Maharashtra Board Class 11 Maths Part 1 Chapter 1 Sets and Relations 1.2 Solutions

Question 1. If \( (x - 1, y + 4) = (1, 2) \), find the values of \( x \) and \( y \).
Answer: Given, \( (x - 1, y + 4) = (1, 2) \).
By the definition of equality of ordered pairs, we can equate the corresponding elements of both pairs to solve for the unknown variables.
This gives us:
\( x - 1 = 1 \) and \( y + 4 = 2 \)

\( \implies x = 1 + 1 \)

\( \implies x = 2 \)
and
\( y = 2 - 4 \)

\( \implies y = -2 \)
Therefore, the values are \( x = 2 \) and \( y = -2 \).
In simple words: Two ordered pairs are equal only when their first parts are equal to each other and their second parts are also equal. By setting the corresponding parts equal, we easily solve for x and y.

🎯 Exam Tip: Always remember to equate the first element with the first element, and the second with the second. Double-check your basic arithmetic signs when moving terms across the equals sign to avoid silly mistakes.

Question 2. If \( \left(x + \frac{1}{3}, \frac{y}{3} - 1\right) = \left(\frac{1}{3}, \frac{3}{2}\right) \), find \( x \) and \( y \).
Answer: Given, \( \left(x + \frac{1}{3}, \frac{y}{3} - 1\right) = \left(\frac{1}{3}, \frac{3}{2}\right) \).
By the definition of equality of ordered pairs, we equate the corresponding components to set up our equations:
\( x + \frac{1}{3} = \frac{1}{3} \) and \( \frac{y}{3} - 1 = \frac{3}{2} \)
For the first equation:
\( x = \frac{1}{3} - \frac{1}{3} \)

\( \implies x = 0 \)
For the second equation:
\( \frac{y}{3} = \frac{3}{2} + 1 \)

\( \implies \frac{y}{3} = \frac{5}{2} \)

\( \implies y = \frac{15}{2} \)
Thus, the values are \( x = 0 \) and \( y = \frac{15}{2} \).
In simple words: Just like the first problem, we set the first terms equal to find x, and the second terms equal to find y. Working carefully with fractions helps us get the correct final values.

🎯 Exam Tip: When dealing with fractions, find a common denominator before adding or subtracting terms to ensure accuracy in your final steps.

Question 2. Find the values of \( x \) and \( y \) if \( \left(x + \frac{1}{3}, \frac{y}{3} - 1\right) = \left(\frac{1}{3}, \frac{3}{2}\right) \).
Answer: Given, \( \left(x + \frac{1}{3}, \frac{y}{3} - 1\right) = \left(\frac{1}{3}, \frac{3}{2}\right) \). By the definition of equality of ordered pairs, we can equate the corresponding elements directly.
\( x + \frac{1}{3} = \frac{1}{3} \) and \( \frac{y}{3} - 1 = \frac{3}{2} \)
\( \implies x = \frac{1}{3} - \frac{1}{3} \) and \( \frac{y}{3} = \frac{3}{2} + 1 \)
\( \implies x = 0 \) and \( \frac{y}{3} = \frac{5}{2} \)
\( \implies y = \frac{15}{2} \)
Thus, the values are \( x = 0 \) and \( y = \frac{15}{2} \).
In simple words: Two ordered pairs are equal only if their corresponding elements are equal. We set the first parts equal to find \( x \), and the second parts equal to find \( y \).

🎯 Exam Tip: Always equate the first element of the first pair to the first element of the second pair, and do the same for the second elements to avoid calculation errors.

Question 3. If \( A = \{a, b, c\} \), \( B = \{x, y\} \), find \( A \times B \), \( B \times A \), \( A \times A \), \( B \times B \).
Answer: Given sets are \( A = \{a, b, c\} \) and \( B = \{x, y\} \). The Cartesian product represents all possible ordered pairs between these sets.
\( A \times B = \{(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)\} \)
\( B \times A = \{(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)\} \)
\( A \times A = \{(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)\} \)
\( B \times B = \{(x, x), (x, y), (y, x), (y, y)\} \)
In simple words: The Cartesian product of two sets is a set of all possible ordered pairs where the first element comes from the first set and the second element comes from the second set.

🎯 Exam Tip: Remember that \( A \times B \) is not equal to \( B \times A \). Double-check the order of elements in each pair to ensure they match the order of the sets.

Question 4. If \( P = \{1, 2, 3\} \) and \( Q = \{6, 4\} \), find the sets \( P \times Q \) and \( Q \times P \).
Answer: Given sets are \( P = \{1, 2, 3\} \) and \( Q = \{6, 4\} \). These sets allow us to form distinct ordered pairs in both directions.
\( P \times Q = \{(1, 6), (1, 4), (2, 6), (2, 4), (3, 6), (3, 4)\} \)
\( Q \times P = \{(6, 1), (6, 2), (6, 3), (4, 1), (4, 2), (4, 3)\} \)
In simple words: To find \( P \times Q \), pair every number in set \( P \) with every number in set \( Q \). For \( Q \times P \), do the reverse by starting with elements from \( Q \).

🎯 Exam Tip: The total number of elements in \( P \times Q \) is always the number of elements in \( P \) multiplied by the number of elements in \( Q \). Use this to verify you haven't missed any pairs.

Question 5. Let \( A = \{1, 2, 3, 4\} \), \( B = \{4, 5, 6\} \), \( C = \{5, 6\} \). Find
(i) \( A \times (B \cap C) \)
(ii) \( (A \times B) \cap (A \times C) \)
(iii) \( A \times (B \cup C) \)
(iv) \( (A \times B) \cup (A \times C) \)
Answer: Given sets are \( A = \{1, 2, 3, 4\} \), \( B = \{4, 5, 6\} \), and \( C = \{5, 6\} \). We will solve each part step-by-step to demonstrate the distributive properties of Cartesian products.
(i) First, find the intersection of \( B \) and \( C \):
\( B \cap C = \{5, 6\} \)
Now, find the Cartesian product:
\( A \times (B \cap C) = \{(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)\} \)

(ii) First, find \( A \times B \) and \( A \times C \):
\( A \times B = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)\} \)
\( A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)\} \)
Now, find their intersection:
\( (A \times B) \cap (A \times C) = \{(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)\} \)

(iii) First, find the union of \( B \) and \( C \):
\( B \cup C = \{4, 5, 6\} \)
Now, find the Cartesian product:
\( A \times (B \cup C) = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)\} \)

(iv) Using the sets \( A \times B \) and \( A \times C \) calculated in part (ii), find their union:
\( (A \times B) \cup (A \times C) = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)\} \)
In simple words: For these problems, always solve the operation inside the parentheses first (like finding the common elements or combining the sets) before finding the Cartesian product.

🎯 Exam Tip: Notice that \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) and \( A \times (B \cup C) = (A \times B) \cup (A \times C) \). These are distributive laws of Cartesian product over intersection and union, which can help you verify your answers.

Question 6. Express \( \{(x, y) \mid x^2 + y^2 = 100, \text{ where } x, y \in W\} \) as a set of ordered pairs.
Answer:
We have the relation \( x^2 + y^2 = 100 \), where \( x \) and \( y \) must be whole numbers (\( W \)).
Let us find the pairs of whole numbers whose squares sum up to 100:
When \( x = 0 \) and \( y = 10 \),
\( x^2 + y^2 = 0^2 + 10^2 = 100 \)
When \( x = 6 \) and \( y = 8 \),
\( x^2 + y^2 = 6^2 + 8^2 = 100 \)
When \( x = 8 \) and \( y = 6 \),
\( x^2 + y^2 = 8^2 + 6^2 = 100 \)
When \( x = 10 \) and \( y = 0 \),
\( x^2 + y^2 = 10^2 + 0^2 = 100 \)
\( \therefore \) Set of ordered pairs = \( \{(0, 10), (6, 8), (8, 6), (10, 0)\} \)
In simple words: We need to find pairs of whole numbers that, when squared and added together, equal 100. The only whole numbers that work are 0, 6, 8, and 10, which give us the four pairs shown above.

🎯 Exam Tip: Remember that whole numbers (\( W \)) start from 0. Do not forget to include the boundary cases \( (0, 10) \) and \( (10, 0) \) in your final set.

Question 7. Write the domain and range of the following relations.
(i) \( \{(a, b) \mid a \in \mathbb{N}, a < 6 \text{ and } b = 4\} \)
(ii) \( \{(a, b) \mid a, b \in \mathbb{N}, a + b = 12\} \)
(iii) \( \{(2, 4), (2, 5), (2, 6), (2, 7)\} \)
Answer:
(i) Let \( R_1 = \{(a, b) \mid a \in \mathbb{N}, a < 6 \text{ and } b = 4\} \)
The set of values of 'a' represents the domain, and the set of values of 'b' represents the range.
Since \( a \in \mathbb{N} \) and \( a < 6 \), the possible values for \( a \) are \( 1, 2, 3, 4, 5 \).
Since \( b = 4 \), the only value for \( b \) is \( 4 \).
\( \therefore a = 1, 2, 3, 4, 5 \) and \( b = 4 \)
Domain\( (R_1) = \{1, 2, 3, 4, 5\} \)
Range\( (R_1) = \{4\} \)

(ii) Let \( R_2 = \{(a, b) \mid a, b \in \mathbb{N}, a + b = 12\} \)
Since \( a \) and \( b \) are natural numbers whose sum is 12, the possible ordered pairs are:
\( R_2 = \{(1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3), (10, 2), (11, 1)\} \)
Domain\( (R_2) = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} \)
Range\( (R_2) = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} \)

(iii) Let \( R_3 = \{(2, 4), (2, 5), (2, 6), (2, 7)\} \)
The domain is the set of all first components of the ordered pairs, and the range is the set of all second components.
Domain\( (R_3) = \{2\} \)
Range\( (R_3) = \{4, 5, 6, 7\} \)
In simple words: The domain is the set of all the first numbers in the pairs, and the range is the set of all the second numbers. We simply list these values inside curly brackets without repeating any numbers.

🎯 Exam Tip: Always write domain and range in set notation using curly brackets \( \{ \} \). Even if an element like 2 repeats in the ordered pairs, write it only once in the domain set.

p>Question. Find the domain and range of the following relations:
(ii) Let \( R_2 = \{(a, b) / a, b \in \mathbb{N} \text{ and } a + b = 12\} \)
(iii) Let \( R_3 = \{(2, 4), (2, 5), (2, 6), (2, 7)\} \)
Answer:
(ii) Given \( R_2 = \{(a, b) / a, b \in \mathbb{N} \text{ and } a + b = 12\} \).
Now, \( a, b \in \mathbb{N} \) and \( a + b = 12 \).
When \( a = 1 \), \( b = 11 \)
When \( a = 2 \), \( b = 10 \)
When \( a = 3 \), \( b = 9 \)
When \( a = 4 \), \( b = 8 \)
When \( a = 5 \), \( b = 7 \)
When \( a = 6 \), \( b = 6 \)
When \( a = 7 \), \( b = 5 \)
When \( a = 8 \), \( b = 4 \)
When \( a = 9 \), \( b = 3 \)
When \( a = 10 \), \( b = 2 \)
When \( a = 11 \), \( b = 1 \)
\( \therefore \text{Domain}(R_2) = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} \)
\( \text{Range}(R_2) = \{11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\} \)

(iii) Given \( R_3 = \{(2, 4), (2, 5), (2, 6), (2, 7)\} \).
\( \text{Domain}(R_3) = \{2\} \)
\( \text{Range}(R_3) = \{4, 5, 6, 7\} \). This represents a relation where a single element in the domain maps to multiple distinct values in the range.
In simple words: To find the domain, we collect all the first numbers from each pair, and for the range, we collect all the second numbers. For the second relation, the number 2 is the only first element, so it is the only number in the domain.

🎯 Exam Tip: Always write the domain and range elements inside curly brackets as sets, and ensure that duplicate elements are written only once.

Question 8. Let \( A = \{6, 8\} \) and \( B = \{1, 3, 5\} \). Let \( R = \{(a, b) / a \in A, b \in B, a - b \text{ is an even number}\} \). Show that \( R \) is an empty relation from \( A \) to \( B \).
Answer:
Given \( A = \{6, 8\} \) and \( B = \{1, 3, 5\} \).
The relation is defined as \( R = \{(a, b) / a \in A, b \in B, a - b \text{ is an even number}\} \).
Here, \( a \in A \)
\( \implies a = 6, 8 \)
and \( b \in B \)
\( \implies b = 1, 3, 5 \).
Let us check the difference \( a - b \) for all possible pairs:
When \( a = 6 \) and \( b = 1 \), \( a - b = 6 - 1 = 5 \), which is odd.
When \( a = 6 \) and \( b = 3 \), \( a - b = 6 - 3 = 3 \), which is odd.
When \( a = 6 \) and \( b = 5 \), \( a - b = 6 - 5 = 1 \), which is odd.
When \( a = 8 \) and \( b = 1 \), \( a - b = 8 - 1 = 7 \), which is odd.
When \( a = 8 \) and \( b = 3 \), \( a - b = 8 - 3 = 5 \), which is odd.
When \( a = 8 \) and \( b = 5 \), \( a - b = 8 - 5 = 3 \), which is odd.
Since the difference \( a - b \) is odd for all possible ordered pairs, no elements satisfy the given condition. Thus, there are no elements in the relation \( R \), which means \( R = \emptyset \). Therefore, \( R \) is an empty relation from \( A \) to \( B \).
In simple words: An empty relation means no pairs from the two sets fit the rule. Here, the rule says subtracting the numbers must give an even result, but every single subtraction gives an odd number, so the relation has no pairs at all.

🎯 Exam Tip: To prove a relation is empty, systematically show that every possible combination of elements fails to satisfy the given condition, and conclude with \( R = \emptyset \).

p>Question 9. Write the relation in the Roster form and hence find its domain and range.
(i) \( R_1 = \{(a, a^2) \mid a \text{ is a prime number less than } 15\} \)
(ii) \( R_2 = \left\{\left(a, \frac{1}{a}\right) \mid 0 < a \le 5, a \in \mathbb{N}\right\} \)
Answer:
(i) \( R_1 = \{(a, a^2) \mid a \text{ is a prime number less than } 15\} \)
The prime numbers less than 15 are 2, 3, 5, 7, 11, and 13.
\( \therefore a = 2, 3, 5, 7, 11, 13 \)
\( \therefore a^2 = 4, 9, 25, 49, 121, 169 \)
\( \therefore R_1 = \{(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)\} \)
\( \therefore \text{Domain}(R_1) = \{a \mid a \text{ is a prime number less than } 15\} = \{2, 3, 5, 7, 11, 13\} \)
\( \text{Range}(R_1) = \{a^2 \mid a \text{ is a prime number less than } 15\} = \{4, 9, 25, 49, 121, 169\} \)

(ii) \( R_2 = \left\{\left(a, \frac{1}{a}\right) \mid 0 < a \le 5, a \in \mathbb{N}\right\} \)
Since \( a \in \mathbb{N} \) and \( 0 < a \le 5 \), the values of \( a \) are 1, 2, 3, 4, and 5.
\( \dots a = 1, 2, 3, 4, 5 \)
\( \therefore \frac{1}{a} = 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \)
\( \therefore R_2 = \left\{(1, 1), \left(2, \frac{1}{2}\right), \left(3, \frac{1}{3}\right), \left(4, \frac{1}{4}\right), \left(5, \frac{1}{5}\right)\right\} \)
\( \dots \text{Domain}(R_2) = \{a \mid 0 < a \le 5, a \in \mathbb{N}\} = \{1, 2, 3, 4, 5\} \)
\( \therefore \text{Range}(R_2) = \left\{\frac{1}{a} \mid 0 < a \le 5, a \in \mathbb{N}\right\} = \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\} \)
These sets represent the inputs and outputs of the given mathematical relations.
In simple words: To write a relation in roster form, we list all the ordered pairs that satisfy the given condition. The domain is the set of all first elements (inputs), and the range is the set of all second elements (outputs).

🎯 Exam Tip: Always list the elements of the domain and range inside curly braces, and ensure you only include unique values without repeating any elements.

Question 10. \( R = \{(a, b) \mid b = a + 1, a \in \mathbb{Z}, 0 < a < 5\} \). Find the range of \( R \).
Answer:
Given relation: \( R = \{(a, b) \mid b = a + 1, a \in \mathbb{Z}, 0 < a < 5\} \)
Since \( a \) is an integer between 0 and 5 (excluding 0 and 5):
\( \therefore a = 1, 2, 3, 4 \)
Now, substitute these values into \( b = a + 1 \):
For \( a = 1 \), \( b = 1 + 1 = 2 \)
For \( a = 2 \), \( b = 2 + 1 = 3 \)
For \( a = 3 \), \( b = 3 + 1 = 4 \)
For \( a = 4 \), \( b = 4 + 1 = 5 \)
\( \therefore b = 2, 3, 4, 5 \)
The range consists of all the second elements of the ordered pairs in the relation.
\( \therefore \text{Range}(R) = \{2, 3, 4, 5\} \)
This shows how each integer input is mapped directly to its consecutive successor.
In simple words: We find the allowed values for \( a \), which are 1, 2, 3, and 4. Then we add 1 to each of these to get the values of \( b \), which form our range: 2, 3, 4, and 5.

🎯 Exam Tip: Pay close attention to the inequality signs; \( 0 < a < 5 \) means 0 and 5 are not included, whereas \( 0 \le a \le 5 \) would include them.

Question 11. Find the following relations as sets of ordered pairs.

Question 1. Find the set of ordered pairs for each of the following relations:
(i) \( \{(x, y) \mid y = 3x, x \in \{1, 2, 3\}, y \in \{3, 6, 9, 12\}\} \)
(ii) \( \{(x, y) \mid y > x + 1, x \in \{1, 2\} \text{ and } y \in \{2, 4, 6\}\} \)
(iii) \( \{(x, y) \mid x + y = 3, x, y \in \{0, 1, 2, 3\}\} \)
Answer:
(i) \( \{(x, y) \mid y = 3x, x \in \{1, 2, 3\}, y \in \{3, 6, 9, 12\}\} \)
Here, \( y = 3x \)
When \( x = 1 \), \( y = 3(1) = 3 \)
When \( x = 2 \), \( y = 3(2) = 6 \)
When \( x = 3 \), \( y = 3(3) = 9 \)
\( \implies \) Ordered pairs are \( \{(1, 3), (2, 6), (3, 9)\} \)

(ii) \( \{(x, y) \mid y > x + 1, x \in \{1, 2\} \text{ and } y \in \{2, 4, 6\}\} \)
Here, \( y > x + 1 \)
When \( x = 1 \) and \( y = 2 \), \( 2 \ngtr 1 + 1 \)
When \( x = 1 \) and \( y = 4 \), \( 4 > 1 + 1 \)
When \( x = 1 \) and \( y = 6 \), \( 6 > 1 + 1 \)
When \( x = 2 \) and \( y = 2 \), \( 2 \ngtr 2 + 1 \)
When \( x = 2 \) and \( y = 4 \), \( 4 > 2 + 1 \)
When \( x = 2 \) and \( y = 6 \), \( 6 > 2 + 1 \)
\( \implies \) Ordered pairs are \( \{(1, 4), (1, 6), (2, 4), (2, 6)\} \)

(iii) \( \{(x, y) \mid x + y = 3, x, y \in \{0, 1, 2, 3\}\} \)
Here, \( x + y = 3 \)
When \( x = 0 \), \( y = 3 \)
When \( x = 1 \), \( y = 2 \)
When \( x = 2 \), \( y = 1 \)
When \( x = 3 \), \( y = 0 \)
\( \implies \) Ordered pairs are \( \{(0, 3), (1, 2), (2, 1), (3, 0)\} \)
In simple words: To find the ordered pairs, we substitute the given values of \( x \) and \( y \) into the condition and check if it holds true. The pairs that satisfy the condition are written together inside curly brackets.

🎯 Exam Tip: Always write down the step-by-step substitution for each value of \( x \) to show the examiner how you arrived at the ordered pairs and avoid calculation errors.