(A – B) ∩ (A – C)
Hint:
Given A, B and C are any three sets.
Now, A – (B ∪ C) = (A – B) ∩ (A – C)

Difference of A and B of B and A


Hint:


A – B will read as difference of A and B of B and A


Ex: Let A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7}


Now, A – B = {2, 4}

(A × B) ∪ (A × C)


Hint:


Given A, B and C are any three sets.


Now, A × (B ∪ C) = (A × B) ∪ (A × C)

[5, 6, 7, 8, 9]


Hint:


Given A = [5, 6, 7] and B = [7, 8, 9]


then A ∪ B = [5, 6, 7, 8, 9]

2 and 3


Hint:


2 and 3 is the null set.

Range R = {1, 3, 5}


Hint:


Given R = {(2, 1),(4, 3),(4, 5)}


then Range(R) = {1, 3, 5}

{0, 1, 4, 9, 16, 25, 36, 49, 64, 81}


Hint:


The set S consists of the square of an integer less than 100


So, S = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}

41


Hint:


The number of students who took at least one of the three subjects can be found by finding out A ∪ B ∪ C, where A is the set of those who took Physics, B the set of those who took Chemistry and C the set of those who opted for Math.


Now, A ∪ B ∪ C = A + B + C – (A ∩ B + B ∩ C + C ∩ A) + (A ∩ B ∩ C)


A is the set of those who opted for Physics = 120/2 = 60 students


B is the set of those who opted for Chemistry = 120/5 = 24


C is the set of those who opted for Math = 120/7 = 17


The 10th, 20th, 30th….. numbered students would have opted for both Physics and Chemistry.


Therefore, A ∩ B = 120/10 = 12


The 14th, 28th, 42nd….. Numbered students would have opted for Physics and Math.


Therefore, C ∩ A = 120/14 = 8


The 35th, 70th…. numbered students would have opted for Chemistry and Math.


Therefore, B ∩ C = 120/35 = 3


And the 70th numbered student would have opted for all three subjects.


Therefore, A ∪ B ∪ C = 60 + 24 + 17 – (12 + 8 + 3) + 1 = 79


Number of students who opted for none of the three subjects = 120 – 79 = 41

symmetric


Hint:


Given {(a, b) : a² + b² = 1} on the set S.


Now a² +b² = b² + a² = 1


So, the given relation is symmetric.

6, 4


Hint:


Let A and B be two sets having m and n numbers of elements respectively


Number of subsets of A = 2m


Number of subsets of B = 2n


Now, according to question


2m – 2n = 48


⇒ 2n(2m – n – 1) = 24(22 – 1)


So, n = 4


and m – n = 2


⇒ m – 4 = 2


⇒ m = 2 + 4


⇒ m = 6

(- ∞, ∞)


Hint:


Let the given function is


y = 3x – 2


⇒ y + 2 = 3x


⇒ x = (y + 2)/3


Now x is saisfied by all values.


So, Range{f(x)} = R = (-∞, ∞)

B = C


Hint:


Given A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C then B = C

B = C

Hint:

Given A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C then B = C

X < 0, Y > 0


Hint: In the second quadrant,





X < 0, Y > 0

Infinite


Hint:


There are infinite many rational and irrational numbers are possible between 0 and 1


This is because between any two numbers, there are infinite numbers.

Finite Set


Hint:


In mathematics, and more specifically set theory, the empty set is the unique set having no elements and its size or cardinality (count of elements in a set) is zero.


So, an empty set is a finite set.

[5, 6, 7, 8, 9]


Hint:


Given A = [5, 6, 7] and B = [7, 8, 9]


then A U B = [5, 6, 7, 8, 9]

A = {1, 2, 3} and B = {2, 1, 3}


Hint:


Two sets are equal if and only if they have the same elements.


So, A = {1, 2, 3} and B = {2, 1, 3} are equal sets.

27


Hint:


Total students = 50


Students who did not opt for math = 10


Students who did not opt for Science = 15


Students who did not opt for either maths or science = 2


Total of 40 students in math and 13 did not opt for science but did for math

= 40 – 13 = 27


So, students of the class opted for both math and science is 27