# RBSE Solutions for Class 11 Maths Chapter 1 Sets Ex 1.1

## Rajasthan Board RBSE Class 11 Maths Chapter 1 Sets Ex 1.1

Question 1.
Fill the symbols ∈ or ∉ in the blanks to make the following statements correct:
(i) 3…. {1, 2, 3, 4, 5}
(ii) 2.5…N
(iii) 0………Q
Solution:
(i) ∈
(ii) ∉
(iii) ∈

Question 2.
Fill the symbols ⊂ or ⊄ in the blanks to make following statements correct:
(i) {2, 3, 4}……….{1, 2, 3, 4, 5}
(ii) {a, e, o}……..{a, b, c}
(iii) {x : x is an equilateral triangle in a plane}……..{x : x is a triangle in a plane}
(iv) {x : x is a natural number}…….{x : x is an odd whole number}
Solution:
(i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}
(ii) {a, e, o} ⊄ {a, b, c}
(iii) Many triangles are possible in a plane.
So, {x: x is an equilateral triangle in a plane} ⊂ (x : x is a triangle in a plane}
(iv) First set in the tabular form is (1, 2, 3, 4,….} and second set in the tabular form is {…… -3, -1, 0, 1, 3, 5,..}.
So, (x : x is a natural number} ⊄ {x : x is an odd whole number}.

Question 3.
Examine the following statements :
(i) {a, b} ⊂ {b, a, c]
(ii) {a, e} ⊂ (x : x is vowel of English alphabet}
(iii) {1, 2, 3} ⊄ {1, 3, 2, 5}
(iv) {x : x is an even natural number less than 6} ⊄ {x : x is a natural number which divide 36}
Solution:
(i) {a, b} ⊂ {b, a, c}
Elements of first set (a, b) is present in second set.
Hence, the statement is true.

(ii) {a, e} ⊂ {x : x is vowel of English alphabet}
Tabular form of seconds set is {a, e, i, o, u}
Elements of first set (a, e is present is second set.
Hence, the statement is true.

(iii) {1, 2, 3} ⊄ {1, 3, 2, 5}
Elements of first set is also the elements of seconds set.
Hence, the statement is false.

(iv) {x : x is an even natural number less then 6} ⊂ (x : x is a natural number which divide 36}
{2, 4} ⊂ {1, 2, 3, 4, 6, 9, 12, 18, 36}
Hence, the statement is true.

Question 4.
Write power sets of the following:
(i) {a}
(ii) {1, 2, 3}
(iii) {a, b}
(iv) Φ
Solution:
(i) {a}
Subsets of {a} are Φ and {a}.
Hence P{a} = {Φ, {a}}

(ii) {1, 2, 3}
Subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}.
Hence, P{1, 2, 3} = {Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

(iii) {a, b}
Subsets of {a, b} are Φ, {a}, {b}, {a, b}
Hence, P{a, b} = {Φ, {a}, {b}, {a, b}}

(iv) Φ
Φ is null set, then P(Φ) = {Φ}

Question 5.
Write the following as intervals :
(i) {x : x ∈ R, -3 < x < 6}
(ii) {x : x ∈ R, -4 ≤ x ≤ 8}
(iii) {x : x ∈ R, 4 < x ≤ 9}
(iv) {x : x ∈ R, -6 ≤ x < -1}
Solution:
(i) {x : x ∈ R, -3 < x < 6}
-3 < x < 6 is open interval where -3 and 6 are not included.
Hence, {x : x ∈ R, -3 < x < 6} has interval (-3, 6).

(ii) {x : x ∈ R, -4 ≤ x ≤ 8}
-4 ≤ x ≤ 8 is closed interval in which (-4 and 8) are included.
Hence, interval is [-4, 8]

(iii) {x : x ∈ R, 4 < x ≤ 9}
4 < x ≤ 9 is semi clsoed interval where the interval is less than 4 and equal to 9.
Hence, {x : x ∈ R, 4 < x ≤ 9} has interval (4, 9].

(iv) {x : x ∈ R, -6 ≤ x < -1}
-6 ≤ x < -1 is semi closed interval where the interval is equal to -6 and less than -1.
Hence, {x : x ∈ R, – 6 ≤ x < -1} has interval [-6, -1)

Question 6.
Write the following set in builder form.
(i) (-4, 0)
(ii) [6, 8]
(iii) [-3, 7)
(iv) (3, 10]
Solution:
(i) (-4, 0) = {x : x ∈ R, -4 < x < 0}
(ii) [6, 8] = {x : x ∈ R, 6 ≤ x ≤ 8}
(iii) [- 3, 7) = {x : x ∈ R, -3 ≤ x < 7}
(iv) (3, 10] = (x : x ∈ R, 3 < x ≤ 10}

Question 7.
A = {1, 3, 5}, B = {1, 4, 6} and C = {2, 4, 6, 8}, then which of the following may be considered as in universal set:
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) (1, 2, 3, 4, 5, 6, 7, 8}
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) Φ
Solution:
For set A = {1, 3, 5}, B = {2, 4, 6} and C = {2, 4, 6, 8}
we have (ii) and (iii) universal set, because it contains all the elements of the given sets A, B and C.
Hence, option (ii) and (iii) are correct.