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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. The relation “is a factor of” on the set N of all natural numbers is not
•  Reflaxive
•  Symmetric
•  Antisymmetric
•  Transitive
Solution
Clearly, 2 is a factor of 6 but 6 is not a factor of 2. So, the relation ‘is factor of’ is not symmetric. However, it is reflexive and transitive

Q2.Which of the following is an equivalence relation?
•  Is father of
•  Is less than
•  Is congruent to
•  Is an uncle of
Solution
No solution available

Q3.  If a N={a x∶x∈N} and b N∩c N=d N, where b,c∈N then
•  d=bc
•  c=bd
•  b=cd
•  NONE OF THESE
Solution
We have, b N={b x│x∈N}= Set of positive integral multiples of b c N={c x│x∈N}= Set of positive integral multiples of c ∴c N={c x | x∈N}= Set of positive integral multiples of b and c both ⇒d=1.c.m.of b and c

Q4. The relation R={(1,3),(3,5)} is defined on the set with minimum number of elements of
natural numbers. The minimum number of elements to be included in R so that R is an
equivalence relation, is
•  5
•  6
•  7
•  8
Solution
To make R a reflexive relation, we must have (1,1),(3,3) and (5,5) in it. In order to make R a symmetric relation, we must inside (3,1) and (5,3) in it. Now, (1,3)∈R and (3,5)∈R. So, to make R a transitive relation, we must have, (1,5)∈R. But, R must be symmetric also. So, it should also contain (5,1). Thus, we have R={(1,1),(3,3),(5,5),(1,3),(3,5),(3,1),(5,3),(1,5),(5,1)} Clearly, it is an equivalence relation on A{1,3,5}

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
•  Reflaxive
•  Symmetric
•  Transitive
•  None of these
Solution
No solution available

Q6. Let R={(a,a)} be a relation on a set A. Then, R is
•  Symmetric
•  Asymmetric
•  Symmetric and Antisymmetric
•  Neither symmetric nor antisymmetric
Solution
No solution available

Q7.The number of elements in the set {(a,b):2a2+3b2=35,a,b∈Z}, where Z is the set of all integers, is
• 2
•  4
•  8
•  12
Solution
The possible sets are {±2,±3} and {±4,±1}; therefore, number of elements in required set is 8. A∩B=ϕ⇒B-A=B

Q8.If A,B and C are three sets such that A∩B=A∩Cand A∪B=A∪C, then
• A=C
•  B=C
•  A∩B=ϕ
•  A=B
Solution
Given, A∩B=A∩C and A∪B=A∪C ⇒B=C

Q9.Let X={1,2,3,4,5} and Y={1,3,5,7,9}. Which of the following is/are not relations from X to Y?
•  R1={(x,y)|y=2+x,x∈X,y∈Y}
•  R2={(1,1),(2,1),(3,3),(4,3),(5,5)}
•  R3={(1,1),(1,3),(3,5),(3,7),(5,7)}
•  R4={(1,3),(2,5),(2,4),(7,9)}
Solution
R4 is not a relation from A to B, because (7,9)∈R4 but (7,9)∉A×B

Q10. In a class of 175 students the following data shows the number of students opting one or
more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30;
Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and
Chemistry 18. How many students have offered Mathematics alone?

•  35
•  48
•  60
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