Let A = {1, 2, 3, ..., 100} and R be a relation on A such that R = {(a, b) : a = 2b + 1}. Let be a sequence of k elements of R such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k, for which such a sequence exists, is equal to : [2025]
8
5
6
7
Let the smallest alue is in A = {1, 2, 3, ..., 100}.
Since,
The sequence is (63, 31), (31, 15), (15, 7), (7, 3), (3, 1)}
k = 5.
Let A = {–3, –2, –1, 0, 1, 2, 3). Let R be a relation on A defined by xRy if and only if . Let be the number of elements in R. Let m be the minimum number of elements required to be added in R to make it reflexive relation. Then + m is equal to [2025]
18
20
17
19
(1)
A = {–3, –2, –1, 0, 1, 2, 3)
xRy if and only if
y = –3
y = –2
y = –1
y = 0
y = 1
y = 2
y = 3
R = {(–3, –3), (3, –3), (–2, –2),(2, –2), (–2, –1), (2, –1), (–2, 0), (–1, 0), (0,0), (1, 0), (2, 0), (–1, 1), (0, 1), (1, 1), (0, 2)}
= 15
To make it reflexive we will add (–1, –1), (2, 2), (3, 3) in R
+ m = 15 + 3 = 18.
Let A = {–2, –1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if y = max{x, 1}. Let be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then + m + n is equal to [2025]
13
12
14
11
(2)
We have, A = {–2, –1, 0, 1, 2, 3} and
R = {(–2, 1), (—1, 1), (0, 1), (0, 0), (1, 1), (2, 2), (3, 3)}.
Now, number of elements in R i.e., = 6
For R to be reflexive,
R = {(–2, –2), (–1, –1), (0, 0), (–2, 1), (–1, 1), (0, 1), (1, 1), (2, 2), (3, 3)}
So, we need to add three elements to make it reflexive.
m = 3
For R to be symmetric,
R = {(–2, 1), (1, –2), (–1, 1), (1, –1), (0, 1), (1, 0), (1, 1), (2, 2), (3, 3)}
So, we need to add three elements to make it symmetric.
n = 3
So, + m + n = 6 + 3 + 3 = 12.
Let A = {–3, –2, –1, 0, 1, 2, 3} and R be a relation on A defined by xRy if and only if 2x – y {0, 1}. Let be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then + m + n is equal to : [2025]
17
16
18
15
(1)
We have, A = {–3, –2, –1, 0, 1, 2, 3}, R is defined on A as xRy such that 2x – y {0, 1}.
i.e., 2x – y = 0 or 2x – y = 1
R = {(0, 0), (–1, –2), (1, 2), (0, –1), (2,3), (1, 1), (–1, –3)} i.e., = 7
For R to be reflexive, i.e., we need 5 more elements {(2, 2), (–1, –1), (3, 3), (–3, –3), (–2, –2)} so m = 5 and for R to be symmetric, we need 5 more elements {(–2, –1), (2, 1), (–1, 0), (3, 2), (–3, –1)}, so n = 5.
= m + n = 7 + 5 + 5 = 17.
Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A defined by (x, y) R if and only if max {x, y} {3, 4}. Then among the statements
() : The number of elements in R is 18, and
() : The relation R is symmetric but neither reflexive nor transitive. [2025]
both are true
only () is true
both are false
only () is true
(2)
R = {(0, 3), (3, 0), (0, 4), (4, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)}
Number of elements in R = 16
is false.
Now, (0, 0), (1, 1), (2, 2), (5, 5) R R is not reflexive.
Again, let (a, b) R then (b, a) R
As max {a, b} = max {b, a} R is symmetric.
Now, R is not transitive as (0, 3), (3, 1) R but (0, 1) R.
is true.
The number of non-empty equivalence relations on the sett {1, 2, 3} is: [2025]
5
6
7
4
(1)
Partitions of set {1, 2, 3} is {{1}}. {2}, {3}}, {{1, 2}, {3}}, {{1, 3}, {2}}, {{2, 3}, {1}}, {{1, 2, 3}}
Number of non-empty equivalence relations on the set {1, 2, 3} = 5.
Let R = {(1, 2), (2, 3), (3, 3)} be a relation defined on the set {1, 2, 3, 4}. Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is: [2025]
7
10
9
8
(1)
Let A = {1, 2, 3, 4}
R = {(1, 2), (2, 3), (3, 3)} [Given]
For an equivalence relation 'R' should be reflexive, symmetric and transitive.
Minimum elements that should be added are
(2, 1), (3, 2), (1, 1), (2, 2), (4, 4), (1, 3), (3, 1)
Minimum number of elements required = 7.
Let and
If , then is: [2025]
18
12
24
15
(2)
We have, and
From given figure, we get C = {(3, 0), (–3, 0), (0, 3), (0, –3)}
Let X = R R. Define a relation R on X as:
.
Statement I: R is an equivalence relation.
Statement II: For some (a, b) X, the set
S = {(x, y) X :(x, y) R (a, b)} represents a line parallel to y = x.
In the light of the above statements, choose the correct answer from the options given below: [2025]
Statement I is true but Statement II is false.
Statement I is false but State II is true.
Both Statement I and Statement II are false.
Both Statement I and Statement II is true.
(1)
State I:
Reflexive :
R is reflexive.
Symmetric :
R is symmetric.
Transitive :
and
Hence, relation R is equivalence relation.
Statement I is true.
For Statement II: (x, y) R (a, b), for some
So, Statement II is false.
The relation R = {(x, y) : x, y Z and x + y is even} is: [2025]
reflexive and transitive but not symmetric.
reflexive and symmetric but not transitive.
symmetric and transtive but not reflexive.
an equivalence relation.
(4)
R = {(x, y) : x, y Z and x + y is even}
For reflexive : x + x = 2x is even
For symmetric : x + y = y + x is even
For transitive : x + y is even and y + z is even then z + x is also even.
So, the relation is an equivalence.