Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 1 Sets Ex 1.1 help students build a strong mathematical foundation.
Ex 1.1 Class 9 Advanced Maths Solutions
Advanced Maths Class 9 Exercise 1.1 Solutions
Exercise 1.1 Class 9 Advanced Maths Solutions
Question 1.
List the elements of the following sets.
(a) {x : x is an integer and x2 = 9}
(b) {x : x is a positive integer less than 5}
(c) {x : x is even natural number divisible by 5}
(d) {x : x ∈ N and x < -1}
Solution:
(a) Given, x is an integer and x2 =9.
Now, x2 = 9 => x = ±3
Hence, the set is {-3, 3}.
(b) Given, x is a positive integer less than 5.
Positive integers less than 5 are 1, 2, 3, 4.
Hence, the set is {1, 2, 3, 4}.
(c) Given, x is an even natural number divisible by 5.
Now, numbers divisible by 5 are 5, 10, 15, 20, …
Among these, even numbers are 10, 20, 30,….
Hence, the set is {10, 20, 30, …}.
(d) Given, x ∈ N and x < -1
But, natural numbers are 1, 2, 3, …. and none is less than -1.
Hence, no such element exists.
Therefore, the set is Φ (empty set).
Question 2.
Determine which elements of the set
A = {-5, -√3, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), π, 13.4, \(\frac{1}{3}\), \(\frac{19}{2}\)} are
(a) natural numbers
(b) whole numbers
(c) integers
(d) rational numbers
(e) real numbers
Solution:
Given, A = {-5, -√3, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), π, 13.4, \(\frac{1}{3}\), \(\frac{19}{2}\)}
(a) Natural numbers are 1, 2, 3, …
From the set, no element is a positive integer.
Hence, natural numbers is Φ.
(b) Whole numbers are 0, 1, 2, 3, …
From the set, only 0 is a whole number.
Hence, whole numbers is {0}.
(c) Integers are…., -2, -1, 0, 1, 2, ….
From the set, integers are -5 and 0.
Hence, integers is {-5, 0}.
(d) Rational numbers are numbers of the form \(\frac{p}{q}\).
From the set,
-5, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), 13.4, \(\frac{1}{3}\), \(\frac{19}{2}\) are rational.
Hence, rational numbers
= { -5, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), 13.4, \(\frac{1}{3}\), \(\frac{19}{2}\)}.
(e) All given numbers are real numbers.
Hence, real numbers
= { -5, -√3, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), 13.4, π \(\frac{1}{3}\), \(\frac{19}{2}\)}.
Question 3.
Write the following sets in roster form.
(a) {x : x is a two-digit number and the sum of the digits is 5}
(b) {x : x is an integer and |x| ≤ 9}
(c) {x : x is letter of the word ‘SWEET’}
(d) {x : x = \(\frac{n+1}{n}\), where n is a natural number and n < 6}
(e) {x : x is a composite number}
Solution:
(a) Given, x is a two-digit number and the sum of its digits is 5.
Possible numbers are 14(1 + 4 = 5), 23, 32, 41, 50.
Hence, the set in roster form is {14, 23, 32, 41, 50}.
(b) Given, x is an integer and |x| ≥ 9.
So, x ≤ -9 or x ≥ 9
Hence, the set is {…, -11, -10, -9, 9, 10, 11,…}.
(c) Given, x is a letter of the word ‘SWEET’.
The letters in the word ‘SWEET’ (without repeating any letter) are S, W, E, T.
Hence, the set is {S, W, E, T}.
(d) Given, x = \(\frac{n+1}{n}\), where n is a natural number and n< 6.
so, n = 1, 2, 3, 4, 5 and x = \(\frac{2}{1}\), \(\frac{3}{2}\), \(\frac{4}{3}\), \(\frac{5}{4}\), \(\frac{6}{5}\).
Hence the set is {2, \(\frac{3}{2}\), \(\frac{4}{3}\), \(\frac{5}{4}\), \(\frac{6}{5}\)}
(e) Given, x is a composite number.
Composite numbers are 4, 6, 8, 9, 10, 12,…
Hence, the set is {4, 6, 8, 9, 10, 12, …}.
Question 4.
Write the following sets in set-builder form.
(i) {2, 4, 6, 8, …}
(ii) {3, 6, 9, 12, 15}
(iii) {1, 4, 9, 16, …}
(iv) {8, 9, 10, 11, …}
(v) {1, 2, 3, 6}
Can two different sets have the same roster form?
Solution:
(i) Let A = {2, 4, 6, 8, …}
Let x represent the elements of given set.
Given, numbers are even natural number.
Thus, A = {x : x = 2n, n ∈ N}.
(ii) Let B = {3, 6, 9, 12, 15}
Let x represents the elements of given set.
Given, numbers are the first five multiple of 5.
Thus, B = {x : x = 3n, n ∈ N, n ≤ 5}.
(iii) Let C = {1, 4, 9, 16, …}
Let x represent the elements of given set.
Given, number are the square of natural numbers.
Thus, C = {x : x = n2,n ∈ N}.
(iv) Let D = {8, 9, 10, 11, …}
Let x represent the element of given set.
Given, numbers are natural numbers greater than or equal to 8.
Thus, D = {x : x ∈ N, x ≥ 8}.
(v) Let E = {1, 2, 3, 6}
Given, numbers are factors of 6.
Thus, E = {x : x ∈ N, x divides 6}
Yes, two different sets can have the same roster form if they are defined differently but contain exactly the same elements.
Question 5.
Which of the following pairs of sets are equal.
(i) {D, E, C, E, N, T} and {C, E, N, T, D]
(ii) {a, b, π, √2} and {a, π, √2, b}
(iii) {x : x is zero of the polynomial x2} and {x : x is the root of the equation, x2 = 0}
(iv) {x : x has numerical value less than or equal to 1) and {x : x is the root of the equation, x2 – 1 = 0}
(v) {5, 10, 15, 20} and {5, 10, 15, 20, …}
(vi) Φ and {Φ}
Solution:
(i) Given, sets are
{D, E, C, E, N, T} = {D, E, C, N, T}
and {C, E, N, T, D} = {D, E, C, N, T}.
Here, we see that both sets have exactly the same elements.
Hence, both sets are equal.
(ii) Given, sets are {a, b, π, √2} and {a, π, √2, b] have exactly the same elements.
Hence, both sets are equal.
(iii) Let A = {x : x is zero of the polynomial x2}
= {0}
and B = {x : x is the root of the equation, x2 = 0}
= {0}
Here, both the sets contain single element 0.
Hence, both sets are equal.
(iv) Let A = {x : x has numerical value less than or equal to 1}
= {x : x ≤ 1}
= {… 4, -3 -1, 0, 1}
and B = {x : x is the root of the equation, x2 – 1 = 0} x2 – 1 = 0
(x – 1)(x + 1) = 0
x = 1, -1
So, B = {-1, 1}.
Here, the elements of both sets are not same.
Hence, both sets are not equal.
(v) Here, {5, 10, 15, 20} has four elements
and {5, 10, 15, 20, …} has infinitely many elements.
Hence, both sets are not equal.
(vi) Here, Φ is the empty set, but {Φ} is a set containing the empty set as an element.
Hence, both sets are not equal.
Question 6.
State which of the following sets are finite or infinite.
(i) {x : x ∈ Z and (x – 1)(x + 2)(x – 3) = 0}
(ii) {x : x and 2 are co-prime}
(iii) {x : x is a rational number between 3 and 4}
(iv) {x : x is an integer and |x| ≥ 5}
Solution:
(i) Given, (x – 1)(x + 2)(x – 3) = 0
⇒ x – 1 = 0 or x + 2= 0 or x – 3 = 0
⇒ x = 1, -2, 3
∴ {1, -2, 3} has 3 elements.
Hence, it is a finite.
(ii) Given, x and 2 are co-prime.
All odd integers are co-prime with 2.
So, infinitely many values of x exist.
Hence, the set is an infinite.
(iii) Given, x is a rational number between 3 and 4.
There are infinitely many rational numbers between any two numbers.
Hence, the set is an infinite.
(iv) Given, x is an integer and |x| ≥ 5.
So, x ≤ -5 or x ≥ 5.
There are infinitely many such integers.
Hence, the set is an infinite.
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