Mathematics: Class 11 — Chapter 2: Relations and Functions

Given two non-empty sets $P$ and $Q$, the Cartesian product $P \times Q$ is defined as:

If either $P$ or $Q$ is the null set $\phi$, then their Cartesian product is also the null set:

• Equality of Ordered Pairs: Two ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$ are equal if and only if their corresponding elements are equal: $$(a_1, b_1) = (a_2, b_2) \iff a_1 = a_2 \text{ and } b_1 = b_2$$
• Cardinality of Product Set: If set $A$ has $p$ elements and set $B$ has $q$ elements, the Cartesian product $A \times B$ has $pq$ elements: $$n(A) = p, n(B) = q \implies n(A \times B) = pq$$
• Infinite Products: If $A$ and $B$ are non-empty sets and at least one of them is an infinite set, then $A \times B$ is also an infinite set.
• Ordered Triplet: The Cartesian product of three sets $A \times A \times A$ forms ordered triplets: $$A \times A \times A = \{(a, b, c) : a, b, c \in A\}$$ Here, $(a, b, c)$ is called an ordered triplet.

Mathematical Problem Statement: If $(x + 1, y - 2) = (3, 1)$, find the values of $x$ and $y$.

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Step-by-Step Algorithmic Solution Trace:

1. Equate the corresponding elements of the equal ordered pairs: $$x + 1 = 3$$ $$y - 2 = 1$$
2. Solve the linear equations independently: $$x = 3 - 1 = 2$$ $$y = 1 + 2 = 3$$

Final Evaluation Output: $x = 2, y = 3$

Mathematical Problem Statement: Let $A = \{1, 2, 3\}$, $B = \{3, 4\}$, and $C = \{4, 5, 6\}$. Find $A \times (B \cap C)$.

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Step-by-Step Algorithmic Solution Trace:

1. Find the intersection of sets $B$ and $C$: $$B \cap C = \{3, 4\} \cap \{4, 5, 6\} = \{4\}$$
2. Compute the Cartesian product of $A$ with the resulting set: $$A \times (B \cap C) = \{1, 2, 3\} \times \{4\}$$ $$A \times (B \cap C) = \{(1, 4), (2, 4), (3, 4)\}$$

Final Evaluation Output: $\{(1, 4), (2, 4), (3, 4)\}$

• Relation ($R$): A relation $R$ from a non-empty set $A$ to a non-empty set $B$ is a subset of the Cartesian product $A \times B$. It is derived by describing a relationship between the first element $x$ and the second element $y$ of the ordered pairs in $A \times B$. $$R \subseteq A \times B$$
• Image: If $(x, y) \in R$, then $y$ is called the image of $x$ under the relation $R$.
• Domain: The set of all first elements of the ordered pairs in a relation $R$ from $A$ to $B$: $$\text{Domain}(R) = \{x \in A : (x, y) \in R \text{ for some } y \in B\}$$
• Range: The set of all second elements of the ordered pairs in a relation $R$ from $A$ to $B$: $$\text{Range}(R) = \{y \in B : (x, y) \in R \text{ for some } x \in A\}$$
• Codomain: The entire set $B$ is called the codomain of the relation $R$. Note that: $$\text{Range}(R) \subseteq \text{Codomain}(R)$$
• Total Number of Relations: If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = pq$. The total number of possible subsets of $A \times B$ is $2^{pq}$. Thus, the total number of relations that can be defined from $A$ to $B$ is: $$\text{Total Relations} = 2^{pq}$$

Relations can be represented in three primary ways:

1. Roster Form: Explicitly listing all ordered pairs belonging to the relation, e.g., $R = \{(1, 2), (2, 3), (3, 4)\}$.
2. Set-builder Form: Defining the algebraic rule or condition that the elements must satisfy, e.g., $R = \{(x, y) : y = x + 1, x, y \in A\}$.
3. Arrow Diagram: A visual representation where arrows are drawn from elements of the domain to their corresponding images in the codomain.

Mathematical Problem Statement: Let $A = \{1, 2, 3, 4, 5, 6\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : y = x + 1\}$. Depict this using roster form and find its domain, codomain, and range.

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Step-by-Step Algorithmic Solution Trace:

1. Evaluate $y = x + 1$ for each $x \in A$:
   • For $x = 1 \implies y = 2 \in A \implies (1, 2) \in R$
   • For $x = 2 \implies y = 3 \in A \implies (2, 3) \in R$
   • For $x = 3 \implies y = 4 \in A \implies (3, 4) \in R$
   • For $x = 4 \implies y = 5 \in A \implies (4, 5) \in R$
   • For $x = 5 \implies y = 6 \in A \implies (5, 6) \in R$
   • For $x = 6 \implies y = 7 \notin A \implies$ No pair formed.
2. Write the relation in roster form: $$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$$
3. Extract the domain (first elements): $$\text{Domain} = \{1, 2, 3, 4, 5\}$$
4. Extract the range (second elements): $$\text{Range} = \{2, 3, 4, 5, 6\}$$
5. Identify the codomain (entire target set): $$\text{Codomain} = A = \{1, 2, 3, 4, 5, 6\}$$

Final Evaluation Output: $R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$, $\text{Domain} = \{1, 2, 3, 4, 5\}$, $\text{Range} = \{2, 3, 4, 5, 6\}$

A relation $f$ from a set $A$ to a set $B$ is said to be a function if every element of set $A$ has one and only one image in set $B$.

In other words, a relation $f$ is a function if:

1. The domain of $f$ is the entire set $A$.
2. No two distinct ordered pairs in $f$ have the same first element. That is, if $(a, b) \in f$ and $(a, c) \in f$, then $b = c$.

We write this as $f : A \to B$, where $f(a) = b$. Here, $b$ is the image of $a$ under $f$, and $a$ is the preimage of $b$ under $f$.

• Real Valued Function: A function whose range is either the set of real numbers $\mathbb{R}$ or a subset of $\mathbb{R}$.
• Real Function: A function whose domain and range are both either the set of real numbers $\mathbb{R}$ or subsets of $\mathbb{R}$.

The following table outlines the core real functions, their mathematical definitions, domains, and ranges:

Mathematical Problem Statement: Examine the relation $R = \{(2,1), (3,1), (4,2)\}$ and state whether it is a function or not, giving reasons.

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Step-by-Step Algorithmic Solution Trace:

1. Identify the domain elements: $\text{Domain} = \{2, 3, 4\}$.
2. Check if each domain element is mapped to a unique image:
   • $2$ is mapped to $1$ (unique).
   • $3$ is mapped to $1$ (unique).
   • $4$ is mapped to $2$ (unique).
3. Since no domain element is associated with more than one image, the uniqueness condition is satisfied.

Final Evaluation Output: The relation $R$ is a function.

Mathematical Problem Statement: Find the domain of the real function $f(x) = \frac{x^2 + 3x + 5}{x^2 - 5x + 4}$.

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Step-by-Step Algorithmic Solution Trace:

1. Identify the condition for a rational function to be defined: The denominator must not be zero. $$x^2 - 5x + 4 \neq 0$$
2. Factorize the quadratic expression in the denominator: $$x^2 - 5x + 4 = (x - 4)(x - 1)$$
3. Find the roots where the denominator becomes zero: $$(x - 4)(x - 1) = 0 \implies x = 4 \text{ or } x = 1$$
4. Exclude these points from the set of real numbers $\mathbb{R}$ to obtain the domain: $$\text{Domain}(f) = \mathbb{R} - \{1, 4\}$$

Final Evaluation Output: $\text{Domain}(f) = \mathbb{R} - \{1, 4\}$

Let $f : X \to \mathbb{R}$ and $g : X \to \mathbb{R}$ be two real functions, where $X \subseteq \mathbb{R}$:

• Addition: $$(f + g)(x) = f(x) + g(x), \quad \forall x \in X$$
• Subtraction: $$(f - g)(x) = f(x) - g(x), \quad \forall x \in X$$
• Multiplication by a Scalar: For any scalar $\alpha \in \mathbb{R}$: $$(\alpha f)(x) = \alpha f(x), \quad \forall x \in X$$
• Multiplication (Pointwise Product): $$(fg)(x) = f(x)g(x), \quad \forall x \in X$$
• Quotient: $$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, \quad \forall x \in X \text{ such that } g(x) \neq 0$$

Mathematical Problem Statement: Let $f(x) = x^2$ and $g(x) = 2x + 1$ be two real functions. Find $(f + g)(x)$, $(f - g)(x)$, $(fg)(x)$, and $\left(\frac{f}{g}\right)(x)$.

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Step-by-Step Algorithmic Solution Trace:

1. Compute addition: $$(f + g)(x) = f(x) + g(x) = x^2 + 2x + 1$$
2. Compute subtraction: $$(f - g)(x) = f(x) - g(x) = x^2 - (2x + 1) = x^2 - 2x - 1$$
3. Compute multiplication: $$(fg)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2$$
4. Compute quotient and specify the domain restriction: $$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1}$$ The denominator must not be zero: $$2x + 1 \neq 0 \implies x \neq -\frac{1}{2}$$

Final Evaluation Output: $(f+g)(x) = x^2+2x+1$, $(f-g)(x) = x^2-2x-1$, $(fg)(x) = 2x^3+x^2$, $\left(\frac{f}{g}\right)(x) = \frac{x^2}{2x+1}$ (where $x \neq -\frac{1}{2}$)