NCERT Solutions for Class 11 Maths
Chapter 1- Sets
Exercise 1.3 (Page No-12)
1. 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}
Solution:
(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}
Solution:
(i) False.
Because each element of {a, b} is an element of {b, c, a}.
(ii) True.
Because a, e are two vowels of the English alphabet.
(iii) False.
2 ∈ {1, 2, 3} but, 2∉ {1, 3, 5}
(iv) True.
Here element of {a} i.e a is also an element of {a, b, c}.
(v) False.
Here Elements of {a, b, c} are a, b, c. Hence, {a} ⊂ {a, b, c}
(vi) True.
{x: x is an even natural number less than 6} = {2, 4}
{x: x is a natural number which divides 36}= {1, 2, 3, 4, 6, 9, 12, 18, 36}
3. 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
Solution:
Here A= {1, 2, {3, 4}, 5}
(i) {3, 4} ⊂ A is incorrect
Because 3 ∈ {3, 4}; where, 3∉A.
(ii) {3, 4} ∈A is correct
Because {3, 4} is an element of A.
(iii) {{3, 4}} ⊂ A is correct
Because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.
(iv) 1∈A is correct
Because 1 is an element of A.
(v) 1⊂ A is incorrect
Because an element of a set can never be a subset of itself.
(vi) {1, 2, 5} ⊂ A is correct
Because each element of {1, 2, 5} is also an element of A.
(vii) {1, 2, 5} ∈ A is incorrect
Because {1, 2, 5} is not an element of A.
(viii) {1, 2, 3} ⊂ A is incorrect
Because 3 ∈ {1, 2, 3} and 3 ∉ A.
(ix) Φ ∈ A is incorrect
Because Φ is not an element of A.
(x) Φ ⊂ A is correct
Because Φ is a subset of every set.
(xi) {Φ} ⊂ A is incorrect
Because Φ ⊂ A.
4. Write down all the subsets of the following sets:
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ
Solution:
(i) Subsets of {a} are
Φ and {a}.
(ii) Subsets of {a, b} are
Φ, {a}, {b}, and {a, b}.
(iii) Subsets of {1, 2, 3} are
Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and {1, 2, 3}.
(iv) Only subset of Φ is Φ.
5. How many elements has P (A), if A = Φ?
Solution:
If A is a set with m elements
n (A) = m then n [P (A)] = 2m
If A = Φ we get n (A) = 0
Hence, n [P(A)] = 20 = 1
Therefore, P (A) has only one element.
6. 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}
Solution:
(i) {x: x ∈ R, –4 < x ≤ 6} = (–4, 6]
(ii) {x: x ∈ R, –12 < x < –10} = (–12, –10)
(iii) {x: x ∈ R, 0 ≤ x < 7} = [0, 7)
(iv) {x: x ∈ R, 3 ≤ x ≤ 4} = [3, 4]
7. Write the following intervals in set-builder form:
(i) (–3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [–23, 5)
Solution:
(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}
8. What universal set (s) would you propose for each of the following?
(i) The set of right triangles
(ii) The set of isosceles triangles
Solution:
(i) Among the set of right triangles, the universal set will be the set of triangles or the set of polygons.
(ii) Among the set of isosceles triangles, the universal set will be the set of triangles or the set of polygons.
9. 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 universals 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:
(i) We can see that A ⊂ {0, 1, 2, 3, 4, 5, 6}
B ⊂ {0, 1, 2, 3, 4, 5, 6}
But C ⊄ {0, 1, 2, 3, 4, 5, 6}
Hence, the set {0, 1, 2, 3, 4, 5, 6} cannot be considered the universal set for the sets A, B, and C.
(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ
Hence, Φ cannot be the universal set for the sets A, B, and C.
(iii) 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}
Hence, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} may be considered as the universal set for the sets A, B, and C.
(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
But C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
Hence, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be considered the universal set for the sets A, B, and C.