Samacheer Kalvi Class 9 Maths Solutions Chapter 1 Set Language Exercise 1.2

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.2

Question 1. Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {y : y = \( \frac{4}{3n} \), n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integer, x ∈ Z and − 5 ≤ x < 5}
(v) S = {x : x is all leap years between 1882 and 1906}.
Answer:
(i) M = {p, q, r, s, t, u}. This set lists six distinct elements.
So, the cardinal number of M is \( n(M) = 6 \).
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}. Here, 'n' is a whole number (starting from 0).
Let's find the values of x:
If \( n = 0, x = 3(0) + 2 = 2 \)
If \( n = 1, x = 3(1) + 2 = 5 \)
If \( n = 2, x = 3(2) + 2 = 8 \)
If \( n = 3, x = 3(3) + 2 = 11 \)
If \( n = 4, x = 3(4) + 2 = 14 \)
If \( n = 5, x = 3(5) + 2 = 17 \). This value is not less than 15, so we stop here.
So, P = {2, 5, 8, 11, 14}. This set contains five distinct elements.
Therefore, the cardinal number of P is \( n(P) = 5 \).
(iii) Q = {y : y = \( \frac{4}{3n} \), n ∈ N and 2 < n ≤ 5}. Here, 'n' is a natural number (starting from 1). The condition 2 < n ≤ 5 means n can be 3, 4, or 5.
Let's find the values of y:
If \( n = 3, y = \frac{4}{3(3)} = \frac{4}{9} \)
If \( n = 4, y = \frac{4}{3(4)} = \frac{4}{12} = \frac{1}{3} \)
If \( n = 5, y = \frac{4}{3(5)} = \frac{4}{15} \)
So, Q = { \( \frac{4}{9}, \frac{1}{3}, \frac{4}{15} \) }. This set contains three distinct elements.
Therefore, the cardinal number of Q is \( n(Q) = 3 \).
(iv) R = {x : x is an integer, x ∈ Z and − 5 ≤ x < 5}. This means 'x' can be any whole number from -5 up to, but not including, 5.
So, R = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4}. If you count them, there are ten distinct integers.
Therefore, the cardinal number of R is \( n(R) = 10 \).
(v) S = {x : x is all leap years between 1882 and 1906}. A leap year happens every four years, except for years ending in 00 that are not divisible by 400. We need to find the leap years in the given range.
The leap years between 1882 and 1906 are: 1884, 1888, 1892, 1896, 1904. (Note: 1900 was not a leap year because it is a century year not divisible by 400).
So, S = {1884, 1888, 1892, 1896, 1904}. This set contains five distinct years.
Therefore, the cardinal number of S is \( n(S) = 5 \).
In simple words: The cardinal number of a set is just how many different items or elements are inside that set. Count each item only once to find the cardinal number.

🎯 Exam Tip: Always list out the elements of a set explicitly before counting them, especially for sets defined by rules, to avoid mistakes.

Question 2. Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x\( ^2 \) – 5x + 6 = 0, x ∈ N}
Answer:
(i) X = The set of all districts in Tamilnadu.
The number of districts in Tamilnadu is a fixed and countable quantity. Although it can change over time, at any given moment, there's a specific count.
Therefore, X is a **Finite set**.
(ii) Y = The set of all straight lines passing through a point.
Through any single point, it is possible to draw an unlimited number of straight lines, each going in a slightly different direction. You can imagine an infinite number of lines radiating out from that point.
Therefore, Y is an **Infinite set**.
(iii) A = {x : x ∈ Z and x < 5}. This set includes all integers (positive, negative, and zero) that are less than 5.
So, A = {..., -2, -1, 0, 1, 2, 3, 4}. The '...' at the beginning means the numbers go on forever in the negative direction, so there is no end to the list.
Therefore, A is an **Infinite set**.
(iv) B = {x : x\( ^2 \) – 5x + 6 = 0, x ∈ N}. First, we need to solve the quadratic equation to find the values of x.
\( x^2 - 5x + 6 = 0 \)
\( (x - 2)(x - 3) = 0 \)
\( \implies x - 2 = 0 \) or \( x - 3 = 0 \)
\( \implies x = 2 \) or \( x = 3 \)
Both 2 and 3 are natural numbers (x ∈ N).
So, B = {2, 3}. This set contains two distinct elements, which is a fixed and countable number.
Therefore, B is a **Finite set**.
In simple words: A set is 'finite' if you can count all its elements and reach an end. A set is 'infinite' if its elements go on forever and you can never finish counting them.

🎯 Exam Tip: When a set is defined by a rule, always list out its elements if possible to confirm whether it has a countable number of items or not.

Question 3. Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word "VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = {x: x is a letter in the word "LIFE"}
Y = {F, I, L, E}
(iv) G = {x: x is a prime number and 3 < x < 23}
H = {1, 2, 3, 6, 9, 18}
Answer:
(i) A = The set of vowels in the English alphabets.
A = {a, e, i, o, u}. So, the cardinal number \( n(A) = 5 \).
B = The set of all letters in the word "VOWEL”.
B = {V, O, W, E, L}. So, the cardinal number \( n(B) = 5 \).
Since \( n(A) = n(B) = 5 \), the sets A and B have the same number of elements. They are **Equivalent sets**.
(ii) C = {2, 3, 4, 5}. So, the cardinal number \( n(C) = 4 \).
D = {x : x ∈ W, 1 < x < 5}. This means x is a whole number (0, 1, 2...) strictly between 1 and 5.
So, D = {2, 3, 4}. The cardinal number \( n(D) = 3 \).
Since \( n(C) \neq n(D) \) and the elements are also different, C and D are **Unequal sets**.
(iii) X = {x: x is a letter in the word "LIFE"}.
X = {L, I, F, E}. So, the cardinal number \( n(X) = 4 \).
Y = {F, I, L, E}. So, the cardinal number \( n(Y) = 4 \).
Since all elements in X are also in Y, and all elements in Y are also in X, these sets contain exactly the same elements. Therefore, X and Y are **Equal sets**.
(iv) G = {x: x is a prime number and 3 < x < 23}. A prime number is only divisible by 1 and itself.
The prime numbers between 3 and 23 are: 5, 7, 11, 13, 17, 19.
So, G = {5, 7, 11, 13, 17, 19}. The cardinal number \( n(G) = 6 \).
H = {1, 2, 3, 6, 9, 18}. The cardinal number \( n(H) = 6 \).
Since \( n(G) = n(H) = 6 \), the sets G and H have the same number of elements. They are **Equivalent sets**.
In simple words: 'Equal sets' have exactly the same items. 'Equivalent sets' have the same *number* of items, even if the items themselves are different. 'Unequal sets' are neither equal nor equivalent.

🎯 Exam Tip: Remember that equal sets are always equivalent, but equivalent sets are not always equal. Always compare both the elements and the cardinal numbers.

Question 4. Identify the following sets as null set or singleton set.
(i) A = {x: x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2
(iii) C = {0}.
(iv) D = The set of all triangles having four sides.
Answer:
(i) A = {x: x ∈ N, 1 < x < 2}. This set asks for natural numbers that are strictly between 1 and 2.
There are no natural numbers between 1 and 2. Natural numbers are whole numbers starting from 1 (1, 2, 3...).
Therefore, A is a **Null set** (or empty set).
(ii) B = The set of all even natural numbers which are not divisible by 2.
By definition, an even number is any integer that is divisible by 2. So, there cannot be an even natural number that is *not* divisible by 2.
Therefore, B is a **Null set**.
(iii) C = {0}. This set contains exactly one element, which is the number 0.
Therefore, C is a **Singleton set**.
(iv) D = The set of all triangles having four sides.
A triangle, by definition, is a polygon with exactly three sides. It is impossible for a triangle to have four sides.
Therefore, D is a **Null set**.
In simple words: A 'null set' is a set with no items inside it at all. A 'singleton set' is a set that has exactly one item inside it.

🎯 Exam Tip: Carefully read the conditions for each set. Sometimes, the conditions describe something that is impossible, resulting in a null set.

Question 5. State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s}
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
(iii) E = {x : x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
Answer:
(i) A = {f, i, a, s} and B = {a, n, f, h, s}.
To check if they are disjoint or overlapping, we look for common elements (their intersection).
The common elements are 'a', 'f', and 's'. So, \( A \cap B = \{a, f, s\} \).
Since the intersection is not empty (it has common elements), A and B are **Overlapping sets**.
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}.
First, let's list the elements of each set:
C = {prime numbers greater than 2} = {3, 5, 7, 11, 13, 17, 19, ...}. All prime numbers greater than 2 are odd.
D = {even prime numbers}. The only even prime number is 2.
So, D = {2}.
Now, let's find the intersection: \( C \cap D = \emptyset \) (there are no common elements).
Since their intersection is the empty set, C and D are **Disjoint sets**.
(iii) E = {x : x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}.
First, let's list the elements of each set:
E = {factors of 24} = {1, 2, 3, 4, 6, 8, 12, 24}.
F = {multiples of 3 less than 30} = {3, 6, 9, 12, 15, 18, 21, 24, 27}.
Now, let's find the intersection: The common elements are 3, 6, 12, 24.
So, \( E \cap F = \{3, 6, 12, 24\} \).
Since the intersection is not empty, E and F are **Overlapping sets**.
In simple words: 'Overlapping sets' share some common items. 'Disjoint sets' have no items in common at all. We check this by finding what items are in both sets.

🎯 Exam Tip: To determine if sets are disjoint or overlapping, always find their intersection. If the intersection is empty, they are disjoint; otherwise, they are overlapping.

Question 6. If S = {square, rectangle, circle, rhombus, triangle}. List the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°.
(iv) The set of shapes which have 5 sides.
Answer:
Given set S = {square, rectangle, circle, rhombus, triangle}.
(i) The set of shapes which have 4 equal sides.
From the set S, a square has 4 equal sides, and a rhombus also has 4 equal sides. A rectangle has 4 sides, but only opposite sides are equal. A triangle has 3 sides, and a circle has no straight sides.
So, the subset is {square, rhombus}.
(ii) The set of shapes which have radius.
From the set S, a circle has a radius. Other shapes like square, rectangle, rhombus, and triangle do not have a radius as a defining feature.
So, the subset is {circle}.
(iii) The set of shapes in which the sum of all interior angles is 180°.
In geometry, the sum of the interior angles of a triangle is always 180°. Other shapes like squares, rectangles, and rhombuses are quadrilaterals, where the sum of interior angles is 360°.
So, the subset is {triangle}.
(iv) The set of shapes which have 5 sides.
From the set S, none of the listed shapes (square, rectangle, circle, rhombus, triangle) have exactly 5 sides. Shapes with 5 sides are called pentagons.
So, the subset is \( \emptyset \) (an empty set, as there are no such shapes in S).
In simple words: A subset is a collection of items taken from a larger set. We pick only the items that fit a certain description from the main list.

🎯 Exam Tip: When forming a subset, ensure every element you include is present in the original set and satisfies the given condition.

Question 7. If A = {a,{a, b}}, write all the subsets of A.
Answer:
Given set A = {a, {a, b}}.
This set A has two elements: the first element is 'a', and the second element is the set '{a, b}'.
Let's consider these two distinct elements as X = 'a' and Y = '{a, b}'. So, A = {X, Y}.
The subsets of any set with 'n' elements are \( 2^n \). Since A has 2 elements, it will have \( 2^2 = 4 \) subsets.
The subsets are:
1. The empty set: \( \emptyset \)
2. The set containing only the first element: {a}
3. The set containing only the second element: {{a, b}}
4. The set containing both elements (which is A itself): {a, {a, b}}
So, the subsets of A are \( \emptyset, \{a\}, \{\{a, b\}\}, \{a, \{a, b\}\} \). A power set contains all possible subsets of a given set.
In simple words: To find all subsets, start with an empty set. Then list sets with one item from the original set, then sets with two items, and so on, until you list the original set itself.

🎯 Exam Tip: Pay close attention when a set contains other sets as elements. Treat each element (whether it's a number, a letter, or another set) as a single, distinct item when forming subsets.

Question 8. write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = \( \emptyset \)
Answer:
(i) A = {a, b}. This set has 2 elements, so its power set will have \( 2^2 = 4 \) subsets.
The power set P(A) = { \( \emptyset \), {a}, {b}, {a, b} }.
(ii) B = {1, 2, 3}. This set has 3 elements, so its power set will have \( 2^3 = 8 \) subsets.
The power set P(B) = { \( \emptyset \), {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }.
(iii) D = {p, q, r, s}. This set has 4 elements, so its power set will have \( 2^4 = 16 \) subsets.
The power set P(D) = { \( \emptyset \), {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {p, q, s}, {p, r, s}, {q, r, s}, {p, q, r, s} }.
(iv) E = \( \emptyset \). This set has 0 elements, so its power set will have \( 2^0 = 1 \) subset.
The power set P(E) = { \( \emptyset \) }.
Note: The empty set is always a subset of every set. Its power set contains only the empty set itself. This means the power set of an empty set is not empty.
In simple words: The power set of any set is a new set that contains ALL possible subsets of the original set. If a set has 'n' items, its power set will always have \( 2^n \) subsets.

🎯 Exam Tip: Always remember that the empty set \( \emptyset \) is a subset of every set, and the set itself is also one of its subsets when forming a power set.

Question 9. Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red, blue, yellow}
(ii) X = {x\( ^2 \) : x ∈ N, x\( ^2 \) ≤ 100}
Answer:
(i) W = {red, blue, yellow}.
The number of elements in W is \( n(W) = 3 \).
Number of subsets = \( 2^{n(W)} = 2^3 = 8 \).
Number of proper subsets = \( 2^{n(W)} - 1 = 2^3 - 1 = 8 - 1 = 7 \).
(ii) X = {x\( ^2 \) : x ∈ N, x\( ^2 \) ≤ 100}.
Here, x is a natural number (1, 2, 3, ...) and its square must be less than or equal to 100.
So, x can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. (Because \( 10^2 = 100 \) and \( 11^2 = 121 \), which is greater than 100).
Therefore, X = { \( 1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2, 10^2 \) } = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}.
The number of elements in X is \( n(X) = 10 \).
Number of subsets = \( 2^{n(X)} = 2^{10} = 1024 \).
Number of proper subsets = \( 2^{n(X)} - 1 = 2^{10} - 1 = 1024 - 1 = 1023 \).
In simple words: The number of subsets tells you how many different groups you can make from the items in a set, including an empty group and the full group. 'Proper subsets' are all these groups *except* for the full group itself.

🎯 Exam Tip: Remember the formulas: Number of subsets = \( 2^n \) and Number of proper subsets = \( 2^n - 1 \), where 'n' is the number of elements in the set.

Question 10.
(i) If n(A) = 4, find n[P(A)]
(ii) If n(A) = 0, find n[P(A)]
(iii) If n[P(A)] = 256, find n(A)
Answer:
(i) If n(A) = 4, find n[P(A)].
The number of elements in set A is 4. The number of elements in the power set P(A) is \( 2^{n(A)} \).
So, \( n[P(A)] = 2^4 = 16 \).
(ii) If n(A) = 0, find n[P(A)].
The number of elements in set A is 0, meaning A is an empty set ( \( A = \emptyset \) ).
The number of elements in the power set P(A) is \( 2^{n(A)} \).
So, \( n[P(A)] = 2^0 = 1 \). The power set of an empty set contains only the empty set itself: \( P(\emptyset) = \{\emptyset\} \).
(iii) If n[P(A)] = 256, find n(A).
We know that \( n[P(A)] = 2^{n(A)} \). We are given \( n[P(A)] = 256 \).
So, \( 2^{n(A)} = 256 \).
To find \( n(A) \), we need to express 256 as a power of 2:
\( 256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8 \). This means 2 multiplied by itself 8 times gives 256.
Thus, \( 2^{n(A)} = 2^8 \).
Therefore, \( n(A) = 8 \).
In simple words: The number of items in the power set is found by raising 2 to the power of how many items are in the original set. If you know the power set's size, you can work backward to find the original set's size.

🎯 Exam Tip: Familiarize yourself with powers of 2 (e.g., \( 2^0=1, 2^1=2, 2^2=4, \dots, 2^{10}=1024 \)) as they are often used in set theory questions involving power sets.