RBSE Solutions Class 11 Maths Chapter 2 Relations and Functions Exercise 2.1

Question 1. If A = {1, 2, 3}, B = {4, 5, 6}, then which of the following is relation from A to B? Justify your answer also.
(i) {(1, 4), (3, 5), (3, 6)}
(ii) {(1, 6), (2, 6), (3, 6)}
(iii) {(1, 5), (3, 4), (5, 1), (3, 6)}
(iv) {(2, 4), (2, 6), (3, 6), (4, 2)}
(v) A x B
Answer: First, let's find the Cartesian product \( A \times B \). This means making all possible pairs where the first element comes from set A and the second element comes from set B. Here, \( A = \{1, 2, 3\} \) and \( B = \{4, 5, 6\} \). So, \( A \times B = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\} \). Now, we check each given set to see if it is a subset of \( A \times B \). A set is a relation from A to B if all its ordered pairs are found in \( A \times B \).
(i) Let \( R_1 = \{(1, 4), (3, 5), (3, 6)\} \) All elements (1, 4), (3, 5), and (3, 6) are present in \( A \times B \).
\( \implies \) Hence, \( R_1 \) is a relation from A to B.
(ii) Let \( R_2 = \{(1, 6), (2, 6), (3, 6)\} \) All elements (1, 6), (2, 6), and (3, 6) are present in \( A \times B \).
\( \implies \) Hence, \( R_2 \) is a relation from A to B.
(iii) Let \( R_3 = \{(1, 5), (3, 4), (5, 1), (3, 6)\} \) Here, (5, 1) is not present in \( A \times B \), because 5 is not an element of set A.
\( \implies \) Hence, \( R_3 \) is not a relation from A to B.
(iv) Let \( R_4 = \{(2, 4), (2, 6), (3, 6), (4, 2)\} \) Here, (4, 2) is not present in \( A \times B \), because 4 is not an element of set A.
\( \implies \) Hence, \( R_4 \) is not a relation from A to B.
(v) \( A \times B \) By definition, the Cartesian product itself is a relation from A to B. It contains all possible pairs, so it is the "largest" possible relation.
\( \implies \) Hence, \( A \times B \) is a relation from A to B.
In simple words: A relation between two sets, A and B, is like a list of pairs where the first number comes from A and the second number comes from B. We checked each list to see if all its pairs were valid, meaning the first number was from A and the second from B. Only the first two lists and the full \( A \times B \) itself met this rule.

🎯 Exam Tip: To verify if a set of ordered pairs is a relation from set A to set B, ensure that for every pair (x, y) in the relation, x must belong to A and y must belong to B.

Question 2. Express the following relations in the rules form defined in N:
(i) {(1, 3), (2, 5), (3, 7), (4, 9), ...}
(ii) {(2, 3), (4, 2), (6, 1)}
(iii) {(2, 1), (3, 2), (4, 3), (5, 4), ...}
Answer: We need to find a rule that connects the first number (x) to the second number (y) in each pair, where x and y are natural numbers (from set N).
(i) Given relation: \( \{(1, 3), (2, 5), (3, 7), (4, 9), ...\} \) Let's look at the pattern: When \( x = 1 \), \( y = 3 \) When \( x = 2 \), \( y = 5 \) When \( x = 3 \), \( y = 7 \) When \( x = 4 \), \( y = 9 \) The y-values (3, 5, 7, 9, ...) form an arithmetic progression (A.P.). The first term \( a = 3 \) and the common difference \( d = 2 \). The nth term formula for an A.P. is \( T_n = a + (n - 1)d \). So, \( T_n = 3 + (n - 1) \times 2 = 3 + 2n - 2 = 2n + 1 \). If we let \( n = x \) and \( T_n = y \), the rule is \( y = 2x + 1 \). So, the required rule is \( \{(x, y) \mid x, y \in N, y = 2x + 1\} \).
(ii) Given relation: \( \{(2, 3), (4, 2), (6, 1)\} \) Let's examine the pattern: When \( x = 2 \), \( y = 3 \) When \( x = 4 \), \( y = 2 \) When \( x = 6 \), \( y = 1 \) Here, as x increases by 2, y decreases by 1. The x-values (2, 4, 6) form an A.P. with \( a_x = 2 \) and \( d_x = 2 \). So \( x = 2 + (n - 1)2 = 2n \). The y-values (3, 2, 1) form an A.P. with \( a_y = 3 \) and \( d_y = -1 \). So \( y = 3 + (n - 1)(-1) = 3 - n + 1 = 4 - n \). From \( x = 2n \), we get \( n = x/2 \). Substitute n into the y-equation: \( y = 4 - (x/2) \). This can also be written as \( 2y = 8 - x \), or \( x + 2y = 8 \). Since x and y are natural numbers, and for the given pairs, y is positive, we also need to consider a condition for y. For \( x = 2, y = 3 \). \( 2 + 2(3) = 8 \). For \( x = 4, y = 2 \). \( 4 + 2(2) = 8 \). For \( x = 6, y = 1 \). \( 6 + 2(1) = 8 \). Also, y values are 3, 2, 1, so \( y < 4 \). So, the required rule is \( \{(x, y) \mid x, y \in N, x + 2y = 8 \text{ and } y < 4\} \).
(iii) Given relation: \( \{(2, 1), (3, 2), (4, 3), (5, 4), ...\} \) Let's observe the pattern: When \( x = 2 \), \( y = 1 \) When \( x = 3 \), \( y = 2 \) When \( x = 4 \), \( y = 3 \) When \( x = 5 \), \( y = 4 \) In each pair, the value of y is one less than the value of x. So, the rule is \( y = x - 1 \), or \( x = y + 1 \). The x-values (2, 3, 4, 5, ...) form an A.P. with \( a = 2 \) and \( d = 1 \). So \( T_n = 2 + (n - 1)1 = n + 1 \). If we let \( n = x \) and \( T_n = y \), this gives \( y = x - 1 \). So, the required rule is \( \{(x, y) \mid x, y \in N, y = x - 1\} \) or \( \{(x, y) \mid x, y \in N, x = y + 1\} \).
In simple words: For each list of pairs, we looked for a simple mathematical connection between the first number (x) and the second number (y). For example, in the first list, y was always double x plus one. In the second, x plus two times y always equalled eight, and y was small. In the third, y was always one less than x.

🎯 Exam Tip: To find the rule for a relation, analyze the pattern in both the x-values and y-values, especially if they form an arithmetic or geometric progression.

Question 3. A relation R from set A = {2, 3, 4, 5} to set B = {3, 6, 7, 10} is defined in such a way that xRy ⇔ x is a prime number related to y. Write relation R in the set form or order pairs and also find domain and range of if.
Answer: Given sets are \( A = \{2, 3, 4, 5\} \) and \( B = \{3, 6, 7, 10\} \). The relation R is defined as \( xRy \Leftrightarrow x \) is co-prime to \( y \). Two numbers are co-prime if their only common factor is 1. We need to find pairs (x, y) where \( x \in A \), \( y \in B \), and x and y are co-prime.
Let's check each element in A with each element in B:
For \( x = 2 \in A \): - Is 2 co-prime to 3? Yes (common factors: 1). So, (2, 3) \( \in R \). - Is 2 co-prime to 6? No (common factors: 1, 2). - Is 2 co-prime to 7? Yes (common factors: 1). So, (2, 7) \( \in R \). - Is 2 co-prime to 10? No (common factors: 1, 2).
For \( x = 3 \in A \): - Is 3 co-prime to 3? No (common factors: 1, 3). - Is 3 co-prime to 6? No (common factors: 1, 3). - Is 3 co-prime to 7? Yes (common factors: 1). So, (3, 7) \( \in R \). - Is 3 co-prime to 10? Yes (common factors: 1). So, (3, 10) \( \in R \).
For \( x = 4 \in A \): - Is 4 co-prime to 3? Yes (common factors: 1). So, (4, 3) \( \in R \). - Is 4 co-prime to 6? No (common factors: 1, 2). - Is 4 co-prime to 7? Yes (common factors: 1). So, (4, 7) \( \in R \). - Is 4 co-prime to 10? No (common factors: 1, 2).
For \( x = 5 \in A \): - Is 5 co-prime to 3? Yes (common factors: 1). So, (5, 3) \( \in R \). - Is 5 co-prime to 6? Yes (common factors: 1). So, (5, 6) \( \in R \). - Is 5 co-prime to 7? Yes (common factors: 1). So, (5, 7) \( \in R \). - Is 5 co-prime to 10? No (common factors: 1, 5).
Therefore, the relation R in set form is: \( R = \{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)\} \)
The domain of R is the set of all first elements in the ordered pairs of R. Domain \( R = \{2, 3, 4, 5\} \).
The range of R is the set of all second elements in the ordered pairs of R. Range \( R = \{3, 6, 7, 10\} \).
In simple words: We are looking for pairs of numbers (x, y) where x is from set A, y is from set B, and x and y share no common factor other than 1. This means they are "co-prime." We checked every possible combination and listed the pairs that fit the rule. The domain is all the 'x' numbers we used, and the range is all the 'y' numbers that appeared.

🎯 Exam Tip: When dealing with co-prime relations, remember that 1 is always a common factor. Co-prime means the *only* common factor is 1. Always list all elements of the domain and range, ensuring no duplicates.

Question 4. A relation R in Z (set of integers) is defined by \( x^2 + y^2 = 25 \). Write the relation R in the form of a set of ordered pairs and also write their domain.
Answer: The relation R is defined on the set of integers Z, such that \( x^2 + y^2 = 25 \). We need to find all integer pairs (x, y) that satisfy this equation. The set of integers \( Z = \{..., -3, -2, -1, 0, 1, 2, 3, ...\} \).
Let's consider possible integer values for x and find corresponding y values:
If \( x = 0 \): \( 0^2 + y^2 = 25 \) \( y^2 = 25 \) \( y = \pm 5 \)
\( \implies \) So, (0, 5) \( \in R \) and (0, -5) \( \in R \).
If \( x = \pm 3 \): \( (\pm 3)^2 + y^2 = 25 \) \( 9 + y^2 = 25 \) \( y^2 = 16 \) \( y = \pm 4 \)
\( \implies \) So, (3, 4) \( \in R \), (3, -4) \( \in R \), (-3, 4) \( \in R \), (-3, -4) \( \in R \).
If \( x = \pm 4 \): \( (\pm 4)^2 + y^2 = 25 \) \( 16 + y^2 = 25 \) \( y^2 = 9 \) \( y = \pm 3 \)
\( \implies \) So, (4, 3) \( \in R \), (4, -3) \( \in R \), (-4, 3) \( \in R \), (-4, -3) \( \in R \).
If \( x = \pm 5 \): \( (\pm 5)^2 + y^2 = 25 \) \( 25 + y^2 = 25 \) \( y^2 = 0 \) \( y = 0 \)
\( \implies \) So, (5, 0) \( \in R \) and (-5, 0) \( \in R \).
If \( x \) is any other integer (e.g., \( x = \pm 1, \pm 2, \pm 6 \)), \( x^2 \) would be 1, 4, or 36 respectively. If \( x^2 = 1 \), \( y^2 = 24 \) (not a perfect square). If \( x^2 = 4 \), \( y^2 = 21 \) (not a perfect square). If \( x^2 = 36 \), \( y^2 = -11 \) (not possible for real y, so not for integer y either).
Hence, the relation R in the form of ordered pairs is: \( R = \{(0, 5), (0, -5), (3, 4), (-3, 4), (3, -4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5, 0), (-5, 0)\} \)
The domain of R is the set of all first elements in these ordered pairs: Domain of \( R = \{0, 3, -3, 4, -4, 5, -5\} \).
The problem also asks for \( R^{-1} \) and its domain. The inverse relation \( R^{-1} \) is formed by swapping the elements of each pair in R. \( R^{-1} = \{(5, 0), (-5, 0), (4, 3), (4, -3), (-4, 3), (-4, -3), (3, 4), (3, -4), (-3, 4), (-3, -4), (0, 5), (0, -5)\} \) The domain of \( R^{-1} \) is the set of all first elements in \( R^{-1} \). This is the same as the range of R. Domain of \( R^{-1} = \{5, -5, 4, -4, 3, -3, 0\} \).
In simple words: We needed to find all pairs of whole numbers (positive, negative, or zero) that, when you square them both and add them up, give you 25. We tested different 'x' values and found the 'y' values that fit. The list of all these pairs is the relation R. The domain is simply all the first numbers (x values) from these pairs. For the inverse relation, we just swap the x and y values in each pair.

🎯 Exam Tip: When finding integer solutions for equations like \( x^2 + y^2 = k \), always test positive and negative values for x (and y), including zero, as squaring hides the sign. Remember that the domain consists of all unique first elements.

Question 5. If a relation \( \Phi \) from set C of complex numbers to a set R of real numbers is so defined that \( x \Phi y \Leftrightarrow |x| = y \forall x \in C, y \in R \)
(i) (1 + i) \( \Phi \) 3
(ii) 3 \( \Phi \) (-3)
(iii) (2 + 3i) \( \Phi \) 13
(iv) (1 + i) \( \Phi \) 1
Answer: The relation \( \Phi \) is defined such that \( x \Phi y \) if and only if \( |x| = y \), where \( x \) is a complex number and \( y \) is a real number. The absolute value (or modulus) of a complex number \( x = a + bi \) is calculated as \( |x| = \sqrt{a^2 + b^2} \).
(i) Check (1 + i) \( \Phi \) 3: Here, \( x = 1 + i \) (so \( a = 1, b = 1 \)) and \( y = 3 \). Calculate \( |x| = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \). Is \( |x| = y \)? Is \( \sqrt{2} = 3 \)? No, \( \sqrt{2} \approx 1.414 \).
\( \implies \) Therefore, (1 + i) \( \Phi \) 3 is false.
(ii) Check 3 \( \Phi \) (-3): Here, \( x = 3 \) (which can be written as \( 3 + 0i \), so \( a = 3, b = 0 \)) and \( y = -3 \). Calculate \( |x| = |3| = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \). Is \( |x| = y \)? Is \( 3 = -3 \)? No.
\( \implies \) Therefore, 3 \( \Phi \) (-3) is false.
(iii) Check (2 + 3i) \( \Phi \) 13: Here, \( x = 2 + 3i \) (so \( a = 2, b = 3 \)) and \( y = 13 \). Calculate \( |x| = |2 + 3i| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \). Is \( |x| = y \)? Is \( \sqrt{13} = 13 \)? No.
\( \implies \) Therefore, (2 + 3i) \( \Phi \) 13 is false.
(iv) Check (1 + i) \( \Phi \) 1: Here, \( x = 1 + i \) (so \( a = 1, b = 1 \)) and \( y = 1 \). Calculate \( |x| = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \). Is \( |x| = y \)? Is \( \sqrt{2} = 1 \)? No.
\( \implies \) Therefore, (1 + i) \( \Phi \) 1 is false.
In simple words: This question asks us to check if the "size" of a complex number (called its absolute value) is equal to a given real number. For each case, we calculated the absolute value of the complex number, which is like its distance from zero on a special graph. Then we compared this size to the real number provided. In all the examples, the absolute value did not match the given real number, so all statements were false.

🎯 Exam Tip: Remember that the absolute value (modulus) of a complex number \( a+bi \) is always a non-negative real number, calculated as \( \sqrt{a^2+b^2} \). The relation \( |x|=y \) requires y to be non-negative. If y is negative, the relation is immediately false.

Question 6. If a relation R from set A = {1, 2, 3, 4, 5} to set B = {1, 4, 5} is defined such that x < y, then write R in the form of set of order pairs. Also find \( R^{-1} \)
Answer: Given set \( A = \{1, 2, 3, 4, 5\} \) and set \( B = \{1, 4, 5\} \). The relation R is defined as \( x < y \), where \( x \in A \) and \( y \in B \). We need to list all pairs (x, y) such that x is from A, y is from B, and x is strictly less than y.
Let's find the pairs:
For \( x = 1 \in A \): - Is \( 1 < 1 \)? No. - Is \( 1 < 4 \)? Yes. So, (1, 4) \( \in R \). - Is \( 1 < 5 \)? Yes. So, (1, 5) \( \in R \).
For \( x = 2 \in A \): - Is \( 2 < 1 \)? No. - Is \( 2 < 4 \)? Yes. So, (2, 4) \( \in R \). - Is \( 2 < 5 \)? Yes. So, (2, 5) \( \in R \).
For \( x = 3 \in A \): - Is \( 3 < 1 \)? No. - Is \( 3 < 4 \)? Yes. So, (3, 4) \( \in R \). - Is \( 3 < 5 \)? Yes. So, (3, 5) \( \in R \).
For \( x = 4 \in A \): - Is \( 4 < 1 \)? No. - Is \( 4 < 4 \)? No. - Is \( 4 < 5 \)? Yes. So, (4, 5) \( \in R \).
For \( x = 5 \in A \): - Is \( 5 < 1 \)? No. - Is \( 5 < 4 \)? No. - Is \( 5 < 5 \)? No.
So, the relation R in set of ordered pairs is: \( R = \{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\} \)
To find the inverse relation, \( R^{-1} \), we swap the elements in each ordered pair of R. \( R^{-1} = \{(4, 1), (5, 1), (4, 2), (5, 2), (4, 3), (5, 3), (5, 4)\} \)
In simple words: We had two groups of numbers, A and B. We needed to make pairs where the first number from A was smaller than the second number from B. We checked every possible combination to make these pairs. After listing all the pairs for R, we found its inverse, \( R^{-1} \), by simply flipping each pair (swapping the first and second numbers).

🎯 Exam Tip: Always be careful with strict inequalities like 'x < y' versus 'x <= y'. For the inverse relation \( R^{-1} \), simply reverse the order of elements in each ordered pair of R.

Question 7. Express the following relations in the form of sets or orders pairs:
(i) R₁ is relation from set A = {1, 2, 3, 4, 5, 6} to set B = {1, 2, 3} such that "x = 2y".
(ii) R₂ is a relation set A = {8, 9, 10, 11} to set B = {5, 6, 7, 8} such that “y = x - 2”.
(iii) R₃ is a relation in set A = {0, 1, 2,..., 10} defined by \( 2x + 3y = 12 \).
(iv) R₄ is a relation from set A = {5, 6, 7, 8} to set B = {10, 12, 15, 16, 18} is defined such that “x is divisor of y"
Answer: We need to list the ordered pairs for each given relation based on its rule and the specified sets.
(i) Relation R₁: from set \( A = \{1, 2, 3, 4, 5, 6\} \) to set \( B = \{1, 2, 3\} \) such that \( x = 2y \). We can rewrite the rule as \( y = \frac{x}{2} \). We take values of \( x \) from set A and find \( y \). If \( y \) is in set B, we include the pair (x, y). - If \( x = 1 \), \( y = \frac{1}{2} \) (not in B). - If \( x = 2 \), \( y = \frac{2}{2} = 1 \) (in B). So, (2, 1) \( \in R_1 \). - If \( x = 3 \), \( y = \frac{3}{2} \) (not in B). - If \( x = 4 \), \( y = \frac{4}{2} = 2 \) (in B). So, (4, 2) \( \in R_1 \). - If \( x = 5 \), \( y = \frac{5}{2} \) (not in B). - If \( x = 6 \), \( y = \frac{6}{2} = 3 \) (in B). So, (6, 3) \( \in R_1 \). Thus, \( R_1 = \{(2, 1), (4, 2), (6, 3)\} \).
(ii) Relation R₂: from set \( A = \{8, 9, 10, 11\} \) to set \( B = \{5, 6, 7, 8\} \) such that \( y = x - 2 \). We take values of \( x \) from set A and calculate \( y \). If \( y \) is in set B, we include the pair (x, y). - If \( x = 8 \), \( y = 8 - 2 = 6 \) (in B). So, (8, 6) \( \in R_2 \). - If \( x = 9 \), \( y = 9 - 2 = 7 \) (in B). So, (9, 7) \( \in R_2 \). - If \( x = 10 \), \( y = 10 - 2 = 8 \) (in B). So, (10, 8) \( \in R_2 \). - If \( x = 11 \), \( y = 11 - 2 = 9 \) (not in B). Thus, \( R_2 = \{(8, 6), (9, 7), (10, 8)\} \).
(iii) Relation R₃: in set \( A = \{0, 1, 2, ..., 10\} \) defined by \( 2x + 3y = 12 \). Here, both x and y must come from set A. We can rewrite the rule as \( 3y = 12 - 2x \), or \( y = \frac{12 - 2x}{3} \). We choose values of \( x \) from A such that \( y \) is also an integer in A. - If \( x = 0 \), \( y = \frac{12 - 2(0)}{3} = \frac{12}{3} = 4 \) (in A). So, (0, 4) \( \in R_3 \). - If \( x = 1 \), \( y = \frac{12 - 2(1)}{3} = \frac{10}{3} \) (not an integer). - If \( x = 2 \), \( y = \frac{12 - 2(2)}{3} = \frac{8}{3} \) (not an integer). - If \( x = 3 \), \( y = \frac{12 - 2(3)}{3} = \frac{6}{3} = 2 \) (in A). So, (3, 2) \( \in R_3 \). - If \( x = 4 \), \( y = \frac{12 - 2(4)}{3} = \frac{4}{3} \) (not an integer). - If \( x = 5 \), \( y = \frac{12 - 2(5)}{3} = \frac{2}{3} \) (not an integer). - If \( x = 6 \), \( y = \frac{12 - 2(6)}{3} = \frac{0}{3} = 0 \) (in A). So, (6, 0) \( \in R_3 \). - If \( x = 7 \), \( y = \frac{12 - 2(7)}{3} = \frac{-2}{3} \) (not an integer). - If \( x = 8 \), \( y = \frac{12 - 2(8)}{3} = \frac{-4}{3} \) (not an integer). - If \( x = 9 \), \( y = \frac{12 - 2(9)}{3} = \frac{-6}{3} = -2 \) (not in A, as A goes from 0 to 10). - If \( x = 10 \), \( y = \frac{12 - 2(10)}{3} = \frac{-8}{3} \) (not an integer). Thus, \( R_3 = \{(0, 4), (3, 2), (6, 0)\} \).
(iv) Relation R₄: from set \( A = \{5, 6, 7, 8\} \) to set \( B = \{10, 12, 15, 16, 18\} \) such that "x is divisor of y". We select pairs (x, y) where \( x \in A \), \( y \in B \), and x divides y evenly. - If \( x = 5 \): - 5 is a divisor of 10. So, (5, 10) \( \in R_4 \). - 5 is a divisor of 15. So, (5, 15) \( \in R_4 \). - 5 is not a divisor of 12, 16, 18. - If \( x = 6 \): - 6 is a divisor of 12. So, (6, 12) \( \in R_4 \). - 6 is a divisor of 18. So, (6, 18) \( \in R_4 \). - 6 is not a divisor of 10, 15, 16. - If \( x = 7 \): - 7 is not a divisor of 10, 12, 15, 16, 18. - If \( x = 8 \): - 8 is a divisor of 16. So, (8, 16) \( \in R_4 \). - 8 is not a divisor of 10, 12, 15, 18. Thus, \( R_4 = \{(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)\} \).
In simple words: For each part, we were given specific rules and groups of numbers. We carefully picked numbers from the first group and used the rule to find matching numbers for the second group. If the calculated second number was in its allowed group, we made a pair. We continued this until all possible pairs were found for each relation.

🎯 Exam Tip: Always clearly define the sets for x and y. When applying a rule, check that both x and the resulting y are valid elements of their respective sets. For "x is a divisor of y", ensure y is perfectly divisible by x.