CBSE Class 12 Mathematics Relations And Functions Assignment Set 18

XII - Mathematics Assignment No-04 - Relations and Functions

Q1. Let * be a binary operation on \(N\) given by \(a * b = \text{l.c.m of } (a, b) \ \forall \ a, b \in N\). Find (i) \(2 * 4\) (ii) \(3 * 5\) (iii) \(7 * 21\)
Answer: The binary operation is defined as \(a * b = \text{l.c.m}(a, b)\).
(i) \(2 * 4 = \text{l.c.m}(2, 4) = 4\)
(ii) \(3 * 5 = \text{l.c.m}(3, 5) = 15\)
(iii) \(7 * 21 = \text{l.c.m}(7, 21) = 21\)
In simple words: Find the least common multiple for each pair of numbers.

Exam Tip: For any two numbers where one divides the other completely, their LCM is always the larger number.

Q2. Let * be a binary operation on \(Z\) defined as \(a * b = a + 3b^2\) \forall \ a, b \in Z \). Find (i) \(1 * 3\) (ii) \(2 * 4\) (iii) \(3 * 2\)
Answer: The binary operation is given by \(a * b = a + 3b^2\).
(i) \(1 * 3 = 1 + 3(3)^2\)
\(= 1 + 3(9)\)
\(= 1 + 27 = 28\)
(ii) \(2 * 4 = 2 + 3(4)^2\)
\(= 2 + 3(16)\)
\(= 2 + 48 = 50\)
(iii) \(3 * 2 = 3 + 3(2)^2\)
\(= 3 + 3(4)\)
\(= 3 + 12 = 15\)
In simple words: Substitute the given values of a and b into the formula \(a + 3b^2\) and calculate.

Exam Tip: Be careful with the order of operations-always evaluate the exponent before multiplying by 3 and adding a.

Q3. Find the number of binary operations on the set \(\{a, b\}\)
Answer: Let \(S = \{a, b\}\). The number of elements in set \(S\) is \(n = 2\).
A binary operation on \(S\) is a function from \(S \times S\) to \(S\).
The number of elements in \(S \times S\) is \(n \times n = 2 \times 2 = 4\).
The total number of binary operations is \(n^{n^2} = 2^{2^2} = 2^4 = 16\).
In simple words: Use the formula \(n^{n^2}\) where n is the number of elements in the set.

Exam Tip: Remember the general formula: for a set with n elements, the total number of binary operations is \(n^{n^2}\).

Q4. Define a binary operation * on the set \(\{0, 1, 2, 3, 4, 5\}\) as a * b =  a + b, & if  a + b < 6 a + b - 6, & if a + b  > 6 . Show that zero is the identity operation and each element 'a' of the set is invertible with \((6-a)\) being the inverse of a.
Answer: Let \(S = \{0, 1, 2, 3, 4, 5\}\).
Identity Element:
An element \(e \in S\) is an identity element if \(a * e = a = e * a\) \forall \ a \in S \).
Let \(e = 0\). For any \(a \in S\):
\(a * 0 = a + 0 = a\) (since \(a + 0 < 6\))
\(0 * a = 0 + a = a\) (since \(0 + a < 6\))
Thus, \(a * 0 = a = 0 * a\) \forall \ a \in S \). Therefore, 0 is the identity element.
Invertible Elements:
An element \(a \in S\) is invertible if there exists an element \(b \in S\) such that \(a * b = e = b * a\), where \(e = 0\).
Let \(b = 6 - a\). Since \(a \in \{0, 1, 2, 3, 4, 5\}\), \(b = 6 - a \in S\).
Consider \(a * (6 - a)\):
Since \(a + (6 - a) = 6 \ge 6\), we use the second branch of the definition:
\(a * (6 - a) = a + (6 - a) - 6 = 6 - 6 = 0 = e\).
Similarly, \((6 - a) * a = (6 - a) + a - 6 = 0 = e\).
Thus, every element \(a\) is invertible and its inverse is \((6 - a)\).
In simple words: Check that operating by 0 leaves any number unchanged (identity), and operating by \((6-a)\) gives a sum of 6 which wraps around to 0 (inverse).

Exam Tip: Always verify that both the identity element and the calculated inverse elements actually belong to the given set.

Q5. Which of the following binary operations on the set \(N\) are associative and which are Commutative?
(i) \(a * b = 1\) \forall \ a, b \in N \)
(ii) \(a * b = \frac{a + b}{2}\) \forall \ a, b \in N \)
Answer: (i) For \(a * b = 1\):
Commutativity: \(a * b = 1\) and \(b * a = 1 \implies a * b = b * a\). Hence, it is Commutative.
Associativity: \((a * b) * c = 1 * c = 1\) and \(a * (b * c) = a * 1 = 1 \implies (a * b) * c = a * (b * c)\). Hence, it is Associative.
(ii) For \(a * b = \frac{a + b}{2}\):
Commutativity: \(a * b = \frac{a + b}{2} = \frac{b + a}{2} = b * a\). Hence, it is Commutative.
Associativity: Let \(a = 1, b = 2, c = 3\).
\((1 * 2) * 3 = \left(\frac{1 + 2}{2}\right) * 3 = \frac{3}{2} * 3 = \frac{\frac{3}{2} + 3}{2} = \frac{9}{4}\)
\(1 * (2 * 3) = 1 * \left(\frac{2 + 3}{2}\right) = 1 * \frac{5}{2} = \frac{1 + \frac{5}{2}}{2} = \frac{7}{4}\)
Since \(\frac{9}{4} \neq \frac{7}{4}\), the operation is not associative.
In simple words: Check if swapping the inputs changes the result (Commutative) and if grouping changes the result (Associative).

Exam Tip: To disprove associativity, a single counterexample is sufficient and faster than a general proof.

Q6. Consider the binary operation * on the set \(A = \{1, 2, 3, 4, 5\}\) defined by \(a * b = \min(a, b)\). Write the operation table of the operation *
Answer: The operation table for \(a * b = \min(a, b)\) on set \(A = \{1, 2, 3, 4, 5\}\) is given below:

In simple words: Fill each cell in the table by taking the smaller of the row and column headers.

Exam Tip: For min and max operations, the table is always symmetric along the main diagonal.

Q7. Let * be a binary operation on the set \(Q\) of rational numbers as follows:
(a) \(a * b = \frac{ab}{4}\)
(b) \(a * b = ab^2\)
Find which of the binary operations are Commutative and which are associative?
Answer: (a) For \(a * b = \frac{ab}{4}\):
Commutativity: \(a * b = \frac{ab}{4} = \frac{ba}{4} = b * a\). It is Commutative.
Associativity: \((a * b) * c = \left(\frac{ab}{4}\right) * c = \frac{\left(\frac{ab}{4}\right)c}{4} = \frac{abc}{16}\)
\(a * (b * c) = a * \left(\frac{bc}{4}\right) = \frac{a\left(\frac{bc}{4}\right)}{4} = \frac{abc}{16}\)
Since \((a * b) * c = a * (b * c)\), it is Associative.
(b) For \(a * b = ab^2\):
Commutativity: Let \(a = 2, b = 3\).
\(2 * 3 = 2(3^2) = 2(9) = 18\)
\(3 * 2 = 3(2^2) = 3(4) = 12\)
Since \(18 \neq 12\), it is not Commutative.
Associativity: Let \(a = 1, b = 2, c = 3\).
\((1 * 2) * 3 = (1 \cdot 2^2) * 3 = 4 * 3 = 4(3^2) = 4(9) = 36\)
\(1 * (2 * 3) = 1 * (2 \cdot 3^2) = 1 * 18 = 1(18^2) = 324\)
Since \(36 \neq 324\), it is not Associative.
In simple words: Test commutativity by swapping variables, and test associativity by checking if different groupings yield the same result.

Exam Tip: Multiplication of rational numbers is commutative and associative, which helps simplify fraction-based operations.

Q8. Let \(A = N \times N\) and let * be a binary operation on \(A\) defined by \((a, b) * (c, d) = (a + c, b + d)\). Show that * on \(A\) is Commutative and Associative. Also find the identity element on \(A\) if exist.
Answer: Commutativity:
Let \((a, b), (c, d) \in A\).
\((a, b) * (c, d) = (a + c, b + d)\)
\(= (c + a, d + b)\) (since addition is commutative in \(N\))
\(= (c, d) * (a, b)\)
Hence, * is Commutative.
Associativity:
Let \((a, b), (c, d), (e, f) \in A\).
\(((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f)\)
\(= (a + (c + e), b + (d + f))\) (since addition is associative in \(N\))
\(= (a, b) * (c + e, d + f) = (a, b) * ((c, d) * (e, f))\)
Hence, * is Associative.
Identity Element:
Let \((e_1, e_2) \in A\) be the identity element such that for any \((a, b) \in A\):
\((a, b) * (e_1, e_2) = (a, b)\)
\((a + e_1, b + e_2) = (a, b)\)
This implies \(a + e_1 = a \implies e_1 = 0\), and \(b + e_2 = b \implies e_2 = 0\).
However, \((0, 0) \notin N \times N\) because 0 is not a natural number.
Therefore, no identity element exists in \(N \times N\).
In simple words: The operation is commutative and associative because vector addition is component-wise. There is no identity because 0 is not a natural number.

Exam Tip: Always check that any candidate for an identity element belongs strictly to the defined domain set.