1. **Problem:** Determine the properties of the relation $R$ on set $G$ where $R = \{(x, y) : x \text{ and } y \text{ have the same mathematics teacher}\}$. **Step 1:** Understand the relation. If $x$ and $y$ have the same teacher, then: - Reflexive: Every student has the same teacher as themselves, so $(x,x) \in R$ for all $x$. - Symmetric: If $x$ has the same teacher as $y$, then $y$ has the same teacher as $x$. - Transitive: If $x$ and $y$ have the same teacher, and $y$ and $z$ have the same teacher, then $x$ and $z$ have the same teacher. **Step 2:** Since all three properties hold, $R$ is an equivalence relation. **Answer:** (D) $R$ is an equivalence relation. 2. **Problem:** For real $x$, $f(x) = x^3$. Determine if $f$ is one-one and/or onto on $\mathbb{R}$. **Step 1:** Check one-one (injective): For $f(x_1) = f(x_2)$, $x_1^3 = x_2^3$ implies $x_1 = x_2$. So $f$ is one-one. **Step 2:** Check onto (surjective): For any $y \in \mathbb{R}$, $x = \sqrt[3]{y}$ satisfies $f(x) = y$. So $f$ is onto. **Answer:** (C) $f$ is one-one and onto on $\mathbb{R}$. 3. **Problem:** Relation $R$ on set $L$ of all straight lines in a plane defined by $lRm$ if and only if $l$ is perpendicular to $m$ and $l \neq m$. Determine properties. **Step 1:** Reflexive? No, because no line is perpendicular to itself. **Step 2:** Symmetric? Yes, if $l$ is perpendicular to $m$, then $m$ is perpendicular to $l$. **Step 3:** Transitive? No, if $l$ is perpendicular to $m$ and $m$ is perpendicular to $n$, $l$ is not necessarily perpendicular to $n$. **Answer:** (B) symmetric. 4. **Problem:** Relation $R$ on $A = \{x \in \mathbb{Z} : 0 \leq x \leq 10\}$ defined by $R = \{(x,y) : x = y\}$ is an equivalence relation. Find number of equivalence classes. **Step 1:** Since $R$ relates only identical elements, each element forms its own equivalence class. **Step 2:** Number of elements in $A$ is $11$ (from $0$ to $10$ inclusive). **Answer:** (D) 11. **Assertion-Reason Questions:** 1. **Assertion:** $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 3x - 5$ is bijective. **Reason:** A function is bijective if it is both surjective and injective. **Explanation:** - Injective: If $f(x_1) = f(x_2)$, then $3x_1 - 5 = 3x_2 - 5 \Rightarrow x_1 = x_2$. - Surjective: For any $y \in \mathbb{Z}$, $x = \frac{y + 5}{3}$ is integer only if $y + 5$ divisible by 3, so not all integers are images. Hence, $f$ is injective but not surjective onto $\mathbb{Z}$. **Answer:** Assertion is false, Reason is true. 2. **Assertion:** Relation $P$ on $X = \{0,2,4,6,8\}$ with $P = \{(0,2), (4,2), (4,6), (8,6), (2,4), (0,4)\}$ is transitive. **Reason:** $P$ has subset $\{(a,b), (b,c), (a,c)\}$. **Check:** For example, $(0,2)$ and $(2,4)$ are in $P$, but $(0,4)$ is also in $P$, so transitive for these elements. However, $(4,2)$ and $(2,4)$ are in $P$, but $(4,4)$ is not in $P$, so not transitive. **Answer:** Assertion is false, Reason is true. 3. **Assertion:** Relation $R = \{(x,y) : x + y \text{ is prime}, x,y \in \mathbb{N}\}$ is not reflexive. **Reason:** $2n$ is composite for all natural numbers $n$. **Check:** For reflexivity, $x + x = 2x$ must be prime for all $x$, but $2x$ is even and greater than 2 for $x > 1$, so not prime. **Answer:** Both assertion and reason are true, and reason correctly explains assertion. 4. **Assertion:** Relation $R$ on $A = \{1,2,3,4,5,6\}$ defined by $R = \{(x,y) : y \text{ divisible by } x\}$ is not an equivalence relation. **Reason:** $R$ is an equivalence relation if it is reflexive, symmetric, and transitive. **Check:** - Reflexive: $x$ divides $x$, so reflexive. - Symmetric: If $y$ divisible by $x$, $x$ divisible by $y$? Not always, so not symmetric. Hence, $R$ is not equivalence relation. **Answer:** Both assertion and reason are true, reason correctly explains assertion. **Very Short Answer Questions:** 1. **Problem:** $A = \{a,b,c\}$, $R = \{(a,a),(b,c),(a,b)\}$. Find minimum ordered pairs to add to make $R$ reflexive and transitive. **Step 1:** Reflexive pairs missing: $(b,b)$ and $(c,c)$. **Step 2:** For transitivity: - $(a,b)$ and $(b,c)$ in $R$, so add $(a,c)$. **Answer:** Add $(b,b), (c,c), (a,c)$. 2. **Problem:** $A=\{1,2,3\}$, $B=\{4,5,6,7\}$, $f=\{(1,4),(2,5),(3,6)\}$. Is $f$ one-one? **Step 1:** Check if different elements in $A$ map to different elements in $B$. All images $4,5,6$ are distinct. **Answer:** Yes, $f$ is one-one (injective).