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Test: Real Analysis- 7 - Mathematics MCQ


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20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Test: Real Analysis- 7

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Test: Real Analysis- 7 - Question 1

A set contains 2n+1 elements. The number of subsets of this set containing more than n elements is equal to

Detailed Solution for Test: Real Analysis- 7 - Question 1

Let the original set contains (2n + 1) elements, then subset of this set containing more than n elements, i.e., subsets containing (n + 1) elements, (n + 2) elements,....... (2n + 1) elements.
Hence, Required number of subsets

Test: Real Analysis- 7 - Question 2

Let R be a relation on the set N of natural numbers defined by nRm <=> n is a factor of m(i.e., n|m). Then, R is

Detailed Solution for Test: Real Analysis- 7 - Question 2

Since, n | n for all n ∈ N, therefore R is reflexive.
Since, 2 | 6 but 6 | 2 , therefore R is not symmetric. Let nRm

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Test: Real Analysis- 7 - Question 3

Assertion (A): Function f: {1, 2, 3 } {a, b, c, d) is defined by f= {(1, a), (2, b), (3, c)} has no inverse.
Reason (R): Functions ‘/ ’ is not one-to-one.

Detailed Solution for Test: Real Analysis- 7 - Question 3


Since, the image of the each element of A is belonging to only one element in B. But d has no pre-image in A Therefore, function is one-one but not onto.
So, its inverse does not exist.

Test: Real Analysis- 7 - Question 4

A class has 175 students.
The following data shows the number of students obtaining 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.
Q. How many students have offered Mathematics alone

Detailed Solution for Test: Real Analysis- 7 - Question 4

We have, n(M alone) = n(M) - n(M ∩ C) -n (M ∩ P ) + n(M ∩ P ∩ C)
=100 - 28 - 30 + 18 = 60

Test: Real Analysis- 7 - Question 5

​If A, B, C are three sets, then what is A - (B - C) equal to

Detailed Solution for Test: Real Analysis- 7 - Question 5

In the above Venn diagram, shaded area shows A - (B - C)


In the above Venn diagram, horizontal lines mean (A - B) and vertical lines mean (A ∩ C).
Total shaded portion = 
Therefore, 

Test: Real Analysis- 7 - Question 6

Consider the following relations
I. A - B = A - ( A ∩ B )
II. A = (A ∩ B ) ∪ (A -B )
III. A - (B ∪ C) = (A - B) ∪ (A - C)
Q. Which of these is/are correct?

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

A - B = A - (A ∩ B) is correct.
A = (A ∩ B) ∩ (A - B) is correct.

III is false
Therefore, I and II are True.

Test: Real Analysis- 7 - Question 7

In an examination out of 100 students, 75 passed in English, 60 passed in Mathematics and 45 passed in both English and Mathematics. What is the number of students passed in exactly one of the two subjects?

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

Since, n(E) = 75, n(M)
= 60 and n(E ∩ M) = 45
Hence, n(E∪M) = n(E) + n(M) - n(E ∩M)
= 75 + 60 - 45 = 90
Required number of students = 90 - 45 = 45

Test: Real Analysis- 7 - Question 8

If, A = {1,2,3,4} and
R = {(1, 1), (1, 3), (2, 2), (3, 1) (3, 4), (4, 3) (4, 4) is a relation on A x A, then which one of the following is correct?

Detailed Solution for Test: Real Analysis- 7 - Question 8

Correct Answer :- c

Explanation : Let A = {1,2,3,4} and R be a relation on A given by R={(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,1),(1,3)}.

Now for all 1,2,3,4 ∈ A, (1,1),(2,2),(3,3),(4,4) ∈ R, this gives the relation R is reflexive.

Again for (1,2) ∈ R

⇒(2,1)∈R and (1,3)∈R

⇒(3,1)∈R for 1,2∈A. 

This gives the relation R is symmetric.

But the relation is not transitive as (2,1),(1,3)∈R but (2,3)does not ∈ R.

Test: Real Analysis- 7 - Question 9

Let U = the set of all triangles, P = the set of all isosceles triangles, Q = the set of all equilateral triangles, R = the set of all right angled triangles. What do the sets P ∩ Q and R- P represents respectively?

Detailed Solution for Test: Real Analysis- 7 - Question 9

μ =the set of all triangles
P =the set of all isosceles triangles
Q =the set of all equilateral triangles
R =the set of all the right angled triangles
Since, P ∩ Q represents the set of isosceles triangles and R - P represents the set of non-isosceles right angled triangles.

Test: Real Analysis- 7 - Question 10

If X = {4n - 3n - 1 : n ∈ N} and Y = {9 ( n -  l) : n ∈ N}, then X ∪ Y is equal to:

Detailed Solution for Test: Real Analysis- 7 - Question 10


So, 4n - 3k - 1 is a multiples of 9 for all n ∈ N
Hence, X contains elements, which are multiples of 9 and clearly Y contains all multiples of 9.

Test: Real Analysis- 7 - Question 11

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

Detailed Solution for Test: Real Analysis- 7 - Question 11

(a) Since, (1 ,1 ); (2 ,2 ); (3 ,3 ) ∈ R
therefore R is reflexive.
(1, 2) ∈ R but (2, 1) ∉ R, therefore R is not symmetric. It can be easily seen that R is transitive.
(c) Here,R{(l, 3), (2,2); (3,2)}, &= {(2,1); (3,2); (3,2); (2 ,3 )}. Then RoS = {(2,3), (3,2); (2,2)}

Test: Real Analysis- 7 - Question 12

​Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S is

Detailed Solution for Test: Real Analysis- 7 - Question 12

Since, 1 + a • a = 1 + a2 > 0,, Therefore, (a, a) ∈ R
So, R is reflexive.
Also, (a, b) ∈ R => 1 + ab > 0 => 1 + ba > 0
=> (b, a) ∈ R,
Hence, R is symmetric.
Since, (a, b) ∈ R and (b, c) ∈ R need
not imply (a, c) ∈ R.
Hence, R is not transitive

Test: Real Analysis- 7 - Question 13

If A - {x: f(x) = 0} and B = { x : g(x) = 0}, then A∩B will be

Test: Real Analysis- 7 - Question 14

Let R be the relation defined on the set of natural number N as aRb; a, b ∈ N, if a divides b.Then, which one of the following is correct?

Detailed Solution for Test: Real Analysis- 7 - Question 14

For reflexive aRa implies a divides a
So, R is reflexive.
For symmetric
aRb impliesa divides b
bRa impliesa divides a
which may not be possible.
So, R is not symmetric.
For transitive
aRb impliesa divides b implies b = ka bRc impliesb divides c implies c = lb Now, c = Ika
or a divides c implies aRc
or bRc implies cRa
Thus, R is transitive.

Test: Real Analysis- 7 - Question 15

If A = {(x, y ) : x2 + y2 = 25} and B = {(x, y ) : x2 + 9y2 = 144}, then A ∩ B contains

Detailed Solution for Test: Real Analysis- 7 - Question 15

A = Set of all values (x, y) :x2 + y2 = 25 = 52

Clearly, A ∩ B consists of four points.

Test: Real Analysis- 7 - Question 16

If A = {4n + 2 | n is a natural number} and B = .{3n | n is a natural number}, then what is (A ∩ B) equal to?

Detailed Solution for Test: Real Analysis- 7 - Question 16

Since, A = {An + 2 | n e N}
= {6,10,14,18,22,26,30,.,.} 
and B = {3n | n ∈ N}
= {3,6,9,12,15,18,21,24,...}
So, A ∩ B = { 6 ,18,30,...}
= {6 + (n - 1)12 | n ∈ N}
= {12n - 6 | n ∈ N}
(c) x = 3(mod 7) => x - 3 = Ip, (p ∈ z) implies x= 7p + 3,p ∈ z i.e., {7p + 3 :p ∈ z}

Test: Real Analysis- 7 - Question 17

If (1,3), (2,5) and (3,3) are three elements of A x B and the total number of elements in A x B is 6, then the remaining elements of A x B are

Detailed Solution for Test: Real Analysis- 7 - Question 17

Since, R is defined over the set of non-negative integers, then R = {(6,0), (0,6)} because the option (b) and (c) does not have integers in all pairs. Also in option (a) and (c), the most appropriate option is (c).

Test: Real Analysis- 7 - Question 18

Solution set of x = 3 (mod 7), p ∈ Z, is given by

Detailed Solution for Test: Real Analysis- 7 - Question 18

 We first find R-1 we have
R-1 = {(5,4); (4 ,1 ); (6 ,4 ); (6 ,7 ); (7,3}.
We now obtain the elements of R-1 OR we first pick the elements of R and then of R-1 Since, (4, 5) ∈ R and 

Test: Real Analysis- 7 - Question 19

Assertion (A): is not a set. Here, R is the set of real numbers.
Reason (R): For every real number x, x2 ≥ 0.

Detailed Solution for Test: Real Analysis- 7 - Question 19

Both (A) and (R) are true and (R) is the correct explanation of (A).
Since, x2 is never negative but it can be positive or zero.

Test: Real Analysis- 7 - Question 20

If X and Y are any two non-empty sets, then what is (X - Y)' equal to?

Detailed Solution for Test: Real Analysis- 7 - Question 20

From the figure, it is clear that

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