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

Question 1. If \( \left( \frac{x}{3} + 1, y - \frac{2}{3} \right) = \left( \frac{5}{3}, \frac{1}{3} \right) \), find the values of x and y?
Answer: Given that \( \left( \frac{x}{3} + 1, y - \frac{2}{3} \right) = \left( \frac{5}{3}, \frac{1}{3} \right) \).
Since two ordered pairs are equal, their corresponding elements must be equal.
So, we get two equations:
\( \frac{x}{3} + 1 = \frac{5}{3} \)
Subtract 1 from both sides:
\( \frac{x}{3} = \frac{5}{3} - 1 \)
\( \frac{x}{3} = \frac{5 - 3}{3} \)
\( \frac{x}{3} = \frac{2}{3} \)
Multiply both sides by 3:
\( x = 2 \)

And for the y-coordinate:
\( y - \frac{2}{3} = \frac{1}{3} \)
Add \( \frac{2}{3} \) to both sides:
\( y = \frac{1}{3} + \frac{2}{3} \)
\( y = \frac{1 + 2}{3} \)
\( y = \frac{3}{3} \)
\( y = 1 \)
Therefore, the values are \( x = 2 \) and \( y = 1 \).
In simple words: When two pairs are equal, the first parts must be equal to each other, and the second parts must be equal to each other. Solve these two simple equations to get the values for x and y.

Exam Tip: Remember that for two ordered pairs to be equal, their corresponding first components must be equal, and their second components must also be equal. This forms a system of two separate equations.

Question 2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in \( (A \times B) \).
Answer: We are given that the number of elements in set A, \( n(A) = 3 \).
The set B is given as \( B = \{3, 4, 5\} \).
We can count the elements in set B to find \( n(B) \). There are 3 elements in B, so \( n(B) = 3 \).
The number of elements in the Cartesian product \( A \times B \) is found by multiplying the number of elements in A by the number of elements in B.
So, \( n(A \times B) = n(A) \times n(B) \).
Substituting the given values: \( n(A \times B) = 3 \times 3 \).
Therefore, \( n(A \times B) = 9 \).
In simple words: To find how many items are in the combined set (A × B), just multiply the number of items in set A by the number of items in set B.

Exam Tip: The number of elements in a Cartesian product \( A \times B \) is always the product of the number of elements in set A and the number of elements in set B, i.e., \( n(A \times B) = n(A) \times n(B) \).

Question 3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G?
Answer: Given sets are \( G = \{7, 8\} \) and \( H = \{5, 4, 2\} \).
(i) To find the Cartesian product \( G \times H \), we form all possible ordered pairs where the first element comes from G and the second element comes from H.
\( G \times H = \{7, 8\} \times \{5, 4, 2\} \)
\( G \times H = \{ (7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2) \} \)
(ii) To find the Cartesian product \( H \times G \), we form all possible ordered pairs where the first element comes from H and the second element comes from G.
\( H \times G = \{5, 4, 2\} \times \{7, 8\} \)
\( H \times G = \{ (5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8) \} \)
In simple words: For \( G \times H \), pair each item from G with every item from H. For \( H \times G \), pair each item from H with every item from G. The order of the sets in the product matters.

Exam Tip: Remember that \( G \times H \) is generally not equal to \( H \times G \) because the order of elements in an ordered pair matters. Make sure to list all possible unique pairs according to the specified order of the sets.

Question 4. State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly?
1. If P = {m, n} and Q = {n, m), then P × Q = {(m, n), (n, m)}
2. If A and B are non empty sets, then A × B is a non empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
3. If A = {1, 2}, B = {3, 4}, then A × (Β ∩ φ) = φ.
Answer:
1. False.
Given \( P = \{m, n\} \) and \( Q = \{n, m\} \). Since sets are determined by their elements and not the order, \( Q \) is also \( \{m, n\} \). Therefore, \( P = Q \).
The Cartesian product \( P \times Q \) would be:
\( P \times Q = \{m, n\} \times \{m, n\} \)
\( P \times Q = \{ (m, m), (m, n), (n, m), (n, n) \} \)
The given statement only lists two pairs, so it is incorrect.
Correct statement: If \( P = \{m, n\} \) and \( Q = \{n, m\} \), then \( P \times Q = \{ (m, m), (m, n), (n, m), (n, n) \} \).
In simple words: The original statement is wrong because it misses some pairs. When you multiply two sets, you must list every possible pair, not just some of them.
2. False.
The definition of the Cartesian product \( A \times B \) states that it is the set of all ordered pairs \( (x, y) \) where the first element \( x \) belongs to set A, and the second element \( y \) belongs to set B.
The given statement incorrectly says \( x \in B \) and \( y \in A \). This describes \( B \times A \), not \( A \times B \).
Correct statement: If A and B are non-empty sets, then \( A \times B \) is a non-empty set of ordered pairs \( (x, y) \) such that \( x \in A \) and \( y \in B \).
In simple words: The first item in a pair from \( A \times B \) must come from A, and the second item must come from B. The original statement mixed up which set the items came from.
3. True.
First, find the intersection of B and the empty set \( \phi \). The intersection of any set with the empty set is always the empty set.
\( B \cap \phi = \phi \)
Next, find the Cartesian product of A with the empty set \( \phi \). The Cartesian product of any set with the empty set is always the empty set.
\( A \times (B \cap \phi) = A \times \phi = \phi \)
This matches the statement, so it is true.
In simple words: When you find what two sets have in common and one of them is empty, the result is always empty. And if you "multiply" any set by an empty set, the answer is always an empty set. So the statement is correct.

Exam Tip: Be very careful with the definitions of Cartesian product and set operations like intersection. Understand that \( A \times B \) requires the first element from A and the second from B, and that any operation involving the empty set often results in the empty set itself.

Question 5. If \( A = \{-1, 1\} \), find \( A \times A \times A \).
Answer: Given set is \( A = \{-1, 1\} \).
First, let's find \( A \times A \):
\( A \times A = \{-1, 1\} \times \{-1, 1\} \)
\( A \times A = \{ (-1, -1), (-1, 1), (1, -1), (1, 1) \} \)
Now, we find \( A \times A \times A \) by taking the Cartesian product of \( (A \times A) \) with A:
\( A \times A \times A = (A \times A) \times A \)
\( A \times A \times A = \{ (-1, -1), (-1, 1), (1, -1), (1, 1) \} \times \{-1, 1\} \)
To form the triplets, we take each ordered pair from \( A \times A \) and pair it with each element from A:
\( A \times A \times A = \{ ((-1, -1), -1), ((-1, -1), 1), ((-1, 1), -1), ((-1, 1), 1), ((1, -1), -1), ((1, -1), 1), ((1, 1), -1), ((1, 1), 1) \} \)
This can be written as triplets directly:
\( A \times A \times A = \{ (-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1) \} \)
In simple words: To find \( A \times A \times A \), first find all pairs from \( A \times A \). Then, for each of those pairs, combine it with every single item from set A to make three-item groups.

Exam Tip: When finding \( A \times A \times A \), methodically pair each element from the first \( A \times A \) result with each element of the last A. This ensures you don't miss any combinations and clearly lists all 2³ = 8 possible triplets.

Question 6. If \( A \times B = \{ (a, x), (a, y), (b, x), (b, y) \} \). Find A and B?
Answer: Given the Cartesian product \( A \times B = \{ (a, x), (a, y), (b, x), (b, y) \} \).
In a Cartesian product \( A \times B \), the first elements of all ordered pairs belong to set A, and the second elements of all ordered pairs belong to set B.
Looking at the first elements of the pairs in \( A \times B \): a, a, b, b.
So, set A contains these distinct first elements: \( A = \{a, b\} \).
Looking at the second elements of the pairs in \( A \times B \): x, y, x, y.
So, set B contains these distinct second elements: \( B = \{x, y\} \).
Therefore, \( A = \{a, b\} \) and \( B = \{x, y\} \).
In simple words: When you have a list of paired items that came from \( A \times B \), the first part of each pair always comes from set A, and the second part always comes from set B. Just collect all the unique first parts for A, and all the unique second parts for B.

Exam Tip: To find the original sets from their Cartesian product, simply collect all the unique first components of the ordered pairs to form the first set, and all the unique second components to form the second set. Remember sets only contain unique elements.

Question 7. Let \( A = \{1, 2\} \), \( B = \{1, 2, 3, 4\} \), \( C = \{5, 6\} \) and \( D = \{5, 6, 7, 8\} \)? Verify that:
1. \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
2. \( A \times C \) is a subset of \( B \times D \).
Answer: Given sets: \( A = \{1, 2\} \), \( B = \{1, 2, 3, 4\} \), \( C = \{5, 6\} \) and \( D = \{5, 6, 7, 8\} \).
1. To verify \( A \times (B \cap C) = (A \times B) \cap (A \times C) \):
First, calculate the Left Hand Side (L.H.S.):
\( B \cap C = \{1, 2, 3, 4\} \cap \{5, 6\} \)
Since there are no common elements between B and C, their intersection is the empty set.
\( B \cap C = \phi \)
Now, find \( A \times (B \cap C) \):
\( A \times (B \cap C) = \{1, 2\} \times \phi \)
The Cartesian product of any set with the empty set is always the empty set.
\( \text{L.H.S.} = \phi \)
Next, calculate the Right Hand Side (R.H.S.):
First, find \( A \times B \):
\( A \times B = \{1, 2\} \times \{1, 2, 3, 4\} \)
\( A \times B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4) \} \)
Next, find \( A \times C \):
\( A \times C = \{1, 2\} \times \{5, 6\} \)
\( A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \)
Now, find the intersection of \( (A \times B) \) and \( (A \times C) \):
\( (A \times B) \cap (A \times C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4) \} \cap \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \)
There are no common ordered pairs between \( A \times B \) and \( A \times C \).
\( \text{R.H.S.} = \phi \)
Since L.H.S. = R.H.S. (\( \phi = \phi \)), the first statement is verified as true.
In simple words: First, find what's common in B and C. Then "multiply" that by A. Next, "multiply" A by B, and A by C separately. Then find what's common in these two multiplied results. Both sides should end up being empty, which means they are equal.
2. To verify \( A \times C \) is a subset of \( B \times D \):
First, find \( A \times C \):
\( A \times C = \{1, 2\} \times \{5, 6\} \)
\( A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \)
Next, find \( B \times D \):
\( B \times D = \{1, 2, 3, 4\} \times \{5, 6, 7, 8\} \)
\( B \times D = \{ (1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8) \} \)
To check if \( A \times C \) is a subset of \( B \times D \), every element (ordered pair) of \( A \times C \) must also be an element of \( B \times D \).
Let's check the elements of \( A \times C \):
\( (1, 5) \in B \times D \)
\( (1, 6) \in B \times D \)
\( (2, 5) \in B \times D \)
\( (2, 6) \in B \times D \)
Since all elements of \( A \times C \) are present in \( B \times D \), we can confirm that \( A \times C \) is a subset of \( B \times D \).
\( (A \times C) \subset (B \times D) \).
This second statement is also verified as true.
In simple words: First, list all the pairs for \( A \times C \) and then all the pairs for \( B \times D \). If every single pair that is in \( A \times C \) can also be found in the list for \( B \times D \), then \( A \times C \) is a subset of \( B \times D \).

Exam Tip: When verifying set identities or subset relationships, always show the step-by-step calculation of each side or set. This demonstrates a complete understanding and helps avoid errors in complex expressions. For subset verification, ensure every element of the first set is explicitly shown to be an element of the second set.

Question 8. Let \( A = \{1, 2\} \) and \( B = \{3, 4\} \). Write \( A \times B \). How many subsets does \( A \times B \) have? List them?
Answer: Given sets: \( A = \{1, 2\} \) and \( B = \{3, 4\} \).
First, find the Cartesian product \( A \times B \):
\( A \times B = \{1, 2\} \times \{3, 4\} \)
\( A \times B = \{ (1, 3), (1, 4), (2, 3), (2, 4) \} \)
Next, find the number of elements in \( A \times B \):
\( n(A) = 2 \) and \( n(B) = 2 \).
So, \( n(A \times B) = n(A) \times n(B) = 2 \times 2 = 4 \).
The number of subsets of a set with \( k \) elements is \( 2^k \). Here, \( k = n(A \times B) = 4 \).
Therefore, the number of subsets of \( A \times B \) is \( 2^4 = 16 \).
Now, list all 16 subsets of \( A \times B \):
1. The empty set: \( \phi \)
2. Subsets with one element:
\( \{ (1, 3) \} \)
\( \{ (1, 4) \} \)
\( \{ (2, 3) \} \)
\( \{ (2, 4) \} \)
3. Subsets with two elements:
\( \{ (1, 3), (1, 4) \} \)
\( \{ (1, 3), (2, 3) \} \)
\( \{ (1, 3), (2, 4) \} \)
\( \{ (1, 4), (2, 3) \} \)
\( \{ (1, 4), (2, 4) \} \)
\( \{ (2, 3), (2, 4) \} \)
4. Subsets with three elements:
\( \{ (1, 3), (1, 4), (2, 3) \} \)
\( \{ (1, 3), (1, 4), (2, 4) \} \)
\( \{ (1, 3), (2, 3), (2, 4) \} \)
\( \{ (1, 4), (2, 3), (2, 4) \} \)
5. Subsets with four elements (the set itself):
\( \{ (1, 3), (1, 4), (2, 3), (2, 4) \} \)
In simple words: First, create all the possible pairs from A and B to get \( A \times B \). Then, count how many pairs are in \( A \times B \). If there are \( k \) pairs, there will be \( 2^k \) total subsets. Finally, write down every possible group you can make from these pairs, including an empty group and the group of all pairs.

Exam Tip: When listing subsets, it's helpful to organize them by the number of elements (zero, one, two, etc.) to ensure you don't miss any. Remember the formula for the number of subsets is \( 2^n \), where \( n \) is the number of elements in the set.

Question 9. Let A and B be two sets such that \( n(A) = 3 \), \( n(B) = 2 \). If \( (x, 1), (y, 2), (z, 1) \) are in \( A \times B \), find A and B, where x, y and z are distinct elements?
Answer: We are given that \( n(A) = 3 \) and \( n(B) = 2 \).
We are also given that the ordered pairs \( (x, 1), (y, 2), (z, 1) \) are elements of \( A \times B \).
In any ordered pair \( (a, b) \) from \( A \times B \), the first element \( a \) must belong to set A, and the second element \( b \) must belong to set B.
From the given pairs, the first elements are x, y, and z.
Since x, y, and z are distinct elements and \( n(A) = 3 \), these must be all the elements of set A.
Therefore, \( A = \{x, y, z\} \).
From the given pairs, the second elements are 1, 2, and 1.
The distinct second elements are 1 and 2.
Since \( n(B) = 2 \), these must be all the elements of set B.
Therefore, \( B = \{1, 2\} \).
In simple words: Since A has 3 items and B has 2 items, and we know some pairs, we can figure out the sets. The first parts of the pairs (x, y, z) are all the items in set A. The second parts of the pairs (1, 2) are all the items in set B.

Exam Tip: Remember that in \( A \times B \), the first components of the ordered pairs form set A, and the second components form set B. Use the given number of elements for each set to confirm that you have identified all the distinct elements correctly.

Question 10. The cartesian product \( A \times A \) has 9 elements among which are found \( (-1, 0) \) and \( (0, 1) \). Find the set A and the remaining elements of \( A \times A \)?
Answer: We are given that the Cartesian product \( A \times A \) has 9 elements, so \( n(A \times A) = 9 \).
This implies that \( n(A) \times n(A) = 9 \), which means \( n(A) = 3 \). So, set A must have 3 distinct elements.
We are also given that \( (-1, 0) \in A \times A \) and \( (0, 1) \in A \times A \).
If \( (a, b) \in A \times A \), then both \( a \in A \) and \( b \in A \).
From \( (-1, 0) \in A \times A \), it follows that \( -1 \in A \) and \( 0 \in A \).
From \( (0, 1) \in A \times A \), it follows that \( 0 \in A \) and \( 1 \in A \).
Combining these elements, we see that \( \{-1, 0, 1\} \) are elements of A.
Since we know \( n(A) = 3 \) and we have found three distinct elements \( \{-1, 0, 1\} \), then set A must be \( A = \{-1, 0, 1\} \).
Now, we need to find all elements of \( A \times A \).
\( A \times A = \{-1, 0, 1\} \times \{-1, 0, 1\} \)
\( A \times A = \{ (a, b) \mid a \in \{-1, 0, 1\}, b \in \{-1, 0, 1\} \} \)
Listing all possible pairs:
\( A \times A = \{ (-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1) \} \)
The question states that \( (-1, 0) \) and \( (0, 1) \) are among these elements, which they are.
The remaining elements of \( A \times A \) are all the elements except \( (-1, 0) \) and \( (0, 1) \).
Remaining elements: \( \{ (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1) \} \)
In simple words: Since \( A \times A \) has 9 elements, set A must have 3 items. We are given two pairs from \( A \times A \), so we can figure out that -1, 0, and 1 are the three items in set A. Then, list all possible pairs you can make using these three items to find all elements of \( A \times A \). The remaining elements are all of them except the two that were given in the question.

Exam Tip: When \( n(A \times A) = k \), then \( n(A) = \sqrt{k} \). Use the components of the given ordered pairs to deduce the elements of set A. Once set A is known, carefully construct all pairs in \( A \times A \) to ensure no elements are missed or duplicated.