Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 1 Sets Ex 1.1 to verify their answers.
Advanced Maths Class 9 Exercise 1.1 Solutions
Class 9 Advanced Maths Ex 1.1 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) x2 = 9 ⇒ x = 3 or x = -3
Hence, the set is {-3, 3}.
(b) The positive integers less than 5 are 1, 2, 3, and 4.
Hence, the set is {1, 2, 3, 4}.
(c) A number divisible by both 2 (even number) and 5 must be divisible by 10.
Hence, the set is {10, 20, 30, 40,…}.
(d) There is no natural number less than -1.
Therefore, the set is the empty set Φ.
Question 2.
Determine which elements of the set
A = {-5, -√3, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), π, 13.4, \(\frac{1}{3}\), \(\frac{19}{2}\)}
(a) natural numbers.
(b) whole numbers.
(c) integers.
(d) rational numbers.
(e) real numbers.
Solution:
| Element | Classification |
| -5 | Negative integer. It is an integer, rational number, real number. |
| \(-\sqrt{3}\) | Irrational (non-terminating, non-recurring decimal). It is a real number only. |
| \(-\frac{1}{2}\) | Negative fraction. It is a rational number and real number. |
| 0 | Zero is a whole number, integer, rational number and real number. (Not a Natural number.) |
| \(\frac{2}{5}\) | Positive fraction = 0.4 (terminating deci-mal). It is a rational number and real num-ber. |
| π | Irrational (non-terminating, non-recur¬ring). It is a real number only. |
| 13.4 | Terminating decimal. It is a rational number and real number. |
| \(\frac{1}{3}\) | Non-terminating recurring decimal (0.333…). It is a rational number and real number. |
| \(\frac{19}{2}\) | \(\frac{19}{2}\) = 9.5 a positive terminating decimal. It is a rational number and real number. |
(a) Φ, none of the element of A is a natural number
(b) {0}
(c) {-5, 0}
(d) -5, –\(\frac{1}{2}\), 0, \(\frac{2}{5}\), 13.4, \(\frac{1}{3}\), \(\frac{19}{2}\)
(e) { -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 the 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) The two-digit numbers whose digits add up ,to 5 are 14, 23, 32, 41, 50.
Hence, the set in roster form is {14, 23, 32, 41, 50}.
(b) |x| ≥ 9 → x ≤ -9 or x ≥ 9, so x can be …, -12,-11, -10, -9, 9,10,11,12,…
Hence, the set in roster form is {…, -12, -11, -10, -9, 9, 10, 11,12,…}.
(c) The letters of the word “SWEET” are S, W, E, E, T. Since repetition is not allowed in a set, the set becomes {S,W,E, T}.
(d) The value of n is less than 6, so the values of n are: 1, 2, 3, 4, 5, x = 2, \(\frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5}\)
For n = 1:
x = \(\frac{n+1}{n}=\frac{1+1}{1}\) = 2
Similarly, for n = 2, 3, 4, 5, the values of x are \(\frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5} \).
Hence, the set in roster form is \(\left\{2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \frac{6}{5}\right\}\)
(e) Composite numbers are positive integers greater than 1 that have factors other than 1 and themselves.
The composite numbers are 4, 6,8, 9,10,12, 14,…
Hence, the set in roster form is {4, 6, 8, 9, 10, 12, 14,…}.
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) This is the set of even natural numbers.
Hence, in set-builder form, {x : x = 2n, n ∈ N}
(ii) These are the first five multiples of 3.
Hence, in set-builder form, {x : x = 3n, n ∈ N, n ≤ 5}
(iii) These are the perfect squares of the natural numbers:
12, 22, 32, 42…
Hence, in set-builder form, {x : x = n2, n ∈ N}
(iv) This is the set of natural numbers greater than or equal to 8.
Hence, in set-builder form, {x : x ∈ N, x ≥ 8}
(v) These are the natural numbers that are the factors of 6. Hence, in set-builder form, {x : x is a natural number and factor of 6}
No, two different sets cannot have the same roster form because a set is completely determined by its elements. If two sets contain the same elements, then they are equal sets.
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) In a set, repeated elements are counted only once.
Therefore, {D, E, C, E, N, T} = {C, D, E, N, T}
and {C, E, N, T, D} = {C, D, E, N, T}
Since both sets contain the same elements,
Hence, the given sets are equal.
(ii) Both sets contain the same elements a, b, √2, π.
The order of elements does not matter in a set.
Hence, the given sets are equal.
(iii) Zero of x2 : x2 = 0 gives x = 0, so the set is {0}.
For root of the equation, x2 = 0 gives x = 0, so the set is {0}.
Hence, both sets are {0}.
Therefore, the two sets are equal.
(iv) The first set represents all numbers satisfying x ≤ 1, which contains infinitely many numbers less than or equal 1: x ≤ 1
For the second set, x2 – 1 = 0
⇒ (x – 1)(x + 1) = 0
⇒ x = ±1
Thus, the second set is {-1,1}.
Since the two sets do not contain the same elements, they are not equal.
(v) The first set contains only four elements, namely 5, 10, 15, and 20. The second set contains infinite multiples of 5. The number of elements in the two sets are different.
Hence, the sets are not equal.
(vi) Φ denotes the empty set, which contains no element.
{Φ} is a set containing one element, namely the empty set itself.
Therefore, the two 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) (x – 1)(x + 2)(x – 3) = 0
⇒ x = 1,-2, 3
Hence, the set is {-2, 1, 3}.
Since the set contains only three elements, it is a finite set.
(ii) A number is coprime with 2 if it has no common factor with 2 other than 1. Thus, all odd integers are coprime with 2.
Hence, the set is {… -5, -3, -1,1, 3,5,7,9,…}.
Since there are infinitely many odd numbers, the given set is infinite.
(iii) We know that between any two distinct real numbers, there are infinitely many rational numbers.
Therefore, the given set contains infinitely many elements. Hence, the given set is infinite.
(iv) |x| ≥ 5 ⇒ x ≤ -5 or x ≥ 5.
Hence, the set can be written as {…, -7, -6, -5, 5, 6,7,…}.
There are infinitely many integers satisfying the given condition. Hence, the set is infinite.