As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
Q1. Consider the following statements:
p∶ Every reflexive relation is symmetric relation
q∶ Every anti-symmetric relation is reflexive
Which of the following is/are true?
p∶ Every reflexive relation is symmetric relation
q∶ Every anti-symmetric relation is reflexive
Which of the following is/are true?
Solution
Clearly, none of the statements is true
Clearly, none of the statements is true
Q2.If R is a relation on a finite set having n elements, then the number of relations on A is
Solution
No solution available
No solution available
Q3. 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
are related if they share exactly one letter”. The relation R is
Solution
Clearly, R={(BHEL,SAIL),(SAIL,BHEL),(BHEL,GAIL), (GAIL,BHEL),(BHEL,IOCL),(IOCL,BHEL)} We observe that R is symmetric only
Clearly, R={(BHEL,SAIL),(SAIL,BHEL),(BHEL,GAIL), (GAIL,BHEL),(BHEL,IOCL),(IOCL,BHEL)} We observe that R is symmetric only
Q4. Let A be a set of compartments in a train. Then the relation R defined on A as aRb iff "a and
b have the link between them”, then which of the following is true for R?
Solution
No solution available
No solution available
Q5.Consider the following statements:
(i) Every reflexive relation is antisymmetric
(ii) Every symmetric relation is antisymmetric
Which one among (i) and (ii) is true?
(i) Every reflexive relation is antisymmetric
(ii) Every symmetric relation is antisymmetric
Which one among (i) and (ii) is true?
Solution
No solution available
No solution available
Q6. Let A={1,2,3,4},B={2,4,6}. Then, the number of sets C such that A∩B⊆C⊆A ∪B is
Solution
A∩B={2,4} {A∩B}⊆{1,2,4},{3,2,4},{6,2,4},{1,3,2,4}, {1,6,2,4},{6,3,2,4},{2,4},{1,3,6,2,4}⊆A∪B ⇒ n(C)=8
A∩B={2,4} {A∩B}⊆{1,2,4},{3,2,4},{6,2,4},{1,3,2,4}, {1,6,2,4},{6,3,2,4},{2,4},{1,3,6,2,4}⊆A∪B ⇒ n(C)=8
Q7.The relation R defined in N as a R b⟺b is divisible by a is
Solution
For any a∈N, we have a|a Therefore R is reflexive R is not symmetric, because a R b does not imply that b R a
For any a∈N, we have a|a Therefore R is reflexive R is not symmetric, because a R b does not imply that b R a
Q8.Given the relation R={(1,2),(2,3)} on the set A={1,2,3}, the minimum number of ordered
pairs which when added to R make it an equivalence relation is
pairs which when added to R make it an equivalence relation is
Solution
R is reflexive if it contains (1,1),(2,2),(3,3) ∵(1,2)∈R,(2,3)∈R ∵R is symmetric, if (2,1),(3,2)∈R Now, R={(1,1),(2,2),(3,3),(2,1),(3,2),(2,3),(1,2)} R will be transitive, if (3,1),(1,3)∈R Thus, R becomes an equivalence relation by adding (1,1)(2,2)(3,3),(2,1)(3,2),(1,3),(3,1). Hence, the total number of ordered pairs is 7
R is reflexive if it contains (1,1),(2,2),(3,3) ∵(1,2)∈R,(2,3)∈R ∵R is symmetric, if (2,1),(3,2)∈R Now, R={(1,1),(2,2),(3,3),(2,1),(3,2),(2,3),(1,2)} R will be transitive, if (3,1),(1,3)∈R Thus, R becomes an equivalence relation by adding (1,1)(2,2)(3,3),(2,1)(3,2),(1,3),(3,1). Hence, the total number of ordered pairs is 7
Q9.Let n(U)=700,n(A)=200,n(B)=300 and n(A∩B)=100. Then, n(Ac∩Bc )=
Solution
We have, n(Ac∩Bc) =n{(A∪B)c } =n(U)-n(A∪B) =n(U)-{n(A)+n(B)-n(A∩B) } =700-(200+300-100)=300
We have, n(Ac∩Bc) =n{(A∪B)c } =n(U)-n(A∪B) =n(U)-{n(A)+n(B)-n(A∩B) } =700-(200+300-100)=300
Q10. The value of (A∪B∪C) ∩(A∩Bc∩Cc )^C ∩Ccis