Exercise 1.3 Class 11 maths 1. Sets - ncert solutions - Toppers Study

Exercise 1.3 

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Q1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces :
(i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 }

(ii) { a, b, c } . . . { b, c, d }
(iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school}
(iv) {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with
radius 1 unit}
(v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} . . . {x : x is an integer}

Solutions: 

(i) { 2, 3, 4 } ⊂ { 1, 2, 3, 4,5 } 

(ii) { a, b, c } ⊄ { b, c, d }

(iii) {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}
(iv) {x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}
(vii) {x : x is an even natural number} ⊂ {x : x is an integer}

2. Examine whether the following statements are true or false:
(i) { a, b } ⊄ { b, c, a }
(ii) { a, e } ⊂ { x : x is a vowel in the English alphabet}
(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }
(iv) { a }⊂ { a, b, c }
(v) { a }∈ { a, b, c }
(vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}

Solutions: 

(i) False, Because Each element of { a, b } is also an element of { b, c, a }. 

(ii) True, Because {a, e} is also vowels of English alphabet. 

(iii) False, Hence 2 ∈ {1, 2, 3}; while, 2 ∉ {1, 3, 5}

(iv) True, Because each elements of set { a } is also element of { a, b, c }
Q3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?
(i) {3, 4} ⊂ A

(ii) {3, 4} ∈ A

(iii) {{3, 4}} ⊂ A
(iv) 1 ∈ A (v) 1 ⊂ A

(vi) {1, 2, 5} ⊂ A
(vii) {1, 2, 5} ∈ A

(viii) {1, 2, 3} ⊂ A

(ix) φ ∈ A
(x) φ ⊂ A

(xi) {φ} ⊂ A

Solutions: 

Given that A = {1, 2, {3, 4}, 5}

(i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; while, 3∉A.

(ii) The statement {3, 4} ∈ A is correct because {3, 4} is an element of A.

(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.

(iv) The statement 1∈A is correct because 1 is an element of A.

(v) The statement 1⊂ A is incorrect because an element of a set can never be a subset of itself.

(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.

(vii)The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an

element of A.

(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A.

(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A.

(x) The statement Φ ⊂ A is correct because Φ is a subset of every set.

(xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.

Q4. Write down all the subsets of the following sets
(i) {a} (ii) {a, b} (iii) {1, 2, 3} (iv) φ

Solutions: 

(i) The subsets of {a} are φ and {a}.

(ii) The subsets of {a, b} are φ, {a}, {b} and {a, b}. 

(iii) The subsets of {1, 2, 3} are φ, {1}, {2}, {3}, {1, 2}, {1,3}, {2, 3} and {1, 2, 3}.

(iv) The subset of φ is φ. 

Q5. How many elements has P(A), if A = φ?

Solution: 

Given that A = φ

Therefore, no. of elements n(A) = 0 

n[P(A)] = 2ⁿ = 2⁰ = 1 

Hence, P(A) has only 1 element. 
Q6. Write the following as intervals :
(i) {x : x ∈ R, – 4 < x ≤ 6}

(ii) {x : x ∈ R, – 12 < x < –10}
(iii) {x : x ∈ R, 0 ≤ x < 7}

(iv) {x : x ∈ R, 3 ≤ x ≤ 4}

Solutions: 

(i) {x: x ∈ R, –4 < x ≤ 6} is an open interval from -4 to 6, including 6 but excluding -4.

Hence interval = (–4, 6]

(ii) {x: x ∈ R, –12 < x < –10} is an open interval from -12 to -10, excluding both -12 and -10.

Hence interval = (–12, –10)

(iii) {x: x ∈ R, 0 ≤ x < 7} is an open interval from 0 to 7, including 0 but excluding 7.

Hence interval = [0, 7)

(iv) {x: x ∈ R, 3 ≤ x ≤ 4} is an close interval from 3 to 4, including both 3 and 4.

Hence interval = [3, 4]

Q7. Write the following intervals in set-builder form :
(i) (– 3, 0)

(ii) [6 , 12]

(iii) (6, 12]

(iv) [–23, 5)

Solutions: 

(i) (–3, 0) = {x: x ∈ R, –3 < x < 0}

(ii) [6, 12] = {x: x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, 12] = {x: x ∈ R, 6 < x ≤ 12}

(iv) [–23, 5) = {x: x ∈ R, –23 ≤ x < 5}

Q8. What universal set(s) would you propose for each of the following :
(i) The set of right triangles.

(ii) The set of isosceles triangles.

Solutions: 

(i) The sets of all possible triangles and polygons can be universal set for the right triangles.

(ii) The sets of all possible triangles and polygons can be universal set for the isosceles triangles.

Q9. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) φ
(iii) {0,1,2,3,4,5,6,7,8,9,10}
(iv) {1,2,3,4,5,6,7,8}

Solution: 

(iii) {0,1,2,3,4,5,6,7,8,9,10} can be universal set (s) for all the three sets A, B and C. 

Because, 

A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the

sets A, B, and C.