CBSE Class 12 Mathematics Relations And Functions Assignment Set 20

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:
(i) \(2 * 4 = \text{l.c.m of } (2, 4) = 4\)
(ii) \(3 * 5 = \text{l.c.m of } (3, 5) = 15\)
(iii) \(7 * 21 = \text{l.c.m of } (7, 21) = 21\)
In simple words: The binary operation finds the least common multiple (LCM) of the two given numbers.

Exam Tip: For prime numbers or when one number is a multiple of the other, finding LCM is straightforward (for multiples, the larger number is the LCM).

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:
(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 values of \(a\) and \(b\) into the given algebraic expression carefully, following the order of operations.

Exam Tip: Always evaluate exponents before multiplication and addition to avoid calculation errors.

Q3. Find the number of binary operations on the set \(\{a, b\}\)
Answer:
Let \(S = \{a, b\}\). The number of elements in \(S\) is \(n = 2\).
A binary operation on \(S\) is a function from \(S \times S \to S\).
Number of elements in \(S \times S = 2 \times 2 = 4\).
Total number of binary operations = \(n^{(n^2)} = 2^{2^2} = 2^4 = 16\).
In simple words: A binary operation is a mapping from the cartesian product to the set itself, so use the formula \(n^{(n^2)}\).

Exam Tip: Remember the general formula: if a set has \(n\) 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 \(S = \{0, 1, 2, 3, 4, 5\}\).
Part 1: Identity Element
Let \(e \in S\) be the identity element. Then for any \(a \in S\):
\(a * e = a \implies a + e = a\) (assuming \(a + e < 6\)) \(\implies e = 0\).
Also, for \(a = 0\), \(0 * 0 = 0 + 0 = 0\). Thus, \(0\) is the identity element in \(S$.
Part 2: Invertibility
Let \(b = 6 - a\) be the inverse of \(a\). Since \(a \in \{0, 1, 2, 3, 4, 5\}\), clearly \(b \in S\).
Consider \(a * (6 - a)\):
Since \(a + (6 - a) = 6 \ge 6\), by definition of the operation:
\(a * (6 - a) = a + (6 - a) - 6 = 6 - 6 = 0\), which is the identity element.
Similarly, \((6 - a) * a = 0\).
Hence, each element \(a\) is invertible with its inverse being \((6 - a)\) (note: for \(a = 0\), inverse is \(6-0 = 6 \equiv 0 \pmod 6\)).
In simple words: Check that operating by 0 leaves elements unchanged, and operating any element with \(6-a\) yields 0 (the identity).

Exam Tip: For modulo-based operations, ensure that the calculated inverse also belongs 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 \(b * a = 1\). Since \(a * b = b * a\), it is commutative.
- Associativity: \((a * b) * c = 1 * c = 1\), and \(a * (b * c) = a * 1 = 1\). Since \((a * b) * c = a * (b * c)\), it is associative.
(ii) For \(a * b = \frac{a + b}{2}\):
- Commutativity: \(a * b = \frac{a + b}{2} = \frac{b + a}{2} = b * a\). 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: Test commutativity by swapping inputs, and test associativity by checking if grouping changes the final result.

Exam Tip: To prove an operation is not associative, providing a single counterexample with numbers is sufficient and fully valid.

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 \(A = \{1, 2, 3, 4, 5\}\) is constructed by taking the minimum of the row and column elements:

In simple words: Fill each cell of the table with the smaller of the two intersecting elements.

Exam Tip: Double-check table symmetry along the main diagonal to verify commutativity.

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: \(a * b = \frac{ab}{4} = \frac{ba}{4} = b * a\). Hence, 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: \(a * b = ab^2\) and \(b * a = ba^2\). Since \(ab^2 \neq ba^2\) in general, it is not commutative.
- Associativity: Let \(a = 2, b = 2, c = 2\).
\((2 * 2) * 2 = (2(2)^2) * 2 = 8 * 2 = 8(2)^2 = 32\)
\(2 * (2 * 2) = 2 * (2(2)^2) = 2 * 8 = 2(8)^2 = 128\)
Since \(32 \neq 128\), it is not associative.
In simple words: Test both properties algebraically using definitions, and use numerical counterexamples when properties fail.

Exam Tip: Multiplication of rational numbers is commutative, but placing variables in exponents like \(ab^2\) breaks symmetry.

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 it exists.
Answer:
1. Commutativity:
For any \((a, b), (c, d) \in N \times N\):
\((a, b) * (c, d) = (a + c, b + d)\)
By commutative property of addition in natural numbers: \((c + a, d + b) = (c, d) * (a, b)\).
Hence, \(*\) is commutative.
2. Associativity:
For any \((a, b), (c, d), (e, f) \in N \times N\):
\(((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f)\)
\((a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + (c + e), b + (d + f))\)
By associativity of addition in natural numbers, both are equal. Hence, \(*\) is associative.
3. Identity Element:
Let \((e_1, e_2) \in N \times N\) be the identity element. Then for any \((a, b) \in N \times N\):
\((a, b) * (e_1, e_2) = (a, b) \implies (a + e_1, b + e_2) = (a, b)\)
\(\implies a + e_1 = a \implies e_1 = 0\)
However, \(0 \notin N\) (since \(N\) is the set of natural numbers).
Thus, no identity element exists in \(N \times N\).
In simple words: Coordinate-wise addition inherits commutativity and associativity from natural numbers, but fails to have an identity because 0 is not a natural number.

Exam Tip: Always verify that the candidate identity element strictly belongs to the specified set \(N\).