In mathematics a set is a collection of distinct
elements. The elements that make up a set can be any kind of things: people, letters of the alphabet,
numbers, points in space, lines, other geometrical shapes, variables, or even other sets. Two sets are equal if
and only if they have precisely the same elements.
Sets are ubiquitous in modern mathematics. Indeed, set theory, more specifically Zermelo–Fraenkel set theory, has
been the standard way to provide rigorous foundations for all branches of mathematics since the first half of the
20th century..

**Q1.**If A_n is the set of first n prime numbers, then⋃_(n=2)^10▒〖A_n=〗

Solution

(b) Clearly, A_2⊂A_3⊂A_4⊂⋯⊂A_10 ∴⋃_(n=2)^10▒〖A_n=A_10={2,3,5,7,11,13,17,19,23,29}〗

(b) Clearly, A_2⊂A_3⊂A_4⊂⋯⊂A_10 ∴⋃_(n=2)^10▒〖A_n=A_10={2,3,5,7,11,13,17,19,23,29}〗

**Q2.**If n(u)=100,n(A)=50,n(B)=20 and n(A∩B)=10, then n{(A∪B)^c }

Solution

(c) n((A∪B)^c }=n(U)-n(A∪B) =n(U)-{n(A)+n(B)-n(A∩B)} =100-(50+20-10)=40

(c) n((A∪B)^c }=n(U)-n(A∪B) =n(U)-{n(A)+n(B)-n(A∩B)} =100-(50+20-10)=40

**Q3.**If A={p:p=(n+2)(2n^5+3n^4+4n^3+5n^2+6)/(n^2+2n),n,p∈Z^+} then the number of elements in the set A, is

Solution

(c) We have, p=((n+2)(2n^5+3n^4+4n^3+5n^2+6))/(n^2+2n) ⇒p=2n^4+3n^3+4n^2+5n+6/n Clearly, p∈Z^+ iff n=1,2,3,6. So, A has 4 elements

(c) We have, p=((n+2)(2n^5+3n^4+4n^3+5n^2+6))/(n^2+2n) ⇒p=2n^4+3n^3+4n^2+5n+6/n Clearly, p∈Z^+ iff n=1,2,3,6. So, A has 4 elements

**Q4.**A relation between two persons is defined as follows: aRb⇔a and b born in different months. Then, R is

Solution

(b) Let (a,b)∈R. Then, a and b are born in different months ⇒(b,a)∈R So, R is symmetric Clearly, R is neither reflexive nor transitive

(b) Let (a,b)∈R. Then, a and b are born in different months ⇒(b,a)∈R So, R is symmetric Clearly, R is neither reflexive nor transitive

**Q5.**Two finite sets have m and n elements. The number of elements in the power set of first set is 48 more than the total number of elements in the power set of the second set. Then, the value of M and N are

Solution

(c) According to question, 2^m-2^n=48 This is possible only if m=6 and n=4.

(c) According to question, 2^m-2^n=48 This is possible only if m=6 and n=4.

**Q6.**Which of the following is true?

Solution

(b) A∩Ï•=Ï• is true.

(b) A∩Ï•=Ï• is true.

**Q7.**If n(A∩B=10,n(B∩C)=20) and n(A∩C)=30, then the greatest possible value of n(A∩B∩C) is

Solution

(c) The greatest possible value of n(A∩B∩C) is the least amongst the values n(A∩B),n(B∩C) and n(A∩C) i.e. 10

(c) The greatest possible value of n(A∩B∩C) is the least amongst the values n(A∩B),n(B∩C) and n(A∩C) i.e. 10

Solution

(c) The set A is the set of all points on the hyperbola xy=1 having its two branches in the first and third quadrants, while the set B is the set of all points on y=-x which lies in second and four quadrants. These two curves do not intersect. Hence, A∩B=Ï•.

(c) The set A is the set of all points on the hyperbola xy=1 having its two branches in the first and third quadrants, while the set B is the set of all points on y=-x which lies in second and four quadrants. These two curves do not intersect. Hence, A∩B=Ï•.

**Q9.**If X={4^n-3n-1∶n∈N} and Y={9(n-1):n∈N}, then X∪Y is equal to

Solution

(b) We have, X= Set of some multiple of 9 and, Y= Set of all multiple of 9 ∴X⊂Y⇒X∪Y=Y

(b) We have, X= Set of some multiple of 9 and, Y= Set of all multiple of 9 ∴X⊂Y⇒X∪Y=Y

**Q10.**If R is a relation from a finite set A having m elements to a finite set B having n elements, then the number of relations from A to B is

Solution

(c) Since, A=B∩C and B=C∩A, Then A≡B

(c) Since, A=B∩C and B=C∩A, Then A≡B