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. For any two sets A and B, A-(A-B) equals
Solution
(c) Now, A-(A-B)=A-(A-B^C) =A〖∩(A∩B^C )〗^C =A∩(A^C∪B) =(A∩A^C )∪(A∩B) =A∩B
(c) Now, A-(A-B)=A-(A-B^C) =A〖∩(A∩B^C )〗^C =A∩(A^C∪B) =(A∩A^C )∪(A∩B) =A∩B
Q2.If A and B are two given sets,
then A∩(A∩B)^c is equal to
Solution
(d) We have, A∩(A∩B)^c=A∩(A^c∪B^c) ⇒A∩(A∩B)^c=(A∩A^c)∪(A∩B^c) ⇒A∩(A∩B)^c=Ï•∪(A∩B^c )=A∩B^c
(d) We have, A∩(A∩B)^c=A∩(A^c∪B^c) ⇒A∩(A∩B)^c=(A∩A^c)∪(A∩B^c) ⇒A∩(A∩B)^c=Ï•∪(A∩B^c )=A∩B^c
Q3. Let Z denote the set of integers, then
{x∈Z:|x-3|<4}n{x∈Z:|x-4|<5}=
Solution
(c) We have, {x∈Z:|x-3|<4}={x∈Z:-1
(c) We have, {x∈Z:|x-3|<4}={x∈Z:-1
Q4. The relation ‘is not equal to’ is defined on R, is
Solution
(b) For any a,b∈R a≠b⇒b≠a⇒R is symmetric Clearly, 2≠-3 and -3≠2, but 2=2. So, R is not transitive. Clearly, R is not reflexive
(b) For any a,b∈R a≠b⇒b≠a⇒R is symmetric Clearly, 2≠-3 and -3≠2, but 2=2. So, R is not transitive. Clearly, R is not reflexive
Q5.
Let L denote the set of all straight lines in a plane. Let a relation R be defined by α R β⇔α⊥β,α,β∈L.
Then R is
Solution
Symmetric
Symmetric
Q6. An investigator interviewed 100 students to determine the
performance of three drinks milk, coffee and tea. The investigator reported that 10 students take all
three drinks milk, coffee and tea; 20 students take milk and coffee, 30 students take coffee and tea, 25
students take mile and tea, 12 students take milk only, 5 students take coffee only and 8 students take
tea only. Then, the number of students who did not take any of the three drinks, is
Solution
B
B
Q7.
If n(A×B)=45, then n(A) cannot be
Solution
(b) We have, n(A×B)=45 ⇒n(A)×n(B)=45 ⇒n(A) and n(B) are factors of 45 such that their product is 45 Hence, n(A) cannot be 17
(b) We have, n(A×B)=45 ⇒n(A)×n(B)=45 ⇒n(A) and n(B) are factors of 45 such that their product is 45 Hence, n(A) cannot be 17
Q8.
Let A and B have 3 and 6 elements respectively. What can be the minimum number of elements in
A∪B?
Solution
(b) A∪B will contain minimum number of elements if A⊂B and in that case, we have n(A∪B)=n(B)=6
(b) A∪B will contain minimum number of elements if A⊂B and in that case, we have n(A∪B)=n(B)=6
Q9.
If A,B and C are three sets such that A⊃B⊃C, then (A∪B∪C)-(A∩B∩C)=
Solution
(c) We have, A⊃B⊃C ∴A∪B∪C=A and A∩B∩C=C ⇒(A∪B∪C)-(A∩B∩C)=A-C
(c) We have, A⊃B⊃C ∴A∪B∪C=A and A∩B∩C=C ⇒(A∪B∪C)-(A∩B∩C)=A-C
Q10. Let
A={ONGC,BHEL,SAIL,GAIL,IOCL} and R be a relation defined as “two elements of A are related if they share
exactly one letter”. The relation R is
Solution
(c) Clearly, R={(BHEL,SAIL),(SAIL,BHEL),(BHEL,GAIL), (GAIL,BHEL),(BHEL,IOCL),(IOCL,BHEL)} We observe that R is symmetric only
(c) Clearly, R={(BHEL,SAIL),(SAIL,BHEL),(BHEL,GAIL), (GAIL,BHEL),(BHEL,IOCL),(IOCL,BHEL)} We observe that R is symmetric only