OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(a)

 

Question 1. x ∈ {2, 4,6,9} and y ∈ {4, 6, 18, 27, 54}. From all ordered pairs (x, y) such that x is a factor of y and x < y.
Answer: We are given two sets: \( x \in \{2, 4, 6, 9\} \) and \( y \in \{4, 6, 18, 27, 54\} \). We need to find all pairs \((x, y)\) where \(x\) divides \(y\) evenly (is a factor of \(y\)) and \(x\) is smaller than \(y\). Both conditions must be met for each pair.
Let's check each element in set \(x\):
For \(x = 2\):
\(2\) is a factor of \(4\), \(6\), \(18\), \(54\).
Also, \(2 < 4\), \(2 < 6\), \(2 < 18\), \(2 < 54\).
So, the possible pairs are: \((2, 4), (2, 6), (2, 18), (2, 54)\).
For \(x = 4\):
\(4\) is a factor of \(4\).
However, \(4\) is not less than \(4\) (\(4 < 4\) is false).
Therefore, there are no pairs starting with \(4\).
For \(x = 6\):
\(6\) is a factor of \(18\), \(54\).
Also, \(6 < 18\), \(6 < 54\).
So, the possible pairs are: \((6, 18), (6, 54)\).
For \(x = 9\):
\(9\) is a factor of \(18\), \(27\), \(54\).
Also, \(9 < 18\), \(9 < 27\), \(9 < 54\).
So, the possible pairs are: \((9, 18), (9, 27), (9, 54)\).
Combining all these pairs, the required possible ordered pairs are:
\( \{(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)\} \)
In simple words: We looked at numbers from the first set (x) and paired them with numbers from the second set (y). The rules were that x must divide y perfectly, and x must be smaller than y. We found all such pairs.

🎯 Exam Tip: When checking conditions for ordered pairs, always make sure every single condition (like being a factor AND being less than) is met for each pair before including it in the final set.

 

Question 2. Find the numbers x and y if (x + 3, y − 5) = (5, 0).
Answer: We are given that two ordered pairs are equal: \( (x + 3, y - 5) = (5, 0) \).
For two ordered pairs to be equal, their corresponding components must be equal. This means the first part of the first pair must equal the first part of the second pair, and the second parts must also be equal.
So, we set up two separate equations:
1. \( x + 3 = 5 \)
2. \( y - 5 = 0 \)
Now, we solve each equation:
From equation 1:
\( x + 3 = 5 \)
\( x = 5 - 3 \)
\( x = 2 \)
From equation 2:
\( y - 5 = 0 \)
\( y = 0 + 5 \)
\( y = 5 \)
Thus, the numbers are \( x = 2 \) and \( y = 5 \). This fundamental property of ordered pairs allows us to solve for unknown variables by comparing components.
In simple words: When two pairs like (a, b) and (c, d) are equal, it means 'a' must be the same as 'c', and 'b' must be the same as 'd'. So we make two small math problems and solve them to find x and y.

🎯 Exam Tip: Remember that ordered pairs are equal if and only if their first components are equal AND their second components are equal. This principle is key for solving such problems.

 

Question 3. If A = {1, 3, 5, 7} and B = {2, 4, 6}, find
(i) A x A
(ii) A x B
(iii) B x A
(iv) B x B
(v) n (A x A)
(vi) n (A x B)
(vii) n (B x A)
(viii) n (B x B)
Answer: We are given the sets \( A = \{1, 3, 5, 7\} \) and \( B = \{2, 4, 6\} \).
The number of elements in set A is \( n(A) = 4 \).
The number of elements in set B is \( n(B) = 3 \).
A Cartesian product combines every element from the first set with every element from the second set to form ordered pairs.
(i) To find \( A \times A \), we pair each element of A with every element of A:
\( A \times A = \{1, 3, 5, 7\} \times \{1, 3, 5, 7\} \)
\( = \{(1, 1), (1, 3), (1, 5), (1, 7), (3, 1), (3, 3), (3, 5), (3, 7), (5, 1), (5, 3), (5, 5), (5, 7), (7, 1), (7, 3), (7, 5), (7, 7)\} \)
(ii) To find \( A \times B \), we pair each element of A with every element of B:
\( A \times B = \{1, 3, 5, 7\} \times \{2, 4, 6\} \)
\( = \{(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6), (7, 2), (7, 4), (7, 6)\} \)
(iii) To find \( B \times A \), we pair each element of B with every element of A:
\( B \times A = \{2, 4, 6\} \times \{1, 3, 5, 7\} \)
\( = \{(2, 1), (2, 3), (2, 5), (2, 7), (4, 1), (4, 3), (4, 5), (4, 7), (6, 1), (6, 3), (6, 5), (6, 7)\} \)
(iv) To find \( B \times B \), we pair each element of B with every element of B:
\( B \times B = \{2, 4, 6\} \times \{2, 4, 6\} \)
\( = \{(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)\} \)
(v) To find \( n(A \times A) \), which is the number of elements in \( A \times A \):
\( n(A \times A) = n(A) \times n(A) = 4 \times 4 = 16 \)
(vi) To find \( n(A \times B) \), which is the number of elements in \( A \times B \):
\( n(A \times B) = n(A) \times n(B) = 4 \times 3 = 12 \)
(vii) To find \( n(B \times A) \), which is the number of elements in \( B \times A \):
\( n(B \times A) = n(B) \times n(A) = 3 \times 4 = 12 \)
(viii) To find \( n(B \times B) \), which is the number of elements in \( B \times B \):
\( n(B \times B) = n(B) \times n(B) = 3 \times 3 = 9 \)
In simple words: The Cartesian product means making all possible pairs by taking the first item from the first set and the second item from the second set. For counting how many pairs there are, you just multiply the number of items in the first set by the number of items in the second set.

🎯 Exam Tip: Remember that \( A \times B \) is generally not the same as \( B \times A \) unless \( A = B \) or one of the sets is empty. However, the number of elements, \( n(A \times B) \), is always equal to \( n(B \times A) \).

 

Question 4. Answer true or false :
(i) If P = {m, n} and Q = {m, n}, then P x Q = {(m, n), (n, m)}
(ii) {(a, x), (a, y), (b, x), (b, y)} is a product set.
(iii) If n (A) = x and n (B) = y and A ∩ B = Φ, then n (A x B) = xy.
(iv) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
Answer:
(i) If \( P = \{m, n\} \) and \( Q = \{m, n\} \), then \( P \times Q = \{(m, m), (m, n), (n, m), (n, n)\} \). The given statement only lists two pairs.
Therefore, the statement is **False**.
(ii) The set \( \{(a, x), (a, y), (b, x), (b, y)\} \) can be formed by taking the Cartesian product of \( A = \{a, b\} \) and \( B = \{x, y\} \). So, it is a product set.
Therefore, the statement is **True**.
(iii) The number of elements in the Cartesian product \( n(A \times B) \) is always equal to \( n(A) \times n(B) \), which is \( x \times y \). The condition \( A \cap B = \Phi \) (A and B are disjoint) does not affect the number of elements in their Cartesian product.
Therefore, the statement is **True**.
(iv) 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 \( y \in B \). The statement incorrectly says \( x \in B \) and \( y \in A \), which describes \( B \times A \) instead.
Therefore, the statement is **False**.
In simple words: We checked each statement about sets and their pairs. For a Cartesian product, the first element of each pair comes from the first set, and the second from the second set. The number of pairs is always found by multiplying the sizes of the two sets, no matter if they overlap or not.

🎯 Exam Tip: Pay close attention to the definition of Cartesian product and how it's formed. The order of sets matters for the elements of the product, but not for the total count of elements.

 

Question 5. Given A = {1, 2}, B = {3}, C = {4, 5}, test whether the following are true
(i) A x (B u C) = (A u B) x (A u C)
(ii) A x (B ∩ C) = (A x B) ∩ (A x C).
Answer: We are given the sets: \( A = \{1, 2\} \), \( B = \{3\} \), and \( C = \{4, 5\} \).
(i) Test whether \( A \times (B \cup C) = (A \cup B) \times (A \cup C) \):
First, let's find the Left Hand Side (L.H.S.):
\( B \cup C = \{3\} \cup \{4, 5\} = \{3, 4, 5\} \)
\( A \times (B \cup C) = \{1, 2\} \times \{3, 4, 5\} \)
\( = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)\} \)
Now, let's find the Right Hand Side (R.H.S.):
\( A \cup B = \{1, 2\} \cup \{3\} = \{1, 2, 3\} \)
\( A \cup C = \{1, 2\} \cup \{4, 5\} = \{1, 2, 4, 5\} \)
\( (A \cup B) \times (A \cup C) = \{1, 2, 3\} \times \{1, 2, 4, 5\} \)
\( = \{(1, 1), (1, 2), (1, 4), (1, 5), (2, 1), (2, 2), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5)\} \)
Comparing L.H.S. and R.H.S., we see that they are not equal: \( L.H.S. \neq R.H.S. \)
Thus, the given statement \( A \times (B \cup C) = (A \cup B) \times (A \cup C) \) is **not true**. This shows that the Cartesian product does not distribute over union in this way.
(ii) Test whether \( A \times (B \cap C) = (A \times B) \cap (A \times C) \):
First, let's find the Left Hand Side (L.H.S.):
\( B \cap C = \{3\} \cap \{4, 5\} = \Phi \) (the empty set, as there are no common elements)
\( A \times (B \cap C) = \{1, 2\} \times \Phi = \Phi \) (the Cartesian product with an empty set is always empty)
Now, let's find the Right Hand Side (R.H.S.):
\( A \times B = \{1, 2\} \times \{3\} = \{(1, 3), (2, 3)\} \)
\( A \times C = \{1, 2\} \times \{4, 5\} = \{(1, 4), (1, 5), (2, 4), (2, 5)\} \)
\( (A \times B) \cap (A \times C) = \{(1, 3), (2, 3)\} \cap \{(1, 4), (1, 5), (2, 4), (2, 5)\} = \Phi \) (as there are no common ordered pairs)
Comparing L.H.S. and R.H.S., we see that they are equal: \( L.H.S. = R.H.S. \)
Thus, the given statement \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is **true**. This illustrates a distributive property of the Cartesian product over intersection.
In simple words: We checked if some rules about making pairs from sets hold true. We found that the rule with "OR" (union) did not work, but the rule with "AND" (intersection) did work. You have to calculate both sides of the equals sign to check.

🎯 Exam Tip: Remember the distributive properties for Cartesian products: \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is true, but \( A \times (B \cup C) \neq (A \times B) \cup (A \times C) \) in general. Always work out both sides to verify.

 

Question 6. If A = {1, 2, 3, 4}, B = {5, 7, 9}, C = {2, 4, 6}, find
(i) A x B
(ii) (B x C)
(iii) C X A and draw their graphs.
Answer: We are given the sets: \( A = \{1, 2, 3, 4\} \), \( B = \{5, 7, 9\} \), and \( C = \{2, 4, 6\} \).
(i) To find \( A \times B \), we list all ordered pairs where the first element is from A and the second is from B:
\( A \times B = \{1, 2, 3, 4\} \times \{5, 7, 9\} \)
\( = \{(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)\} \)
The graph for \( A \times B \) is shown below, where X-axis values are from A and Y-axis values are from B. Each ordered pair is a point on this grid.

(ii) To find \( B \times C \), we list all ordered pairs where the first element is from B and the second is from C:
\( B \times C = \{5, 7, 9\} \times \{2, 4, 6\} \)
\( = \{(5, 2), (5, 4), (5, 6), (7, 2), (7, 4), (7, 6), (9, 2), (9, 4), (9, 6)\} \)
The graph for \( B \times C \) is shown below, where X-axis values are from B and Y-axis values are from C.
(iii) To find \( C \times A \), we list all ordered pairs where the first element is from C and the second is from A:
\( C \times A = \{2, 4, 6\} \times \{1, 2, 3, 4\} \)
\( = \{(2, 1), (2, 2), (2, 3), (2, 4), (4, 1), (4, 2), (4, 3), (4, 4), (6, 1), (6, 2), (6, 3), (6, 4)\} \)
The graph for \( C \times A \) is shown below, where X-axis values are from C and Y-axis values are from A. Each ordered pair corresponds to a unique point on this grid.
In simple words: We made all possible pairs for three different combinations of sets. Then, for each combination, we drew a picture (a grid) to show where these pairs would be if we thought of them as points. Each box in the grid that has a pair shows that pair as a spot.

🎯 Exam Tip: When graphing Cartesian products, represent the first set's elements on the x-axis and the second set's elements on the y-axis. Each ordered pair corresponds to a specific point on this grid, often represented as a cell or a dot.

 

Question 7. Some elements of A x B are (a, x), (c, y), (d, z). If A = {a, b, c, d], find the remaining elements of A x B such that n (A x B) is least.
Answer: We are given the set \( A = \{a, b, c, d\} \) and some elements of the Cartesian product \( A \times B \): \( (a, x), (c, y), (d, z) \).
For \( n(A \times B) \) to be the least possible, the set \( B \) must contain only the unique second components from the given ordered pairs.
The second components in the given pairs are \( x, y, \) and \( z \).
Therefore, we take \( B = \{x, y, z\} \).
Now we can find the complete Cartesian product \( A \times B \):
\( A \times B = \{a, b, c, d\} \times \{x, y, z\} \)
\( = \{(a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z), (d, x), (d, y), (d, z)\} \)
The elements given in the question were \( (a, x), (c, y), (d, z) \).
To find the *remaining* elements, we remove these from the complete list of \( A \times B \):
Remaining elements \( = \{(a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, z), (d, x), (d, y), (d, z)\} \)
In simple words: We knew some pairs in A x B. To make the total number of pairs as small as possible, we had to choose the smallest possible set B. This means B should only include the second parts (x, y, z) that we already saw. Then we made all possible pairs and removed the ones we already knew to find the rest.

🎯 Exam Tip: When asked to find the "least" number of elements in a Cartesian product, define the second set (or first, if applicable) using only the unique elements observed in the given partial product. This minimizes the size of the set and thus the product.

 

Question 8. The ordered pairs (1, 1), (2, 2), (3, 3) are among the elements in the set A x B. If A and B have 3 elements each, how many elements in all does the set A x B have ? Also find the remaining elements.
Answer: We are given that \( n(A) = 3 \) and \( n(B) = 3 \).
The total number of elements in the set \( A \times B \) is found by multiplying the number of elements in A by the number of elements in B:
\( n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9 \)
So, the set \( A \times B \) has \( 9 \) elements in total.
We are also given that \( (1, 1), (2, 2), (3, 3) \) are elements of \( A \times B \).
For any ordered pair \( (p, q) \) in \( A \times B \), \( p \) must be an element of \( A \) and \( q \) must be an element of \( B \).
Looking at the given pairs:
First components: \( 1, 2, 3 \). These must be in set \( A \).
Second components: \( 1, 2, 3 \). These must be in set \( B \).
Since \( n(A) = 3 \) and we have found three distinct elements \( \{1, 2, 3\} \) that must be in A, this means \( A = \{1, 2, 3\} \).
Similarly, since \( n(B) = 3 \) and we have found three distinct elements \( \{1, 2, 3\} \) that must be in B, this means \( B = \{1, 2, 3\} \).
Now we can find all elements of \( A \times B \):
\( A \times B = \{1, 2, 3\} \times \{1, 2, 3\} \)
\( = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \)
The given elements were \( (1, 1), (2, 2), (3, 3) \).
To find the remaining elements, we remove these given pairs from the complete set of \( A \times B \):
Remaining elements \( = \{(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)\} \)
In simple words: We first found the total number of pairs by multiplying the number of items in set A and set B. Since we knew some pairs and how many items were in A and B, we could figure out exactly what was in set A and set B. Then we listed all possible pairs and removed the ones we already knew to find the others.

🎯 Exam Tip: The knowledge of \( n(A) \) and \( n(B) \) combined with specific elements of \( A \times B \) can often uniquely determine the sets A and B themselves. Always confirm that the deduced sets match the given cardinalities.

 

Question 9. If A = {1, 4}, B = {2, 3, 6} and C = {2, 3, 7}, then verify that
(i) A x (B u C) = (A x B) u (A x C)
(ii) A x (B ∩ C) = (A x B) ∩ (A x C).
Answer: We are given the sets: \( A = \{1, 4\} \), \( B = \{2, 3, 6\} \), and \( C = \{2, 3, 7\} \).
(i) To verify \( A \times (B \cup C) = (A \times B) \cup (A \times C) \):
First, calculate the Left Hand Side (L.H.S.):
\( B \cup C = \{2, 3, 6\} \cup \{2, 3, 7\} = \{2, 3, 6, 7\} \)
\( L.H.S. = A \times (B \cup C) = \{1, 4\} \times \{2, 3, 6, 7\} \)
\( = \{(1, 2), (1, 3), (1, 6), (1, 7), (4, 2), (4, 3), (4, 6), (4, 7)\} \)
Now, calculate the Right Hand Side (R.H.S.):
\( A \times B = \{1, 4\} \times \{2, 3, 6\} = \{(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)\} \)
\( A \times C = \{1, 4\} \times \{2, 3, 7\} = \{(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)\} \)
\( R.H.S. = (A \times B) \cup (A \times C) \)
\( = \{(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)\} \cup \{(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)\} \)
\( = \{(1, 2), (1, 3), (1, 6), (1, 7), (4, 2), (4, 3), (4, 6), (4, 7)\} \)
Comparing L.H.S. and R.H.S., we see that \( L.H.S. = R.H.S. \)
Thus, the statement \( A \times (B \cup C) = (A \times B) \cup (A \times C) \) is **verified as true**.
(ii) To verify \( A \times (B \cap C) = (A \times B) \cap (A \times C) \):
First, calculate the Left Hand Side (L.H.S.):
\( B \cap C = \{2, 3, 6\} \cap \{2, 3, 7\} = \{2, 3\} \) (common elements are 2 and 3)
\( L.H.S. = A \times (B \cap C) = \{1, 4\} \times \{2, 3\} \)
\( = \{(1, 2), (1, 3), (4, 2), (4, 3)\} \)
Now, calculate the Right Hand Side (R.H.S.):
\( A \times B = \{1, 4\} \times \{2, 3, 6\} = \{(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)\} \)
\( A \times C = \{1, 4\} \times \{2, 3, 7\} = \{(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)\} \)
\( R.H.S. = (A \times B) \cap (A \times C) \)
\( = \{(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)\} \cap \{(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)\} \)
\( = \{(1, 2), (1, 3), (4, 2), (4, 3)\} \)
Comparing L.H.S. and R.H.S., we see that \( L.H.S. = R.H.S. \)
Thus, the statement \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is **verified as true**. The Cartesian product distributes over both union and intersection.
In simple words: We checked two math rules for sets. We calculated both sides of the equals sign for each rule. Both times, we found that the two sides matched perfectly, so both rules are correct for these sets.

🎯 Exam Tip: These two identities demonstrate the distributive properties of the Cartesian product over set union and intersection. It's important to remember that these properties hold true for any sets A, B, and C.

 

Question 10. If A = {2, 3}, B = {1, 2, 3}, C = {2, 3, 4} show that A x A = (B x B) ∩ (C x C).
Answer: We are given the sets: \( A = \{2, 3\} \), \( B = \{1, 2, 3\} \), and \( C = \{2, 3, 4\} \).
We need to show that \( A \times A = (B \times B) \cap (C \times C) \).
First, calculate the Left Hand Side (L.H.S.):
\( L.H.S. = A \times A = \{2, 3\} \times \{2, 3\} \)
\( = \{(2, 2), (2, 3), (3, 2), (3, 3)\} \)
Now, calculate the Right Hand Side (R.H.S.):
First, find \( B \times B \):
\( B \times B = \{1, 2, 3\} \times \{1, 2, 3\} \)
\( = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \)
Next, find \( C \times C \):
\( C \times C = \{2, 3, 4\} \times \{2, 3, 4\} \)
\( = \{(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)\} \)
Finally, find the intersection of \( (B \times B) \) and \( (C \times C) \):
\( R.H.S. = (B \times B) \cap (C \times C) \)
\( = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\} \cap \{(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)\} \)
\( = \{(2, 2), (2, 3), (3, 2), (3, 3)\} \) (These are the ordered pairs common to both products)
Comparing the L.H.S. and R.H.S., we see that:
\( L.H.S. = \{(2, 2), (2, 3), (3, 2), (3, 3)\} \)
\( R.H.S. = \{(2, 2), (2, 3), (3, 2), (3, 3)\} \)
Therefore, \( L.H.S. = R.H.S. \), which means \( A \times A = (B \times B) \cap (C \times C) \) is shown to be true. This demonstrates how the intersection of Cartesian products can result in a new Cartesian product.
In simple words: We wanted to show that one set of pairs (A x A) is the same as finding common pairs from two other sets (B x B) and (C x C). We made all the pairs for each side, and then found that they ended up being exactly the same.

🎯 Exam Tip: When showing set equalities, always calculate each side of the equation completely and then compare the resulting sets. Remember that \( (P \times Q) \cap (R \times S) \) consists of pairs \( (x, y) \) where \( x \in P \cap R \) and \( y \in Q \cap S \).

 

Question 11. If A and B be two sets such that n (A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y, z are distinct elements.
Answer: We are given that \( n(A) = 3 \) and \( n(B) = 2 \).
We are also given three ordered pairs that are elements of \( A \times B \): \( (x, 1), (y, 2), (z, 1) \).
A fundamental rule of Cartesian products is that if \( (a, b) \) is an element of \( A \times B \), then \( a \) must be an element of set \( A \), and \( b \) must be an element of set \( B \).
Let's look at the first components of the given pairs: \( x, y, z \).
Since \( x, y, z \) are given as distinct elements, and they are all first components of pairs in \( A \times B \), they must belong to set \( A \).
We know that \( n(A) = 3 \). Since we have found three distinct elements \( \{x, y, z\} \) that must be in A, this means:
\( A = \{x, y, z\} \)
Now, let's look at the second components of the given pairs: \( 1, 2, 1 \).
These second components must belong to set \( B \). The unique values among them are \( 1 \) and \( 2 \).
We know that \( n(B) = 2 \). Since we have found two distinct elements \( \{1, 2\} \) that must be in B, this means:
\( B = \{1, 2\} \)
So, set A is \( \{x, y, z\} \) and set B is \( \{1, 2\} \). These sets satisfy the given conditions for the number of elements.
In simple words: We were told how many items were in set A and set B, and given some example pairs from A x B. We used the first part of each example pair to find the items in set A, and the second part to find the items in set B. Because the example parts were unique and matched the given counts, we could figure out both sets.

🎯 Exam Tip: When given elements of \( A \times B \) and the cardinalities of A and B, remember that all first components must belong to A, and all second components must belong to B. Use the "distinct elements" clue to correctly identify the unique members of each set.

 

Question 12. The Cartesian product A x A has 9 elements among which are found (- 1, 0) and (0, 1). Find the set A are the remaining elements of A x A.
Answer: We are given that the Cartesian product \( A \times A \) has 9 elements.
Since \( n(A \times A) = n(A) \times n(A) \), we can write:
\( n(A) \times n(A) = 9 \)
\( \implies n(A) = 3 \) (because \( 3 \times 3 = 9 \))
This means set A must contain exactly 3 distinct elements.
We are also given that two elements of \( A \times A \) are \( (-1, 0) \) and \( (0, 1) \).
If an ordered pair \( (p, q) \) is in \( A \times A \), then both \( p \) and \( q \) must be elements of set \( A \).
From the pair \( (-1, 0) \), we know that \( -1 \in A \) and \( 0 \in A \).
From the pair \( (0, 1) \), we know that \( 0 \in A \) and \( 1 \in A \).
Combining these, the elements that must be in A are \( \{-1, 0, 1\} \).
Since \( n(A) \) must be 3, and we have found exactly three distinct elements, we can conclude that:
\( A = \{-1, 0, 1\} \)
Now, we can find all elements of the Cartesian product \( A \times A \):
\( A \times A = \{-1, 0, 1\} \times \{-1, 0, 1\} \)
\( = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\} \)
The given elements from \( A \times A \) were \( (-1, 0) \) and \( (0, 1) \).
To find the *remaining* elements, we remove these given pairs from the complete set of \( A \times A \):
Remaining elements \( = \{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\} \)
In simple words: We knew how many pairs were in A x A, so we figured out how many items were in set A itself. Then, using the example pairs provided, we found all the items that must be in A. Since this matched the total count for A, we knew exactly what set A was. Finally, we listed all pairs in A x A and took out the ones we were already given to find the rest.

🎯 Exam Tip: When \( A \times A \) is involved, remember that both components of any ordered pair must come from the same set A. This helps to quickly deduce the elements of A from partial information about \( A \times A \).