# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.4 | Set 1

• Last Updated : 11 Feb, 2021

### Question 1. Which of the following statements are true? Give a reason to support your answer.

(i) For any two sets A and B either A B or B A.

(ii) Every subset of an infinite set is infinite.

(iii) Every subset of a finite set is finite.

(iv) Every set has a proper subset.

(v) {a, b, a, b, a, b,….} is an infinite set.

(vi) {a, b, c} and {1, 2, 3} are equivalent sets.

(vii) A set can have infinitely many subsets.

Solution:

(i) False

It is not mandatory for any two set A and B to be either A B or B A.

(ii) False

Let us consider a set, A = {2,3,4}.

It is finite subset with infinite set N of natural numbers.

(iii) True

A finite set can never have an infinite subset.

Therefore, every subset of a finite set is finite.

(iv) False

Null set, also known as empty set doesn’t have a proper subset.

(v) False

A set can never have duplicate entries.

Removing duplications, the set {a, b} becomes a finite set.

(vi) True

Equivalent sets have the same number of elements. Both the sets have three elements, therefore, are equivalent.

(vii) False

Let us consider, a set A = {1}

The subsets of this set A can be ϕ and {1} which are finite.

### Question 2. State whether the following statements are true or false:

(i) 1 ∈ {1,2,3}

(ii) a ⊂ {b,c,a}

(iii) {a} ∈ {a,b,c}

(iv) {a, b} = {a, a, b, b, a}

(v) The set {x: x + 8 = 8} is the null set.

Solution:

(i) True

1 is a part of the given set, therefore it belongs to this set {1, 2, 3} .

(ii) False

Since, a is an element and not a subset of a set {b, c, a}, therefore this statement is false.

(iii) False

Since, {a} is not an element but a subset of set {b, c, a}.

(iv) True

A set cannot have duplicate entries. Therefore, removing duplicate entries from RHS, the sets become equivalent.

(v) False

Given, x+8 = 8

Solving we get, x = 0

Therefore, the given set becomes a single ton set with the only element being {0}. Where, it is not a null set.

### A = {x: x satisfies x2 – 8x + 12 = 0}, B = {2,4,6}, C = {2,4,6,8,….}, D = {6}

Solution:

We have,

A = x2 – 8x + 12=0

Solving we get,

⇒ (x–6) (x–2) =0

Solving for x,

⇒ x = 2 or x = 6

Therefore,

A = {2, 6}

Given,

B = {2, 4, 6}

C = {2, 4, 6, 8}

D = {6}

Therefore,

D ⊂ A ⊂ B ⊂ C

### Question 4. Write which of the following statements are true? Justify your answer.

(i) The set of all integers is contained in the set of all rational numbers.

(ii) The set of all crows is contained in the set of all birds.

(iii) The set of all rectangles is contained in the set of all squares.

(iv) The set of all rectangle is contained in the set of all squares.

(v) The sets P = {a} and B = {{a}} are equal.

(vi) The sets A={x: x is a letter of word “LITTLE”} AND, b = {x: x is a letter of the word “TITLE”} are equal.

Solution:

(i) True

A rational number is a fractional number which is represented by the form p/q where p and q are integers where q is not equal to 0. Substituting, q = 1, we get p=q, which is an integer.

(ii) True

Crows are also birds, so all the crows are contained in the set of all birds.

(iii) False

Every square can be a rectangle, where the length and breadth of the rectangle are same. But, the reverse is not true, that is, not every rectangle cannot be a square.

(iv) False

Every square can be a rectangle, where the length and breadth of the rectangle are same. But, the reverse is not true, that is, not every rectangle cannot be a square.

(v) False

We have,

P = {a}

B = {{a}}

But, {a} = P

B = {P}

Hence, they are not equivalent.

(vi) True

We have,

A = For “LITTLE”

A = {L, I, T, E} = {E, I, L, T}

B = For “TITLE”

B = {T, I, L, E} = {E, I, L, T}

Therefore,

A = B

### Question 5. Which of the following statements are correct? Write a correct form of each of the incorrect statements.

(i) a ⊂ {a, b, c}

(ii) {a} {a, b, c}

(iii) a {{a}, b}

(iv) {a} ⊂ {{a}, b}

(v) {b, c} ⊂ {a,{b, c}}

(vi) {a, b} ⊂ {a,{b, c}}

(vii) ϕ {a, b}

(viii) ϕ ⊂ {a, b, c}

(ix) {x: x + 3 = 3}= ϕ

Solution:

(i) False

a is not a subset of given set but belongs to the given set.It is just an element.

Correct form is – a ∈ {a, b, c}

(ii) In this {a} is subset of {a, b, c}

Correct form is – {a} ⊂ {a, b, c}

(iii) False

‘a’ is not the element of the set.

Correct form is – {a} ∈ {{a}, b}

(iv) False

{a} is not a subset of given set.

Correct form is – {a} ∈ {{a}, b}

(v) {b, c} is not a subset of given set. But it belongs to the given set.

Correct form is – {b, c} ∈ {a,{b, c}}

(vi) {a, b} is not a subset of given set.

Correct form is – {a, b}⊄{a,{b, c}}

(vii) ϕ does not belong to the given set but it is subset.

Correct form is – ϕ ⊂ {a, b}

(viii) True

It is the correct form. ϕ is subset of every set.

(ix) x + 3 = 3

Evaluating, we get,

x = 0 = {0}, which is not ϕ

Correct form is – {x: x + 3 = 3} ≠ ϕ

### Question 6. Let A = {a, b,{c, d}, e}. Which of the following statements are false and why?

(i) {c, d} ⊂ A

(ii) {c, d} ∈ A

(iii) {{c, d}} ⊂ A

(iv) a ∈ A

(v) a ⊂ A.

(vi) {a, b, e} ⊂ A

(vii) {a, b, e}∈ A

(viii) {a, b, c} ⊂ A

(ix) ϕ∈ A

(x) {ϕ} ⊂ A

Solution:

(i) False

{c, d} is not a subset of A but it belongs to the set A.

Therefore,

{c, d} ∈ A

(ii) True

{c, d} ∈ A

(iii) True

{c, d} is a subset of A.

(iv) It is true that a belongs to A.

(v) False

The element a is not a subset of A but it belongs to the set A.

(vi) True

{a, b, e} is a subset of A.

(vii) False

{a, b, e} does not belong to A, {a, b, e} ⊂ A this is the correct form.

(viii) False

{a, b, c} is not a subset of A

(ix) False

ϕ is a subset of A.

ϕ ⊂ A.

(x) False

{ϕ} is not subset of A, ϕ is a subset of A. Therefore, it is false.

### Question 7. Let A = {{1, 2, 3}, {4, 5}, {6, 7, 8}}. Determine which of the following is true or false:

(i) 1 ∈ A

(ii) {1, 2, 3} ⊂ A

(iii) {6, 7, 8} ∈ A

(iv) {4, 5} ⊂ A

(v) ϕ ∈ A

(vi) ϕ ⊂ A

Solution:

(i) False

1 is not an element of the set A.

(ii) True

{1,2,3} ∈ A. this is correct form.

(iii) True.

The set {6, 7, 8} is an element of A, that is {6, 7, 8} ∈ A.

(iv) True

{{4, 5}} is a subset of the specified set A={4, 5}.

(v) False

Φ is a subset of the given set A, not an element of A.

(vi) True

Φ is a subset of every set, so it is also a subset of the given set A.

### Question 8. Let A = {ϕ, {ϕ}, 1, {1, ϕ}, 2}. Which of the following are true?

(i) ϕ ∈ A

(ii) {ϕ} ∈ A

(iii) {1} ∈ A

(iv) {2, ϕ} ⊂ A

(v) 2 ⊂ A

(vi) {2, {1}} ⊄A

(vii) {{2}, {1}} ⊄ A

(viii) {ϕ, {ϕ}, {1, ϕ}} ⊂ A

(ix) {{ϕ}} ⊂ A

Solution:

(i) True

Φ is an element of set A, therefore it belongs to set A. Hence, the given statement is true.

(ii) True

{Φ} is an element of set A, and not a subset. Hence, the given statement is true.

(iii) False

The subset 1 is not an element of A. Hence, the given statement is false

(iv) True

{2, Φ} is a subset of the given set A. Hence, the given statement is true.

(v) False

2 is not a subset of set A, it is an element of set A. Hence, the given statement is false.

(vi) True

{2, {1}} is not a subset of the given set A. Hence, the given statement is true.

(vii) True

Neither {2} and nor {1} is a subset of set A. Hence, the given statement is true.

(viii) True

All three {ϕ, {ϕ}, {1, ϕ}} are subset of set A. Hence, the given statement is true.

(ix) True

{{ϕ}} is a subset of set A. Hence, the given statement is true.

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