Real Number System MCQ Quiz - Objective Question with Answer for Real Number System - Download Free PDF

Explanation:

1. L partitions S into equivalence classes.

Let the equivalence classes be   , where k is the number of equivalence classes.

2. Each a ∈ D  belongs to a distinct equivalence class since no two elements in D are related under L .

Maximum Elements in    :

1. Suppose D = S

Since D contains all n elements, and we take the union of all  ,

we include all pairs (a, x) such that a ≠  x and a L x .

2. For an equivalence relation on S ,

the total number of pairs (a, x) such that a ≠ x is equal to the number of ways to pick ordered pairs in S ,

excluding the cases where a = x

This gives: Total pairs  = n(n - 1).
   
3. Since each pair (a, x) is counted twice (once as (a, x) and once as (x, a) ),

the maximum number of distinct elements in the union is = 

Hence Option(2) is the correct answer.

Explanation:

We know that sum of two convex set is convex set. Also, intersection of two convex set is convex set.

So, (1), (2), (4) are correct.

But union of two convex set may not be convex set.

Option (3) is incorrect.

Explanation:

The number of cosets of H in G is the index of H in G , denoted [G : H]

This is given by:,

if G and H are finite

Every element of H is of the form:

Here, z can be any complex number, so H is infinite, as    is uncountable

Each element of G is of the form:

Since F has exactly 2020 elements and z can take any value in   , G is also infinite

The cosets of H in G are determined by the different values of    in F

For each F , the matrix 
   represents a unique left coset of H

Since    takes exactly 2020 distinct values in F , there are exactly 2020 cosets of H in G

The number of cosets of H in G is 2020.

Hence, Option(3) is the correct answer.

Explanation:   

A = {a ∈ R : x2 = a(a + 1)(a + 2) has a real root  

Since x² ≥ 0 

a(a + 1)(a + 2) ≥ 0 

a = 0 , a = -1 , a = -2 

a ∈ [-2 , -1] ∪ [ 0, ∞) 

⇒The number of connected components of A is 2

Hence 2 is the correct answer.

Explanation:  

  :  is infinite } ,

where    is the circle group under multiplication, and  

Understanding   :

The set consists of all complex numbers of the form:

 ,   

This describes the unit circle in the complex plane, where each z has modulus 1

Meaning of    :

The exponential form    maps    in    to a point on the unit circle

For any integer n ,   , implying that the mapping is periodic with period 1

The set    collects all real numbers    such that    is well-defined

Since    is a complex number of modulus 1 for all    ,

it is well-defined for every real number

The set    is the entire real line   

The real numbers are uncountable

Hence Option(4) is the Correct Answer.

Explanation:

x is a real number

(1): n2 sin  → ∞ as n → ∞

So, there exists an integer n ≥ 1 such that n2 sin ≥ x.

(1) is correct

(2): n2 cos  → ∞ as n → ∞

So, there exists an integer n ≥ 1 such that n2 cos ≥ x.

(2) is correct

(3): ne-n → 0 as n → ∞

So, For large n, ne-n will not be greater than x

For, n = 1000, ne-n is not greater than x

(3) is false 

(4): n(log n)-1 = n/(log n) → → ∞ as n → ∞

Hence there exists an integer n ≥ 2 such that n(log n)-1 ≥ x.

(4) is correct

Explanation:

The number of cosets of H in G is the index of H in G , denoted [G : H]

This is given by:,

if G and H are finite

Every element of H is of the form:

Here, z can be any complex number, so H is infinite, as    is uncountable

Each element of G is of the form:

Since F has exactly 2020 elements and z can take any value in   , G is also infinite

The cosets of H in G are determined by the different values of    in F

For each F , the matrix 
   represents a unique left coset of H

Since    takes exactly 2020 distinct values in F , there are exactly 2020 cosets of H in G

The number of cosets of H in G is 2020.

Hence, Option(3) is the correct answer.

Explanation:

Option (2)

f2(1, 2) = 1.2 + 1 + 2 = 5

f2(2, 1) = 2.1 + 2 + 1 = 5

then f2(1, 2) = f2(2, 1) ⇒ f2(m, n) is not injective.

option (2) - False.

Option (3)

f3(8, 1) = 82 + 13 = 65 = f3(1, 4) = 12 + 43

⇒ f3(8, 1) ± f3(1, 4) ⇒ f3(m, n) is not injective.

option (3) - False.

Option (4)

f4(1, 4) = 12 ⋅ 43 = 64

f4(8, 1) = 82 ⋅ 13 = 64

⇒ f4(1, 4) = f4(8, 1) ⇒ f4(m, n) is not infective.

option (4) - False.

Option (1)

f1(m, n) = 2m 3n

Let f1(m1, n1) = f1(m2, n2)

⇒ 2m1 × 3n1 = 2m2 × 3n2

⇒ m1 = m2 and n1 = n2

⇒ (m1, n1) = (m2, n2)

So, f1(m, n) = 2m 3n is injective.

Option (1) - True.

Explanation:  

  :  is infinite } ,

where    is the circle group under multiplication, and  

Understanding   :

The set consists of all complex numbers of the form:

 ,   

This describes the unit circle in the complex plane, where each z has modulus 1

Meaning of    :

The exponential form    maps    in    to a point on the unit circle

For any integer n ,   , implying that the mapping is periodic with period 1

The set    collects all real numbers    such that    is well-defined

Since    is a complex number of modulus 1 for all    ,

it is well-defined for every real number

The set    is the entire real line   

The real numbers are uncountable

Hence Option(4) is the Correct Answer.

Explanation:

The Archimedean property states that for any two real numbers x > 0 and y,

there exists a positive integer n such that nx > y

This means that for any positive real number x, you can always find an integer

n that, when multiplied by x, will exceed any given number y.

The correct option is option 2.

Explanation:

We have   

Let's use the conjugate:

= 

Hence Option (2) is correct.

Explanation:

Given |x|, |y| ≤ 2 where x, y are real and |x| ≠ lyl, 

now taking    and   

now   

and  = 

and     so by here statement C and D are false 

and as  so

 ⇔  ⇔ 

option A is true 

and  as 0\)  so

 ⇔ 

option B  is also true  

Therefore  correct answer is  option: 3

Explanation:

1. L partitions S into equivalence classes.

Let the equivalence classes be   , where k is the number of equivalence classes.

2. Each a ∈ D  belongs to a distinct equivalence class since no two elements in D are related under L .

Maximum Elements in    :

1. Suppose D = S

Since D contains all n elements, and we take the union of all  ,

we include all pairs (a, x) such that a ≠  x and a L x .

2. For an equivalence relation on S ,

the total number of pairs (a, x) such that a ≠ x is equal to the number of ways to pick ordered pairs in S ,

excluding the cases where a = x

This gives: Total pairs  = n(n - 1).
   
3. Since each pair (a, x) is counted twice (once as (a, x) and once as (x, a) ),

the maximum number of distinct elements in the union is = 

Hence Option(2) is the correct answer.

Explanation:

We know that sum of two convex set is convex set. Also, intersection of two convex set is convex set.

So, (1), (2), (4) are correct.

But union of two convex set may not be convex set.

Option (3) is incorrect.