CBSE Class 11 Maths Relations and Functions Ex 2.2: Concepts & Solutions

Understanding Relations, Domain, Codomain, and Range

A relation $R$ from a non-empty set $A$ to a non-empty set $B$ is defined as a subset of the Cartesian product $A \times B$. This subset is derived by describing a specific relationship between the first element $x$ and the second element $y$ of the ordered pairs $(x, y)$. The second element $y$ is called the image of the first element $x$.

When working on Exercise 2.2, you must identify three critical sets:

1. Domain: The set of all first elements in the ordered pairs belonging to the relation $R$.
2. Range: The set of all second elements (images) in the ordered pairs of $R$.
3. Codomain: The entire set $B$ from which the second elements are drawn. Note that the $\text{Range} \subseteq \text{Codomain}$.

You can represent any relation in three ways: Set-builder form (using a mathematical rule), Roster form (by explicitly listing ordered pairs), and Arrow Diagrams (drawing visual maps from domain elements to range elements).

Step-by-Step Approach to Solve Exercise 2.2 Problems

1. Identify the Base Sets and Rule — Start by identifying the given sets $A$ and $B$, along with the logical or algebraic rule connecting $x$ and $y$ (e.g., $y = x + 1$ or $y = x^2$).
2. Generate Valid Ordered Pairs (Roster Form) — Substitute elements of domain-set $A$ into the relation's rule. Check if the resulting output $y$ belongs to set $B$. Keep only those pairs $(x, y)$ where both $x \in A$ and $y \in B$ hold true.
3. Extract the Domain and Range — Compile the first components of your generated pairs into a single set to find the Domain. Compile the second components (unique values only) to form the Range.
4. State the Codomain — Directly write down the entire target set $B$. The codomain does not depend on which elements are actually mapped.

Step-by-Step Worked NCERT Examples

• Example 1: Let $A = {1, 2, 3, 4, 5, 6}$. Define a relation $R$ from $A$ to $A$ by $R = {(x, y) : y = x + 1}$. Write this relation in roster form and find its domain, range, and codomain. Solution Steps: - Step 1: The relation rule is $y = x + 1$ where both $x, y \in A$. - Step 2: Plug each element of $A$ into the rule: For $x=1 \implies y=2 \implies (1, 2)$ For $x=2 \implies y=3 \implies (2, 3)$ For $x=3 \implies y=4 \implies (3, 4)$ For $x=4 \implies y=5 \implies (4, 5)$ For $x=5 \implies y=6 \implies (5, 6)$ For $x=6 \implies y=7 \notin A$ (cannot be included). - Step 3: Roster Form: $R = {(1,2), (2,3), (3,4), (4,5), (5,6)}$. - Step 4: Domain (first elements) = ${1, 2, 3, 4, 5}$. - Step 5: Range (second elements) = ${2, 3, 4, 5, 6}$. - Step 6: Codomain (entire second set $A$) = ${1, 2, 3, 4, 5, 6}$.
• Example 2: A relation $R$ is defined on the set of natural numbers $\mathbb{N}$ by $R = {(x, y) : y = x + 5, x \text{ is a natural number less than } 4}$. List the relation and write the domain and range. Solution Steps: - Step 1: Identify inputs $x$. Since $x < 4$ and $x \in \mathbb{N}$, we have $x = {1, 2, 3}$. - Step 2: Calculate corresponding $y$ values using $y = x + 5$: For $x=1 \implies y=6 \implies (1, 6)$ For $x=2 \implies y=7 \implies (2, 7)$ For $x=3 \implies y=8 \implies (3, 8)$ - Step 3: Roster Form: $R = {(1, 6), (2, 7), (3, 8)}$. - Step 4: Domain = ${1, 2, 3}$. - Step 5: Range = ${6, 7, 8}$.

Exam Traps to Avoid in Exercise 2.2

1. Confusing Codomain with Range: Students often write down the codomain as the range. Remember, the Range only contains elements that actually have an incoming arrow (outputs), whereas the Codomain is the entire target set $B$.
2. Out of Set Outputs: When evaluating an algebraic relation, ensure that both $x$ and $y$ lie in the domain and target sets respectively. For example, if $A = {1, 2, 3}$, the pair $(3, 4)$ is invalid for a relation on set $A$ because $4 \notin A$.
3. Total Relations Formula: Keep in mind that if set $A$ has $n$ elements and set $B$ has $m$ elements, the total number of relations from $A$ to $B$ is $2^{nm}$.

Practice Questions with Solutions

• Q: Let $A = {1, 2, 3, 5}$ and $B = {4, 6, 9}$. Define a relation $R$ from $A$ to $B$ by $R = {(x, y) : \text{the difference between } x \text{ and } y \text{ is odd}; x \in A, y \in B}$. Write $R$ in roster form. A: Step 1: Understand that the difference $|x - y|$ is odd if and only if one number is even and the other is odd. Step 2: Test pairs of $x \in A$ and $y \in B$: - For $x=1$ (odd): odd difference with $y=4$ (even) and $y=6$ (even). Pairs: $(1,4), (1,6)$. - For $x=2$ (even): odd difference with $y=9$ (odd). Pairs: $(2,9)$. - For $x=3$ (odd): odd difference with $y=4$ (even) and $y=6$ (even). Pairs: $(3,4), (3,6)$. - For $x=5$ (odd): odd difference with $y=4$ (even) and $y=6$ (even). Pairs: $(5,4), (5,6)$. Step 3: Collect all valid pairs into the roster set. Final answer: $R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}$
• Q: Find the domain and range of the relation $R$ defined by $R = {(x, x+5) : x \in {0, 1, 2, 3, 4, 5}}$. A: Step 1: Identify all values of $x$: $0, 1, 2, 3, 4, 5$. Step 2: Calculate the corresponding second elements $(x+5)$ for each value of $x$: - For $x=0 \implies 0+5=5$ - For $x=1 \implies 1+5=6$ - For $x=2 \implies 2+5=7$ - For $x=3 \implies 3+5=8$ - For $x=4 \implies 4+5=9$ - For $x=5 \implies 5+5=10$ Step 3: Compile the domain from the first inputs and range from the calculated outputs. Final answer: Domain = ${0, 1, 2, 3, 4, 5}$, Range = ${5, 6, 7, 8, 9, 10}$
• Q: Let $A = {1, 2, 3, ..., 14}$. Define a relation $R$ from $A$ to $A$ by $R = {(x, y) : 3x - y = 0, \text{ where } x, y \in A}$. Write down its domain, codomain and range. A: Step 1: Express the relation condition as $y = 3x$. Step 2: Find pairs $(x, y)$ where both elements belong to $A = {1, 2, 3, ..., 14}$: - $x = 1 \implies y = 3 \in A \implies (1,3)$ - $x = 2 \implies y = 6 \in A \implies (2,6)$ - $x = 3 \implies y = 9 \in A \implies (3,9)$ - $x = 4 \implies y = 12 \in A \implies (4,12)$ - $x = 5 \implies y = 15 \notin A$ (exclude this pair). Step 3: State the Domain, Range, and Codomain from these pairs. - Domain = first components = ${1, 2, 3, 4}$ - Range = second components = ${3, 6, 9, 12}$ - Codomain = the entire set $A = {1, 2, 3, ..., 14}$ Final answer: Domain = ${1, 2, 3, 4}$, Range = ${3, 6, 9, 12}$, Codomain = ${1, 2, 3, ..., 14}$
• Q: Write the relation $R = {(x, x^3) : x \text{ is a prime number less than } 10}$ in roster form. A: Step 1: Identify all prime numbers less than 10. These are $x \in {2, 3, 5, 7}$. Step 2: Find $x^3$ for each prime number: - For $x=2 \implies 2^3 = 8 \implies (2, 8)$ - For $x=3 \implies 3^3 = 27 \implies (3, 27)$ - For $x=5 \implies 5^3 = 125 \implies (5, 125)$ - For $x=7 \implies 7^3 = 343 \implies (7, 343)$ Step 3: Combine these into the roster form set. Final answer: $R = {(2, 8), (3, 27), (5, 125), (7, 343)}$

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