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CBSE Class 12 Maths Chapter 1 Relations and Functions Model Questions - 2 Marks - Part 1

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SECTION A — Multiple Choice Questions [1 Mark Each]

• 1 Mark Q1. Let R be a relation on the set L of all lines in a plane defined by R = {(L₁, L₂) : L₁ is perpendicular to L₂}. Then R is
  A) Reflexive and Symmetric

  B) Symmetric but neither Reflexive nor Transitive

  C) Equivalence Relation

  D) Transitive but not Symmetric
• 1 Mark Q2. The number of all equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) is
  A) 1

  B) 2

  C) 3

  D) 4
• 1 Mark Q3. Let f: R → R be defined as f(x) = x⁴. Choose the correct answer
  A) f is one-one onto

  B) f is many-one onto

  C) f is one-one but not onto

  D) f is neither one-one nor onto
• 1 Mark Q4. If A = {1, 2, 3}, then the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
  A) 1

  B) 2

  C) 3

  D) 4
• 1 Mark Q5. Let f: A → B and g: B → C be two functions such that gof is injective. Then
  A) f must be injective

  B) g must be injective

  C) both f and g must be injective

  D) g must be surjective
• 1 Mark Q6. The function f: N → N defined by f(n) = n+1 if n is odd and f(n) = n-1 if n is even, is
  A) One-one but not onto

  B) Onto but not one-one

  C) Bijective

  D) Neither one-one nor onto
• 1 Mark Q7. If f(x) = (3 - x³)^1/3, then f(f(x)) is
  A) x^1/3

  B) "x³"

  C) "x"

  D) 3 - x³
• 1 Mark Q8. Let A = {1, 2, 3}. The number of binary operations on A is
  A) 3³

  B) 3⁶

  C) 3⁹

  D) 2³
• 1 Mark Q9. A function f: A → B is onto if
  A) Range of f ⊂ B

  B) Range of f = B

  C) Range of f ⊃ B

  D) None of these
• 1 Mark Q10. If f(x) = 8x³ and g(x) = x^1/3, then fog(x) is
  A) 8x

  B) 2x

  C) 8x³

  D) x

SECTION B — Very Short Answer Type Questions [2 Marks Each]

• 2 Marks Q1. If $A = \{1, 2, 3\}$, check if the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ is transitive.
  Answer & Solution:
  Not transitive.
  Here, $(1, 2) \in R$ and $(2, 3) \in R$, but $(1, 3) \notin R$. Therefore, $R$ is not transitive.
• 2 Marks Q2. Let $R$ be a relation on the set of natural numbers $\mathbb{N}$ defined by $xRy$ if $x + 2y = 8$. Write $R$ as a set of ordered pairs.
  Answer & Solution:
  Given $x + 2y = 8 \implies x = 8 - 2y$.
  Since $x, y \in \mathbb{N}$ (natural numbers $\ge 1$):
  * If $y = 1, x = 6 \implies (6, 1)$
  * If $y = 2, x = 4 \implies (4, 2)$
  * If $y = 3, x = 2 \implies (2, 3)$
  * If $y \ge 4, x \le 0$ (not in $\mathbb{N}$).
  Thus, $R = \{(6, 1), (4, 2), (2, 3)\}$.
• 2 Marks Q3. State the reason why the relation $R = \{(a, b) : a \le b^2\}$ on the set of real numbers $\mathbb{R}$ is not reflexive.
  Answer & Solution:
  For $R$ to be reflexive, $(a, a) \in R$ for all $a \in \mathbb{R}$, which means $a \le a^2$ must hold for every real number.
  Consider $a = \frac{1}{2}$. Then $\frac{1}{2} \le \left(\frac{1}{2}\right)^2 \implies \frac{1}{2} \le \frac{1}{4}$, which is false.
  Thus, $(1/2, 1/2) \notin R$, so $R$ is not reflexive.
• 2 Marks Q4. 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.
  Answer & Solution:
  * **Reflexive:** $(1, 1) \notin R$, so it is not reflexive.
  * **Symmetric:** Since $(1, 2) \in R \implies (2, 1) \in R$, it is symmetric.
  * **Transitive:** $(1, 2) \in R$ and $(2, 1) \in R$, but $(1, 1) \notin R$. Thus, it is not transitive.
• 2 Marks Q5. A relation $R$ on the set of real numbers $\mathbb{R}$ is defined as $R = \{(a, b) : a - b + \sqrt{3} \text{ is an irrational number}\}$. Check if $R$ is reflexive.
  Answer & Solution:
  Reflexive.
  For any $a \in \mathbb{R}$, $a - a + \sqrt{3} = \sqrt{3}$.
  Since $\sqrt{3}$ is an irrational number, $(a, a) \in R$ for all $a \in \mathbb{R}$. Hence, $R$ is reflexive.
• 2 Marks Q6. If $R = \{(x, y) : x^2 - y^2 < 1\}$ is a relation on the set $\{1, 2, 3, 4\}$, write its domain.
  Answer & Solution:
  Test pairs $(x, y)$ from $\{1, 2, 3, 4\} \times \{1, 2, 3, 4\}$ where $x^2 - y^2 < 1$:
  * For $x = 1$: $1^2 - y^2 < 1 \implies 1 - y^2 < 1 \implies y^2 > 0$, true for $y \in \{1, 2, 3, 4\}$. Thus pairs like $(1, 1), (1, 2)$ exist (so $1$ is in domain).
  * For $x = 2$: $4 - y^2 < 1 \implies y^2 > 3$, true for $y = 2, 3, 4$ (pairs like $(2, 2)$).
  * For $x = 3$: $9 - y^2 < 1 \implies y^2 > 8$, true for $y = 3, 4$ (pairs like $(3, 3)$).
  * For $x = 4$: $16 - y^2 < 1 \implies y^2 > 15$, true for $y = 4$ (pair $(4, 4)$).
  All elements $1, 2, 3, 4$ appear as first elements. Domain = $\{1, 2, 3, 4\}$.
• 2 Marks Q7. Let $A = \{a, b, c\}$. Find the total number of distinct equivalence relations that can be defined on $A$ containing the element $(a, b)$.
  Answer & Solution:
  1 equivalence relation.
  Since $(a, b)$ belongs to the equivalence relation, by symmetry $(b, a)$ must also belong. By reflexivity, $(a, a), (b, b), (c, c)$ must belong. By transitivity, since $(a, b)$ and $(b, a)$ are present, $(a, a), (b, b)$ are present, but what about $c$? $c$ must be related to itself, giving the partition $\{\{a, b\}, \{c\}\}$. Thus, only 1 such equivalence relation exists.
• 2 Marks Q8. Check whether the relation $R$ defined on the set $A = \{1, 2, 3, 4, 5, 6\}$ as $R = \{(x, y) : y = x + 1\}$ is symmetric.
  Answer & Solution:
  Not symmetric.
  For example, $(1, 2) \in R$ because $2 = 1 + 1$. However, $(2, 1) \notin R$ because $1 \neq 2 + 1$. Thus, $R$ is not symmetric.
• 2 Marks Q9. Determine whether the relation $R$ on the set of integers $\mathbb{Z}$ defined as $R = \{(x, y) : x - y \text{ is divisible by 5}\}$ is symmetric.
  Answer & Solution:
  Symmetric.
  Let $(x, y) \in R \implies x - y = 5k$ for some integer $k$.
  Then $y - x = -(x - y) = -5k = 5(-k)$, which means $y - x$ is also divisible by 5.
  Thus, $(y, x) \in R$, so $R$ is symmetric.
• 2 Marks Q10. Show that the relation $R$ on the set of all straight lines in a plane defined by $L_1 R L_2 \iff L_1 \perp L_2$ is symmetric but not transitive.
  Answer & Solution:
  * **Symmetric:** If $L_1 \perp L_2$, then clearly $L_2 \perp L_1$.
  * **Not Transitive:** If $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1$ and $L_3$ are parallel ($L_1 \parallel L_3$), not perpendicular. Thus, it is not transitive.
• 2 Marks Q11. Test the transitivity of the relation $R$ on $\mathbb{R}$ defined by $aRb \iff 1 + ab > 0$.
  Answer & Solution:
  Not transitive.
  Counterexample: Let $a = 2, b = -1, c = -2$.
  * $1 + ab = 1 + (2)(-1) = -1 \not> 0$? Wait, let's pick numbers where $1+ab>0$ and $1+bc>0$:
  Let $a = 2, b = 0, c = -3$.
  * $1 + ab = 1 + 0 = 1 > 0 \implies (2, 0) \in R$.
  * $1 + bc = 1 + 0 = 1 > 0 \implies (0, -3) \in R$.
  * For $(a, c) = (2, -3)$, $1 + ac = 1 + (2)(-3) = -5 \not> 0$.
  Thus, $R$ is not transitive.
• 2 Marks Q12. Let $R$ be a relation on the set $A = \{1, 2, 3, 4\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1)\}$. Is $R$ an equivalence relation? Justify.
  Answer & Solution:
  Not an equivalence relation.
  It is reflexive and symmetric, but not transitive because $(3, 1) \in R$ and $(1, 2) \in R$, but $(3, 2) \notin R$.
• 2 Marks Q13. If $R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$, check if $R$ is transitive.
  Answer & Solution:
  Not transitive.
  Counterexample: Take $a = 16, b = 4, c = 2$.
  * $16 = 4^2 \implies (16, 4) \in R$.
  * $4 = 2^2 \implies (4, 2) \in R$.
  * But $16 \neq 2^2$ ($16 \neq 4$), so $(16, 2) \notin R$. Thus, $R$ is not transitive.
• 2 Marks Q14. Let $R$ be an equivalence relation on $\mathbb{Z}$ defined by $aRb$ if $a - b$ is an even integer. Find the equivalence class $[0]$.
• 2 Marks Q15. For the relation $R = \{(x, y) : x, y \in \mathbb{Z}, x \equiv y \pmod 3\}$, find the equivalence class $[2]$.
  Answer & Solution:
  $[2] = \{x \in \mathbb{Z} : x \equiv 2 \pmod 3\}$
  This means $x - 2$ is a multiple of 3, or $x = 3k + 2$ for $k \in \mathbb{Z}$.
  $[2] = \{\dots, -4, -1, 2, 5, 8, \dots\}$.
• 2 Marks Q16. If $A = \{1, 2, 3\}$, what is the smallest equivalence relation containing $(1, 2)$?
  Answer & Solution:
  To be reflexive: $\{(1, 1), (2, 2), (3, 3)\}$.
  To contain $(1, 2)$ and be symmetric: add $(2, 1)$.
  Smallest equivalence relation = $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.
• 2 Marks Q17. If $A = \{1, 2, 3\}$, what is the largest equivalence relation that can be formed on $A$?
  Answer & Solution:
  The universal relation, which is the Cartesian product $A \times A$:
  $R = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$.
• 2 Marks Q18. Let $R$ be a relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a, b) R (c, d) \iff a+d = b+c$. Show that $(1, 2) R (3, 4)$.
  Answer & Solution:
  Here $a = 1, b = 2$ and $c = 3, d = 4$.
  Check condition $a + d = b + c$:
  * LHS: $1 + 4 = 5$
  * RHS: $2 + 3 = 5$
  Since $5 = 5$, $(1, 2) R (3, 4)$ is verified.
• 2 Marks Q19. Prove that the empty relation on a non-empty set $A$ is symmetric and transitive but not reflexive.
  Answer & Solution:
  * **Reflexive:** If $a \in A$, $(a, a) \notin \emptyset$, so not reflexive.
  * **Symmetric:** The condition $(x, y) \in \emptyset \implies (y, x) \in \emptyset$ is vacuously true.
  * **Transitive:** The condition $((x, y) \in \emptyset \land (y, z) \in \emptyset) \implies (x, z) \in \emptyset$ is vacuously true.
• 2 Marks Q20. If $R_1$ and $R_2$ are two equivalence relations on a set $A$, prove that $R_1 \cap R_2$ is also an equivalence relation.
  Answer & Solution:
  * **Reflexive:** $(a, a) \in R_1$ and $(a, a) \in R_2 \implies (a, a) \in R_1 \cap R_2$.
  * **Symmetric:** If $(a, b) \in R_1 \cap R_2$, then $(a, b) \in R_1 \implies (b, a) \in R_1$, and similarly in $R_2$. Thus $(b, a) \in R_1 \cap R_2$.
  * **Transitive:** Follows similarly by intersection properties. Hence, $R_1 \cap R_2$ is an equivalence relation.
• 2 Marks Q21. Give an example of a relation which is reflexive and transitive but not symmetric
  Answer & Solution:
  On the set of real numbers $\mathbb{R}$, the relation $R = \{(a, b) : a \le b\}$.
  * $a \le a$ (Reflexive)
  * $a \le b$ and $b \le c \implies a \le c$ (Transitive)
  * $2 \le 3$ is true, but $3 \le 2$ is false (Not symmetric).
• 2 Marks Q22. Give an example of a relation which is symmetric and transitive but not reflexive.
  Answer & Solution:
  Let $A = \{1, 2, 3\}$. The relation $R = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$ is symmetric and transitive, but not reflexive because $(3, 3) \notin R$.
• 2 Marks Q23. Let $R = \{(a, b) : a \le b\}$ be a relation on $\mathbb{R}$. Prove that $R$ is transitive.
  Answer & Solution:
  Let $(a, b) \in R$ and $(b, c) \in R$.
  This means $a \le b$ and $b \le c$.
  By the transitive property of inequality in real numbers, $a \le c$.
  Therefore, $(a, c) \in R$, proving $R$ is transitive.
• 2 Marks Q24. Find the number of all possible relations on the set $A = \{1, 2, 3\}$.
  Answer & Solution:
  Number of elements in $A$, $n = 3$.
  Number of elements in $A \times A$ is $n^2 = 3^2 = 9$.
  Total number of relations is $2^{n^2} = 2^9 = 512$.
• 2 Marks Q25. If $R$ is a relation on the set $A = \{1, 2, 3, 4\}$ defined by $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$, find its range.
  Answer & Solution:
  The range of a relation is the set of all second elements of the ordered pairs.
  Second elements: $\{3, 2, 4, 3, 1\}$.
  Range = $\{1, 2, 3, 4\}$.
• 2 Marks Q26. Check the injectivity of the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x^2$.
  Answer & Solution:
  Injective (One-one).
  Let $f(x_1) = f(x_2) \implies x_1^2 = x_2^2$.
  Since $x_1, x_2 \in \mathbb{N}$ (positive integers), $x_1 = x_2$. Thus, $f$ is injective.
• 2 Marks Q27. Check the injectivity of the function $f: \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = x^2$.
  Answer & Solution:
  Not injective.
  Counterexample: $f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$.
  Since $f(1) = f(-1)$ but $1 \neq -1$, $f$ is not injective.
• 2 Marks Q28. Show that the constant function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = c$ (where $c$ is a constant) is neither one-one nor onto if $\mathbb{R}$ has more than one element.
  Answer & Solution:
  * **Not one-one:** For any distinct $x_1, x_2 \in \mathbb{R}$, $f(x_1) = c = f(x_2)$.
  * **Not onto:** The range of $f$ is only $\{c\}$, whereas the codomain is $\mathbb{R}$ (which contains elements other than $c$). Thus, elements like $c+1$ have no pre-image.
• 2 Marks Q29. Check if the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = |x|$ is injective.
  Answer & Solution:
  Not injective.
  Counterexample: $f(1) = |1| = 1$ and $f(-1) = |-1| = 1$.
  Since $f(1) = f(-1)$ with $1 \neq -1$, $f$ is not injective.
• 2 Marks Q30. Show that the Modulus function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = |x|$, is not surjective.
  Answer & Solution:
  For $f$ to be surjective, every element in the codomain $\mathbb{R}$ must have a pre-image.
  However, for any negative real number (e.g., $y = -1$), there is no $x \in \mathbb{R}$ such that $|x| = -1$ (since absolute values are always non-negative). Thus, it is not surjective.
• 2 Marks Q31. Check whether the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3 - 4x$ is surjective.
  Answer & Solution:
  Surjective (Onto).
  Let $y \in \mathbb{R}$. Then $y = 3 - 4x \implies 4x = 3 - y \implies x = \frac{3-y}{4}$.
  For any real number $y$, $x$ is a well-defined real number, meaning every element in the codomain has a pre-image. Hence, $f$ is surjective.
• 2 Marks Q32. Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^3$. Prove that $f$ is injective.
  Answer & Solution:
  Let $f(x_1) = f(x_2) \implies x_1^3 = x_2^3$.
  Taking the cube root on both sides yields $x_1 = x_2$.
  Therefore, $f$ is injective.
• 2 Marks Q33. Examine if the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x + 1$ is onto.
  Answer & Solution:
  Not onto.
  Consider the element $1 \in \mathbb{N}$ (codomain). There is no $x \in \mathbb{N}$ such that $x + 1 = 1$, which would require $x = 0 \notin \mathbb{N}$. Thus, $1$ has no pre-image.
• 2 Marks Q34. Show that the Signum function $f: \mathbb{R} \to \mathbb{R}$ is neither one-one nor onto.
  Answer & Solution:
  * **Not one-one:** $f(2) = 1$ and $f(5) = 1$, but $2 \neq 5$.
  * **Not onto:** The range of the signum function is $\{-1, 0, 1\}$, which does not cover the entire codomain $\mathbb{R}$ (e.g., $2$ has no pre-image).
• 2 Marks Q35. Check the bijectivity of the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 5$.
  Answer & Solution:
  Bijective.
  * **Injective:** $2x_1 + 5 = 2x_2 + 5 \implies x_1 = x_2$.
  * **Surjective:** For any $y \in \mathbb{R}$, $x = \frac{y-5}{2} \in \mathbb{R}$ such that $f(x) = y$.
  Since it is both one-one and onto, it is bijective.
• 2 Marks Q36. Let $A = \{1, 2, 3\}$ and $B = \{4, 5, 6, 7\}$. Let $f = \{(1, 4), (2, 5), (3, 6)\}$ be a function from $A$ to $B$. Show that $f$ is one-one but not onto.
  Answer & Solution:
  * **One-one:** Distinct elements in $A$ have distinct images in $B$.
  * **Not onto:** The element $7 \in B$ has no pre-image in $A$, as the range is $\{4, 5, 6\} \neq B$.
• 2 Marks Q37. If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = \frac{x}{x^2+1}$, find whether the function is one-one.
  Answer & Solution:
  Not one-one.
  Counterexample: Let $x_1 = 2$ and $x_2 = \frac{1}{2}$.
  * $f(2) = \frac{2}{2^2+1} = \frac{2}{5}$
  * $f\left(\frac{1}{2}\right) = \frac{1/2}{(1/2)^2+1} = \frac{1/2}{5/4} = \frac{2}{5}$
  Since $f(2) = f(1/2)$ but $2 \neq 1/2$, $f$ is not one-one.
• 2 Marks Q38. Prove that the Greatest Integer Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = [x]$, is neither one-one nor onto.
  Answer & Solution:
  * **Not one-one:** $f(1.2) = 1$ and $f(1.8) = 1$, but $1.2 \neq 1.8$.
  * **Not onto:** The range of $f$ is the set of integers $\mathbb{Z}$, whereas the codomain is $\mathbb{R}$. Non-integers (e.g., $1.5$) have no pre-image.
• 2 Marks Q39. Let $f: \mathbb{R} - \{3\} \to \mathbb{R} - \{1\}$ be defined by $f(x) = \frac{x-2}{x-3}$. Show that $f$ is one-one.
  Answer & Solution:
  Let $f(x_1) = f(x_2) \implies \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}$.
  Cross-multiplying: $(x_1-2)(x_2-3) = (x_2-2)(x_1-3)$
  $x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$
  $-3x_1 - 2x_2 = -3x_2 - 2x_1 \implies x_1 = x_2$. Thus, $f$ is one-one.
• 2 Marks Q40. Using the function in Q39, check if $f$ is onto.
  Answer & Solution:
  Onto.
  Let $y \in \mathbb{R} - \{1\}$. Set $y = \frac{x-2}{x-3} \implies y(x-3) = x-2 \implies yx - 3y = x - 2$.
  $x(y-1) = 3y-2 \implies x = \frac{3y-2}{y-1}$.
  Since $y \neq 1$, $x$ is well-defined and $x \in \mathbb{R}-\{3\}$. Thus, every element in the codomain has a pre-image.
• 2 Marks Q41. If a set $A$ contains 3 elements and set $B$ contains 4 elements, find the number of one-one functions from $A$ to $B$.
  Answer & Solution:
  Number of one-one functions from a set of size $m$ to a set of size $n$ (where $m \le n$) is given by $P(n, m)$.
  Here $n = 4, m = 3 \implies P(4, 3) = \frac{4!}{(4-3)!} = 4 \times 3 \times 2 = 24$.
• 2 Marks Q42. If a set $A$ contains 4 elements and set $B$ contains 3 elements, find the number of one-one functions from $A$ to $B$.
  Answer & Solution:
  0.
  Since the size of the domain ($4$) is greater than the size of the codomain ($3$), by the Pigeonhole Principle, no one-one function can exist.
• 2 Marks Q43. Find the total number of onto functions from a set $A = \{1, 2, 3\}$ to itself.
  Answer & Solution:
  For a finite set to itself, a function is onto if and only if it is one-one (bijection).
  Total number of bijections from a set of size $n$ to itself is $n!$.
  Here $n = 3 \implies 3! = 6$.
• 2 Marks Q44. If $f: A \to B$ is a bijection where $n(A) = 5$, what must be the value of $n(B)$?
  Answer & Solution:
  Since $f$ is a bijection, it is both one-one and onto, meaning every element of $B$ is uniquely paired with an element of $A$. Therefore, $n(B) = n(A) = 5$.
• 2 Marks Q45. Let $f: [0, \infty) \to [0, \infty)$ be defined by $f(x) = x^2$. Show that $f$ is a bijection.
  Answer & Solution:
  * **Injective:** $x_1^2 = x_2^2 \implies x_1 = x_2$ (since domain is restricted to non-negative numbers).
  * **Surjective:** For every $y \in [0, \infty)$, $x = \sqrt{y} \in [0, \infty)$ such that $f(x) = y$.
  Thus, $f$ is a bijection.
• 2 Marks Q46. State the condition under which a linear function $f(x) = mx + c$ (where $m \neq 0$) from $\mathbb{R} \to \mathbb{R}$ is a bijective function.
  Answer & Solution:
  The function $f(x) = mx + c$ is bijective for any $m \neq 0$ over $\mathbb{R}$, as it is strictly monotonic (either increasing or decreasing) and covers all real numbers.
• 2 Marks Q47. Let $f: \mathbb{N} \to \mathbb{Z}$ be defined by: $$f(n) = \begin{cases} \frac{n}{2}, & \text{if } n \text{ is even} \\ -\frac{n-1}{2}, & \text{if } n \text{ is odd} \end{cases}$$ Find the value of $f(4) + f(5)$.
  Answer & Solution:
  * For $n = 4$ (even): $f(4) = \frac{4}{2} = 2$.
  * For $n = 5$ (odd): $f(5) = -\frac{5-1}{2} = -\frac{4}{2} = -2$.
  Sum: $f(4) + f(5) = 2 + (-2) = 0$.
• 2 Marks Q48. If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = \sin x$, find the range of $f$. Is it onto?
  Answer & Solution:
  * **Range:** $[-1, 1]$.
  * **Onto:** No, because the range $[-1, 1]$ is a proper subset of the codomain $\mathbb{R}$ (e.g., $2 \in \mathbb{R}$ has no pre-image).
• 2 Marks Q49. If $A = \{-1, 1\}$, find the number of all onto functions from $A$ to $A$.
  Answer & Solution:
  For a finite set to itself, the number of onto functions equals the number of bijections, which is $n!$.
  Here $n = 2 \implies 2! = 2$.
• 2 Marks Q50. Give an example of a function $f: \mathbb{N} \to \mathbb{N}$ which is onto but not one-one.
  Answer & Solution:
  Define $f: \mathbb{N} \to \mathbb{N}$ as:
  $$f(x) = \begin{cases} 1, & \text{if } x = 1 \text{ or } x = 2 \\ x - 1, & \text{if } x > 2 \end{cases}$$
  * **Not one-one:** $f(1) = 1$ and $f(2) = 1$, but $1 \neq 2$.
  * **Onto:** Every natural number $y \in \mathbb{N}$ has a pre-image (e.g., $y$ has pre-image $y+1$).

SECTION C — Short Answer Type Questions [3 Marks Each]

• 3 Marks Q1. 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.
  Answer & Solution:
  * **Reflexive:** For any $a \in A$, $|a - a| = 0$, which is an even integer. Thus, $(a, a) \in R$.
  * **Symmetric:** If $(a, b) \in R$, then $|a - b|$ is even $\implies |b - a|$ is even $\implies (b, a) \in R$.
  * **Transitive:** If $(a, b) \in R$ and $(b, c) \in R$, then $|a - b|$ and $|b - c|$ are even. Since the sum of two even integers is even, $(a - b) + (b - c) = a - c$ is even, so $|a - c|$ is even $\implies (a, c) \in R$.
  Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.
• 3 Marks Q2. Prove that the relation $R$ on the set $\mathbb{Z}$ of all integers defined by $(x, y) \in R \iff (x - y)$ is divisible by 5 is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** For any $x \in \mathbb{Z}$, $x - x = 0 = 5(0)$, which is divisible by 5. Thus, $(x, x) \in R$.
  * **Symmetric:** If $(x, y) \in R$, $x - y = 5k$ for some integer $k$. Then $y - x = -(x - y) = 5(-k)$, so $(y - x)$ is divisible by 5 $\implies (y, x) \in R$.
  * **Transitive:** If $(x, y) \in R$ and $(y, z) \in R$, then $x - y = 5k$ and $y - z = 5m$. Adding them gives $x - z = 5(k + m)$, which is divisible by 5 $\implies (x, z) \in R$.
  Hence, $R$ is an equivalence relation.
• 3 Marks Q3. Let $A = \{x \in \mathbb{Z} : 0 \le x \le 12\}$. Show that the relation $R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$ is an equivalence relation. Find the equivalence class [1].
  Answer & Solution:
  * **Reflexive, Symmetric, and Transitive:** Proved similarly to multiples/differences (difference of an element with itself is $0$, symmetric by absolute value property, and transitive by sum of multiples of 4).
  * **Equivalence Class $[1]$:** $[1] = \{x \in A : |x - 1| \text{ is a multiple of } 4\}$.
  Checking elements in $A$: $1-1=0$ ($0$ is a multiple of $4$), $5-1=4$, $9-1=8$.
  Thus, $[1] = \{1, 5, 9\}$.
• 3 Marks Q4. Show that the relation $R$ defined on 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.
  Answer & Solution:
  * **Reflexive:** Every polygon $P_1$ has the same number of sides as itself $\implies (P_1, P_1) \in R$.
  * **Symmetric:** If $(P_1, P_2) \in R$, then $P_1$ and $P_2$ have the same number of sides $\implies P_2$ and $P_1$ have the same number of sides $\implies (P_2, P_1) \in R$.
  * **Transitive:** If $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$, then $P_1, P_2$ have equal sides and $P_2, P_3$ have equal sides $\implies P_1$ and $P_3$ have the same number of sides $\implies (P_1, P_3) \in R$.
  Thus, $R$ is an equivalence relation.
• 3 Marks Q5. Let $L$ be the set of all lines in a 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.
  Answer & Solution:
  * **Reflexive:** Every line is parallel to itself ($L_1 \parallel L_1$) $\implies (L_1, L_1) \in R$.
  * **Symmetric:** If $L_1 \parallel L_2$, then $L_2 \parallel L_1 \implies (L_2, L_1) \in R$.
  * **Transitive:** If $L_1 \parallel L_2$ and $L_2 \parallel L_3$, then $L_1 \parallel L_3 \implies (L_1, L_3) \in R$.
  Hence, $R$ is an equivalence relation.
• 3 Marks Q6. Prove that the relation $R$ on the set $\mathbb{N} \times \mathbb{N}$ defined by $(a, b) R (c, d) \iff a + d = b + c$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** For any $(a, b) \in \mathbb{N} \times \mathbb{N}$, $a + b = b + a \implies (a, b) R (a, b)$.
  * **Symmetric:** If $(a, b) R (c, d) \implies a + d = b + c \implies c + b = d + a \implies (c, d) R (a, b)$.
  * **Transitive:** If $(a, b) R (c, d)$ and $(c, d) R (e, f)$, then $a + d = b + c$ and $c + f = d + e$. Adding both equations: $a + d + c + f = b + c + d + e \implies a + f = b + e \implies (a, b) R (e, f)$.
  Thus, it is an equivalence relation.
• 3 Marks Q7. Let $R$ be a relation on the set $\mathbb{N} \times \mathbb{N}$ defined by $(a, b) R (c, d) \iff ad = bc$. Show that $R$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** $ab = ba \implies (a, b) R (a, b)$.
  * **Symmetric:** If $ad = bc \implies cb = da \implies (c, d) R (a, b)$.
  * **Transitive:** If $(a, b) R (c, d)$ and $(c, d) R (e, f)$, then $ad = bc$ and $cf = de$. Multiplying gives $adcf = bcee \implies af = be$ (dividing by non-zero $c$) $\implies (a, b) R (e, f)$.
  Hence, $R$ is an equivalence relation.
• 3 Marks Q8. Show that the relation $R$ on the set $A = \mathbb{Z}$ defined by $R = \{(a, b) : 2 \text{ divides } (a - b)\}$ is an equivalence relation.
  Answer & Solution:
  Identical to divisibility by 5 (Q2):
  * Reflexive: $a - a = 0$ is divisible by 2.
  * Symmetric: If $2 \mid (a - b)$, then $2 \mid (b - a)$ as $b - a = -(a - b)$.
  * Transitive: If $2 \mid (a - b)$ and $2 \mid (b - c)$, then their sum $(a - c)$ is also divisible by 2.
  Thus, $R$ is an equivalence relation.
• 3 Marks Q9. Let $A$ be the set of all books in a library of a college. $R$ is a relation on $A$ given by $R = \{(x, y) : x \text{ and } y \text{ have the same number of pages}\}$. Check if $R$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** Every book $x$ has the same number of pages as itself $\implies (x, x) \in R$.
  * **Symmetric:** If book $x$ and book $y$ have the same pages, then $y$ and $x$ have the same pages $\implies (y, x) \in R$.
  * **Transitive:** If $x, y$ have same pages and $y, z$ have same pages, then $x, z$ have same pages $\implies (x, z) \in R$.
  Therefore, $R$ is an equivalence relation.
• 3 Marks Q10. Show that the relation $R$ in the set $A = \{1, 2, 3\}$ given by $R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** Contains $(1,1), (2,2), (3,3)$, so it is reflexive.
  * **Symmetric:** Contains $(1,2)$ and its reverse $(2,1)$, so it is symmetric.
  * **Transitive:** Checking all combinations of pairs (e.g., $(1,2)$ and $(2,1) \implies (1,1)$), transitivity holds.
  Hence, $R$ is an equivalence relation.
• 3 Marks Q11. Show that the relation $R$ in the set $\mathbb{R}$ of real numbers, defined as $R = \{(a, b) : a \le b^2\}$, is neither reflexive nor symmetric nor transitive.
  Answer & Solution:
  * **Not Reflexive:** For $a = 1/2$, $1/2 \le (1/2)^2 = 1/4$ is false.
  * **Not Symmetric:** For $(1, 2)$, $1 \le 2^2$ ($1 \le 4$) is true, but $2 \le 1^2$ ($2 \le 1$) is false.
  * **Not Transitive:** For $3, 2, 1.5$: $3 \le 2^2$ ($3 \le 4$) and $2 \le (1.5)^2$ ($2 \le 2.25$), but $3 \le (1.5)^2$ ($3 \le 2.25$) is false.
• 3 Marks Q12. Check whether the relation $R$ defined in the set $\{1, 2, 3, 4, 5, 6\}$ as $R = \{(a, b) : b = a + 1\}$ is reflexive, symmetric, or transitive.
  Answer & Solution:
  * **Reflexive:** False, since $a = a + 1$ is never true for any element.
  * **Symmetric:** False, since $(1, 2) \in R$ ($2 = 1+1$), but $(2, 1) \notin R$ ($1 \neq 2+1$).
  * **Transitive:** False, since $(1, 2) \in R$ and $(2, 3) \in R$, but $(1, 3) \notin R$ ($3 \neq 1+1$).
• 3 Marks Q13. Show that the relation $R$ in the set $\mathbb{R}$ of real numbers, defined as $R = \{(a, b) : a \le b^3\}$, is neither reflexive nor symmetric nor transitive.
  Answer & Solution:
  * **Not Reflexive:** For $a = 1/2$, $1/2 \le (1/2)^3 = 1/8$ is false.
  * **Not Symmetric:** For $(1, 2)$, $1 \le 2^3$ ($1 \le 8$) holds, but $2 \le 1^3$ ($2 \le 1$) fails.
  * **Not Transitive:** Counterexample can be found with fractions like $3, 2, 1.5$.
• 3 Marks Q14. Examine if the relation $R$ on the set $\mathbb{R}$ defined by $R = \{(a, b) : 1 + ab > 0\}$ is reflexive, symmetric, and transitive.
  Answer & Solution:
  * **Reflexive:** $1 + a^2 > 0$ for all real $a$. True (Reflexive).
  * **Symmetric:** $1 + ab > 0 \implies 1 + ba > 0$. True (Symmetric).
  * **Transitive:** False. Counterexample: $a = 2, b = 0, c = -3$. $1 + ab = 1 > 0$ and $1 + bc = 1 > 0$, but $1 + ac = 1 + (2)(-3) = -5 \not> 0$. Not transitive.
• 3 Marks Q15. Let $R$ be a relation on the set $A = \{1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$. Is $R$ an equivalence relation? Justify.
  Answer & Solution:
  Not an equivalence relation.
  Although it contains $(1, 1), (2, 2), (3, 3), (4, 4)$ (reflexive), it is not symmetric because $(1, 2) \in R$ but $(2, 1) \notin R$.
• 3 Marks Q16. Determine whether the relation $R$ in the set $A$ of human beings in a town at a particular time given by $R = \{(x, y) : x \text{ is exactly 7 cm taller than } y\}$ is reflexive, symmetric, or transitive.
  Answer & Solution:
  * **Reflexive:** No person is 7 cm taller than themselves.
  * **Symmetric:** If $x$ is 7 cm taller than $y$, then $y$ is 7 cm shorter than $x$, not taller.
  * **Transitive:** If $x$ is 7 cm taller than $y$, and $y$ is 7 cm taller than $z$, then $x$ is 14 cm taller than $z$, not 7 cm. Thus, it is not transitive.
• 3 Marks Q17. Show that the relation $R$ on the set $\mathbb{R}$ of real numbers defined as $R = \{(a, b) : a \le b\}$ is reflexive and transitive but not symmetric.
  Answer & Solution:
  * **Reflexive:** $a \le a$ is always true.
  * **Transitive:** If $a \le b$ and $b \le c$, then $a \le c$.
  * **Not Symmetric:** $2 \le 3$ is true, but $3 \le 2$ is false.
• 3 Marks Q18. Let $R$ be a relation on the set of natural numbers $\mathbb{N}$ defined by $R = \{(x, y) : x + 4y = 10\}$. Find the domain and range of $R$. Is it reflexive?
  Answer & Solution:
  Given $x + 4y = 10 \implies x = 10 - 4y$. For $x, y \in \mathbb{N}$:
  * If $y = 1, x = 6 \implies (6, 1)$
  * If $y = 2, x = 2 \implies (2, 2)$
  * For $y \ge 3, x \le -2$ (not in $\mathbb{N}$).
  * **Domain:** $\{2, 6\}$, **Range:** $\{1, 2\}$.
  * **Reflexive:** No, because $(1, 1) \notin R$ ($1 + 4(1) = 5 \neq 10$).
• 3 Marks Q19. Prove that the perpendicular relation $R = \{(L_1, L_2) : L_1 \perp L_2\}$ on the set of all lines in a plane is symmetric but neither reflexive nor transitive.
  Answer & Solution:
  * **Not Reflexive:** A line cannot be perpendicular to itself ($L_1 \not\perp L_1$).
  * **Symmetric:** If $L_1 \perp L_2$, then $L_2 \perp L_1$.
  * **Not Transitive:** If $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1 \parallel L_3$, not perpendicular.
• 3 Marks Q20. Let $A = \{1, 2, 3\}$. Write the smallest equivalence relation containing $(1, 2)$ and the largest equivalence relation on $A$.
  Answer & Solution:
  * **Smallest equivalence relation containing $(1, 2)$:**
  Include reflexivity $\{(1,1), (2,2), (3,3)\}$, symmetry for $(1,2)$ which adds $(2,1)$, and check transitivity. Result: $\{(1,1), (2,2), (3,3), (1,2), (2,1)\}$.
  * **Largest equivalence relation:** The universal relation $A \times A$ (all 9 pairs).
• 3 Marks Q21. If $R = \{(a, b) : a^2 - b^2 \text{ is divisible by 3}\}$ is an equivalence relation on integers $\mathbb{Z}$, find all the distinct equivalence classes.
  Answer & Solution:
  The condition $a^2 \equiv b^2 \pmod 3$ partitions the integers into 3 equivalence classes based on their remainders modulo 3:
  * $[0] = \{3k : k \in \mathbb{Z}\}$
  * $[1] = \{3k + 1 : k \in \mathbb{Z}\}$
  * $[2] = \{3k + 2 : k \in \mathbb{Z}\}$
• 3 Marks Q22. Let $R$ be a relation on the set $A = \{x \in \mathbb{Z} : 0 \le x \le 10\}$ given by $R = \{(a, b) : |a - b| \text{ is a multiple of 3}\}$. Find the equivalence class of $[2]$.
  Answer & Solution:
  $[2] = \{x \in A : |x - 2| \text{ is a multiple of } 3\}$.
  Checking $0 \le x \le 10$:
  * $|2 - 2| = 0$ (multiple of 3) $\implies 2$
  * $|5 - 2| = 3$ (multiple of 3) $\implies 5$
  * $|8 - 2| = 6$ (multiple of 3) $\implies 8$
  Thus, $[2] = \{2, 5, 8\}$.
• 3 Marks Q23. Show that the relation $R = \{(a, b) : a \text{ divides } b\}$ on the set $\mathbb{N}$ of natural numbers is reflexive and transitive but not symmetric.
  Answer & Solution:
  * **Reflexive:** Every number divides itself ($a \mid a$).
  * **Transitive:** If $a \mid b$ and $b \mid c$, then $a \mid c$.
  * **Not Symmetric:** $2 \mid 4$ is true, but $4 \mid 2$ is false.
• 3 Marks Q24. Prove that the intersection of two equivalence relations on a set $A$ is also an equivalence relation on $A$.
  Answer & Solution:
  Let $R_1$ and $R_2$ be equivalence relations on $A$, and let $R = R_1 \cap R_2$.
  * **Reflexive:** $(a, a) \in R_1$ and $(a, a) \in R_2 \implies (a, a) \in R$.
  * **Symmetric:** If $(a, b) \in R \implies (a, b) \in R_1$ and $(a, b) \in R_2 \implies (b, a) \in R_1$ and $(b, a) \in R_2 \implies (b, a) \in R$.
  * **Transitive:** Follows similarly through the intersection property. Hence, $R_1 \cap R_2$ is an equivalence relation.
• 3 Marks Q25. Is the union of two equivalence relations on a set $A$ necessarily an equivalence relation? Give an example to justify your answer.
  Answer & Solution:
  No, not necessarily.
  Example: Let $A = \{1, 2, 3\}$. Let $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$ and $R_2 = \{(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)\}$. Both are equivalence relations.
  Their union is $R_1 \cup R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}$.
  Here, $(1, 2)$ and $(2, 3)$ belong to the union, but $(1, 3)$ does not. Thus, transitivity fails, showing the union is not necessarily an equivalence relation.
• 3 Marks Q26. Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3 - 4x$ is bijective.
  Answer & Solution:
  * **Injectivity:** Let $f(x_1) = f(x_2) \implies 3 - 4x_1 = 3 - 4x_2 \implies -4x_1 = -4x_2 \implies x_1 = x_2$. Thus, $f$ is one-one.
  * **Surjectivity:** Let $y \in \mathbb{R}$. Then $y = 3 - 4x \implies 4x = 3 - y \implies x = \frac{3-y}{4}$. For any $y \in \mathbb{R}$, $x \in \mathbb{R}$ and $f(x) = y$. Thus, $f$ is onto.
  Since $f$ is both one-one and onto, it is bijective.
• 3 Marks Q27. Check the injectivity and surjectivity of the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x^2$.
  Answer & Solution:
  * **Injectivity:** Let $f(x_1) = f(x_2) \implies x_1^2 = x_2^2$. Since $x_1, x_2 \in \mathbb{N}$, $x_1 = x_2$. Thus, $f$ is injective.
  * **Surjectivity:** Consider $2 \in \mathbb{N}$ (codomain). There is no $x \in \mathbb{N}$ such that $x^2 = 2$ (since $\sqrt{2} \notin \mathbb{N}$). Thus, $f$ is not surjective.
• 3 Marks Q28. Check the injectivity and surjectivity of the function $f: \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = x^2$.
  Answer & Solution:
  * **Injectivity:** Not injective, because $f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$, but $1 \neq -1$.
  * **Surjectivity:** Not surjective, because negative integers in $\mathbb{Z}$ (e.g., $-1$) have no pre-image since squares of integers are non-negative.
• 3 Marks Q29. Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is a bijection.
  Answer & Solution:
  * **Injectivity:** Let $f(x_1) = f(x_2) \implies x_1^3 = x_2^3 \implies x_1 = x_2$. Hence, $f$ is injective.
  * **Surjectivity:** For any $y \in \mathbb{R}$, there exists $x = y^{1/3} \in \mathbb{R}$ such that $f(x) = (y^{1/3})^3 = y$. Hence, $f$ is surjective.
  Therefore, $f$ is a bijection.
• 3 Marks Q30. Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Consider the function $f: A \to B$ defined by $f(x) = \frac{x - 2}{x - 3}$. Is $f$ one-one and onto? Justify.
  Answer & Solution:
  * **One-one:** Let $f(x_1) = f(x_2) \implies \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3} \implies (x_1-2)(x_2-3) = (x_2-2)(x_1-3) \implies x_1 = x_2$. Thus, $f$ is one-one.
  * **Onto:** Let $y \in B$ ($y \neq 1$). Then $y = \frac{x-2}{x-3} \implies y(x-3) = x-2 \implies x = \frac{3y-2}{y-1}$. Since $y \neq 1$, $x \in A$ is well-defined. Thus, every element has a pre-image, making $f$ onto.
• 3 Marks Q31. Show that the Modulus Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = |x|$, is neither one-one nor onto.
  Answer & Solution:
  * **Not one-one:** $f(1) = |1| = 1$ and $f(-1) = |-1| = 1$, but $1 \neq -1$.
  * **Not onto:** The range of $f$ is $[0, \infty)$, which does not equal the codomain $\mathbb{R}$ (negative numbers like $-5$ have no pre-image).
• 3 Marks Q32. Show that the Greatest Integer Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = [x]$, is neither one-one nor onto.
  Answer & Solution:
  * **Not one-one:** $f(1.2) = 1$ and $f(1.8) = 1$, but $1.2 \neq 1.8$.
  * **Not onto:** The range of $f$ is $\mathbb{Z}$, which is a proper subset of $\mathbb{R}$ (non-integers have no pre-images).
• 3 Marks Q33. Show that the Signum Function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = 1$ if $x > 0$, $0$ if $x = 0$, and $-1$ if $x < 0$, is neither one-one nor onto.
  Answer & Solution:
  * **Not one-one:** $f(2) = 1$ and $f(5) = 1$, but $2 \neq 5$.
  * **Not onto:** The range is $\{-1, 0, 1\} \neq \mathbb{R}$ (e.g., $2$ has no pre-image).
• 3 Marks Q34. Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x}{x^2 + 1}$ is neither one-one nor onto.
  Answer & Solution:
  * **Not one-one:** $f(2) = \frac{2}{5}$ and $f\left(\frac{1}{2}\right) = \frac{2}{5}$, but $2 \neq \frac{1}{2}$.
  * **Not onto:** The range is $[-1/2, 1/2]$, which is a subset of $\mathbb{R}$ (elements like $2$ have no pre-image).
• 3 Marks Q35. Let $f: \mathbb{N} \to \mathbb{N}$ be defined by $f(n) = \frac{n+1}{2}$ if $n$ is odd, and $\frac{n}{2}$ if $n$ is even. State whether the function $f$ is bijective. Justify your answer.
  Answer & Solution:
  Not bijective (not one-one).
  For $n = 1$ (odd), $f(1) = \frac{1+1}{2} = 1$.
  For $n = 2$ (even), $f(2) = \frac{2}{2} = 1$.
  Since $f(1) = f(2) = 1$ but $1 \neq 2$, the function is not injective, and hence not bijective.
• 3 Marks Q36. Let $A = \{-1, 0, 1, 2\}$, $B = \{-4, -2, 0, 2\}$ and $f, g: A \to B$ be functions defined by $f(x) = x^2 - x$ and $g(x) = 2\left|x - \frac{1}{2}\right| - 1$. Are $f$ and $g$ equal? Justify.
  Answer & Solution:
  Evaluate both functions for each element in $A$:
  * $f(-1) = 2, g(-1) = 2$
  * $f(0) = 0, g(0) = 0$
  * $f(1) = 0, g(1) = 0$
  * $f(2) = 2, g(2) = 2$
  Since $f(x) = g(x)$ for all $x \in A$ and they share the same domain and codomain, yes, $f$ and $g$ are equal.
• 3 Marks Q37. Show that an injective function $f: \{1, 2, 3\} \to \{1, 2, 3\}$ must be onto.
  Answer & Solution:
  Since $f$ is injective, 3 distinct elements of the domain must map to 3 distinct elements of the codomain. Because the codomain also contains exactly 3 elements, every element in the codomain must be mapped to. Hence, $f$ is onto.
• 3 Marks Q38. Show that a surjective function $f: \{1, 2, 3\} \to \{1, 2, 3\}$ must be one-one.
  Answer & Solution:
  Since $f$ is surjective on finite sets of equal cardinality ($3$ elements), all elements of the codomain are covered. If two domain elements mapped to the same codomain element, another codomain element would be left without a pre-image, violating surjectivity. Thus, it must be one-one.
• 3 Marks Q39. Examine whether the function $f: [0, \infty) \to \mathbb{R}$ defined by $f(x) = \sqrt{x}$ is one-one and onto.
  Answer & Solution:
  * **One-one:** Let $\sqrt{x_1} = \sqrt{x_2} \implies x_1 = x_2$. Thus, it is one-one.
  * **Not onto:** The range is $[0, \infty)$, whereas the codomain is $\mathbb{R}$. Negative real numbers have no pre-image. Thus, it is not onto.
• 3 Marks Q40. Consider a function $f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \sin x$. Is this function bijective? If not, how can you restrict the domain to make it bijective?
  Answer & Solution:
  * **Bijective:** No, it is not one-one (e.g., $\sin(0) = \sin(\pi) = 0$).
  * **Domain Restriction:** By restricting the domain to $[-\pi/2, \pi/2]$ (or any principal branch interval like $[0, \pi]$), the function becomes strictly monotonic and bijective.
• 3 Marks Q41. Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = 2x^3 - 5$. Prove that $f$ is a bijective function.
  Answer & Solution:
  * **Injective:** $2x_1^3 - 5 = 2x_2^3 - 5 \implies x_1^3 = x_2^3 \implies x_1 = x_2$.
  * **Surjective:** For any $y \in \mathbb{R}$, $x = \left(\frac{y+5}{2}\right)^{1/3} \in \mathbb{R}$ such that $f(x) = y$.
  Thus, $f$ is both one-one and onto, hence bijective.
• 3 Marks Q42. Let $A = \mathbb{R} - \{-4/3\}$. Show that the function $f: A \to \mathbb{R}$ defined as $f(x) = \frac{4x}{3x + 4}$ is one-one. Find its range to make it an onto function.
  Answer & Solution:
  * **One-one:** $\frac{4x_1}{3x_1+4} = \frac{4x_2}{3x_2+4} \implies 12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2 \implies x_1 = x_2$.
  * **Range:** Let $y = \frac{4x}{3x+4} \implies x = \frac{4y}{4 - 3y}$. For $x$ to be real, $4 - 3y \neq 0 \implies y \neq 4/3$. Thus, the range is $\mathbb{R} - \{4/3\}$.
• 3 Marks Q43. Check the injectivity and surjectivity of the function $f: \mathbb{Q} \to \mathbb{Q}$ defined by $f(x) = 3x + 5$.
  Answer & Solution:
  * **Injective:** $3x_1 + 5 = 3x_2 + 5 \implies x_1 = x_2$. (Injective)
  * **Surjective:** For any $y \in \mathbb{Q}$, $x = \frac{y-5}{3} \in \mathbb{Q}$. (Surjective)
• 3 Marks Q44. Show that the function $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x + 1$ if $x$ is odd, and $f(x) = x - 1$ if $x$ is even, is a bijection.
  Answer & Solution:
  * **Injective & Surjective:** This function pairs each odd number $2k-1$ with $2k$, and each even number $2k$ with $2k-1$. Every natural number has a unique pre-image and image, making it a bijection.
• 3 Marks Q45. Prove that the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = ax + b$ where $a, b \in \mathbb{R}$ and $a \neq 0$ is a bijective function.
  Answer & Solution:
  * **Injective:** $ax_1 + b = ax_2 + b \implies x_1 = x_2$.
  * **Surjective:** For any $y \in \mathbb{R}$, $x = \frac{y-b}{a} \in \mathbb{R}$ (since $a \neq 0$).
  Thus, it is bijective.
• 3 Marks Q46. Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^4$. Check if the function is injective, surjective, or both.
  Answer & Solution:
  * **Not injective:** $f(1) = 1$ and $f(-1) = 1$, but $1 \neq -1$.
  * **Not surjective:** Range is $[0, \infty)$, so negative numbers in codomain $\mathbb{R}$ have no pre-images.
• 3 Marks Q47. Show that the function $f: (-1, 1) \to \mathbb{R}$ defined by $f(x) = \frac{x}{1 - |x|}$ is one-one and onto.
  Answer & Solution:
  * **One-one:** Analyzing cases for positive and negative $x$ shows that $f$ is strictly increasing, hence injective.
  * **Onto:** For any $y \in \mathbb{R}$, we can solve for $x = \frac{y}{1 + |y|} \in (-1, 1)$, ensuring every real number has a valid pre-image.
• 3 Marks Q48. If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = \frac{3x^2 + x - 1}{x^2 + 1}$, find whether the function is many-one or one-one.
  Answer & Solution:
  Many-one.
  Setting $f(x_1) = f(x_2)$ or evaluating specific points reveals that multiple distinct domain values can map to the same image, proving the function is not one-one.
• 3 Marks Q49. Let $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c\}$. Find the total number of functions from $A$ to $B$. How many of them are one-one?
  Answer & Solution:
  * **Total functions:** $3^4 = 81$.
  * **One-one functions:** $0$, because the size of the domain ($4$) exceeds the size of the codomain ($3$).
• 3 Marks Q50. Prove that the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3 + x$ is an injective function.
  Answer & Solution:
  Derivative method: $f'(x) = 3x^2 + 1 > 0$ for all $x \in \mathbb{R}$.
  Since the derivative is strictly positive everywhere, the function is strictly increasing and therefore injective.

SECTION D — Case-Based/Source-Based Integrated Questions [4 Marks Each]

• 4 Marks Q1. Case Study: Amusement Park Rides
  An amusement park designer is mapping out the relationship between different rides using coordinates. Let $A = \{1, 2, 3\}$ be the set representing three major rides (Roller Coaster, Giant Wheel, and Bumper Cars). A relation $R$ is defined on set $A$ as: $$R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$$

  a) Is the relation $R$ reflexive? Justify. (1 Marks)

  b) Is the relation $R$ symmetric? Justify. (1 Marks)

  c) Check whether $R$ is transitive. (1 Marks)

  d) Is $R$ an equivalence relation? If not, what minimum ordered pair should be added to make it one? (1 Marks)

• 4 Marks Q2. Case Study: Cellular Network Connections
  A telecom company analyzes signal interference between four towers represented by the set $A = \{a, b, c, d\}$. The relation $R$ on $A$ is defined as: $$R = \{(a, a), (b, b), (c, c), (d, d), (a, b), (b, a), (b, c), (c, b)\}$$

  a) Find whether the relation is symmetric. (1 Marks)

  b) Show that $R$ is not transitive by giving a counterexample. (1 Marks)

  c) If we remove $(b, c)$ and $(c, b)$ from $R$, does the new relation become an equivalence relation? (1 Marks)

  d) Find the number of elements in the power set of $R$. (1 Marks)

• 4 Marks Q3. Case Study: Blood Group Compatibility
  In a medical study, researchers define a relation $R$ on the set of all human beings such that $xRy$ if and only if $x$ and $y$ have the exact same blood group.

  a) Prove that $R$ is a reflexive relation. (1 Marks)

  b) Prove that $R$ is a symmetric relation. (1 Marks)

  c) Prove that $R$ is a transitive relation. (1 Marks)

  d) Based on your answers, what special type of relation is $R$? (1 Marks)

• 4 Marks Q4. Case Study: E-Commerce Delivery Logistics
  An e-commerce company routes packages using a function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 3x + 5$, where $x$ represents the weight of the package in kg and $f(x)$ represents the shipping cost in dollars.

  a) Show that the function $f(x)$ is injective (one-one). (1.5 Marks)

  b) Show that the function $f(x)$ is surjective (onto). (1.5 Marks)

  c) Is this function bijective? What does this mean for the company regarding tracking costs back to weights? (1 Marks)

• 4 Marks Q5. Case Study: The Architectural Arch
  An architect designs a parabolic arch for a monument. The height of the arch is modeled by a function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x^2$. f(x) | . | / -----+---/----+--> x | / | .

  a) Check if the function is one-one. Provide a counterexample if it is not. (1 Marks)

  b) Check if the function is onto. Provide a counterexample if it is not. (1 Marks)

  c) If the domain and co-domain are restricted to the set of natural numbers $\mathbb{N}$, check the injectivity of $f(x) = x^2$. (2 Marks)

• 4 Marks Q6. Case Study: School Student Database
  A school database stores student records. Let $A$ be the set of all students in Class XII. A relation $R$ in $A$ is given by $R = \{(a, b) : \text{height of } a \text{ is exactly } 5 \text{ cm more than height of } b\}$.

  a) Is $R$ reflexive? Why or why not? (1 Marks)

  b) Is $R$ symmetric? Why or why not? (1 Marks)

  c) If $(a, b) \in R$ and $(b, c) \in R$, find the relation between the heights of $a$ and $c$. Is $R$ transitive? (2 Marks)

• 4 Marks Q7. Case Study: Coding and Cryptography
  A software engineer creates an encryption algorithm using a function $f: \mathbb{N} \rightarrow \mathbb{N}$ defined by: $$f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases}$$

  a) Find $f(1)$ and $f(2)$. (1 Marks)

  b) Based on Q1, is the function one-one? Explain. (1 Marks)

  c) Is the function onto? Justify your answer. (2 Marks)

• 4 Marks Q8. Case Study: Digital Image Processing
  In digital image processing, a pixel scaling function $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined as $f(x) = [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$ (Floor function).

  a) Find the values of $f(2.3)$ and $f(2.9)$. (1 Marks)

  b) Is this function one-one? Explain using the values from Q1. (1 Marks)

  c) What is the range of this function? (1 Marks)

  d) Is this function onto? Justify. (1 Marks)

• 4 Marks Q9. Case Study: Aviation Altitude Control
  An aircraft's autopilot system utilizes a cubic function to adjust altitude smoothing. The function $f: \mathbb{R} \rightarrow \mathbb{R}$ is given by $f(x) = x^3$.

  a) Prove mathematically that $f(x) = x^3$ is a one-one function. (2 Marks)

  b) Prove that $f(x)$ is an onto function. (1.5 Marks)

  c) What is the term used for functions that are both one-one and onto? (0.5 Marks)

• 4 Marks Q10. Case Study: Library Catalog System
  A university library defines a relation $R$ on the set $B$ of all books in the library given by $R = \{(x, y) : x \text{ and } y \text{ have the same number of pages}\}$.

  a) Check if $R$ is reflexive. (1 Marks)

  b) Check if $R$ is symmetric. (1 Marks)

  c) Check if $R$ is transitive. (1 Marks)

  d) If Book $X$ has 350 pages, what does the equivalence class $[X]$ represent? (1 Marks)

• 4 Marks Q11. Case Study: Absolute Value Sensors
  A thermal sensor records temperature fluctuations using an absolute value function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = |x|$.

  a) Evaluate $f(-5)$ and $f(5)$. (1 Marks)

  b) Is the sensor function one-one? Justify based on Q1. (1 Marks)

  c) Can the output of this sensor ever be negative? What is the range of $f(x)$? (1 Marks)

  d) Is the function onto? Explain (1 Marks)

• 4 Marks Q12. Case Study: Real Estate Plot Subdivisions
  A developer divides a plot of land. Let $L$ be the set of all straight lines in a XY-plane. Let $R$ be a relation in $L$ defined as $R = \{(L_1, L_2) : L_1 \text{ is parallel to } L_2\}$.

  a) Prove that $R$ is reflexive. (1 Marks)

  b) Prove that $R$ is symmetric. (1 Marks)

  c) Prove that $R$ is transitive. (1 Marks)

  d) Find the set of all lines related to the line $y = 2x + 4$. (1 Marks)

• 4 Marks Q13. Case Study: Perpendicular Structural Beams
  In a construction blueprint, $L$ is the set of all lines in a 2D plane. A relation $R$ is defined as $R = \{(L_1, L_2) : L_1 \text{ is perpendicular to } L_2\}$.

  a) Can a line be perpendicular to itself? Is $R$ reflexive? (1 Marks)

  b) If $L_1 \perp L_2$, is $L_2 \perp L_1$? Is $R$ symmetric? (1 Marks)

  c) If $L_1 \perp L_2$ and $L_2 \perp L_3$, what is the geometric relationship between $L_1$ and $L_3$? Is $R$ transitive? (2 Marks)

• 4 Marks Q14. Case Study: Social Media Networks
  On a social media application, a relation $R$ is defined on the set of all active users. User $A$ is related to User $B$ ($(A, B) \in R$) if $A$ follows $B$.

  a) If user $A$ follows themselves, the relation is reflexive. Is it universally true that everyone follows themselves on social media? (1 Marks)

  b) If $A$ follows $B$, does it imply $B$ must follow $A$? What does this say about symmetry? (1 Marks)

  c) If $A$ follows $B$ and $B$ follows $C$, does it mean $A$ follows $C$? What does this say about transitivity? (1 Marks)

  d) Is this relation an equivalence relation? (1 Marks)

• 4 Marks Q15. Case Study: Set Theory Operations
  Let $A = \{1, 2, 3\}$. A relation $R$ on set $A$ is given by $R = \{(1, 2), (2, 3)\}$.

  a) Write the minimum number of ordered pairs to be added to $R$ so that it becomes reflexive. (1 Marks)

  b) Write the ordered pairs to be added to the original $R$ to make it symmetric. (1 Marks)

  c) What ordered pair must be added to $R$ to satisfy the transitivity condition for the existing elements? (1 Marks)

  d) Find the total number of possible relations on set $A$. (1 Marks)

• 4 Marks Q16. Case Study: Mathematical Mapping
  Consider a function $f: \mathbb{R} - \{3\} \rightarrow \mathbb{R} - \{1\}$ defined by: $$f(x) = \frac{x-2}{x-3}$$

  a) Show that $f(x)$ is a one-one function. (2 Marks)

  b) Show that $f(x)$ is an onto function. (2 Marks)

• 4 Marks Q17. Case Study: Divisibility in Inventory Systems
  An inventory tracking system assigns unique identification numbers to items. Let $\mathbb{Z}$ be the set of integers. A relation $R$ on $\mathbb{Z}$ is defined by $R = \{(a, b) : 2 \text{ divides } (a - b)\}$.

  a) Show that $R$ is reflexive. (1 Marks)

  b) Show that $R$ is symmetric. (1 Marks)

  c) Show that $R$ is transitive. (1.5 Marks)

  d) How many distinct equivalence classes are formed by this relation? (0.5 Marks)

• 4 Marks Q18. Case Study: Linear Economic Projections
  An economic forecast tool uses a linear function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 4x - 7$.

  a) Check if the function is injective. (1.5 Marks)

  b) Check if the function is surjective. (1.5 Marks)

  c) Find the value of $x$ if $f(x) = 5$. (1 Marks)

• 4 Marks Q19. Case Study: Human Genetics Mapping
  Let $X$ be the set of all human beings alive in the world at a given time. A relation $R$ on $X$ is defined as $R = \{(a, b) : a \text{ is the brother of } b\}$.

  a) Is $R$ reflexive? Explain with a reason involving female individuals. (1 Marks)

  b) Is $R$ symmetric? (Hint: Consider if $a$ is a brother of a female $b$). (1.5 Marks)

  c) Is $R$ transitive? Explain carefully. (1.5 Marks)

• 4 Marks Q20. Case Study: The Signum Signal
  An electrical engineer uses a signum function $f: \mathbb{R} \rightarrow \mathbb{R}$ to switch control states in a circuit. The function is defined as: $$f(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases}$$

  a) Find $f(5)$ and $f(10)$. Is the function one-one? (1.5 Marks)

  b) What is the range of this function? (1 Marks)

  b) Is this function onto? Explain based on its co-domain. (1.5 Marks)

SECTION E — Long Answer Type Questions [5 Marks Each]

• 5 Marks Q1. Show that the relation $R$ in the set $A = \{x \in \mathbb{Z} : 0 \le x \le 12\}$, given by $R = \{(a, b) : |a - b| \text{ is a multiple of } 4\}$, is an equivalence relation. Find the set of all elements related to 1.
  Answer & Solution:
  * **Reflexive:** For any $a \in A$, $|a - a| = 0$, which is a multiple of $4$. So, $(a, a) \in R$.
  * **Symmetric:** If $(a, b) \in R$, then $|a - b|$ is a multiple of $4$ $\implies |b - a|$ is also a multiple of $4$ $\implies (b, a) \in R$.
  * **Transitive:** If $(a, b) \in R$ and $(b, c) \in R$, then $|a - b|$ and $|b - c|$ are multiples of $4$. Thus, their sum/difference implies $|a - c|$ is a multiple of $4$ $\implies (a, c) \in R$.
  Since $R$ is reflexive, symmetric, and transitive, it is an **equivalence relation**.
  * **Elements related to 1:** Let $x \in A$ such that $|x - 1|$ is a multiple of $4$. The possible values in $A = \{0, 1, \dots, 12\}$ are $\{1, 5, 9\}$.
• 5 Marks Q2. Let $\mathbb{N}$ be the set of all natural numbers and $R$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a, b) R (c, d) \iff ad(b + c) = bc(a + d)$. Show that $R$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** For any $(a, b) \in \mathbb{N} \times \mathbb{N}$, $ab(b + a) = ba(a + b)$, so $(a, b) R (a, b)$. Reflexive.
  * **Symmetric:** If $(a, b) R (c, d)$, then $ad(b + c) = bc(a + d) \implies cb(d + a) = da(c + b) \implies (c, d) R (a, b)$. Symmetric.
  * **Transitive:** Let $(a, b) R (c, d)$ and $(c, d) R (e, f)$. Translating to fractions $\frac{a}{b} = \frac{c}{d}$ and $\frac{c}{d} = \frac{e}{f}$, we get $\frac{a}{b} = \frac{e}{f}$, which means $af(b + e) = be(a + f) \implies (a, b) R (e, f)$. Transitive.
  Thus, $R$ is an **equivalence relation**.
• 5 Marks Q3. Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Consider the function $f: A \to B$ defined by $f(x) = \frac{x - 2}{x - 3}$. Is $f$ one-to-one and onto? Justify your answer.
  Answer & Solution:
  * **One-to-one:** Let $f(x_1) = f(x_2) \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \implies (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \implies x_1 = x_2$. Thus, $f$ is **one-to-one**.
  * **Onto:** For any $y \in B$ ($y \neq 1$), let $y = \frac{x - 2}{x - 3} \implies x = \frac{3y - 2}{y - 1}$. Since $y \neq 1$, $x \in A$ ($x \neq 3$) is well-defined, and $f(x) = y$. Thus, $f$ is **onto**.
• 5 Marks Q4. Show that the relation $R$ defined on 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-angled triangle $T$ with sides 3, 4, and 5?
  Answer & Solution:
  * **Reflexive:** Every polygon $P_1$ has the same number of sides as itself $\implies (P_1, P_1) \in R$.
  * **Symmetric:** If $(P_1, P_2) \in R$, $P_1$ and $P_2$ have the same number of sides $\implies P_2$ and $P_1$ have the same number of sides $\implies (P_2, P_1) \in R$.
  * **Transitive:** If $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$, then $P_1, P_2, P_3$ all have the same number of sides $\implies (P_1, P_3) \in R$.
  Thus, $R$ is an **equivalence relation**.
  * **Set related to $T$:** Triangle $T$ has $3$ sides. Therefore, the set of all elements related to $T$ is the **set of all triangles** in $A$.
• 5 Marks Q5. Prove that the relation $R$ on the set $\mathbb{Z}$ of all integers defined by $(x, y) \in R \iff (x - y)$ is divisible by $n$ (where $n$ is a fixed positive integer) is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** For any $x \in \mathbb{Z}$, $x - x = 0$, which is divisible by $n$. So, $(x, x) \in R$.
  * **Symmetric:** If $(x, y) \in R$, then $x - y = kn$ for some integer $k$. Then $y - x = -kn$, which is also divisible by $n$. So, $(y, x) \in R$.
  * **Transitive:** If $(x, y) \in R$ and $(y, z) \in R$, then $x - y = k_1n$ and $y - z = k_2n$. Adding them gives $x - z = (k_1 + k_2)n$, which is divisible by $n$. So, $(x, z) \in R$.
  Hence, $R$ is an **equivalence relation**.
• 5 Marks Q6. 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 $y = 2x + 4$.
  Answer & Solution:
  * **Reflexive:** Every line is parallel to itself $\implies (L_1, L_1) \in R$.
  * **Symmetric:** If $L_1 \parallel L_2$, then $L_2 \parallel L_1 \implies (L_2, L_1) \in R$.
  * **Transitive:** If $L_1 \parallel L_2$ and $L_2 \parallel L_3$, then $L_1 \parallel L_3 \implies (L_1, L_3) \in R$.
  Thus, $R$ is an **equivalence relation**.
  * **Lines related to $y = 2x + 4$:** Parallel lines share the same slope. Thus, the set of all lines related to $y = 2x + 4$ is given by **$y = 2x + c$**, where $c \in \mathbb{R}$.
• 5 Marks Q7. Let $A = \{1, 2, 3, \dots, 9\}$ and $R$ be the relation in $A \times A$ defined by $(a, b) R (c, d)$ if $a + d = b + c$ for $(a, b), (c, d)$ in $A \times A$. Prove that $R$ is an equivalence relation. Also, obtain the equivalence class $[(2, 5)]$.
  Answer & Solution:
  * **Reflexive:** $a + b = b + a \implies (a, b) R (a, b)$.
  * **Symmetric:** If $a + d = b + c$, then $c + b = d + a \implies (c, d) R (a, b)$.
  * **Transitive:** If $a + d = b + c$ and $c + f = d + e$, adding them yields $a + f = b + e \implies (a, b) R (e, f)$.
  Thus, $R$ is an **equivalence relation**.
  * **Equivalence class $[(2, 5)]gart$:** $(c, d)$ such that $2 + d = 5 + c \implies d - c = 3$. The pairs are **$\{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\}$**.
• 5 Marks Q8. Show that the relation $R$ on the set $A = \mathbb{Z}$ of integers given by $R = \{(a, b) : 2 \text{ divides } a - b\}$ is an equivalence relation. Write all its equivalence classes.
  Answer & Solution:
  * **Reflexive:** $a - a = 0$ is divisible by $2$.
  * **Symmetric:** If $a - b$ is divisible by $2$, then $b - a$ is also divisible by $2$.
  * **Transitive:** If $a - b$ and $b - c$ are divisible by $2$, their sum $a - c$ is divisible by $2$.
  Thus, $R$ is an **equivalence relation**.
  * **Equivalence Classes:** There are two equivalence classes: * $[0] = \{2k : k \in \mathbb{Z}\}$ (set of all even integers) * $[1] = \{2k + 1 : k \in \mathbb{Z}\}$ (set of all odd integers)
• 5 Marks Q9. Let $R$ be a relation on the set $A$ of ordered pairs of positive integers defined by $(x, y) R (u, v) \iff xv = yu$. Show that $R$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** $xy = yx \implies (x, y) R (x, y)$.
  * **Symmetric:** If $(x, y) R (u, v)$, then $xv = yu \implies uy = vx \implies (u, v) R (x, y)$.
  * **Transitive:** If $xv = yu$ and $ub = vc$, working with fractions $\frac{x}{y} = \frac{u}{v}$ and $\frac{u}{v} = \frac{c}{b}$ gives $\frac{x}{y} = \frac{c}{b} \implies xb = yc \implies (x, y) R (c, b)$.
  Thus, $R$ is an **equivalence relation**.
• 5 Marks Q10. Let $f: X \to Y$ be a function. Define a relation $R$ in $X$ given by $R = \{(a, b) : f(a) = f(b)\}$. Examine if $R$ is an equivalence relation.
  Answer & Solution:
  * **Reflexive:** For any $a \in X$, $f(a) = f(a) \implies (a, a) \in R$.
  * **Symmetric:** If $(a, b) \in R$, then $f(a) = f(b) \implies f(b) = f(a) \implies (b, a) \in R$.
  * **Transitive:** If $(a, b) \in R$ and $(b, c) \in R$, then $f(a) = f(b)$ and $f(b) = f(c) \implies f(a) = f(c) \implies (a, c) \in R$.
  Therefore, $R$ is an **equivalence relation**.
• 5 Marks Q11. Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 4x + 3$ is a bijective function.
  Answer & Solution:
  * **Injective:** $4x_1 + 3 = 4x_2 + 3 \implies 4x_1 = 4x_2 \implies x_1 = x_2$ (One-one).
  * **Surjective:** For any $y \in \mathbb{R}$, $y = 4x + 3 \implies x = \frac{y - 3}{4} \in \mathbb{R}$ such that $f(x) = y$ (Onto).
  Since it is both one-one and onto, $f$ is **bijective**.
• 5 Marks Q12. Consider $f: \mathbb{R}_+ \to [4, \infty)$ given by $f(x) = x^2 + 4$. Show that $f$ is bijective, where $\mathbb{R}_+$ is the set of all non-negative real numbers.
  Answer & Solution:
  * **Injective:** Let $f(x_1) = f(x_2) \implies x_1^2 + 4 = x_2^2 + 4 \implies x_1^2 = x_2^2$. Since domain is $\mathbb{R}_+$ ($x \ge 0$), $x_1 = x_2$.
  * **Surjective:** For any $y \in [4, \infty)$, $y = x^2 + 4 \implies x = \sqrt{y - 4}$. Since $y \ge 4$, $x \in \mathbb{R}_+$, and $f(x) = y$.
  Thus, $f$ is **bijective**.
• 5 Marks Q13. Let $f: \mathbb{N} \to \mathbb{N}$ be defined by $f(n) = \frac{n + 1}{2}$ if $n$ is odd, and $\frac{n}{2}$ if $n$ is even. State whether the function $f$ is bijective. Justify your answer.
  Answer & Solution:
  Not bijective (not one-one).
  For $n = 1$ (odd), $f(1) = \frac{1+1}{2} = 1$.
  For $n = 2$ (even), $f(2) = \frac{2}{2} = 1$.
  Since $f(1) = f(2)$ but $1 \neq 2$, the function is not injective, and hence not bijective.
• 5 Marks Q14. Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is a bijection.
  Answer & Solution:
  * **Injective:** $x_1^3 = x_2^3 \implies x_1 = x_2$. (One-one)
  * **Surjective:** For any $y \in \mathbb{R}$, $x = y^{1/3} \in \mathbb{R}$ such that $f(x) = y$. (Onto)
  Thus, $f$ is a **bijection**.
• 5 Marks Q15. Consider $f: \mathbb{R} - \left\{-\frac{4}{3}\right\} \to \mathbb{R}$ defined as $f(x) = \frac{4x}{3x + 4}$. Show that $f$ is a one-to-one function. Is it onto if the codomain is changed to the range of $f$?
  Answer & Solution:
  * **One-to-one:** $\frac{4x_1}{3x_1+4} = \frac{4x_2}{3x_2+4} \implies x_1 = x_2$. Thus, $f$ is **one-to-one**.
  * **Onto check:** The range is found by setting $y = \frac{4x}{3x+4} \implies x = \frac{4y}{4-3y}$, which shows $y \neq 4/3$. If the codomain is changed to the range $\mathbb{R} - \{4/3\}$, every element has a pre-image, making it **onto**.
• 5 Marks Q16. Show that the Signum function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = 1$ if $x > 0$, $0$ if $x = 0$, and $-1$ if $x < 0$, is neither one-to-one nor onto.
  Answer & Solution:
  * **Not one-to-one:** $f(2) = 1$ and $f(5) = 1$, but $2 \neq 5$.
  * **Not onto:** The range of $f$ is $\{-1, 0, 1\}$, which is a proper subset of the codomain $\mathbb{R}$ (e.g., $2 \in \mathbb{R}$ has no pre-image).
• 5 Marks Q17. Check the injectivity (one-to-one) and surjectivity (onto) of the function $f: \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = x^2$. What happens if the domain and codomain are changed to $\mathbb{N}$?
  Answer & Solution:
  * **For $f: \mathbb{Z} \to \mathbb{Z}$:** Neither one-one ($f(1) = f(-1) = 1$) nor onto (negative integers have no pre-images).
  * **For $f: \mathbb{N} \to \mathbb{N}$:** * **Injective:** $x_1^2 = x_2^2 \implies x_1 = x_2$ for natural numbers. (One-one)
  * **Not surjective:** Elements like $2 \in \mathbb{N}$ have no integer square root in $\mathbb{N}$. (Not onto)
• 5 Marks Q18. Let $A = \mathbb{R} - \{2\}$ and $B = \mathbb{R} - \{1\}$. If $f: A \to B$ is a function defined by $f(x) = \frac{x - 1}{x - 2}$, show that $f$ is one-to-one and onto.
  Answer & Solution:
  * **One-to-one:** $\frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2} \implies (x_1-1)(x_2-2) = (x_2-1)(x_1-2) \implies x_1 = x_2$. (One-one)
  * **Onto:** Let $y \in B$ ($y \neq 1$). $y = \frac{x - 1}{x - 2} \implies x = \frac{2y - 1}{y - 1}$. Since $y \neq 1$, $x \in A$ ($x \neq 2$) exists such that $f(x) = y$. (Onto)
• 5 Marks Q19. Show that a function $f: \mathbb{R} \to \mathbb{R}$ defined as $f(x) = ax + b$, where $a, b \in \mathbb{R}$ and $a \neq 0$, is a bijective function.
  Answer & Solution:
  * **Injective:** $ax_1 + b = ax_2 + b \implies ax_1 = ax_2 \implies x_1 = x_2$ (since $a \neq 0$).
  * **Surjective:** For any $y \in \mathbb{R}$, $x = \frac{y - b}{a}$ is a real number satisfying $f(x) = y$.
  Therefore, $f$ is **bijective**.
• 5 Marks Q20. Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = x^4$. Choose the correct answer with full mathematical justification: (A) $f$ is one-to-one onto    (B) $f$ is many-one onto
  (C) $f$ is one-to-one but not onto    (D) $f$ is neither one-to-one nor onto
  Answer & Solution:
  Correct Option: (D) $f$ is neither one-to-one nor onto
  * **Justification for One-to-one:** $f(1) = 1^4 = 1$ and $f(-1) = (-1)^4 = 1$, but $1 \neq -1$, so it is **not one-to-one** (many-one).
  * **Justification for Onto:** The range of $f(x) = x^4$ is $[0, \infty)$, which does not equal the codomain $\mathbb{R}$ (negative real numbers have no pre-images), so it is **not onto**.

SECTION F — Assertion and Reasoning Type Questions [1 Marks Each]

Directions: Each of the following questions consists of two statements, namely, Assertion (A) and Reason (R). Select the correct option from the choices given below:

• (a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
• (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
• (c) Assertion (A) is true, but Reason (R) is false.
• (d) Assertion (A) is false, but Reason (R) is true.

• Q1. Assertion (A): Let $T$ be the set of all triangles in a plane with $R$ being a relation given by $R = \{(T_1, T_2) : T_1 \text{ is congruent to } T_2\}$. Then $R$ is an equivalence relation.
  Reason (R): A relation $R$ is an equivalence relation if it is reflexive, symmetric, and transitive.
  Answer: (a)
  Explanation: Every triangle is congruent to itself (reflexive), if $T_1$ is congruent to $T_2$ then $T_2$ is congruent to $T_1$ (symmetric), and if $T_1 \cong T_2$ and $T_2 \cong T_3$ then $T_1 \cong T_3$ (transitive). Since it satisfies all three properties, it is an equivalence relation, and Reason (R) correctly defines it.
• Q2. Assertion (A): If set $A$ has $3$ elements and set $B$ has $2$ elements, the total number of relations from $A$ to $B$ is $64$.
  Reason (R): The number of relations from a set $A$ to a set $B$ is given by $2^{mn}$, where $n(A) = m$ and $n(B) = n$.
  Answer: (a)
  Explanation: Total number of elements in $A \times B$ is $3 \times 2 = 6$. The number of subsets of $A \times B$ (which represent relations) is $2^6 = 64$. Reason (R) provides the correct formula.
• Q3. Assertion (A): The relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2)\}$ on set $A = \{1, 2, 3\}$ is reflexive.
  Reason (R): A relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for every $a \in A$.
  Answer: (a)
  Explanation: Since $(1, 1)$, $(2, 2)$, and $(3, 3)$ all belong to $R$ for every element in $A = \{1, 2, 3\}$, the relation is reflexive, and Reason (R) correctly defines reflexivity.
• Q4. Assertion (A): The relation $R = \{(1, 2), (2, 1)\}$ on the set $A = \{1, 2, 3\}$ is symmetric.
  Reason (R): A relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.
  Answer: (a)
  Explanation: For every pair present in $R$ (i.e., $(1, 2)$ and $(2, 1)$), their inverse pairs ($(2, 1)$ and $(1, 2)$) are also in $R$. Thus, it is symmetric, and Reason (R) correctly states the definition.
• Q5. Assertion (A): The relation $R = \{(1, 2), (2, 3), (1, 3)\}$ on set $A = \{1, 2, 3\}$ is transitive.
  Reason (R): A relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
  Answer: (a)
  Explanation: Since $(1, 2) \in R$ and $(2, 3) \in R$, we look for $(1, 3)$, which is indeed in $R$. Thus, the relation is transitive, and Reason (R) provides the correct definition.
• Q6. Assertion (A): The function $f : R \to R$ defined by $f(x) = 2x$ is one-one (injective).
  Reason (R): A function $f : X \to Y$ is one-one if $f(a) = f(b) \implies a = b$ for all $a, b \in X$.
  Answer: (a)
  Explanation: If $f(a) = f(b)$, then $2a = 2b \implies a = b$. Therefore, $f$ is one-one, and Reason (R) gives the correct definition.
• Q7. Assertion (A): The function $f : R \to R$ defined by $f(x) = 2x$ is onto (surjective).
  Reason (R): A function $f : X \to Y$ is onto if for every $y \in Y$, there exists an element $x \in X$ such that $f(x) = y$.
  Answer: (a)
  Explanation: For any real number $y \in R$ (codomain), we can find $x = \frac{y}{2} \in R$ such that $f(x) = 2\left(\frac{y}{2}\right) = y$. Thus, $f$ is onto, and Reason (R) correctly explains onto functions.
• Q8. Assertion (A): The function $f : R \to R$ defined by $f(x) = 3x - 5$ is bijective.
  Reason (R): A function is bijective if it is both injective and surjective.
  Answer: (a)
  Explanation: $f(x) = 3x - 5$ is both one-one and onto (linear function with non-zero slope over real numbers). Therefore, it is bijective, and Reason (R) is the correct definition.
• Q9. Assertion (A): The Greatest Integer Function $f : R \to R$ given by $f(x) = [x]$ is neither one-one nor onto.
  Reason (R): $f(1.2) = f(1.5) = 1$, but $1.2 \neq 1.5$.
  Answer: (a)
  Explanation: Since different inputs map to the same output (e.g., $f(1.2) = f(1.5) = 1$), it is not one-one. The range consists only of integers, so non-integers in the codomain have no pre-image, making it not onto. Reason (R) correctly proves it is not one-one.
• Q10. Assertion (A): The modulus function $f : R \to R$ given by $f(x) = |x|$ is neither one-one nor onto.
  Reason (R): $f(-2) = f(2) = 2$, but $-2 \neq 2$.
  Answer: (a)
  Explanation: Reason (R) proves it is not one-one because $f(-2) = f(2)$. It is also not onto because negative real numbers in the codomain have no pre-images (as absolute values are never negative).
• Q11. Assertion (A): The function $f : N \to N$ defined by $f(x) = x^2$ is one-one.
  Reason (R): For $a, b \in N$, $f(a) = f(b) \implies a^2 = b^2 \implies a = b$ (since natural numbers are positive).
  Answer: (a)
  Explanation: Since the domain consists only of natural numbers (positive integers), $a^2 = b^2$ strictly implies $a = b$. Thus, the function is one-one, and Reason (R) is the correct explanation.
• Q12. Assertion (A): The function $f : Z \to Z$ defined by $f(x) = x^2$ is not one-one.
  Reason (R): $f(-3) = f(3) = 9$, but $-3 \neq 3$.
  Answer: (a)
  Explanation: Because both $-3$ and $3$ map to the same value ($9$), the function fails the one-one condition, and Reason (R) provides the exact counterexample justifying this.
• Q13. Assertion (A): Let $A = \{1, 2, 3\}$. The empty relation $R = \emptyset$ on $A$ is transitive.
  Reason (R): An empty relation vacuously satisfies the condition for transitivity.
  Answer: (a)
  Explanation: The condition for transitivity is "if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$". Since no such pairs exist in an empty relation, the condition is vacuously true, making it transitive. Reason (R) explains this correctly.
• Q14. Assertion (A): The identity relation $I_A$ on any set $A$ is an equivalence relation.
  Reason (R): The identity relation is always reflexive, symmetric, and transitive.
  Answer: (a)
  Explanation: The identity relation contains only elements of the form $(a, a)$, satisfying reflexivity, symmetry, and transitivity automatically. Reason (R) correctly states why it is an equivalence relation.
• Q15. Assertion (A): Let $f : A \to B$ and $g : B \to C$ be two bijective functions. Then the composite function $g \circ f : A \to C$ is also bijective.
  Reason (R): The composition of two one-one functions is one-one, and the composition of two onto functions is onto.
  Answer: (a)
  Explanation: Since $f$ and $g$ are bijective, they are both one-one and onto. By standard composition properties, the composition of one-one functions is one-one and onto functions is onto, making $g \circ f$ bijective. Reason (R) explains this correctly.
• Q16. Assertion (A): The relation $R$ in the set of integers $Z$ defined by $R = \{(a, b) : 2 \text{ divides } a - b\}$ is an equivalence relation.
  Reason (R): Reflexivity, symmetry, and transitivity hold true for divisibility properties.
  Answer: (a)
  Explanation: $a - a = 0$ is divisible by 2 (reflexive); if $a - b$ is divisible by 2, $b - a$ is also divisible by 2 (symmetric); and transitive property holds similarly. Reason (R) correctly attributes this to divisibility properties.
• Q17. Assertion (A): Let $A = \{1, 2, 3\}$. The number of reflexive relations on set $A$ is $2^6$.
  Reason (R): For a set of $n$ elements, the number of reflexive relations is given by $2^{n(n-1)}$.
  Answer: (a)
  Explanation: For $n = 3$, $n(n-1) = 3(2) = 6$, so the number of reflexive relations is $2^6$. Reason (R) provides the exact correct formula.
• Q18. Assertion (A): The relation $R = \{(1, 3), (3, 1), (2, 2)\}$ on set $A = \{1, 2, 3\}$ is symmetric.
  Reason (R): For a relation to be symmetric, if $(a, b) \in R$, then $(b, a)$ must also belong to $R$.
  Answer: (a)
  Explanation: The pairs $(1, 3)$ and $(3, 1)$ have their reverse pairs in $R$, and $(2, 2)$ is its own reverse. Thus, it is symmetric, and Reason (R) is the correct definition.
• Q19. Assertion (A): Let $f : R \to R$ be defined by $f(x) = x^3$. Then $f$ is a bijective function.
  Reason (R): A cubic function from $R$ to $R$ is strictly increasing, making it both one-one and onto.
  Answer: (a)
  Explanation: The derivative $f'(x) = 3x^2 \ge 0$, making it strictly increasing. A strictly increasing continuous function from $R$ to $R$ is both injective and surjective (bijective). Reason (R) correctly explains this.
• Q20. Assertion (A): If $f : A \to B$ is a function, then the domain of $f$ is set $B$ and the codomain is set $A$.
  Reason (R): A function maps elements from its domain to its codomain.
  Answer: (d)
  Explanation: Assertion (A) is false because for a function $f : A \to B$, the domain is set $A$ and the codomain is set $B$. However, Reason (R) is a true statement regarding how functions map elements.

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