CBSE Class 11 Mathematics Relations and Functions Assignment Set 07

Relations

Question 1. Let \(R\) be a relation on the set \(N\) of natural numbers defined by \(a + 3b = 15\). Find (i) \(R\) (ii) domain of \(R\) (iii) range of \(R\).
Answer: Given relation: \(a + 3b = 15\), where \(a, b \in N\)
\(a = 15 - 3b\)
Since \(a, b \in N\), we substitute values for \(b\):
For \(b = 1\), \(a = 15 - 3(1) = 12\) (since \(12 \in N\))
For \(b = 2\), \(a = 15 - 3(2) = 9\) (since \(9 \in N\))
For \(b = 3\), \(a = 15 - 3(3) = 6\) (since \(6 \in N\))
For \(b = 4\), \(a = 15 - 3(4) = 3\) (since \(3 \in N\))
For \(b = 5\), \(a = 15 - 3(5) = 0\) (not in \(N\), since natural numbers start from 1)
(i) \(R = \{(12, 1), (9, 2), (6, 3), (3, 4)\}\)
(ii) Domain of \(R = \{12, 9, 6, 3\}\)
(iii) Range of \(R = \{1, 2, 3, 4\}\)
In simple words: We find pairs of natural numbers that satisfy the given equation and then collect their first and second elements for domain and range.

Exam Tip: Always check that elements of the domain and range belong to the specified set, which is \(N\) in this case.

Question 2. Let \(A = \{3, 5\}\) and \(B = \{7, 11\}\). Let \(R = \{(a, b) : a \in A, b \in B, a + b \text{ is odd}\}\). Show that \(R\) is an empty relation from \(A\) into \(B\).
Answer: Let us find all possible sums of elements \(a \in A\) and \(b \in B\):
\(3 + 7 = 10\) (even)
\(3 + 11 = 14\) (even)
\(5 + 7 = 12\) (even)
\(5 + 11 = 16\) (even)
Since no sum is odd, no ordered pair satisfies the condition of relation \(R\).
Therefore, \(R = \emptyset\). Thus, \(R\) is an empty relation from \(A\) into \(B\).
In simple words: Since adding any number from set A to any number from set B gives an even number, no pairs can be formed, making the relation empty.

Exam Tip: An empty relation contains no elements from the Cartesian product.

Question 3. Let \(A = \{a, b\}\), \(B = \{c, d\}\). Find the number of relations from \(A\) to \(B\).
Answer: Number of elements in set \(A\), \(n(A) = 2\)
Number of elements in set \(B\), \(n(B) = 2\)
Number of elements in Cartesian product \(A \times B\) is \(n(A \times B) = n(A) \times n(B) = 2 \times 2 = 4\)
Total number of subsets of \(A \times B\) is \(2^{n(A \times B)} = 2^4 = 16\)
Since every relation is a subset of \(A \times B\), the number of relations from \(A\) to \(B\) is 16.
In simple words: The total number of relations from set A to set B is equal to 2 raised to the power of the product of the number of elements in both sets.

Exam Tip: Remember the formula: If \(n(A) = p\) and \(n(B) = q\), total relations = \(2^{pq}\).

Question 4. Let \(A = \{1, 2, 3\}\), \(B = \{7, 8\}\). Represent the relation from set \(A\) to set \(B\) by arrow diagram.
Answer: Any arbitrary relation from \(A\) to \(B\) can be represented by selecting a subset of \(A \times B\). Let us define a relation \(R = \{(1, 7), (2, 7), (3, 8)\}\) as an example of a relation from \(A\) to \(B\).
Arrow diagram representation:
In simple words: An arrow diagram maps elements of set A to set B using directed lines based on the relation pairs.

Exam Tip: Draw two ellipses for sets A and B, place elements inside them, and draw arrows from domain elements to corresponding range elements.

Question 5. Let \(A = \{1, 2, 3, 4, 5, 6, 7\}\). Define a relation \(R\) from \(A\) to \(A\) by \(R = \{(x, y) : y = x + 2\}\). (i) Write the relation from \(A\) to \(A\) (ii) Write its domain, co-domain, and range of \(R\) (iii) Show the relation by arrow diagram.
Answer: Given \(A = \{1, 2, 3, 4, 5, 6, 7\}\) and \(y = x + 2\).
For \(x = 1, y = 1 + 2 = 3\)
For \(x = 2, y = 2 + 2 = 4\)
For \(x = 3, y = 3 + 2 = 5\)
For \(x = 4, y = 4 + 2 = 6\)
For \(x = 5, y = 5 + 2 = 7\)
For \(x = 6, y = 6 + 2 = 8\) (not in \(A\))
For \(x = 7, y = 7 + 2 = 9\) (not in \(A\))
(i) \(R = \{(1, 3), (2, 4), (3, 5), (4, 6), (5, 7)\}\)
(ii) Domain \(= \{1, 2, 3, 4, 5\}\), Range \(= \{3, 4, 5, 6, 7\}\), Co-domain \(= \{1, 2, 3, 4, 5, 6, 7\}\)
(iii) Arrow diagram representation:
In simple words: Each element x in set A is paired with y which is 2 greater than x, provided y is also inside set A.

Exam Tip: Co-domain is always the entire second set, while range is only the set of second elements that actually get mapped.

Question 6. If \(A = \{1, 2, 3, \dots, 15\}\) and \(R\) is a relation on \(A\) defined by \(R = \{(a, b) : 3a - b = 0, a, b \in A\}\). Write \(R\) in roster form.
Answer: Given \(3a - b = 0 \implies b = 3a\), where \(a, b \in \{1, 2, 3, \dots, 15\}\).
For \(a = 1, b = 3(1) = 3\)
For \(a = 2, b = 3(2) = 6\)
For \(a = 3, b = 3(3) = 9\)
For \(a = 4, b = 3(4) = 12\)
For \(a = 5, b = 3(5) = 15\)
For \(a = 6, b = 3(6) = 18\) (not in set \(A\))\
Therefore, roster form of \(R\) is:
\(R = \{(1, 3), (2, 6), (3, 9), (4, 12), (5, 15)\}\)
In simple words: We calculate b as 3 times a for each element in set A until b exceeds 15.

Exam Tip: Always make sure both coordinates of the ordered pair lie within the specified set A.

Question 7. Find the domain and the range of the relation \(R\) defined by \(R = \{(x + 2, x + 4) : x \in N, x < 10\}\).
Answer: Given \(x \in N\) and \(x < 10\), so \(x \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\).
The relation elements are generated by ordered pairs \((x + 2, x + 4)\):
For \(x = 1: (1+2, 1+4) = (3, 5)\)
For \(x = 2: (2+2, 2+4) = (4, 6)\)
For \(x = 3: (3+2, 3+4) = (5, 7)\)
For \(x = 4: (4+2, 4+4) = (6, 8)\)
For \(x = 5: (5+2, 5+4) = (7, 9)\)
For \(x = 6: (6+2, 6+4) = (8, 10)\)
For \(x = 7: (7+2, 7+4) = (9, 11)\)
For \(x = 8: (8+2, 8+4) = (10, 12)\)
For \(x = 9: (9+2, 9+4) = (11, 13)\)
Domain \(= \{3, 4, 5, 6, 7, 8, 9, 10, 11\}\)
Range \(= \{5, 6, 7, 8, 9, 10, 11, 12, 13\}\)
In simple words: Substitute each natural number less than 10 into the coordinate expressions to find all ordered pairs, then extract all first and second elements.

Exam Tip: Be careful with strict inequalities like \(x < 10\); it means values stop at 9.

Question 8. If \(A = \{1, 2\}\), \(B = \{3, 4, 5\}\). Find number of relations from (i) \(A\) to \(A\) (ii) \(B\) to \(B\) (iii) \(A\) to \(B\).
Answer: Given \(n(A) = 2\) and \(n(B) = 3\).
(i) Number of relations from \(A\) to \(A\):
\(n(A \times A) = n(A) \times n(A) = 2 \times 2 = 4\)
Total relations = \(2^4\)
(ii) Number of relations from \(B\) to \(B\):
\(n(B \times B) = n(B) \times n(B) = 3 \times 3 = 9\)
Total relations = \(2^9\)
(iii) Number of relations from \(A\) to \(B\):
\(n(A \times B) = n(A) \times n(B) = 2 \times 3 = 6\)
Total relations = \(2^6\)
In simple words: Calculate the number of elements in the Cartesian product of the two sets and use it as the exponent of 2.

Exam Tip: Total relations from set X to set Y is always \(2^{n(X) \times n(Y)}\).

Question 9. Let \(A = \{1, 2, 3, 4, 5\}\) and \(B = \{2, 4, 6, 8, 10\}\). Let \(R = \{(a, b) : a \in A, b \in B, a \text{ divides } b\}\). Find a relation from \(A\) into \(B\) and show that domain of \(R\) is \(A\) and range of \(R\) is \(B\).
Answer: Given \(A = \{1, 2, 3, 4, 5\}\) and \(B = \{2, 4, 6, 8, 10\}\).
Condition: \(a\) divides \(b\) (i.e., \(b\) is a multiple of \(a\)).
\(R = \{(1, 2), (1, 4), (1, 6), (1, 8), (1, 10), (2, 2), (2, 4), (2, 6), (2, 8), (2, 10), (3, 6), (4, 4), (4, 8), (5, 10)\}\)
Domain of \(R = \{1, 2, 3, 4, 5\} = A\)
Range of \(R = \{2, 4, 6, 8, 10\} = B\)
In simple words: We pair each number in set A with every number in set B that it can divide evenly.

Exam Tip: "a divides b" means \(\frac{b}{a}\) is an integer.

Question 10. This figure shows a relationship between the sets \(A\) and \(B\). Write this relation (i) in set-builder form (ii) in roster form. Find also domain and range.
Answer: From the given arrow diagram, the elements mapped are: \(1 \rightarrow 1\), \(2 \rightarrow 4\), \(3 \rightarrow 9\), \(4 \rightarrow 16\).
(i) Set-builder form:
\(R = \{(x, y) : y = x^2, x, y \in N, x \le 4\}\)
(ii) Roster form:
\(R = \{(1, 1), (2, 4), (3, 9), (4, 16)\}\)
Domain \(= \{1, 2, 3, 4\}\)
Range \(= \{1, 4, 9, 16\}\)
In simple words: Each number in set A is squared to get its corresponding paired number in set B.

Exam Tip: Always look for algebraic patterns (like squares, cubes, or linear functions) between input and output elements in arrow diagrams.