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All questions of Sets, Relations & Functions for A Level Exam
The number of proper subsets of the set { 1, 2 , 3 } is :- a)7
- b)5
- c)6
- d)8
Correct answer is 'A'. Can you explain this answer?
The number of proper subsets of the set { 1, 2 , 3 } is :
a)
7
b)
5
c)
6
d)
8
|
Kritika Bajaj answered |
Number of subsets of a given set = 2n
The Shaded region in the following figure illustrates
- a)A ∩ ( B ∪ C)
- b)A ∩ B ∩ C
- c)A ∪ B ∪ C
- d)(A ∩ B) ∪ (A ∩ C)
Correct answer is option 'D'. Can you explain this answer?
The Shaded region in the following figure illustrates
a)
A ∩ ( B ∪ C)
b)
A ∩ B ∩ C
c)
A ∪ B ∪ C
d)
(A ∩ B) ∪ (A ∩ C)
|
New Words answered |
First which region is over which region Then We will see that A is on the B so A intersection B and after C is on the A so, A intersection C after that we have to take all intersection part so A intersection B is Union with A intersection C.
The shaded region represents (A ∩ B) ∪ (A ∩ C).
If A = {a, b, c} then the number of proper subsets of A are:
- a)3
- b)10
- c)7
- d)8
Correct answer is option 'C'. Can you explain this answer?
If A = {a, b, c} then the number of proper subsets of A are:
a)
3
b)
10
c)
7
d)
8
|
Pooja Nair answered |
- Number of proper subsets of a given set = 2m - 1, where m is the number of elements.
- Here the number of elements is 3. So the number of proper subsets of A = 23 - 1 = 7.
If Q={x:x=1/y,where y∈N}, then- a)0∈Q
- b)1∈Q
- c)2∈Q
- d)2/3∈Q
Correct answer is option 'B'. Can you explain this answer?
If Q={x:x=1/y,where y∈N}, then
a)
0∈Q
b)
1∈Q
c)
2∈Q
d)
2/3∈Q
|
Rohit Jain answered |
As y ∈ N, y can be 1, 2, 3, 4...
∴ x will be 1, 1/2, 1/3, 1/4...
⇒ 1 ∈ Q
∴ x will be 1, 1/2, 1/3, 1/4...
⇒ 1 ∈ Q
Can you explain the answer of this question below:
If U= set of all whole numbers less than 12, A=set of all whole numbers less than 10, B= Set of all odd natural numbers less than 10, then what is (A∩B)’?
A:{3,5,7,9}
B:{1,3,5,7,9}
C:{0,2,4,6,8,10,11}
D:{1,3,5,7}
The answer is a.
|
Hansa Sharma answered |
U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}
Can you explain the answer of this question below:If A = {5, 10, 15}, B = ϕ, then B – A is
- A:
5
- B:
{5,10}
- C:
ϕ
- D:
{5,10, 15}
The answer is c.
If A = {5, 10, 15}, B = ϕ, then B – A is
5
{5,10}
ϕ
{5,10, 15}
|
Om Desai answered |
If A = {5, 10, 15}, B = ϕ
B - A will have those elements which are in B but not in A.
B - A = ϕ
B - A will have those elements which are in B but not in A.
B - A = ϕ
Which of the following is not an empty set?- a){x : x is a multiple of 7, x < 7, x ∈ N}
- b)Set of common points of two parallel lines in a plane
- c){x : 6 + 2x > 5x + 3, x ∈ N}
- d)Set of smallest whole number
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not an empty set?
a)
{x : x is a multiple of 7, x < 7, x ∈ N}
b)
Set of common points of two parallel lines in a plane
c)
{x : 6 + 2x > 5x + 3, x ∈ N}
d)
Set of smallest whole number
|
Ayush Joshi answered |
As, the set of Whole numbers is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...};
So, according to the set of whole numbers above, the smallest whole number should be "0" and therefore the set of smallest whole number is not empty.
For real number x and y, we writeis an irrationalnumber. Then the relation R is:a)Reflexiveb)Symmetricc)Transitived)EquivalenceCorrect answer is option 'A'. Can you explain this answer?
|
Avantika Dasgupta answered |
xRy => x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
Given, xRy => x - y + √2 is irrational ............1
and yRz => y - z + √2 is irrational ............2
Add equation 1 and 2, we get
(x - y + √2) + (y - z + √2) is irrational
= x - z + √2 is irrational
= xRz is irrational
So, the relation R is transitive.
If f(x) = x2 and g(x) = cosx, which of the following is true?- a)f + g is even function
- b)f – g is an odd function
- c)f + g is not defined
- d)f + g is an odd function
Correct answer is option 'A'. Can you explain this answer?
If f(x) = x2 and g(x) = cosx, which of the following is true?
a)
f + g is even function
b)
f – g is an odd function
c)
f + g is not defined
d)
f + g is an odd function
|
|
Raghav Bansal answered |
if f(x) is an odd function
So, f(−x)=−f(x)
F(−x)=cos(f(−x))
=cos(−f(x))
=cos(f(x))
=F(x)
So cos(f(x)) is an even function
So, f(x) and g(x) is an even function
The set A = {1,4,9,16,25—} in set builder form is written as- a)A = {x:x is a prime number}
- b)A ={x:x is the cube of a natural number}
- c)A = {x:x is the square of a natural number}
- d)A = {x:x is an even natural number}
Correct answer is 'C'. Can you explain this answer?
The set A = {1,4,9,16,25—} in set builder form is written as
a)
A = {x:x is a prime number}
b)
A ={x:x is the cube of a natural number}
c)
A = {x:x is the square of a natural number}
d)
A = {x:x is an even natural number}
|
|
Krishna Iyer answered |
- We know that, 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25
- Therefore the set A = {1, 4, 9, 16, 25...} can be written in set builder form as:
A = {x: x is the square of a natural number}
If ordered pair (a + 2b, 9) = (7, 3a + 2b), then the values of a and b are- a)9, 7
- b)1, 3
- c)7, 9
- d)3, 1
Correct answer is option 'B'. Can you explain this answer?
If ordered pair (a + 2b, 9) = (7, 3a + 2b), then the values of a and b are
a)
9, 7
b)
1, 3
c)
7, 9
d)
3, 1
|
Learners Habitat answered |
A+2B=7------(1)
3A+2B=9------(2)
Subtracting (2) from (1)
we get, 2A=2
A=1
Subsitute A=1 in (1)
1+2B=7
2B=6
B=3
therefore A = 1, B = 3
3A+2B=9------(2)
Subtracting (2) from (1)
we get, 2A=2
A=1
Subsitute A=1 in (1)
1+2B=7
2B=6
B=3
therefore A = 1, B = 3
For the set of all natural numbers the universal set can be- a)Set of all odd numbers
- b)Set of all even numbers
- c)Set of all integers
- d)Set of all prime numbers
Correct answer is option 'C'. Can you explain this answer?
For the set of all natural numbers the universal set can be
a)
Set of all odd numbers
b)
Set of all even numbers
c)
Set of all integers
d)
Set of all prime numbers
|
|
Krishna Iyer answered |
Integers contain all the natural numbers. So it can be a universal set for natural numbers. In other options, there are only some of the elements of natural numbers.
If A = {1, 2, 3 }, B = {4, 5, 6 }, which of the following is not a relation from A to Ba) R1 = {(1, 4), (1, 5),(1, 6)}b) R2 = {(1, 5), (2, 4), (3, 6)}c) R3 = {(4, 2), (2, 6), (5, 1), (2, 4)}d) R4 = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}Correct answer is option 'C'. Can you explain this answer?
|
Poonam Reddy answered |
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Can you explain the answer of this question below: Identify the null set from the following- A:Odd natural numbers divisible by 2
- B:The set which contains only zero as its element
- C:The set of even prime numbers
- D:The set of points common to two intersecting lines
The answer is a.
Identify the null set from the following
A:
Odd natural numbers divisible by 2
B:
The set which contains only zero as its element
C:
The set of even prime numbers
D:
The set of points common to two intersecting lines
|
Geetika Shah answered |
Since none of the odd number is divisible by 2, so it would be a null set.
If A = { (1, 2, 3}, then the relation R = {(2, 3)} in A isa)symmetric and transitive
|
|
Pallavi Ahuja answered |
Symmetric
The relation R is said to be symmetric if for every element (a, b) in R, the element (b, a) is also in R. In other words, if (a, b) ∈ R, then (b, a) ∈ R.
In this case, the relation R = {(2, 3)} has only one element. To check if it is symmetric, we need to check if (3, 2) is also in R. Since (3, 2) is not in R, the relation R is not symmetric.
Transitive
The relation R is said to be transitive if for every elements (a, b) and (b, c) in R, the element (a, c) is also in R. In other words, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
In this case, we need to check if (2, 3) ∈ R and (3, 2) ∈ R imply that (2, 2) ∈ R. Since (2, 3) ∈ R and (3, 2) is not in R, we cannot determine if (2, 2) ∈ R or not. Therefore, the relation R is not transitive.
Summary
- The relation R = {(2, 3)} is not symmetric because it does not contain the element (3, 2).
- The relation R is not transitive because we cannot determine if (2, 2) ∈ R or not based on the given relation R.
Therefore, the relation R = {(2, 3)} is neither symmetric nor transitive.
The relation R is said to be symmetric if for every element (a, b) in R, the element (b, a) is also in R. In other words, if (a, b) ∈ R, then (b, a) ∈ R.
In this case, the relation R = {(2, 3)} has only one element. To check if it is symmetric, we need to check if (3, 2) is also in R. Since (3, 2) is not in R, the relation R is not symmetric.
Transitive
The relation R is said to be transitive if for every elements (a, b) and (b, c) in R, the element (a, c) is also in R. In other words, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
In this case, we need to check if (2, 3) ∈ R and (3, 2) ∈ R imply that (2, 2) ∈ R. Since (2, 3) ∈ R and (3, 2) is not in R, we cannot determine if (2, 2) ∈ R or not. Therefore, the relation R is not transitive.
Summary
- The relation R = {(2, 3)} is not symmetric because it does not contain the element (3, 2).
- The relation R is not transitive because we cannot determine if (2, 2) ∈ R or not based on the given relation R.
Therefore, the relation R = {(2, 3)} is neither symmetric nor transitive.
If A = {1, 3, 6}, B = {4, 8} then A x A is- a){(1, 3), (1, 6), (3, 1) {(3, 6), (6, 1), (6, 3)}
- b){(1, 1), (1, 3), (1, 6), {(3, 1), (3, 3), (3, 6), {(6, 1), (6, 3), (6, 6)}
- c){(1, 4), (1, 8), (3, 4), {(3, 8), (6, 4), (6, 8)}
- d){(1, 1), (3, 3), (6, 6)}
Correct answer is option 'B'. Can you explain this answer?
If A = {1, 3, 6}, B = {4, 8} then A x A is
a)
{(1, 3), (1, 6), (3, 1) {(3, 6), (6, 1), (6, 3)}
b)
{(1, 1), (1, 3), (1, 6), {(3, 1), (3, 3), (3, 6), {(6, 1), (6, 3), (6, 6)}
c)
{(1, 4), (1, 8), (3, 4), {(3, 8), (6, 4), (6, 8)}
d)
{(1, 1), (3, 3), (6, 6)}
|
|
Raghav Bansal answered |
It is distributive law.
If in class XI of a certain school, 24 students got distinction in mathematics, 19 students got distinction in physics and 15 students got distinction in both the subjects,then the number of students in class XI is :- a)43
- b)58
- c)28
- d)39
Correct answer is option 'C'. Can you explain this answer?
If in class XI of a certain school, 24 students got distinction in mathematics, 19 students got distinction in physics and 15 students got distinction in both the subjects,then the number of students in class XI is :
a)
43
b)
58
c)
28
d)
39
|
Lohit Matani answered |
r :- c
Explanation:- Students got distinction in mathematics = 24
Student got distinction in physics = 19
Students got distinction in both= 15
Total number of students = (24+19)-15
= 28
Explanation:- Students got distinction in mathematics = 24
Student got distinction in physics = 19
Students got distinction in both= 15
Total number of students = (24+19)-15
= 28
U= set of all teachers in a schoolB= Set of all Mathematics teachers in a schoolSo B’=?- a)set of all non mathematics teachers
- b)set of all language teachers
- c)set of all non science teachers
- d)Set of all science teachers
Correct answer is option 'A'. Can you explain this answer?
U= set of all teachers in a schoolB= Set of all Mathematics teachers in a schoolSo B’=?
a)
set of all non mathematics teachers
b)
set of all language teachers
c)
set of all non science teachers
d)
Set of all science teachers
|
Nandini Iyer answered |
U = Set of all teachers in a school
B = Set of all Mathematics teachers in a school
B’ = U – B = Set of all Non-Mathematics teachers in a school
B = Set of all Mathematics teachers in a school
B’ = U – B = Set of all Non-Mathematics teachers in a school
Can you explain the answer of this question below: If P = {1, 2, 3, 4}, then the number of subsets in its power set will be- A:8
- B:16
- C:4
- D:6
The answer is b.
If P = {1, 2, 3, 4}, then the number of subsets in its power set will be
A:
8
B:
16
C:
4
D:
6
|
|
Om Desai answered |
Number of elements in a power set = 2n, where n = number of elements in the set P.
Hence, 24 = 16.
Hence, 24 = 16.
If n(A) = p and n(B) = q, then how many relations are there from A to B.- a)pq
- b)3pq
- c)2pq
- d)(pq)2
Correct answer is option 'C'. Can you explain this answer?
If n(A) = p and n(B) = q, then how many relations are there from A to B.
a)
pq
b)
3pq
c)
2pq
d)
(pq)2
|
|
Preeti Khanna answered |
Number of relation from A to B = 2n(A)×n(B)
n(A) = p, n(B) = q
⇒ Number of relation from A to B = 2pq
⇒ Number of relation from A to B = 2pq
Given the sets A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, then A ∪ (B ∩ C) is
- a){1, 2, 3, 4}
- b){1, 2, 3, 4, 5, 6}
- c){1, 2, 4, 5}
- d){3}
Correct answer is option 'A'. Can you explain this answer?
Given the sets A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, then A ∪ (B ∩ C) is
a)
{1, 2, 3, 4}
b)
{1, 2, 3, 4, 5, 6}
c)
{1, 2, 4, 5}
d)
{3}
|
|
Pooja Shah answered |
B ∩ C = {4}, as only 4 is common between B and C
∴ A ∪ (B ∩ C) = {1, 2, 3, 4}.
∴ A ∪ (B ∩ C) = {1, 2, 3, 4}.
Number of binary sets on the set {p, q,r} is:- a)39
- b)16
- c)18
- d)36
Correct answer is option 'A'. Can you explain this answer?
Number of binary sets on the set {p, q,r} is:
a)
39
b)
16
c)
18
d)
36
|
Guru Randhawa answered |
It os must to remember these basic results the answer is 1 at 8 corners 1/8 sphere is present :8*1/8=1 that means overall( effective) one atom is present in ssc
Which of the following is a finite set?- a)Set of roots of the equation x2 – 25 = 0
- b)Set of natural numbers
- c)The set of lines parallel to y – axis
- d)The set of numbers which are multiples of 5
Correct answer is 'A'. Can you explain this answer?
Which of the following is a finite set?
a)
Set of roots of the equation x2 – 25 = 0
b)
Set of natural numbers
c)
The set of lines parallel to y – axis
d)
The set of numbers which are multiples of 5
|
|
Hansa Sharma answered |
- A finite set is a set that has a finite number of elements.
- Since x2 – 25 = 0 has finite number of elements. So it is a finite set.
Choose the incorrect statement- a)If a set has only one element, we call it a singleton set.
- b)Set of all even prime numbers is a subset of set of all natural numbers.
- c)Φ is not a subset of any set.
- d)Every set is a subset of itself.
Correct answer is option 'C'. Can you explain this answer?
Choose the incorrect statement
a)
If a set has only one element, we call it a singleton set.
b)
Set of all even prime numbers is a subset of set of all natural numbers.
c)
Φ is not a subset of any set.
d)
Every set is a subset of itself.
|
Rohit Joshi answered |
A set A is a proper subset of a set B if A is a subset of B and there is at least one element of B that's not an element of A. Thus, the void set is a subset of all sets, and it's a proper subset of every set except itself
If A = {1, 2, 3}, and B = {3, 6} then the number of relations from A to B is
- a)32
- b)23
- c)23
- d)26
Correct answer is option 'D'. Can you explain this answer?
If A = {1, 2, 3}, and B = {3, 6} then the number of relations from A to B is
a)
32
b)
23
c)
23
d)
26
|
|
Preeti Iyer answered |
The number of relations between sets can be calculated using 2mn where m and n represent the number of members in each set.
So, number of relations from A to B is 26.
So, number of relations from A to B is 26.
A is a set with 6 elements. So, the number of subsets is:- a)12
- b)32
- c)24
- d)64
Correct answer is option 'D'. Can you explain this answer?
A is a set with 6 elements. So, the number of subsets is:
a)
12
b)
32
c)
24
d)
64
|
|
Neha Joshi answered |
{1, 2, 3, 4, 5, 6} is a set of 6 elements, so it has 26 = 64 subsets
Let R be a relation defined on the set of N natural numbers as R = {(x, y): y is a factor of x, x, y∈ N} then,- a)(2, 4) ∈ R
- b)(9, 5) ∈ R
- c)(4, 2) ∈ R
- d)(3, 9) ∈ R
Correct answer is option 'C'. Can you explain this answer?
Let R be a relation defined on the set of N natural numbers as R = {(x, y): y is a factor of x, x, y∈ N} then,
a)
(2, 4) ∈ R
b)
(9, 5) ∈ R
c)
(4, 2) ∈ R
d)
(3, 9) ∈ R
|
|
Om Desai answered |
R = {(x, y): y is a factor of x, x, y∈ N}
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R
Choose the correct statementa){2} ⊆ {1,4,6}b){2,4,6} ⊂ {2}c){2} ⊂ {2,4,6}d){2} ⊇ {2,4,6}Correct answer is option 'C'. Can you explain this answer?
|
|
Leelu Bhai answered |
Yes option C is correct as {2} is a subset of {2, 4, 6}.
In rule method the null set is represented by- a){}
- b){x: x ≠ x}
- c){x : x = x}
- d)φ
Correct answer is option 'B'. Can you explain this answer?
In rule method the null set is represented by
a)
{}
b)
{x: x ≠ x}
c)
{x : x = x}
d)
φ
|
Aarav Sharma answered |
It is fundamental concept.
Let A = {1, 2, 3, …, 50} and R be the relation “is square of” on A. Then R as a subset of A x A is written as:
- a){(1, 1), (2, 2), (3, 3), (4, 4) (5, 5), (6, 6), (7, 7)}
- b){(1, 1), (2, 5), (3, 9), (4, 15) (5, 25), (6, 36), (7, 49)}
- c){(1, 1), (4, 4), (9, 9), (16, 16) (25, 25), (36, 36), (49, 49)}
- d){(1, 1), (4, 2), (9, 3), (16, 4) (25, 5), (36, 6), (49, 7)}
Correct answer is option 'D'. Can you explain this answer?
Let A = {1, 2, 3, …, 50} and R be the relation “is square of” on A. Then R as a subset of A x A is written as:
a)
{(1, 1), (2, 2), (3, 3), (4, 4) (5, 5), (6, 6), (7, 7)}
b)
{(1, 1), (2, 5), (3, 9), (4, 15) (5, 25), (6, 36), (7, 49)}
c)
{(1, 1), (4, 4), (9, 9), (16, 16) (25, 25), (36, 36), (49, 49)}
d)
{(1, 1), (4, 2), (9, 3), (16, 4) (25, 5), (36, 6), (49, 7)}
|
Tejas Verma answered |
Since x is the square of y in relation R.
i.e. option D is correct.
i.e. option D is correct.
If 
= 22 then x and y =- a)x=3,y=1
- b)x = 3, y=6
- c)x=1 ,y=2
- d)
x = 2, y = 1
Correct answer is option 'D'. Can you explain this answer?
If
= 22 then x and y =
a)
x=3,y=1
b)
x = 3, y=6
c)
x=1 ,y=2
d)
x = 2, y = 1
|
Lavanya Menon answered |
R × ( P' ∪ Q' )'=R × [( P' )' ∩ ( Q' )' ] =R × (P ∩ Q) = (R × P) ∩ (R × Q)
If U= set of all whole numbers less than 12, A=set of all whole numbers less than 10, B= Set of all odd natural numbers less than 10, then what is (A∩B)’?- a){3,5,7,9}
- b){0,1,3,5,7,9}
- c){0,2,4,6,8,10,11}
- d){1,3,5,7}
Correct answer is option 'A'. Can you explain this answer?
If U= set of all whole numbers less than 12, A=set of all whole numbers less than 10, B= Set of all odd natural numbers less than 10, then what is (A∩B)’?
a)
{3,5,7,9}
b)
{0,1,3,5,7,9}
c)
{0,2,4,6,8,10,11}
d)
{1,3,5,7}
|
|
Hansa Sharma answered |
U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}
A = set of all trianglesB = set of all right triangles
A – B=?- a)set of all equilateral triangles
- b)set of all triangles which have all three angles different
- c)set of all isosceles triangles
- d)set of all triangles which do not have one right angle
Correct answer is option 'D'. Can you explain this answer?
A = set of all triangles
B = set of all right triangles
A – B=?
A – B=?
a)
set of all equilateral triangles
b)
set of all triangles which have all three angles different
c)
set of all isosceles triangles
d)
set of all triangles which do not have one right angle
|
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Geetika Shah answered |
A = Set of all triangles
B = Set of all right triangles
A – B = Set of triangles after removing all right triangles from A
= Set of all triangles which do not have right angle
B = Set of all right triangles
A – B = Set of triangles after removing all right triangles from A
= Set of all triangles which do not have right angle
Can you explain the answer of this question below:Let f: Q→Q be a function given by f(x) = x2,then f -1(9) =
- A:
{3}
- B:
f
- C:
{-3, 3}
- D:
{-3}
The answer is c.
Let f: Q→Q be a function given by f(x) = x2,then f -1(9) =
{3}
f
{-3, 3}
{-3}
|
Manish Suthar answered |
Function has only one image not 2 . is it right So the answer should be 3
If A = {2, 4, 6, 8} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9,}, then A’=- a){1,3,5,7,9}
- b){1,3,5,7}
- c){1,3,5,6,9}
- d){1,3,5,8,9}
Correct answer is 'A'. Can you explain this answer?
If A = {2, 4, 6, 8} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9,}, then A’=
a)
{1,3,5,7,9}
b)
{1,3,5,7}
c)
{1,3,5,6,9}
d)
{1,3,5,8,9}
|
|
Pooja Shah answered |
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8}
A’ = U – A
A’ = {1, 3, 5, 7, 9}
A = {2, 4, 6, 8}
A’ = U – A
A’ = {1, 3, 5, 7, 9}
Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R is- a)An Equivalence Relation
- b)Only Reflexive
- c)Symmetric and reflexive.
- d)Only Transitive
Correct answer is option 'C'. Can you explain this answer?
Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R is
a)
An Equivalence Relation
b)
Only Reflexive
c)
Symmetric and reflexive.
d)
Only Transitive
|
|
Anaya Patel answered |
(m, n) R (p, q) <=> mq(n + p) = np(m + q)
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.
A—B is read as ?- a) Difference of B and A
- b) Difference of A and B of B and A
- c) None of the above
- d) Both a and b
Correct answer is option 'B'. Can you explain this answer?
A—B is read as ?
a)
Difference of B and A
b)
Difference of A and B of B and A
c)
None of the above
d)
Both a and b
|
|
Gaurav Kumar answered |
x2 = 16 ⇒ x = ±4 and 2x = 6 ⇒x = 3
There is no value of x which satisfies both the above equations. Thus, A = φ
Given 
1 and B = {0, 1, 2, 3, 4, 5, 6} then A x B is- a){ (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}
- b)Φ
- c){(3, 5), (3, 6), (4, 5), (4, 6)}
- d)Infinite
Correct answer is option 'B'. Can you explain this answer?
Given 1 and B = {0, 1, 2, 3, 4, 5, 6} then A x B is
1 and B = {0, 1, 2, 3, 4, 5, 6} then A x B isa)
{ (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}
b)
Φ
c)
{(3, 5), (3, 6), (4, 5), (4, 6)}
d)
Infinite
|
|
Krishna Iyer answered |
A = Φ
B = {0, 1, 2, 3, 4, 5, 6}
A × B = Φ (Cartesian product with an empty set is an empty set)
B = {0, 1, 2, 3, 4, 5, 6}
A × B = Φ (Cartesian product with an empty set is an empty set)
Which of the following statement is false?- a)If R is a relation from A to B, then domain of R is a subset of A and range of R is a subset of B.
- b)If A = {1, 2, 3, 4} and B = {5, 6}, then the number of relations from A to B is 256.
- c)A subset of A × B is called a relation from the set A to set B.
- d)If A = {x, y, z} and B = {1, 2} then the number of relations from A to B is 22.
Correct answer is option 'D'. Can you explain this answer?
Which of the following statement is false?
a)
If R is a relation from A to B, then domain of R is a subset of A and range of R is a subset of B.
b)
If A = {1, 2, 3, 4} and B = {5, 6}, then the number of relations from A to B is 256.
c)
A subset of A × B is called a relation from the set A to set B.
d)
If A = {x, y, z} and B = {1, 2} then the number of relations from A to B is 22.
|
|
Riya Banerjee answered |
Option D: The number of relations from A to B is 23*2= 26
, where a, b are the elements of a set of non-zero rational numbers, isFrom the sets given below, select equal sets :A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4},
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}- a)A and C
- b)A and B
- c)B and D
- d)B and C
Correct answer is option 'C'. Can you explain this answer?
From the sets given below, select equal sets :
A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4},
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}
a)
A and C
b)
A and B
c)
B and D
d)
B and C
|
Mansi Chopra answered |
The sets are equal, if they have the exact same elements in them. Since option B & D have exactly same number of elements in them So, B & D are equal sets.
If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to- a)f(c) + g(a)
- b)f(a) – g(c)
- c)f(d) – g(b)
- d)f(b) – g(b)
Correct answer is option 'C'. Can you explain this answer?
If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to
a)
f(c) + g(a)
b)
f(a) – g(c)
c)
f(d) – g(b)
d)
f(b) – g(b)
|
|
Leelu Bhai answered |
F(g(x)) - g(f(x)) = f(cx + d) - g(ax + b)= a(cx + d) + b - c(ax + b) - d= acx + ad + b - acx - bc - d= ad + b - bc - dnow by putting each
If the number of outcomes in collection are 2 and the distinct outcomes are 4 then the count value according to permutations is- a)2
- b)12
- c)24
- d)4
Correct answer is option 'C'. Can you explain this answer?
If the number of outcomes in collection are 2 and the distinct outcomes are 4 then the count value according to permutations is
a)
2
b)
12
c)
24
d)
4
|
|
Naina Sharma answered |
The number of non- empty subsets =
2n − 1 = 24 − 1 = 16 − 1 = 15
2n − 1 = 24 − 1 = 16 − 1 = 15
Can you explain the answer of this question below: If A = {-1, 1, 2}, then n(A x A x A) =
- A:
9
- B:
16
- C:
27
- D:
3
The answer is c.
If A = {-1, 1, 2}, then n(A x A x A) =
9
16
27
3
|
|
Om Desai answered |
n(A) = 3
n(A × A × A) = n(A) * n(A) * n(A)
n(A × A × A) = 3 * 3 * 3 = 27
n(A × A × A) = n(A) * n(A) * n(A)
n(A × A × A) = 3 * 3 * 3 = 27
Consider the set A of all divisors of 30. How many subsets of A contains even divisors only?- a)2
- b)16
- c)28
- d)4
Correct answer is 'B'. Can you explain this answer?
Consider the set A of all divisors of 30. How many subsets of A contains even divisors only?
a)
2
b)
16
c)
28
d)
4
|
|
Himaja Ammu answered |
Set of divisors of 30={1,2,3,6,10,15,30} in these the even divisors r={2,6,10,30} we know no.of subsets to any set=2^n so answer is 2^4=16
The number of elements in the Power set P(S) of the set S = [ [ Φ] , 1, [ 2, 3 ]] is- a)2
- b)4
- c)8
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The number of elements in the Power set P(S) of the set S = [ [ Φ] , 1, [ 2, 3 ]] is
a)
2
b)
4
c)
8
d)
None of these
|
|
Preeti Iyer answered |
There’s a result in mathematics used for this. It says that a power set B of any set A is a set of all the subsets of A and the number of elements of B will be 2^n where n is the number of elements of A.
So taking your question as an example;
A = {1,2,3}
B : set of all subsets of A
List out all the subsets of A - {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}
Number of elements in A (n) = 3
so 23 = 8
So, B = {{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}}
and the number of elements are 8.
So taking your question as an example;
A = {1,2,3}
B : set of all subsets of A
List out all the subsets of A - {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}
Number of elements in A (n) = 3
so 23 = 8
So, B = {{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}}
and the number of elements are 8.
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