Class 12 : Maths (English) – Chapter 1: Relations and Functions

EXPLANATION & SUMMARY

✳️ Explanation Section (~1700 words)
🔵 1. Introduction
Relations and Functions form the foundational concept of modern mathematics. They help us understand how two quantities are connected.
In Class 12, you expand the knowledge of relations and functions studied in Class 11, learning about new types like invertible functions, composition, and binary operations.

🟢 2. Relation – Basic Concept
✏️ Definition:
A relation is a connection between elements of two sets.
If there are two non-empty sets A and B, then any subset of their Cartesian product A × B is called a relation from A to B.
💡 Notation:
If (a, b) ∈ R, then we say “a is related to b under relation R” → a R b
📘 Example:
Let A = {1, 2, 3}, B = {4, 5, 6},
Define R = {(1, 4), (2, 5), (3, 6)}
⇒ R is a relation from A to B.

🟡 3. Types of Relations
Let R be a relation on a set A (i.e. R ⊆ A × A):
🔹 (i) Reflexive Relation
A relation R is reflexive if every element is related to itself.
➡️ (a, a) ∈ R for every a ∈ A
📘 Example: A = {1, 2, 3}, R = {(1,1), (2,2), (3,3)} is reflexive.
🔹 (ii) Symmetric Relation
R is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R
📘 Example: R = {(1,2), (2,1), (2,3), (3,2)} is symmetric.
🔹 (iii) Transitive Relation
R is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
📘 Example: R = {(1,2), (2,3), (1,3)} is transitive.
🔹 (iv) Equivalence Relation
If R is reflexive, symmetric, and transitive,
then R is called an equivalence relation.
💡 Equivalence relation partitions a set into disjoint equivalence classes.

🔴 4. Function – Definition and Basics
✏️ Definition:
A function is a special relation where each element of set A is associated with exactly one element of set B.
A function from A to B is denoted by f: A → B.
💡 Important: Every function is a relation, but not every relation is a function.

🟢 5. Types of Functions
Let f: A → B, and A = Domain, B = Co-domain, Range = image of A under f.
🔹 (i) One-One Function (Injective)
If distinct elements in A map to distinct elements in B:
a₁ ≠ a₂ ⇒ f(a₁) ≠ f(a₂)
🔹 (ii) Onto Function (Surjective)
If Range = Co-domain,
every element of B has a pre-image in A.
🔹 (iii) Bijective Function
If function is both One-One and Onto,
then it is called Bijective.
💡 Bijective functions are invertible.

🟡 6. Composition of Functions
If f: A → B and g: B → C,
then composition g ∘ f: A → C is defined by
➡️ (g ∘ f)(x) = g(f(x))
📘 Example:
Let f(x) = 2x + 3, g(x) = x²
Then (g ∘ f)(x) = g(f(x)) = (2x + 3)²
💡 Composition is associative:
h ∘ (g ∘ f) = (h ∘ g) ∘ f

🔴 7. Invertible Function
A function f: A → B is invertible if there exists a function g: B → A such that
➡️ g ∘ f = I_A and f ∘ g = I_B
(where I_A and I_B are identity functions)
✏️ Condition:
A function is invertible if and only if it is bijective.
💡 Identity Function: I_A(x) = x for all x ∈ A.

🟢 8. Some Standard Functions

🔹 Identity Function
f(x) = x
Graph: Straight line through origin at 45°.
🔹 Constant Function
f(x) = c (constant)
Graph: Parallel to x-axis.
🔹 Linear Function
f(x) = mx + c
Graph: Straight line (slope = m).
🔹 Quadratic Function
f(x) = ax² + bx + c
Graph: Parabola.
🔹 Polynomial Function
f(x) = a₀ + a₁x + a₂x² + … + aₙxⁿ
🔹 Rational Function
f(x) = P(x)/Q(x), Q(x) ≠ 0
🔹 Modulus Function
f(x) = |x|
🔹 Signum Function
f(x) =
-1, x < 0
0, x = 0
1, x > 0

🟡 9. Graphical Representation
Graph helps in visualizing domain, range, and nature of functions.
💡 For one-one functions: horizontal line cuts graph at most once.
💡 For onto functions: every y has at least one x.

🔴 10. Binary Operation
A binary operation on a set A is a function
➡️ * : A × A → A
📘 Example: Addition (+) on R:
(a, b) → a + b
💡 Properties:
Closure: a * b ∈ A
Associativity: (a * b) * c = a * (b * c)
Commutativity: a * b = b * a
Identity: ∃ e ∈ A, a * e = e * a = a
Inverse: ∃ a⁻¹ ∈ A, a * a⁻¹ = e

🟢 11. Equivalence Class (Extra Insight)
If R is equivalence on A,
the set [a] = {x ∈ A | x R a} is called equivalence class.
All such classes partition A.

💡 12. Important Results
f is bijective ⇔ f⁻¹ exists
Composition of bijective functions is bijective
Inverse of inverse is the function itself: (f⁻¹)⁻¹ = f
Composition respects inverse: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹

🧠 13. Real-life Applications
One-one relations model unique mappings (e.g., roll number → student).
Functions represent dependencies (e.g., temperature → pressure).
Inverse functions used in solving equations, transformations.

✳️ Summary Section (~300 words)
🔹 A relation is a subset of A × B; a function is a relation with unique mapping.
🔹 Reflexive, symmetric, and transitive relations combine to form equivalence relations.
🔹 Function types:
One-one (Injective)
Onto (Surjective)
Bijective = One-one + Onto
🔹 Composition: (g ∘ f)(x) = g(f(x))
🔹 Invertible function exists only for bijective functions.
🔹 Binary operation: mapping from A × A to A, following closure and other properties.
🔹 Common functions:
Identity: f(x) = x
Constant: f(x) = c
Linear: f(x) = mx + c
Quadratic: f(x) = ax² + bx + c
🔹 Graph analysis helps identify one-one/onto nature.
🔹 Function inverse reverses input-output pairing.

📝 Quick Recap
🧠 Relation = subset of Cartesian product
✳️ Function = unique output for each input
🔢 One-one ⇔ unique mapping
🎯 Onto ⇔ all elements of co-domain covered
🧭 Bijective ⇔ invertible
🔄 Composition: g ∘ f = g(f(x))
🔁 Inverse: f⁻¹ exists ⇔ f bijective
⚙️ Binary operation: A × A → A

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QUESTIONS FROM TEXTBOOK

Exercise 1.1

🔵 Question 1:
Determine whether each of the following relations are reflexive, symmetric, and transitive:

✳️ (i)
Relation R in the set A = {1, 2, 3, …, 13, 14} defined as
➡️ R = {(x, y) : 3x − y = 0}
🟢 Answer:
💡 Given condition: 3x − y = 0 ⇒ y = 3x
✔️ Reflexive:
We check if (x, x) ∈ R
For (x, x),
➡️ 3x − x = 2x ≠ 0
Hence (x, x) ∉ R
❌ Not reflexive.
✔️ Symmetric:
If (x, y) ∈ R ⇒ y = 3x
For symmetry, need (y, x) ∈ R ⇒ x = 3y
But if y = 3x, then x = y/3 ≠ 3y
❌ Not symmetric.
✔️ Transitive:
Suppose (x, y) ∈ R and (y, z) ∈ R
⇒ y = 3x, z = 3y = 9x
Check if (x, z) ∈ R
⇒ z = 3x ? → No, because z = 9x
❌ Not transitive.
✅ Therefore, R is neither reflexive, nor symmetric, nor transitive.

✳️ (ii)
Relation R in the set N of natural numbers defined as
➡️ R = {(x, y) : y = x + 5 and x < 4}
🟢 Answer:
Given: y = x + 5, valid only for x = 1, 2, 3
✔️ Reflexive:
For (x, x) ⇒ x = x + 5 ❌ False
Hence not reflexive.
✔️ Symmetric:
If (x, y) ∈ R ⇒ y = x + 5
To have (y, x) ∈ R, we need x = y + 5 ❌ Not true
Hence not symmetric.
✔️ Transitive:
If (x, y) ∈ R and (y, z) ∈ R,
then y = x + 5 and z = y + 5 = x + 10
To have (x, z) ∈ R, we need z = x + 5, but z = x + 10 ❌
Hence not transitive.
✅ Therefore, R is neither reflexive, nor symmetric, nor transitive.

✳️ (iii)
Relation R in the set A = {1, 2, 3, 4, 5, 6} as
➡️ R = {(x, y) : y is divisible by x}
🟢 Answer:
✔️ Reflexive:
For every x, x divides x ✔️
✅ Reflexive.
✔️ Symmetric:
If (x, y) ∈ R ⇒ y divisible by x
But x may not be divisible by y
E.g. (2, 4) ∈ R but (4, 2) ∉ R
❌ Not symmetric.
✔️ Transitive:
If (x, y) ∈ R and (y, z) ∈ R
⇒ y = k₁x, z = k₂y = k₁k₂x
⇒ z divisible by x ✔️
✅ Transitive.
✅ Therefore, R is reflexive and transitive, but not symmetric.

✳️ (iv)
Relation R in the set Z of all integers defined as
➡️ R = {(x, y) : x − y is an integer}
🟢 Answer:
For x, y ∈ Z, x − y ∈ Z always true.
✔️ Reflexive: x − x = 0 ∈ Z ✔️ ✅
✔️ Symmetric: If (x, y) ∈ R, then x − y ∈ Z ⇒ y − x = −(x − y) ∈ Z ✔️ ✅
✔️ Transitive: If (x, y) ∈ R and (y, z) ∈ R
⇒ x − y ∈ Z, y − z ∈ Z
Add ⇒ x − z ∈ Z ✔️ ✅
✅ Therefore, R is reflexive, symmetric, and transitive.
Hence, R is an equivalence relation.

✳️ (v)
Relation R in the set A of human beings given by:
(a) R = {(x, y) : x and y work at the same place}
✅ Reflexive (same person works at same place)
✅ Symmetric (if x works with y, y works with x)
✅ Transitive (if x with y, y with z ⇒ x with z)
✔️ Equivalence relation
(b) R = {(x, y) : x and y live in same locality}
✅ Reflexive ✔️ Symmetric ✔️ Transitive
✔️ Equivalence relation
(c) R = {(x, y) : x is exactly 7 cm taller than y}
❌ Not reflexive
❌ Not symmetric
❌ Not transitive
✖️ Not equivalence
(d) R = {(x, y) : x is wife of y}
❌ Not reflexive
❌ Not symmetric (wife of y ≠ y wife of x)
❌ Not transitive
✖️ Not equivalence
(e) R = {(x, y) : x is father of y}
❌ Not reflexive
❌ Not symmetric
❌ Not transitive
✖️ Not equivalence

🔵 Question 2:
Show that the relation R in the set ℝ of real numbers, defined as
➡️ R = {(a, b) : a ≤ b²}
is neither reflexive, nor symmetric, nor transitive.
🟢 Answer:
💡 Given R = {(a, b) : a ≤ b²}
✔️ Reflexive:
For R to be reflexive, (a, a) ∈ R ∀ a ∈ ℝ
Check: a ≤ a²
⮞ True for a ≥ 1 and a ≤ 0, but false for 0 < a < 1 (e.g. ½ ≤ ¼ ❌)
∴ R is not reflexive.
✔️ Symmetric:
If (a, b) ∈ R ⇒ a ≤ b²
For symmetry, (b, a) ∈ R ⇒ b ≤ a² must hold.
Counterexample: a = 2, b = 1
→ a ≤ b² ⇒ 2 ≤ 1² ❌
So R is not symmetric.
✔️ Transitive:
Take a = 3, b = 2, c = 1
a ≤ b² ⇒ 3 ≤ 4 ✔️
b ≤ c² ⇒ 2 ≤ 1² ❌ fails
Hence no guarantee of a ≤ c²
∴ R is not transitive.
✅ Therefore, R is neither reflexive, nor symmetric, nor transitive.

🔵 Question 3:
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:
💡 Pairs: (1,2), (2,3), (3,4), (4,5), (5,6)
✔️ Reflexive:
(a, a) not present since b = a + 1 ≠ a ❌
∴ Not reflexive.
✔️ Symmetric:
If (a, b) ∈ R ⇒ b = a + 1
Then (b, a) requires a = b + 1, which is false ❌
∴ Not symmetric.
✔️ Transitive:
If (a, b) ∈ R and (b, c) ∈ R
Then b = a + 1, c = b + 1 = a + 2
Check if (a, c) ∈ R ⇒ c = a + 1 ❌
∴ Not transitive.
✅ Hence, R is neither reflexive, nor symmetric, nor transitive.

🔵 Question 4:
Show that the relation R in ℝ defined as
➡️ R = {(a, b) : a ≤ b}
is reflexive and transitive, but not symmetric.
🟢 Answer:
✔️ Reflexive:
For all a ∈ ℝ, a ≤ a ✔️
∴ Reflexive ✅
✔️ Symmetric:
If (a, b) ∈ R ⇒ a ≤ b
Then (b, a) requires b ≤ a, which may not hold
Counterexample: a = 2, b = 3
→ 2 ≤ 3 ✔️ but 3 ≤ 2 ❌
∴ Not symmetric ❌
✔️ Transitive:
If (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c ⇒ a ≤ c ✔️
∴ Transitive ✅
✅ Hence, R is reflexive and transitive, but not symmetric.

🔵 Question 5:
Check whether the relation R in ℝ defined by
➡️ R = {(a, b) : a ≤ b³}
is reflexive, symmetric, or transitive.
🟢 Answer:
💡 Given a ≤ b³
✔️ Reflexive:
Check (a, a) ⇒ a ≤ a³
True for a ≥ 1 and a ≤ 0, but false for 0 < a < 1
e.g. ½ ≤ (½)³ = 1/8 ❌
∴ Not reflexive ❌
✔️ Symmetric:
If a ≤ b³, we need b ≤ a³
Take a = 2, b = 1:
2 ≤ 1³ ❌, 1 ≤ 2³ ✔️
∴ Not symmetric ❌
✔️ Transitive:
Take a = 2, b = 3, c = 1
2 ≤ 3³ = 27 ✔️, 3 ≤ 1³ = 1 ❌
Transitivity fails ❌
✅ Hence, R is neither reflexive, nor symmetric, nor transitive.

🔵 Question 6:
Show that the relation R in the set {1, 2, 3} given by
➡️ R = {(1, 2), (2, 1)}
is symmetric but neither reflexive nor transitive.
🟢 Answer:
💡 Given R = {(1, 2), (2, 1)}
✔️ Reflexive:
For reflexivity, (a, a) must belong to R for all a.
But (1, 1), (2, 2), (3, 3) are not in R.
❌ Not reflexive.
✔️ Symmetric:
If (a, b) ∈ R, then (b, a) ∈ R.
→ (1, 2) ∈ R ⇒ (2, 1) ∈ R ✔️
→ (2, 1) ∈ R ⇒ (1, 2) ∈ R ✔️
✅ Symmetric.
✔️ Transitive:
If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
Here (1, 2) and (2, 1) ∈ R, but (1, 1) ∉ R
❌ Not transitive.
✅ Therefore, R is symmetric, but neither reflexive nor transitive.

🔵 Question 7:
Show that the relation R in the set A of all the books in a library of a college,
given by
➡️ R = {(x, y) : x and y have the same number of pages}
is an equivalence relation.
🟢 Answer:
Let x and y be any books.
✔️ Reflexive:
Every book has the same number of pages as itself.
(x, x) ∈ R ✔️
✔️ Symmetric:
If x and y have the same number of pages,
then y and x also have the same number of pages ✔️
✔️ Transitive:
If x and y have the same number of pages, and y and z have the same number of pages,
then x and z also have the same number of pages ✔️
✅ Hence, R is reflexive, symmetric, and transitive,
∴ R is an equivalence relation.

🔵 Question 8:
Show that the relation R in the set A = {1, 2, 3, 4, 5}
defined by
➡️ R = {(a, b) : |a − b| is even}
is an equivalence relation.
Show that
– all the elements of {1, 3, 5} are related to each other,
– all the elements of {2, 4} are related to each other,
– no element of {1, 3, 5} is related to any element of {2, 4}.
🟢 Answer:
💡 Given R = {(a, b) : |a − b| is even}
✔️ Reflexive:
For all a ∈ A, |a − a| = 0 (even)
∴ (a, a) ∈ R ✔️
✔️ Symmetric:
If |a − b| is even, then |b − a| = |a − b| is also even ✔️
✔️ Transitive:
If |a − b| and |b − c| are even,
then |a − c| = |(a − b) + (b − c)|
Sum of two even numbers is even ✔️
∴ Transitive.
✅ Hence, R is equivalence relation.
🧠 Equivalence classes:
For 1: {1, 3, 5}
For 2: {2, 4}
For 3: {1, 3, 5}
For 4: {2, 4}
For 5: {1, 3, 5}

🔵 Question 9:
Show that each of the relation R in the set
A = {x ∈ ℤ : 0 ≤ x ≤ 12} given by
(i) R = {(a, b) : |a − b| is a multiple of 4}
(ii) R = {(a, b) : a = b}
is an equivalence relation.
Find the set of all elements related to 1 in each case.
🟢 Answer:
(i)
💡 R = {(a, b) : |a − b| is multiple of 4}
✔️ Reflexive:
|a − a| = 0, multiple of 4 ✔️
✔️ Symmetric:
If |a − b| multiple of 4, then |b − a| same ✔️
✔️ Transitive:
If |a − b| = 4m, |b − c| = 4n,
then |a − c| = 4(m ± n), also multiple of 4 ✔️
✅ Equivalence relation.
📘 Elements related to 1:
All b such that |1 − b| multiple of 4
⇒ b = 1, 5, 9
✅ Related set = {1, 5, 9}

(ii)
💡 R = {(a, b) : a = b}
✔️ Reflexive: (a, a) ✔️
✔️ Symmetric: if a = b ⇒ b = a ✔️
✔️ Transitive: if a = b and b = c ⇒ a = c ✔️
✅ Equivalence relation.
📘 Elements related to 1: only {1}

🔵 Question 10:
Give an example of a relation which is:
(i) Symmetric but neither reflexive nor transitive
🟢 Example: R = {(1, 2), (2, 1)} on A = {1, 2, 3}
✔️ Symmetric ✅
❌ Not reflexive
❌ Not transitive
(ii) Transitive but neither reflexive nor symmetric
🟢 Example: R = {(1, 2), (2, 3), (1, 3)} on A = {1, 2, 3}
✔️ Transitive ✅
❌ Not reflexive
❌ Not symmetric
(iii) Reflexive and symmetric but not transitive
🟢 Example: R = {(1, 1), (2, 2), (1, 2), (2, 1)} on A = {1, 2, 3}
✔️ Reflexive, Symmetric ✅
❌ Not transitive (since (1,2), (2,1) ⇒ need (1,1) ✔️, (2,2) ✔️, but if we omit some, fails)
(iv) Reflexive and transitive but not symmetric
🟢 Example: R = {(1, 1), (2, 2), (3, 3), (1, 2)}
✔️ Reflexive ✅
✔️ Transitive ✅
❌ Not symmetric
(v) Symmetric and transitive but not reflexive
🟢 Example: R = {(1, 2), (2, 1)}
✔️ Symmetric ✅
✔️ Transitive (no triplets violating) ✅
❌ Not reflexive

🔵 Question 11:
Show that the relation R in the set A of points in a plane given by
➡️ R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}
is an equivalence relation.
Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
🟢 Answer:
Let P(x₁, y₁) and Q(x₂, y₂) be two points in the plane.
Distance from origin:
➡️ OP = √(x₁² + y₁²), OQ = √(x₂² + y₂²)
Given condition:
P R Q ⇔ OP = OQ
✔️ Reflexive:
For every P, OP = OP
∴ (P, P) ∈ R ✅
✔️ Symmetric:
If (P, Q) ∈ R ⇒ OP = OQ
Then OQ = OP ⇒ (Q, P) ∈ R ✅
✔️ Transitive:
If (P, Q) ∈ R and (Q, R) ∈ R,
then OP = OQ and OQ = OR ⇒ OP = OR
⇒ (P, R) ∈ R ✅
✅ Hence, R is reflexive, symmetric, and transitive.
∴ R is an equivalence relation.
💡 Set of all points related to P(x₁, y₁):
All Q(x, y) such that √(x² + y²) = √(x₁² + y₁²)
⇒ x² + y² = x₁² + y₁²
➡️ This represents a circle with centre at origin and radius = √(x₁² + y₁²), passing through P. ✅

🔵 Question 12:
Show that the relation R defined in the set A of all triangles as
➡️ R = {(T₁, T₂) : T₁ is similar to T₂}
is an equivalence relation.
Consider three right-angled triangles T₁ with sides 3, 4, 5; T₂ with sides 5, 12, 13; T₃ with sides 6, 8, 10.
Which triangles among T₁, T₂, T₃ are related?
🟢 Answer:
✔️ Reflexive:
Every triangle is similar to itself.
∴ (T₁, T₁) ∈ R ✅
✔️ Symmetric:
If T₁ ∼ T₂, then T₂ ∼ T₁ ✅
✔️ Transitive:
If T₁ ∼ T₂ and T₂ ∼ T₃, then T₁ ∼ T₃ ✅
✅ Hence, R is equivalence relation.
💡 Now check similarity:
T₁ sides 3, 4, 5
T₂ sides 5, 12, 13
T₃ sides 6, 8, 10
Check ratios:
T₁: 3:4:5
T₃: 6:8:10 = 3:4:5 ✅ → T₁ ∼ T₃
T₂: 5:12:13 ≠ 3:4:5 ❌
✅ Therefore, T₁ and T₃ are related (similar), T₂ is not related to them.

🔵 Question 13:
Show that the relation R defined in the set A of all polygons as
➡️ R = {(P₁, P₂) : P₁ and P₂ 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, 5?
🟢 Answer:
✔️ Reflexive:
Each polygon has same number of sides as itself. ✅
✔️ Symmetric:
If P₁ and P₂ have same number of sides,
then P₂ and P₁ also have same number of sides. ✅
✔️ Transitive:
If P₁ and P₂ have same number of sides and P₂ and P₃ also have same number,
then P₁ and P₃ have same number of sides. ✅
✅ Hence R is an equivalence relation.
💡 A right-angled triangle has 3 sides,
So, the set of all elements related to T is
➡️ All triangles (3-sided polygons) in A.

🔵 Question 14:
Let L be the set of all lines in XY-plane and R the relation in L defined as
➡️ R = {(L₁, L₂) : L₁ is parallel to L₂}
Show that R is an equivalence relation.
Find the set of all lines related to the line y = 2x + 4.
🟢 Answer:
✔️ Reflexive:
Every line is parallel to itself. ✅
✔️ Symmetric:
If L₁ ∥ L₂, then L₂ ∥ L₁. ✅
✔️ Transitive:
If L₁ ∥ L₂ and L₂ ∥ L₃, then L₁ ∥ L₃. ✅
✅ Hence, R is an equivalence relation.
💡 Line y = 2x + 4 has slope m = 2.
All lines parallel to it have same slope 2.
➡️ Set of all related lines:
{y = 2x + c | c ∈ ℝ}

Question 15
Let R be the relation in the set {1, 2, 3, 4} given by
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer:
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Answer
✏️ Reflexive check: need (1,1), (2,2), (3,3), (4,4) ∈ R → all four are present ✔️
✏️ Symmetric check: (1,2) ∈ R but (2,1) ∉ R → not symmetric ❌
✏️ Transitive check: verify chains:
• (1,2) and (2,2) ⇒ (1,2) ∈ R ✔️
• (1,3) and (3,2) ⇒ (1,2) ∈ R ✔️
• (3,2) and (2,2) ⇒ (3,2) ∈ R ✔️
• Self-pairs keep transitivity intact ✔️
Hence transitive ✔️
✅ Correct option: (B) R is reflexive and transitive but not symmetric.

Question 16
Let R be the relation in the set ℕ given by R = {(a, b) : a = b − 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R
(B) (3, 8) ∈ R
(C) (6, 8) ∈ R
(D) (8, 7) ∈ R
Answer
✏️ Condition: a = b − 2 and b > 6
• (2,4): 2 = 4 − 2 ✔️ but b = 4 ≯ 6 ❌ → not in R
• (3,8): 3 = 8 − 2 ⇒ 3 = 6 ❌ → not in R
• (6,8): 6 = 8 − 2 ✔️ and b = 8 > 6 ✔️ → in R
• (8,7): 8 = 7 − 2 ⇒ 8 = 5 ❌ → not in R
✅ Correct option: (C) (6, 8) ∈ R

Exercise 1.2

🔵 Question 1:
Show that the function f : R* → R*, defined by f(x) = 1/x is one-one and onto, where R* is the set of all non-zero real numbers.
Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
🟢 Answer:
✏️ Let f(x) = 1/x
➤ Step 1: Show f is one-one
Take any x₁, x₂ ∈ R*, and suppose
f(x₁) = f(x₂)
➡️ 1/x₁ = 1/x₂
➡️ Cross multiply: x₂ = x₁
✔️ Hence, f is one-one.

➤ Step 2: Show f is onto
Let y ∈ R* be any element.
We need to find x ∈ R* such that f(x) = y.
➡️ f(x) = y
➡️ 1/x = y
➡️ x = 1/y
Since y ≠ 0, 1/y ∈ R*
✔️ Therefore, f is onto.
✅ So, f is one-one and onto on R*.

➤ Step 3: When domain = N
If domain = N, co-domain = R*:
f(x) = 1/x ∈ (0,1] ⊂ R*
Here,
✔️ f is one-one (no repetition)
❌ Not onto, since negative and >1 values in R* are not covered.
✅ Final: f is one-one but not onto when domain is N.

🔵 Question 2:
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : R → R given by f(x) = x²
(iv) f : N → N given by f(x) = x³
(v) f : Z → Z given by f(x) = x³
🟢 Answer:

✳️ (i) f : N → N, f(x) = x²
➡️ f(1)=1, f(2)=4, f(3)=9 (distinct)
✔️ One-one (injective)
🧠 Every natural number is not a perfect square
❌ Not onto
✅ One-one but not onto

✳️ (ii) f : Z → Z, f(x) = x²
➡️ f(2)=4, f(-2)=4
❌ Not one-one
🧠 Negative integers not obtained as f(x)
❌ Not onto
✅ Neither one-one nor onto

✳️ (iii) f : R → R, f(x) = x²
➡️ f(2)=4, f(-2)=4
❌ Not one-one
🧠 Negative y not possible since x² ≥ 0
❌ Not onto
✅ Neither one-one nor onto

✳️ (iv) f : N → N, f(x) = x³
➡️ Distinct inputs → distinct outputs
✔️ One-one
🧠 Not all natural numbers are cubes
❌ Not onto
✅ One-one but not onto

✳️ (v) f : Z → Z, f(x) = x³
➡️ f(a)=f(b) ⇒ a³=b³ ⇒ a=b
✔️ One-one
🧠 For every y ∈ Z, take x = ∛y ∈ Z
✔️ Onto
✅ One-one and onto

🔵 Question 3:
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
🟢 Answer:
✏️ Definition: [x] = greatest integer ≤ x
Examples: [2.3] = 2, [−1.2] = −2

➤ Step 1: Check one-one
Take x₁ = 1.2 and x₂ = 1.5
➡️ f(1.2) = [1.2] = 1
➡️ f(1.5) = [1.5] = 1
❌ f(x₁) = f(x₂) though x₁ ≠ x₂
✅ Hence not one-one

➤ Step 2: Check onto
For any y ∈ R, can we find x such that [x] = y?
Let y = 2.5
No x exists such that [x] = 2.5, since [x] is always integer.
❌ Hence not onto
✅ Therefore, f is neither one-one nor onto.

🔵 Question 4:
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto,
where |x| is x, if x is positive or 0, and |x| = –x, if x is negative.
🟢 Answer:
✏️ Let f(x) = |x|
➡️ For x = 2, f(2) = |2| = 2
➡️ For x = –2, f(–2) = |–2| = 2
💡 Since f(2) = f(–2) but 2 ≠ –2, the function is not one-one.
➡️ Co-domain = R
➡️ Range = R⁺ ∪ {0}
⚡ Range ≠ Co-domain ⇒ function is not onto.
✔️ Therefore, f(x) = |x| is neither one-one nor onto.

🔵 Question 5:
Show that the Signum Function f : R → R, given by
f(x) =
1, if x > 0
0, if x = 0
–1, if x < 0
is neither one-one nor onto.
🟢 Answer:
✏️ Take f(2) = 1 and f(5) = 1
💡 Different inputs give same output ⇒ not one-one.
➡️ Range = {–1, 0, 1}
➡️ Co-domain = R
💡 Some elements like 2, 3 ∈ R are not in range ⇒ not onto.
✔️ Hence, the function is neither one-one nor onto.

🔵 Question 6:
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let
f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
Show that f is one-one.
🟢 Answer:
✏️ Each element of A has a distinct image in B:
f(1) = 4, f(2) = 5, f(3) = 6
💡 No two elements of A have the same image ⇒ f is one-one.
➡️ Range = {4, 5, 6}
➡️ Co-domain = {4, 5, 6, 7}
⚡ Range ≠ Co-domain ⇒ not onto.
✔️ Therefore, f is one-one but not onto.

🔵 Question 7:
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x²
🟢 Answer (i):
✏️ Let f(x₁) = f(x₂)
⇒ 3 – 4x₁ = 3 – 4x₂
⇒ x₁ = x₂
✔️ So, f is one-one.
➡️ For any y ∈ R,
y = 3 – 4x ⇒ x = (3 – y)/4 ∈ R
✔️ Hence, f is onto.
💡 Therefore, f is bijective (both one-one and onto).
🟢 Answer (ii):
✏️ f(x) = 1 + x²
For x₁ = 1 and x₂ = –1,
f(1) = 2 and f(–1) = 2
💡 Different inputs same output ⇒ not one-one.
➡️ Range = [1, ∞)
➡️ Co-domain = R
⚡ Range ≠ Co-domain ⇒ not onto.
✔️ Therefore, f(x) = 1 + x² is neither one-one nor onto.

🔵 Question 8:
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
🟢 Answer:
✏️ Let f(a, b) = (b, a)
➡️ One-one:
Assume f(a₁, b₁) = f(a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂)
⇒ b₁ = b₂ and a₁ = a₂
✔️ Hence (a₁, b₁) = (a₂, b₂)
💡 So f is one-one.
➡️ Onto:
Let (b, a) ∈ B × A.
We can find (a, b) ∈ A × B such that f(a, b) = (b, a).
✔️ So every element in co-domain has a pre-image.
✅ Therefore, f is bijective.

🔵 Question 9:
Let f : N → N be defined by
f(n) = (n + 1)/2, if n is odd
f(n) = n/2, if n is even
for all n ∈ N.
State whether the function f is bijective.
🟢 Answer:
➡️ For odd n: f(1)=1, f(3)=2, f(5)=3, …
➡️ For even n: f(2)=1, f(4)=2, f(6)=3, …
💡 Range = N (since every natural number appears)
✔️ Each value in N has exactly one pre-image (either odd or even number).
So f is one-one and onto.
✅ Hence f is bijective.

🔵 Question 10:
Let A = R – {3} and B = R – {1}.
Consider f : A → B defined by f(x) = (x – 2)/(x – 3).
Is f one-one and onto?
🟢 Answer:
✏️ Let f(x₁) = f(x₂)
⇒ (x₁ – 2)/(x₁ – 3) = (x₂ – 2)/(x₂ – 3)
Cross multiply: (x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
Simplify ⇒ x₁ = x₂
✔️ f is one-one.
➡️ To check onto:
Let y ∈ B.
y = (x – 2)/(x – 3)
⇒ y(x – 3) = x – 2
⇒ yx – 3y = x – 2
⇒ x(y – 1) = 3y – 2
⇒ x = (3y – 2)/(y – 1)
Since y ≠ 1 ⇒ x ≠ 3 ⇒ x ∈ A.
✔️ So f is onto.
✅ Hence f is bijective.

🔵 Question 11:
Let f : R → R be defined as f(x) = x^4.
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
🟢 Answer:
➡️ f(1) = 1 and f(–1) = 1
💡 Different inputs same output ⇒ not one-one.
➡️ Range = [0, ∞), Co-domain = R
💡 Range ≠ Co-domain ⇒ not onto.
✔️ Hence correct option: (D) f is neither one-one nor onto.

🔵 Question 12:
Let f : R → R be defined as f(x) = 3x.
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
🟢 Answer:
✏️ Let f(x₁) = f(x₂)
⇒ 3x₁ = 3x₂
⇒ x₁ = x₂
✔️ So f is one-one.
➡️ For any y ∈ R, let y = 3x ⇒ x = y/3 ∈ R.
✔️ So f is onto.
✅ Correct option: (A) f is one-one onto.

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

OTHER IMPORTANT QUESTIONS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

Questions 1 to 18 (Section A – MCQs)

Question 1:
If A = {1, 2, 3}, the total number of relations on A is
🔵 (A) 16
🟢 (B) 64
🟠 (C) 512
🔴 (D) 8
Answer: (C) 512

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

Question 3:
If R = {(1,2), (2,3), (1,3)}, then R is
🔵 (A) Symmetric
🟢 (B) Transitive
🟠 (C) Reflexive
🔴 (D) None
Answer: (B) Transitive

Question 4:
Number of reflexive relations on a set with 3 elements =
🔵 (A) 2⁶
🟢 (B) 2³
🟠 (C) 2⁹
🔴 (D) 3
Answer: (A) 2⁶

Question 5:
If f(x) = 3x + 2, then f(2) =
🔵 (A) 6
🟢 (B) 8
🟠 (C) 7
🔴 (D) 10
Answer: (B) 8

Question 6:
The range of f(x) = x² + 1 is
🔵 (A) [1, ∞)
🟢 (B) (−∞, 1]
🟠 (C) ℝ
🔴 (D) None
Answer: (A) [1, ∞)

Question 7:
If f(x) = 2x + 3, g(x) = x², then (g ∘ f)(x) =
🔵 (A) 4x² + 4x + 1
🟢 (B) (2x + 3)²
🟠 (C) 2x² + 3
🔴 (D) None
Answer: (B) (2x + 3)²

Question 8:
If f(x) = 3x − 5, then f⁻¹(x) =
🔵 (A) (x + 5)/3
🟢 (B) (x − 5)/3
🟠 (C) 3x + 5
🔴 (D) None
Answer: (A) (x + 5)/3

Question 9:
A function f: A → B is onto if
🔵 (A) Range ⊆ Co-domain
🟢 (B) Range = Co-domain
🟠 (C) Range ⊂ Co-domain
🔴 (D) None
Answer: (B) Range = Co-domain

Question 10:
A constant function is always
🔵 (A) One-one
🟢 (B) Many-one
🟠 (C) Bijective
🔴 (D) Onto
Answer: (B) Many-one

Question 11:
If f(x) = 1/x, then f(f(x)) =
🔵 (A) x
🟢 (B) 1/x
🟠 (C) x²
🔴 (D) None
Answer: (A) x

Question 12:
The inverse of the identity function is
🔵 (A) Itself
🟢 (B) Constant
🟠 (C) Reciprocal
🔴 (D) None
Answer: (A) Itself

Question 13:
If f(x) = x³, then f⁻¹(x) =
🔵 (A) x³
🟢 (B) x¹ᐟ³
🟠 (C) 1/x³
🔴 (D) None
Answer: (B) x¹ᐟ³

Question 14:
The binary operation * on ℝ defined by a * b = a + b + 2 is
🔵 (A) Commutative
🟢 (B) Not associative
🟠 (C) Not closed
🔴 (D) None
Answer: (A) Commutative

Question 15:
Identity element for * defined above is
🔵 (A) 0
🟢 (B) −2
🟠 (C) 2
🔴 (D) None
Answer: (B) −2

Question 16:
If f(x) = |x|, range is
🔵 (A) [0, ∞)
🟢 (B) (−∞, 0]
🟠 (C) ℝ
🔴 (D) None
Answer: (A) [0, ∞)

Question 17:
If f(x) = 1 − x, then f(f(x)) =
🔵 (A) x
🟢 (B) 1 − x
🟠 (C) 1
🔴 (D) None
Answer: (A) x

Question 18:
The number of equivalence relations on {1, 2, 3} is
🔵 (A) 5
🟢 (B) 3
🟠 (C) 6
🔴 (D) 4
Answer: (A) 5

Q19 to Q27 (Section B – 2 marks each)

🔵 Question 19:
Define equivalence relation with an example.
🟢 Answer:
A relation R on a set A is equivalence if it is
✳️ Reflexive ✳️ Symmetric ✳️ Transitive
💡 Example:
Let A = ℝ, R = {(a,b) | a − b is an integer}
✔️ Reflexive → a − a = 0 ∈ ℤ
✔️ Symmetric → if a − b ∈ ℤ ⇒ b − a = −(a − b) ∈ ℤ
✔️ Transitive → if a − b, b − c ∈ ℤ ⇒ a − c ∈ ℤ
✅ Hence, R is an equivalence relation.

🔵 Question 20:
Find f ∘ g and g ∘ f if f(x) = x + 1, g(x) = 2x.
🟢 Answer:
✏️ (g ∘ f)(x) = g(f(x)) = g(x + 1) = 2(x + 1) = 2x + 2
✏️ (f ∘ g)(x) = f(g(x)) = f(2x) = 2x + 1
➡️ ✔️ (g ∘ f)(x) = 2x + 2, (f ∘ g)(x) = 2x + 1

🔵 Question 21:
Show that f : ℝ → ℝ given by f(x) = 3x + 2 is one-one and onto.
🟢 Answer:
✏️ Let f(x₁) = f(x₂)
➡️ 3x₁ + 2 = 3x₂ + 2
➡️ x₁ = x₂ ✔️ one-one
💡 For onto: y = 3x + 2 ⇒ x = (y − 2)/3 ∈ ℝ
➡️ Range = Co-domain = ℝ
✅ Hence f is bijective.

🔵 Question 22:
Find f⁻¹ if f : ℝ → ℝ, f(x) = 2x − 3.
🟢 Answer:
✏️ y = 2x − 3
✏️ Swap ⇒ x = 2y − 3
✏️ Solve ⇒ y = (x + 3)/2
✔️ f⁻¹(x) = (x + 3)/2

🔵 Question 23:
If f(x) = x² + 2, find the range of f over ℝ.
🟢 Answer:
✳️ Minimum value of x² = 0
➡️ Minimum f(x) = 0 + 2 = 2
✔️ Range = [2, ∞)

🔵 Question 24:
Verify that the operation * on ℝ defined by a * b = a + b + 2
is commutative and associative.
🟢 Answer:
✨ Commutative:
a * b = a + b + 2 = b + a + 2 = b * a ✔️
✨ Associative:
(a * b) * c = (a + b + 2) + c + 2 = a + b + c + 4
a * (b * c) = a + (b + c + 2) + 2 = a + b + c + 4 ✔️
✅ Hence * is commutative and associative.

🔵 Question 25:
Find the identity element for the operation * above.
🟢 Answer:
✏️ Let e be identity: a * e = a
➡️ a + e + 2 = a
➡️ e = −2
✔️ Identity = −2

🔵 Question 26:
Find inverse of element a under * (a * b = a + b + 2).
🟢 Answer:
✏️ Let b = inverse of a: a * b = e = −2
➡️ a + b + 2 = −2
➡️ b = −a − 4
✔️ Inverse = −a − 4

🔵 Question 27:
Check whether f : ℝ → ℝ, f(x) = x³ is bijective.
🟢 Answer:
✏️ f(x₁) = f(x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂ ✔️ one-one
✏️ For any y ∈ ℝ, x = y¹ᐟ³ ∈ ℝ ✔️ onto
✅ Hence f is bijective.

🔴 Section D – Long Answer Questions (4 marks each)

🔵 Question 28:
Let A = {1, 2, 3} and B = {a, b, c}. Define a relation R from A to B as R = {(1, a), (2, b), (3, c)}.
✳️ Show that R is a function. Also find f⁻¹.
🟢 Answer:
✔️ Since each element of A is related to exactly one element of B,
➡️ R is a function.
💡 Let f: A → B be defined by f(1) = a, f(2) = b, f(3) = c.
Then, the inverse function f⁻¹: B → A is
➡️ f⁻¹(a) = 1, f⁻¹(b) = 2, f⁻¹(c) = 3.
Thus,
✔️ f⁻¹ = {(a, 1), (b, 2), (c, 3)}

🔵 Question 29:
Prove that the function f(x) = 2x + 3 is invertible and find f⁻¹.
🟢 Answer:
✳️ Given f: ℝ → ℝ, f(x) = 2x + 3
➤ Step 1: Check one-one
Let f(x₁) = f(x₂)
➡️ 2x₁ + 3 = 2x₂ + 3
➡️ x₁ = x₂
✔️ Hence one-one.
➤ Step 2: Check onto
Let y ∈ ℝ,
We need x such that y = 2x + 3
➡️ x = (y − 3)/2
Since x ∈ ℝ, ✔️ f is onto.
Therefore, f is bijective, hence invertible.
➤ Step 3: To find inverse
Put y = f(x) = 2x + 3
Swap x and y:
➡️ x = 2y + 3
Solve for y:
➡️ y = (x − 3)/2
✔️ ∴ f⁻¹(x) = (x − 3)/2

🔵 Question 30:
Find the range of the function f(x) = x² + 4x + 5, x ∈ ℝ.
🟢 Answer:
✳️ Given f(x) = x² + 4x + 5
➤ Step 1: Convert to vertex form
f(x) = (x² + 4x + 4) + 1 = (x + 2)² + 1
Since (x + 2)² ≥ 0,
➡️ f(x) ≥ 1
➤ Step 2: Minimum value = 1
∴ Range = [1, ∞)
✔️ Range of f(x) = [1, ∞)

🔵 Question 31:
Let A = {1, 2, 3, 4} and define a relation R on A by
R = {(a, b) | a − b is even}
Prove that R is an equivalence relation.
🟢 Answer:
✳️ To prove R is equivalence, check:
➤ Reflexive:
a − a = 0, which is even ⇒ (a, a) ∈ R ∀ a ∈ A ✔️
➤ Symmetric:
If (a, b) ∈ R ⇒ a − b is even ⇒ (b − a) is also even
∴ (b, a) ∈ R ✔️
➤ Transitive:
If (a, b) ∈ R and (b, c) ∈ R
Then a − b and b − c both even
⇒ (a − c) = (a − b) + (b − c) even ✔️
Hence (a, c) ∈ R
✔️ ∴ R is reflexive, symmetric, transitive, so equivalence relation.

🟣 Section E – Case/Application Based Questions (5 marks each)

🔵 Question 32:
A company defines salary increment f(x) = 5x + 2000, where x is the grade number (x ∈ {1, 2, 3, 4}).
Find
(i) Salary increment for each grade.
(ii) Whether f is one-one and onto from set A = {1, 2, 3, 4} to B = {2005, 2010, 2015, 2020}.
🟢 Answer:
➤ (i) f(1) = 2005, f(2) = 2010, f(3) = 2015, f(4) = 2020
✔️ Set B = {2005, 2010, 2015, 2020}
➤ (ii)
Each element of A has a unique image → one-one ✔️
Every element of B is covered → onto ✔️
✔️ ∴ f is bijective.

🔵 Question 33:
Consider f: ℝ → ℝ, f(x) = x³ + x
Show that f is one-one and onto, hence invertible. Find f⁻¹.
🟢 Answer:
✳️ Given f(x) = x³ + x
➤ One-one:
Let f(x₁) = f(x₂)
⇒ x₁³ + x₁ = x₂³ + x₂
⇒ (x₁ − x₂)(x₁² + x₁x₂ + x₂² + 1) = 0
Since (x₁² + x₁x₂ + x₂² + 1) > 0,
⇒ x₁ = x₂ ✔️
➤ Onto:
Let y ∈ ℝ, we need x such that y = x³ + x
The cubic x³ + x − y = 0 has at least one real root ✔️
∴ For every y, ∃ x ⇒ onto.
Hence f is bijective ⇒ invertible.
Inverse f⁻¹ exists (no simple algebraic form).
✔️ f is invertible.

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