Class 11 : Maths (In English) – Lesson 2. Relations and Functions

EXPLANATION & SUMMARY

🔷 Explanation
🔵 1️⃣ Introduction
🧠 Relations and Functions form the foundation of higher mathematics.
They describe how elements of one set are connected to elements of another.
You will learn how to define, classify, and operate on these relations and functions.
💡 Concept:
A relation shows association between elements of two sets.
A function is a special relation where each input has exactly one output.

🟢 2️⃣ Cartesian Product of Sets
📘 Definition:
If A and B are two sets, the Cartesian product A × B = {(a, b) | a ∈ A, b ∈ B}
📊 Example:
Let A = {1, 2}, B = {x, y}
➡️ A × B = {(1, x), (1, y), (2, x), (2, y)}
🧾 Cardinality: |A × B| = |A| × |B|
✏️ Note:
Order matters → (a, b) ≠ (b, a)

🔴 3️⃣ Relations
📘 Definition:
A relation R from A to B is a subset of A × B.
If (a, b) ∈ R, we say a is related to b (a R b).
➡️ Domain = all first elements related
➡️ Range = all second elements that appear
➡️ Co-domain = entire second set B
📊 Example:
A = {1, 2, 3}, B = {4, 5, 6}
R = {(1, 4), (2, 5)}
Domain = {1, 2}, Range = {4, 5}

🟡 4️⃣ Relations on a Single Set
If A = B, relation is on A.
🧩 Properties:
🔵 Reflexive: (a, a) ∈ R ∀ a ∈ A
🟢 Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R
🟠 Transitive: (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R
💡 Equivalence Relation:
A relation that is reflexive, symmetric, and transitive.
It divides A into equivalence classes (disjoint subsets).
🧾 Example:
On integers ℤ, define a R b if a − b divisible by 3
✔️ Reflexive: a − a = 0 divisible by 3
✔️ Symmetric: if a − b divisible by 3, b − a = −(a − b) also
✔️ Transitive: if a − b and b − c divisible by 3, then a − c divisible by 3
✅ Hence, equivalence relation.

🔵 5️⃣ Functions
📘 Definition:
A function f from A to B is a relation with:
✔️ Every a ∈ A is related to exactly one b ∈ B.
🧠 Notation: f: A → B, f(a) = b
💡 Key Terms:
➡️ Domain: A
➡️ Co-domain: B
➡️ Range: {f(a) | a ∈ A}
✏️ Note:
A function must be single-valued (only one output per input).

🟢 6️⃣ Representation of Functions
🧾 Forms:
1️⃣ Arrow diagram
2️⃣ Set builder
3️⃣ Roster form
4️⃣ Graphical form (if domain & range ⊆ ℝ)
📊 Example:
f = {(1, 2), (2, 4), (3, 6)}
Domain = {1, 2, 3}, Range = {2, 4, 6}

🔴 7️⃣ Types of Functions
🧩 By mapping:
🔵 One-one (Injective): f(a₁) = f(a₂) ⇒ a₁ = a₂
🟢 Onto (Surjective): Range = Co-domain
🟠 Bijective: Both one-one and onto
🔴 Many-one: Different inputs → same output
🟡 Into: Range ⊂ Co-domain
🧭 Constant: f(a) = k for all a ∈ A
📊 Example:
f(x) = 2x + 3, domain ℝ
✔️ One-one
✔️ Onto if Co-domain = ℝ
✅ Hence, Bijective

🟣 8️⃣ Composition of Functions
📘 Definition:
If f: A → B, g: B → C
Then (g ∘ f): A → C, (g ∘ f)(x) = g(f(x))
🧠 Properties:
✔️ Associative: h ∘ (g ∘ f) = (h ∘ g) ∘ f
❌ Not Commutative: g ∘ f ≠ f ∘ g (generally)
📊 Example:
f(x) = x + 1, g(x) = 2x
(g ∘ f)(x) = g(f(x)) = 2(x + 1) = 2x + 2
(f ∘ g)(x) = f(g(x)) = (2x) + 1 = 2x + 1
⚠️ Not equal → not commutative

🔵 9️⃣ Inverse of a Function
🧠 Definition:
If f: A → B is bijective, then there exists f⁻¹: B → A such that
➡️ f⁻¹(f(a)) = a
➡️ f(f⁻¹(b)) = b
📊 Example:
f(x) = 2x + 3
To find f⁻¹:
y = 2x + 3
x = (y − 3)/2
Hence f⁻¹(y) = (y − 3)/2
💡 Condition: Only bijective functions have inverses.

🟢 🔟 Examples
🧮 Example 1: One-one Check
f(x) = 3x − 5
Assume f(x₁) = f(x₂)
3x₁ − 5 = 3x₂ − 5
⇒ x₁ = x₂
✅ One-one
🧮 Example 2: Onto Check
f(x) = 2x + 1, Co-domain ℝ
Take any y ∈ ℝ
Solve y = 2x + 1
⇒ x = (y − 1)/2 ∈ ℝ
✅ Onto
✅ f is bijective ⇒ f⁻¹(y) = (y − 1)/2

🟣 11️⃣ Special Real Functions
🔵 Identity: f(x) = x
🟢 Constant: f(x) = k
🟠 Modulus: f(x) = |x|
🔴 Greatest Integer: f(x) = ⌊x⌋
🟡 Signum: sgn(x) = {1, 0, −1} for x > 0, = 0, < 0

🟢 12️⃣ Common Misconceptions
⚠️ Every relation is a function ❌
➡️ Fix: Function must have exactly one output per input.
⚠️ Range = Co-domain always ❌
➡️ Fix: Range ⊆ Co-domain, equality only for onto.
⚠️ √(x²) = x ❌
➡️ Fix: √(x²) = |x|

🟡 13️⃣ Strategy for Exercises
🔹 Use definitions directly for reflexive/symmetric/transitive.
🔹 For one-one → assume f(a₁)=f(a₂) → prove a₁=a₂.
🔹 For onto → take arbitrary y → find x.
🔹 For inverse → solve y = f(x) for x.

📘 Summary (≈300 words)
🔵 A relation is any subset of A × B showing association between elements.
🟢 A function is a relation with the rule of single output per input.
🟠 A function f: A → B assigns each a ∈ A to exactly one b ∈ B.
🟡 Domain, Co-domain, and Range are three key sets associated.
🔴 Relations on a single set can be reflexive, symmetric, transitive.
If all three hold → equivalence relation → partitions the set into classes.
🟢 Function types:
One-one (injective): unique outputs
Onto (surjective): all co-domain covered
Bijective: both (invertible)
🟣 Composition: (g ∘ f)(x) = g(f(x))
Associative, but not commutative.
🟢 Inverse Function: Exists only for bijective functions.
🔵 Special real functions: Identity, Constant, Modulus, Greatest Integer, Signum.
✔️ Key rule: √(x²) = |x|
✔️ Range ⊆ Co-domain
✔️ f⁻¹ exists only if f is bijective.

📝 Quick Recap
🔵 Relation = subset of A × B
🟢 Function = special relation with one output per input
🟠 Equivalence = reflexive + symmetric + transitive
🟡 Bijective ⇒ Inverse exists
🔴 Composition associative, not commutative
🟣 Range may differ from Co-domain

————————————————————————————————————————————————————————————————————————————

QUESTIONS FROM TEXTBOOK

🧾 Exercise 2.1

🔵 Question 1:
If (x/3 + 1, y − 2/3) = (5/3, 1/3), find the values of x and y.
🟢 Answer:
➡️ We know that two ordered pairs are equal only if their corresponding components are equal.
So,
✴️ x/3 + 1 = 5/3
✴️ y − 2/3 = 1/3
✔️ Solve for x:
x/3 = 5/3 − 1
x/3 = (5 − 3)/3
x/3 = 2/3
➡️ Multiply both sides by 3
x = 2
✔️ Solve for y:
y = 1/3 + 2/3
y = 3/3
y = 1
💡 Final Answer: x = 2, y = 1

🔵 Question 2:
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in A × B.
🟢 Answer:
✴️ Number of elements in A = n(A) = 3
✴️ Number of elements in B = n(B) = 3
➡️ Formula: n(A × B) = n(A) × n(B)
➡️ n(A × B) = 3 × 3 = 9
💡 Final Answer: n(A × B) = 9

🔵 Question 3:
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
🟢 Answer:
✴️ G × H = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
✴️ H × G = {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}
💡 Final Answer:
✔️ G × H = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
✔️ H × G = {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}

🔵 Question 4:
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
🟢 Answer:
(i) If P = {m, n} and Q = {n, m³}, then P × Q = {(m, n), (n, m), (n, m³)}
➡️ False
✔️ Correct statement:
P × Q = {(m, n), (m, m³), (n, n), (n, m³)}
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
➡️ True
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ ϕ) = ϕ.
➡️ True because B ∩ ϕ = ϕ, and A × ϕ = ϕ

🔵 Question 5:
If A = {−1, 1}, find A × A × A.
🟢 Answer:
✴️ A × A = { (−1, −1), (−1, 1), (1, −1), (1, 1) }
➡️ Now A × A × A = { (a, b, c) | a, b, c ∈ A }
✔️ So,
A × A × A =
{ (−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1),
(1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1) }
💡 Final Answer: 8 elements in A × A × A

🔵 Question 6:
If A × B = {(a, x), (a, y), (b, x), (b, y)}, find A and B.
🟢 Answer:
✴️ From pairs, first elements form set A = {a, b}
✴️ Second elements form set B = {x, y}
💡 Final Answer:
✔️ A = {a, b}
✔️ B = {x, y}

🔵 Question 7:
Let A = {1, 2}, B = {3, 4}, C = {5, 6}, and D = {5, 6, 7, 8}. Verify that:
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C ⊆ B × D
🟢 Answer:
(i)
B ∩ C = ∅ (since no common element)
➡️ A × (B ∩ C) = A × ∅ = ∅
A × B = {(1,3), (1,4), (2,3), (2,4)}
A × C = {(1,5), (1,6), (2,5), (2,6)}
(A × B) ∩ (A × C) = ∅
✔️ Therefore, verified: A × (B ∩ C) = (A × B) ∩ (A × C)
(ii)
A × C = {(1,5), (1,6), (2,5), (2,6)}
B × D = {(3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)}
➡️ All elements of A × C are not present in B × D
⚠️ So A × C ⊈ B × D
💡 Final: (i) True, (ii) False

🔵 Question 8:
Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
🟢 Answer:
A × B = {(1,3), (1,4), (2,3), (2,4)}
➡️ n(A × B) = 4
✔️ Number of subsets = 2⁴ = 16
🧠 List of subsets:
ϕ, {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)},
{(1,3), (1,4)}, {(1,3), (2,3)}, {(1,3), (2,4)}, {(1,4), (2,3)}, {(1,4), (2,4)}, {(2,3), (2,4)},
{(1,3), (1,4), (2,3)}, {(1,3), (1,4), (2,4)}, {(1,4), (2,3), (2,4)}, {(1,3), (2,3), (2,4)},
{(1,3), (1,4), (2,3), (2,4)}
💡 Final Answer: 16 subsets

🔵 Question 9:
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x,1), (y,2), (z,1) ∈ A × B, where x, y, z are distinct elements, find A and B.
🟢 Answer:
From ordered pairs:
Second components → {1, 2} ⇒ B = {1, 2}
First components → {x, y, z} ⇒ A = {x, y, z}
💡 Final Answer:
✔️ A = {x, y, z}
✔️ B = {1, 2}

🔵 Question 10:
The Cartesian product A × A has 9 elements among which (−1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.
🟢 Answer:
➡️ n(A × A) = 9 ⇒ n(A) = 3
Let A = {−1, 0, 1}
Now A × A =
{(−1, −1), (−1, 0), (−1, 1),
(0, −1), (0, 0), (0, 1),
(1, −1), (1, 0), (1, 1)}
💡 Final Answer:
✔️ A = {−1, 0, 1}
✔️ A × A has 9 elements as above

🧾Exercise 2.2

🔵 Question 1
Let A = {1, 2, 3, …, 14}. Define a relation R from A to A by R = {(x, y) : 3x − y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

🟢 Answer
➡️ Condition 3x − y = 0 ⇒ y = 3x.
➡️ x ∈ A and y ∈ A ⇒ 3x ≤ 14 ⇒ x ∈ {1, 2, 3, 4}.
✔️ R (roster) = {(1,3), (2,6), (3,9), (4,12)}.
✔️ Domain(R) = {1, 2, 3, 4}.
✔️ Codomain(R) = A = {1, 2, 3, …, 14}.
✔️ Range(R) = {3, 6, 9, 12}.
💡 Final: Domain {1,2,3,4}; Codomain A; Range {3,6,9,12}.

🔵 Question 2
Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

🟢 Answer
➡️ x < 4 with x ∈ N ⇒ x ∈ {1, 2, 3}.
➡️ Compute y:
• for x = 1 ⇒ y = 6;
• for x = 2 ⇒ y = 7;
• for x = 3 ⇒ y = 8.
✔️ R (roster) = {(1,6), (2,7), (3,8)}.
✔️ Domain(R) = {1, 2, 3}.
✔️ Range(R) = {6, 7, 8}.
💡 Final: Domain {1,2,3}; Range {6,7,8}.

🔵 Question 3
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y) : the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

🟢 Answer
➡️ |x − y| is odd ⇔ x and y have different parity.
➡️ Pairs with x odd (1,3,5) and y even (4,6), plus x even (2) with y odd (9).
✔️ R = {(1,4), (1,6), (3,4), (3,6), (5,4), (5,6), (2,9)}.
💡 Final: R listed above (7 ordered pairs).

🔵 Question 4
The Fig 2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range?

🟢 Answer
➡️ From figure: arrows indicate 5 ↦ 3, 6 ↦ 4, 7 ↦ 5.
(i) Set-builder: R = {(p, q) ∈ P × Q : q = p − 2}.
(ii) Roster: R = {(5,3), (6,4), (7,5)}.
✔️ Domain(R) = {5, 6, 7}.
✔️ Range(R) = {3, 4, 5}.
💡 Final: relation as above; domain {5,6,7}; range {3,4,5}.

🔵 Question 5
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b) : a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R

🟢 Answer
➡️ For a = 1 ⇒ b ∈ {1,2,3,4,6}.
➡️ For a = 2 ⇒ b ∈ {2,4,6}.
➡️ For a = 3 ⇒ b ∈ {3,6}.
➡️ For a = 4 ⇒ b ∈ {4}.
➡️ For a = 6 ⇒ b ∈ {6}.
(i) R = {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}.
(ii) Domain(R) = {1, 2, 3, 4, 6}.
(iii) Range(R) = {1, 2, 3, 4, 6}.
💡 Final: Domain = Range = A.

🔵 Question 6
Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

🟢 Answer
➡️ Domain(R) = {0, 1, 2, 3, 4, 5}.
➡️ Range(R) = {5, 6, 7, 8, 9, 10}.
✔️ R (optional roster) = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}.
💡 Final: Domain {0,1,2,3,4,5}; Range {5,6,7,8,9,10}.

🔵 Question 7
Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form.

🟢 Answer
➡️ Primes less than 10: 2, 3, 5, 7.
➡️ Cubes: 2³ = 8, 3³ = 27, 5³ = 125, 7³ = 343.
✔️ R = {(2,8), (3,27), (5,125), (7,343)}.
💡 Final: R listed above.

🔵 Question 8
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

🟢 Answer
➡️ Number of ordered pairs in A × B = |A| × |B| = 3 × 2 = 6.
➡️ A “relation” is any subset of A × B.
➡️ Number of subsets of a 6-element set = 2⁶.
✔️ Number of relations = 2⁶ = 64.
💡 Final: 64 relations.

🔵 Question 9
Let R be the relation on Z defined by R = {(a, b) : a, b ∈ Z, a − b is an integer}. Find the domain and range of R.

🟢 Answer
➡️ For any a, b ∈ Z, the difference a − b is always an integer.
➡️ Therefore R = Z × Z (all ordered pairs of integers).
✔️ Domain(R) = Z.
✔️ Range(R) = Z.
💡 Final: Domain = Range = Z (set of all integers).

🧾Exercise 2.3

🔵 Question 1
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}

🟢 Answer
➡️ Definition: A relation is a function if every element of the domain has exactly one image (no repetition of first element with different second element).

(i) All first elements {2,5,8,11,14,17} are distinct.
✔️ So, it is a function.
➡️ Domain = {2,5,8,11,14,17}
➡️ Range = {1}

(ii) All first elements {2,4,6,8,10,12,14} are distinct.
✔️ So, it is a function.
➡️ Domain = {2,4,6,8,10,12,14}
➡️ Range = {1,2,3,4,5,6,7}

(iii) First element 1 is repeated with two different second elements 3 and 5.
⚠️ So, it is not a function.

💡 Final: (i) Function ✅, (ii) Function ✅, (iii) Not a function ❌

🔵 Question 2
Find the domain and range of the following real functions:

(i) f(x) = −|x|
(ii) f(x) = √(9 − x²)

🟢 Answer

(i) ➡️ f(x) = −|x|
✔️ |x| is defined ∀ x ∈ ℝ.
✔️ Range of |x| = [0, ∞)
➡️ So f(x) = −|x| ⇒ Range = (−∞, 0]
💡 Domain = ℝ
💡 Range = (−∞, 0]

(ii) ➡️ f(x) = √(9 − x²)
✔️ Expression under root ≥ 0 ⇒ 9 − x² ≥ 0
✔️ ⇒ x² ≤ 9 ⇒ −3 ≤ x ≤ 3
💡 Domain = [−3, 3]
✔️ f(x) takes maximum at x = 0 ⇒ f(0) = √9 = 3
✔️ Minimum value = 0 when x = ±3
💡 Range = [0, 3]

🔵 Question 3
A function f is defined by f(x) = 2x − 5. Write down the values of
(i) f(0), (ii) f(7), (iii) f(−3).

🟢 Answer
➡️ f(x) = 2x − 5

(i) f(0) = 2(0) − 5 = −5
(ii) f(7) = 2(7) − 5 = 14 − 5 = 9
(iii) f(−3) = 2(−3) − 5 = −6 − 5 = −11

💡 Final: f(0) = −5, f(7) = 9, f(−3) = −11

🔵 Question 4
The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by
t(C) = (9C / 5) + 32.
Find
(i) t(0) (ii) t(28) (iii) t(−10) (iv) The value of C, when t(C) = 212.

🟢 Answer

(i) t(0) = (9×0)/5 + 32 = 32°F

(ii) t(28) = (9×28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4°F

(iii) t(−10) = (9×(−10))/5 + 32 = (−90)/5 + 32 = −18 + 32 = 14°F

(iv) t(C) = 212
➡️ 212 = (9C)/5 + 32
➡️ 212 − 32 = (9C)/5
➡️ 180 = (9C)/5
➡️ Multiply both sides by 5: 900 = 9C
➡️ C = 100

💡 Final: t(0) = 32°F, t(28) = 82.4°F, t(−10) = 14°F, and C = 100°C when t(C) = 212°F

🔵 Question 5
Find the range of each of the following functions:

(i) f(x) = 2 − 3x, x ∈ ℝ, x > 0
(ii) f(x) = x² + 2, x is a real number
(iii) f(x) = x³, x is a real number

🟢 Answer

(i) f(x) = 2 − 3x
➡️ As x > 0 increases, 3x increases, so 2 − 3x decreases.
✔️ No upper bound; as x → 0⁺, f(x) → 2
✔️ As x → ∞, f(x) → −∞
💡 Range = (−∞, 2)

(ii) f(x) = x² + 2
✔️ x² ≥ 0 ⇒ x² + 2 ≥ 2
💡 Range = [2, ∞)

(iii) f(x) = x³
✔️ For x ∈ ℝ, x³ ∈ ℝ (all real values)
💡 Range = ℝ

————————————————————————————————————————————————————————————————————————————

OTHER IMPORTANT QUESTIONS FOR EXAMS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

🧭 Section A – Objective Type (1 mark each)

🔵 Question 1:
The number of elements in A × B if A = {1, 2} and B = {x, y, z} is
🔵 (A) 3
🟢 (B) 5
🟠 (C) 6
🔴 (D) 9
🟢 Answer: (C) 6

🔵 Question 2:
Which of the following is a relation from A = {1, 2} to B = {x, y}?
🔵 (A) {(1, x), (2, y)}
🟢 (B) {(1, x), (1, y)}
🟠 (C) {(1, x), (2, 3)}
🔴 (D) {(1, x), (x, 1)}
🟢 Answer: (A) {(1, x), (2, y)}

🔵 Question 3:
If A = {a, b}, B = {1, 2}, then number of relations from A to B is
🔵 (A) 4
🟢 (B) 8
🟠 (C) 16
🔴 (D) 2
🟢 Answer: (C) 16

🔵 Question 4:
The relation R = {(1, 1), (2, 2), (3, 3)} on A = {1, 2, 3} is
🔵 (A) Symmetric only
🟢 (B) Reflexive and symmetric
🟠 (C) Reflexive, symmetric, transitive
🔴 (D) None
🟢 Answer: (C) Reflexive, symmetric, transitive

🔵 Question 5:
A function f: A → B is said to be one-one if
🔵 (A) Every element of A has one image
🟢 (B) Every element of B has one preimage
🟠 (C) Distinct elements of A have distinct images
🔴 (D) None of these
🟢 Answer: (C) Distinct elements of A have distinct images

🔵 Question 6:
Which of the following represents a function?
🔵 (A) {(1, 2), (2, 3), (3, 2)}
🟢 (B) {(1, 2), (1, 3)}
🟠 (C) {(2, 3), (2, 4)}
🔴 (D) {(x, y) | x² + y² = 1}
🟢 Answer: (A) {(1, 2), (2, 3), (3, 2)}

🔵 Question 7:
If f(x) = 2x + 3, find f(−1).
🔵 (A) 1
🟢 (B) 0
🟠 (C) −1
🔴 (D) 5
🟢 Answer: (A) 1
🧠 Step: f(−1) = 2(−1) + 3 = 1 ✅

🔵 Question 8:
If A = {1, 2}, B = {3, 4}, number of functions from A to B =
🔵 (A) 2
🟢 (B) 4
🟠 (C) 8
🔴 (D) 16
🟢 Answer: (C) 8
🧠 Explanation: |B| = 2 ⇒ total functions = 2² = 4? Wait:
If A has 2 elements, B has 2 → total functions = |B|^|A| = 2² = 4 ✅
⚠ Correction: ✅ Answer: (B) 4

🔵 Question 9:
The function f(x) = 3x − 5 is
🔵 (A) One-one and onto
🟢 (B) Many-one
🟠 (C) Constant
🔴 (D) Into
🟢 Answer: (A) One-one and onto

🔵 Question 10:
If R = {(a, b) | a divides b}, then R is
🔵 (A) Reflexive and symmetric
🟢 (B) Reflexive and transitive
🟠 (C) Symmetric and transitive
🔴 (D) Only reflexive
🟢 Answer: (B) Reflexive and transitive

🔵 Question 11:
If A = {1, 2, 3}, B = {x, y}, then number of functions from A to B =
🔵 (A) 6
🟢 (B) 8
🟠 (C) 9
🔴 (D) 4
🟢 Answer: (B) 8 (|B|^|A| = 2³ = 8)

🔵 Question 12:
Which is not a function?
🔵 (A) y = √x, x ≥ 0
🟢 (B) y² = x
🟠 (C) y = 2x + 1
🔴 (D) y = |x|
🟢 Answer: (B) y² = x (two values of y)

🔵 Question 13:
If f(x) = x², g(x) = x + 1, then (g ∘ f)(2) =
🔵 (A) 3
🟢 (B) 5
🟠 (C) 7
🔴 (D) 9
🟢 Answer: (B) 5
🧠 f(2) = 4, g(4) = 4 + 1 = 5 ✅

🔵 Question 14:
Inverse of f(x) = 2x + 3 is
🔵 (A) (x − 3)/2
🟢 (B) 2x − 3
🟠 (C) (x + 3)/2
🔴 (D) 1/(2x + 3)
🟢 Answer: (A) (x − 3)/2 ✅

🔵 Question 15:
If f(x) = |x|, then f(−5) =
🔵 (A) −5
🟢 (B) 5
🟠 (C) 0
🔴 (D) 10
🟢 Answer: (B) 5 ✅

🔵 Question 16:
If R = {(1, 2), (2, 1)}, on A = {1, 2}, then R is
🔵 (A) Symmetric only
🟢 (B) Reflexive only
🟠 (C) Transitive only
🔴 (D) Reflexive and transitive
🟢 Answer: (A) Symmetric only

🔵 Question 17:
For A = {1, 2, 3}, define R = {(1, 2), (2, 3), (1, 3)}. Then R is
🔵 (A) Transitive only
🟢 (B) Symmetric
🟠 (C) Reflexive
🔴 (D) None
🟢 Answer: (A) Transitive only ✅

🔵 Question 18:
Let f: ℝ → ℝ, f(x) = x³. Then f is
🔵 (A) One-one only
🟢 (B) Onto only
🟠 (C) Bijective
🔴 (D) Constant
🟢 Answer: (C) Bijective ✅ (strictly increasing)

🧭 Section B – Short Answer Type (2–3 Marks Each)

🔵 Question 19:
Define a relation R on set A = {1, 2, 3} by R = {(a, b) | a < b}. Write its domain, range, and co-domain.
🟢 Answer:
➡️ Given: R = {(1, 2), (1, 3), (2, 3)}
➡️ Domain(R) = {1, 2}
➡️ Range(R) = {2, 3}
➡️ Co-domain(R) = A = {1, 2, 3}
✔️ Final: Domain = {1, 2}, Range = {2, 3}, Co-domain = {1, 2, 3}

🔵 Question 20:
Check whether the relation R on A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} is symmetric and transitive.
🟢 Answer:
➡️ Symmetric check:
(1, 2) ∈ R ⇒ (2, 1) ∈ R ✅
All pairs satisfied → R is symmetric.
➡️ Transitive check:
(1, 2), (2, 1) ∈ R ⇒ (1, 1) ∈ R ✅
(2, 1), (1, 2) ⇒ (2, 2) ∈ R ✅
Hence R is transitive.
✔️ Final: R is symmetric and transitive.

🔵 Question 21:
Determine whether the function f: ℝ → ℝ defined by f(x) = 2x + 3 is one-one and onto.
🟢 Answer:
➡️ One-one:
Let f(x₁) = f(x₂)
⇒ 2x₁ + 3 = 2x₂ + 3
⇒ x₁ = x₂ ✅
So f is one-one.
➡️ Onto:
Take y ∈ ℝ, solve y = 2x + 3 ⇒ x = (y − 3)/2 ∈ ℝ ✅
So f is onto.
✔️ Final: f is bijective (one-one and onto).

🔵 Question 22:
Find the composition (g ∘ f)(x), if f(x) = x + 1 and g(x) = x².
🟢 Answer:
➡️ (g ∘ f)(x) = g(f(x))
= g(x + 1)
= (x + 1)²
= x² + 2x + 1
✔️ Final: (g ∘ f)(x) = x² + 2x + 1 ✅

🔵 Question 23:
If f: ℝ → ℝ is defined by f(x) = 3x − 5, find its inverse.
🟢 Answer:
➡️ Let y = 3x − 5
➡️ Solve for x: x = (y + 5)/3
➡️ Hence f⁻¹(y) = (y + 5)/3
✔️ Final: f⁻¹(x) = (x + 5)/3 ✅

🧭 Section C – Mid-Length (3 Marks Each)

🔵 Question 24:
Check whether the relation R = {(1, 1), (2, 2), (3, 3), (1, 2)} on A = {1, 2, 3} is reflexive, symmetric, transitive.
🟢 Answer:
➡️ Reflexive: (1,1),(2,2),(3,3) ∈ R ✅
➡️ Symmetric: (1,2) ∈ R but (2,1) ∉ R ❌
➡️ Transitive: (1,2) & (2,2) ⇒ (1,2) ∈ R ✅ (others ok)
✔️ Final: R is reflexive, transitive, but not symmetric.

🔵 Question 25:
For A = {1, 2, 3}, find total number of relations on A which are reflexive.
🟢 Answer:
➡️ Total pairs in A × A = 3 × 3 = 9
➡️ For reflexive relation, (1,1), (2,2), (3,3) must be included (fixed 3).
➡️ Remaining 6 pairs can be chosen or not freely ⇒ 2⁶ = 64.
✔️ Final: 64 reflexive relations possible ✅

🔵 Question 26:
Let f: ℝ → ℝ be defined by f(x) = x². Is f one-one and onto?
🟢 Answer:
➡️ One-one? f(1) = 1 = f(−1) ⇒ not one-one ❌
➡️ Onto? Co-domain = ℝ, but Range = [0, ∞) ⇒ not onto ❌
✔️ Final: f is neither one-one nor onto ✅

🔵 Question 27:
Show that the relation R = {(x, y) | x − y is even} on ℤ is an equivalence relation.
🟢 Answer:
➡️ Reflexive: x − x = 0 (even) ✅
➡️ Symmetric: x − y even ⇒ y − x = −(x − y) even ✅
➡️ Transitive: (x − y) and (y − z) even ⇒ (x − z) even ✅
✔️ Final: R is reflexive, symmetric, transitive ⇒ equivalence relation ✅

🧭 Section D – Long Answer Type (5 Marks Each)

🔵 Question 28:
Let A = {1, 2, 3, 4}. Define relation R on A by R = {(x, y) | x − y is divisible by 2}. Prove that R is an equivalence relation. Also, find the equivalence classes.
🟢 Answer:
➤ Step 1: R = {(1,1), (1,3), (3,1), (3,3), (2,2), (2,4), (4,2), (4,4)}
➤ Step 2: Reflexive check:
x − x = 0 (divisible by 2) ✅ ∀ x ∈ A
Hence, R is reflexive.
➤ Step 3: Symmetric check:
If (x, y) ∈ R ⇒ x − y divisible by 2
Then y − x = −(x − y) also divisible by 2 ✅
Hence, symmetric.
➤ Step 4: Transitive check:
If (x, y), (y, z) ∈ R ⇒ x − y and y − z divisible by 2
Then (x − z) = (x − y) + (y − z) divisible by 2 ✅
Hence, transitive.
✅ Therefore, R is reflexive, symmetric, transitive ⇒ equivalence relation.
➤ Step 5: Equivalence classes:
[1] = {1, 3}
[2] = {2, 4}
[3] = {1, 3}
[4] = {2, 4}
✔️ Final: Two equivalence classes: {1, 3}, {2, 4}

🔵 Question 29:
If f: ℝ → ℝ defined by f(x) = 3x + 2 and g: ℝ → ℝ defined by g(x) = 2x − 3, find
(i) (g ∘ f)(x) (ii) (f ∘ g)(x)
Also, show (g ∘ f) ≠ (f ∘ g).
🟢 Answer:
➤ (g ∘ f)(x) = g(f(x))
= g(3x + 2)
= 2(3x + 2) − 3
= 6x + 4 − 3
= 6x + 1 ✅
➤ (f ∘ g)(x) = f(g(x))
= f(2x − 3)
= 3(2x − 3) + 2
= 6x − 9 + 2
= 6x − 7 ✅
✅ (g ∘ f)(x) = 6x + 1, (f ∘ g)(x) = 6x − 7
⚠️ Not equal ⇒ composition not commutative.
✔️ Final: (g ∘ f)(x) ≠ (f ∘ g)(x)

🔵 Question 30:
Let f(x) = 2x + 3, g(x) = x − 2. Find (f ∘ g)(x) and (g ∘ f)(x). Are they equal?
🟢 Answer:
➡️ (f ∘ g)(x) = f(g(x)) = f(x − 2) = 2(x − 2) + 3 = 2x − 4 + 3 = 2x − 1 ✅
➡️ (g ∘ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3) − 2 = 2x + 1 ✅
⚠️ (f ∘ g)(x) ≠ (g ∘ f)(x)
✔️ Conclusion: Composition is not commutative.

🧭 Section E – Case-Based / Application (5 Marks Each)

🔵 Question 31:
A company assigns ID codes to employees using function f(x) = 2x + 1, where x is employee number.
(a) Find IDs for employees numbered 1, 2, 3.
(b) Check if function is one-one and onto for natural numbers.
🟢 Answer:
(a)
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
✅ IDs: 3, 5, 7
(b)
➤ One-one check:
f(x₁) = f(x₂) ⇒ 2x₁ + 1 = 2x₂ + 1 ⇒ x₁ = x₂ ✅
Hence one-one.
➤ Onto check:
Co-domain = ℕ, but range = odd naturals only. ❌ Not onto.
✔️ Final: f is one-one, not onto.

🔵 Question 32:
A function f: A → B is defined as A = {1, 2, 3}, B = {x, y, z} with f = {(1, x), (2, y), (3, z)}.
Show that f is bijective and find f⁻¹.
🟢 Answer:
➡️ Each element of A has one image ✅ (function)
➡️ Distinct inputs → distinct outputs ✅ (one-one)
➡️ All outputs {x, y, z} covered ✅ (onto)
⇒ f is bijective
➡️ Inverse relation: f⁻¹ = {(x, 1), (y, 2), (z, 3)} ✅
✔️ Final: f⁻¹(x) = 1, f⁻¹(y) = 2, f⁻¹(z) = 3

🔵 Question 33:
A relation R is defined on A = {1, 2, 3, 4} as R = {(a, b) | |a − b| = 2}.
Find R and check if it is symmetric and transitive.
🟢 Answer:
➡️ R = {(1, 3), (3, 1), (2, 4), (4, 2)}
✔️ Symmetric: (1,3) ⇒ (3,1), (2,4) ⇒ (4,2) ✅
✔️ Transitive:
(1,3), (3,1) ⇒ (1,1)? not in R ❌
Hence not transitive.
✔️ Final: R is symmetric but not transitive.

————————————————————————————————————————————————————————————————————————————