Rbse Solutions Class 12 Maths Chapter 1: Exercise 1.1 (Relations and Functions)

If you’ve just opened your NCERT Maths book, Chapter 1 might look a bit intimidating with all those $x, y$ and set brackets. But here’s a secret from someone who has analyzed the RBSE Ajmer Board patterns for years: Exercise 1.1 is actually a scoring goldmine if you understand the logic instead of memorizing the steps.

For the 2025-26 academic session, the Rajasthan Board is leaning heavily into “Application-based” questions. This means you won’t just be asked for definitions; you’ll need to prove them. Let’s break down the solutions for Exercise 1.1 with the exact logic your board examiner wants to see.

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The Big Three: Reflexive, Symmetric, & Transitive

Before we touch the pen to paper, you need to have these three concepts crystal clear in your head. Think of a relation $R$ on a set $A$ like a set of rules:

Transitive (संक्रामक): The “bridge” rule. If $1 \to 2$ and $2 \to 3$, then $1$ must jump directly to $3$.

Reflexive (स्वतुल्य): Every element is related to itself. If $(1, 1)$ isn’t there, it’s not reflexive.

Symmetric (सममित): It’s a two-way street. If $1$ is related to $2$, then $2$ must be related to $1$.

Question 1: Determine Reflexivity, Symmetry, and Transitivity

Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set $A = \{1, 2, 3 \dots 13, 14\}$ defined as $R = \{(x, y): 3x – y = 0\}$

Solution:

The relation $R$ can be written in roster form by choosing $x \in A$ such that $y = 3x$ is also in $A$.

$$R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$$

• Reflexive: A relation $R$ on set $A$ is reflexive if $(x, x) \in R$ for every $x \in A$.
  • Since $(1, 1)$, $(2, 2), \dots (14, 14) \notin R$ (e.g., $3(1) – 1 = 2 \neq 0$), R is not reflexive.
• Symmetric: A relation $R$ is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$.
  • We have $(1, 3) \in R$, but $(3, 1) \notin R$ (since $3(3) – 1 = 8 \neq 0$). R is not symmetric.
• Transitive: A relation $R$ is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
  • We have $(1, 3) \in R$ and $(3, 9) \in R$, but $(1, 9) \notin R$ (since $3(1) – 9 = -6 \neq 0$). R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

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(ii) Relation R in the set $N$ of natural numbers defined as $R = \{(x, y): y = x + 5 \text{ and } x < 4\}$

Solution:

The relation $R$ can be written in roster form:

$$R = \{(1, 6), (2, 7), (3, 8)\}$$

• Reflexive:
  • Since $(1, 1)$, $(2, 2)$, etc., are not in $R$, R is not reflexive.
• Symmetric:
  • We have $(1, 6) \in R$, but $(6, 1) \notin R$ (since $1 \neq 6 + 5$). R is not symmetric.
• Transitive:
  • There are no pairs in $R$ of the form $(x, y)$ and $(y, z)$. Thus, the condition for transitivity is vacuously true.
  • Correction/Detailed Check: The definition of transitive only requires that if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. Since there are no elements $(x, y)$ and $(y, z)$ where $y$ is common, the relation is transitive.

Conclusion: R is transitive but neither reflexive nor symmetric.

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(iii) Relation R in the set $A = \{1, 2, 3, 4, 5, 6\}$ as $R = \{(x, y): y \text{ is divisible by } x\}$

Solution:

• Reflexive:
  • For every $x \in A$, $x$ is always divisible by $x$. Thus, $(x, x) \in R$ for all $x \in A$. R is reflexive.
• Symmetric:
  • We have $(2, 4) \in R$ (since 4 is divisible by 2).
  • But, $(4, 2) \notin R$ (since 2 is not divisible by 4). R is not symmetric.
• Transitive:
  • Let $(x, y) \in R$ and $(y, z) \in R$.
  • This means $y = k_1 x$ and $z = k_2 y$ for some integers $k_1, k_2$.
  • Substituting $y$, we get $z = k_2 (k_1 x) = (k_1 k_2) x$.
  • Since $k_1 k_2$ is an integer, $z$ is divisible by $x$, so $(x, z) \in R$. R is transitive.

Conclusion: R is reflexive and transitive but not symmetric.

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(iv) Relation R in the set $Z$ of all integers defined as $R = \{(x, y): x – y \text{ is an integer}\}$

Solution:

The difference between any two integers is always an integer.

• Reflexive:
  • For any $x \in Z$, $x – x = 0$, which is an integer. Thus, $(x, x) \in R$. R is reflexive.
• Symmetric:
  • If $(x, y) \in R$, then $x – y = k$ (an integer).
  • Then $y – x = -(x – y) = -k$. Since $k$ is an integer, $-k$ is also an integer.
  • Thus, $(y, x) \in R$. R is symmetric.
• Transitive:
  • If $(x, y) \in R$ and $(y, z) \in R$, then $x – y = k_1$ and $y – z = k_2$ (where $k_1, k_2$ are integers).
  • $x – z = (x – y) + (y – z) = k_1 + k_2$.
  • Since the sum of two integers is an integer, $x – z$ is an integer. Thus, $(x, z) \in R$. R is transitive.

Conclusion: R is reflexive, symmetric, and transitive (an equivalence relation).

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(v) Relation R in the set A of human beings in a town at a particular time given by:

(a) $R = \{(x, y): x \text{ and } y \text{ work at the same place}\}$

• Reflexive: $x$ and $x$ always work at the same place. R is reflexive.
• Symmetric: If $x$ and $y$ work at the same place, then $y$ and $x$ work at the same place. R is symmetric.
• Transitive: If $x$ and $y$ work at the same place, and $y$ and $z$ work at the same place, then $x$ and $z$ must work at the same place. R is transitive.

Conclusion: R is an equivalence relation.

(b) $R = \{(x, y): x \text{ and } y \text{ live in the same locality}\}$

• Reflexive: $x$ and $x$ always live in the same locality. R is reflexive.
• Symmetric: If $x$ and $y$ live in the same locality, then $y$ and $x$ live in the same locality. R is symmetric.
• Transitive: If $x$ and $y$ live in the same locality, and $y$ and $z$ live in the same locality, then $x$ and $z$ live in the same locality. R is transitive.

Conclusion: R is an equivalence relation.

(c) $R = \{(x, y): x \text{ is exactly 7 cm taller than } y\}$

• Reflexive: $x$ cannot be exactly 7 cm taller than $x$. R is not reflexive.
• Symmetric: If $x$ is 7 cm taller than $y$, then $y$ is 7 cm shorter than $x$, so $y$ is not 7 cm taller than $x$. R is not symmetric.
• Transitive: If $x$ is 7 cm taller than $y$, and $y$ is 7 cm taller than $z$, then $x$ is $7+7=14$ cm taller than $z$. Thus $(x, z) \notin R$. R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

(d) $R = \{(x, y): x \text{ is wife of } y\}$

• Reflexive: $x$ cannot be the wife of $x$. R is not reflexive.
• Symmetric: If $x$ is the wife of $y$, then $y$ is the husband of $x$, not the wife of $x$. R is not symmetric.
• Transitive: If $x$ is the wife of $y$, and $(y, z) \in R$, then $y$ would have to be the wife of $z$, which is impossible (as $y$ is a male husband). The condition for transitivity (having $(x, y)$ and $(y, z)$) is never met, so it is vacuously transitive.
  • Self-Correction/Standard Answer: While technically transitive, the common context focuses on the practical failure. It is often cited as not transitive because the scenario $(x, y)$ and $(y, z)$ cannot occur.

Conclusion: R is not reflexive, not symmetric, and usually considered not transitive in the context of human relationships, though technically transitive based on formal logic.

(e) $R = \{(x, y): x \text{ is father of } y\}$

• Reflexive: $x$ cannot be the father of $x$. R is not reflexive.
• Symmetric: If $x$ is the father of $y$, then $y$ is the son/daughter of $x$, not the father of $x$. R is not symmetric.
• Transitive: If $x$ is the father of $y$, and $y$ is the father of $z$, then $x$ is the grandfather of $z$, not the father of $z$. R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Question 2. Show that the relation $R$ in the set $\mathbb{R}$ of real numbers, defined as $R = \{(a, b): a \leq b^2\}$ is neither reflexive nor symmetric nor transitive.

Solution: $R = \{(a, b): a \leq b^2\}$ defined on $\mathbb{R}$.

1. Reflexive: We check if $a \leq a^2$ for all $a \in \mathbb{R}$.
   • Take $a = \frac{1}{2}$. $\frac{1}{2} \leq \left(\frac{1}{2}\right)^2 \implies 0.5 \leq 0.25$, which is False.
   • $\therefore R$ is not reflexive.
2. Symmetric: We check if $(a, b) \in R \implies (b, a) \in R$.
   • Take $(1, 4) \in R$ since $1 \leq 4^2 = 16$ (True).
   • Check $(4, 1)$. Is $4 \leq 1^2 = 1$? False.
   • $\therefore R$ is not symmetric.
3. Transitive: We check if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
   • Take $(3, 2) \in R$ since $3 \leq 2^2 = 4$ (True).
   • Take $(2, 1.5) \in R$ since $2 \leq (1.5)^2 = 2.25$ (True).
   • Check $(3, 1.5)$. Is $3 \leq (1.5)^2 = 2.25$? False.
   • $\therefore R$ is not transitive.

Conclusion: $R$ is neither reflexive, nor symmetric, nor transitive.

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Question 3. Check whether the relation $R$ defined in the set $A = \{1, 2, 3, 4, 5, 6\}$ as $R = \{(a, b): b = a + 1\}$ is reflexive, symmetric or transitive.

Solution: Set $A = \{1, 2, 3, 4, 5, 6\}$. $R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$.

1. Reflexive: $(a, a) \notin R$. E.g., $(1, 1) \notin R$ (since $1 \neq 1 + 1$).
   • $\therefore R$ is not reflexive.
2. Symmetric: $(1, 2) \in R$, but $(2, 1) \notin R$ (since $1 \neq 2 + 1$).
   • $\therefore R$ is not symmetric.
3. Transitive: $(1, 2) \in R$ and $(2, 3) \in R$, but $(1, 3) \notin R$ (since $3 \neq 1 + 1$).
   • $\therefore R$ is not transitive.

Conclusion: $R$ is neither reflexive, nor symmetric, nor transitive.

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Question 4. Show that the relation $R$ in $\mathbb{R}$ defined as $R = \{(a, b): a \leq b\}$, is reflexive and transitive but not symmetric.

Solution: $R = \{(a, b): a \leq b\}$ defined on $\mathbb{R}$.

1. Reflexive: $a \leq a$ is always true (since $a = a$).
   • $\therefore R$ is reflexive.
2. Symmetric: Take $(2, 4) \in R$ since $2 \leq 4$ (True).
   • Check $(4, 2)$. Is $4 \leq 2$? False.
   • $\therefore R$ is not symmetric.
3. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $a \leq b$ and $b \leq c$.
   • By transitivity of inequality, $a \leq c$. $\therefore (a, c) \in R$.
   • $\therefore R$ is transitive.

Conclusion: $R$ is reflexive and transitive but not symmetric.

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Question 5. Check whether the relation $R$ in $\mathbb{R}$ defined as $R = \{(a, b): a \leq b^3\}$ is reflexive, symmetric or transitive.

Solution: This is the same as Question 2 but with $b^3$.

1. Reflexive: Take $a = \frac{1}{2}$. Is $\frac{1}{2} \leq \left(\frac{1}{2}\right)^3$? $0.5 \leq 0.125$ is False.
   • $\therefore R$ is not reflexive.
2. Symmetric: Take $(1, 2) \in R$ since $1 \leq 2^3 = 8$ (True).
   • Check $(2, 1)$. Is $2 \leq 1^3 = 1$? False.
   • $\therefore R$ is not symmetric.
3. Transitive: Take $(10, 3) \in R$ since $10 \leq 3^3 = 27$.
   • Take $(3, 1.5) \in R$ since $3 \leq (1.5)^3 = 3.375$.
   • Check $(10, 1.5)$. Is $10 \leq (1.5)^3 = 3.375$? False.
   • $\therefore R$ is not transitive.

Conclusion: $R$ is neither reflexive, nor symmetric, nor transitive.

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Question 6. Show that the relation $R$ in the set $A = \{1, 2, 3\}$ given by $R = \{(1, 2), (2, 1)\}$ is symmetric but neither reflexive nor transitive.

Solution: Set $A = \{1, 2, 3\}$, $R = \{(1, 2), (2, 1)\}$.

1. Reflexive: $(1, 1), (2, 2), (3, 3) \notin R$.
   • $\therefore R$ is not reflexive.
2. Symmetric: $(1, 2) \in R$ and $(2, 1) \in R$.
   • $\therefore R$ is symmetric.
3. Transitive: $(1, 2) \in R$ and $(2, 1) \in R$. For transitivity, we need $(1, 1) \in R$.
   • Since $(1, 1) \notin R$.
   • $\therefore R$ is not transitive.

Conclusion: $R$ is symmetric but neither reflexive nor transitive.

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Question 7. Show that the relation $R$ in the set $A$ of all the books in a library, given by $R = \{(x, y): x \text{ and } y \text{ have same number of pages}\}$ is an equivalence relation.

Solution:

1. Reflexive: $(x, x) \in R$ since book $x$ has the same number of pages as itself. (Holds)
2. Symmetric: If $(x, y) \in R$, then $x$ and $y$ have the same number of pages. Thus, $y$ and $x$ have the same number of pages, so $(y, x) \in R$. (Holds)
3. Transitive: If $(x, y) \in R$ and $(y, z) \in R$, then $x$ and $y$ have the same pages, and $y$ and $z$ have the same pages. Thus, $x$ and $z$ have the same number of pages, so $(x, z) \in R$. (Holds)

Conclusion: $R$ is reflexive, symmetric, and transitive, hence it is an equivalence relation.

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Question 8. Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5\}$ given by $R = \{(a, b): |a – b| \text{ is even}\}$ is an equivalence relation. Show that all the elements of $\{1, 3, 5\}$ are related to each other and no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

Solution: $A = \{1, 2, 3, 4, 5\}$, $R = \{(a, b): |a – b| \text{ is even}\}$.

1. Equivalence Relation Proof:
   • Reflexive: $|a – a| = 0$, which is even. (Holds)
   • Symmetric: $|a – b|$ is even $\implies |b – a| = |-(a – b)| = |a – b|$ is even. (Holds)
   • Transitive: If $|a – b|$ is even and $|b – c|$ is even, then $(a – b)$ and $(b – c)$ are both even integers. Their sum $(a – c) = (a – b) + (b – c)$ is also even. $\therefore |a – c|$ is even. (Holds)
   • $R$ is an equivalence relation.
2. Related Sets Proof:
   • Set $\{1, 3, 5\}$: All elements are odd. The difference between any two odd numbers (Odd – Odd) is always Even. Thus, all elements are related.
   • Sets $\{1, 3, 5\}$ and $\{2, 4\}$: $\{1, 3, 5\}$ are odd, $\{2, 4\}$ are even. The difference between an odd and an even number (Odd – Even or Even – Odd) is always Odd. Thus, $|a – b|$ is not even, and no element from $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

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Question 9. Show that the relation $R$ in the set $A = \{x \in \mathbb{Z}: 0 \leq x \leq 12\}$, given by: (i) $R = \{(a, b): |a – b| \text{ is a multiple of } 4\}$ and (ii) $R = \{(a, b): a = b\}$ are equivalence relations. Find the set of all elements related to 1 in each case.

Solution (i): $R = \{(a, b): |a – b| \text{ is a multiple of } 4\}$

1. Equivalence Proof:
   • Reflexive: $|a – a| = 0$, which is $4 \times 0$. (Holds)
   • Symmetric: $|a – b|$ is a multiple of 4 $\implies |b – a|$ is a multiple of 4. (Holds)
   • Transitive: If $a – b = 4k_1$ and $b – c = 4k_2$, then $a – c = 4(k_1 + k_2)$. (Holds)
   • $R$ is an equivalence relation.
2. Elements related to 1: We need $b \in A$ such that $|1 – b|$ is a multiple of 4, i.e., $|1 – b| = 0, 4, 8, 12$.
   • $|1 – b| = 0 \implies b = 1$
   • $|1 – b| = 4 \implies b = 5$ (since $-3 \notin A$)
   • $|1 – b| = 8 \implies b = 9$ (since $-7 \notin A$)
   • $|1 – b| = 12 \implies b = 13$ (not in $A$)
   • Set of elements related to 1 is $\{1, 5, 9\}$.

Solution (ii): $R = \{(a, b): a = b\}$

1. Equivalence Proof:
   • Reflexive: $a = a$. (Holds)
   • Symmetric: $a = b \implies b = a$. (Holds)
   • Transitive: $a = b$ and $b = c \implies a = c$. (Holds)
   • $R$ is an equivalence relation.
2. Elements related to 1: We need $b \in A$ such that $a = b$, i.e., $1 = b$.
   • Set of elements related to 1 is $\{1\}$.

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Question 10. Give an example of a relation which is…

(i) Symmetric but neither reflexive nor transitive.

• Example: On $A = \{5, 6, 7\}$, let $R = \{(5, 6), (6, 5)\}$.
  • Not reflexive (e.g., $(5, 5) \notin R$). Symmetric (holds). Not transitive (e.g., $(5, 6) \in R$ and $(6, 5) \in R$, but $(5, 5) \notin R$).

(ii) Transitive but neither reflexive nor symmetric.

• Example: On $\mathbb{R}$, let $R = \{(a, b): a < b\}$.
  • Not reflexive ($a < a$ is false). Not symmetric (e.g., $1 < 2$ but $2 < 1$ is false). Transitive (If $a < b$ and $b < c$, then $a < c$).

(iii) Reflexive and symmetric but not transitive.

• Example: On $A = \{4, 6, 8\}$, let $R = \{(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)\}$.
  • Reflexive (holds). Symmetric (holds). Not transitive (e.g., $(4, 6) \in R$ and $(6, 8) \in R$, but $(4, 8) \notin R$).

(iv) Reflexive and transitive but not symmetric.

• Example: On $\mathbb{R}$, let $R = \{(a, b): a \geq b\}$ (This is the inverse of Q4).
  • Reflexive ($a \geq a$). Not symmetric (e.g., $2 \geq 1$ but $1 \geq 2$ is false). Transitive (If $a \geq b$ and $b \geq c$, then $a \geq c$).

(v) Symmetric and transitive but not reflexive.

• Example: On $A = \{1, 2\}$, let $R = \{(1, 1)\}$.
  • Not reflexive (e.g., $(2, 2) \notin R$). Symmetric (vacuously true, since only $(1, 1)$ requires checking). Transitive (vacuously true, since the only case is $(1, 1)$ and $(1, 1)$, which implies $(1, 1)$).
  • Alternative: On $A = \{-5, -6\}$, let $R = \{(-5, -6), (-6, -5), (-5, -5)\}$. Not reflexive (since $(-6, -6) \notin R$). Symmetric (holds). Transitive (holds).

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Question 11. Show that the relation $R$ in the set $A$ of points in a plane given by $R = \{(P, Q): \text{distance of } P \text{ from origin is same as the distance of } Q \text{ from the origin}\}$, is an equivalence relation. Further, show that the set of all point related to a point $P \neq (0, 0)$ is the circle passing through $P$ with origin as centre.

Solution: Let $OP$ denote the distance of point $P$ from the origin $O$. $R = \{(P, Q): OP = OQ\}$.

1. Equivalence Proof:
   • Reflexive: $OP = OP$. (Holds)
   • Symmetric: $OP = OQ \implies OQ = OP$. (Holds)
   • Transitive: $OP = OQ$ and $OQ = OS \implies OP = OS$. (Holds)
   • $R$ is an equivalence relation.
2. Related Set: The set of all points $Q$ related to $P$ are points $Q$ such that $OQ = OP$. This means $Q$ is at a fixed distance $OP$ from the origin. This set of points forms a circle with the origin as centre and radius $OP$.

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Question 12. Show that the relation $R$ defined in the set $A$ of all triangles as $R = \{(T_1, T_2): T_1 \text{ is similar to } T_2\}$, is equivalence relation. Consider three right angle triangles $T_1$ with sides $3, 4, 5$, $T_2$ with sides $5, 12, 13$ and $T_3$ with sides $6, 8, 10$. Which triangles among $T_1, T_2$ and $T_3$ are related?

Solution: $R = \{(T_1, T_2): T_1 \sim T_2\}$ (similar).

1. Equivalence Proof:
   • Reflexive: $T_1 \sim T_1$. (Holds)
   • Symmetric: $T_1 \sim T_2 \implies T_2 \sim T_1$. (Holds)
   • Transitive: $T_1 \sim T_2$ and $T_2 \sim T_3 \implies T_1 \sim T_3$. (Holds)
   • $R$ is an equivalence relation.
2. Related Triangles: Triangles are related if they are similar, i.e., their corresponding sides are proportional.
   • Check $T_1 (3, 4, 5)$ and $T_3 (6, 8, 10)$:$$\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$$
   • Since the sides are proportional, $T_1$ is similar to $T_3$.
   • $T_1$ and $T_2$: $\frac{3}{5} \neq \frac{4}{12}$. Not related.
   • $T_2$ and $T_3$: $\frac{5}{6} \neq \frac{12}{8}$. Not related.

Conclusion: $\mathbf{T_1}$ and $\mathbf{T_3}$ are related.

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Question 13. Show that the relation $R$ defined in the set $A$ of all polygons as $R = \{(P_1, P_2): P_1 \text{ and } P_2 \text{ have same number of sides}\}$, is an equivalence relation. What is the set of all elements in $A$ related to the right angle triangle $T$ with sides $3, 4$ and $5$?

Solution:

1. Equivalence Proof:
   • Reflexive: $P_1$ has the same number of sides as itself. (Holds)
   • Symmetric: $P_1$ and $P_2$ have the same number of sides $\implies P_2$ and $P_1$ have the same number of sides. (Holds)
   • Transitive: If $P_1$ and $P_2$ have $n$ sides, and $P_2$ and $P_3$ have $n$ sides, then $P_1$ and $P_3$ have $n$ sides. (Holds)
   • $R$ is an equivalence relation.
2. Related Set: The triangle $T$ has 3 sides. The set of all elements in $A$ related to $T$ are all polygons that also have 3 sides.
   • The set of all elements related to $T$ is the set of all triangles.

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Question 14. Let $L$ be the set of all lines in XY plane and $R$ be the relation in $L$ defined as $R = \{(L_1, L_2): L_1 \text{ is parallel to } L_2\}$. Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y = 2x + 4$.

Solution:

1. Equivalence Proof:
   • Reflexive: $L_1$ is parallel to $L_1$. (Holds)
   • Symmetric: $L_1 \parallel L_2 \implies L_2 \parallel L_1$. (Holds)
   • Transitive: $L_1 \parallel L_2$ and $L_2 \parallel L_3 \implies L_1 \parallel L_3$. (Holds)
   • $R$ is an equivalence relation.
2. Related Lines: The line $y = 2x + 4$ has a slope $m = 2$. Lines related to it must be parallel, meaning they must also have a slope of 2.
   • The set of all lines related to $y = 2x + 4$ is given by: $\mathbf{y = 2x + c, \text{ where } c \in \mathbb{R}}$.

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Question 15. Let $R$ be the relation in the set $\{1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Solution:

1. Reflexive: $(1, 1), (2, 2), (3, 3), (4, 4) \in R$. (Holds)
2. Symmetric: $(1, 2) \in R$, but $(2, 1) \notin R$. (Fails)
3. Transitive: Check $(3, 2) \in R$ and there are no other pairs starting with 2 to check. Check $(1, 3) \in R$ and $(3, 2) \in R \implies (1, 2) \in R$. (Holds)
4. Conclusion: $R$ is reflexive and transitive but not symmetric.

Correct Answer: (B)

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Question 16. Let $R$ be the relation in the set $N$ given by $R = \{(a, b): a = b – 2, b > 6\}$. Choose the correct answer.

(A) $(2, 4) \in R$

(B) $(3, 8) \in R$

(C) $(6, 8) \in R$

(D) $(8, 7) \in R$

Solution: Check each option against both conditions: $b > 6$ and $a = b – 2$.

• (A) $(2, 4)$: $4 \not> 6$. (Incorrect)
• (B) $(3, 8)$: $8 > 6$. Check: $a = 8 – 2 = 6$. Since $3 \neq 6$. (Incorrect)
• (C) $(6, 8)$: $8 > 6$. Check: $a = 8 – 2 = 6$. Since $6 = 6$. (Correct)
• (D) $(8, 7)$: $7 > 6$. Check: $a = 7 – 2 = 5$. Since $8 \neq 5$. (Incorrect)

Correct Answer: (C)

💡 Topper’s Tips for 2026 RBSE Exams

1. Don’t skip the ‘Given’ part: Always write “Given: Set A = {…}” at the start. It fills the page and shows the examiner you are organized.
2. The Equivalence Relation Jackpot: If a question asks you to “Show that it’s an Equivalence Relation,” you already know the answer is YES for all three. Your job is just to provide the formal proof!
3. Use Math symbols: Instead of writing “for all,” use the symbol $\forall$. It makes your answer sheet look professional.

Frequently Asked Questions (Maths Ch 1)