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.**If A={1,2,3},B={a,b}, then A×B mapped A to Bis

Solution

Given, A={1,2,3},B={a,b} ∴ A×B={(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)}

Given, A={1,2,3},B={a,b} ∴ A×B={(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)}

**Q2.**If P is the set of all parallelograms, and T is the set of all trapeziums, then P∩T is

Solution

Clearly, P⊂T ∴P∩T=P

Clearly, P⊂T ∴P∩T=P

**Q3.**If A={4,6,10,12} and R is a relation defined on A as “two elements are related iff they have exactly one common factor other than 1”. Then the relation R is

Solution

Clearly, R={(4,6),(4,10),(6,4),(10,4)(6,10),(10,6),(10,12),(12,10)} Clearly, R is symmetric (6,10)∈R and (10,12)∈R but (6,12)∉R So, R is not transitive Also, R is not reflexive

Clearly, R={(4,6),(4,10),(6,4),(10,4)(6,10),(10,6),(10,12),(12,10)} Clearly, R is symmetric (6,10)∈R and (10,12)∈R but (6,12)∉R So, R is not transitive Also, R is not reflexive

**Q5.**If A is a non-empty set, then which of the following is false?

p∶ There is at least one reflexive relation on A

q∶ There is at least one symmetric relation on A

Solution

The identity relation on a set A is reflexive and symmetric both. So, there is always a reflexive and symmetric relation on a set

The identity relation on a set A is reflexive and symmetric both. So, there is always a reflexive and symmetric relation on a set

**Q8.**Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset

of A consisting of all determinants with value 1. Let C be the subset of the set of all

determinants with value -1. Then

Solution

Since the value of a determinant charges by minus sign by interchanging any two rows or columns. Therefore, corresponding to every element ∆ of B there is an element ∆' in C obtained by interchanging two adjacent rows (or columns) in ∆. It follows from this that n(B)≤n(C) Similarly, we have n(C)≤n(B) Hence, n(B)=n(C) ⇒B=C

Since the value of a determinant charges by minus sign by interchanging any two rows or columns. Therefore, corresponding to every element ∆ of B there is an element ∆' in C obtained by interchanging two adjacent rows (or columns) in ∆. It follows from this that n(B)≤n(C) Similarly, we have n(C)≤n(B) Hence, n(B)=n(C) ⇒B=C

**Q9.**Let R be a reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then,

Solution

Since R is reflexive relation on A ∴(a,a)∈R for all a∈A ⇒ The minimum number of ordered pairs in R is n Hence, m≥n

Since R is reflexive relation on A ∴(a,a)∈R for all a∈A ⇒ The minimum number of ordered pairs in R is n Hence, m≥n

**Q10.**For any two sets A and B, A-(A-B) equals