Maharashtra Board Class 11 Maths Part 1 Chapter 1 Sets and Relations Miscellaneous Solutions

Question 1. Write the following sets in set builder form:
(i) \( \{10, 20, 30, 40, 50\} \)
(ii) \( \{a, e, i, o, u\} \)
(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Answer:
(i) \( \{x \mid x = 10n, n \in \mathbb{N} \text{ and } n \le 5\} \)
(ii) \( \{x \mid x \text{ is a vowel in the English alphabet}\} \)
(iii) \( \{x \mid x \text{ is a day of the week}\} \)
Set-builder notation is a highly efficient mathematical shorthand used to describe a set by specifying the exact properties that its members must satisfy.
In simple words: Set-builder form is just a way of writing a set by describing its common rule or property instead of listing all its elements. For example, instead of listing all the vowels, we just write a rule that says 'x is a vowel'.

🎯 Exam Tip: When writing sets in set-builder form, always clearly define the variable (like \(x\)) and state the exact condition or formula it must satisfy, including the set of numbers it belongs to (like natural numbers \( \mathbb{N} \)).

Question 1. Write the following sets in set-builder form:
(i) Let A = {10, 20, 30, 40, 50}
(ii) Let B = {a, e, i, o, u}
(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Answer:
(i) \( \therefore A = \{x \mid x = 10n, n \in \mathbb{N} \text{ and } n \le 5\} \)
(ii) \( \dots B = \{x \mid x \text{ is a vowel of English alphabets}\} \)
(iii) \( \dots C = \{x \mid x \text{ represents days of a week}\} \)
This method is highly efficient for representing infinite sets or sets with a clear mathematical pattern.
In simple words: Set-builder form describes the common property of all the elements in a set instead of listing them one by one.

🎯 Exam Tip: Always define the domain of the variable (like \( n \in \mathbb{N} \)) and its limits clearly when writing in set-builder form.

Question 2. If \( U = \{x \mid x \in \mathbb{N}, 1 \le x \le 12\} \), \( A = \{1, 4, 7, 10\} \), \( B = \{2, 4, 6, 7, 11\} \), \( C = \{3, 5, 8, 9, 12\} \). Write the sets:
(i) \( A \cup B \)
(ii) \( B \cap C \)
(iii) \( A - B \)
(iv) \( B - C \)
(v) \( A \cup B \cup C \)
(vi) \( A \cap (B \cup C) \)
Answer:
Given:
\( U = \{x \mid x \in \mathbb{N}, 1 \le x \le 12\} = \{1, 2, 3, \dots, 12\} \)
\( A = \{1, 4, 7, 10\} \)
\( B = \{2, 4, 6, 7, 11\} \)
\( C = \{3, 5, 8, 9, 12\} \)

(i) \( A \cup B = \{1, 2, 4, 6, 7, 10, 11\} \)
(ii) \( B \cap C = \{ \} \)
(iii) \( A - B = \{1, 10\} \)
(iv) \( B - C = \{2, 4, 6, 7, 11\} \)
(v) \( A \cup B \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \)
(vi) \( B \cup C = \{2, 3, 4, 5, 6, 7, 8, 9, 11, 12\} \)
\( \therefore A \cap (B \cup C) = \{4, 7\} \)
These operations form the foundational algebra of sets used across modern mathematics.
In simple words: Union (\( \cup \)) combines all elements from both sets, intersection (\( \cap \)) finds only the common elements, and subtraction (\( - \)) removes elements of the second set from the first.

🎯 Exam Tip: Double-check each element carefully when performing set operations to avoid missing any numbers or including duplicates in unions.

Question 3. In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice and 80 drink both apple as well as orange juice. Find how many students drink neither apple juice nor orange juice.
Answer:
Let \( U \) be the set of surveyed students, \( A \) be the set of students who drink apple juice, and \( B \) be the set of students who drink orange juice.
\( n(U) = 425 \)
\( n(A) = 115 \)
\( n(B) = 160 \)
\( n(A \cap B) = 80 \)
We need to find the number of students who drink neither apple juice nor orange juice, which is \( n(A' \cap B') \).
By De Morgan's Law:
\( n(A' \cap B') = n((A \cup B)') = n(U) - n(A \cup B) \)
First, find \( n(A \cup B) \):
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)

\( \implies n(A \cup B) = 115 + 160 - 80 = 195 \)
Now, find \( n(A' \cap B') \):
\( n(A' \cap B') = 425 - 195 = 230 \)
Thus, 230 students drink neither apple juice nor orange juice. Using Venn diagrams can also help visualize these disjoint and overlapping groups of students easily.
In simple words: Out of 425 students, 195 drink at least one type of juice. Subtracting this from the total gives 230 students who do not drink either juice.

🎯 Exam Tip: Always state what each set represents (like \( U, A, B \)) before writing down the numerical values to keep your steps clear and logical.

Question 3. ... drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?
Answer:
Let \( A \) = set of students who drink apple juice
\( B \) = set of students who drink orange juice
\( X \) = set of all students

\( \therefore n(X) = 425, n(A) = 115, n(B) = 160, n(A \cap B) = 80 \)

Diagram representation:

• Set A (only): 35
• Intersection of A and B: 80
• Set B (only): 80
• Universal Set X

No. of students who neither drink apple juice nor orange juice:
\( n(A' \cap B') = n(A \cup B)' \)
\( = n(X) - n(A \cup B) \)
\( = 425 - [n(A) + n(B) - n(A \cap B)] \)
\( = 425 - (115 + 160 - 80) \)
\( = 230 \)
Thus, there are 230 students who do not drink either of the two juices.
In simple words: To find the students who drink neither juice, we first find how many drink at least one juice (which is 195) and subtract this from the total number of students (425), giving us 230.

🎯 Exam Tip: Always define your sets clearly at the beginning and use De Morgan's Laws \( n(A' \cap B') = n(X) - n(A \cup B) \) to solve "neither/nor" questions easily.

Question 4. In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?
Answer:
Let \( A \) = set of teachers who teach Mathematics
\( B \) = set of teachers who teach Physics

\( n(A \cup B) = 20, n(A) = 12, n(A \cap B) = 4 \)

Diagram representation:

• Set A (only): 8
• Intersection of A and B: 4
• Set B

Since, \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
\( \therefore 20 = 12 + n(B) - 4 \)
\( \dots n(B) = 12 \)
\( \therefore \) Number of teachers who teach physics = 12. This shows that exactly 12 teachers in the school are qualified to teach physics.
In simple words: Out of 20 teachers, 12 teach math. Since 4 teach both, the remaining 8 teach only math. This leaves 12 teachers who must teach physics to make up the total of 20.

🎯 Exam Tip: Remember the standard formula \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \) and substitute the given values carefully to find the unknown set.

Question 5.
(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A
Answer:
(i) The elements of the Cartesian products are:
\( A \times A = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\} \)
\( A \times B = \{(1,2), (1,4), (2,2), (2,4), (3,2), (3,4)\} \)
\( B \times A = \{(2,1), (2,2), (2,3), (4,1), (4,2), (4,3)\} \)
\( B \times B = \{(2,2), (2,4), (4,2), (4,4)\} \)
These Cartesian products represent all possible ordered pairs formed by taking the first element from the first set and the second element from the second set.
In simple words: Cartesian product means pairing every element of the first set with every element of the second set to form pairs like (x, y).

🎯 Exam Tip: When writing Cartesian products, always ensure the number of elements in \( A \times B \) equals \( n(A) \times n(B) \) to double-check your work.

Question 5. (i) Find \( (A \times B) \cap (B \times A) \) if \( A = \{1, 2, 3\} \) and \( B = \{2, 4\} \).
(ii) If \( A = \{-1, 1\} \), find \( A \times A \times A \).
Answer:
(i) Given \( A = \{1, 2, 3\} \) and \( B = \{2, 4\} \).
\( A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \)
\( A \times B = \{(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)\} \)
\( B \times A = \{(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)\} \)
\( B \times B = \{(2, 2), (2, 4), (4, 2), (4, 4)\} \)
\( \implies (A \times B) \cap (B \times A) = \{(2, 2)\} \)

(ii) Given \( A = \{-1, 1\} \).
\( \implies A \times A \times A = \{(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)\} \)
In simple words: To find the intersection of two Cartesian products, we look for the ordered pairs that are common to both sets. For the triple product, we list all possible combinations of three elements chosen from the set.

🎯 Exam Tip: When writing Cartesian products, ensure you list all ordered pairs systematically to avoid missing any combinations, and double-check the total number of elements using the formula \( n(A \times B) = n(A) \times n(B) \).

Question 6. If \( A = \{1, 2, 3\} \), \( B = \{4, 5, 6\} \), which of the following are relations from \( A \) to \( B \)?
(i) \( R_1 = \{(1, 4), (1, 5), (1, 6)\} \)
(ii) \( R_2 = \{(1, 5), (2, 4), (3, 6)\} \)
(iii) \( R_3 = \{(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)\} \)
(iv) \( R_4 = \{(4, 2), (2, 6), (5, 1), (2, 4)\} \)
Answer:
Given \( A = \{1, 2, 3\} \) and \( B = \{4, 5, 6\} \).
The Cartesian product is:
\( A \times B = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \)

(i) \( R_1 = \{(1, 4), (1, 5), (1, 6)\} \)
Since \( R_1 \subseteq A \times B \),
\( \implies R_1 \) is a relation from \( A \) to \( B \).

(ii) \( R_2 = \{(1, 5), (2, 4), (3, 6)\} \)
Since \( R_2 \subseteq A \times B \),
\( \implies R_2 \) is a relation from \( A \) to \( B \).

(iii) \( R_3 = \{(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)\} \)
Since \( R_3 \subseteq A \times B \),
\( \implies R_3 \) is a relation from \( A \) to \( B \).

(iv) \( R_4 = \{(4, 2), (2, 6), (5, 1), (2, 4)\} \)
Since \( (4, 2) \in R_4 \), but \( (4, 2) \notin A \times B \),
\( \implies R_4 \) is not a relation from \( A \) to \( B \).
In simple words: A relation from set A to set B is simply any subset of the Cartesian product \( A \times B \). If a set contains even one ordered pair that is not in \( A \times B \), it cannot be a relation from A to B.

🎯 Exam Tip: To prove a subset is a relation, show that every ordered pair in it belongs to the Cartesian product \( A \times B \). If even one element like \( (4, 2) \) has its first element not in \( A \), it is not a relation.

Question 7. Determine the domain and range of the following relation.
\( R = \{(a, b) / a \in N, a < 5, b = 4\} \)
Answer:
\( R = \{(a, b) / a \in N, a < 5, b = 4\} \)
\( \therefore \text{Domain } (R) = \{a / a \in N, a < 5\} = \{1, 2, 3, 4\} \)
\( \text{Range } (R) = \{b / b = 4\} = \{4\} \). This relation maps the first four natural numbers to the constant value of 4.
In simple words: The domain is the set of all first numbers in the pairs, which are natural numbers less than 5. The range is the set of all second numbers, which is only 4.

🎯 Exam Tip: Remember that natural numbers (\( N \)) start from 1, so 0 is not included in the domain.