Class 9 Advanced Maths Chapter 1 Sets Ex 1.2 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 1 Sets Ex 1.2 to verify their answers.

Advanced Maths Class 9 Exercise 1.2 Solutions

Class 9 Advanced Maths Ex 1.2 Solutions

Question 1.
Fill in the blanks with symbol c or cz.
(i) {2, 3, 4} {1, 2, 3, 4, 5}
(ii) {x : x are triangles in a plane} ………….. {x : x are polygons in a plane}
(iii) {x is an integer} ………….. {x : x is a multiple of 4}
(iv) Φ …………. {Φ}
(v) {x : x = \(\frac{m-1}{m}\), where m is a non-zero integer} …………….. {x : x is a rational number}.
(vi) {x : x = n2} ………….. {x : x = n3}, (where n is a
natural number)
(vii) {x : x ∈ R} ………….. {x : x = 2n}, (where, R is a real number)
Solution:
(i) Every element of {2, 3, 4} belongs to {1, 2, 3, 4, 5}.
Hence, {2, 3,4} ⊂ {1,2, 3,4, 5}.

(ii) Every triangle is a polygon.
Hence, {x | x are triangles in a plane} ⊂ {x | x are polygons in a plane}.

(iii) Not every integer is a multiple of 4.
For example, 3 is an integer but not a multiple of 4.
Hence, {x is an integer} ⊄ {x : x is a multiple of 4}.

(iv) The empty set is a subset of every set.
Hence, Φ ⊂ {Φ}.

(v) Since \(\frac{m-1}{m}\) is always a rational number for every non-zero integer m, every element of the first set is rational.
Hence, {x | x = \(\frac{m-1}{m}\), where m is a non-zero integer} ⊄ {x | x is a rational number}.

(vi) Not every perfect square is a perfect cube.
For example, 4 = 2² but 4 is not a perfect cube.
Hence, {x | x = n²} ⊄ {x | x = n³}.

(vii) The set {x | x = 2ⁿ} represents powers of 2, whereas {x | x ∈ R} represents all real numbers.
Not every real number is a power of 2. For example, 3 ∈ R but 3 ≠ 2ⁿ for any natural number n.
Hence, {x | x ∈ R} ⊄ {x | x = 2ⁿ}.

Question 2.
Determine whether the following statements are true or false.
(i) 1 ∈ {1}
(ii) {2} ∈ {2}
(iii) {2} ∈ {{2}}
(iv) Φ ∈ {1, 2, 3}
(v) Φ ⊆ {1, 2, 3}
Solution:
(i) The set {1} contains the element 1. Hence, the statement is True.
(ii) The set {2} contains the number 2, not the set {2}. Therefore, {2} is not an element of {2}. Hence, the statement is False.

(iii) The set {{2}} contains one element, namely {2}. Therefore, {2} is an element of {{2}}.
Hence, the statement is True.

(iv) The set {1, 2, 3} does not contain the empty set as an element.
Hence, the statement is False.

(v) The empty set is a subset of every set. Hence, the statement is True.

Question 3.
Write the power set of the following sets:
(i) {1}
(ii) {p, q}
(iii) {1, 2, 5}
(iv) {Φ, {Φ}}
Solution:
(i) Let A = {1}
The subsets of A are Φ and {1}.
Since the power set of a set is the set containing all its subsets.
Therefore, the power set of A is P(A) = {Φ , {1}}.

(ii) Let B = {p, q}
The subsets of B are Φ, {p}, {q} and {p, q}.
Therefore, the power set of B is P(B) = {Φ, {p}, {q}, {p, q}}.

(iii) Let C = {1, 2, 5}
The subsets of C are Φ, {1}, {2}, {5}, {1, 2}, {2, 5}, {1, 5} and {1, 2, 5}.
Therefore, the power set of C is P(C) = {Φ, {1}, {2}, {5}, {1, 2}, {2, 5}, {1, 5}, {1, 2, 5}}.

(iv) Let D = {Φ, {Φ}}
The subsets of D are Φ, {Φ}, {{Φ}}, and {Φ, {Φ}}.
Therefore, the power set of D is P(D) = {Φ, {Φ}, {{Φ}}, {Φ,{Φ}}}.

Question 4.
What is the cardinality of the following sets?
(i) {a}
(ii) {a, {a}}
(iii) {Φ, 1, 2, {1, 2}}
(iv) {1, {1}, {1, {1}}}
(v) {Φ, {Φ}, {Φ{Φ}}}
Solution:
(i) Let A = {a}
The set A contains only one element, namely a.
Therefore, the cardinality of the set A is n(A) = 1.

(ii) Let B = {a, {a}}
The set B contains two elements, namely a and {a}.
Therefore, the cardinality of the set B is n(B) = 2.

(iii) Let C= {Φ, 1, 2, {1, 2}}
The set C contains 4 elements, namely Φ, 1, 2, {1, 2}.
Therefore, the cardinality of the set C is n(C) = 4.

(iv) Let D = {1, {1}, {1, {1}}}
The set D contains 3 elements, namely 1, {1}, {1, {1}}.
Therefore, the cardinality of the set D is n(D) = 3.

(v) Let E = {Φ, {Φ}, {Φ, {Φ}}}
The set E contains 3 elements, namely Φ, {Φ}, {Φ, {Φ}}.
Therefore, the cardinality of the set E is n(E) = 3.

Question 5.
Let A be a set and n(A) = 10, then find the value of n[P(A)]? What it A has 100 elements?
Solution:
Let A be a set such that n(A) = 10
We know that if a set contains n elements, then the number of elements in its power set is given by
n[P(A)] = 2ⁿ
Therefore, n[P(A)] = 2¹⁰ = 1024
Hence, the power set of A contains 1024 elements.
If the set A has 100 elements, then n[P(A)] = 2¹⁰⁰
= 1.27 × 10³⁰.