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. Three sets A,B,Care such that A=B∩C and B=C∩A, then
Solution
Since, A=B∩C and B=C∩A, Then A≡B
Since, A=B∩C and B=C∩A, Then A≡B
Q2.Which of the following cannot be the number of elements in the power set of any finite set?
Solution
The power set of a set containing n elements has 2n elements. Clearly, 2n cannot be equal to 26
The power set of a set containing n elements has 2n elements. Clearly, 2n cannot be equal to 26
Q3. Let P={(x,y)│x2+y2=1,x,y∈R}. Then, P is
Solution
Obviously the relation is not reflexive and transitive but it is symmetric, because x2+y2=1⇒y2+x2=1
Obviously the relation is not reflexive and transitive but it is symmetric, because x2+y2=1⇒y2+x2=1
Q4. Let S={1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to
Solution
Required number = (34+1)/2=41
Required number = (34+1)/2=41
Q5.The finite sets A and B have m and n elements respectively. if the total number of subsets of
A is 112 more than the total number of subsets of B, then the volume of m is
A is 112 more than the total number of subsets of B, then the volume of m is
Solution
According to the given condition, 2m=112+2n ⇒ 2m-2n=112 ⇒ m=7,n=4
According to the given condition, 2m=112+2n ⇒ 2m-2n=112 ⇒ m=7,n=4
Q6. If S is a set with 10 elements and A={(x,y):x,y∈S,x≠y}, then the number of elements in A is
Solution
Number of element is S=10 And A={(x,y);x,y∈S,x≠y} ∴ Number of element in A=10×9=90
Number of element is S=10 And A={(x,y);x,y∈S,x≠y} ∴ Number of element in A=10×9=90
Q7.The void relation on a set A is
Solution
The void relation R on A is not reflexive as (a,a)∉R for any a∈A. The void relation is symmetric and transitive
The void relation R on A is not reflexive as (a,a)∉R for any a∈A. The void relation is symmetric and transitive
Q8.If A={1,3,5,7,9,11,13,15,17},B={2,4…,18} and N is the universal set, then A'∪((A∪B)∩B') is
Solution
We have, (A∪B)∩B^'=A ∴((A∪B)∩B^' )∪A^'=A∪A^'=N
We have, (A∪B)∩B^'=A ∴((A∪B)∩B^' )∪A^'=A∪A^'=N
Q9.Let A={1,2,3,4}, and let R={(2,2),(3,3),(4,4),(1,2)} be a relation on A. Then, R is
Solution
Since (1,1)∉R. So, R is not reflexive Now, (1,2)∈R but, (2,1)∉R. Therefore, R is not symmetric. Clearly, R is transitive
Since (1,1)∉R. So, R is not reflexive Now, (1,2)∈R but, (2,1)∉R. Therefore, R is not symmetric. Clearly, R is transitive
Q10. For real numbers x and y, we write x Ry⇔x-y+√2 is an irrational number. Then, the relation R is
Solution
For any x∈R, we have x-x+√2=√2 an irrational number ⇒x R x for all x So, R is reflexive R is not symmetric, because √2 R 1 but 1 √2 R is not transitive also because √2 R 1 and 1 R 2 √2 but √2 2√2
For any x∈R, we have x-x+√2=√2 an irrational number ⇒x R x for all x So, R is reflexive R is not symmetric, because √2 R 1 but 1 √2 R is not transitive also because √2 R 1 and 1 R 2 √2 but √2 2√2