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={a x∶x∈N} and b N ∩c N=d N, where b,c∈N are relatively prime, then

Solution

(a) We have, b N={b x│x∈N}= Set of positive integral multiples of b c N={c x│x∈N}= Set positive integral multiples of c ∴bN∩cN= Set of positive integral multiples of bc ⇒bN∩cN=bc N [∵b and c are prime] Hence, d=bc

(a) We have, b N={b x│x∈N}= Set of positive integral multiples of b c N={c x│x∈N}= Set positive integral multiples of c ∴bN∩cN= Set of positive integral multiples of bc ⇒bN∩cN=bc N [∵b and c are prime] Hence, d=bc

**Q2.**Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n are

Solution

(b) It is given that 2^m-2^n=56 Obviously, m=6,n=3 satisfy the equation

(b) It is given that 2^m-2^n=56 Obviously, m=6,n=3 satisfy the equation

**Q3.**{n(n+1)(2n+1):n∈Z}⊂

Solution

(a) Let A={n(n+1)(2n+1):n∈Z} Putting n=±1,±2,…., we get A={…-30,-6,0,6,30,…} ⇒ {n(n+1)(2n+1):n∈Z}⊂{6k:k∈Z}

(a) Let A={n(n+1)(2n+1):n∈Z} Putting n=±1,±2,…., we get A={…-30,-6,0,6,30,…} ⇒ {n(n+1)(2n+1):n∈Z}⊂{6k:k∈Z}

**Q4.**If a set A contains n elements, then which of the following cannot be the number of reflexive relations on the set A?

Solution

(d) A relation on set A is a subset of A×A Let A={a_1,a_2,…,a_n }. Then, a reflexive relation on A must contain at least n elements (a_1,a_1 ),(a_2,a_2 ),…,(a_n,a_n) ∴ Number of reflexive relations on A is 2^(n^2-n) Clearly, n^2-n=n,n^2-n=n-1,n^2-n=n^2-1 have solutions in N but n^2-n=n+1 is not solvable in N. So, 2^(n+1) cannot be the number of reflexive relations on A

(d) A relation on set A is a subset of A×A Let A={a_1,a_2,…,a_n }. Then, a reflexive relation on A must contain at least n elements (a_1,a_1 ),(a_2,a_2 ),…,(a_n,a_n) ∴ Number of reflexive relations on A is 2^(n^2-n) Clearly, n^2-n=n,n^2-n=n-1,n^2-n=n^2-1 have solutions in N but n^2-n=n+1 is not solvable in N. So, 2^(n+1) cannot be the number of reflexive relations on A

**Q5.**15. Universal set, U={x:x^5-6x^4+11x^3-6x^2=0} And A={x:x^2-5x+6=0} B={x:x^2-3x+2=0} Then, (A∩B)' is equal to

Solution

(c) U={x:x^5+6x^4+11x^3-6x^2=0}={0,1,2,3} A={x:x^2-5x+6=0}={2,3} And B={x:x^2-3x+2=0}={2,1} ∴ (A∩B)^'=U-(A∩B) ={0,1,2,3}-{2}={0,1,3}

(c) U={x:x^5+6x^4+11x^3-6x^2=0}={0,1,2,3} A={x:x^2-5x+6=0}={2,3} And B={x:x^2-3x+2=0}={2,1} ∴ (A∩B)^'=U-(A∩B) ={0,1,2,3}-{2}={0,1,3}

**Q6.**16. If S is the set of squares and R is the set of rectangles, then (S∪R)-(S∩S) is

Solution

(d) Clearly, S⊂R ∴S∪R=R and S∩R=S ⇒(S∩R)-(S∩R)= Set of rectangles which are not squares

(d) Clearly, S⊂R ∴S∪R=R and S∩R=S ⇒(S∩R)-(S∩R)= Set of rectangles which are not squares

**Q7.**In rule method the null set is represented by

Solution

C

C

**Q8.**A,B and C are three non-empty sets. If A⊂B and B⊂C, then which of the following is true? Let A and B have 3 and 6 elements respectively. What can be the minimum number of elements in A∪B?

Solution

(c) We have, A⊂B and B⊂C ∴A∪B=B and B∩C=B ⇒A∪B=B∩C

(c) We have, A⊂B and B⊂C ∴A∪B=B and B∩C=B ⇒A∪B=B∩C

**Q9.**If A and B are two sets, then A∩(A∪B) equals

Solution

(a) We have, A⊂A∪B ⇒A∩(A∪B)=A

(a) We have, A⊂A∪B ⇒A∩(A∪B)=A

**Q10.**20. If A={(x,y):y^2=x;x,y∈R} and B={(x,y):y=|x|;x,y∈R}, then

Solution

(d) Clearly, y^2=x and y=|x| intersect at (0,0),(1,1) and (-1,-1). Hence, option (d) is correct

(d) Clearly, y^2=x and y=|x| intersect at (0,0),(1,1) and (-1,-1). Hence, option (d) is correct