Mathematics Exam  >  Mathematics Tests  >  Topic-wise Tests & Solved Examples for Mathematics  >  Test: Real Analysis - 6 - Mathematics MCQ

Test: Real Analysis - 6 - Mathematics MCQ


Test Description

20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Test: Real Analysis - 6

Test: Real Analysis - 6 for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation. The Test: Real Analysis - 6 questions and answers have been prepared according to the Mathematics exam syllabus.The Test: Real Analysis - 6 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Real Analysis - 6 below.
Solutions of Test: Real Analysis - 6 questions in English are available as part of our Topic-wise Tests & Solved Examples for Mathematics for Mathematics & Test: Real Analysis - 6 solutions in Hindi for Topic-wise Tests & Solved Examples for Mathematics course. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free. Attempt Test: Real Analysis - 6 | 20 questions in 60 minutes | Mock test for Mathematics preparation | Free important questions MCQ to study Topic-wise Tests & Solved Examples for Mathematics for Mathematics Exam | Download free PDF with solutions
Test: Real Analysis - 6 - Question 1

Let A and B be two sets such that n(A) = 0.16, n(B) = 0.14, n(A ∪ B) = 0.25. Then n(A ∩ B) is equal to

Detailed Solution for Test: Real Analysis - 6 - Question 1

Hence,
n(A ∪ B) = n(A) + n(B)
0.25 = 0.16 + 0.14 - n(A ∩ B)
implies n(A ∪ B) = 0.30 - 0.25 = 0.05

Test: Real Analysis - 6 - Question 2

​20 teachers of a school either teach Mathematics of Physics. 12 of them teach Mathematics while 4 teach both the subjects. Then, the number of teachers teaching Physics only is

Detailed Solution for Test: Real Analysis - 6 - Question 2

Let n(P) = Number of teachers in Physics.
n(M) = Number of teachers in Maths Then, n(P ∪ M ) = n(P) + n(M) - n(P∩M)
20 = n(P) + 12 - 4 => n(P) = 12

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Real Analysis - 6 - Question 3

If U is the universal set and P is a subset of U, then what is P ∩ {(P - U ) ∪ ( U - P ) } equal to?

Detailed Solution for Test: Real Analysis - 6 - Question 3

Test: Real Analysis - 6 - Question 4

Two Unite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n are

Detailed Solution for Test: Real Analysis - 6 - Question 4

Since,

Test: Real Analysis - 6 - Question 5

In the set A = {1,2,3,4, 5}, a relation R is defined by R = {(x , y ) | x, y ∈ A and x < y } . Then,R is

Detailed Solution for Test: Real Analysis - 6 - Question 5

Since, xx, therefore R is not reflexive. Also, x<y does not imply that
y < x. So, R is not symetric.
Let xRy and yRz.
Then, x < y and y < z implies, x < z i. e.. xRz.
Hence, R is transitive.

Test: Real Analysis - 6 - Question 6

If A and B are two subsets of a set X, then what is A ∩ (A ∪ B)' equal to?

Detailed Solution for Test: Real Analysis - 6 - Question 6

Here,

Test: Real Analysis - 6 - Question 7

What is the inverse of the function y = 5log x?

Detailed Solution for Test: Real Analysis - 6 - Question 7

Given that,

Test: Real Analysis - 6 - Question 8

For non-empty subsets A, B and C of a set X such that A ∪ B = B ∩ C, which one of the following is the strongest inference that can be derived?

Detailed Solution for Test: Real Analysis - 6 - Question 8

Since, A ∪ B = B ∩ C
From above strong inference is
AB C
e.g., A= {a},B= {a,b},C= {a,b.c}
A ∪ B = {a, b}, B ∩ C = { a, b }

Test: Real Analysis - 6 - Question 9

The relation R = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)} on a set A = { 1 ,2 , 3} is

Detailed Solution for Test: Real Analysis - 6 - Question 9

Since, R= {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}
Reflexive
Since,  1R1, 2R23R3
Hence, R is a reflexive relation.
Symmetric
Since, 1R1, 2R3
Hence, R is not a symmetric relation.
Transitive
Since,1R2, 2R3 implies, 1R3
Hence, R is a transitive relation.

Test: Real Analysis - 6 - Question 10

In a class of 100 students, 55 students have passed in Mathematics and 67 students have passed in Physics. Then, the number of student who have passed in Physics only, is

Detailed Solution for Test: Real Analysis - 6 - Question 10

Test: Real Analysis - 6 - Question 11

Let Rbe a relation defined by R1 = {(a, b ) | a ≥ b, a, b ∈ R}. Then R1 is

Detailed Solution for Test: Real Analysis - 6 - Question 11

For any a ∈ R, we haye a ≥ a. Therefore, the relation R is reflexive but it is hot symmetric as (2, 1) ∈ R but (1, 2) ∉ R. The relation R is transitive also, because (a, b) ∈ R, (b, c) ∈ R imply that a ≥ b and b ≥. c which is turn imply that a ≥ c.

Test: Real Analysis - 6 - Question 12

Of the members of three athletic teams in a school 21 are in the cricket team, 26 are in the hockey team and 29 are in the football team. Among them, 14 play hockey and cricket, 15 play hockey and football, and 12 play football and cricket. 8 play all the three games. The total number of members in the three athletic teams is

Detailed Solution for Test: Real Analysis - 6 - Question 12

Let B, H, F denote the set of members who are the basketball team, hockey team and football team, respectively. Then , we are given n (B) = 21, n(H) = 26, n(F) = 29

Thus, these are 43 members in all

Test: Real Analysis - 6 - Question 13

In a class of 30 pupils, 12 take needle work, 16 take Physics and 18 take History. If all the 30 students take atleast one subject and no one takes all three, then the number of pupils taking 2 subjects is

Detailed Solution for Test: Real Analysis - 6 - Question 13

Given, n(N) = 12, n(P) = 16, n(H)

Therefore, 
Now, number of pupils taking two subjects
= 16 - 0 =16

Test: Real Analysis - 6 - Question 14

For a set A, consider the following statements
1. A ∪ P(A) = P(A)
2. {A} ∩ P(A)=A
3. P(A) - {A} = P{A}
where P denotes power set.
Which of the statements given above is/are correct?

Detailed Solution for Test: Real Analysis - 6 - Question 14

 Let A = {1,2} and {A} = {(1,2)} implies,P(A) = {{1}, {2},.{1,2}, φ}
Then, A ∪ P(A) ≠ P(A)
and{A} ∩ P(A) = {{ l , 2}} ∩ {{1}, {2}, {1,2},φ}
= {1,2}=A

Test: Real Analysis - 6 - Question 15

The number of elements in the set
{(a, b) : 2a2 + 3b2 = 35, a b ∈ Z}, where Z is the set of all integer, is

Detailed Solution for Test: Real Analysis - 6 - Question 15

Given set is {(a, b) :2a2 + 3b2-35,a,b ∈ Z}
We can that, 2(±2)2 + 3(±3)2 = 35
and 2(±4)2 + 3(±l)1 = 35
So, (2,3), (2,-3), (-2, -3), (2,3), (4.1), (4, 1), ( 4, 1), (-A, 1) are 8 elements of the set.
Hence, n = 8

Test: Real Analysis - 6 - Question 16

If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets A x B and B x A are

Detailed Solution for Test: Real Analysis - 6 - Question 16

Test: Real Analysis - 6 - Question 17

Assertion (A): y = 2x + 3 is a one-to-one real-valued function.
Reason (R): x1 ≠ x2 ==> y≠ y2, y1 = 2 x1 + 3, y2 = 2x2 + 3, for any two real x1, and x2

Detailed Solution for Test: Real Analysis - 6 - Question 17

Given,


So, function is one-to-one

Test: Real Analysis - 6 - Question 18

If A, B and C are three finite sets, then what is [(A ∪ B) ∩ C]' equal to

Detailed Solution for Test: Real Analysis - 6 - Question 18

We know that,

Test: Real Analysis - 6 - Question 19

If Q =then 

Detailed Solution for Test: Real Analysis - 6 - Question 19

Since 
[Since, y ∈ N]
Therefore, 1/y can be 1. [Since, y can be 1]

Test: Real Analysis - 6 - Question 20

Consider the following statements
1. φ ∈ {φ}
2. {φ} φ

Which of the statements given above is/are correct?

27 docs|150 tests
Information about Test: Real Analysis - 6 Page
In this test you can find the Exam questions for Test: Real Analysis - 6 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Real Analysis - 6, EduRev gives you an ample number of Online tests for practice
Download as PDF