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  \( C_1, C_2, \dots, C_k \) , 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  \(\bigcup_{a ∈ D} S_a \)  :

1. Suppose D = S

Since D contains all n elements, and we take the union of all  \(S_a \),

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 = \(\frac{n(n - 1)}{2} \)

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:\( [G : H] = \frac{|G|}{|H|} \),

if G and H are finite

Every element of H is of the form:

\(\begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}, \quad z \in \mathbb{C} \) 

Here, z can be any complex number, so H is infinite, as  \(\mathbb{C} \)  is uncountable

Each element of G is of the form:

\(\begin{bmatrix} \omega & z \\ 0 & 1 \end{bmatrix}, \quad \omega \in F, \, z \in \mathbb{C} \) 

Since F has exactly 2020 elements and z can take any value in  \(\mathbb{C} \) , G is also infinite

The cosets of H in G are determined by the different values of  \(\omega \)  in F

For each \(\omega \in \) F , the matrix 
\(\begin{bmatrix} \omega & z \\ 0 & 1 \end{bmatrix} \)   represents a unique left coset of H

Since  \(\omega \)  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:  

\(\Theta = \{\theta \in \mathbb{R} \)  : \(e^{i 2 \pi \theta} \) is infinite } ,

where  \(S^1 = \{z \in \mathbb{C} : |z| = 1\} \)  is the circle group under multiplication, and  \(i = \sqrt{-1} \)

Understanding \(S^1 \)  :

The set \(S^1 \) consists of all complex numbers of the form:

\(z = e^{i \phi} \) ,   \( \phi \in \mathbb{R} \)

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

Meaning of  \(e^{i 2 \pi \theta} \)  :

The exponential form  \( e^{i 2 \pi \theta} \)  maps  \(\theta \)  in  \(\mathbb{R} \)  to a point on the unit circle

For any integer n ,  \(e^{i 2 \pi (\theta + n)} = e^{i 2 \pi \theta} \) , implying that the mapping is periodic with period 1

The set  \(\Theta \)  collects all real numbers  \(\theta \)  such that  \(e^{i 2 \pi \theta} \)  is well-defined

Since  \(e^{i 2 \pi \theta} \)  is a complex number of modulus 1 for all  \(\theta \in \mathbb{R} \)  ,

it is well-defined for every real number

The set  \(\Theta \)  is the entire real line \(\mathbb{R} \)  

The real numbers are uncountable

Hence Option(4) is the Correct Answer.

Explanation:

x is a real number

(1): n2 sin\(\frac{1}{n}\)  → ∞ as n → ∞

So, there exists an integer n ≥ 1 such that n2 sin\(\frac{1}{n}\) ≥ x.

(1) is correct

(2): n2 cos\(\frac{1}{n}\)  → ∞ as n → ∞

So, there exists an integer n ≥ 1 such that n2 cos\(\frac{1}{n}\) ≥ 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:\( [G : H] = \frac{|G|}{|H|} \),

if G and H are finite

Every element of H is of the form:

\(\begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}, \quad z \in \mathbb{C} \) 

Here, z can be any complex number, so H is infinite, as  \(\mathbb{C} \)  is uncountable

Each element of G is of the form:

\(\begin{bmatrix} \omega & z \\ 0 & 1 \end{bmatrix}, \quad \omega \in F, \, z \in \mathbb{C} \) 

Since F has exactly 2020 elements and z can take any value in  \(\mathbb{C} \) , G is also infinite

The cosets of H in G are determined by the different values of  \(\omega \)  in F

For each \(\omega \in \) F , the matrix 
\(\begin{bmatrix} \omega & z \\ 0 & 1 \end{bmatrix} \)   represents a unique left coset of H

Since  \(\omega \)  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:  

\(\Theta = \{\theta \in \mathbb{R} \)  : \(e^{i 2 \pi \theta} \) is infinite } ,

where  \(S^1 = \{z \in \mathbb{C} : |z| = 1\} \)  is the circle group under multiplication, and  \(i = \sqrt{-1} \)

Understanding \(S^1 \)  :

The set \(S^1 \) consists of all complex numbers of the form:

\(z = e^{i \phi} \) ,   \( \phi \in \mathbb{R} \)

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

Meaning of  \(e^{i 2 \pi \theta} \)  :

The exponential form  \( e^{i 2 \pi \theta} \)  maps  \(\theta \)  in  \(\mathbb{R} \)  to a point on the unit circle

For any integer n ,  \(e^{i 2 \pi (\theta + n)} = e^{i 2 \pi \theta} \) , implying that the mapping is periodic with period 1

The set  \(\Theta \)  collects all real numbers  \(\theta \)  such that  \(e^{i 2 \pi \theta} \)  is well-defined

Since  \(e^{i 2 \pi \theta} \)  is a complex number of modulus 1 for all  \(\theta \in \mathbb{R} \)  ,

it is well-defined for every real number

The set  \(\Theta \)  is the entire real line \(\mathbb{R} \)  

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.

Solution:

Given:

The notation (a)* is defined as : 

\(\left.\begin{array}{rl} (a)^* & =a \text { if } a>0 \\ & =0 \text { if } a \leq 0 \end{array}\right\}\)

We are asked to find the necessary condition for two real numbers x and y such that (xy)* = (x)* (y)*.

Step 1: Analyzing (xy)*

According to the definition, (xy)* will be:

• xy if xy > 0
• 0 if xy ≤ 0

Step 2: Analyzing (x)* (y)*

(x)* and (y)* will depend on the values of x and y:

• (x)* = x if x > 0, otherwise (x)* = 0
• (y)* = y if y > 0, otherwise (y)* = 0

Therefore:

• If both x > 0 and y > 0, then (x)* (y)* = xy.
• If either x ≤ 0 or y ≤ 0, then (x)* (y)* = 0.

Step 3: Condition for (xy)* = (x)* (y)*

For the equality (xy)* = (x)* (y)* to hold:

1. Case 1: If x ≥ 0 and y ≥ 0:
   • When x > 0 and y > 0, (xy)* = xy and (x)* (y)* = xy, so the equality holds.
   • When x = 0 or y = 0, (xy)* = 0 and (x)* (y)* = 0, so the equality holds.
2. Case 2: If x ≥ 0 or y ≥ 0 (one of them is non-negative):
   • When one of x or y is zero or positive, (xy)* will equal (x)* (y)* due to the nature of the zero multiplication rule, which also satisfies the condition.

The correct answer is 4) 

Explanation:

We have  \(√{n+1}-√{n} \) 

Let's use the conjugate:

= \(\frac{(√{n+1}-√{n})(√{n+1}+√{n}) }{(√{n+1}+√{n})}\)

\(=\frac{n+1-n}{(√{n+1}+√{n})}\)

\(=\frac{1}{(√{n+1}+√{n})} \le \frac{1}{2√{n}} \le \frac{1}{√{n}} \ \ \forall n\)

Hence Option (2) is correct.

Explanation:

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

now taking   \(x=\frac{4}{5}\) and  \(y=\frac{-3}{5}\) 

now \(|x-y|=|\frac{4}{5}-(\frac{-3}{5})|=\frac{7}{5}\)  

and  \(\frac{1}{|x|+|y|}\)= \(\frac{1}{|\frac{4}{5}|+|\frac{-3}{5}|}=\frac{5}{7}\)

and  \( \frac{1}{|x+y|}\) \(= \frac{1}{|\frac{4}{5}+\frac{-3}{5}|}=5\)  so by here statement C and D are false 

and as \(|ab|=|a|\times|b| \) so

\(\left|x^2-y^2\right| \geq 1\) ⇔ \(|(x+y)(x-y)| \geq 1\) ⇔ \(|x+y||x-y| \geq 1\)

option A is true 

and  as \(|a| \neq |b|\implies|a+b|>0\)  so

\(|x+y||x-y| \geq 1\) ⇔ \(|x-y| \geq \frac{1}{|x+y|}\)

option B  is also true  

Therefore  correct answer is  option: 3

Explanation:

1. L partitions S into equivalence classes.

Let the equivalence classes be  \( C_1, C_2, \dots, C_k \) , 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  \(\bigcup_{a ∈ D} S_a \)  :

1. Suppose D = S

Since D contains all n elements, and we take the union of all  \(S_a \),

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 = \(\frac{n(n - 1)}{2} \)

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.