RBSE Solutions Class 12 Maths Chapter 1 Composite Functions Exercise 1.3

Question 1. Determine whether each of the following operation define a binary operation on the given set or not. Also, Justify your answer.
(i) a*b = a, on N
(ii) a*b = a + b - 3, on N
(iii) a*b = a + 3b, on N
(iv) a*b = a/b, on Q
(v) a*b = a - b, on R
Answer:
(i) The operation \( a*b = a \) is a binary operation on natural numbers (N). If you select any two natural numbers, \( a \) and \( b \), the result \( a*b = a \) will always be a natural number. For instance, if \( a=7 \) and \( b=12 \), then \( a*b=7 \), which is a natural number. This confirms the operation keeps results within the set of natural numbers.
(ii) The operation \( a*b = a + b - 3 \) is not a binary operation on natural numbers (N). For example, if we take \( a=1 \) and \( b=1 \), then \( a*b = 1 + 1 - 3 = -1 \). Since \( -1 \) is not a natural number, the operation does not always keep the result within the set N.
(iii) The operation \( a*b = a + 3b \) is a binary operation on natural numbers (N). If you pick any two natural numbers, such as \( a=1 \) and \( b=2 \), then \( a*b = 1 + 3 \times 2 = 1 + 6 = 7 \). Since \( 7 \) is a natural number, this operation consistently results in a natural number when performed on any two natural numbers.
(iv) The operation \( a*b = \frac{a}{b} \) is not always a binary operation on rational numbers (Q). For instance, if we take \( a=22 \) and \( b=7 \), both of which are rational numbers. The given context considers the result \( \frac{22}{7} \) as a value that does not belong to the set of rational numbers. Thus, the operation is not closed within Q for all possible pairs.
(v) The operation \( a*b = a - b \) is a binary operation on real numbers (R). This means if you take any two real numbers, \( a \) and \( b \), their difference \( a - b \) will always be another real number. For example, \( 8 - \sqrt{3} \) is still a real number. The set of real numbers is closed under subtraction, ensuring results stay within the set.
In simple words: An operation is binary if its result always stays within the original set of numbers you are using. You need to check if the rule for combining numbers keeps the answer in the same type of number (like natural, rational, or real).

🎯 Exam Tip: To prove an operation is *not* binary, a single counterexample (a pair of numbers from the set whose result is *not* in the set) is enough. For "is binary", you need to show it holds for all cases, often by logical reasoning.

Question 2. Determine which of the following binary operation is commutative and which is associative :
(i) * on N defined as a*b = \( 2^{ab} \)
(ii) * on N defined as a*b = a + b + \( a^{2}b \)
(iii) * on Z defined as a*b = a - b
(iv) * on Q defined as a*b = ab + 1
(v) * on R defined as a*b = a + b - 7
Answer:
(i) Given \( a*b = 2^{ab} \) on N.
* **Commutativity:** To check if \( a*b = b*a \). We have \( a*b = 2^{ab} \). Since multiplication of natural numbers is commutative (\( ab = ba \)), we can write \( 2^{ab} = 2^{ba} \). This means \( a*b = b*a \). So, the operation is commutative.
* **Associativity:** To check if \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (2^{ab})*c = 2^{((2^{ab})c)} \)
\( a*(b*c) = a*(2^{bc}) = 2^{(a(2^{bc}))} \)
These two expressions are generally not equal for all natural numbers \( a, b, c \). For instance, if \( a=1, b=2, c=3 \), then \( (1*2)*3 = 2^{12} \) and \( 1*(2*3) = 2^{64} \). Since \( 2^{12} \neq 2^{64} \), the operation is not associative.
* **Conclusion:** The operation \( a*b = 2^{ab} \) is commutative but not associative.

(ii) Given \( a*b = a + b + a^2b \) on N.
* **Commutativity:** To check if \( a*b = b*a \).
\( a*b = a + b + a^2b \)
\( b*a = b + a + b^2a \)
Since \( a^2b \) is generally not equal to \( b^2a \) (e.g., \( 1*2 = 1+2+1^2 \cdot 2 = 5 \), while \( 2*1 = 2+1+2^2 \cdot 1 = 7 \)), \( a*b \neq b*a \). So, the operation is not commutative.
* **Associativity:** To check if \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (a + b + a^2b)*c = (a + b + a^2b) + c + (a + b + a^2b)^2c \)
\( a*(b*c) = a*(b + c + b^2c) = a + (b + c + b^2c) + a^2(b + c + b^2c) \)
These two expressions are generally not equal. Therefore, the operation is not associative.
* **Conclusion:** The operation \( a*b = a + b + a^2b \) is neither commutative nor associative.

(iii) Given \( a*b = a - b \) on Z (integers).
* **Commutativity:** To check if \( a*b = b*a \).
\( a*b = a - b \)
\( b*a = b - a \)
Since \( a - b \neq b - a \) (unless \( a=b \)), the operation is not commutative. For example, \( 5*2 = 3 \) but \( 2*5 = -3 \).
* **Associativity:** To check if \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (a - b)*c = (a - b) - c = a - b - c \)
\( a*(b*c) = a*(b - c) = a - (b - c) = a - b + c \)
Since \( a - b - c \neq a - b + c \) (unless \( c=0 \)), the operation is not associative.
* **Conclusion:** The operation \( a*b = a - b \) is neither commutative nor associative.

(iv) Given \( a*b = ab + 1 \) on Q (rational numbers).
* **Commutativity:** To check if \( a*b = b*a \).
\( a*b = ab + 1 \)
\( b*a = ba + 1 \)
Since multiplication of rational numbers is commutative (\( ab = ba \)), we have \( ab + 1 = ba + 1 \). So, the operation is commutative.
* **Associativity:** To check if \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1 \)
\( a*(b*c) = a*(bc + 1) = a(bc + 1) + 1 = abc + a + 1 \)
Since \( abc + c + 1 \neq abc + a + 1 \) (unless \( a=c \)), the operation is not associative.
* **Conclusion:** The operation \( a*b = ab + 1 \) is commutative but not associative.

(v) Given \( a*b = a + b - 7 \) on R (real numbers).
* **Commutativity:** To check if \( a*b = b*a \).
\( a*b = a + b - 7 \)
\( b*a = b + a - 7 \)
Since addition of real numbers is commutative (\( a+b = b+a \)), we have \( a+b-7 = b+a-7 \). So, the operation is commutative.
* **Associativity:** To check if \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (a + b - 7)*c = (a + b - 7) + c - 7 = a + b + c - 14 \)
\( a*(b*c) = a*(b + c - 7) = a + (b + c - 7) - 7 = a + b + c - 14 \)
Since both sides are equal, the operation is associative.
* **Conclusion:** The operation \( a*b = a + b - 7 \) is both commutative and associative.
In simple words: Commutativity means you can swap the order of the two numbers being operated on and get the same result. Associativity means you can change how you group three or more numbers (which pair you calculate first) and still get the same final result. You must test these properties for each operation.

🎯 Exam Tip: For each property (commutativity, associativity), write down the definition, apply it to the given operation, and then show the steps clearly. For disproving a property, a single counterexample with specific numbers is usually sufficient and easier to demonstrate.

Question 3. If * be an operation on Z, defined as a*b = a + b + 1, \( \forall \) a, b \( \in \) Z then prove that * is commutative and associative, find its identity element. Also, find inverse element of any integer in Z.
Answer:
Given the operation \( a*b = a + b + 1 \) on the set of integers (Z).
* **Commutativity:** To prove commutativity, we need to show \( a*b = b*a \).
\( a*b = a + b + 1 \)
\( b*a = b + a + 1 \)
Since addition is commutative (\( a + b = b + a \)), it is clear that \( a + b + 1 = b + a + 1 \). Thus, the operation is commutative.
* **Associativity:** To prove associativity, we need to show \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (a + b + 1)*c = (a + b + 1) + c + 1 = a + b + c + 2 \)
\( a*(b*c) = a*(b + c + 1) = a + (b + c + 1) + 1 = a + b + c + 2 \)
Since both sides are equal, the operation is associative.
* **Identity Element:** An identity element \( e \) is an integer such that \( a*e = a \) for all \( a \in Z \).
\( a*e = a \)
\( \implies a + e + 1 = a \)
Subtract \( a \) from both sides: \( e + 1 = 0 \)
So, \( e = -1 \).
Thus, \( -1 \) is the identity element in Z for this operation.
* **Inverse Element:** The inverse of an element \( a \in Z \), denoted \( x \), is an integer such that \( a*x = e \), where \( e \) is the identity element.
We found \( e = -1 \). So, we set \( a*x = -1 \).
\( a + x + 1 = -1 \)
Rearrange to solve for \( x \): \( x + 1 = -1 - a \)
\( x = -a - 2 \)
Therefore, the inverse of any integer \( a \) is \( -(a+2) \). For example, the inverse of 3 would be \(-(3+2) = -5\).
In simple words: First, we prove that swapping numbers or changing their grouping doesn't change the outcome. Then, we find a special number (identity element) that doesn't change any other number when operated with it. Finally, we find for each number another number (its inverse) that, when operated together, gives the special identity element.

🎯 Exam Tip: Clearly state the definitions for commutativity, associativity, identity, and inverse before showing your work. Use proper algebraic steps to solve for the identity and inverse elements.

Question 4. If * be a binary operation defined on R – {1}. as a*b = a + b - ab, \( \forall \) a, b \( \in \) R – {1} prove that * is commutative and associative. Find identity element and also find inverse of a.
Answer:
Given the operation \( a*b = a + b - ab \) on the set \( R - \{1\} \) (real numbers excluding 1).
* **Commutativity:** To prove commutativity, we need to show \( a*b = b*a \).
\( a*b = a + b - ab \)
\( b*a = b + a - ba \)
Since addition and multiplication of real numbers are commutative, \( a + b - ab = b + a - ba \). Thus, the operation is commutative.
* **Associativity:** To prove associativity, we need to show \( (a*b)*c = a*(b*c) \).
\( (a*b)*c = (a + b - ab)*c \)
\( = (a + b - ab) + c - (a + b - ab)c \)
\( = a + b - ab + c - ac - bc + abc \)
\( a*(b*c) = a*(b + c - bc) \)
\( = a + (b + c - bc) - a(b + c - bc) \)
\( = a + b + c - bc - ab - ac + abc \)
Since both expressions are equal, the operation is associative.
* **Identity Element:** An identity element \( e \) is a real number such that \( a*e = a \) for all \( a \in R - \{1\} \).
\( a*e = a \)
\( \implies a + e - ae = a \)
Subtract \( a \) from both sides: \( e - ae = 0 \)
Factor out \( e \): \( e(1 - a) = 0 \)
Since \( a \in R - \{1\} \), we know that \( a \neq 1 \), which means \( (1 - a) \neq 0 \).
Therefore, we can divide by \( (1 - a) \), giving \( e = 0 \).
Thus, \( 0 \) is the identity element.
* **Inverse Element:** The inverse of an element \( a \in R - \{1\} \), denoted \( x \), is a real number such that \( a*x = e \), where \( e \) is the identity element.
We found \( e = 0 \). So, we set \( a*x = 0 \).
\( a + x - ax = 0 \)
Rearrange to solve for \( x \): \( x - ax = -a \)
Factor out \( x \): \( x(1 - a) = -a \)
Since \( a \neq 1 \), \( (1 - a) \neq 0 \), so we can divide:
\( x = \frac{-a}{1 - a} \)
This can also be written as \( x = \frac{a}{a - 1} \).
Therefore, the inverse of any element \( a \) is \( \frac{a}{a - 1} \). For example, the inverse of 2 would be \(\frac{2}{2-1} = 2\).
In simple words: This question asks us to prove that the order of numbers and how they are grouped does not matter for this operation. Then, we find the special number that leaves other numbers unchanged, and finally, for each number, we find its partner that combines with it to give that special unchanged number.

🎯 Exam Tip: When finding the identity or inverse element, remember to clearly state the conditions (e.g., \( a \neq 1 \) when dividing by \( 1-a \)) that ensure your solution is valid within the specified set.

Question 5. Four functions are defined on set \( R_0 \), Such that, \( f_1(x) = x, f_2(x) = -x, f_3(x) = 1/x, f_4(x) = -1/x \) Construct the composition table for the composition of functions \( f_1, f_2, f_3, f_4 \). Also, find identity element and inverse of every element.
Answer:
Given four functions defined on the set \( R_0 \) (all real numbers except zero):
\( f_1(x) = x \)
\( f_2(x) = -x \)
\( f_3(x) = \frac{1}{x} \)
\( f_4(x) = -\frac{1}{x} \)

We need to construct a composition table for these functions and find the identity and inverse elements. The composition of two functions \( f_i \) and \( f_j \) is denoted by \( (f_i \circ f_j)(x) = f_i(f_j(x)) \).

Let's calculate each composition:
* \( (f_1 \circ f_1)(x) = f_1(f_1(x)) = f_1(x) = x = f_1 \)
* \( (f_1 \circ f_2)(x) = f_1(f_2(x)) = f_1(-x) = -x = f_2 \)
* \( (f_1 \circ f_3)(x) = f_1(f_3(x)) = f_1(\frac{1}{x}) = \frac{1}{x} = f_3 \)
* \( (f_1 \circ f_4)(x) = f_1(f_4(x)) = f_1(-\frac{1}{x}) = -\frac{1}{x} = f_4 \)

* \( (f_2 \circ f_1)(x) = f_2(f_1(x)) = f_2(x) = -x = f_2 \)
* \( (f_2 \circ f_2)(x) = f_2(f_2(x)) = f_2(-x) = -(-x) = x = f_1 \)
* \( (f_2 \circ f_3)(x) = f_2(f_3(x)) = f_2(\frac{1}{x}) = -\frac{1}{x} = f_4 \)
* \( (f_2 \circ f_4)(x) = f_2(f_4(x)) = f_2(-\frac{1}{x}) = -(-\frac{1}{x}) = \frac{1}{x} = f_3 \)

* \( (f_3 \circ f_1)(x) = f_3(f_1(x)) = f_3(x) = \frac{1}{x} = f_3 \)
* \( (f_3 \circ f_2)(x) = f_3(f_2(x)) = f_3(-x) = \frac{1}{-x} = -\frac{1}{x} = f_4 \)
* \( (f_3 \circ f_3)(x) = f_3(f_3(x)) = f_3(\frac{1}{x}) = \frac{1}{(\frac{1}{x})} = x = f_1 \)
* \( (f_3 \circ f_4)(x) = f_3(f_4(x)) = f_3(-\frac{1}{x}) = \frac{1}{(-\frac{1}{x})} = -x = f_2 \)

* \( (f_4 \circ f_1)(x) = f_4(f_1(x)) = f_4(x) = -\frac{1}{x} = f_4 \)
* \( (f_4 \circ f_2)(x) = f_4(f_2(x)) = f_4(-x) = -\frac{1}{(-x)} = \frac{1}{x} = f_3 \)
* \( (f_4 \circ f_3)(x) = f_4(f_3(x)) = f_4(\frac{1}{x}) = -\frac{1}{(\frac{1}{x})} = -x = f_2 \)
* \( (f_4 \circ f_4)(x) = f_4(f_4(x)) = f_4(-\frac{1}{x}) = -\frac{1}{(-\frac{1}{x})} = x = f_1 \)

**Composition Table:**

* **Identity Element:** By observing the table, \( f_1 \) acts as the identity element. When any function is composed with \( f_1 \), the function itself is returned (e.g., \( f_i \circ f_1 = f_i \) and \( f_1 \circ f_i = f_i \)). This is similar to multiplying by 1 in arithmetic.
* **Inverse Element:** An inverse element \( f_j \) for \( f_i \) is one such that their composition results in the identity element, \( f_1 \).
* The inverse of \( f_1 \) is \( f_1 \), as \( f_1 \circ f_1 = f_1 \).
* The inverse of \( f_2 \) is \( f_2 \), as \( f_2 \circ f_2 = f_1 \).
* The inverse of \( f_3 \) is \( f_3 \), as \( f_3 \circ f_3 = f_1 \).
* The inverse of \( f_4 \) is \( f_4 \), as \( f_4 \circ f_4 = f_1 \).
In this specific set of functions, each function is its own inverse.
In simple words: We create a chart showing what happens when you combine any two of these four functions. The identity function is like a special "do-nothing" function, and each function's inverse is the one that "undoes" its effect, bringing you back to the identity function.
🎯 Exam Tip: When constructing a composition table, systematically calculate each entry. \( (f_i \circ f_j)(x) \) means applying \( f_j \) first, then \( f_i \). The identity element is the one that leaves other functions unchanged when composed with them. For inverses, look for pairs that result in the identity element.