Chapter 16
Relations
1. Consider the following relations : 5. Let R = {(P, Q) | P and Q are at the same
R = {(x, y) | x, y are real numbers and x = wy for distance from the origin} be a relation, then the
some rational number w}; equivalence class of (1, –1) is the set :
[JEE (Main)-2021]
m p
S , m, n, p and q are integers such that (1) S = {(x, y) | x2 + y2 = 2}
n q
n, q 0 and qm = pn}. Then [AIEEE-2010] (2) S = {(x, y) | x2 + y2 = 1}
(1) R is an equivalence relation but S is not an (3) S = {(x, y) | x2 + y2 = 2}
equivalence relation
(4) S = {(x, y) | x2 + y2 = 4}
(2) Neither R nor S is an equivalence relation
6. Let A = {2, 3, 4, 5, ..., 30} and ' ' be an
(3) S is an equivalence relation but R is not an equivalence relation on A × A, defined by
equivalence relation
(a, b) (c, d) , if and only if ad = bc. Then the
(4) R and S both are equivalence relations
number of ordered pairs which satisfy this
2. If R = {(x, y) ; x, y Z, x2 + 3y2 8} is a relation equivalence relation with ordered pair (4, 3) is equal
on the set of integers Z, then the domain of to : [JEE (Main)-2021]
R–1 is [JEE (Main)-2020]
(1) 7 (2) 8
(1) {0, 1}
(3) 5 (4) 6
(2) {–2, –1, 1, 2}
(3) {–1, 0, 1} 7. Let N be the set of natural numbers and a relation
R on N be defined by
(4) {–2, –1, 0, 1, 2}
3. Let R1 and R2 be two relation defined as follows : R x, y N N : x 3 3x 2y xy 2 3y 3 0.
R1 = {(a, b) R2 : a2 + b2 Q} and
Then the relation R is [JEE (Main)-2021]
R2 = {(a, b) R2 : a2 + b2 Q}, where Q is the
(1) An equivalence relation
set of all rational numbers. Then
(2) Reflexive and symmetric, but not transitive
[JEE (Main)-2020]
(3) Reflexive but neither symmetric nor transitive
(1) Neither R1 nor R2 is transitive.
(2) R2 is transitive but R1 is not transitive. (4) Symmetric but neither reflexive nor transitive
(3) R1 and R2 are both transitive. 8. Let be the set of all integers,
(4) R1 is transitive but R2 is not transitive.
A ( x, y ) : ( x 2)2 y 2 4 ,
4. Let f : R R be defined as f(x) = 2x – 1 and
1
x
B ( x, y ) : x 2 y 2 4 and
g : R – {1} R be defined as g x 2.
x 1
Then the composition function f(g(x)) is :
C ( x, y ) : ( x 2)2 ( y 2 2)2 4
[JEE (Main)-2021] If the total number of relations from A B to
(1) neither one-one nor onto A C is 2p, then the value of p is
(2) onto but not one-one [JEE (Main)-2021]
(3) both one-one and onto (1) 16 (2) 49
(4) one-one but not onto (3) 25 (4) 9
9. Which of the following is not correct for relation R [JEE (Main)-2022]
on the set of real numbers?
(1) reflexive, symmetric but not transitive
[JEE (Main)-2021]
(1) (x, y) R |x – y| 1 is reflexive and (2) reflexive, transitive but not symmetric
symmetric.
(3) reflexive but not symmetric and transitive
(2) (x, y) R 0 |x| – |y| 1 is neither transitive
nor symmetric (4) an equivalence relation
(3) (x, y) R 0 < |x – y| 1 is symmetric and 13. Let R1 and R2 be two relations defined on by a R1
transitive b ab 0 and a R2 b a b. Then,
(4) (x, y) R |x| – |y| 1 is reflexive but not
[JEE (Main)-2022]
symmetric
(1) R1 is an equivalence relation but not R2
10. Let R and R be relations on the set {1, 2, ....., 50}
1 2
such that (2) R2 is an equivalence relation but not R1
R = {(p, pn) : p is a prime and n 0 is an integer} and (3) Both R1 and R2 are equivalence relations
1
R = {(p, pn) : p is a prime and n = 0 or 1}.
2 (4) Neither R1 nor R2 is an equivalence relation
Then, the number of elements in R – R is ______. 14. For N, consider a relation R on N given by
1 2
R = {(x, y) : 3x + y is a multiple of 7}. The relation R
[JEE (Main)-2022] is an equivalence relation if and only if
11. Let R1 = {(a, b) N × N : |a – b| 13} and [JEE (Main)-2022]
R2 = {(a, b) N × N : |a – b| ‘“ 13}. Then on N: (1) = 14
[JEE (Main)-2022] (2) is a multiple of 4
(1) Both R1 and R2 are equivalence relations (3) 4 is the remainder when is divided by 10
(2) Neither R1 nor R2 is an equivalence relation (4) 4 is the remainder when is divided by 7
(3) R1 is an equivalence relation but R2 is not 15. Let R be a relation from the set {1, 2, 3, ....., 60} to
itself such that R = {(a, b) : b = pq, where p, q 3
(4) R2 is an equivalence relation but R1 is not
are prime numbers}. Then, the number of elements
12. Let a set A = A1 A2 … Ak, where Ai Aj = in R is : [JEE (Main)-2022]
for i j, 1 i, j k. Define the relation R from A to
(1) 600 (2) 660
A by R = {(x, y) : y Ai if and only if x Ai, 1 i
k}. Then, R is : (3) 540 (4) 720
Chapter 16
Relations
1. Answer (3) So clearly it is one-one but not onto
R is not an equivalence relation because 0 R 1 but
1 R 0 , S is an equivalence relation.
2. Answer (3)
Given R = {(x, y) : x, yZ, x2 + 3y2 d 8} (1, 1)
So R = {(1, 1), (2, 1), (1, –1), (0, 1), (1, 0)} 0
So DR 1 ^1, 0, 1`
3. Answer (1)
5. Answer (1)
(I) If (a, b) R1 and (b, c) R1
? R = {(P, Q) | P and Q are at the same distance
a2 + b2 Q and b2 + c2 Q
from the origin}.
then a2 + 2b 2 + c 2 Q but we cannot say
anything about a2 + c2, that it is rational or not. Then equivalence class of (1, –1) will contain
all such points which lies on circumference of
So R1 is not transitive
the circle of centre at origin and passing
(II) If (a, b) R2 and (b, c) R2
through point (1, –1).
a2 + b2 Q and b2 + c2 Q
but we can’t say anything about a2 + c2, that i.e., radius of circle 12 12 2
it is rational or irrational.
? Required equivalence class of (S)
So R2 is not transitive
= {(x, y) | x2 + y2 = 2}.
4. Answer (4)
6. Answer (1)
Here f : R o R, f(x) = 2x – 1
Let (4, 3) (c, d)
1
x
and g : R – {1} o R g(x) 2
x 1 c d
4d = 3c k say
4 3
So, f(g(x)) = 2 g(x) – 1
For c, d A, k = 1, 2, 3, ...., 7
§ 1·
¨x2¸ 7. Answer (3)
2¨ ¸ 1
¨¨ x 1 ¸¸ x2(x – 3y) – y2(x – 3y) = 0
© ¹
(x – y) (x + y) (x – 3y) = 0 ...(i)
2x 1 x 1 x 1 1 ? (i) holds for all (x, x) ? R is reflexive
x 1 x 1
if (x, y) holds then (y, x) may or may not holds for
1 factors (x + y), (x – 3y) ? R is NOT symmetric
1
x 1 Similarly (x – 3y) factor doesn’t hold for transitive
8. Answer (3) ? (2, 11) R1 and (11, 19) R1 but
The set A and set B are represented as : (2, 19) R1
? R1 is not transitive
(0, 2) (A B) Hence R1 is not equivalence
In R2 : (13, 3) R2 and (3, 26) R2 but
(–2, 4) (0, 0) (2, 0) (4, 0) (13, 26) R2 (' |13 – 26| = 13)
? R2 is not transitive
Hence R2 is not equivalence.
12. Answer (4)
? A B = {(0, 0), (1, 0), (2, 0), (1, 1), (1, –1)}
The set A and set C are represented as : R = {(x, y) : yAi, iff xAi, 1 d i d k}
(1) Reflexive
(a, a) aAi iff aAi
C
(0, 2) (2) Symmetric
A (a, b) aAi iff bAi
(2, 0)
(b, a) R as bAi iff aAi
(A C) (3) Transitive
(a, b)R & (b, c) R.
? A C = {(1, 1), (2, 0), (2, 1), (2, 2), (3, 2)} aAi iff bAi & bAi iff cAi
? Total number relations from A B to A C = 25×5
aAi iff cAi
? p = 25
9. Answer (3) (a, c)R.
(x, y) R 0 < |x – y| d 1. Relation is equivalence
R is symmetric because |x – y| = |y – x| 13. Answer (4)
But R is not transitive a R1 b ab t 0
For example
So, definitely (a, a) R1 as a2 t 0
x = 0.2, y = 0.9, z = 1.5
If (a, b) R1 (b, a) R1
0 d |x – y| = 0.7 d 1
But if (a, b) R1, (b, c) R1
0 d |y – z| = 0.6 d 1
Then (a, c) may or may not belong to R1
But |x – z| = 1.3 > 1
{Consider a = –5, b = 0, c = 5 so (a, b) and (b, c)
10. Answer (8)
R1 but ac < 0}
R1 – R2 = {(2, 22), (2, 23), (2, 24), (2, 25), (3, 32)
(3, 33), (5, 52), (7, 72)} So, R1 is not equivalence relation
So number of elements = 8 a R2 b a t b
11. Answer (2) (a, a) R2 so reflexive relation
R1 = {(a, b) N × N : |a – b| d 13} and If (a, b) R2 then (b, a) may or may not belong to R2
R2 = {(a, b) N × N : |a – b| d 13} So not symmetric
Hence it is not equivalence relation
In R1: ' |2 – 11| = 9 d13
14. Answer (4) 3a + (7k – 3)b = 7k1 and
R = {(x, y) : 3x + Dy is multiple of 7}, Now R to be an 3b + (7k2 – 3) c = 7k3
equivalence relation
Adding 3a + 7kb + (7k2 – 3) c = 7 (k1 + k3)
(1) R should be reflexive : (a, a)R aN
3a + (7k2 – 3) c = 7 m
? 3a + aD = 7k
?(a, c)R
? (3 + D) a = 7k
?R is transitive
?3 + D = 7k1 D = 7k1 –3
?D = 7k – 3 = 7k + 4
= 7k1 + 4
(2) R should be symmetric : aRb bRa 15. Answer (2)
aRb : 3a + (7k –3) b = 7 m b can take its values as 9, 15, 21, 33, 39, 51, 57,
3(a – b) + 7kb = 7 m 25, 35, 55, 49
3(b – a) + 7 ka = 7 m b can take these 11 values
So, aRb bRa and a can take any of 60 values
? R will be symmetric for a = 7k1 –3
So, number of elements in R = 60 × 11
(3) Transitive : Let (a, b)R, (b, c)R
= 660
