−2,b>6}. Choose the correct answer.
Answer:
R={(a,b):a=b−2,b>6}
(A) Since b<6, so (2,4)∉R
(B) Since 3≠8−2, so (3,8)∉R
(C) Since, 8>6 and 6=8−2 so (6,8)∈R
(D) Since 8≠7−2, so (8,7)∉R
The correct answer is option C.
NCERT Relations and Functions Class 12 Solutions: Exercise: 1.2 Page Number: 10-11 Total Questions: 12 |
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:
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:
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:
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:
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:
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.
Answer:
f:R⟶R
f(x)=[x]
One-one:
For 1.5,1.7∈R then f(1.5)=[1.5]=1 and f(1.7)=[1.7]=1
but 1.5≠1.7
∴ f is not one-one, i.e. not injective.
For 0.6∈R there is no x in R such that f(x)=[0.6]
∴ f is not onto, i.e. not surjective.
Hence, f is not injective but not surjective.
Answer:
f: R→R
f(x)=|x|
f(x)=|x|=xifx≥0and−xifx<0
One-one:
For −1,1∈R then f(−1)=|−1|=1
f(1)=|1|=1
−1≠1
∴ f is not one-one, i.e. not injective.
For −2∈R,
We know f(x)=|x| is always positive there is no x in R such that f(x)=|x|=−2
∴ f is not onto, i.e. not surjective.
Hence, f(x)=|x| is neither one-one nor onto.
Question 5: Show that the Signum Function f: R→R, given by f(x)={1 if x>0 0 if x=0 −1 if x<0, is neither one-one nor onto.
Answer:
f:R→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.
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.
(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.
(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
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.
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.
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,
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 are an integral part of mathematics, and the NCERT Class 12 textbook discusses the following Maths topics.
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}
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.
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.
Concept Name |
JEE |
NCERT |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
✅ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
❌ | |
✅ |
❌ | |
f(x) = min{g1(x), g2(x).......... } or max{g1(x), g2(x).......... } |
✅ |
❌ |
Students can access all NCERT Class 12 Maths solutions from the links below all-in-one place.
Also read,
Students can check the following links for more in-depth learning.
Students can check the following links for more in-depth learning.
Frequently Asked Questions (FAQs)
In NCERT Class 12 Maths Chapter 1, the composition of functions means applying one function to the result of another. If we have two functions f(x) and g(x), their composition is written as (f o g)(x), which means f(g(x)).
Steps to find composition:
1. First, find g(x) (solve for x in g ).
2. Substitute g(x) into f(x).
3. Simplify the expression.
For example, if f(x)=x2 and g(x)=x+1, then:
(f o g)(x)=f(g(x))=f(x+1)=(x+1)2
In NCERT Class 12 Maths Chapter 1, the domain and range of a function describe its input and output values:
In NCERT Class 12 Maths, one-one (injective) and onto (surjective) functions are two important types of functions:
A function can be both one-one and onto (bijective).
NCERT Class 12 Maths Chapter 1 "Relations and Functions" covers essential standards of set principles and mappings. Key subjects encompass:
In NCERT Class 12 Maths Chapter 1, relations and functions are different concepts:
Thus, every function is a relation, but not every relation is a function.
On Question asked by student community
Hello
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%.
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE