Week 1 - Graded Assignment 1

Course: Jan 2026 - Mathematics I

Week 1 - Graded Assignment 1

Question 1

                              

Define a relation on the set of 525 ponds such that two ponds are related if both are polluted by fertilisers and pharmaceutical products. Which of the following is/are true?

• Relation is reflexive.
• Relation is transitive.
• Relation is symmetric.
• This is an equivalence relation.

Solution

[!solution]+ Abstract Solution (Strategy)

1. [Relation Property Logic]: Let $S$ be the subset of ponds polluted by $\{F, Ph\}$. A pond $x$ relates to $y$ ($xRy$) if and only if $x \in S$ and $y \in S$.
2. [Symmetry]: Does $xRy \implies yRx$? Yes, because the condition $x, y \in S$ is commutative.
3. [Transitivity]: If $x, y \in S$ and $y, z \in S$, then $x, z \in S$ holds inherently by set membership.
4. [Reflexivity Constraint]: For reflexivity, every pond in the universe $U$ (525 ponds) must be in $S$.

Procedure

• Step 1: Identify that not all 525 ponds are in the set $S$ ($F \cap Ph$). Thus, if pond $k \notin S$, then $(k, k) \notin R$. Reflexivity fails.
• Step 2: If Pond A and Pond B are in the polluted list, then Pond B and Pond A are in the polluted list. Symmetry holds.
• Step 3: If A & B are polluted, and B & C are polluted, then A & C are polluted. Transitivity holds.
• Step 4: Since it is not reflexive, it is not an equivalence relation.
• Result: Transitive and Symmetric.

Question 2

• $R$ is a transitive relation.
• $R$ is a function.
• $R$ is not an equivalence relation.
• $R$ is a reflexive relation.
• $R$ is a symmetric relation.

Solution

[!solution]+ Abstract Solution (Strategy)

1. [Simplification]: The condition $x - y = 0$ simplifies to $y = x$.
2. [Axiom Testing]: Test the identity rule $x=y$ against the definitions of Reflexivity, Symmetry, and Transitivity.
3. [Function Mapping]: For each $x$, determine if there is exactly one $y$.

Procedure

• Step 1 (Reflexive): $x = x$ is true for all $x \in \mathbb{R}$.
• Step 2 (Symmetric): If $x = y$, then $y = x$.
• Step 3 (Transitive): If $x = y$ and $y = z$, then $x = z$.
• Step 4 (Function): The relation $y=x$ assigns a unique output for every input.
• Result: Reflexive, Symmetric, Transitive (Equivalence) and Function.

Question 3

• $R_1 = \{(x, y) \mid x, y \in \mathbb{R}, x+y>2 \}$
• $R_2 = \{(x, y) \mid x, y \in \mathbb{R}, x>y \}$
• $R_3 = \{(x,y) \mid x,y \in \mathbb{R},\ x + y = 12\}$
• $R_4 = \{(x, y) \mid x, y \in \mathbb{R}, y=x^2 \}$

Solution

[!solution]+ Abstract Solution (Strategy)

1. [Eliminate Non-Functions]: Relations involving inequalities ($>, <$) assign multiple $y$ values to a single $x$. These are not functions.
2. [Injectivity Test]: A function $y=f(x)$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$.

Procedure

• Step 1: $R_1, R_2$ are inequalities. For a fixed $x$, multiple $y$'s satisfy the relation. Fails function test.
• Step 2: Evaluate $R_4$: $y=x^2$. If $y=4$, $x$ can be $2$ or $-2$. Not one-one.
• Step 3: Evaluate $R_3$: $y = 12 - x$. This is a linear function; every $x$ yields a unique $y$, and every $y$ yields a unique $x$.
• Result: $R_3$ is a one-one function.

Question 4

• Function $f: \mathbb{N} \to \mathbb{R}$ such that $f(x) = x^2$ is not onto.
• Relation $R = \{(1,1),(1,2), (3,1) \}$ on a set $A = \{1, 2, 3\}$ is a function.
• Function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x) = |x|$ is not one-one.
• Function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x) = (x-1)^2$ is one- one and onto.

Solution

[!solution]+ Abstract Solution (Strategy)

1. [Onto Test]: Does the set of outputs (Range) cover the entire target set (Codomain)?
2. [One-one Test]: Does different inputs produce different outputs?

Procedure

• Step 1: $f(x)=x^2$ on $\mathbb{N} \to \mathbb{R}$. The range is $\{1, 4, 9, \dots\}$ which does not include negative numbers or fractions in $\mathbb{R}$. Not onto (True).
• Step 2: $R$ maps $1 \to 1$ and $1 \to 2$. Fails function test (unique image).
• Step 3: $f(x)=|x|$. $f(1) = 1$ and $f(-1) = 1$. Fails one-one test. Not one-one (True).
• Step 4: $(x-1)^2$ fails both (similar to $x^2$).
• Result: Options 1 and 3 are correct.

Question 5

Solution

Abstract Solution (Strategy)

1. [Calculate Integer Overlap]: Find Natural numbers within the bounds of $B$.
2. [Execute Subtraction]: Remove any Natural numbers falling within the range specified by $C$.

Procedure

• Step 1: $A \cap B = \{0, 1, 2, \dots, 104\}$.
• Step 2: $C$ contains all rational numbers in $(10, 80]$. This includes the integers $\{11, 12, \dots, 80\}$.
• Step 3: Remove those integers from the set.
• Step 4: Integers in $[0, 10]$ = 11. Integers in $[81, 104]$ = $104 - 81 + 1 = 24$.
• Step 5: Total remaining = $11 + 24 = 35$.
• Result: 35

Question 6

Solution

Abstract Solution (Strategy)

1. [Principle of Inclusion-Exclusion]: For three overlapping sets $F, P, Ph$, the total union is given by: $n(F \cup P \cup Ph) = \sum n(S_i) - \sum n(S_i \cap S_j) + n(F \cap P \cap Ph)$
2. [Variable Isolation]: Solve for the unknown intersection $x = n(F \cap P \cap Ph)$.

Procedure

• Step 1: Identify Union $n(U) = 525$.
• Step 2: Individual Counts: $n(F)=230, n(P)=245, n(Ph)=257$.
• Step 3: Intersection Pairs: $n(F \cap P)=100, n(F \cap Ph)=82, n(P \cap Ph)=77$.
• Step 4: Substitute: $525 = (230 + 245 + 257) - (100 + 82 + 77) + x$ $525 = 732 - 259 + x$ $525 = 473 + x$
• Step 5**: $x = 525 - 473 = 52$.
• Result: 52

Question 7

• $f: P \to Q$ is a function, where $P = \{\text{Yashubh, Navrtna, Rathi, Rakesh, Mahesh}\}$ and $Q=\{\text{ Shubh, Rabi, Mahendra, Rajat}\}$.
• If $f: P \to Q$ is a function, where $P=\{\text{ Yashubh, Navrtna, Rathi, Rakesh}\}$ and $Q=\{\text{Shubh, Rabi, Mahendra, Rajat}\}$, then $f$ is one-one.
• If $f: P \to Q$ is a function, where $P=\{\text{Yashubh, Navrtna, Rathi, Rakesh}\}$ and $Q=\{ \text{ Shubh, Rabi}\}$, then $f$ is onto.
• If $f: P \to Q$ is a function, where $P=\{\text{Yashubh, Rathi}\}$ and $Q=\{\text{Shubh, Rabi}\}$, then $f$ is bijective.

Solution

Abstract Solution (Strategy)

1. [Functional Mapping]: A relation is a function if every element in the domain maps to exactly one element in the codomain.
2. [Injectivity (One-One)]: Does each son in $P$ have a unique father in $Q$? (In a standard tree, Yes).
3. [Surjectivity (Onto)]: Is every father in $Q$ linked to at least one son in $P$?

Procedure

• Step 1: Test $P=\{\text{... Mahesh}\}$. Mahesh has no father in the set $Q$. Thus, he has no image. Not a function.
• Step 2: If $P$ is the set of 4 grandsons, each has exactly one parent in $Q$. Since no two grandsons in $P$ share the same father across the different families mapping to $Q$, we check: Yashubh & Navrtna map to Shubh. Rathi & Rakesh map to Rabi.
• Step 3 (Onto): If $Q=\{\text{Shubh, Rabi}\}$, both are mapped to by the grandsons. Onto holds.
• Step 4 (Bijection): If $P=\{\text{Yashubh, Rathi}\}$ and $Q=\{\text{Shubh, Rabi}\}$, then Yashubh $\to$ Shubh and Rathi $\to$ Rabi. This is a one-to-one correspondence. Bijective holds.
• Result: Options 2, 3, and 4 are correct.

Question 8

• $f$ is neither one to one nor onto.
• $f$ is one to one but not onto.
• $f$ is onto but not one to one.
• $f$ is bijective.

Solution

Abstract Solution (Strategy)

1. [Direct Comparison]: Compare the cardinality of the domain and the range.
2. [Correspondence Check]: Determine if each material maps to a unique, distinct value.

Procedure

• Step 1: Count elements in Material set ($n=6$) and Value set ($n=6$).
• Step 2: Each material is assigned exactly one constant from the table.
• Step 3: No dielectric constant repeats for different materials in this specific set.
• Step 4: Since $f$ is both injective and surjective ($Range = Codomain$), it is bijective.
• Result: Bijective.

Question 9

• $A = \{x \in \mathbb{N} \mid x \equiv 0 \pmod 2, 1 \leq x \leq 10\}$
• $B = \{x \in \mathbb{N} \mid x \equiv 0 \pmod 5, 6 \leq x \leq 25\}$
• $C = \{x \in \mathbb{N} \mid x \equiv 0 \pmod 7, 7 \leq x \leq 29\}$

Solution

Abstract Solution (Strategy)

1. [Enumerate Sets]: Explicitly list all elements in $A, B, \text{ and } C$.
2. [Exclusive Regions]: Calculate the "Only A", "Only B", and "Only C" parts of the Venn diagram and sum their sizes.

Procedure

• Step 1: $A = \{2, 4, 6, 8, 10\}$.
• Step 2: $B = \{10, 15, 20, 25\}$.
• Step 3: $C = \{7, 14, 21, 28\}$.
• Step 4: Find overlaps: $A \cap B = \{10\}$, $A \cap C = \emptyset$, $B \cap C = \emptyset$.
• Step 5:
  • $A \setminus (B \cup C) = \{2, 4, 6, 8\}$ (Count = 4)
  • $B \setminus (C \cup A) = \{15, 20, 25\}$ (Count = 3)
  • $C \setminus (B \cup A) = \{7, 14, 21, 28\}$ (Count = 4)
• Step 6: Total = $4 + 3 + 4 = 11$.
• Result: 11

Question 10

• Dabangg ($D$): 95
• Avatar ($A$): 100
• RRR ($R$): 100
• $D \cap A = 50$, $A \cap R = 40$, $D \cap R = 55$. Find the number of people who watched only RRR and Avatar.

Solution

Abstract Solution (Strategy)

1. [Triple Intersection]: Solve for $n(D \cap A \cap R)$ using the PIE for the total union $n(D \cup A \cup R)$.
2. [Exclude shared regions]: The "Only RRR and Avatar" region is $n(A \cap R) - n(D \cap A \cap R)$.

Procedure

• Step 1: Union formula: $180 = (95+100+100) - (50+40+55) + x$.
• Step 2: $180 = 295 - 145 + x \implies 180 = 150 + x \implies x = 30$.
• Step 3: Intersection of Avatar and RRR is 40.
• Step 4: "Only" intersection = $40 - 30 = 10$.
• Result: 10

Question 11

• $R$ is an equivalence relation.
• (Rathi, Navrtna) $\in R$
• $S$ is an equivalence relation.
• (Mahesh, Rajat) $\in S$.
• (Yashubh, Rathi) $\in R$ and (Rathi, Yashubh) $\in R$.
• $S$ is symmetric relation.

Solution

Abstract Solution (Strategy)

1. [Membership Test]: Check specific relations against the provided tree.
2. [Property Verification]: Test $R$ and $S$ for Reflexivity/Symmetry/Transitivity.

Procedure

• Step 1: $R$ (Cousins) is not reflexive (A is not a cousin of itself). Not equivalence.
• Step 2: Rathi's parent is Rabi. Navrtna's parent is Shubh. Rabi and Shubh are siblings. Cousins holds.
• Step 3: (Mahesh, Rajat) is Father-Son, not Son-Father. Incorrect.
• Step 4: Cousin relation is inherently symmetric. If Yashubh is Rathi's cousin, Rathi is Yashubh's.
• Step 5: $S$ is not symmetric (if A is son of B, B is father of A).
• Result: Options 2 and 5 are correct.

Question 12

Solution

Abstract Solution (Strategy)

1. [Identify Pairs in S]: Count all (child, parent) edges.
2. [Identify Pairs in R]: Count distinct (cousin A, cousin B) pairs.

Procedure

• Step 1 (n = |S|):
  • Mahesh has 4 sons (Shubh, Rabi, Mahendra, Rajat). (4 pairs)
  • Shubh has 2 sons. (2 pairs)
  • Rabi has 2 sons. (2 pairs)
  • Total $n = 4 + 2 + 2 = 8$.
• Step 2 (m = |R|):
  • Cousins exist between the 4 grandsons.
  • Yashubh/Navrtna (Shubh's sons) are cousins to Rathi/Rakesh (Rabi's sons).
  • Each grandson in Shubh's family has 2 cousins in Rabi's. (2 sons $\times$ 2 cousins = 4 pairs).
  • Symmetry doubles this (A $\to$ B and B $\to$ A). Total $m = 4 \times 2 = 8$.
• Step 3: $m + n = 8 + 8 = 16$.
• Result: 16

Question 13

• $f$ is one to one but not onto
• $f$ is neither one to one nor onto.
• $f$ is onto but not one to one.
• $f$ is a bijective function.

Solution

Abstract Solution (Strategy)

1. [Surjectivity]: For any integer $k$, can we find $p, q$ such that $p - q = k$?
2. [Injectivity]: Do different fractions map to the same integer difference?

Procedure

• Step 1 (Onto): Let $k \in \mathbb{Z}$. Choose $q = 1, p = k + 1$. Then $f((k+1)/1) = (k+1) - 1 = k$. Since every integer $k$ has a preimage, it is Onto.
• Step 2 (One-one): Consider $f(3/2) = 3 - 2 = 1$. Consider $f(5/4) = 5 - 4 = 1$.
• Step 3: Since $3/2 \neq 5/4$ but $f(3/2) = f(5/4)$, the function is not one-to-one.
• Result: Onto but not one-to-one.

Question 14

• $a\rightarrow iii$, $b\rightarrow iv$, $c\rightarrow i$, $d\rightarrow ii$
• $a\rightarrow iii$, $b\rightarrow i$, $c\rightarrow iv$, $d\rightarrow ii$
• $a\rightarrow iii$, $b\rightarrow i$, $c\rightarrow ii$, $d\rightarrow iv$

Solution

Abstract Solution (Strategy)

1. [Direct Evaluation]: Compute the elements for each set identity.
2. [Matching]: Correlate computed sets with provided lists.

Procedure

• Step 1: $A \setminus B$ (In A, not in B) = $\{3, 4\}$. (Mates with $iii$)
• Step 2: $A \cap B$ (Common) = $\{5, 6, 7, 8\}$. (Mates with $i$)
• Step 3: $A \cup B$ (All) = $\{3, 4, 5, 6, 7, 8, 9, 10\}$. (Mates with $ii$)
• Step 4: $B \setminus A$ (In B, not in A) = $\{9, 10\}$. (Mates with $iv$)
• Result: Option 3.

Question 15

• $D_1 \cup D_2$
• $D_1 \setminus D_2$
• $D_2 \setminus D_1$
• $D_1 \cap D_2$

Solution

Abstract Solution (Strategy)

1. [Definition Search]: For a sum $h(x) = f(x) + g(x)$ to be defined, $x$ must satisfy the constraints of both functions simultaneously.

Procedure

• Step 1: If $x \notin D_1$, $f_1(x)$ is undefined.
• Step 2: If $x \notin D_2$, $f_2(x)$ is undefined.
• Step 3: Therefore, $x$ must belong to the set of values present in both $D_1$ and $D_2$.
• Result: Intersection $D_1 \cap D_2$.

Question 16

• $f$ is one to one but not onto
• $f$ is neither one to one nor onto.
• $g$ is onto but not one to one.
• $g$ is a bijective function.

Solution

Abstract Solution (Strategy)

1. [Range Verification]: Check if the codomain ($1, 2, 3...$) is fully covered by the range.
2. [Inverse Check]: If $g$ maps back uniquely and covers $1$, test bijectivity.

Procedure

• Step 1 (f): $f(x)=x+1$. The range is $\{2, 3, 4...\}$. Since $1$ is never produced, it is not onto.
• Step 2 (f): Distinct inputs always yield distinct successors. One-to-one.
• Step 3 (g): $g(1)=1, g(2)=1, g(3)=2...$ wait. $g(x) = f(x)-1$ for $x>1$.
• Wait! Let's re-read g: $g(1)=1, g(2)=2, g(3)=3$. $g(x) = x$ essentially.
• Step 4: If $g(x)$ maps $1 \to 1$ and $x \to x$, it is the identity function. Identity is bijective.
• Result: Options 1 and 4.

Question 17

• $R_1$ is reflexive but not symmetric.
• $R_2$ is both reflexive and symmetric.
• $R_1$ is neither symmetric nor transitive.
• $R_2$ is both reflexive and transitive.

Solution

Abstract Solution (Strategy)

1. [Property Definitions]:
   • Reflexive: $aRa$
   • Symmetric: $aRb \implies bRa$
   • Transitive: $aRb \text{ and } bRc \implies aRc$

Procedure

• Step 1 ($R_1$): $(a, a+1)$. Not reflexive ($a \neq a+1$). Not symmetric ($2=1+1$ but $1 \neq 2+1$). Not transitive ($1, 2$ and $2, 3$ doesn't imply $1, 3$).
• Step 2 ($R_2$): $b \ge a$. $a \ge a$ is true (Reflexive). $b \ge a$ doesn't imply $a \ge b$ (Not symmetric). $c \ge b$ and $b \ge a \implies c \ge a$ (Transitive).
• Result: Options 3 and 4.

Question 18

• $f$ is one to one but not onto.
• $f$ is neither one to one nor onto.
• $f$ is onto but not one to one
• $f$ is a bijective function.

Solution

Abstract Solution (Strategy)

1. [Range Check]: Compare the range of $f(n)$ (multiples of 6) with the codomain $A$ (multiples of 3).

Procedure

• Step 1 (One-one): $6n_1 = 6n_2 \implies n_1 = n_2$. One-to-one.
• Step 2 (Onto): Range = $\{6, 12, 18, \dots\}$.
• Step 3: Codomain $A = \{3, 6, 9, 12, \dots\}$.
• Step 4: Elements like $3, 9, 15$ are in $A$ but are not in the range ($6n$).
• Step 5: Since the range does not equal the codomain, it is not onto.
• Result: One-to-one but not onto.