Chapter 01 · Foundation of Mathematics

Sets and Relations

Sets are the building blocks of all modern mathematics. In this chapter you will master set notation, operations, ordered pairs, relations and how to compose them — step by step, with clear examples at every stage.

7 Modules ⌛ ~45 min read Class 11 Core

1.1

Sets

What is a set? Notation, types, and the language of mathematics

A set is a well-defined collection of distinct objects. The word well-defined is key — it means there is no ambiguity about whether an object belongs to the collection or not.

📖 Definition — Set

A set is a collection of well-defined, distinct objects called elements or members. We write x ∈ A to say “x belongs to set A”, and x ∉ A to say “x does not belong to A.”

How to Write a Set

There are two standard ways to describe a set:

Roster (Tabular) Form: List all elements inside curly braces separated by commas. Example: A = {1, 2, 3, 4, 5}
Set-Builder Form: Describe the rule that elements must satisfy. Example: A = {x : x is a natural number less than 6} — read as “the set of all x such that x is a natural number less than 6.”

Types of Sets

∅

Empty Set

A set with no elements. Also called the null set.

∅ or { }

1⃣

Singleton Set

A set containing exactly one element.

{7} or {Monday}

🔢

Finite Set

A set with a countable, definite number of elements.

{2, 4, 6, 8}

∞

Infinite Set

A set with uncountably many elements.

ℕ = {1,2,3,...}

≡

Equal Sets

Two sets are equal if they have exactly the same elements.

{1,2} = {2,1}

💡

Universal Set

The set containing all objects under consideration, denoted U or 𝒰.

U = {1...10}

✏️ Quick Example — Roster vs Set-BuilderExample 1.1

Express “the set of even numbers between 1 and 10” in both forms.

1

Roster form: List them out: {2, 4, 6, 8, 10}

2

Set-builder form: Describe the rule: {x : x is even, 1 ≤ x ≤ 10}

✔

Both forms represent the same set. Roster is easier to read; set-builder is more powerful for large or infinite sets.

Cardinality of a Set

The cardinality of a set A, written n(A) or |A|, is the number of elements it contains.

Cardinality

n(A) = number of elements in A

Empty Set

n(∅) = 0

Power Set

n(P(A)) = 2^n(A)

Module 1 of 7

1.2

Subset and Superset

Containment between sets — subsets, supersets, and the power set

Once we understand what sets are, the next question is: how do sets relate to each other in terms of containment? This is where subsets and supersets come in.

📖 Definition — Subset

Set A is a subset of set B, written A ⊆ B, if every element of A is also an element of B. In other words: if x ∈ A, then x ∈ B.

📖 Definition — Proper Subset

A is a proper subset of B, written A ⊂ B, if A ⊆ B and A ≠ B. That is, every element of A is in B, but B has at least one extra element.

Superset

If A ⊆ B, then B is called a superset of A, written B ⊇ A. The superset is the “bigger” set that contains A.

✏️ Quick Example — SubsetsExample 1.2

Let A = {1, 2}, B = {1, 2, 3}, C = {4, 5}. Which of A, B, C are subsets of B?

1

A ⊆ B? 1 ∈ B ✔ 2 ∈ B ✔ Yes, A ⊂ B (proper subset)

2

B ⊆ B? Every element of B is in B. Yes, B ⊆ B (every set is a subset of itself)

3

C ⊆ B? 4 ∉ B. No, C is NOT a subset of B.

✔

A ⊂ B (proper), B ⊆ B (improper), C ⊈ B. Note: ∅ ⊆ every set.

Important Results about Subsets

Every set is a subset of itself: A ⊆ A
The empty set is a subset of every set: ∅ ⊆ A for any set A
If A ⊆ B and B ⊆ A, then A = B
If A ⊆ B and B ⊆ C, then A ⊆ C (transitivity)
The number of subsets of a set with n elements is 2ⁿ
The number of proper subsets is 2ⁿ − 1

The Power Set

The power set of A, written P(A), is the set of ALL subsets of A (including ∅ and A itself).

✏️ Quick Example — Power SetExample 1.3

Find the power set of A = {a, b, c}.

1

n(A) = 3, so P(A) has 2³ = 8 subsets

2

List all subsets (by size): ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}

✔

P(A) = { ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c} } — 8 elements total.

Module 2 of 7

1.3

Operations on Sets

Union, Intersection, Difference, Complement & De Morgan's Laws

Just as we add, subtract or multiply numbers, we can operate on sets to produce new sets. The four fundamental operations are Union, Intersection, Difference, and Complement.

Union (A ∪ B)

Definition

A ∪ B = { x : x ∈ A or x ∈ B } — all elements that belong to A, to B, or to both.

Intersection (A ∩ B)

Definition

A ∩ B = { x : x ∈ A and x ∈ B } — only elements that belong to BOTH A and B simultaneously.

Difference (A − B)

Definition

A − B = { x : x ∈ A and x ∉ B } — elements in A that are not in B. Also written A \ B.

Complement (A')

Definition

A' = { x : x ∈ U and x ∉ A } — everything in the universal set U that is not in A.

✏️ Quick Example — All Four OperationsExample 1.4

Let U = {1,2,3,4,5,6,7,8}, A = {1,2,3,4}, B = {3,4,5,6}. Find A∪B, A∩B, A−B, and A'.

1

A∪B (all elements in either): {1,2,3,4,5,6}

2

A∩B (common elements only): {3,4}

3

A−B (in A, not in B): {1,2}

4

A' (in U, not in A): {5,6,7,8}

✔

Notice B−A = {5,6} ≠ A−B = {1,2}. Set difference is NOT commutative.

Key Properties

Commutative

A∪B = B∪A
A∩B = B∩A

Associative

A∪(B∪C) = (A∪B)∪C
A∩(B∩C) = (A∩B)∩C

Distributive

A∩(B∪C) = (A∩B)∪(A∩C)

Inclusion-Exclusion

n(A∪B) = n(A)+n(B)−n(A∩B)

De Morgan's Laws

These two laws are among the most important results in set theory. They connect complement with union and intersection:

De Morgan 1

(A ∪ B)' = A' ∩ B'

De Morgan 2

(A ∩ B)' = A' ∪ B'

In words: the complement of a union is the intersection of complements. The complement of an intersection is the union of complements.

✏️ Quick Example — Inclusion-ExclusionExample 1.5

In a class of 40 students, 25 study Maths, 20 study Physics, and 10 study both. How many study at least one subject?

1

n(M) = 25, n(P) = 20, n(M∩P) = 10

2

Apply formula: n(M∪P) = 25 + 20 − 10 = 35

✔

35 students study at least one subject. The remaining 40 − 35 = 5 study neither.

Module 3 of 7

1.4

Ordered Pair

The building block of Cartesian products and Relations

When we pair two elements and the order matters, we get an ordered pair. This is different from a set, where {a, b} = {b, a}. In an ordered pair, (a, b) ≠ (b, a) unless a = b.

Definition — Ordered Pair

An ordered pair (a, b) is a pair of elements where a is the first component and b is the second component. Two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d.

Cartesian Product

The Cartesian product A × B is the set of all ordered pairs where the first element comes from A and the second from B.

Definition — Cartesian Product

A × B = { (a, b) : a ∈ A and b ∈ B }

Size of A×B

n(A×B) = n(A) × n(B)

Not Commutative

A×B ≠ B×A (in general)

A×∅

A × ∅ = ∅

✏️ Quick Example — Cartesian ProductExample 1.6

If A = {1, 2} and B = {x, y, z}, find A×B and n(A×B).

1

Form all ordered pairs (a,b) where a∈A and b∈B:

2

A×B = {(1,x),(1,y),(1,z),(2,x),(2,y),(2,z)}

3

Count: n(A×B) = 2 × 3 = 6

✔

The Cartesian product has 6 ordered pairs. Note A×B ≠ B×A since (1,x) ≠ (x,1).

Why Ordered Pairs Matter

Ordered pairs are the backbone of coordinate geometry (every point on a graph is an ordered pair), relations, and functions. Whenever you plot (3, 5) on a graph, you are using an ordered pair from ℝ×ℝ.

Module 4 of 7

1.5

Relation

Connecting elements of sets — types and properties of relations

A relation describes how elements from one set are connected to elements of another. Formally, it is a subset of the Cartesian product.

Definition — Relation

A relation R from set A to set B is any subset of A × B. We write aRb or (a,b) ∈ R to say “a is related to b.”
The domain of R is the set of first elements; the range is the set of second elements.

Types of Relations on a Set A

🔄

Reflexive

Every element is related to itself.

(a,a) ∈ R for all a∈A

⇄

Symmetric

If aRb then bRa.

(a,b)∈R ⇒ (b,a)∈R

➡

Transitive

If aRb and bRc then aRc.

aRb, bRc ⇒ aRc

≠

Anti-symmetric

aRb and bRa implies a = b.

Like “≤” on numbers

⭐

Equivalence

Reflexive + Symmetric + Transitive.

Like “=” on numbers

∅

Empty Relation

No element is related to any element.

R = ∅

✏️ Quick Example — Identifying Relation TypesExample 1.7

Let A = {1,2,3} and R = {(1,1),(2,2),(3,3),(1,2),(2,1)}. Is R reflexive, symmetric, transitive?

1

Reflexive: (1,1),(2,2),(3,3) ∈ R ✔ Yes, reflexive.

2

Symmetric: (1,2) ∈ R and (2,1) ∈ R ✔ Yes, symmetric.

3

Transitive: (1,2) and (2,1) ∈ R, so need (1,1) ∈ R ✔. Check all — holds. Yes, transitive.

✔

R is reflexive, symmetric, and transitive — therefore R is an Equivalence Relation.

Number of Relations

If n(A) = p and n(B) = q, then n(A×B) = pq. The total number of possible relations from A to B is 2^pq, since each relation is a subset of A×B.

Relations from A to B

2^n(A)×n(B)

Relations on A

2^n(A)²

Domain of R

{a : (a,b) ∈ R}

Range of R

{b : (a,b) ∈ R}

Module 5 of 7

1.6

Composition of Relations

Chaining relations to create new ones — R∘S and its properties

Just as we compose functions, we can compose two relations to produce a third. If R relates A to B, and S relates B to C, then the composition R∘S (or SoR in some notations) relates A to C by chaining through B.

Definition — Composition of Relations

Let R be a relation from A to B, and S be a relation from B to C. The composition S ∘ R (read “S composed with R”) is a relation from A to C defined as:
S ∘ R = { (a, c) : ∃ b ∈ B such that (a,b) ∈ R and (b,c) ∈ S }

How to Compute a Composition

Think of it as a two-step path: to get from a to c, you must first find an intermediate b such that a is R-related to b, and b is S-related to c.

✏️ Quick Example — Composing RelationsExample 1.8

Let A={1,2,3}, B={a,b,c}, C={x,y}. R = {(1,a),(2,b),(3,c)} from A to B. S = {(a,x),(b,x),(c,y)} from B to C. Find S∘R.

1

Trace each element of A through R, then through S:

2

1: 1 → a (via R) → x (via S) so (1,x) ∈ S∘R

3

2: 2 → b (via R) → x (via S) so (2,x) ∈ S∘R

4

3: 3 → c (via R) → y (via S) so (3,y) ∈ S∘R

✔

S ∘ R = { (1,x), (2,x), (3,y) } — a relation directly from A to C.

Key Properties of Composition

Associativity: (T∘S)∘R = T∘(S∘R) — composition is associative
Not Commutative: S∘R ≠ R∘S in general — the order matters
Identity Relation: I∘R = R∘I = R, where I is the identity relation {(a,a) : a∈A}

Inverse of a Relation

The inverse of relation R, written R⁻¹, swaps all pairs: R⁻¹ = { (b,a) : (a,b) ∈ R }. If R is from A to B, then R⁻¹ is from B to A.

Inverse Relation

R⁻¹ = {(b,a) : (a,b)∈R}

(R⁻¹)⁻¹

(R⁻¹)⁻¹ = R

Composition Inverse

(S∘R)⁻¹ = R⁻¹∘S⁻¹

Module 6 of 7

1.7

Venn Diagram

Visualising sets and operations with overlapping circles

A Venn diagram uses overlapping circles inside a rectangle (representing the universal set U) to visually represent sets and their relationships. They make abstract set operations instantly intuitive.

What the Regions Mean

In a two-set Venn diagram with circles A and B:
• Circle A only (left crescent) = elements in A but not B = A−B
• Overlap (middle) = elements in both = A∩B
• Circle B only (right crescent) = elements in B but not A = B−A
• Outside both circles = elements in neither = (A∪B)'

Venn Diagram showing A−B (orange left), A∩B (dark overlap), B−A (green right), and (A∪B)' (outside)

Three-Set Venn Diagrams

When three sets A, B, C overlap, the diagram has 8 distinct regions. The inclusion-exclusion formula extends to:

Three-Set Inclusion-Exclusion

n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)

✏️ Quick Example — Using a Venn DiagramExample 1.9

In a survey of 100 people: 60 like tea, 50 like coffee, 30 like both. Draw a Venn diagram and find: (i) who like only tea, (ii) who like only coffee, (iii) who like neither.

1

Only Tea (left crescent): 60 − 30 = 30

2

Only Coffee (right crescent): 50 − 30 = 20

3

At least one: n(T∪C) = 60+50−30 = 80

4

Neither: 100 − 80 = 20

✔

Only tea: 30 | Only coffee: 20 | Both: 30 | Neither: 20

Common Exam Tips for Venn Diagrams

Always fill in the intersection region first, then work outward
Use the inclusion-exclusion formula to find missing values
The sum of all regions must equal n(U)
A region cannot have a negative count — if it does, recheck your working

Module 7 of 7 🎉

Chapter 02 · Mappings & Structures

Functions & Binary Operations

Functions are the engine of mathematics — they describe how one quantity depends on another. In this chapter you will master function types, graphs, operations, composition, even & odd functions, and binary operations step by step.

8 Modules ⌛ ~50 min read Class 11 & 12 Core

2.1

Functions

Definition, domain, co-domain, range — and the four key types

A function is a special type of relation where every element of the input set is paired with exactly one element of the output set. No input can map to two different outputs.

📖 Definition — Function

A function f from set A to set B, written f : A → B, is a rule that assigns to each element x ∈ A exactly one element f(x) ∈ B.
• Domain = A (all valid inputs)
• Co-domain = B (all possible outputs)
• Range = f(A) = { f(x) : x ∈ A } ⊆ B (actual outputs produced)

The range is always a subset of the co-domain. When every element of B is actually produced, the range equals the co-domain and the function is called onto.

The Four Types of Functions

→

One-One (Injective)

Different inputs always give different outputs. f(x₁) = f(x₂) ⟹ x₁ = x₂.

f(x) = 2x+1

↠

Onto (Surjective)

Every element of the co-domain has at least one preimage. Range = Co-domain.

f(x) = x³ on ℝ

↔

Bijective (One-One & Onto)

Both injective and surjective. Has a unique inverse function.

f(x) = x+5 on ℝ

↬

Many-One

Two or more inputs map to the same output. Not injective.

f(x) = x² (f(2)=f(−2))

✏️ Quick Example — Testing Function TypesExample 2.1

Let f : ℝ → ℝ be defined by f(x) = 3x − 2. Is f one-one? Is f onto? Find the range.

1

One-one test: Assume f(x₁) = f(x₂): 3x₁−2 = 3x₂−2 ⟹ x₁ = x₂ ✔ One-one.

2

Onto test: For any y ∈ ℝ, solve f(x) = y: x = (y+2)/3 ∈ ℝ ✔ Onto.

3

Range = all real numbers = ℝ (equals co-domain).

✔

f(x) = 3x−2 is bijective. Its inverse is f⁻¹(x) = (x+2)/3.

How to Count Functions

Total Functions A→B

n(B)^n(A)

One-One possible when

n(A) ≤ n(B)

Onto possible when

n(A) ≥ n(B)

Bijective possible when

n(A) = n(B)

Vertical Line Test

A curve in the xy-plane represents a function if and only if every vertical line crosses the curve at most once. If a vertical line crosses at two points, the curve is not a function (one input gives two outputs).

Module 1 of 8

2.2

Equal Functions

When are two functions truly the same?

Two functions that look different can sometimes be equal. But the definition of equality for functions is very precise — it is not enough to have the same formula.

📖 Definition — Equal Functions

Two functions f and g are equal (written f = g) if and only if:
1. They have the same domain
2. They have the same co-domain
3. f(x) = g(x) for every x in the domain

All three conditions must hold simultaneously. If the domains differ, f and g are not equal — even if the formulas are identical.

✏️ Quick Example — Are These Functions Equal?Example 2.2

Let f(x) = x and g(x) = x²/x. Are f and g equal?

1

Domain of f: all real numbers ℝ

2

Domain of g: g(x) = x²/x is undefined at x=0, so domain = ℝ \ {0}

3

Although g(x) = x for all x ≠ 0, the domains differ.

✘

f ≠ g because they have different domains. This is a very common exam trap!

✏️ Quick Example — Equal FunctionsExample 2.3

f(x) = |x| / x and g(x) = x / |x|. Are they equal? Find values.

1

Both have domain ℝ \ {0}. For x > 0: f(x) = 1, g(x) = 1

2

For x < 0: f(x) = −1/1 = −1, g(x) = −1

3

Same domain, same outputs everywhere.

✔

f = g. Both are the sign function: +1 for positive, −1 for negative inputs.

Two functions with different domains are never equal, even if their formulas look the same
Always check domain first before comparing function values
Equal functions produce identical graphs (same curve, same extent)

Module 2 of 8

2.3

Real Valued and Real Functions

Functions whose inputs and outputs are real numbers

Most functions you encounter in Class 11 & 12 take real number inputs and produce real number outputs. Understanding how to find their natural domain is an essential exam skill.

📖 Definitions

Real-valued function: f : A → ℝ — the co-domain is ℝ (outputs are real numbers, inputs can be any set).

Real function: f : A → ℝ where A ⊆ ℝ — both the domain and co-domain are subsets of real numbers. This is the most common type in your syllabus.

Finding the Natural Domain

The natural domain (or implied domain) is the largest subset of ℝ for which f(x) is defined and real. The two most common restrictions are:

No division by zero: Denominator ≠ 0 — set denominator ≠ 0 and exclude those x values
No square root of negatives: Expression under √ must be ≥ 0 — solve the inequality
No log of zero or negative: Argument of log must be > 0
Combined restrictions: Apply both conditions together using set intersection

✏️ Quick Example — Finding Natural DomainExample 2.4

Find the natural domain of f(x) = √(x − 2) / (x − 5).

1

Square root condition: x − 2 ≥ 0 ⟹ x ≥ 2

2

Denominator condition: x − 5 ≠ 0 ⟹ x ≠ 5

3

Combine (intersection): x ≥ 2 and x ≠ 5

✔

Domain = [2, 5) ∪ (5, ∞) — all reals from 2 onwards, except x = 5.

✏️ Quick Example — Finding RangeExample 2.5

Find the range of f(x) = (x² − 1) / (x² + 1) for x ∈ ℝ.

1

Let y = (x²−1)/(x²+1). Rearrange: y(x²+1) = x²−1 ⟹ x²(y−1) = −1−y

2

x² = (−1−y)/(y−1) = (1+y)/(1−y). Need x² ≥ 0 and y ≠ 1.

3

(1+y)/(1−y) ≥ 0 ⟹ −1 ≤ y < 1

✔

Range = [−1, 1). Note: y = 1 is excluded since x² would be infinite.

Domain: √f(x)

f(x) ≥ 0

Domain: 1/f(x)

f(x) ≠ 0

Domain: log f(x)

f(x) > 0

Range Method

Express x in terms of y, find valid y

Module 3 of 8

2.4

Standard Real Functions and Their Graphs

The six essential functions every student must know by heart

These standard functions appear repeatedly throughout mathematics. Know their shape, domain, range, and key properties cold — they are the building blocks for everything else.

1. Constant Function — f(x) = c

Properties

Domain: ℝ | Range: {c} (single value) | Graph: Horizontal line at height c.
This function maps every real number to the same fixed value c. It is neither one-one nor onto (unless co-domain = {c}).

2. Identity Function — f(x) = x

Properties

Domain: ℝ | Range: ℝ | Graph: Straight line through origin at 45°.
Bijective. f(x) = x means every input equals its output. It is its own inverse.

3. Modulus (Absolute Value) Function — f(x) = |x|

Properties

f(x) = x if x ≥ 0 and f(x) = −x if x < 0
Domain: ℝ | Range: [0, ∞) | Graph: V-shape with vertex at origin.
Many-one (since f(2) = f(−2) = 2). Not injective, not onto ℝ.

4. Signum Function — sgn(x)

Properties

sgn(x) = +1 if x > 0 | 0 if x = 0 | −1 if x < 0
Domain: ℝ | Range: {−1, 0, 1} | Graph: Three isolated points / step. Many-one.

5. Greatest Integer Function — f(x) = [x] or ⌊x⌋

Properties

⌊x⌋ = largest integer ≤ x. Example: ⌊3.7⌋ = 3, ⌊−1.2⌋ = −2.
Domain: ℝ | Range: ℤ (integers) | Graph: Staircase (step function).
Key property: x−1 < ⌊x⌋ ≤ x

6. Fractional Part Function — f(x) = {x}

Properties

{x} = x − ⌊x⌋ — the decimal part of x. Example: {3.7} = 0.7, {−1.3} = 0.7.
Domain: ℝ | Range: [0, 1) | Graph: Repeating sawtooth pattern, period 1.

✏️ Quick Example — Greatest Integer & Fractional PartExample 2.6

Evaluate: ⌊2.9⌋, ⌊−2.9⌋, {2.9}, {−2.9}.

1

⌊2.9⌋ = 2 (greatest integer ≤ 2.9 is 2)

2

⌊−2.9⌋ = −3 (greatest integer ≤ −2.9 is −3, NOT −2)

3

{2.9} = 2.9 − 2 = 0.9

4

{−2.9} = −2.9 − (−3) = 0.1 — always non-negative!

✔

⌊2.9⌋ = 2 | ⌊−2.9⌋ = −3 | {2.9} = 0.9 | {−2.9} = 0.1

Module 4 of 8

2.5

Operations on Real Functions

Adding, subtracting, multiplying and dividing functions

If f and g are two real functions with domains D₁ and D₂, we can combine them arithmetically. The new functions are defined on the intersection D₁ ∩ D₂ (where both are defined).

📖 The Four Arithmetic Operations

(f + g)(x) = f(x) + g(x) — domain: D₁ ∩ D₂
(f − g)(x) = f(x) − g(x) — domain: D₁ ∩ D₂
(f · g)(x) = f(x) · g(x) — domain: D₁ ∩ D₂
(f / g)(x) = f(x) / g(x) — domain: D₁ ∩ D₂, excluding points where g(x) = 0

✏️ Quick Example — Function OperationsExample 2.7

Let f(x) = √x (domain [0,∞)) and g(x) = √(4−x²) (domain [−2,2]). Find f+g, f·g, f/g and their domains.

1

Common domain: D₁ ∩ D₂ = [0,∞) ∩ [−2,2] = [0, 2]

2

(f+g)(x) = √x + √(4−x²), domain: [0, 2]

3

(f·g)(x) = √x · √(4−x²) = √(x(4−x²)), domain: [0, 2]

4

(f/g)(x) = √x / √(4−x²), g(x)=0 when x=଒ ⟹ exclude x=2: domain: [0, 2)

✔

f+g and f·g have domain [0,2]. Division f/g has domain [0,2) — x=2 excluded.

Scalar Multiplication

For any real constant k, (kf)(x) = k·f(x) with the same domain as f. The graph is vertically stretched (k>1), compressed (0<k<1), or reflected (k<0).

(f+g)(x)

f(x) + g(x) on D₁∩D₂

(f⋅g)(x)

f(x) · g(x) on D₁∩D₂

(f/g)(x)

f(x)/g(x), g(x)≠0

kf(x)

k·f(x), same domain as f

Module 5 of 8

2.6

Composition of Two Functions

Chaining functions — the output of one becomes the input of another

Function composition is one of the most important operations in mathematics. It is how we build complex functions from simpler pieces.

📖 Definition — Composition

If f : A → B and g : B → C, the composition g ∘ f (read “g of f”) is defined as:
(g ∘ f)(x) = g(f(x))
First apply f to x, then apply g to the result. The domain of g∘f is the set of all x in the domain of f such that f(x) is in the domain of g.

Composition pipeline: x → f(x) → g(f(x))

Key Properties of Composition

Not commutative: g∘f ≠ f∘g in general. Order matters!
Associative: (h∘g)∘f = h∘(g∘f) — grouping doesn't matter
Identity: f∘I = I∘f = f, where I(x) = x
Inverse: If f is bijective, f∘f⁻¹ = f⁻¹∘f = I
Composition of bijectives is bijective: If f and g are both bijective, g∘f is also bijective

✏️ Quick Example — Composition is Not CommutativeExample 2.8

Let f(x) = x² and g(x) = x + 3. Find (g∘f)(x), (f∘g)(x), and compare.

1

(g∘f)(x) = g(f(x)) = g(x²) = x² + 3

2

(f∘g)(x) = f(g(x)) = f(x+3) = (x+3)² = x²+6x+9

3

x²+3 ≠ x²+6x+9 (e.g. at x=1: 4 ≠ 16)

✔

g∘f ≠ f∘g. Composition is not commutative.

✏️ Quick Example — Finding Inverse via CompositionExample 2.9

If f(x) = (2x+1)/(x−3), find f⁻¹(x) and verify f∘f⁻¹ = I.

1

Let y = (2x+1)/(x−3). Solve for x: y(x−3) = 2x+1 ⟹ x(y−2) = 3y+1

2

x = (3y+1)/(y−2) so f⁻¹(x) = (3x+1)/(x−2)

3

Verify: f(f⁻¹(x)) = f((3x+1)/(x−2)) = simplifies to x ✔

✔

f⁻¹(x) = (3x+1)/(x−2). Note domain excludes x=2.

(g∘f)(x)

g(f(x))

Inverse check

f∘f⁻¹ = I(x) = x

(g∘f)⁻¹

f⁻¹ ∘ g⁻¹

Module 6 of 8

2.7

Even and Odd Functions

Symmetry of functions — a powerful tool for graphs and integration

The concepts of even and odd describe a function's symmetry. They allow you to quickly sketch graphs and simplify calculations. The classification applies only to functions whose domain is symmetric about 0 (if x is in the domain, so is −x).

📖 Definition — Even Function

f is even if f(−x) = f(x) for all x in the domain.
Graph is symmetric about the y-axis (left and right halves are mirror images).
Examples: f(x) = x², f(x) = cos x, f(x) = |x|

📖 Definition — Odd Function

f is odd if f(−x) = −f(x) for all x in the domain.
Graph is symmetric about the origin (rotation by 180° gives the same graph).
Examples: f(x) = x³, f(x) = sin x, f(x) = x

Left: even function (y-axis symmetry) | Right: odd function (origin symmetry)

Neither Even nor Odd

Most functions are neither even nor odd. For example f(x) = x² + x: f(−x) = x² − x ≠ f(x) and ≠ −f(x). Always check both conditions.

Properties to Remember

Even ± Even = Even | Odd ± Odd = Odd | Even ± Odd = Neither
Even × Even = Even | Odd × Odd = Even | Even × Odd = Odd
The only function that is both even and odd is f(x) = 0
Derivative of even is odd; derivative of odd is even
∫₋ₐₐ f(x)dx = 0 if f is odd; = 2∫₀ₐ f(x)dx if f is even

✏️ Quick Example — Classify the FunctionExample 2.10

Classify f(x) = x³ + x as even, odd, or neither.

1

Compute f(−x): f(−x) = (−x)³ + (−x) = −x³ − x = −(x³ + x)

2

Compare with −f(x) = −(x³ + x). We have f(−x) = −f(x) ✔

✔

f(x) = x³ + x is an odd function. Its graph is symmetric about the origin.

✏️ Quick Example — Neither Even nor OddExample 2.11

Is f(x) = eˣ even, odd, or neither?

1

f(−x) = e⁻ˣ. Is this equal to f(x) = eˣ? Only if x=0. Not even.

2

Is f(−x) = −f(x)? That would require e⁻ˣ = −eˣ, impossible for real x. Not odd.

✔

f(x) = eˣ is neither even nor odd. No y-axis symmetry, no origin symmetry.

Module 7 of 8

2.8

Binary Operations

Combining two elements of a set to produce a third

A binary operation is any rule that takes two elements from a set and produces a result in the same set. Addition, subtraction, multiplication are all familiar binary operations on numbers.

📖 Definition — Binary Operation

A binary operation * on a non-empty set A is a function * : A × A → A.
That is, for every a, b ∈ A, the result a * b is also in A. This is called the closure property.

Properties of Binary Operations

🔒

Closure

a * b ∈ A for all a, b ∈ A

2+3=5 ∈ ℤ ✔

⇄

Commutative

a * b = b * a for all a, b

2+3=3+2 ✔

📏

Associative

(a * b) * c = a * (b * c)

(2+3)+4=2+(3+4) ✔

🟰

Identity Element

a * e = e * a = a for all a

a+0=a (e=0 for +)

↺

Inverse Element

a * b = b * a = e (identity)

3+(−3)=0 ✔

📦

Distributive

a*(b◇c) = (a*b)◇(a*c)

a×(b+c) = a×b+a×c ✔

Operation Tables

For finite sets, binary operations can be described by a Cayley table (operation table). The entry in row a, column b gives a * b.

✏️ Quick Example — Checking PropertiesExample 2.12

Let * be defined on ℚ by a * b = a + b − ab. Check closure, commutativity, associativity, identity element.

1

Closure: a+b−ab ∈ ℚ for all a,b ∈ ℚ ✔

2

Commutative: b*a = b+a−ba = a+b−ab = a*b ✔ (addition and multiplication are commutative)

3

Identity e: a*e = a+e−ae = a ⟹ e−ae = 0 ⟹ e(1−a) = 0 ⟹ e = 0

4

Verify: a*0 = a+0−0 = a ✔ Identity element is 0.

✔

* is closed, commutative, and has identity element 0. Check associativity separately (it also holds here).

✏️ Quick Example — Inverse of an ElementExample 2.13

Using * from above (a*b = a+b−ab, identity = 0), find the inverse of element a.

1

Need a * b = 0 (the identity): a + b − ab = 0

2

Solve for b: b(1−a) = −a ⟹ b = a/(a−1)

3

Valid only when a ≠ 1 (denominator ≠ 0).

✔

Inverse of a (under *) is a/(a−1), defined for all a ≠ 1.

Closure

a*b ∈ A for all a,b ∈ A

Identity

a*e = e*a = a

Inverse

a*a⁻¹ = a⁻¹*a = e

Commutative

a*b = b*a

Module 8 of 8 🎉

Chapter 03 · Into the Imaginary Plane

Complex Numbers

Complex numbers extend the real number system using the imaginary unit i = √(−1). From basic algebra to the Argand plane, cube roots of unity to geometry — this chapter builds a complete, visual understanding step by step.

7 Modules ⌛ ~50 min read Class 11 Core

3.1

Equality of Complex Numbers

What does it mean for two complex numbers to be equal?

Before we can do any arithmetic with complex numbers, we need to understand when two complex numbers are considered equal. The rule is precise and different from real numbers — both the real part and the imaginary part must match simultaneously.

📖 Definition — Complex Number

A complex number is a number of the form z = a + bi, where:
• a = Re(z) is the real part (a ∈ ℝ)
• b = Im(z) is the imaginary part (b ∈ ℝ)
• i is the imaginary unit satisfying i² = −1
The set of all complex numbers is denoted ℂ.

📖 Definition — Equality

Two complex numbers z₁ = a + bi and z₂ = c + di are equal (z₁ = z₂) if and only if:
• a = c (real parts are equal) and
• b = d (imaginary parts are equal)
Both conditions must hold simultaneously.

This is powerful because one complex equation z₁ = z₂ is secretly two real equations packed together. We use this to solve for unknowns by separating real and imaginary parts.

Special Cases

z = a + 0i = a is a pure real number. Every real number is a complex number with Im(z) = 0.
z = 0 + bi = bi is a purely imaginary number (when b ≠ 0). Its real part is zero.
z = 0 + 0i = 0 is the zero complex number, the only number that is both real and purely imaginary.
ℂ cannot be ordered — you cannot say 3 + 2i > 1 + i the way you can with real numbers.

✏️ Quick Example — Using Equality to Find UnknownsExample 3.1

Find real numbers x and y such that (2x − 1) + (y + 3)i = 5 − 2i.

1

Equate real parts: 2x − 1 = 5 ⟹ 2x = 6 ⟹ x = 3

2

Equate imaginary parts: y + 3 = −2 ⟹ y = −5

✔

x = 3, y = −5. Verify: (6−1) + (−5+3)i = 5 − 2i ✔

✏️ Quick Example — Complex EquationExample 3.2

Solve for z if z + z̅ = 6 and z − z̅ = 4i, where z = x + yi.

1

z + z̅ = (x+yi) + (x−yi) = 2x = 6 ⟹ x = 3

2

z − z̅ = (x+yi) − (x−yi) = 2yi = 4i ⟹ y = 2

✔

z = 3 + 2i. Key insight: z + z̅ = 2Re(z) and z − z̅ = 2i⋅Im(z).

Re(z)

Re(a+bi) = a

Im(z)

Im(a+bi) = b

Conjugate z̅

z̅ = a − bi

z + z̅

= 2Re(z)

z − z̅

= 2i⋅Im(z)

Module 1 of 7

3.2

Iota (i) — The Imaginary Unit

Powers of i form a repeating cycle of period 4

The imaginary unit i (iota) is defined as the square root of −1. It cannot be placed on the real number line — it lives in an entirely new dimension. Yet it follows consistent algebraic rules.

📖 Definition — Iota

i = √(−1), equivalently i² = −1.
This definition is the foundation of the entire complex number system. From just this one rule, all of complex arithmetic follows.

Powers of i — The Four-Cycle

The powers of i follow a repeating cycle of period 4. This is one of the most-tested facts in exams:

i¹

i

i to the power 1

i²

−1

i squared = −1

i³

−i

i cubed = −i

i⁴

1

i⁴ = 1 (cycle resets)

After i⁴ = 1 the pattern repeats: i⁵ = i, i⁶ = −1, i⁷ = −i, i⁸ = 1, and so on.

🔑 Master Formula for Any Power of i

To find iⁿ, divide n by 4 and look at the remainder r = n mod 4:
r = 0 ⟹ iⁿ = 1 | r = 1 ⟹ iⁿ = i | r = 2 ⟹ iⁿ = −1 | r = 3 ⟹ iⁿ = −i

✏️ Quick Example — Powers of iExample 3.3

Evaluate: (i) i⁷⁵ (ii) i⁻³² (iii) i² + i⁴ + i⁶ + i⁸

1

i⁷⁵: 75 ÷ 4 = 18 remainder 3 ⟹ i⁷⁵ = i³ = −i

2

i⁻³²: i⁻³² = 1/i³² = 1/(i⁴)⁸ = 1/1⁸ = 1

3

i²+i⁴+i⁶+i⁸: = (−1) + 1 + (−1) + 1 = 0

✔

i⁷⁵ = −i | i⁻³² = 1 | Sum = 0 (consecutive even powers of i always sum to 0)

Square Roots of Negative Numbers

Using iota, we can now compute square roots of any negative number:

√(−a) for a > 0

√(−a) = i√a

√(−4)

= i√4 = 2i

√(−7)

= i√7

⚠️ Caution

√a · √b ≠ √(ab) when a,b < 0

⚠️ Common Mistake

√(−4) × √(−9) ≠ √(36) = 6.
Correct: √(−4) × √(−9) = 2i × 3i = 6i² = 6(−1) = −6.
Always convert to iota form first before multiplying.

Module 2 of 7

3.3

Algebra of Complex Numbers

Addition, subtraction, multiplication, division, modulus, and conjugate

Complex numbers form a field — we can add, subtract, multiply, and divide them (except by zero). All operations follow naturally from treating i as a variable with the extra rule i² = −1.

Addition and Subtraction

Rule

Add or subtract the real parts together and the imaginary parts together:
(a+bi) ± (c+di) = (a±c) + (b±d)i

Multiplication

Rule — use FOIL then substitute i² = −1

(a+bi)(c+di) = ac + adi + bci + bdi² = ac + (ad+bc)i + bd(−1)
= (ac−bd) + (ad+bc)i

Conjugate and Modulus

Conjugate & Modulus

The conjugate of z = a+bi is z̅ = a−bi (flip the sign of the imaginary part).
The modulus of z is |z| = √(a²+b²) — the distance from the origin in the Argand plane.
Key property: z · z̅ = |z|² = a² + b² (always a non-negative real number!)

Division

To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator. This makes the denominator a real number.

✏️ Quick Example — All Four OperationsExample 3.4

Let z₁ = 3 + 4i and z₂ = 1 − 2i. Find z₁+z₂, z₁z₂, z₁/z₂, |z₁|, and z̅₁.

1

Sum: (3+1) + (4−2)i = 4 + 2i

2

Product: (3+4i)(1−2i) = 3−6i+4i−8i² = 3−2i+8 = 11 − 2i

3

Division: z₁/z₂ = (3+4i)(1+2i)/[(1−2i)(1+2i)] = (3+6i+4i−8)/(1+4) = −1 + 2i

4

Modulus: |z₁| = √(9+16) = √25 = 5

5

Conjugate: z̅₁ = 3 − 4i

✔

Sum: 4+2i | Product: 11−2i | Quotient: −1+2i (= (−5+10i)/5) | |z₁|=5 | z̅₁=3−4i

Properties of Modulus and Conjugate

|z|

√(a²+b²)

z·z̅

|z|² = a²+b²

|z₁z₂|

= |z₁| × |z₂|

|z₁/z₂|

= |z₁| / |z₂|

z̅̅

= z (double conjugate)

Triangle △

|z₁+z₂| ≤ |z₁|+|z₂|

Polar z⁻¹

= z̅ / |z|²

|z|=0

⟺ z = 0

Module 3 of 7

3.4

Argand Plane and Argument of a Complex Number

Visualising complex numbers as points — polar form and the angle θ

Every complex number z = a + bi can be plotted as the point (a, b) in a 2D coordinate plane. When we use this geometric interpretation, the plane is called the Argand plane or complex plane.

📖 The Argand Plane

• x-axis = Real axis (Re) — plots the real part
• y-axis = Imaginary axis (Im) — plots the imaginary part
• z = a + bi is the point P(a, b)
• |z| = √(a²+b²) is the distance OP from origin O to point P

z = 3+4i plotted as point (3,4). Distance |z| = 5. Angle θ = arg(z) ≈ 53.13°

Argument of a Complex Number

📖 Definition — Argument

The argument of z = a+bi, written arg(z) or θ, is the angle the line OP makes with the positive real axis, measured anti-clockwise.
The principal argument Arg(z) is the unique value of θ in the interval (−π, π].

Q1: a>0, b>0

θ = tan⁻¹(b/a)

0 < θ < π/2

Q2: a<0, b>0

θ = π − tan⁻¹(|b/a|)

π/2 < θ < π

Q3: a<0, b<0

θ = −π + tan⁻¹(|b/a|)

−π < θ < −π/2

Q4: a>0, b<0

θ = −tan⁻¹(|b/a|)

−π/2 < θ < 0

Polar Form and Euler's Formula

Polar Form

z = r(cosθ + i sinθ) where r = |z| and θ = arg(z).
Euler's Formula: e^iθ = cosθ + i sinθ, so z = re^iθ.
De Moivre's Theorem: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)

✏️ Quick Example — Polar FormExample 3.5

Express z = −1 + √3 i in polar form and find arg(z).

1

a = −1, b = √3. r = |z| = √(1+3) = 2

2

Since a < 0, b > 0 (Quadrant 2): θ = π − tan⁻¹(√3/1) = π − π/3 = 2π/3

3

Polar form: z = 2(cos 2π/3 + i sin 2π/3)

✔

r = 2, arg(z) = 2π/3 (= 120°), z = 2e^i𢱒π/3

Modulus r

r = |z| = √(a²+b²)

arg(z₁z₂)

= arg(z₁) + arg(z₂)

arg(z₁/z₂)

= arg(z₁) − arg(z₂)

De Moivre

(re^iθ)ⁿ = rⁿe^inθ

Module 4 of 7

3.5

Cube Roots of Unity

The three solutions of z³ = 1 and their elegant properties

The equation z³ = 1 has three solutions in ℂ. One is the obvious real root z = 1, and the other two are complex conjugates. These are called the cube roots of unity.

📖 The Three Cube Roots of Unity

Solving z³ − 1 = 0: factor as (z−1)(z²+z+1) = 0.
• z = 1 (real root)
• z = ω = (−1 + √3 i) / 2
• z = ω² = (−1 − √3 i) / 2

1, ω, and ω² lie on the unit circle, equally spaced at 120° apart

The Golden Properties of ω

ω³ = 1

Fundamental property

1 + ω + ω² = 0

Sum of all three roots

ω² = ω̅

They are conjugates

ω = ω⁻²

Inverse of ω

ω³ⁿ = 1

ω to any multiple of 3

ω²+ω+1 = 0

Quadratic satisfied by ω

✏️ Quick Example — Using ω PropertiesExample 3.6

Evaluate: (i) 1 + ω² + ω⁴ (ii) (1+ω)(1+ω²) (iii) ω²¹ + ω²²

1

ω⁴ = ω³⋅ω = 1⋅ω = ω. So 1+ω²+ω⁴ = 1+ω²+ω = 0

2

(1+ω)(1+ω²) = 1+ω²+ω+ω³ = (1+ω+ω²)+ω³ = 0 + 1 = 1

3

ω²¹ = (ω³)⁷ = 1. ω²² = ω²¹⋅ω = ω. So ω²¹+ω²² = 1+ω = −ω² (since 1+ω+ω²=0)

✔

(i) 0 | (ii) 1 | (iii) −ω² = (1+√3 i)/2

Module 5 of 7

3.6

nth Roots of Unity

The n solutions of zⁿ = 1 — equally spaced on the unit circle

The cube roots of unity generalise beautifully. The equation zⁿ = 1 has exactly n distinct complex roots, all lying on the unit circle |z| = 1 in the Argand plane, equally spaced at angles of 2π/n.

📖 nth Roots of Unity

The n solutions of zⁿ = 1 are:
zₖ = e^i𢱒πk/n = cos(2πk/n) + i sin(2πk/n) for k = 0, 1, 2, …, n−1

Let α = e^i2π/n (the primitive nth root). Then all roots are:
1, α, α², α³, …, αⁿ⁻¹

Key Properties

Sum of all nth roots of unity = 0 (for n ≥ 2): 1 + α + α² + … + αⁿ⁻¹ = 0
Product of all nth roots of unity = (−1)ⁿ⁻¹ = +1 if n is odd, −1 if n is even
All roots lie on the unit circle (each has modulus 1)
The roots form a regular n-gon in the Argand plane
Non-real roots come in conjugate pairs: if α is a root, so is α̅

nth roots of zⁿ=1

e^i2πk/n, k=0,1,…,n−1

Sum of roots

= 0 (for n≥2)

Product of roots

= (−1)ⁿ⁻¹

Each root |zₖ|

= 1 (unit circle)

✏️ Quick Example — 4th Roots of UnityExample 3.7

Find all 4th roots of unity and plot them on the Argand plane.

1

k=0: e⁰ = 1 (angle 0°)

2

k=1: e^iπ/2 = cos90°+i sin90° = i (angle 90°)

3

k=2: e^iπ = cos180°+i sin180° = −1 (angle 180°)

4

k=3: e^i3π/2 = cos270°+i sin270° = −i (angle 270°)

✔

Roots: {1, i, −1, −i}. Sum = 1+i−1−i = 0 ✔. They form a square on the unit circle.

✏️ Quick Example — nth Root of any Complex NumberExample 3.8

Find the square roots of i (i.e. solve z² = i).

1

Write i in polar form: i = e^iπ/2 (r=1, θ=π/2)

2

Square roots: z = e^{i(π/2+2πk)/2} for k=0,1

3

k=0: e^iπ/4 = cos45°+i sin45° = (1+i)/√2

4

k=1: e^i5π/4 = −(1+i)/√2

✔

√i = ±(1+i)/√2 = ±(√2/2 + √2/2 i)

Module 6 of 7

3.7

Geometry of Complex Numbers

Distance, midpoints, loci, and geometric interpretation of operations

The Argand plane connects algebra and geometry beautifully. Every algebraic operation on complex numbers has a precise geometric meaning — addition is vector translation, multiplication is rotation and scaling.

Distance and Midpoint

Geometric Formulas

If z₁ = a₁+b₁i and z₂ = a₂+b₂i represent points P and Q:
• Distance PQ = |z₂ − z₁| = √[(a₂−a₁)² + (b₂−b₁)²]
• Midpoint = (z₁ + z₂) / 2
• Section formula (ratio m:n): z = (mz₂ + nz₁) / (m+n)

Geometric Meaning of Operations

➕

Addition z₁+z₂

Vector sum (parallelogram law). Translates point z₁ by the vector z₂.

➖

Subtraction z₂−z₁

Vector from z₁ to z₂. Its modulus = distance between points.

✖️

Multiply by r e^iθ

Scales modulus by r and rotates by angle θ anti-clockwise.

🔃

Multiply by i

Rotation by exactly 90° anti-clockwise (since arg(i) = π/2).

🗺️

Conjugate z̅

Reflection about the real axis (x-axis).

◎

|z − z₀| = r

Locus is a circle centred at z₀ with radius r.

Standard Loci in the Complex Plane

A locus is the set of all points z satisfying a given condition. These appear frequently in exams:

|z − z₀| = r — Circle with centre z₀ and radius r
|z − z₁| = |z − z₂| — Perpendicular bisector of segment z₁z₂
|z − z₁| / |z − z₂| = k (k≠1) — Circle (Apollonius circle)
arg(z − z₀) = θ — Ray from z₀ at angle θ
Re(z) = c — Vertical line x = c
Im(z) = c — Horizontal line y = c

✏️ Quick Example — Distance and LocusExample 3.9

Find the locus of z if |z − (3+2i)| = 4. Also find the distance between z₁ = 2+3i and z₂ = 5−i.

1

Locus: |z − (3+2i)| = 4 is a circle with centre (3,2) and radius 4.

2

Distance: |z₂−z₁| = |(5−i)−(2+3i)| = |3−4i|

3

= √(3² + (−4)²) = √(9+16) = 5

✔

Locus: circle, centre (3,2), radius 4. | Distance PQ = 5 units.

✏️ Quick Example — RotationExample 3.10

If z = 3+i, find iz and interpret geometrically.

1

iz = i(3+i) = 3i + i² = 3i − 1 = −1 + 3i

2

Original z = (3,1). New iz = (−1,3). Check: (−1)²+3² = 10 = 3²+1² ✔ same modulus

✔

Multiplying by i rotates z by 90° anti-clockwise. The modulus is preserved: |z| = |iz| = √10.

Distance |z₂−z₁|

√[(a₂−a₁)²+(b₂−b₁)²]

Midpoint

(z₁+z₂)/2

× by i

Rotation 90° CCW

× by −1

Rotation 180°

Circle locus

|z−z₀| = r

Perp. bisector

|z−z₁|=|z−z₂|

Module 7 of 7 🎉

Chapter 04 · Roots, Discriminants & Inequalities

Theory of Equations & Inequations

From polynomials and quadratic equations to the powerful discriminant and linear inequalities — this chapter gives you the tools to find, analyse and verify solutions with complete confidence.

6 Modules ⌛ ~45 min read Class 11 Core

4.1

Polynomial

Definition, degree, types — and two key theorems

A polynomial is the most fundamental algebraic expression — a finite sum of terms, each being a non-negative integer power of a variable multiplied by a real coefficient. Understanding polynomials is the foundation for everything in this chapter.

📖 Definition — Polynomial

A polynomial in variable x is an expression of the form:
p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
where n is a non-negative integer, aₙ, aₙ₋₁, …, a₀ ∈ ℝ, and aₙ ≠ 0.
• n is the degree (highest power with non-zero coefficient)
• aₙ is the leading coefficient
• a₀ is the constant term

Types of Polynomials by Degree

0️⃣

Constant

Degree 0. Value never changes.

p(x) = 7

1️⃣

Linear

Degree 1. Straight-line graph.

p(x) = 3x + 5

2️⃣

Quadratic

Degree 2. Parabola graph.

p(x) = 2x²−x+1

3️⃣

Cubic

Degree 3. S-curve graph.

p(x) = x³−2x

4️⃣

Quartic

Degree 4. W or M shaped.

p(x) = x⁴−1

n️⃣

nth Degree

Degree n. Has at most n real roots.

p(x) = aₙxⁿ+…

Types by Number of Terms

Monomial: one term — e.g. 5x³
Binomial: two terms — e.g. x² − 4
Trinomial: three terms — e.g. x² + 3x + 2
Zero Polynomial: p(x) = 0. Its degree is undefined (sometimes defined as −∞ or −1)

Remainder Theorem

📖 Remainder Theorem

When a polynomial p(x) is divided by (x − a), the remainder is p(a).
No long division needed — just substitute x = a into the polynomial!

Factor Theorem

📖 Factor Theorem

(x − a) is a factor of p(x) if and only if p(a) = 0.
Equivalently: a is a root of p(x) ⟺ (x − a) divides p(x) exactly.

✏️ Quick Example — Remainder & Factor TheoremExample 4.1

Let p(x) = x³ − 4x² + x + 6. (i) Find the remainder when divided by (x−3). (ii) Is (x−2) a factor?

1

Remainder by (x−3): p(3) = 27 − 36 + 3 + 6 = 0

2

Since p(3) = 0, (x−3) is actually a factor with zero remainder.

3

(x−2) factor test: p(2) = 8 − 16 + 2 + 6 = 0 ✔

✔

Remainder by (x−3) = 0; (x−3) is a factor. (x−2) is also a factor. p(x) = (x−2)(x−3)(x+1).

Degree rule

deg[p·q] = deg[p] + deg[q]

Remainder Thm

p(x)÷(x−a) leaves p(a)

Factor Thm

(x−a) | p(x) ⟺ p(a)=0

Max real roots

At most n roots for degree-n poly

Module 1 of 6

4.2

Quadratic Equation

ax² + bx + c = 0 — roots, discriminant, Vieta's formulas, and forming equations

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0 where a ≠ 0. It is the most important equation type in all of Class 11 & 12 mathematics.

📖 Standard Form

ax² + bx + c = 0, where a, b, c ∈ ℝ and a ≠ 0.
• a = leading coefficient (coefficient of x²)
• b = middle coefficient (coefficient of x)
• c = constant term

The Quadratic Formula

✨ The Master Formula

The roots of ax² + bx + c = 0 are:
x = (−b ± √(b² − 4ac)) / 2a
The expression under the square root, D = b² − 4ac, is the discriminant. It determines everything about the nature of the roots.

Nature of Roots — The Discriminant D

✅

D > 0

Two distinct real roots. The parabola crosses the x-axis at two points.

x = (−b±√D)/2a

✏️

D = 0

Two equal (repeated) real roots. The parabola just touches the x-axis.

x = −b/2a (both roots)

❌

D < 0

No real roots. Two complex conjugate roots. Parabola doesn't cross x-axis.

x = (−b±i√|D|)/2a

📐

D is perfect square

Roots are rational (when a, b, c are integers). Can factorise.

D=25: √D=5 (rational)

Three cases of the discriminant and their parabola shapes

Vieta's Formulas — Roots Without Solving

If α and β are the roots of ax² + bx + c = 0, then:

Sum of roots

α + β = −b/a

Product of roots

α · β = c/a

Difference of roots

|α−β| = √D / |a|

α² + β²

= (α+β)² − 2αβ

α³ + β³

= (α+β)³ − 3αβ(α+β)

Form equation

x²−(α+β)x+αβ=0

✏️ Quick Example — Full Quadratic AnalysisExample 4.2

For 2x² − 5x + 3 = 0: find D, state nature of roots, find roots, verify with Vieta's, and form equation with roots 2α and 2β.

1

D = (−5)² − 4(2)(3) = 25 − 24 = 1 > 0 ⟹ two distinct real roots

2

Roots: x = (5±1)/4 ⟹ x = 3/2 or x = 1

3

Vieta's check: α+β = 3/2+1 = 5/2 = −(−5)/2 ✔ αβ = 3/2×1 = 3/2 = 3/2 ✔

4

New equation (roots 2α=3, 2β=2): sum=5, product=6 ⟹ x² − 5x + 6 = 0

✔

D=1, roots: 3/2 and 1, Vieta's verified. New equation: x²−5x+6=0 (factors as (x−2)(x−3)=0).

✏️ Quick Example — Forming a Quadratic from RootsExample 4.3

Form the quadratic equation whose roots are (3+√2) and (3−√2).

1

Sum: (3+√2)+(3−√2) = 6

2

Product: (3+√2)(3−√2) = 9−2 = 7

3

Equation: x² − (sum)x + (product) = 0 ⟹ x² − 6x + 7 = 0

✔

x² − 6x + 7 = 0. Irrational roots always come in conjugate pairs: if (3+√2) is a root, so is (3−√2).

Module 2 of 6

4.3

Quadratic Expression

f(x) = ax² + bx + c as a function — sign, graph, and range analysis

A quadratic expression (or quadratic function) is f(x) = ax² + bx + c viewed as a function of x — not just an equation set to zero. We study where it is positive, negative, zero, and what its graph looks like.

📖 Quadratic Expression

f(x) = ax² + bx + c (a ≠ 0)
Its graph is a parabola:
• Opens upward (⋃-shape) if a > 0
• Opens downward (⋂-shape) if a < 0
The lowest/highest point is the vertex at x = −b/2a.

Completing the Square — Vertex Form

Any quadratic can be rewritten as f(x) = a(x − h)² + k, called the vertex form, where (h, k) is the vertex.

Vertex: h = −b/2a and k = f(h) = −D/4a
If a > 0: minimum value = k at x = h
If a < 0: maximum value = k at x = h
Axis of symmetry: the vertical line x = −b/2a

✏️ Quick Example — Completing the SquareExample 4.4

Express f(x) = 2x² − 8x + 5 in vertex form and find the minimum value.

1

Factor out 2: f(x) = 2(x² − 4x) + 5

2

Complete the square inside: x² − 4x = (x−2)² − 4

3

Substitute back: f(x) = 2[(x−2)² − 4] + 5 = 2(x−2)² − 3

4

Vertex = (2, −3). Since a = 2 > 0, this is the minimum.

✔

f(x) = 2(x−2)² − 3. Minimum value = −3 at x = 2. Axis of symmetry: x = 2.

Sign of a Quadratic Expression

Knowing when f(x) > 0 or f(x) < 0 is crucial for solving inequalities. The sign depends on both the coefficient a and the discriminant D:

a > 0, D < 0

f(x) > 0 for ALL real x. Expression is always positive.

x²+x+1 > 0 always

a > 0, D = 0

f(x) ≥ 0. Zero only at x = −b/2a (the vertex touches axis).

x²−2x+1 ≥ 0

a > 0, D > 0

f(x) < 0 between roots α and β; positive outside.

f(x)<0 for α<x<β

a < 0, D > 0

f(x) > 0 between roots; negative outside.

f(x)>0 for α<x<β

✏️ Quick Example — Range of a QuadraticExample 4.5

Find the range of f(x) = −x² + 4x − 3 for x ∈ ℝ.

1

a = −1 < 0, so parabola opens downward ⟹ maximum exists.

2

Vertex x = −b/2a = −4/(−2) = 2. Maximum = f(2) = −4+8−3 = 1

3

Since a < 0, f(x) ≤ 1 for all x ⟹ Range = (−∞, 1]

✔

Range = (−∞, 1]. Maximum value is 1, achieved at x = 2.

Vertex x-coord

h = −b / 2a

Vertex y-coord

k = −D / 4a

Min (a>0)

f_min = −D/4a

Max (a<0)

f_max = −D/4a

Module 3 of 6

4.4

Inequality

Types, properties, and the wavy curve (sign chart) method

An inequality compares two expressions using <, >, ≤, or ≥. Unlike equations (which give specific points), inequalities give intervals as their solution. Mastering inequalities is essential for every topic ahead.

📖 Types of Inequality

• Strict: f(x) < 0 or f(x) > 0 (excludes the boundary)
• Non-strict: f(x) ≤ 0 or f(x) ≥ 0 (includes the boundary)
• Absolute value: |f(x)| < k or |f(x)| > k
• Polynomial: p(x) > 0, < 0, etc. for higher degree polynomials

Fundamental Properties of Inequalities

Addition/Subtraction: Adding the same quantity to both sides preserves the inequality: a < b ⟹ a+c < b+c
Multiplication by positive: Preserves the sign: a < b and c > 0 ⟹ ac < bc
Multiplication by negative: Reverses the inequality: a < b and c < 0 ⟹ ac > bc
Reciprocal: Taking reciprocals reverses the inequality (when both sides have the same sign)
Transitivity: a < b and b < c ⟹ a < c

⚠️ Critical Rule — Never Forget This!

When you multiply or divide both sides of an inequality by a negative number, you MUST flip the inequality sign.
Example: −2x < 6 ⟹ x > −3 (sign flips when dividing by −2)

The Wavy Curve (Sign Chart) Method

For polynomial inequalities, the wavy curve method is the fastest and most reliable approach. It works for any factored polynomial.

Step 1: Factorise p(x) completely and find all roots (zeros)
Step 2: Mark roots on the number line in increasing order
Step 3: Check the sign of p(x) in the rightmost interval (x very large)
Step 4: Alternate signs between consecutive roots (for simple roots)
Step 5: Read off which intervals satisfy the inequality

✏️ Quick Example — Wavy Curve MethodExample 4.6

Solve (x−1)(x−3)(x+2) > 0.

1

Roots: x = −2, 1, 3. Mark on number line: −2 < 1 < 3

2

For x > 3 (e.g. x=4): all three factors positive ⟹ product > 0 ✔

3

Alternate: 1<x<3 ⟹ negative. −2<x<1 ⟹ positive. x<−2 ⟹ negative.

✔

Solution: x ∈ (−2, 1) ∪ (3, ∞). The positive intervals.

Absolute Value Inequalities

|x| < a (a>0)

−a < x < a

|x| > a (a>0)

x < −a or x > a

|x| ≤ a

−a ≤ x ≤ a

|x| ≥ a

x ≤ −a or x ≥ a

✏️ Quick Example — Absolute Value InequalityExample 4.7

Solve |2x − 3| ≤ 5.

1

Apply |f(x)| ≤ a ⟹ −a ≤ f(x) ≤ a: −5 ≤ 2x−3 ≤ 5

2

Add 3: −2 ≤ 2x ≤ 8

3

Divide by 2: −1 ≤ x ≤ 4

✔

Solution: x ∈ [−1, 4].

Module 4 of 6

4.5

Linear Inequality

One variable and two variables — graphing, solving, and half-planes

A linear inequality is an inequality involving a linear (degree 1) expression. They appear in optimisation, resource allocation, and everywhere numbers need to satisfy constraints rather than exact values.

📖 Linear Inequality in One Variable

Standard forms: ax + b < 0, ax + b > 0, ax + b ≤ 0, ax + b ≥ 0
where a ≠ 0 and a, b ∈ ℝ.
Solution is always an interval on the real number line.

Solving Linear Inequalities in One Variable

Treat exactly like a linear equation, with one extra rule: flip the inequality when multiplying or dividing by a negative number.

✏️ Quick Example — One VariableExample 4.8

Solve and graph on a number line: 3x − 7 > −1 and 2 − 5x ≥ −8.

1

First: 3x > 6 ⟹ x > 2 ⟹ solution: (2, ∞)

2

Second: −5x ≥ −10 ⟹ x ≤ 2 (flip!) ⟹ x ≤ 2 ⟹ (−∞, 2]

3

Intersection (both must hold): (2,∞) ∩ (−∞,2] = ∅

✔

No common solution. The two inequalities have no x satisfying both simultaneously.

Linear Inequality in Two Variables

📖 Two-Variable Linear Inequality

Forms: ax + by < c, ax + by > c, ax + by ≤ c, ax + by ≥ c
Solution is a half-plane: one side of the line ax + by = c.
The line ax + by = c is called the boundary line.

How to Graph a Two-Variable Inequality

Step 1: Draw the boundary line ax + by = c. Use a solid line for ≤ or ≥; dashed line for < or >
Step 2: Choose a test point NOT on the line. The origin (0,0) is easiest (if not on the line)
Step 3: Substitute the test point into the inequality. If it satisfies the inequality, shade that side. If not, shade the other side

✏️ Quick Example — Two VariablesExample 4.9

Graph the solution of 2x + 3y ≤ 12.

1

Boundary line: 2x+3y = 12. Find intercepts: (6,0) and (0,4). Draw a solid line (≤ includes boundary).

2

Test (0,0): 2(0)+3(0) = 0 ≤ 12 ✔ Origin satisfies the inequality.

3

Shade the side containing the origin (below-left of the line).

✔

Solution is the half-plane on and below the line 2x+3y=12, including the line itself.

System of Linear Inequalities

When multiple inequalities must hold simultaneously, the solution is the intersection of all the individual half-planes. This intersection region is called the feasible region and is the basis of Linear Programming.

Solve ax<b (a>0)

x < b/a

Solve ax<b (a<0)

x > b/a (flip!)

AND inequalities

Intersection of solutions

OR inequalities

Union of solutions

Module 5 of 6

4.6

Solution Set

Writing, representing, and interpreting complete solutions using intervals and set notation

The solution set is the complete collection of all values of x that satisfy an equation or inequality. Writing it correctly — using proper notation — is as important as finding it.

📖 Definition — Solution Set

The solution set of an equation or inequality is the set of all real numbers x that make it true.
• For equations: usually a finite set {x₁, x₂, …}
• For inequalities: usually an interval or union of intervals

Interval Notation — The Standard Way

(a, b)

Open interval. Excludes both endpoints. a < x < b.

x ∈ (2, 5)

[a, b]

Closed interval. Includes both endpoints. a ≤ x ≤ b.

x ∈ [−1, 3]

[a, b)

Half-open. Includes a, excludes b. a ≤ x < b.

x ∈ [0, 7)

(a, b]

Half-open. Excludes a, includes b. a < x ≤ b.

x ∈ (−3, 1]

(a, ∞)

All x greater than a. ∞ is never included (always open).

x > 4

(−∞, ∞)

All real numbers. The entire real line ℝ.

x ∈ ℝ

Number Line Representation

On a number line: use a filled dot ● to show a value IS included (≤ or ≥), and an open dot ○ to show a value is NOT included (< or >).

Number line representation: open dot ○ = excluded, filled dot ● = included

✏️ Quick Example — Full Solution with Solution SetExample 4.10

Find the solution set of x² − 5x + 6 < 0 and represent it on a number line.

1

Factorise: x²−5x+6 = (x−2)(x−3)

2

Roots: x = 2 and x = 3. Leading coefficient a = 1 > 0.

3

For a > 0, quadratic < 0 between the roots: 2 < x < 3

4

Strict inequality (<) so endpoints excluded: open dots at 2 and 3.

✔

Solution set = (2, 3). On the number line: open dots at x=2 and x=3 with the segment between them shaded.

✏️ Quick Example — Union of IntervalsExample 4.11

Find the solution set of (x+1)(x−2)(x−4) ≥ 0.

1

Roots: x = −1, 2, 4. Apply wavy curve (positive for x > 4).

2

Sign chart: −∞ to −1: negative. −1 to 2: positive. 2 to 4: negative. 4 to ∞: positive.

3

Need ≥ 0: include positive intervals AND the zeros (≥ includes roots).

✔

Solution set = [−1, 2] ∪ [4, ∞). Closed brackets include the roots.

a<0, D>0: f<0

x ∈ (−∞,α) ∪ (β,∞)

a>0, D>0: f<0

x ∈ (α, β)

a>0, D>0: f>0

x ∈ (−∞,α) ∪ (β,∞)

a>0, D<0

f(x)>0 for all x ∈ ℝ

Module 6 of 6 🎉

Chapter 05 · Patterns in Numbers

Sequence and Series

From the ordered lists of numbers that form sequences to the sums that create series, this chapter unlocks the elegant patterns of AP, GP, HP, and their hybrid AGP — with every formula derived from first principles.

6 Modules ⌛ ~50 min read Class 11 Core

5.1

Sequence

Ordered lists with a pattern — finite, infinite, and special sequences

A sequence is a list of numbers arranged in a definite order according to a rule. Each number in the list is called a term. The order matters — changing it gives a different sequence.

📖 Definition — Sequence

A sequence is a function f : ℕ → ℝ that assigns a real number to each positive integer. The value f(n) = aₙ is called the nth term (or general term) of the sequence.
Notation: {aₙ} or a₁, a₂, a₃, …

Types of Sequences

📏

Finite Sequence

Has a last term. Terminates after a definite number of terms.

2, 4, 6, 8, 10

∞

Infinite Sequence

Continues without end. No last term.

1, 3, 5, 7, 9, …

↗

Increasing

Each term is greater than the previous: aₙ₊₁ > aₙ

1, 2, 4, 8, 16, …

↘

Decreasing

Each term is less than the previous: aₙ₊₁ < aₙ

100, 50, 25, …

🌀

Periodic

Terms repeat after a fixed period p: aₙ₊ₚ = aₙ

1,2,3,1,2,3,…

🔢

Alternating

Signs alternate + and − due to (−1)ⁿ factor.

1,−2,3,−4,…

Finding the General Term

The general term aₙ is a formula that gives the nth term directly without listing all previous terms. For example, aₙ = 2n−1 gives the sequence 1, 3, 5, 7, … (odd numbers).

✏️ Quick Example — Finding the General TermExample 5.1

Find the general term of: (i) 2, 5, 8, 11, … (ii) 1, 4, 9, 16, 25, … (iii) 1, −1, 1, −1, …

1

(i) Differences = 3 each time. Starts at 2. aₙ = 3n − 1

2

(ii) Terms are perfect squares. aₙ = n²

3

(iii) Alternating ±1. aₙ = (−1)ⁿ⁻¹

✔

aₙ = 3n−1 | aₙ = n² | aₙ = (−1)ⁿ⁻¹

Important Special Sums

∑ 1 to n

n(n+1)/2

∑ n²

n(n+1)(2n+1)/6

∑ n³

[n(n+1)/2]²

∑ (2n−1)

n² (sum of n odd nos.)

Module 1 of 6

5.2

Series

The sum of a sequence — partial sums and convergence

When we add the terms of a sequence, we get a series. The key question with any series is: what is the total sum? For finite sequences this is always defined; for infinite sequences, the sum may or may not converge to a finite value.

📖 Definition — Series

If a₁, a₂, a₃, … is a sequence, then the expression a₁ + a₂ + a₃ + … is a series.
The nth partial sum is: Sₙ = a₁ + a₂ + … + aₙ = ∑𝓈₌₁ⁿ a𝓈
For a finite series of n terms, Sₙ is the total sum. For an infinite series, we study whether Sₙ approaches a limit as n → ∞.

Convergent vs Divergent Series

Convergent: The partial sums approach a finite limit L. Then L is the sum to infinity. Example: 1 + 1/2 + 1/4 + … converges to 2.
Divergent: The partial sums grow without bound. Example: 1 + 2 + 3 + … diverges.
The general term test: if aₙ does not approach 0, the series diverges.

✏️ Quick Example — Partial SumExample 5.2

Find S₃ and S₅ for the series with aₙ = n². Also find the general formula for Sₙ.

1

S₃ = 1²+2²+3² = 1+4+9 = 14

2

S₅ = 1+4+9+16+25 = 55

3

General: Sₙ = ∑k² = n(n+1)(2n+1)/6

4

Check: S₃ = 3(4)(7)/6 = 84/6 = 14 ✔ S₅ = 5(6)(11)/6 = 330/6 = 55 ✔

✔

Sₙ = n(n+1)(2n+1)/6. Verified for n=3 and n=5.

Partial sum

Sₙ = ∑𝓈₌₁ⁿ a𝓈

nth term from Sₙ

aₙ = Sₙ − Sₙ₋₁ (n≥2)

a₁ from Sₙ

a₁ = S₁

Module 2 of 6

5.3

Arithmetic Progression (AP)

Constant difference between consecutive terms — the most common progression

An Arithmetic Progression is a sequence where each term is obtained from the previous one by adding a fixed constant called the common difference. It is the simplest and most frequently tested progression.

📖 Definition — AP

A sequence a, a+d, a+2d, a+3d, … is an AP where:
• a = first term • d = common difference = aₙ₊₁ − aₙ
• d can be positive (increasing), negative (decreasing), or zero (constant)

AP: each term is the previous term plus d (common difference)

Key Formulas for AP

nth term

aₙ = a + (n−1)d

Last term l

l = a + (n−1)d

Sum Sₙ

Sₙ = n/2 [2a+(n−1)d]

Sum (using l)

Sₙ = n/2 (a+l)

Common diff d

d = (l−a)/(n−1)

Middle term

(a+l)/2 = AM

AM of a,b

A = (a+b)/2

n AMs between a,b

d = (b−a)/(n+1)

Properties of AP

If a constant is added to (or subtracted from) every term of an AP, the result is also an AP with the same d
If every term is multiplied by a constant k, the result is an AP with common difference kd
Sum of terms equidistant from beginning and end = a₁ + aₙ = a₂ + aₙ₋₁ = … = a + l
If three numbers are in AP: take them as a−d, a, a+d (middle is the AM)
If four numbers are in AP: take them as a−3d, a−d, a+d, a+3d

✏️ Quick Example — Complete AP ProblemExample 5.3

The 7th term of an AP is 20 and the 13th term is 32. Find the AP, the sum of its first 20 terms, and the first term exceeding 50.

1

a+6d = 20 …(i) and a+12d = 32 …(ii). Subtract: 6d = 12 ⟹ d = 2

2

From (i): a = 20−12 = 8. AP: 8, 10, 12, 14, …

3

S₂₀ = 20/2[2(8)+19(2)] = 10[16+38] = 10×54 = 540

4

aₙ > 50: 8+(n−1)2 > 50 ⟹ 2n > 44 ⟹ n > 22. First term: n=23, a₂₃ = 8+44 = 52

✔

AP: 8, 10, 12, … | S₂₀ = 540 | First term > 50 is 52 (23rd term)

Module 3 of 6

5.4

Geometric Progression (GP)

Constant ratio between consecutive terms — exponential growth and decay

A Geometric Progression is a sequence where each term is obtained by multiplying the previous term by a fixed constant called the common ratio. GPs model compound interest, population growth, and radioactive decay.

📖 Definition — GP

A sequence a, ar, ar², ar³, … is a GP where:
• a = first term (a ≠ 0) • r = common ratio = aₙ₊₁ / aₙ (r ≠ 0)
• r > 1: increasing • 0 < r < 1: decreasing towards 0 • r < 0: alternating signs

Key Formulas for GP

nth term

aₙ = a·rⁿ⁻¹

Sum Sₙ (r≠1)

a(rⁿ−1)/(r−1)

Sum Sₙ (r=1)

na

Sum to ∞ |r|<1

S∞ = a/(1−r)

GM of a,b

G = √(ab)

n GMs between a,b

r = (b/a)¹/ⁿ⁺¹

Product of n terms

P = aⁿ · rⁿ(ⁿ⁻¹)/²

AM ≥ GM

(a+b)/2 ≥ √(ab)

Properties of GP

If every term is multiplied by or divided by a non-zero constant, the result is still a GP with the same r
The reciprocals of a GP form another GP with ratio 1/r
If three terms are in GP: take them as a/r, a, ar (middle is the GM)
Product of terms equidistant from ends is constant: a₁·aₙ = a₂·aₙ₋₁ = a²
Sum to infinity exists only when |r| < 1

✏️ Quick Example — GP Sum & InfinityExample 5.4

A GP has first term 3 and common ratio 1/2. Find the 6th term, S₆, and S∞.

1

a₆ = 3×(1/2)⁵ = 3/32 = 3/32

2

S₆ = 3(1−(1/2)⁶)/(1−1/2) = 3(1−1/64)/(1/2) = 6×(63/64) = 189/32

3

|r| = 1/2 < 1, so S∞ = a/(1−r) = 3/(1−1/2) = 6

✔

a₆ = 3/32 | S₆ = 189/32 ≈ 5.906 | S∞ = 6

✏️ Quick Example — AM-GM InequalityExample 5.5

Prove that for positive reals x and y, x/y + y/x ≥ 2.

1

Let a = x/y and b = y/x. Then ab = (x/y)(y/x) = 1 ⟹ GM = √(ab) = 1

2

By AM ≥ GM: (a+b)/2 ≥ √(ab) = 1

3

So a + b ≥ 2, i.e. x/y + y/x ≥ 2. Equality when x = y.

✔

Proved using AM ≥ GM. Equality holds when x = y.

Module 4 of 6

5.5

Harmonic Progression (HP)

Reciprocals of an AP — and the AM-GM-HM inequality chain

A Harmonic Progression is a sequence whose reciprocals form an Arithmetic Progression. HPs appear in optics, music (harmonics), and electrical circuits. There is no direct sum formula for HP — we always convert to AP first.

📖 Definition — HP

A sequence a₁, a₂, a₃, … is a Harmonic Progression if 1/a₁, 1/a₂, 1/a₃, … is an Arithmetic Progression.
General form: 1/a, 1/(a+d), 1/(a+2d), … where a and d are the AP parameters.

Key Results for HP

nth term of HP

aₙ = 1/[a+(n−1)d]

HM of a,b

H = 2ab/(a+b)

a,b,c in HP iff

1/a, 1/b, 1/c in AP

b in HP between a,c

b = 2ac/(a+c)

The AM ≥ GM ≥ HM Chain

For any two positive real numbers a and b, the three classical means satisfy a beautiful inequality chain:

✨ The Most Important Mean Inequality

For positive reals a and b:
AM ≥ GM ≥ HM
That is: (a+b)/2 ≥ √(ab) ≥ 2ab/(a+b)
Equality holds if and only if a = b.
Also: AM × HM = GM² (a beautiful connecting identity)

✏️ Quick Example — HP Terms & HMExample 5.6

If 1/5, 1/10, 1/15, … is an HP, find the 8th term. Also find HM of 4 and 12.

1

Reciprocals: 5, 10, 15, … are in AP with a=5, d=5.

2

8th AP term = 5+7(5) = 40. So 8th HP term = 1/40.

3

HM of 4 and 12: H = 2(4)(12)/(4+12) = 96/16 = 6

4

Verify AM×HM = GM²: AM=8, GM=√48, HM=6. AM×HM = 48 = GM² = 48 ✔

✔

8th HP term = 1/40. HM(4,12) = 6. Verified: AM×HM = GM² = 48.

✏️ Quick Example — Three Numbers in HPExample 5.7

If a, 12, b are in HP and a+b = 22, find a and b.

1

12 is HM of a and b: 12 = 2ab/(a+b) = 2ab/22 = ab/11

2

So ab = 132. With a+b = 22, solve the quadratic:

3

x²−22x+132 = 0 ⟹ (x−11)² = 121−132 < 0. Hmm — adjust: ab=132, sum=22 ⟹ roots via formula: x = [22±√(484−528)]/2. Since D<0, try a+b=26: ab = 12×13 = 156, so a=12 and b=13 (or vice versa) needs rechecking.

4

Correct setup: 1/a + 1/b = 2/12 = 1/6. With a+b=22: (a+b)/ab = 1/6 ⟹ ab = 132. Roots: 11±√(121−132) ⟹ a=6, b=16 (checking: 1/6+1/16=8/48+3/48=11/48≠1/6). Use a=6,b=16: sum=22, product=96, 1/6+1/16≠1/6. a=8, b=14: product=112, 1/8+1/14=7/56+4/56=11/56≠1/6. Correct a=6, b=12 doesn't fit. Answer: a=6, b=18 satisfies: sum=24. Restart with a+b=24: ab=12×12=144 ⟹ a=12=b. Use general formula instead.

✔

Key method: 1/a, 1/12, 1/b in AP ⟹ 2/12 = 1/a+1/b ⟹ (a+b)/ab = 1/6 ⟹ ab = 6(a+b). Substitute a+b=22: ab=132. Roots of t²−22t+132=0 are t=(22±√(484−528))/2 ⟹ complex, so the given data is inconsistent. In well-posed HP problems, always verify the data first.

Module 5 of 6

5.6

Arithmetic-Geometric Progression (AGP)

Products of AP terms with GP terms — summed using the subtraction method

An Arithmetic-Geometric Progression is a sequence formed by multiplying corresponding terms of an AP and a GP. These arise naturally in probability and finance, and their sum requires a special technique.

📖 Definition — AGP

If {aₙ} = a, a+d, a+2d, … is an AP and {bₙ} = 1, r, r², … is a GP, then the sequence:
a, (a+d)r, (a+2d)r², (a+3d)r³, …
is an AGP. Its general term is Tₙ = [a+(n−1)d] · rⁿ⁻¹

Summing an AGP — The Subtraction Method

There is no simple closed-form formula to memorise. Instead, use the multiply-and-subtract (S − rS) technique:

Step 1: Write Sₙ = a + (a+d)r + (a+2d)r² + … + [a+(n−1)d]rⁿ⁻¹
Step 2: Write rSₙ = ar + (a+d)r² + … + [a+(n−2)d]rⁿ⁻¹ + [a+(n−1)d]rⁿ
Step 3: Subtract: Sₙ − rSₙ = a + dr + dr² + … + drⁿ⁻¹ − [a+(n−1)d]rⁿ
Step 4: The remaining terms form a GP. Sum it and solve for Sₙ.

S∞ (|r|<1)

a/(1−r) + dr/(1−r)²

General term Tₙ

[a+(n−1)d]·rⁿ⁻¹

Key technique

Compute S − rS

✏️ Quick Example — Summing an AGPExample 5.8

Find S = 1 + 2×(1/2) + 3×(1/4) + 4×(1/8) + … to infinity.

1

This is an AGP with a=1, d=1, r=1/2. Write:
S = 1 + 2(1/2) + 3(1/4) + 4(1/8) + …

2

(1/2)S = 1(1/2) + 2(1/4) + 3(1/8) + …

3

Subtract: S−(1/2)S = 1 + (1/2) + (1/4) + (1/8) + … = 1/(1−1/2) = 2

4

(1/2)S = 2 ⟹ S = 4

✔

S = 4. Verify using the formula: a/(1−r) + dr/(1−r)² = 1/(1/2) + 1×(1/2)/(1/2)² = 2 + 2 = 4 ✔

Module 6 of 6 🎉

Chapter 06 · The Art of Counting

Permutation and Combination

How many ways can we arrange, select, or organise a group of objects? From the fundamental counting principle to circular arrangements and selections, this chapter builds powerful tools for counting systematically.

5 Modules ⌛ ~45 min read Class 11 Core

6.1

Fundamental Principles of Counting

Multiplication and addition principles — the foundation of all counting

Before we count arrangements or selections, we need two fundamental rules that all of combinatorics rests on. These are so important that every P&C problem reduces to applying these two principles.

📖 Multiplication Principle (AND rule)

If one task can be done in m ways AND a second independent task can be done in n ways, then both tasks together can be done in m × n ways.
Key word: AND ⟹ multiply.
Extends: if k tasks can be done in n₁, n₂, …, n𝓈 ways respectively, total = n₁ × n₂ × … × n𝓈

📖 Addition Principle (OR rule)

If one task can be done in m ways OR a second task can be done in n ways, and the two tasks are mutually exclusive (cannot be done simultaneously), then either task can be done in m + n ways.
Key word: OR ⟹ add.

✏️ Quick Example — Applying Both PrinciplesExample 6.1

A student can travel from city A to B by 3 trains or 2 buses. From B to C by 4 trains or 5 buses. (i) Ways to travel A→C. (ii) Ways to travel from A to C choosing different modes for each leg.

1

A to B: 3+2 = 5 ways (OR principle, different modes)

2

B to C: 4+5 = 9 ways

3

A to C total: 5 × 9 = 45 ways (AND principle — both legs)

✔

45 ways total. Each choice of A→B mode can pair with each B→C mode.

✏️ Quick Example — Number of OutcomesExample 6.2

How many 3-digit numbers can be formed using digits 1–9 (i) with repetition allowed (ii) without repetition?

1

With repetition: Hundreds: 9 choices, Tens: 9, Units: 9 ⟹ 9×9×9 = 729

2

Without repetition: Hundreds: 9, Tens: 8 (one used), Units: 7 ⟹ 9×8×7 = 504

✔

With repetition: 729. Without repetition: 504.

AND ⟹ × (multiply): consecutive decisions, both must happen
OR ⟹ + (add): alternative choices, exactly one happens
Always check: are the events mutually exclusive before adding?
Always check: are the tasks independent before multiplying?

Module 1 of 5

6.2

Factorial

n! — the product of all integers from 1 to n, and why it matters

The factorial of a non-negative integer n is the product of all positive integers from 1 to n. It is the cornerstone of permutation and combination calculations.

📖 Definition — Factorial

n! = n × (n−1) × (n−2) × … × 2 × 1
Special cases: 0! = 1 and 1! = 1 (by definition)
Recursive definition: n! = n × (n−1)! for n ≥ 1

Factorial Values to Memorise

0! = 1

By definition

1! = 1

1

2! = 2

2×1

3! = 6

3×2×1

4! = 24

4×6

5! = 120

5×24

6! = 720

6×120

10! = 3628800

Large quickly!

✏️ Quick Example — Factorial SimplificationExample 6.3

Simplify: (i) 8!/6! (ii) 10!/(7!×3!) (iii) (n+1)!/n!

1

8!/6! = (8×7×6!)/(6!) = 8×7 = 56

2

10!/(7!×3!) = (10×9×8×7!)/(7!×6) = (720/6) = 120

3

(n+1)!/n! = (n+1)×n!/n! = n+1

✔

56 | 120 | n+1. Always cancel common factorial terms before multiplying out!

n!

n × (n−1)!

n!/(n−r)!

n(n−1)…(n−r+1)

0!

= 1 (by convention)

Trailing zeros in n!

⌊n/5⌋+⌊n/25⌋+…

Module 2 of 5

6.3

Permutation

Ordered arrangements — when the order of selection matters

A permutation is an ordered arrangement of objects. The key question is: in how many ways can we arrange r objects chosen from n distinct objects, when the order matters? Swapping two objects gives a different permutation.

📖 Definition — ⁿPᵣ

The number of permutations of n distinct objects taken r at a time (r ≤ n):
ⁿPᵣ = n! / (n−r)! = n(n−1)(n−2)…(n−r+1)
Special case: all n objects ⟹ ⁿPₙ = n!

Fill each slot: 5 choices for 1st, 4 for 2nd (one used), 3 for 3rd

Special Permutation Cases

ⁿPᵣ

n!/(n−r)!

All n objects

ⁿPₙ = n!

With repetition

nᵣ arrangements

n objects, p alike

n!/p!

p alike, q alike

n!/(p!q!)

Always together

Treat as 1 unit

✏️ Quick Example — Permutation with RestrictionsExample 6.4

In how many ways can 5 boys and 3 girls be arranged in a row so that (i) all girls are together (ii) no two girls are adjacent?

1

All girls together: Treat 3 girls as 1 block. 6 units (5+1) arranged in 6! ways. Girls in block: 3! arrangements. Total = 6! × 3! = 720 × 6 = 4320

2

No two girls adjacent: Arrange 5 boys first: 5! = 120 ways. This creates 6 gaps: _B_B_B_B_B_. Choose 3 gaps for girls: ⁶P₃ = 6×5×4 = 120. Total = 5! × ⁶P₃ = 120 × 120 = 14400

✔

All girls together: 4320 | No two girls adjacent: 14400

✏️ Quick Example — Repeated LettersExample 6.5

How many distinct arrangements are there of the letters in MISSISSIPPI?

1

Total letters = 11. Repeated: M=1, I=4, S=4, P=2.

2

Arrangements = 11! / (4! × 4! × 2!) = 39916800 / (24×24×2) = 39916800/1152 = 34650

✔

34,650 distinct arrangements. Divide by factorials of repeated-letter counts.

Module 3 of 5

6.4

Circular Permutation

Arranging objects around a circle — when position is relative

When objects are arranged in a circle, there is no fixed starting point. Rotating all objects by one position gives the same arrangement. This fundamentally changes the count compared to a linear arrangement.

📖 Key Principle

In a circle, one object is fixed to eliminate equivalent rotations. The remaining (n−1) objects are arranged in:
(n−1)! ways for n distinct objects in a circle.
This is because fixing one object leaves n−1 positions to fill freely.

Linear vs Circular: fixing one element eliminates identical rotations

Necklace and Garland (Clockwise = Anti-clockwise)

Necklace Rule

When the arrangement can be flipped (like a necklace or garland), clockwise and anti-clockwise arrangements are identical. In this case:
Number of arrangements = (n−1)! / 2

Circle (distinct)

(n−1)!

Necklace/garland

(n−1)!/2

r from n in circle

n/r × (r−1)! = n!/(r×(n−r)!)

✏️ Quick Example — Circular ArrangementsExample 6.6

In how many ways can 8 people sit at a round table? How many if 2 specific people must always sit together?

1

Unrestricted: (8−1)! = 7! = 5040

2

2 people together: Treat them as one unit ⟹ 7 units around table = (7−1)! = 6! ways. The 2 people within the unit: 2! = 2 ways.

3

Total = 6! × 2! = 720 × 2 = 1440

✔

Unrestricted: 5040 | Two together: 1440

Module 4 of 5

6.5

Combination

Unordered selections — when only which objects matter, not their order

A combination is a selection of objects where order does not matter. Choosing {A, B, C} and {B, C, A} is the same combination. This is how we count committees, teams, and choices.

📖 Definition — ⁿCᵣ

The number of ways to choose r objects from n distinct objects (order doesn't matter):
ⁿCᵣ = n! / [r! × (n−r)!]
Also written as C(n,r) or (ⁿ ᵣ).
Relation to permutation: ⁿCᵣ = ⁿPᵣ / r! (divide by r! to remove order)

Key Properties of ⁿCᵣ

ⁿCᵣ

n!/[r!(n−r)!]

Symmetry

ⁿCᵣ = ⁿCₙ₋ᵣ

ⁿC₀ = ⁿCₙ

= 1

ⁿC₁

= n

Pascal's Identity

ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

Sum of row

∑ ⁿCᵣ = 2ⁿ

Odd-even split

∑(even r)=∑(odd r)=2ⁿ⁻¹

ⁿCᵣ = ⁿCˢ

⟹ r=s or r+s=n

Permutation vs Combination — The Key Difference

🔃

Permutation ⁿPᵣ

Order matters. ABC ≠ BAC ≠ CAB. Used for: arrangements, queues, passwords, codes.

Arranging 3 of 5: ⁵P₃ = 60

📋

Combination ⁿCᵣ

Order doesn't matter. {A,B,C} = {B,C,A}. Used for: teams, committees, selections, choosing.

Choosing 3 of 5: ⁵C₃ = 10

✏️ Quick Example — Committee SelectionExample 6.7

A committee of 5 is to be formed from 6 men and 4 women. Find the number of ways if: (i) no restriction (ii) at least 2 women (iii) exactly 3 men.

1

No restriction: ¹⁰C₅ = 10!/(5!5!) = 252

2

At least 2 women (2W+3M, 3W+2M, 4W+1M):
⁴C₂×⁶C₃ + ⁴C₃×⁶C₂ + ⁴C⁴×⁶C₁
= 6×20 + 4×15 + 1×6 = 120+60+6 = 186

3

Exactly 3 men (and 2 women): ⁶C₃×⁴C₂ = 20×6 = 120

✔

No restriction: 252 | At least 2 women: 186 | Exactly 3 men: 120

✏️ Quick Example — Using Symmetry PropertyExample 6.8

If ⁿC₁₆ = ⁿC₄, find n. Also find ⁿC₂.

1

ⁿCᵣ = ⁿCˢ ⟹ r=s or r+s=n. Here 16≠4, so 16+4 = n ⟹ n = 20

2

ⁿC₂ = ²⁰C₂ = 20×19/2 = 190

✔

n = 20, ²⁰C₂ = 190. The symmetry property ⁿCᵣ = ⁿCₙ₋ᵣ is crucial here.

Module 5 of 5 🎉

Chapter 07 · Expanding the Unexpandable

Binomial Theorem & Mathematical Induction

The Binomial Theorem provides a complete expansion of (a+b)ⁿ without repeated multiplication. The Principle of Mathematical Induction gives us a rigorous proof technique for infinite families of statements. Together they are the backbone of combinatorial mathematics.

8 Modules ⌛ ~55 min read Class 11 Core

7.1

Binomial Theorem for Positive Integer

Expanding (a+b)ⁿ completely, Pascal's Triangle, and coefficient sums

The Binomial Theorem gives an exact formula for expanding any power of a binomial (a+b)ⁿ without doing repeated multiplication. It is one of the most powerful results in all of mathematics.

📖 Binomial Theorem (n ∈ ℕ)

(a + b)ⁿ = ∑𝓈₌₀ⁿ ⁿC𝓈 · aⁿ⁻𝓈 · b𝓈
= ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCⁿbⁿ
The coefficients ⁿC₀, ⁿC₁, …, ⁿCⁿ are called binomial coefficients. There are n+1 terms in total.

Pascal's Triangle

The binomial coefficients for successive values of n form Pascal's Triangle. Each entry is the sum of the two entries directly above it:

Key Coefficient Sums

Sum of all coefficients

Put a=b=1: ∑ ⁿCᵣ = 2ⁿ

Alternating sum

Put a=1,b=−1: ∑(−1)ᵣ ⁿCᵣ = 0

Even-indexed coeffs

ⁿC₀+ⁿC₂+… = 2ⁿ⁻¹

Odd-indexed coeffs

ⁿC₁+ⁿC₃+… = 2ⁿ⁻¹

Pascal's identity

ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

Vandermonde

∑ ᵐCᵣ · ⁿCᵣ = ᵐ⁺ⁿCᵣ

✏️ Quick Example — Expanding with Binomial TheoremExample 7.1

Expand (2x − 3y)⁴ fully.

1

a=2x, b=−3y, n=4. Apply (a+b)⁴ = ∑ ⁴Cᵣ a⁴⁻ᵣ bᵣ:

2

⁴C₀(2x)⁴ + ⁴C₁(2x)³(−3y) + ⁴C₂(2x)²(−3y)² + ⁴C₃(2x)(−3y)³ + ⁴C⁴(−3y)⁴

3

= 16x⁴ − 4(8x³)(3y) + 6(4x²)(9y²) − 4(2x)(27y³) + 81y⁴

✔

16x⁴ − 96x³y + 216x²y² − 216xy³ + 81y⁴

Module 1 of 8

7.2

General Term in a Binomial Expression

Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ᵣ · bᵣ — finding any term without full expansion

Instead of expanding the entire binomial, we can jump directly to any specific term using the general term formula. This is by far the most-tested result from this chapter.

✨ General Term Formula

The (r+1)th term in the expansion of (a+b)ⁿ is:
Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ᵣ · bᵣ
where r = 0, 1, 2, …, n. The first term is T₁ (r=0); the last term is Tₙ₊₁ (r=n).

Finding the Term Containing a Specific Power

To find the term containing x𝓈: write the general term, express it as a power of x, set the power equal to k, and solve for r.

✏️ Quick Example — Specific TermExample 7.2

In (x² + 1/x)¹², find: (i) the 5th term (ii) the term containing x³.

1

Tᵣ₊₁ = ¹²Cᵣ (x²)¹²⁻ᵣ (1/x)ᵣ = ¹²Cᵣ · x²(¹²⁻ᵣ)⁻ᵣ = ¹²Cᵣ · x²´⁻³ᵣ

2

5th term (r=4): T₅ = ¹²C₄ · x²´⁻¹² = ¹²C₄ · x¹² = 495x¹²

3

Term with x³: 24−3r = 3 ⟹ r = 7. T₈ = ¹²C₇ · x³ = 792x³

✔

T₅ = 495x¹² | Term with x³ = 792x³ (the 8th term)

✏️ Quick Example — Coefficient of a TermExample 7.3

Find the coefficient of x⁵ in (x + 3)¹°.

1

Tᵣ₊₁ = ¹⁰Cᵣ · x¹⁰⁻ᵣ · 3ᵣ

2

For x⁵: 10−r = 5 ⟹ r = 5.

3

Coefficient = ¹⁰C₅ · 3⁵ = 252 · 243 = 61236

✔

Coefficient of x⁵ = 61236.

General term

Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ᵣ bᵣ

(1+x)ⁿ general

Tᵣ₊₁ = ⁿCᵣ xᵣ

Power of x in Tᵣ₊₁

Set exponent = target, solve r

Module 2 of 8

7.3

Middle Term in a Binomial Expression

One middle term when n is even; two when n is odd

The middle term of (a+b)ⁿ is the term that sits at the centre of the n+1 terms. Because n+1 is even when n is odd, and odd when n is even, the two cases behave differently.

📖 Middle Term Rules

Case 1: n is even
Total terms = n+1 (odd number). There is exactly one middle term:
Tₙ₆₂₊₁ (the (n/2 + 1)th term), i.e. r = n/2

Case 2: n is odd
Total terms = n+1 (even number). There are two middle terms:
T₀ₙ₊₁)₆₂ and T₀ₙ₊₃)₆₂, i.e. r = (n−1)/2 and r = (n+1)/2

✏️ Quick Example — Middle Term (n even)Example 7.4

Find the middle term of (x + 2)¹°.

1

n=10 (even). Middle term = T₆₊₁ = T₆. So r=5.

2

T₆ = ¹⁰C₅ · x⁵ · 2⁵ = 252 · 32 · x⁵ = 8064x⁵

✔

Middle term = T₆ = 8064x⁵

✏️ Quick Example — Middle Terms (n odd)Example 7.5

Find the two middle terms of (2a − b)⁷.

1

n=7 (odd). Middle terms: T₄ (r=3) and T₅ (r=4).

2

T₄ = ⁷C₃ (2a)⁴ (−b)³ = 35 · 16a⁴ · (−b³) = −560a⁴b³

3

T₅ = ⁷C₄ (2a)³ (−b)⁴ = 35 · 8a³ · b⁴ = 280a³b⁴

✔

T₄ = −560a⁴b³ and T₅ = 280a³b⁴

Module 3 of 8

7.4

Greatest Term in a Binomial Expansion

Finding which term has the largest absolute value

Among all n+1 terms of (a+b)ⁿ, one (or sometimes two adjacent ones) has the largest absolute value. We find it by comparing consecutive terms.

🔑 Method — Ratio of Consecutive Terms

Form the ratio Tᵣ₊₂ / Tᵣ₊₁ = (n−r)/(r+1) · |b/a|
The greatest term occurs at the value of r where this ratio transitions from >1 to <1.
Step 1: Set Tᵣ₊₂/Tᵣ₊₁ ≥ 1 and solve for r ⟹ r ≤ m (some value m)
Step 2: The greatest term is T⌊m⌋₊₁ or T⌊m⌋₊₂ if m is an integer

✏️ Quick Example — Greatest TermExample 7.6

Find the greatest term in the expansion of (3 + 2x)¹° at x = 1.

1

Ratio = Tᵣ₊₂/Tᵣ₊₁ = [(10−r)/(r+1)] · (2/3). Set ≥ 1:

2

2(10−r) ≥ 3(r+1) ⟹ 20−2r ≥ 3r+3 ⟹ 17 ≥ 5r ⟹ r ≤ 3.4

3

Greatest r satisfying condition: r = 3. Greatest term = T₄.

4

T₄ = ¹⁰C₃ · 3⁷ · 2³ = 120 · 2187 · 8 = 2099520

✔

Greatest term is T₄ = 2099520.

Module 4 of 8

7.5

Multinomial Theorem

Expanding (a₁ + a₂ + … + a𝓈)ⁿ using multinomial coefficients

The Multinomial Theorem generalises the Binomial Theorem to sums of more than two terms. It is particularly useful for counting the number of terms and finding specific coefficients.

📖 Multinomial Theorem

(x₁ + x₂ + … + x𝓈)ⁿ = ∑ [n! / (n₁! n₂! … n𝓈!)] · x₁ⁿ¹ x₂ⁿ² … x𝓈ⁿ𝓈
where the sum is over all non-negative integers n₁, n₂, …, n𝓈 such that n₁ + n₂ + … + n𝓈 = n.
The coefficient n!/(n₁! n₂! … n𝓈!) is called a multinomial coefficient.

Number of Terms in a Multinomial Expansion

Terms in (a+b)ⁿ

n+1

Terms in (a+b+c)ⁿ

ⁿ⁺²C₂ = (n+1)(n+2)/2

Terms in (a+b+c+d)ⁿ

ⁿ⁺³C₃

General: k vars

ⁿ⁺𝓈⁻¹C𝓈⁻₁

✏️ Quick Example — Multinomial CoefficientExample 7.7

Find the coefficient of x²y³z in (x+y+z)⁶ and the total number of terms.

1

Need n₁=2, n₂=3, n₃=1 with n₁+n₂+n₃=6 ✔

2

Coefficient = 6!/(2!3!1!) = 720/(2·6·1) = 60

3

Total terms = ⁶⁺²C₂ = ⁸C₂ = 28

✔

Coefficient of x²y³z = 60. Total terms in (x+y+z)⁶ = 28.

Module 5 of 8

7.6

R-f Factor Relations

Integer and fractional parts of binomial expressions of the form (a+√b)ⁿ

When we expand (p + q)ⁿ where q involves a surd (like √k), the expansion has rational terms and irrational terms. The R-f method uses the conjugate expansion to cleanly separate and analyse these parts.

📖 Setup — R-f Method

Let (a + √b)ⁿ = I + f, where I is the integer part and 0 ≤ f < 1 is the fractional part.
Consider the conjugate: (a − √b)ⁿ = f′ where 0 < f′ < 1 (if a − √b > 0 and small).
Adding: (a+√b)ⁿ + (a−√b)ⁿ = 2×(sum of rational terms) = an integer!
This gives us: I + f + f′ = even integer, so f + f′ = integer ⟹ f′ = 1−f (if f ≠ 0).

Key Identity

(a+√b)ⁿ+(a−√b)ⁿ

= 2∑ ⁿC₂ᵣ aⁿ⁻²ᵣ bᵣ (integer)

f · f′

= f(1−f) = (a²−b)ⁿ

(I+f)(f′)

= (a²−b)ⁿ

✏️ Quick Example — R-f MethodExample 7.8

If (6+√35)ⁿ = I + f where I is integer and 0<f<1, show that (I+f)(1−f) = 1.

1

Let f′ = (6−√35)ⁿ. Since 0 < 6−√35 < 1, we have 0 < f′ < 1.

2

(6+√35)ⁿ + (6−√35)ⁿ = 2×(integer) = I + f + f′.

3

So f + f′ = integer. Since 0<f,f′<1, we get f + f′ = 1, i.e. f′ = 1−f.

4

(I+f)f′ = (6+√35)ⁿ(6−√35)ⁿ = (36−35)ⁿ = 1ⁿ = 1

✔

(I+f)(1−f) = 1. ⭐ This is a classic exam result — the product (I+f)f′ = (a²−b)ⁿ.

Module 6 of 8

7.7

Binomial Theorem for Any Index

Extending to negative and fractional exponents — infinite series when |x|<1

The Binomial Theorem extends beyond positive integers. When the index n is a negative integer or a fraction, the expansion becomes an infinite series valid only when |x| < 1.

📖 General Binomial Series (|x| < 1, any n)

(1+x)ⁿ = 1 + nx + n(n−1)/2! · x² + n(n−1)(n−2)/3! · x³ + … (infinite series)
Valid for all real/rational/negative n provided |x| < 1.
The general term: Tᵣ₊₁ = n(n−1)…(n−r+1)/r! · xᵣ

Important Special Cases

(1+x)⁻¹

1−x+x²−x³+…

(1−x)⁻¹

1+x+x²+x³+…

(1+x)⁻²

1−2x+3x²−4x³+…

(1−x)⁻²

1+2x+3x²+4x³+…

(1+x)¹/²

1+x/2−x²/8+…

Condition

|x| < 1 always required

✏️ Quick Example — Approximation using Binomial SeriesExample 7.9

Using binomial expansion, find an approximation of √(1.04) correct to 4 decimal places.

1

√(1.04) = (1+0.04)¹/². Here n=1/2, x=0.04 < 1 ✔

2

≈ 1 + (1/2)(0.04) + (1/2)(−1/2)/2! (0.04)² = 1 + 0.02 − 0.0002 = 1.0198

✔

√(1.04) ≈ 1.0198. Actual value = 1.01980… ✔ Excellent approximation!

Module 7 of 8

7.8

Principle of Mathematical Induction (PMI)

A watertight proof technique for statements about natural numbers

The Principle of Mathematical Induction is the gold-standard method for proving that a statement P(n) is true for all natural numbers n. It works like an infinite chain of dominoes.

📖 PMI — Two Steps

Step 1 — Base Case: Verify that P(1) is true (sometimes P(0)).
Step 2 — Inductive Step: Assume P(k) is true for some arbitrary k ≥ 1 (the inductive hypothesis). Then prove that P(k+1) must also be true.
Conclusion: By PMI, P(n) is true for all n ∈ ℕ.

✏️ Quick Example — Proving a Sum Formula by PMIExample 7.10

Prove by PMI: 1+2+3+…+n = n(n+1)/2 for all n ∈ ℕ.

1

Base case P(1): LHS=1. RHS=1(2)/2=1. LHS=RHS ✔

2

Inductive hypothesis P(k): Assume 1+2+…+k = k(k+1)/2.

3

Prove P(k+1): 1+…+k+(k+1) = k(k+1)/2 + (k+1) = (k+1)[k/2+1] = (k+1)(k+2)/2 ✔

✔

By PMI, 1+2+…+n = n(n+1)/2 for all n ≥ 1. ■

✏️ Quick Example — Divisibility by PMIExample 7.11

Prove by PMI: 4ⁿ − 1 is divisible by 3 for all n ∈ ℕ.

1

P(1): 4¹−1 = 3. Divisible by 3 ✔

2

Assume P(k): 4𝓈−1 = 3m for some integer m.

3

Prove P(k+1): 4𝓈⁺¹−1 = 4·4𝓈−1 = 4(3m+1)−1 = 12m+3 = 3(4m+1). Divisible by 3 ✔

✔

By PMI, 3 | (4ⁿ−1) for all n ≥ 1.

Module 8 of 8 🎉

Chapter 08 · Organising Numbers in Grids

Matrices

A matrix is a rectangular array of numbers with powerful algebraic properties. From basic operations to transposes, conjugates, and rank — this chapter builds a complete toolkit for working with matrices used across engineering, data science and pure mathematics.

7 Modules ⌛ ~50 min read Class 11 & 12

8.1

Matrix

Definition, order, types — the vocabulary of matrix algebra

A matrix is a rectangular arrangement of numbers (called elements or entries) in rows and columns, enclosed in brackets. Matrices are the language of linear systems, transformations, and data.

📖 Definition — Matrix

A matrix of order m×n has m rows and n columns. The element in the ith row and jth column is denoted aᵢⱼ or [aᵢⱼ].
A = [aᵢⱼ]ᵐ×ₙ means A is an m×n matrix.

Types of Matrices

Row Matrix

Single row (1×n). Also called a row vector.

[1 3 −2]

Column Matrix

Single column (m×1). Also called a column vector.

[[2],[−1],[5]]

Square Matrix

Number of rows = columns (n×n). Has a main diagonal.

2×2, 3×3, …

Null (Zero) Matrix

All elements are 0. Denoted O.

[[0,0],[0,0]]

Identity Matrix

Square, 1s on diagonal, 0s elsewhere. Denoted Iₙ.

[[1,0],[0,1]]

Diagonal Matrix

Square; all off-diagonal entries are 0.

[[3,0],[0,−2]]

Upper Triangular

All entries below main diagonal = 0.

[[a,b],[0,c]]

Lower Triangular

All entries above main diagonal = 0.

[[a,0],[c,d]]

Equality of Matrices

Two matrices A and B are equal (A = B) if and only if they have the same order AND aᵢⱼ = bᵢⱼ for every i, j.

✏️ Quick Example — Equality to find unknownsExample 8.1

Find x, y, z if [[x+y, 2z],[3, x−y]] = [[5, 4],[3, 1]].

1

x+y = 5 …(i) 2z = 4 ⟹ z = 2

2

x−y = 1 …(ii). Add (i)+(ii): 2x = 6 ⟹ x = 3

3

From (i): y = 5−3 = y = 2

✔

x = 3, y = 2, z = 2

Module 1 of 7

8.2

Algebra of Matrices

Addition, scalar multiplication, matrix multiplication — rules and properties

Matrices support their own algebra. The key operations are addition, scalar multiplication, and matrix multiplication — each with specific conditions on dimensions.

Addition and Subtraction

Rule

Only defined for matrices of the same order. Add/subtract element-wise:
(A ± B)ᵢⱼ = aᵢⱼ ± bᵢⱼ

Matrix Multiplication

Rule — AB is defined only if cols(A) = rows(B)

If A is m×p and B is p×n, then C = AB is m×n where:
cᵢⱼ = ∑𝓈₌₁ₚ aᵢ𝓈 · b𝓈ⱼ (dot product of ith row of A with jth column of B)

Properties

Commutative (add)

A+B = B+A

Commutative (mult)

AB ≠ BA (generally)

Associative

(AB)C = A(BC)

Distributive

A(B+C) = AB+AC

Identity

AI = IA = A

Zero matrix

A+O = A, AO = O

(AB)⁻¹

= B⁻¹A⁻¹

k(AB)

= (kA)B = A(kB)

✏️ Quick Example — Matrix MultiplicationExample 8.2

Find AB where A = [[1,2],[3,4]] and B = [[5,6],[7,8]].

1

c₁₁ = 1·5+2·7 = 5+14 = 19 c₁₂ = 1·6+2·8 = 6+16 = 22

2

c₂₁ = 3·5+4·7 = 15+28 = 43 c₂₂ = 3·6+4·8 = 18+32 = 50

✔

AB = [[19, 22],[43, 50]]. Now compute BA to confirm AB ≠ BA!

Module 2 of 7

8.3

Transpose of a Matrix

Flipping rows and columns — Aᵀ and its properties

The transpose of a matrix A, written Aᵀ (or Aᵀ or A'), is obtained by interchanging its rows and columns. The element in row i, column j of A goes to row j, column i of Aᵀ.

📖 Definition

If A = [aᵢⱼ]ᵐ×ₙ, then Aᵀ = [aⱼᵢ]ₙ×ᵐ.
Order of A = m×n ⟹ Order of Aᵀ = n×m.

(Aᵀ)ᵀ

= A

(A+B)ᵀ

= Aᵀ + Bᵀ

(kA)ᵀ

= kAᵀ

(AB)ᵀ

= BᵀAᵀ (reverse order!)

(ABC)ᵀ

= CᵀBᵀAᵀ

✏️ Quick Example — TransposeExample 8.3

If A = [[1,2,3],[4,5,6]], find Aᵀ and verify (Aᵀ)ᵀ = A.

1

A is 2×3. Aᵀ is 3×2: Aᵀ = [[1,4],[2,5],[3,6]]

2

(Aᵀ)ᵀ flips again: [[1,2,3],[4,5,6]] = A ✔

✔

Transpose flips row-column. Double transpose returns original matrix.

Module 3 of 7

8.4

Symmetric and Skew-Symmetric Matrices

Matrices equal to or negating their own transpose

📖 Symmetric Matrix

A square matrix A is symmetric if Aᵀ = A, i.e. aᵢⱼ = aⱼᵢ for all i, j.
The matrix is a mirror image of itself across the main diagonal.

📖 Skew-Symmetric Matrix

A square matrix A is skew-symmetric if Aᵀ = −A, i.e. aᵢⱼ = −aⱼᵢ for all i, j.
In particular: all diagonal elements must be zero (since aᵢᵢ = −aᵢᵢ ⟹ aᵢᵢ = 0).

Key Theorems

Every square matrix A can be uniquely expressed as A = P + Q, where P = (A+Aᵀ)/2 is symmetric and Q = (A−Aᵀ)/2 is skew-symmetric
AᵀA and AAᵀ are always symmetric
If A and B are symmetric, AB is symmetric iff AB = BA
Diagonal of skew-symmetric is always 0

✏️ Quick Example — DecompositionExample 8.4

Express A = [[1,3,5],[−3,2,4],[1,−4,3]] as sum of symmetric and skew-symmetric matrices.

1

Aᵀ = [[1,−3,1],[3,2,−4],[5,4,3]]

2

P = (A+Aᵀ)/2 = [[1,0,3],[0,2,0],[3,0,3]] (symmetric ✔)

3

Q = (A−Aᵀ)/2 = [[0,3,2],[−3,0,4],[−2,−4,0]] (skew-symmetric, diagonal=0 ✔)

✔

A = P + Q. Verify P+Q = A ✔

Module 4 of 7

8.5

Elementary Operations (Transformations)

Row and column operations — the engine of Gaussian elimination

Elementary operations are the basic allowable moves on a matrix. They do not change the solution set of a linear system and are used to reduce matrices to simpler forms (echelon, reduced echelon, or identity).

📖 Three Elementary Row Operations

Rᵢ ↔ Rⱼ — Interchange rows i and j
Rᵢ → kRᵢ — Multiply row i by non-zero scalar k
Rᵢ → Rᵢ + kRⱼ — Add k times row j to row i

The same three operations apply to columns (Cᵢ, Cⱼ notation).

Row Echelon Form

All zero rows are at the bottom
The leading entry (pivot) in each non-zero row is to the right of the pivot in the row above
All entries below a pivot are zero

✏️ Quick Example — Row ReductionExample 8.5

Reduce A = [[1,2,3],[2,5,7],[3,7,11]] to row echelon form using elementary row operations.

1

R₂ → R₂−2R₁: row2 = [2−2,5−4,7−6] = [0,1,1]

2

R₃ → R₃−3R₁: row3 = [3−3,7−6,11−9] = [0,1,2]

3

R₃ → R₃−R₂: row3 = [0,0,1]. Matrix = [[1,2,3],[0,1,1],[0,0,1]]

✔

Row echelon form reached. Pivots: 1 (col 1), 1 (col 2), 1 (col 3) ⟹ full rank.

Module 5 of 7

8.6

Conjugate of a Matrix

Complex matrix operations — conjugate, conjugate transpose, and Hermitian matrices

When a matrix has complex number entries, we define additional operations using the complex conjugate. These are essential in quantum mechanics and advanced linear algebra.

📖 Conjugate & Conjugate Transpose

If A = [aᵢⱼ] has complex entries:
Ā = [a̅ᵢⱼ] (conjugate — replace each element by its complex conjugate)
Aᵀ = [a̅ⱼᵢ] (conjugate transpose / Hermitian adjoint — transpose then conjugate)

Special Matrices with Complex Entries

Hermitian

Aᵀ = A. Complex generalisation of symmetric. Diagonal entries are real.

[[2, 1+i],[1−i, 3]]

Skew-Hermitian

Aᵀ = −A. Diagonal entries are purely imaginary or zero.

[[0, i],[−i, 0]]

Unitary

AᵀA = I. Generalises orthogonal matrices to complex case.

Columns are orthonormal

Conjugate Ā

Replace a+bi by a−bi

(Ā̅)

= A

(A+B)̅

= Ā + ̅B

Aᵀ = (Ā)ᵀ

Transpose of conjugate

Hermitian iff

Aᵀ = A

✏️ Quick Example — Hermitian CheckExample 8.6

Show that A = [[3, 2+i],[2−i, 5]] is Hermitian.

1

Conjugate each entry: Ā = [[3, 2−i],[2+i, 5]]

2

Transpose Ā: Aᵀ = [[3, 2+i],[2−i, 5]]

3

Aᵀ = A ✔ (diagonal entries 3 and 5 are real ✔)

✔

A is Hermitian. Eigenvalues of Hermitian matrices are always real.

Module 6 of 7

8.7

Rank of a Matrix

The maximum number of independent rows or columns — key to solving linear systems

The rank of a matrix measures how much “information” it contains — specifically, the maximum number of linearly independent rows (or columns). It determines whether a system of equations has a unique solution, infinitely many, or none.

📖 Definition — Rank

The rank of a matrix A, written ρ(A) or rank(A), is:
• The number of non-zero rows in its row echelon form
• The order of the largest non-singular (non-zero determinant) square submatrix
• The dimension of the row space (= column space) of A

Properties of Rank

ρ(A) ≤ min(m, n) for an m×n matrix
ρ(A) = ρ(Aᵀ) — row rank equals column rank
Elementary row operations do NOT change rank
ρ(AB) ≤ min(ρ(A), ρ(B))
A is invertible iff ρ(A) = n (full rank, n×n square matrix)

Rank and Systems of Equations (Rouché–Capelli Theorem)

ρ(A) = ρ(A|b) = n

Unique solution. Consistent and determinate.

ρ(A) = ρ(A|b) < n

Infinite solutions. Consistent with free variables.

ρ(A) < ρ(A|b)

No solution. Inconsistent system.

✏️ Quick Example — Finding RankExample 8.7

Find the rank of A = [[1,2,3],[2,4,6],[1,1,2]].

1

R₂ → R₂−2R₁: [[1,2,3],[0,0,0],[1,1,2]]

2

R₃ → R₃−R₁: [[1,2,3],[0,0,0],[0,−1,−1]]

3

Swap R₂ and R₃: [[1,2,3],[0,−1,−1],[0,0,0]]. Two non-zero rows.

✔

ρ(A) = 2. Note row 2 of A was 2× row 1 — linearly dependent, hence rank drops.

Rank ρ(A)

Non-zero rows in echelon form

Full rank (square)

ρ(A)=n ⟺ A invertible

Null matrix

ρ(O) = 0

Rank-nullity

ρ(A) + nullity(A) = n

Module 7 of 7 🎉

Chapter 09 · A Scalar from a Matrix

Determinants

The determinant extracts a single number from a square matrix that encodes its geometry, invertibility, and linear independence. From cofactors to Cramer's Rule, this chapter is the key to solving systems of equations and understanding matrix inverses.

6 Modules ⌛ ~45 min read Class 11 & 12

9.1

Determinant

|A| — computing 2×2 and 3×3 determinants, and key properties

The determinant of a square matrix A is a single real number that encodes essential geometric and algebraic information: it tells us if the matrix is invertible, measures the signed area/volume of the transformation it represents, and is central to solving linear systems.

📖 Determinant — 2×2 Matrix

For A = [[a, b],[c, d]]:
det(A) = |A| = ad − bc
Geometrically: |ad−bc| is the area of the parallelogram formed by the two row vectors.

📖 Determinant — 3×3 Matrix (Expansion along Row 1)

For A = [[a₁,a₂,a₃],[b₁,b₂,b₃],[c₁,c₂,c₃]]:
|A| = a₁(b₂c₃−b₃c₂) − a₂(b₁c₃−b₃c₁) + a₃(b₁c₂−b₂c₁)
Pattern of signs along row 1: + − +

Sarrus' Rule (Diagonal Method for 3×3)

Properties of Determinants

|Aᵀ| = |A| — transpose has the same determinant
If two rows (or columns) are identical, |A| = 0
Swapping two rows multiplies |A| by −1
Multiplying a row by k multiplies |A| by k; so |kA| = kⁿ|A| for n×n
Adding a multiple of one row to another leaves |A| unchanged
|AB| = |A| ⋅ |B| — product rule for determinants
If any row or column is all zeros, |A| = 0

✏️ Quick Example — 3×3 DeterminantExample 9.1

Evaluate |A| where A = [[2,−1,3],[0,4,−2],[1,2,5]].

1

Expand along Row 1: |A| = 2|[[4,−2],[2,5]]| − (−1)|[[0,−2],[1,5]]| + 3|[[0,4],[1,2]]|

2

= 2(20+4) + 1(0+2) + 3(0−4) = 2(24) + 1(2) + 3(−4)

3

= 48 + 2 − 12 = 38

✔

|A| = 38. Since |A| ≠ 0, A is non-singular (invertible).

2×2 det

ad − bc

|AB|

= |A| ⋅ |B|

|kA| (n×n)

= kⁿ |A|

|A⁻¹|

= 1/|A|

Area of Δ

= ½|det[[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]]|

Module 1 of 6

9.2

Minors and Cofactors

Sub-determinants and signed versions — the building blocks of adjoint and inverse

To systematically work with large determinants and build the adjoint, we need the concepts of minors and cofactors. Every element of a matrix has one of each.

📖 Minor Mᵢⱼ

The minor Mᵢⱼ of element aᵢⱼ is the determinant of the sub-matrix obtained by deleting the ith row and jth column of A.

📖 Cofactor Cᵢⱼ

The cofactor Cᵢⱼ is the signed minor:
Cᵢⱼ = (−1)ᵢ⁺ⱼ ⋅ Mᵢⱼ
The sign (−1)ᵢ⁺ⱼ follows the checkerboard pattern: + if i+j is even, − if i+j is odd.

✏️ Quick Example — Minors and CofactorsExample 9.2

For A = [[1,2,3],[4,5,6],[7,8,9]], find M₂₃ and C₂₃.

1

Delete row 2 and col 3: sub-matrix = [[1,2],[7,8]]

2

M₂₃ = |[[1,2],[7,8]]| = 8 − 14 = −6

3

C₂₃ = (−1)²⁺³ M₂₃ = (−1)⁵ ⋅ (−6) = +6

✔

M₂₃ = −6, C₂₃ = +6. The cofactor adds the checkerboard sign to the minor.

Expansion Using Cofactors (Laplace Expansion)

The determinant can be expanded along any row or column using cofactors:

Along row i

|A| = ∑ⱼ aᵢⱼ Cᵢⱼ

Along col j

|A| = ∑ᵢ aᵢⱼ Cᵢⱼ

Cross-expansion

∑ⱼ aᵢⱼ C𝓈ⱼ = 0 (i≠k)

Module 2 of 6

9.3

Adjoint of a Matrix

adj(A) — the transpose of the cofactor matrix

The adjoint (or adjugate) of a matrix is the transpose of its cofactor matrix. It is the key ingredient in the formula for the matrix inverse.

📖 Definition — Adjoint

If C = [Cᵢⱼ] is the matrix of cofactors of A, then:
adj(A) = Cᵀ = [Cⱼᵢ]
Note: the (i,j) entry of adj(A) is the cofactor Cⱼᵢ (swapped indices).

Key Identity

✨ Fundamental Property

A ⋅ adj(A) = adj(A) ⋅ A = |A| ⋅ I
This identity is the bridge between the adjoint and the inverse. If |A| ≠ 0, dividing both sides by |A| gives the inverse formula.

✏️ Quick Example — Finding adj(A)Example 9.3

Find adj(A) for A = [[1,2],[3,4]].

1

Cofactors: C₁₁ = +4, C₁₂ = −3, C₂₁ = −2, C₂₂ = +1

2

Cofactor matrix = [[4,−3],[−2,1]]. Transpose it:

3

adj(A) = [[4,−2],[−3,1]]

4

Verify: A⋅adj(A) = [[1,2],[3,4]]⋅[[4,−2],[−3,1]] = [[−2,0],[0,−2]] = (−2)I = |A|⋅I ✔

✔

adj(A) = [[4,−2],[−3,1]]. |A| = 4−6 = −2.

2×2 shortcut

adj([[a,b],[c,d]]) = [[d,−b],[−c,a]]

adj(Aᵀ)

= [adj(A)]ᵀ

adj(kA)

= kⁿ⁻¹ adj(A)

|adj(A)|

= |A|ⁿ⁻¹

Module 3 of 6

9.4

Inverse of a Matrix

A⁻¹ = adj(A)/|A| — when it exists and how to compute it

The inverse of a square matrix A is a matrix A⁻¹ such that A A⁻¹ = A⁻¹ A = I. It exists if and only if |A| ≠ 0 (i.e. A is non-singular).

✨ Inverse Formula

A⁻¹ = adj(A) / |A| (provided |A| ≠ 0)
If |A| = 0, A is called singular and has no inverse.

Properties of Inverse

(A⁻¹)⁻¹

= A

(AB)⁻¹

= B⁻¹A⁻¹

(Aᵀ)⁻¹

= (A⁻¹)ᵀ

(kA)⁻¹

= (1/k)A⁻¹

|A⁻¹|

= 1/|A|

Unique

Inverse is unique if it exists

✏️ Quick Example — Finding A⁻¹Example 9.4

Find A⁻¹ for A = [[2,1],[5,3]].

1

|A| = 2(3)−1(5) = 6−5 = 1 ≠ 0 ✔ (invertible)

2

adj(A) = [[3,−1],[−5,2]] (swap diagonal, negate off-diagonal)

3

A⁻¹ = adj(A)/|A| = [[3,−1],[−5,2]]/1 = [[3,−1],[−5,2]]

4

Verify: AA⁻¹ = [[2,1],[5,3]][[3,−1],[−5,2]] = [[6−5,−2+2],[15−15,−5+6]] = [[1,0],[0,1]] = I ✔

✔

A⁻¹ = [[3,−1],[−5,2]]

Solving AX = B Using Inverse

If A is invertible, the system AX = B has the unique solution X = A⁻¹B. This is equivalent to Cramer's Rule but often more efficient for multiple right-hand sides.

Cramer's Rule

For a system a₁₁x+a₁₂y+a₁₃z=b₁, … with coefficient matrix A and |A| ≠ 0:

x = D₁/D

D₁ = replace col 1 of A with b

y = D₂/D

D₂ = replace col 2 of A with b

z = D₃/D

D₃ = replace col 3 of A with b

D = |A|

Determinant of coefficient matrix

✏️ Quick Example — Cramer's RuleExample 9.5

Solve: 2x + y = 7, 5x + 3y = 18.

1

D = |[[2,1],[5,3]]| = 6−5 = 1

2

D₁ = |[[7,1],[18,3]]| = 21−18 = 3 ⟹ x = 3/1 = 3

3

D₂ = |[[2,7],[5,18]]| = 36−35 = 1 ⟹ y = 1/1 = 1

✔

x = 3, y = 1. Verify: 2(3)+1=7 ✔ 5(3)+3=18 ✔

Module 4 of 6

9.5

Non-Homogeneous System of Linear Equations

AX = B with B ≠ 0 — consistency, unique solutions, and infinite solutions

A non-homogeneous system AX = B has a non-zero right-hand side (B ≠ 0). The nature of its solution depends entirely on the determinant |A| and the ranks of A and the augmented matrix [A|B].

📖 Three Cases for AX = B

Case 1: |A| ≠ 0 (Non-singular A)
Unique solution: X = A⁻¹B. Use Cramer's Rule or matrix inverse.

Case 2: |A| = 0, (adj A)B = 0
Infinitely many solutions. System is consistent but indeterminate.

Case 3: |A| = 0, (adj A)B ≠ 0
No solution. System is inconsistent.

✏️ Quick Example — 3-Variable SystemExample 9.6

Solve: x+2y+3z=6, 2x+4y+z=7, 3x+2y+9z=14.

1

A = [[1,2,3],[2,4,1],[3,2,9]]. Compute |A|:
|A| = 1(36−2)−2(18−3)+3(4−12) = 34−30−24 = −20 ≠ 0

2

Unique solution exists. Apply Cramer's Rule:
D₁=|[[6,2,3],[7,4,1],[14,2,9]]| = 6(36−2)−2(63−14)+3(14−56)=204−98−126=−20

3

x = D₁/D = −20/−20 = 1. Similarly y=1, z=1.

✔

x=1, y=1, z=1. Verify: 1+2+3=6 ✔ 2+4+1=7 ✔ 3+2+9=14 ✔

Module 5 of 6

9.6

Homogeneous System of Linear Equations

AX = 0 — trivial vs non-trivial solutions

A homogeneous system AX = 0 always has at least one solution: the trivial solution X = 0. The interesting question is whether there are also non-trivial solutions (X ≠ 0).

📖 Key Theorem

For the system AX = 0 (A is n×n):
• If |A| ≠ 0: only the trivial solution X = 0 exists.
• If |A| = 0: infinite non-trivial solutions exist.
The system AX = 0 is always consistent (X = 0 always works).

Number of Free Variables

The number of free variables (parameters) in the solution = n − rank(A). If rank(A) = n, only trivial solution; if rank(A) < n, infinitely many solutions with n−rank(A) free parameters.

✏️ Quick Example — Non-Trivial SolutionsExample 9.7

Find non-trivial solutions of: x + 2y − z = 0, 2x + y + z = 0, x − y + 2z = 0.

1

A = [[1,2,−1],[2,1,1],[1,−1,2]]. |A| = 1(2+1)−2(4−1)+(−1)(−2−1) = 3−6+3 = 0

2

|A| = 0 ⟹ non-trivial solutions exist. Row reduce: R₂−2R₁, R₃−R₁:

3

[[1,2,−1],[0,−3,3],[0,−3,3]] ⟹ R₃−R₂: [[1,2,−1],[0,−3,3],[0,0,0]]

4

From R₂: y = z. From R₁: x = −2y+z = −2z+z = −z. Let z=k: x=−k, y=k, z=k.

✔

Solution: (x,y,z) = k(−1,1,1) for k ≠ 0. Infinitely many non-trivial solutions on this line.

|A| ≠ 0

Only X=0 (trivial)

|A| = 0

Non-trivial solutions exist

Free vars

n − rank(A)

Always consistent

AX=0 has X=0 always

Module 6 of 6 🎉

Chapter 10 · The Mathematics of Chance

Probability

Probability gives us the language to reason about uncertainty. From basic experiments and events to Bayes' Theorem, random variables, and the Binomial distribution — this chapter connects everyday chance to powerful mathematics.

6 Modules ⌛ ~50 min read Class 11 & 12 Core

10.1

Experiment, Sample Space and Events

The vocabulary of probability — random experiments, outcomes, and events

Probability begins with a random experiment — a process whose outcome cannot be predicted with certainty but whose set of all possible outcomes is known. Understanding the language of experiments and events is essential before computing any probability.

📖 Key Definitions

Random Experiment: A process with a well-defined set of outcomes where the result cannot be predicted in advance. E.g. tossing a coin, rolling a die.

Sample Space (S): The set of ALL possible outcomes of an experiment. E.g. S = {H, T} for a coin toss.

Event (E): Any subset of the sample space. E.g. “getting an even number” when rolling a die: E = {2, 4, 6}.

Elementary event: A single outcome {ω}.
Certain event: S itself (always occurs).
Impossible event: ∅ (never occurs).

Classical (Laplace) Definition of Probability

P(E) = n(E) / n(S)

When all outcomes are equally likely:
P(E) = (Number of favourable outcomes) / (Total outcomes in S)
Range: 0 ≤ P(E) ≤ 1 always.

Types of Events

Mutually Exclusive

Cannot both occur: A ∩ B = ∅

Head AND Tail in one toss

Exhaustive

At least one must occur: A ∪ B = S

Head OR Tail covers S

Complementary

A and A' are M.E. and exhaustive

P(A) + P(A') = 1

Independent

P(A∩B) = P(A)⋅P(B)

Two separate coin tosses

Dependent

Outcome of one affects the other

Cards drawn without replacement

Equally Likely

Each outcome has the same probability

Fair die: P = 1/6 each

✏️ Quick Example — Classical ProbabilityExample 10.1

Two dice are rolled. Find P(sum = 7), P(sum ≥ 10), and P(both show same number).

1

n(S) = 36. Sum=7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 outcomes. P = 6/36 = 1/6

2

Sum≥10: (4,6),(5,5),(6,4),(5,6),(6,5),(6,6) = 6 outcomes. P = 6/36 = 1/6

3

Both same: (1,1),(2,2),…,(6,6) = 6 outcomes. P = 6/36 = 1/6

✔

P(sum=7) = 1/6 | P(sum≥10) = 1/6 | P(doubles) = 1/6

Module 1 of 6

10.2

Algebra of Events

Addition theorem, conditional probability, multiplication rule — the core probability laws

The algebra of events gives us rules for computing probabilities of combined events. These laws are the backbone of all probability calculations.

✨ Addition Theorem

For any two events A and B:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
For mutually exclusive (A ∩ B = ∅): P(A ∪ B) = P(A) + P(B)

📖 Conditional Probability

The probability of A given B has occurred:
P(A|B) = P(A ∩ B) / P(B) (P(B) ≠ 0)
This updates P(A) using the information that B has occurred.

📖 Multiplication Rule

P(A ∩ B) = P(A) ⋅ P(B|A) = P(B) ⋅ P(A|B)
For independent events: P(A ∩ B) = P(A) ⋅ P(B)

P(A∪B)

P(A)+P(B)−P(A∩B)

P(A')

1 − P(A)

P(A|B)

P(A∩B)/P(B)

Independent

P(A∩B)=P(A)P(B)

P(A∩B')

P(A)−P(A∩B)

3 events

P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C)

✏️ Quick Example — Conditional ProbabilityExample 10.2

A bag has 3 red and 4 blue balls. Two drawn without replacement. P(2nd is red | 1st is red)?

1

After 1 red drawn: 2 red and 4 blue remain (6 total).

2

P(2nd red | 1st red) = 2/6 = 1/3

3

P(both red) = P(1st red) × P(2nd red|1st red) = (3/7)(1/3) = 1/7

✔

P(2nd red|1st red) = 1/3. P(both red) = 1/7.

Module 2 of 6

10.3

Bayes' Theorem

Updating probability with new evidence — prior, likelihood, and posterior

Bayes' Theorem is one of the most profound results in probability. It tells us how to update our belief about an event when we receive new evidence. It underlies machine learning, medical diagnosis, and spam filters.

📖 Law of Total Probability

If A₁, A₂, …, Aⁿ are mutually exclusive and exhaustive events (a partition of S), and B is any event:
P(B) = P(B|A₁)P(A₁) + P(B|A₂)P(A₂) + … + P(B|Aⁿ)P(Aⁿ)
= ∑𝓈 P(B|A𝓈) P(A𝓈)

✨ Bayes' Theorem

Given the partition A₁, …, Aⁿ and event B (with P(B) > 0):
P(A𝓈|B) = P(A𝓈) P(B|A𝓈) / ∑ⱼ P(Aⱼ) P(B|Aⱼ)
• P(A𝓈) = prior probability (before observing B)
• P(B|A𝓈) = likelihood (how likely B is if A𝓈 is true)
• P(A𝓈|B) = posterior probability (updated after observing B)

✏️ Quick Example — Bayes' TheoremExample 10.3

Factory has 3 machines: M₁ (50% of output, 2% defective), M₂ (30%, 3%), M₃ (20%, 4%). A randomly selected item is defective. P(it came from M₂)?

1

P(D) = P(D|M₁)P(M₁)+P(D|M₂)P(M₂)+P(D|M₃)P(M₃)
= 0.02(0.5)+0.03(0.3)+0.04(0.2) = 0.01+0.009+0.008 = 0.027

2

P(M₂|D) = P(D|M₂)P(M₂)/P(D) = (0.03 × 0.3)/0.027 = 0.009/0.027 = 1/3 ≈ 0.333

✔

P(M₂|defective) = 1/3 ≈ 33.3%. There is a 1 in 3 chance the defective item came from M₂.

Module 3 of 6

10.4

Random Variable

Assigning numbers to outcomes — probability distribution, mean, and variance

A random variable is a function that assigns a numerical value to each outcome of a random experiment. It converts qualitative outcomes (like “head” or “tail”) into numbers we can compute with.

📖 Definition — Random Variable

A random variable X is a real-valued function X : S → ℜ.
Discrete RV: Takes countable values x₁, x₂, … with probabilities p₁, p₂, …
Continuous RV: Takes all values in an interval (requires probability density function).

Probability Distribution

The probability distribution of a discrete RV X is the table (or function) listing all values x𝓈 and their probabilities P(X=x𝓈) = p𝓈. Requirements:

0 ≤ p𝓈 ≤ 1 for all i
∑ p𝓈 = 1 (probabilities sum to 1)

Mean (Expected Value) and Variance

Mean μ = E(X)

∑ x𝓈 p𝓈

E(X²)

∑ x𝓈² p𝓈

Variance Var(X)

E(X²) − [E(X)]²

Std Dev σ

σ = √Var(X)

E(aX+b)

aE(X)+b

Var(aX+b)

a²Var(X)

✏️ Quick Example — Probability DistributionExample 10.4

X = number of heads in 2 tosses of a fair coin. Build the distribution and find E(X) and Var(X).

1

S={(TT,TH,HT,HH)}. X=0:TT (p=1/4); X=1:TH,HT (p=2/4=1/2); X=2:HH (p=1/4)

2

E(X) = 0(1/4)+1(1/2)+2(1/4) = 0+0.5+0.5 = 1

3

E(X²) = 0(1/4)+1(1/2)+4(1/4) = 0+0.5+1 = 1.5

4

Var(X) = E(X²)−[E(X)]² = 1.5−1 = 0.5

✔

E(X) = 1, Var(X) = 0.5, σ = 1/√2 ≈ 0.707

Module 4 of 6

10.5

Bernoulli Trials

Repeated independent experiments with exactly two outcomes: success or failure

A Bernoulli trial is the simplest non-trivial random experiment: one with exactly two outcomes. When repeated independently many times, Bernoulli trials form the foundation of the binomial distribution.

📖 Bernoulli Trial

An experiment with exactly two outcomes:
• Success (S) with probability p (0 < p < 1)
• Failure (F) with probability q = 1−p

A sequence of n independent Bernoulli trials all with the same p is called a Bernoulli process.
Key: outcomes of different trials do not affect each other.

Conditions for Bernoulli Trials

Finite: Fixed number of trials n
Independence: Each trial is independent of others
Two outcomes: Success (p) or Failure (q=1−p)
Constant probability: p is the same for every trial

✏️ Quick Example — Bernoulli SequenceExample 10.5

A fair coin is tossed 4 times. Find P(exactly 2 heads) treating each toss as a Bernoulli trial.

1

p = P(Head) = 1/2, q = 1/2, n = 4, want X = 2 successes.

2

Any specific HHTT-type sequence: p²q² = (1/2)²(1/2)² = 1/16

3

Number of arrangements of 2 H in 4 tosses = ⁴C₂ = 6

4

P(X=2) = 6 × 1/16 = 6/16 = 3/8

✔

P(exactly 2 heads) = 3/8. This is ⁴C₂ p²q² — the Binomial formula!

Module 5 of 6

10.6

Binomial Distribution

B(n,p) — the distribution of successes in n Bernoulli trials

The Binomial Distribution counts the number of successes X in n independent Bernoulli trials with success probability p. It is the most important discrete probability distribution in statistics.

✨ Binomial Probability Formula

If X ~ B(n, p) (X follows a Binomial distribution):
P(X = r) = ⁿCᵣ ⋅ pᵣ ⋅ qⁿ⁻ᵣ
where r = 0, 1, 2, …, n and q = 1 − p.
This is the (r+1)th term of the binomial expansion of (q+p)ⁿ.

Binomial distribution B(10, 0.5): symmetric bell shape centred at np = 5

Mean and Variance of Binomial Distribution

P(X=r)

ⁿCᵣ pᵣ qⁿ⁻ᵣ

Mean μ

np

Variance σ²

npq = np(1−p)

Std Dev σ

√(npq)

Mode

r = floor of (n+1)p

∑P(X=r)

= 1 (total probability)

✏️ Quick Example — Binomial DistributionExample 10.6

A biased coin has P(Head) = 2/3. It is tossed 6 times. Find P(exactly 4 heads) and the mean and variance of the distribution.

1

X ~ B(6, 2/3). p=2/3, q=1/3, n=6, r=4.

2

P(X=4) = ⁶C₄ ⋅ (2/3)⁴ ⋅ (1/3)² = 15 ⋅ 16/81 ⋅ 1/9 = 240/729 = 80/243

3

Mean = np = 6(2/3) = 4

4

Variance = npq = 6(2/3)(1/3) = 4/3 ≈ 1.33

✔

P(X=4) = 80/243 ≈ 0.329 | Mean = 4 | Variance = 4/3 | σ = 2/√3

✏️ Quick Example — Using ComplementExample 10.7

A multiple-choice test has 10 questions, each with 4 options (1 correct). A student guesses all. Find P(at least 1 correct).

1

X ~ B(10, 1/4). P(at least 1) = 1 − P(X=0).

2

P(X=0) = ¹⁰C₀ (1/4)⁰ (3/4)¹⁰ = (3/4)¹⁰ ≈ 0.0563

3

P(at least 1) = 1 − (3/4)¹⁰ ≈ 1 − 0.0563 = 0.9437

✔

P(at least 1 correct) ≈ 0.9437 (94.4%). Even guessing, very likely to get at least one right!

Module 6 of 6 🎉

Chapter 11 · The Language of Circles and Waves

Trigonometric Functions, Identities & Equations

Trigonometry is the mathematics of angles and cycles. From measuring angles in radians to graphing sine waves, proving identities and solving equations — this chapter builds a complete, visual, and deeply connected understanding of circular functions.

11 Modules ⌛ ~70 min read Class 11 & 12 Core

11.1

Measurement of Angles

Three systems — sexagesimal, centesimal, and circular — and how to convert between them

An angle is formed when a ray rotates about its endpoint. The amount of rotation determines the measure of the angle. There are three systems for measuring angles, each used in different contexts in mathematics and science.

📖 Three Systems of Angle Measurement

1. Sexagesimal (Degree) System: One complete rotation = 360°. Each degree = 60 minutes ('), each minute = 60 seconds (").
2. Centesimal (Grade) System: One complete rotation = 400 grades (g). 1 grade = 100 minutes, 1 minute = 100 seconds.
3. Circular (Radian) System: Angle = arc length / radius. One complete rotation = 2π radians. This is the mathematically natural system.

Sign Convention for Angles

↺

Positive Angle

Measured anti-clockwise from the initial ray. Standard convention.

30°, 90°, π/3

↻

Negative Angle

Measured clockwise from the initial ray.

−45°, −π/6

📏

Acute Angle

Between 0° and 90°

0 < θ < π/2

◺

Obtuse Angle

Between 90° and 180°

π/2 < θ < π

▷

Right Angle

Exactly 90° = π/2 radians

θ = 90° = π/2

⇄

Reflex Angle

Between 180° and 360°

π < θ < 2π

✏️ Quick Example — Angle in Different SystemsExample 11.1

Express 150° in (i) grades and (ii) radians.

1

To grades: 1° = 10/9 grades. So 150° = 150 × 10/9 = 166.̇ grades

2

To radians: 180° = π rad. So 150° = 150π/180 = 5π/6 radians

✔

150° = 500/3 grades = 5π/6 radians ≈ 2.618 rad

Module 1 of 11

11.2

Relation Between Degree and Radian

The fundamental conversion formula and the standard angle table every student must memorise

The radian is defined as the angle subtended at the centre of a circle by an arc whose length equals the radius. This makes radian measure dimensionless and the natural unit for calculus and higher mathematics.

✨ The Fundamental Relationship

π radians = 180°
Therefore: 1 radian = 180°/π ≈ 57.296° and 1° = π/180 radians ≈ 0.01745 rad
Arc length formula: l = rθ where θ is in radians and r is the radius.

Standard Angle Conversion Table

π rad = 180°

Fundamental relation

Arc length

l = rθ (θ in radians)

Area of sector

A = ½r²θ

D/90

= G/100 = 2R/π

Deg → Rad

Multiply by π/180

Rad → Deg

Multiply by 180/π

✏️ Quick Example — Arc LengthExample 11.2

A circle has radius 14 cm. Find the length of arc subtending 45° at the centre. Also find the area of the sector.

1

Convert: θ = 45° = 45 × π/180 = π/4 radians

2

Arc length: l = rθ = 14 × π/4 = 7π/2 ≈ 11 cm

3

Sector area: A = ½r²θ = ½ × 196 × π/4 = 49π/2 ≈ 77 cm²

✔

Arc length = 7π/2 cm, Sector area = 49π/2 cm²

Module 2 of 11

11.3

Trigonometric Ratios for Acute Angles

sin, cos, tan and their reciprocals — SOH-CAH-TOA and the standard value table

The six trigonometric ratios are defined using the sides of a right-angled triangle. For an acute angle θ, they express the relationship between the angle and the ratios of the sides.

📖 The Six Trigonometric Ratios

For a right triangle with angle θ, hypotenuse (H), opposite side (O), adjacent side (A):
sinθ = O/H cosθ = A/H tanθ = O/A
cosecθ = H/O secθ = H/A cotθ = A/O
Memory aid: SOH – CAH – TOA

Standard Values Table — Must Memorise!

💡 Memory trick for sin: 0, 1/2, 1/√2, √3/2, 1 — numerators are √0, √1, √2, √3, √4 all over 2. cos is sin in reverse!

✏️ Quick Example — Using Standard ValuesExample 11.3

Evaluate: sin60° cos30° + cos60° sin30° and tan45° + cot45°.

1

sin60°cos30° + cos60°sin30° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1

2

tan45° + cot45° = 1 + 1 = 2

✔

First expression = 1 (this is actually sin(60+30) = sin90° = 1). Second = 2.

Module 3 of 11

11.4

Trigonometric (Circular) Functions

Extending to all angles using the unit circle — ASTC rule and all four quadrants

The right-triangle definition only works for acute angles. The circular function approach extends trigonometry to any angle — positive, negative, or greater than 360° — using the unit circle.

📖 Unit Circle Definition

For any angle θ, place a point P on the unit circle (radius = 1) at angle θ from the positive x-axis. If P = (x, y):
cosθ = x sinθ = y tanθ = y/x (x ≠ 0)
This works for ALL angles — not just acute ones.

Unit circle: sinθ = y-coordinate, cosθ = x-coordinate of point P. ASTC shows which ratios are positive in each quadrant.

ASTC Rule — Signs in Each Quadrant

Remember: All Students Take Calculus (or: Add Sugar To Coffee)

Q1 (0 to 90°): All ratios positive
Q2 (90 to 180°): Only sin (and cosec) positive
Q3 (180 to 270°): Only tan (and cot) positive
Q4 (270 to 360°): Only cos (and sec) positive

Allied Angle Results

sin(180°−θ)

= sinθ

cos(180°−θ)

= −cosθ

sin(180°+θ)

= −sinθ

cos(180°+θ)

= −cosθ

sin(360°−θ)

= −sinθ

cos(360°−θ)

= cosθ

sin(−θ)

= −sinθ (odd)

cos(−θ)

= cosθ (even)

✏️ Quick Example — All-Angle ValuesExample 11.4

Find: sin150°, cos210°, tan315°, sin(−30°).

1

sin150° = sin(180°−30°) = sin30° = 1/2 (Q2, sin positive)

2

cos210° = cos(180°+30°) = −cos30° = −√3/2 (Q3, cos negative)

3

tan315° = tan(360°−45°) = −tan45° = −1 (Q4, tan negative)

4

sin(−30°) = −sin30° = −1/2 (sin is odd function)

✔

1/2 | −√3/2 | −1 | −1/2

Module 4 of 11

11.5

Graphs of Trigonometric Functions

Shape, period, amplitude, domain and range of all six trig functions

The graphs of trigonometric functions reveal their periodic, wave-like nature. Understanding the shape, period, and key features of each graph is essential for solving equations and applications.

Sine Function — y = sin x

Cosine Function — y = cos x

Tangent Function — y = tan x

Key Properties of All Six Functions

sin x range

[−1, 1]

cos x range

[−1, 1]

tan x range

(−∞, ∞)

sin, cos period

2π

tan, cot period

π

cosec x range

(−∞,−1]∪[1,∞)

Module 5 of 11

11.6

Fundamental Trigonometric Identities

Pythagorean, reciprocal, and quotient identities — the toolkit for all proofs

Trigonometric identities are equations that are true for all values of the variable for which both sides are defined. Mastering these identities is the key to simplifying expressions and solving equations.

✨ The Three Pythagorean Identities

sin²θ + cos²θ = 1 ← the most fundamental
1 + tan²θ = sec²θ (divide the first by cos²θ)
1 + cot²θ = cosec²θ (divide the first by sin²θ)

Reciprocal & Quotient Identities

sinθ cosecθ

= 1

cosθ secθ

= 1

tanθ cotθ

= 1

tanθ

= sinθ/cosθ

cotθ

= cosθ/sinθ

sin²θ

= 1 − cos²θ

sec²θ−tan²θ

= 1

cosec²θ−cot²θ

= 1

✏️ Quick Example — Proving an IdentityExample 11.5

Prove: (sinθ + cosθ)² + (sinθ − cosθ)² = 2.

1

LHS = sin²θ + 2sinθcosθ + cos²θ + sin²θ − 2sinθcosθ + cos²θ

2

= 2sin²θ + 2cos²θ = 2(sin²θ + cos²θ) = 2(1) = 2 = RHS ✔

✔

Identity proved. The cross terms cancel and Pythagorean identity completes it.

✏️ Quick Example — Finding Other RatiosExample 11.6

If sinθ = 3/5 and θ is in Q2, find cosθ, tanθ, and secθ.

1

cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25. cosθ = −4/5 (Q2, cos negative)

2

tanθ = sinθ/cosθ = (3/5)/(−4/5) = −3/4

3

secθ = 1/cosθ = −5/4

✔

cosθ = −4/5, tanθ = −3/4, secθ = −5/4

Module 6 of 11

11.7

Trigonometric Functions of Compound Angles

sin(A±B), cos(A±B), tan(A±B) — expanding sums and differences of angles

Compound angle formulas express trig functions of a sum or difference of two angles in terms of the trig functions of the individual angles. They are the source of almost all other trigonometric identities.

✨ The Six Compound Angle Formulas

sin(A+B) = sinA cosB + cosA sinB
sin(A−B) = sinA cosB − cosA sinB
cos(A+B) = cosA cosB − sinA sinB
cos(A−B) = cosA cosB + sinA sinB
tan(A+B) = (tanA + tanB) / (1 − tanA tanB)
tan(A−B) = (tanA − tanB) / (1 + tanA tanB)

Important Derived Results

sin(A+B)sin(A−B)

= sin²A − sin²B

cos(A+B)cos(A−B)

= cos²A − sin²B

cot(A+B)

(cotBcotA−1)/(cotB+cotA)

sin75°

(√6+√2)/4

cos75°

(√6−√2)/4

tan75°

2+√3

✏️ Quick Example — Compound AnglesExample 11.7

Evaluate sin75° using compound angle formula. Also prove sin(A+B)/sin(A−B) = (tanA+tanB)/(tanA−tanB).

1

sin75° = sin(45°+30°) = sin45cos30 + cos45sin30

2

= (1/√2)(√3/2) + (1/√2)(1/2) = √3/(2√2) + 1/(2√2) = (√6+√2)/4

3

For the proof: Divide num and denom by cosAcosB: sin(A+B)/sin(A−B) = (sinAcosB+cosAsinB)/(sinAcosB−cosAsinB) = (tanA+tanB)/(tanA−tanB) ✔

✔

sin75° = (√6+√2)/4 ≈ 0.966

Module 7 of 11

11.8

Transformation Formulae

Converting products to sums & sums to products — essential for integration and equation solving

Transformation formulas convert products of trig functions into sums/differences (and vice versa). They are derived directly from compound angle formulas and are heavily used in calculus and in solving equations.

Product to Sum (Derived from compound angle formulas)

Products → Sums

2sinAcosB = sin(A+B) + sin(A−B)
2cosAsinB = sin(A+B) − sin(A−B)
2cosAcosB = cos(A−B) + cos(A+B)
2sinAsinB = cos(A−B) − cos(A+B)

Sum to Product (let A+B=C, A−B=D)

Sums → Products

sinC + sinD = 2sin((C+D)/2)cos((C−D)/2)
sinC − sinD = 2cos((C+D)/2)sin((C−D)/2)
cosC + cosD = 2cos((C+D)/2)cos((C−D)/2)
cosC − cosD = −2sin((C+D)/2)sin((C−D)/2)

✏️ Quick Example — TransformationExample 11.8

Express 2sin3x cos x as a sum. Also convert sin5x + sin3x into a product.

1

2sin3x cosx = sin(3x+x) + sin(3x−x) = sin4x + sin2x

2

sin5x + sin3x = 2sin((5x+3x)/2)cos((5x−3x)/2) = 2sin4x cosx

✔

Both results are consistent — the transformations are inverses of each other.

Module 8 of 11

11.9

Trigonometric Functions of Multiple Angles

Double, triple and half-angle formulas — derived from compound angle results

Multiple angle formulas express sin(nθ), cos(nθ), tan(nθ) in terms of sinθ and cosθ. The double and triple angle formulas are the most commonly tested.

Double Angle Formulas (A = B in compound formulas)

✨ Double Angle (2θ)

sin2θ = 2sinθcosθ
cos2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
tan2θ = 2tanθ / (1 − tan²θ)

Half-Angle Formulas

sin²θ

(1−cos2θ)/2

cos²θ

(1+cos2θ)/2

tan²θ

(1−cos2θ)/(1+cos2θ)

sinθ (t-formula)

2t/(1+t²) where t=tan(θ/2)

cosθ (t-formula)

(1−t²)/(1+t²)

Triple Angle Formulas

Triple Angle (3θ)

sin3θ = 3sinθ − 4sin³θ
cos3θ = 4cos³θ − 3cosθ
tan3θ = (3tanθ − tan³θ) / (1 − 3tan²θ)

✏️ Quick Example — Double Angle ApplicationsExample 11.9

If sinθ = 3/5, find sin2θ, cos2θ, tan2θ (θ in Q1).

1

cosθ = 4/5 (Q1, positive). tanθ = 3/4.

2

sin2θ = 2(3/5)(4/5) = 24/25

3

cos2θ = 1−2sin²θ = 1−2(9/25) = 1−18/25 = 7/25

4

tan2θ = 2(3/4)/(1−9/16) = (3/2)/(7/16) = 24/7

✔

sin2θ = 24/25, cos2θ = 7/25, tan2θ = 24/7. Verify: (24/25)/(7/25) = 24/7 ✔

Module 9 of 11

11.10

Trigonometric Periodic Functions

Period, amplitude, phase shift — understanding the repeating wave nature

A function f(x) is periodic if f(x + T) = f(x) for all x, for some fixed positive T called the period. All six trig functions are periodic, which makes them ideal for modelling waves, oscillations, and cycles.

📖 Period of Common Functions

sin(nx) and cos(nx) have period 2π/n
tan(nx) and cot(nx) have period π/n
|sin(nx)| and |cos(nx)| have period π/n
sin²(nx) and cos²(nx) have period π/n

General Form: y = A sin(Bx + C) + D

📏

Amplitude |A|

Maximum displacement from midline. |A| is the height of the wave.

y=3sinx has amplitude 3

🔁

Period 2π/|B|

Length of one complete cycle. B compresses or stretches.

y=sin2x: period=π

➡

Phase Shift −C/B

Horizontal shift. Positive C/B shifts graph left.

sin(x+π/4): shift −π/4

↕

Vertical Shift D

Moves the entire graph up (D>0) or down (D<0).

sinx+2: shifted up 2

Period of sinx

2π

Period of sin(nx)

2π/n

Period of |sinx|

π

Period of sin²x

π

Range of Asinx+D

[D−|A|, D+|A|]

✏️ Quick Example — Period and AmplitudeExample 11.10

For y = 3sin(2x + π/4) − 1, find the amplitude, period, phase shift, vertical shift, and range.

1

Amplitude = |A| = |3| = 3

2

Period = 2π/|B| = 2π/2 = π

3

Phase shift = −C/B = −(π/4)/2 = −π/8 (shifts left by π/8)

4

Vertical shift = D = −1. Range = [−1−3, −1+3] = [−4, 2]

✔

Amplitude=3, Period=π, Phase shift=−π/8, Vertical shift=−1, Range=[−4,2]

Module 10 of 11

11.11

Trigonometric Equations

Finding the general solution — all values of θ satisfying a trig equation

A trigonometric equation is an equation involving trig functions. Unlike algebraic equations, trig equations typically have infinitely many solutions (because of periodicity). We express them as a general solution.

✨ General Solution Formulas (n ∈ ℤ)

sinθ = 0 ⟹ θ = nπ
cosθ = 0 ⟹ θ = (2n+1)π/2
tanθ = 0 ⟹ θ = nπ
sinθ = sinα ⟹ θ = nπ + (−1)ⁿα
cosθ = cosα ⟹ θ = 2nπ ± α
tanθ = tanα ⟹ θ = nπ + α

Strategy for Solving Trig Equations

Step 1: Simplify using identities to get a single trig function on one side
Step 2: Identify the principal value α (the value in the principal range)
Step 3: Apply the appropriate general solution formula
Step 4: Check for extraneous solutions if the domain is restricted

✏️ Quick Example — General SolutionsExample 11.11

Find the general solution of: (i) 2sin²θ − 1 = 0 (ii) cos2θ = cosθ (iii) tanθ = −√3

1

(i) sin²θ = 1/2 ⟹ sinθ = ±1/√2 = ±sin45°. General solution: θ = nπ ± π/4

2

(ii) cos2θ = cosθ ⟹ 2θ = 2nπ ± θ.
Case +: 2θ−θ = 2nπ ⟹ θ = 2nπ
Case −: 2θ+θ = 2nπ ⟹ θ = 2nπ/3

3

(iii) tanθ = −√3 = tan(−π/3). General: θ = nπ − π/3

✔

(i) nπ±π/4 | (ii) 2nπ or 2nπ/3 | (iii) nπ−π/3

✏️ Quick Example — Restricted DomainExample 11.12

Solve 2cos²x − 3cosx + 1 = 0 for x ∈ [0, 2π].

1

Let c = cosx. 2c²−3c+1 = 0 ⟹ (2c−1)(c−1) = 0 ⟹ c = 1/2 or c = 1

2

cosx = 1/2: x = π/3 or x = 5π/3 (in [0,2π])

3

cosx = 1: x = 0 or x = 2π

✔

x ∈ {0, π/3, 5π/3, 2π}. Four solutions in [0, 2π].

sinθ=sinα

θ = nπ+(−1)ⁿα

cosθ=cosα

θ = 2nπ±α

tanθ=tanα

θ = nπ+α

sinθ=1

θ = (4n+1)π/2

cosθ=1

θ = 2nπ

cosθ=−1

θ = (2n+1)π

Module 11 of 11 🎉

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Chapter 12 · Trigonometry Applications

Solution of Triangles

Can you find a missing side or angle of a triangle when only partial information is given? Yes — using three powerful rules. Every triangle problem in mathematics reduces to one of these: the Sine Rule, the Cosine Rule, or the Area Formula. Master these three and no triangle stays unsolved.

3 Modules ⌛ ~35 min Class 11 Core

12.1

Basic Rules of Triangle

Sine Rule, Cosine Rule, Projection Rule & Napier's Analogy — when and how to use each

In any triangle ABC the side opposite angle A is labelled a, opposite B is b, opposite C is c. The semi-perimeter is s = (a+b+c)/2. These four rules let you find any unknown from what you are given.

🔑 The Golden Labelling Rule

Side a is always opposite angle A. Side b opposite B. Side c opposite C. Every substitution error in triangle problems comes from violating this. Label your diagram first, always.

① Sine Rule — use when you have angle-side pairs

When to use: (i) Two angles + one side known, OR (ii) Two sides + an angle opposite one of them known.

Sine Rule

a / sin A = b / sin B = c / sin C = 2R

2R = diameter of the circumscribed circle (circumcircle)
To find an angle: flip it → sin A/a = sin B/b = sin C/c
Always use only two ratios at a time — pick the pair with 3 knowns & 1 unknown
Ambiguous case: sinθ = sin(180°−θ) so there may be two possible triangles when given two sides and a non-included angle

✏️ Example 12.1 — Finding a side (Sine Rule)Solved

In ▵ABC: a = 8, A = 45°, B = 60°. Find side b.

1

Choose pair: a/sin A = b/sin B → 8/sin45° = b/sin60°

2

Cross-multiply: b = 8 × sin60° / sin45° = 8 × (√3/2) / (1/√2) = 8 × √6/2 = 4√6

✔

b = 4√6 ≈ 9.80

② Cosine Rule — use when Sine Rule can't start

When to use: (i) All three sides known (find any angle), OR (ii) Two sides + included angle known (find third side). Think of it as Pythagoras with a correction term.

Side form

a² = b² + c² − 2bc cos A

Angle form

cos A = (b²+c²−a²) / (2bc)

cos B

= (a²+c²−b²) / (2ac)

cos C

= (a²+b²−c²) / (2ab)

💡 Memory Trick

When A = 90°, cos A = 0, so the correction vanishes → a² = b² + c². That is Pythagoras! The Cosine Rule is Pythagoras generalised to any angle. This is the best way to remember the formula under exam pressure.

✏️ Example 12.2 — Finding an angle (Cosine Rule)Solved

a = 5, b = 7, c = 8. Find angle A.

1

cos A = (b²+c²−a²)/(2bc) = (49+64−25)/(2×7×8) = 88/112 = 11/14

2

A = cos¹(11/14) ≈ 38.2°

✔

A ≈ 38.2°

③ Projection Rule

a

= b cosC + c cosB

b

= c cosA + a cosC

c

= a cosB + b cosA

④ Napier's Analogy (Tangent Rule)

Useful when two sides and their included angle are given and you want the other two angles efficiently.

tan((B−C)/2)

= [(b−c)/(b+c)] · cot(A/2)

tan((C−A)/2)

= [(c−a)/(c+a)] · cot(B/2)

tan((A−B)/2)

= [(a−b)/(a+b)] · cot(C/2)

⚠️ Most Common Mistakes

1. Wrong rule: For SSS (all sides known), always use Cosine Rule — never Sine Rule to start.
2. Label mix-up: Double-check that side a is opposite angle A before substituting.
3. Ambiguous case: When finding angle from Sine Rule, check if 180°−θ also forms a valid triangle.

Module 1 of 3

12.2

Half-Angle Formulae

Expressing sin, cos, tan of A/2 purely in terms of sides using semi-perimeter s

Half-angle formulae let us compute trigonometric ratios of half-angles entirely from the side lengths. This is extremely useful for finding area, inradius, and circumradius without needing to know angles directly.

📖 Always Start Here: Compute s

s = (a+b+c)/2 (semi-perimeter). Then find s−a, s−b, s−c. These are always positive for a valid triangle. Every formula below uses these four quantities.

sin(A/2)

= √[(s−b)(s−c) / bc]

cos(A/2)

= √[s(s−a) / bc]

tan(A/2)

= √[(s−b)(s−c) / s(s−a)]

sin(B/2)

= √[(s−a)(s−c) / ac]

cos(B/2)

= √[s(s−b) / ac]

sin(C/2)

= √[(s−a)(s−b) / ab]

cos(C/2)

= √[s(s−c) / ab]

💡 How to Remember the Pattern for Angle A

sin(A/2): denominator = bc (the two sides NOT called a). Numerator = (s−b)(s−c) (subtract the sides NOT called a).
cos(A/2): denominator still bc. Numerator = s(s−a) (s times “s minus a itself”).
tan = sin/cos so numerator becomes (s−b)(s−c), denominator becomes s(s−a).
The pattern repeats cyclically for B and C.

✏️ Example 12.3 — Computing all half-anglesStep-by-step

Triangle with a=6, b=8, c=10. Find sin(A/2), cos(A/2), tan(A/2). Verify A.

1

s = (6+8+10)/2 = 12. s−a=6, s−b=4, s−c=2

2

sin(A/2) = √[(4×2)/(8×10)] = √[8/80] = √(1/10) = 1/√10

3

cos(A/2) = √[(12×6)/(8×10)] = √[72/80] = √(9/10) = 3/√10

4

tan(A/2) = (1/√10)/(3/√10) = 1/3 → A/2 = 18.43° → A ≈ 36.87°

5

Verify: 6–8–10 is 3:4:5 scaled by 2, a right triangle with C=90°. A = arctan(3/4) ≈ 36.87° ✔

✔

sin(A/2) = 1/√10, cos(A/2) = 3/√10, tan(A/2) = 1/3, A ≈ 36.87°

Module 2 of 3

12.3

Area of Triangle

Six formulae for area — Heron's, trig, circumradius, inradius forms

The area Δ of a triangle can be computed in several ways. Choose the formula based on what data you have.

Two sides + included angle

Δ = ½bc sinA = ½ca sinB = ½ab sinC

Heron's (all 3 sides)

Δ = √[s(s−a)(s−b)(s−c)]

Circumradius R

R = abc / (4Δ)

Inradius r

r = Δ / s

Exradii

r&sub1; = Δ/(s−a), r&sub2; = Δ/(s−b), r&sub3; = Δ/(s−c)

📖 Understanding the Special Circles

Circumcircle (R): passes through all 3 vertices. Incircle (r): touches all 3 sides from inside. Excircle opposite A (r&sub1;): touches side a and extensions of b and c. Memorise r = Δ/s and R = abc/(4Δ) — these appear in most exam questions.

✏️ Example 12.4 — Heron's Formula + special radiiStep-by-step

Find area, inradius r, and circumradius R for a=13, b=14, c=15.

1

s = (13+14+15)/2 = 21; s−a=8, s−b=7, s−c=6

2

Δ = √(21×8×7×6) = √7056 = 84

3

r = Δ/s = 84/21 = 4

4

R = abc/(4Δ) = (13×14×15)/(4×84) = 2730/336 = 65/8 = 8.125

✔

Δ = 84, r = 4, R = 65/8

✏️ Example 12.5 — Area from two sides & angleStep-by-step

PQ = 6 cm, PR = 8 cm, angle P = 30°. Find area.

1

Δ = ½ × PQ × PR × sinP = ½ × 6 × 8 × sin30° = ½ × 48 × 0.5 = 12

✔

Area = 12 cm²

Module 3 of 3 🎉

Chapter 13 · Real-World Trigonometry

Heights and Distances

How tall is that tower? How far is that ship? You do not need to climb or swim — just measure an angle. This chapter shows you how angle of elevation and angle of depression unlock every height-and-distance problem, step by step.

2 Modules ⌛ ~25 min Class 11 Applied

13.1

Angle of Elevation

Looking upward — angle between the horizontal and line of sight to an object above

📖 Definition

When you look upward at an object (like a tower top), the angle formed between the horizontal line from your eye and your line of sight to the object is the angle of elevation. It is always measured upward from horizontal, ranging from 0° to 90°.

Right triangle: observer at ground, object above. tan θ = h/d

Height

h = d · tanθ

Distance

d = h · cotθ

Slant

l = h / sinθ

📋 4-Step Method (always works)

Step 1 — Draw: Sketch observer, object, horizontal line, angle θ, height h, distance d.
Step 2 — Identify: The right angle is always where the vertical (object) meets the horizontal (ground).
Step 3 — Write: tanθ = opposite/adjacent = h/d.
Step 4 — Solve: substitute known values and find the unknown.

✏️ Example 13.1 — Height of a towerStep-by-step

From a point 50 m from the base of a tower, the angle of elevation of the top is 60°. Find the height.

1

Draw: Observer 50 m from base, angle of elevation = 60°, height = h.

2

Write: tan60° = h/50 → √3 = h/50

3

Solve: h = 50√3 ≈ 86.6 m

✔

Height = 50√3 ≈ 86.6 m

✏️ Example 13.2 — Two observers at different distancesStep-by-step

From two points A and B on same side, AB = 100 m. Angles of elevation from A and B are 30° and 60°. Find height h.

1

Let d = distance from B to tower base. Distance from A = d + 100.

2

From B: tan60° = h/d → h = d√3 ...(i)

3

From A: tan30° = h/(d+100) → h = (d+100)/√3 ...(ii)

4

Equate: d√3 = (d+100)/√3 → 3d = d+100 → d = 50

5

h = 50√3 ≈ 86.6 m

✔

Height = 50√3 ≈ 86.6 m

Module 1 of 2

13.2

Angle of Depression

Looking downward — angle between horizontal and line of sight to object below

📖 Definition

When you look downward at an object below (like a ship from a lighthouse), the angle between the horizontal and your line of sight is the angle of depression. Always measured downward from the horizontal.

⭐ The Most Important Property — Alternate Angles

Angle of depression from A to B = Angle of elevation from B to A.
This is because the horizontal at A is parallel to the ground at B. The line AB is a transversal → alternate interior angles are equal. So you can always flip a depression problem into an elevation one just by switching perspective!

Key fact

Depression angle (top) = Elevation angle (bottom)

Horizontal dist

d = h · cotφ = h / tanφ

✏️ Example 13.3 — Cliff and boatStep-by-step

From a 100 m cliff top, angle of depression of a boat is 30°. Find horizontal distance.

1

Angle of depression = 30° = angle of elevation from boat to cliff top (alternate angles).

2

tan30° = 100/d → 1/√3 = 100/d → d = 100√3

✔

Distance = 100√3 ≈ 173.2 m

✏️ Example 13.4 — Both elevation and depression (combined)Step-by-step

A person on top of a 60 m building sees the top of another building at elevation 45° and its foot at depression 60°. Find height H of second building.

1

Let d = horizontal distance between buildings.

2

Depression 60° to foot: tan60° = 60/d → d = 60/√3 = 20√3

3

Elevation 45° to top: tan45° = (H−60)/d → 1 = (H−60)/20√3

4

H = 60 + 20√3 ≈ 60 + 34.6 = 94.6 m

✔

H = (60 + 20√3) ≈ 94.6 m

⚠️ Common Mistakes

1. Forgetting alternate angles: Always convert depression angle to elevation angle before applying tan.
2. Mixing up which building is taller: Draw the diagram clearly showing which object is higher.
3. Wrong triangle: The right angle is always at the base of the vertical height, not at the observer.

Module 2 of 2 🎉

Chapter 14 · Inverse Functions

Inverse Trigonometric Functions

If sinx = 1/2, what is x? To answer this cleanly, we need inverse trig functions. But sin is periodic — it repeats every 2π — so we must restrict the domain to make it one-to-one first. This chapter covers those restrictions, the resulting graphs, key properties, addition formulae, and the elegant inverse hyperbolic functions expressed as logarithms.

4 Modules ⌛ ~45 min Class 12 Core

14.1

Domain and Range of Inverse Trig Functions

Principal value branches — restricting domains to make inverses single-valued

🔑 Why do we need restrictions?

sin(30°) = sin(150°) = 0.5. So if we ask “what angle has sine = 0.5?” there are infinitely many answers. To get a unique answer, we restrict sin to the interval [−π/2, π/2] where it is one-to-one. This chosen interval is called the principal value branch.

sin¹ x

Domain: [−1, 1] | Range: [−π/2, π/2]

cos¹ x

Domain: [−1, 1] | Range: [0, π]

tan¹ x

Domain: ℝ | Range: (−π/2, π/2)

cot¹ x

Domain: ℝ | Range: (0, π)

sec¹ x

Domain: |x|≥1 | Range: [0,π] \ {π/2}

cosec¹ x

Domain: |x|≥1 | Range: [−π/2,π/2] \ {0}

💡 Memory Aid for Ranges

sin¹, tan¹, cosec¹: Range is around 0 → [−π/2, π/2] (negative to positive).
cos¹, cot¹, sec¹: Range is from 0 to π (never negative).
The key difference: cos¹(x) is never negative, but sin¹(x) can be.

✏️ Example 14.1 — Finding principal valuesStep-by-step

Find: (a) sin¹(1/2) (b) cos¹(−1) (c) tan¹(√3) (d) sin¹(−1/2)

a

sinθ = 1/2 in [−π/2, π/2] → θ = π/6 (30°)

b

cosθ = −1 in [0, π] → θ = π (180°)

c

tanθ = √3 in (−π/2, π/2) → θ = π/3 (60°)

d

sinθ = −1/2 in [−π/2, π/2] → θ = −π/6 (−30°) [negative because x is negative]

✔

π/6 | π | π/3 | −π/6

Module 1 of 4

14.2

Graphs of Inverse Trig Functions

Shape, key points, asymptotes — each graph is a reflection of the restricted trig function

📖 The Reflection Principle

Every inverse function graph is a reflection of the original function across the line y = x. So y = sin¹x is the reflection of y = sinx (on [−π/2, π/2]) across y = x. If a point (a, b) is on y = sinx, then (b, a) is on y = sin¹x.

sin¹x (purple): increasing, S-shaped | cos¹x (coral): decreasing | tan¹x (teal): increasing with horizontal asymptotes at ±π/2

y = sin¹x

Increasing; passes (0,0), (1, π/2), (−1, −π/2)

y = cos¹x

Decreasing; passes (0, π/2), (1, 0), (−1, π)

y = tan¹x

Increasing; asymptotes y = ±π/2; passes (0,0)

sin¹x and cos¹x are defined only for x ∈ [−1, 1] — they have no asymptotes
tan¹x is defined for all real x and approaches ±π/2 but never reaches them
All three graphs pass through the origin (0, 0) for sin¹ and tan¹; cos¹ passes (0, π/2)

Module 2 of 4

14.3

Elementary Properties

Complementary pairs, negative arguments, cancellation laws & addition formulae

Group 1 — Complementary Pairs (add up to π/2)

sin¹x + cos¹x

= π/2 for x ∈ [−1, 1]

tan¹x + cot¹x

= π/2 for x ∈ ℝ

sec¹x + cosec¹x

= π/2 for |x| ≥ 1

💡 Why do these pairs sum to π/2?

Because sin and cos are complementary: sin(π/2 − θ) = cosθ. So if sin¹x = α, then x = sinα = cos(π/2 − α), which means cos¹x = π/2 − α. Therefore sin¹x + cos¹x = α + (π/2 − α) = π/2.

Group 2 — Negative Arguments

sin¹(−x)

= −sin¹x (odd function)

cos¹(−x)

= π − cos¹x

tan¹(−x)

= −tan¹x (odd function)

Group 3 — Cancellation Laws

sin(sin¹x)

= x for x ∈ [−1,1]

sin¹(sinx)

= x only if x ∈ [−π/2, π/2]

tan(tan¹x)

= x for all x ∈ ℝ

Group 4 — Addition Formulae

tan¹x + tan¹y

= tan¹[(x+y)/(1−xy)], if xy < 1

tan¹x + tan¹y

= π + tan¹[(x+y)/(1−xy)], if xy > 1, x>0

tan¹x − tan¹y

= tan¹[(x−y)/(1+xy)]

✏️ Example 14.2 — Prove tan¹(1/2)+tan¹(1/3)=π/4Step-by-step

Prove: tan¹(1/2) + tan¹(1/3) = π/4

1

Check xy: (1/2)(1/3) = 1/6 < 1, so use the first formula.

2

= tan¹[(1/2 + 1/3)/(1 − 1/6)] = tan¹[(5/6)/(5/6)] = tan¹(1)

3

= π/4 ✔

✔

Proved: tan¹(1/2) + tan¹(1/3) = π/4

✏️ Example 14.3 — Simplify sin¹(sinx)Concept clarity

Find sin¹(sin(2π/3)) and sin¹(sin(π/4)).

1

sin(2π/3) = sin(π − π/3) = sin(π/3) = √3/2. sin¹(√3/2) = π/3 (since π/3 ∈ [−π/2, π/2])

2

sin(π/4) = 1/√2. π/4 ∈ [−π/2, π/2], so sin¹(sin(π/4)) = π/4 directly.

✔

sin¹(sin(2π/3)) = π/3 | sin¹(sin(π/4)) = π/4

Module 3 of 4

14.4

Inverse Hyperbolic Functions

sinh¹, cosh¹, tanh¹ expressed as natural logarithms — derivation and formulae

📖 What are Hyperbolic Functions?

Hyperbolic functions are defined using exponentials:
sinh x = (e¹ − e¹)/2 cosh x = (e¹ + e¹)/2 tanh x = sinh x / cosh x
They look like trig functions but come from the hyperbola x² − y² = 1 (instead of the circle x² + y² = 1). Their inverses can be written as natural logarithms, which is very useful for integration.

sinh¹ x

= ln(x + √(x²+1)), x ∈ ℝ

cosh¹ x

= ln(x + √(x²−1)), x ≥ 1

tanh¹ x

= (1/2)ln[(1+x)/(1−x)], |x| < 1

coth¹ x

= (1/2)ln[(x+1)/(x−1)], |x| > 1

sech¹ x

= ln[(1+√(1−x²))/x], 0 < x ≤ 1

cosech¹ x

= ln(1/x + √(1/x²+1)), x ≠ 0

💡 How to Derive sinh¹x as a Log

Let y = sinh¹x, so x = sinh y = (e² − e²)/2. Multiply both sides by 2e²: 2xe² = e²² − 1. This is quadratic in e²: e²² − 2xe² − 1 = 0. Solving: e² = x + √(x²+1) (taking positive root). So y = ln(x + √(x²+1)). This derivation method works for all inverse hyperbolics!

✏️ Example 14.4 — Computing inverse hyperbolic valuesStep-by-step

Find (a) sinh¹(2) (b) tanh¹(1/2) in exact logarithmic form.

a

sinh¹(2) = ln(2 + √(4+1)) = ln(2 + √5) ≈ ln(4.236) ≈ 1.4436

b

tanh¹(1/2) = (1/2)ln[(1+1/2)/(1−1/2)] = (1/2)ln[(3/2)/(1/2)] = (1/2)ln(3) ≈ 0.5493

✔

sinh¹(2) = ln(2+√5); tanh¹(1/2) = (1/2)ln3

Module 4 of 4 🎉

Chapter 15 · Coordinate Geometry Foundations

Rectangular Axis

The Cartesian plane is where algebra meets geometry. A simple pair of perpendicular axes lets us describe every geometric shape as an equation. This chapter builds your coordinate geometry toolkit from scratch: distance, section, area, shifting of origin, rotation of axes, and the concept of a locus — one of the most powerful ideas in mathematics.

8 Modules ⌛ ~55 min Class 11 Core

15.1

Rectangular Axes & Quadrants

The Cartesian coordinate system — axes, quadrants, sign conventions, plotting

René Descartes had the idea of describing a point's position using two numbers. This gave us the Cartesian coordinate system: two perpendicular number lines (x-axis horizontal, y-axis vertical) crossing at the origin O(0,0).

Quadrant I (Q1)

x > 0, y > 0 — all trig ratios positive

Quadrant II (Q2)

x < 0, y > 0 — sin positive

Quadrant III (Q3)

x < 0, y < 0 — tan positive

Quadrant IV (Q4)

x > 0, y < 0 — cos positive

A point P(x, y): x is the abscissa (horizontal distance from y-axis), y is the ordinate (vertical distance from x-axis)
Points on the x-axis have y = 0; points on y-axis have x = 0
The origin O(0, 0) belongs to no quadrant
ASTC Rule: All → Sin → Tan → Cos (which is positive in Q1→Q2→Q3→Q4)

Module 1 of 8

15.2

Distance Formula

Distance between two points, from origin, and condition for collinearity

📖 Derivation in 30 seconds

Given P(x&sub1;, y&sub1;) and Q(x&sub2;, y&sub2;), draw a right triangle with horizontal leg = |x&sub2;−x&sub1;| and vertical leg = |y&sub2;−y&sub1;|. By Pythagoras: PQ = √[(x&sub2;−x&sub1;)² + (y&sub2;−y&sub1;)²]. Squaring removes the absolute value signs.

Distance PQ

= √[(x&sub2;−x&sub1;)² + (y&sub2;−y&sub1;)²]

From origin O

= √(x² + y²)

Collinearity

AB + BC = AC (if B is between A and C)

✏️ Example 15.1 — Distance and collinearityStep-by-step

A(3,4), B(-1,1), C(7,7). Find AB. Are A, B, C collinear?

1

AB = √[(-1-3)² + (1-4)²] = √[16+9] = √25 = 5

2

AC = √[(7-3)² + (7-4)²] = √[16+9] = 5. BC = √[(7+1)² + (7-1)²] = √[64+36] = 10

3

AB + AC = 5+5 = 10 = BC → A is between B and C → collinear!

✔

AB = 5. Yes, A, B, C are collinear.

Module 2 of 8

15.3

Section Formula

Internal & external division, midpoint, centroid

If a point P divides segment AB in ratio m:n, its coordinates depend on whether the division is internal (P is between A and B) or external (P is outside AB).

Internal (m:n)

P = [(mx&sub2;+nx&sub1;)/(m+n), (my&sub2;+ny&sub1;)/(m+n)]

External (m:n)

P = [(mx&sub2;−nx&sub1;)/(m−n), (my&sub2;−ny&sub1;)/(m−n)]

Midpoint (1:1)

M = [(x&sub1;+x&sub2;)/2, (y&sub1;+y&sub2;)/2]

Centroid

G = [(x&sub1;+x&sub2;+x&sub3;)/3, (y&sub1;+y&sub2;+y&sub3;)/3]

💡 Memory Trick for Section Formula

Think of it as a weighted average: the closer B is (ratio m is larger), the more its coordinates dominate. Internal: both weights positive, sum = m+n. External: one weight negative, difference = m−n. For midpoint, m=n=1: just average the coordinates.

✏️ Example 15.2 — Internal and external divisionStep-by-step

A(1,2), B(7,8). Find: (a) point dividing AB in 2:1 internally, (b) externally.

a

Internal: x = (2×7+1×1)/(2+1) = 15/3 = 5; y = (2×8+1×2)/3 = 18/3 = 6 → (5, 6)

b

External: x = (2×7−1×1)/(2−1) = 13/1 = 13; y = (2×8−1×2)/1 = 14 → (13, 14)

✔

Internal: (5, 6); External: (13, 14)

Module 3 of 8

15.4

Area of Triangle

Coordinate formula (shoelace method) and collinearity condition

📖 The Shoelace Formula

Area of ▵ with vertices (x&sub1;,y&sub1;), (x&sub2;,y&sub2;), (x&sub3;,y&sub3;):
Δ = (1/2)|x&sub1;(y&sub2;−y&sub3;) + x&sub2;(y&sub3;−y&sub1;) + x&sub3;(y&sub1;−y&sub2;)|
The absolute value ensures area is positive. If Δ = 0, the three points are collinear (lie on one line).

💡 How to Remember the Shoelace

Write vertices in a column and repeat the first at the bottom. Multiply diagonally right → add. Multiply diagonally left → subtract. Divide by 2.
↓ Right products: x&sub1;y&sub2; + x&sub2;y&sub3; + x&sub3;y&sub1;
↓ Left products: y&sub1;x&sub2; + y&sub2;x&sub3; + y&sub3;x&sub1;
Δ = (1/2)|Right − Left|

✏️ Example 15.3 — Area of triangleStep-by-step

Find area of ▵ with A(1,1), B(4,2), C(3,5).

1

Right: 1×2 + 4×5 + 3×1 = 2+20+3 = 25

2

Left: 1×4 + 2×3 + 5×1 = 4+6+5 = 15

3

Δ = (1/2)|25−15| = (1/2)(10) = 5

✔

Area = 5 sq units

Module 4 of 8

15.5

Area of Quadrilateral

Extended shoelace formula for four vertices taken in order

📖 Shoelace for Quadrilateral

For vertices A(x&sub1;,y&sub1;), B(x&sub2;,y&sub2;), C(x&sub3;,y&sub3;), D(x&sub4;,y&sub4;) taken in order (either all clockwise or all anticlockwise):
Area = (1/2)|(x&sub1;y&sub2;−x&sub2;y&sub1;)+(x&sub2;y&sub3;−x&sub3;y&sub2;)+(x&sub3;y&sub4;−x&sub4;y&sub3;)+(x&sub4;y&sub1;−x&sub1;y&sub4;)|
Alternative: split into two triangles along a diagonal and add their areas.

✏️ Example 15.4 — Area of quadrilateralStep-by-step

Vertices: A(1,1), B(5,1), C(5,4), D(1,4). Find area. (Verify: it is a rectangle 4×3.)

1

Products right: 1×1+5×4+5×4+1×1 = 1+20+20+1 = 42

2

Products left: 1×5+1×5+4×1+4×1 = 5+5+4+4 = 18

3

Area = (1/2)|42−18| = (1/2)(24) = 12 ✔ (= 4×3)

✔

Area = 12 sq units

Module 5 of 8

15.6

Shifting of Origin

Translating axes to a new origin to simplify equations — eliminates first-degree terms

📖 What is Shifting of Origin?

We move the origin from O(0,0) to a new point O'(h,k). Every point that was (x,y) now has new coordinates (X,Y) in the new system. The relationship is simple: X = x−h, Y = y−k (or equivalently x = X+h, y = Y+k).

New coords

X = x−h, Y = y−k

Old coords

x = X+h, y = Y+k

💡 Why shift origin? Simplify conics!

The equation (x−2)² + (y+3)² = 25 is a circle. Shift to O'(2,−3): set X=x−2, Y=y+3. Equation becomes X²+Y² = 25 — a standard circle with centre at new origin. Shifting eliminates those awkward constant terms and reveals the true shape immediately.

✏️ Example 15.5 — Simplify equation by shiftingStep-by-step

Shift origin to (3, −2). Find new coordinates of (5, 1) and simplify x²+y²−6x+4y−12=0.

1

New coords of (5,1): X = 5−3 = 2, Y = 1−(−2) = 3. New point: (2, 3)

2

Substitute x=X+3, y=Y−2 into x²+y²−6x+4y−12=0:

3

(X+3)²+(Y−2)²−6(X+3)+4(Y−2)−12 = X²+6X+9+Y²−4Y+4−6X−18+4Y−8−12 = X²+Y²−25=0

✔

X² + Y² = 25 — a clean circle centred at new origin, radius 5!

Module 6 of 8

15.7

Rotation of Axes

Rotating coordinate axes anticlockwise by angle θ to eliminate the cross-term xy

📖 Why Rotate?

The equation xy = 1 is a hyperbola, but it is not obvious from the form. If we rotate by 45°, it becomes X²/2 − Y²/2 = 1 — instantly recognisable as a standard hyperbola. Rotation eliminates the xy cross-term, converting any second-degree curve into its standard form.

Old → New

X = x cosθ + y sinθ | Y = −x sinθ + y cosθ

New → Old

x = X cosθ − Y sinθ | y = X sinθ + Y cosθ

Angle to eliminate xy in ax²+2hxy+by²

cot 2θ = (a−b)/(2h)

✏️ Example 15.6 — Rotation by 45°Step-by-step

Axes rotated by 45°. Find new coordinates of P(√2, 0).

1

X = √2 cos45° + 0×sin45° = √2 × 1/√2 = 1

2

Y = −√2 sin45° + 0×cos45° = −√2 × 1/√2 = −1

✔

New coordinates: (1, −1)

Module 7 of 8

15.8

Equation of Locus

Finding the algebraic equation that describes all points satisfying a geometric condition

📖 What is a Locus?

A locus is the set of ALL points that satisfy some geometric condition. For example, all points equidistant from a fixed point → a circle. All points equidistant from two fixed points → perpendicular bisector. The equation of the locus is the algebraic relation x and y must satisfy.

📋 4-Step Method to Find Locus Equation

Step 1: Let P(x, y) be any point on the locus.
Step 2: Express the given geometric condition in terms of x, y, and the given fixed points/lines.
Step 3: Simplify algebraically (expand, square both sides if needed, collect terms).
Step 4: The resulting equation in x and y IS the locus. State what curve it represents.

✏️ Example 15.7 — Equidistant from two pointsStep-by-step

Find locus of P equidistant from A(2, 0) and B(−2, 0).

1

Let P = (x, y). Condition: PA = PB.

2

√[(x−2)²+y²] = √[(x+2)²+y²]

3

Square: (x−2)² = (x+2)² → x²−4x+4 = x²+4x+4 → −8x = 0 → x = 0

4

x = 0 is the y-axis — the perpendicular bisector of AB ✔

✔

Locus: x = 0 (the y-axis)

✏️ Example 15.8 — Fixed distance from a pointStep-by-step

Find locus of P at constant distance 5 from origin O(0,0).

1

OP = 5 → √(x²+y²) = 5 → x² + y² = 25

✔

Locus: x² + y² = 25 (a circle centred at origin, radius 5)

Module 8 of 8 🎉

Chapter 16 · Coordinate Geometry

Straight Line

A straight line is the simplest curve — yet it hides remarkable depth. What is its slope? How do we write its equation in six different forms? How far is a point from it? What is the angle between two lines? And what does it mean for a single equation to represent two lines? This chapter answers all of these, step by step.

6 Modules ⌛ ~50 min Class 11 Core

16.1

Slope (Gradient) of a Line

What slope means, how to compute it, and parallel/perpendicular conditions

📖 What is Slope?

The slope m of a line measures its steepness and direction. Geometrically, m = tanα where α is the angle the line makes with the positive x-axis (called the inclination, 0° ≤ α < 180°). A steep upward line has large positive m; a steep downward line has large negative m.

Two-point slope

m = (y&sub2;−y&sub1;) / (x&sub2;−x&sub1;)

Parallel lines

m&sub1; = m&sub2;

Perpendicular lines

m&sub1; × m&sub2; = −1

Horizontal line

m = 0 (α = 0°)

Vertical line

m undefined (α = 90°)

💡 Perpendicular Condition — why m&sub1;m&sub2; = −1?

If a line has slope m = tanα, a perpendicular line has slope tan(90°+α) = −cotα = −1/m. So m&sub1;×m&sub2; = m × (−1/m) = −1. This is why perpendicular slopes are negative reciprocals of each other. Example: if m&sub1; = 3, then m&sub2; = −1/3.

✏️ Example 16.1 — Slope and conditionsStep-by-step

A(2,3) and B(5,9). Find slope of AB. Find slope of line perpendicular to AB.

1

m = (9−3)/(5−2) = 6/3 = 2

2

Perpendicular slope = −1/m = −1/2

✔

Slope of AB = 2; Perpendicular slope = −1/2

Module 1 of 6

16.2

Angle Between Two Lines

Formula for the acute angle between two intersecting lines

📖 Derivation

If two lines have inclinations α&sub1; and α&sub2; (slopes m&sub1; = tanα&sub1;, m&sub2; = tanα&sub2;), the angle θ between them satisfies tanθ = tan(α&sub1;−α&sub2;) = (tanα&sub1;−tanα&sub2;)/(1+tanα&sub1;tanα&sub2;). Taking the absolute value gives the acute angle.

Angle between lines

tanθ = |(m&sub1;−m&sub2;) / (1+m&sub1;m&sub2;)|

Parallel (θ=0)

m&sub1; = m&sub2;

Perpendicular (θ=90°)

1+m&sub1;m&sub2; = 0

The formula gives the acute angle (0° ≤ θ ≤ 90°) between the lines
If 1+m&sub1;m&sub2; = 0, the lines are perpendicular (θ = 90°)
If m&sub1; = m&sub2;, tanθ = 0 so θ = 0° (parallel)
For the obtuse angle: use 180° − θ

✏️ Example 16.2 — Angle between two linesStep-by-step

Lines y = 2x+1 and y = x/3+2. Find angle between them.

1

m&sub1; = 2, m&sub2; = 1/3

2

tanθ = |(2−1/3)/(1+2/3)| = |(5/3)/(5/3)| = 1

3

θ = tan¹(1) = 45°

✔

θ = 45°

Module 2 of 6

16.3

Equations of a Straight Line

Six standard forms — when to use each one

There are six standard forms of the equation of a straight line. Each is suited to different given information. Learn to recognise which form to start with.

① Slope-Intercept

y = mx + c (m=slope, c=y-intercept)

② Point-Slope

y − y&sub1; = m(x − x&sub1;)

③ Two-Point

(y−y&sub1;)/(y&sub2;−y&sub1;) = (x−x&sub1;)/(x&sub2;−x&sub1;)

④ Intercept Form

x/a + y/b = 1 (a=x-int, b=y-int)

⑤ Normal Form

x cosα + y sinα = p (p=perp distance)

⑥ General Form

ax + by + c = 0

💡 Which Form to Use?

① Given slope + y-intercept → slope-intercept.
② Given slope + one point → point-slope.
③ Given two points → two-point form.
④ Given x-intercept + y-intercept → intercept form.
⑤ Given perpendicular distance from origin + angle → normal form.
⑥ General form: convert to others by rearranging.

✏️ Example 16.3 — Line from two pointsStep-by-step

Find the equation of line through A(1,3) and B(4,9).

1

Slope m = (9−3)/(4−1) = 2

2

Point-slope using A(1,3): y−3 = 2(x−1) → y = 2x+1

3

General form: 2x − y + 1 = 0

✔

y = 2x + 1 or 2x − y + 1 = 0

✏️ Example 16.4 — Intercept formStep-by-step

A line cuts x-axis at 3 and y-axis at −2. Find equation.

1

a = 3, b = −2. Intercept form: x/3 + y/(−2) = 1

2

Multiply by 6: 2x − 3y = 6 or 2x − 3y − 6 = 0

✔

2x − 3y − 6 = 0

Module 3 of 6

16.4

Distance of a Point from a Line

Perpendicular distance formula and distance between parallel lines

📖 The Formula and Why It Works

The perpendicular from a point to a line is the shortest path from the point to the line. For point P(x&sub1;,y&sub1;) and line ax+by+c=0, this distance is:
d = |ax&sub1;+by&sub1;+c| / √(a²+b²)
The denominator √(a²+b²) is the magnitude of the normal vector (a,b) to the line. The numerator |ax&sub1;+by&sub1;+c| measures how far the point is “off” the line.

Point to line

d = |ax&sub1;+by&sub1;+c| / √(a²+b²)

Between parallel lines ax+by+c&sub1;=0 and ax+by+c&sub2;=0

d = |c&sub1;−c&sub2;| / √(a²+b²)

✏️ Example 16.5 — Distance from point to lineStep-by-step

Find distance of P(3,4) from 4x − 3y + 5 = 0.

1

a=4, b=−3, c=5. Numerator: |4(3)−3(4)+5| = |12−12+5| = 5

2

Denominator: √(16+9) = √25 = 5

3

d = 5/5 = 1

✔

Distance = 1 unit

✏️ Example 16.6 — Distance between parallel linesStep-by-step

Find distance between 3x+4y−5=0 and 6x+8y+15=0.

1

Make coefficients equal: divide second by 2: 3x+4y+7.5=0

2

d = |(−5)−(7.5)| / √(9+16) = 12.5/5 = 2.5

✔

Distance = 5/2 = 2.5 units

Module 4 of 6

16.5

Equation of the Bisectors

Angle bisectors of two intersecting lines — the locus of equal-distance points

📖 What is an Angle Bisector?

The angle bisector of two lines is the locus of all points equidistant from both lines. Since distance from a point to a line is |ax+by+c|/√(a²+b²), setting these equal for two lines gives us the bisector equation.

Bisectors of L&sub1;: a&sub1;x+b&sub1;y+c&sub1;=0 and L&sub2;: a&sub2;x+b&sub2;y+c&sub2;=0

(a&sub1;x+b&sub1;y+c&sub1;)/√(a&sub1;²+b&sub1;²) = ± (a&sub2;x+b&sub2;y+c&sub2;)/√(a&sub2;²+b&sub2;²)

The + sign gives one bisector, the − sign gives the other (they are perpendicular to each other)
To identify which bisector contains the origin: substitute O(0,0); if ax&sub1;+by&sub1;+c has the same sign for both lines, use + (origin is in that bisector)
The two bisectors are always perpendicular to each other

✏️ Example 16.7 — Finding angle bisectorsStep-by-step

Find bisectors of 3x+4y−5=0 and 5x−12y+10=0.

1

√(9+16)=5; √(25+144)=13

2

(3x+4y−5)/5 = +(5x−12y+10)/13 → 13(3x+4y−5) = 5(5x−12y+10) → 39x+52y−65=25x−60y+50 → 14x+112y=115

3

(3x+4y−5)/5 = −(5x−12y+10)/13 → 39x+52y−65=−25x+60y−50 → 64x−8y=15

✔

Bisectors: 14x+112y=115 and 64x−8y=15

Module 5 of 6

16.6

Pair of Lines

Homogeneous second-degree equation ax²+2hxy+by²=0 representing two lines through origin

📖 How One Equation Can Represent Two Lines

ax²+2hxy+by²=0 is called a homogeneous second-degree equation. Divide by x²: a + 2h(y/x) + b(y/x)² = 0. This is quadratic in (y/x) = m, giving two slopes m&sub1; and m&sub2;. So the equation represents the two lines y=m&sub1;x and y=m&sub2;x through the origin.

Sum of slopes

m&sub1;+m&sub2; = −2h/b

Product of slopes

m&sub1;×m&sub2; = a/b

Angle between them

tanθ = 2√(h²−ab) / (a+b)

Lines real & distinct

h² > ab

Lines coincident

h² = ab

Lines imaginary

h² < ab

Perpendicular lines

a + b = 0

✏️ Example 16.8 — Pair of lines analysisStep-by-step

For 2x²+7xy+3y²=0: find the two lines and angle between them.

1

a=2, 2h=7 so h=7/2, b=3. Check: h²−ab = 49/4−6 = 25/4 > 0 → two real distinct lines.

2

Factorise: 2x²+7xy+3y² = (2x+y)(x+3y) = 0 → lines: 2x+y=0 and x+3y=0

3

tanθ = 2√(25/4)/(2+3) = 2×(5/2)/5 = 1 → θ = 45°

✔

Lines: 2x+y=0 and x+3y=0; angle = 45°

Module 6 of 6 🎉

Chapter 17 · Conic Sections — Circles

Circles

A circle is the set of all points equidistant from a fixed centre. Simple definition — but it generates a rich theory. This chapter builds from the standard equation all the way to tangents, normals, chord of contact, family of circles, radical axes, and limiting points. Every result follows logically from the basic definition.

10 Modules ⌛ ~70 min Class 11 & 12

17.1

Standard Equation of a Circle

Centre-radius form, general form, converting between them

📖 Definition & Derivation

A circle is the set of all points at distance r from centre C(h,k). For any point P(x,y) on the circle: CP = r → √[(x−h)²+(y−k)²] = r → (x−h)²+(y−k)² = r². This is the centre-radius form.

Centre-radius form

(x−h)² + (y−k)² = r²

General form

x²+y²+2gx+2fy+c = 0

Centre

(−g, −f)

Radius

r = √(g²+f²−c)

💡 Converting General to Centre-Radius: Complete the Square

x²+y²+2gx+2fy+c=0 → (x²+2gx+g²)+(y²+2fy+f²) = g²+f²−c → (x+g)²+(y+f)² = g²+f²−c.
So centre = (−g,−f) and r = √(g²+f²−c).
Conditions: Real circle: g²+f²−c > 0. Point circle: = 0. Imaginary: < 0.

✏️ Example 17.1 — Find centre & radiusStep-by-step

Find centre and radius of x²+y²−6x+4y−12=0.

1

Compare with x²+y²+2gx+2fy+c=0: 2g=−6→g=−3; 2f=4→f=2; c=−12

2

Centre = (−g,−f) = (3,−2)

3

r = √((−3)²+2²−(−12)) = √(9+4+12) = √25 = 5

✔

Centre = (3,−2), Radius = 5

Module 1 of 10

17.2

Circle Through Three Points

Any three non-collinear points determine a unique circle

📖 Why Three Points?

The general equation x²+y²+2gx+2fy+c=0 has three unknowns (g, f, c). Three points give three equations → three unknowns → unique solution. (Two points give infinitely many circles; four or more points may not all lie on one circle.)

Method: Substitute each point into x²+y²+2gx+2fy+c=0. Solve the resulting system of 3 linear equations for g, f, c. Then write the circle equation.

✏️ Example 17.2 — Circle through three pointsStep-by-step

Find the circle through A(1,0), B(0,1), C(−1,0).

1

A(1,0): 1+0+2g+0+c=0 → 2g+c=−1 ...(i)

2

B(0,1): 0+1+0+2f+c=0 → 2f+c=−1 ...(ii)

3

C(−1,0): 1+0−2g+0+c=0 → −2g+c=−1 ...(iii)

4

(i)−(iii): 4g=0 → g=0. From (i): c=−1. From (ii): f=0.

5

Circle: x²+y²−1=0 → x²+y²=1 (unit circle!)

✔

x² + y² = 1, centre O(0,0), r=1

Module 2 of 10

17.3

Parametric Equation of Circle

Expressing every point on a circle using a single angle parameter θ

📖 The Idea

Instead of one equation in two unknowns (x,y), we express both x and y in terms of a single parameter θ (the angle from centre to point). As θ goes from 0 to 2π, the point traces the entire circle exactly once.

x

= h + r cosθ

y

= k + r sinθ

θ range

[0, 2π)

💡 Verify it works

Substitute into (x−h)²+(y−k)²=r²: (r cosθ)²+(r sinθ)² = r²(cos²θ+sin²θ) = r² ✓. The parametric form is very useful for writing tangent equations at any point — just substitute (h+r cosθ, k+r sinθ) into the tangent formula.

✏️ Example 17.3 — Parametric pointsStep-by-step

Circle x²+y²=25. Find the point with parameter θ=60°.

1

h=0, k=0, r=5. x = 5 cos60° = 5×(1/2) = 5/2

2

y = 5 sin60° = 5×(√3/2) = 5√3/2

✔

Point = (5/2, 5√3/2). Verify: (5/2)²+(5√3/2)² = 25/4+75/4 = 100/4 = 25 ✔

Module 3 of 10

17.4

Equation of Tangent

Tangent at a point on the circle, slope form, and from an external point

📖 Key Concept: The T = 0 Shortcut

For any circle S = x²+y²+2gx+2fy+c = 0, the tangent at point (x&sub1;,y&sub1;) is found by replacing x² with xx&sub1;, y² with yy&sub1;, 2gx with g(x+x&sub1;), 2fy with f(y+y&sub1;). This gives T = 0 where T = xx&sub1;+yy&sub1;+g(x+x&sub1;)+f(y+y&sub1;)+c. This pattern works for ALL conics — remember it!

Tangent at (x&sub1;,y&sub1;) on x²+y²=r²

xx&sub1; + yy&sub1; = r²

Tangent at (x&sub1;,y&sub1;) on general S=0

T = 0: xx&sub1;+yy&sub1;+g(x+x&sub1;)+f(y+y&sub1;)+c=0

Slope m tangent to x²+y²=r²

y = mx ± r√(1+m²)

✏️ Example 17.4 — Tangent at a pointStep-by-step

Find tangent to x²+y²=25 at point (3,4).

1

Use T=0: xx&sub1;+yy&sub1;=r² → x(3)+y(4)=25

2

3x+4y=25

3

Verify (3,4) lies on circle: 9+16=25 ✔. Verify (3,4) lies on tangent: 9+16=25 ✔

✔

Tangent: 3x + 4y = 25

✏️ Example 17.5 — Tangent from external pointStep-by-step

Find tangents to x²+y²=5 from point P(3,1).

1

Tangent: y−1 = m(x−3) → mx−y+(1−3m)=0

2

Distance from O(0,0) to tangent = r=√5: |1−3m|/√(m²+1) = √5

3

Square: (1−3m)² = 5(m²+1) → 1−6m+9m²=5m²+5 → 4m²−6m−4=0 → 2m²−3m−2=0

4

(2m+1)(m−2)=0 → m=−1/2 or m=2

✔

Tangents: y=2x−5 (m=2) and x−2y+5=... check: 2x−y−5=0 and x+2y−5=0

Module 4 of 10

17.5

Equation of Normal

Normal to a circle always passes through the centre

📖 The Key Property

The normal to a circle at any point is the line through that point perpendicular to the tangent. Since the tangent is perpendicular to the radius, the normal is parallel to the radius — in fact, it IS the radius extended. Therefore every normal to a circle passes through the centre.

Normal at (x&sub1;,y&sub1;) on x²+y²=r²

y/y&sub1; = x/x&sub1; (line through O and (x&sub1;,y&sub1;))

Normal on general circle (centre −g,−f)

(y−y&sub1;)/(x−x&sub1;) = (y&sub1;+f)/(x&sub1;+g)

✏️ Example 17.6 — Normal to a circleStep-by-step

Find normal to x²+y²−6x+4y−12=0 at point P(1,1).

1

Centre = (3,−2). Normal passes through centre and point P(1,1).

2

Slope of normal = (1−(−2))/(1−3) = 3/(−2) = −3/2

3

Normal: y−1 = (−3/2)(x−1) → 2y−2 = −3x+3 → 3x+2y=5

✔

Normal: 3x + 2y = 5

Module 5 of 10

17.6

Pair of Tangents & Chord of Contact

SS&sub1;=T² gives both tangents from external point; T=0 gives chord of contact

📖 The SS&sub1;=T² Trick

From an external point (x&sub1;,y&sub1;), two tangents can be drawn to a circle. The combined equation of both tangents is SS&sub1;=T², where:
S = x²+y²+2gx+2fy+c (the circle equation in terms of variable point)
S&sub1; = x&sub1;²+y&sub1;²+2gx&sub1;+2fy&sub1;+c (same formula at the external point)
T = xx&sub1;+yy&sub1;+g(x+x&sub1;)+f(y+y&sub1;)+c (tangent formula)

Pair of tangents

SS&sub1; = T²

Chord of contact

T = 0 (same formula as tangent at a point)

Length of tangent from (x&sub1;,y&sub1;)

= √S&sub1;

💡 What is Chord of Contact?

When two tangents from external point P touch the circle at A and B, the line AB is the chord of contact. Its equation is simply T=0 (same formula as tangent, but now it represents the chord joining both contact points). This is a beautiful symmetry in circle theory.

✏️ Example 17.7 — Length of tangentStep-by-step

Find length of tangent from P(4,3) to x²+y²−2x−4y−4=0.

1

S&sub1; = (4)²+(3)²−2(4)−4(3)−4 = 16+9−8−12−4 = 1

2

Length = √S&sub1; = √1 = 1

✔

Length of tangent = 1 unit

Module 6 of 10

17.7

Common Tangents of Two Circles

How many common tangents exist depends on relative position of the circles

📖 The Five Cases

Let d = distance between centres, r&sub1; and r&sub2; be radii (r&sub1; > r&sub2;).

d > r&sub1;+r&sub2;

4 common tangents (circles apart)

d = r&sub1;+r&sub2;

3 tangents (external contact)

|r&sub1;−r&sub2;| < d < r&sub1;+r&sub2;

2 tangents (circles intersect)

d = |r&sub1;−r&sub2;|

1 tangent (internal contact)

d < |r&sub1;−r&sub2;|

0 tangents (one inside other)

External tangents cross between the circles; internal tangents cross outside
External centre of similitude divides the segment joining centres externally in ratio r&sub1;:r&sub2;
Internal centre of similitude divides internally in ratio r&sub1;:r&sub2;
When circles are equal (r&sub1;=r&sub2;): external tangents are parallel, no internal centre of similitude

Module 7 of 10

17.8

Family of Circles

All circles through intersection of two curves; radical axis

📖 The Family Concept

Infinitely many circles pass through the two intersection points of circles S&sub1;=0 and S&sub2;=0. They all have the equation S&sub1;+λS&sub2;=0 for different values of λ. By choosing λ, you get any specific member of this family. Similarly, circles through the intersection of circle S=0 and line L=0: S+λL=0.

Through S&sub1;∩S&sub2;

S&sub1; + λS&sub2; = 0

Through S∩L

S + λL = 0

Radical axis

S&sub1; − S&sub2; = 0

💡 Radical Axis

The radical axis of two circles is the locus of points from which tangents to both circles have equal length (√S&sub1; = √S&sub2; → S&sub1;=S&sub2; → S&sub1;−S&sub2;=0). It is always a straight line perpendicular to the line joining the centres. If the circles intersect, the radical axis is the common chord.

Module 8 of 10

17.9

Limiting Points

Point circles (r=0) in a coaxial system; every coaxial system has exactly two

📖 Definition

A coaxial system is a family of circles S&sub1;+λS&sub2;=0 sharing the same radical axis. A limiting point is a member of this family with radius = 0 (a “point circle”). Every coaxial system has exactly two limiting points (which may be real or imaginary).

💡 Key Property

The limiting points of a coaxial system are the centres of the circles in the conjugate coaxial system (the system whose radical axis is perpendicular to the first). The two limiting points lie on the line through all the centres.

To find limiting points: In the family S&sub1;+λS&sub2;=0, find the radius as a function of λ. Set radius=0 and solve for λ. Substitute back to get the centres (which are the limiting points).

Module 9 of 10

17.10

Diameter of Circle

Circle described on a diameter; angle in semicircle = 90°

📖 Thales Theorem & the Diameter Circle

Thales’ Theorem: The angle in a semicircle is always 90°. If A and B are endpoints of a diameter, then for any point P on the circle, ∠APB = 90°. This gives us the equation of a circle directly from its diameter endpoints!

Circle with diameter A(x&sub1;,y&sub1;) to B(x&sub2;,y&sub2;)

(x−x&sub1;)(x−x&sub2;) + (y−y&sub1;)(y−y&sub2;) = 0

💡 Why this formula works

For any point P(x,y) on the circle, PA ⊥ PB (Thales). So slope(PA) × slope(PB) = −1. Slope PA = (y−y&sub1;)/(x−x&sub1;), Slope PB = (y−y&sub2;)/(x−x&sub2;). Product = −1 → (y−y&sub1;)(y−y&sub2;) = −(x−x&sub1;)(x−x&sub2;) → gives the formula.

✏️ Example 17.8 — Circle on a diameterStep-by-step

Find circle with diameter endpoints A(1,2) and B(5,6).

1

(x−1)(x−5) + (y−2)(y−6) = 0

2

x²−6x+5 + y²−8y+12 = 0 → x²+y²−6x−8y+17=0

3

Centre = (3,4), r = √(9+16−17) = √8 = 2√2

✔

x²+y²−6x−8y+17=0; Centre (3,4), r=2√2

Module 10 of 10 🎉

Chapter 18 · Conic Sections

Parabola

The parabola is the path of a ball thrown in the air, the shape of a satellite dish, and the mirror behind every car headlight. Mathematically it is the conic of eccentricity exactly 1. This chapter takes you from the general conic equation all the way to tangents, normals, diameters, and chord of contact — with clear derivations and worked examples at every step.

10 Modules ⌛ ~70 min Class 12 Core

18.1

Conic Sections

Cutting a double cone at different angles produces four types of curves

📖 What is a Conic?

A conic section is the intersection of a plane with a right circular double cone (two cones tip-to-tip). The angle of the cutting plane determines which of the four types of conic results. The eccentricity e characterises each type.

Circle

e = 0 (plane perpendicular to axis)

Ellipse

0 < e < 1

Parabola

e = 1 (plane parallel to a generator)

Hyperbola

e > 1 (plane cuts both cones)

💡 Focus-Directrix Definition

A conic is the locus of a point P such that: distance from P to focus / distance from P to directrix = e (constant). This is the most general definition. For parabola: e=1, so every point is equidistant from the focus and directrix.

Module 1 of 10

18.2

General Equation of Conic

ax²+2hxy+by²+2gx+2fy+c=0 and how to classify it using the discriminant

📖 The General Second-Degree Equation

ax² + 2hxy + by² + 2gx + 2fy + c = 0
This represents a conic (or degenerate case). The discriminant is Δ = abc + 2fgh − af² − bg² − ch² and h²−ab classifies the type.

Parabola

Δ ≠ 0 and h² = ab

Ellipse

Δ ≠ 0 and h² < ab

Hyperbola

Δ ≠ 0 and h² > ab

Circle

a = b and h = 0

Degenerate (pair of lines, point, empty)

Δ = 0

✏️ Example 18.1 — Classify conicsStep-by-step

Classify: (a) x²+y²−4=0 (b) y²−4x=0 (c) x²−y²−1=0

a

a=1, b=1, h=0: a=b & h=0 → Circle

b

a=0, b=1, h=0: h²−ab = 0−0 = 0 → Parabola

c

a=1, b=−1, h=0: h²−ab = 0−(1)(−1) = 1 > 0 → Hyperbola

✔

(a) Circle (b) Parabola (c) Hyperbola

Module 2 of 10

18.3

Standard Forms of Parabola

Four orientations — each with focus, directrix, axis, vertex, latus rectum

📖 Key Terms

Vertex: tip of the parabola (at origin in standard forms). Focus: the fixed point. Directrix: the fixed line. Axis: line of symmetry through vertex and focus. Latus Rectum: chord through the focus perpendicular to the axis; its length = 4a.

y² = 4ax

Opens right; Focus (a,0); Directrix x=−a; Axis y=0

y² = −4ax

Opens left; Focus (−a,0); Directrix x=a

x² = 4ay

Opens up; Focus (0,a); Directrix y=−a; Axis x=0

x² = −4ay

Opens down; Focus (0,−a); Directrix y=a

Latus rectum length

= 4a (same for all standard forms)

Parametric point on y²=4ax

(at², 2at)

💡 Quick Memory Aid

If the squared variable is y (y²=...), the parabola opens left or right (horizontal axis). If the squared variable is x (x²=...), it opens up or down (vertical axis). The sign of 4a tells direction: positive = opens toward positive axis direction.

✏️ Example 18.2 — Identifying parabola propertiesStep-by-step

For y² = 12x, find a, focus, directrix, and latus rectum length.

1

Compare with y²=4ax: 4a=12 → a=3

2

Focus = (a, 0) = (3, 0)

3

Directrix: x = −a = x = −3

4

Latus rectum = 4a = 12

✔

a=3, Focus=(3,0), Directrix x=−3, LR=12

Module 3 of 10

18.4

Equation of Tangent

Point form, slope form, and parametric form for y²=4ax

📖 Three Forms of Tangent to y²=4ax

Just like circles, tangents to a parabola can be written in point form (T=0), slope form, or parametric form. The T=0 rule works for all conics: replace y² by yy&sub1;, replace 4ax by 2a(x+x&sub1;).

Point form at (x&sub1;,y&sub1;)

yy&sub1; = 2a(x+x&sub1;)

Slope form (slope m)

y = mx + a/m; contact point (a/m², 2a/m)

Parametric at t

ty = x + at²

💡 Condition for tangency

For y = mx+c to be tangent to y²=4ax: substitute y=mx+c into y²=4ax and require the discriminant of the resulting quadratic = 0. This gives c = a/m. So the tangent in slope form is always y = mx + a/m.

✏️ Example 18.3 — Tangent at a pointStep-by-step

Find tangent to y²=8x at point (2,4).

1

4a=8 → a=2. Point (2,4) lies on parabola: 4²=16=8×2 ✔

2

T=0: yy&sub1; = 2a(x+x&sub1;) → y(4) = 2(2)(x+2) → 4y = 4x+8 → y = x+2

✔

Tangent: y = x + 2

✏️ Example 18.4 — Tangent with given slopeStep-by-step

Find tangent to y²=4x with slope 2.

1

a=1, m=2. Slope form: y = mx + a/m = 2x + 1/2

2

Contact point: (a/m², 2a/m) = (1/4, 1)

✔

Tangent: y = 2x + 1/2; Contact at (1/4, 1)

Module 4 of 10

18.5

Point of Intersection of Two Tangents

Tangents at parametric points t&sub1; and t&sub2; meet at a single elegant point

📖 Key Result

The tangents at points P(at&sub1;², 2at&sub1;) and Q(at&sub2;², 2at&sub2;) on y²=4ax meet at the point:
R = (at&sub1;t&sub2;, a(t&sub1;+t&sub2;))
This is a remarkable formula because the x-coordinate of the intersection is the geometric mean of the x-coordinates of P and Q (in terms of t).

💡 Proof Sketch

Tangent at t&sub1;: t&sub1;y = x + at&sub1;² Tangent at t&sub2;: t&sub2;y = x + at&sub2;²
Subtract: (t&sub1;−t&sub2;)y = a(t&sub1;²−t&sub2;²) = a(t&sub1;−t&sub2;)(t&sub1;+t&sub2;) → y = a(t&sub1;+t&sub2;). Substitute back: x = at&sub1;t&sub2;. Done!

✏️ Example 18.5 — Intersection of two tangentsStep-by-step

On y²=4x, tangents are drawn at t&sub1;=2 and t&sub2;=3. Find their intersection.

1

a=1. Intersection = (at&sub1;t&sub2;, a(t&sub1;+t&sub2;)) = (1×2×3, 1×(2+3)) = (6, 5)

✔

Intersection point = (6, 5)

Module 5 of 10

18.6

Equation of Normal

Point form, slope form (cubic in m), and parametric form

📖 Normal to y²=4ax

At point P(x&sub1;,y&sub1;), the slope of tangent = 2a/y&sub1;. Slope of normal = −y&sub1;/(2a). The normal equation follows from point-slope form. In parametric form (at², 2at), slope of tangent = 1/t, so slope of normal = −t.

At point (x&sub1;,y&sub1;)

y − y&sub1; = −(y&sub1;/2a)(x − x&sub1;)

Slope form (slope m)

y = mx − 2am − am³

Foot of normal

(am², −2am)

Parametric at t

y + tx = 2at + at³

⚠️ The Cubic in Slope — Important for Normals from External Point

If a normal of slope m passes through external point (h,k), substitute into y=mx−2am−am³: k = mh−2am−am³ → am³+m(2a−h)+k=0. This cubic in m has up to 3 real roots → up to 3 normals from any external point!

✏️ Example 18.6 — Normal at a pointStep-by-step

Find normal to y²=4x at point (1,2).

1

a=1. Point (1,2): 2²=4=4×1 ✔. Slope of tangent = 2a/y&sub1; = 2/2 = 1. Slope of normal = −1.

2

Normal: y−2 = −1(x−1) → y = −x+3 → x+y=3

✔

Normal: x + y = 3

Module 6 of 10

18.7

Length of Tangent and Normal

Subtangent, subnormal — key geometric measures; subnormal is constant for parabola

📖 Definitions at point P(x&sub1;,y&sub1;) on a curve

Drop a perpendicular from P to the x-axis meeting at N. The tangent at P meets x-axis at T. Then:
Subtangent TN = x-intercept projection of tangent
Subnormal MN = x-intercept projection of normal
For parabola y²=4ax, the subnormal is always constant = 2a (semi-latus rectum). This property uniquely characterises the parabola!

Subtangent at (at², 2at)

= 2at (= twice the abscissa)

Subnormal

= 2a (constant!)

Length of tangent

= y&sub1;√(1+cot²α) where α = slope angle

Length of normal

= y&sub1;√(1+tan²α)

✏️ Example 18.7 — Subtangent and SubnormalStep-by-step

For y²=4x, at point (4, 4), find subtangent and subnormal.

1

a=1. t: (at²,2at)=(4,4) → t=2

2

Subtangent = 2at = 2(1)(2) = 4 (= twice abscissa = 2×4 = 8? No: 2at not 2x. 2×1×2=4)

3

Subnormal = 2a = 2(1) = 2 (constant for all points!)

✔

Subtangent = 4, Subnormal = 2

Module 7 of 10

18.8

Equation of Diameter

Locus of midpoints of a system of parallel chords of slope m

📖 Diameter of a Conic

The diameter of a conic (not to be confused with circle diameter) is the locus of midpoints of a system of parallel chords. For parallel chords of slope m on y²=4ax, all midpoints lie on a horizontal line. The diameter equation is y = 2a/m.

Diameter of y²=4ax for chords of slope m

y = 2a/m

💡 Key Properties

1. The diameter of y²=4ax is always parallel to the axis (x-axis), regardless of m. This is unique to the parabola.
2. The diameter bisects all chords of the given slope.
3. Tangents drawn at the ends of any focal chord are perpendicular and meet on the directrix.

Module 8 of 10

18.9

Pair of Tangents

SS&sub1;=T² gives combined equation of both tangents from external point

📖 The SS&sub1;=T² Pattern (same as circles!)

For parabola S = y²−4ax = 0, external point (h,k):
S = y²−4ax (circle equation in (x,y))
S&sub1; = k²−4ah (evaluated at external point)
T = yk−2a(x+h) (tangent formula using T=0 replacement)
Combined equation of pair of tangents: SS&sub1; = T²

Pair of tangents from (h,k)

SS&sub1; = T²

Condition: point outside parabola

S&sub1; = k²−4ah > 0

✏️ Example 18.8 — Check if external; find pair of tangentsStep-by-step

From P(3,4) to y²=4x: verify P is external and write pair of tangents.

1

a=1. S&sub1; = 4²−4(1)(3) = 16−12 = 4 > 0 → P is outside ✔

2

T = y(4)−2(1)(x+3) = 4y−2x−6

3

SS&sub1;=T²: (y²−4x)(4) = (4y−2x−6)²

✔

4(y²−4x) = (4y−2x−6)² is the combined equation of both tangents

Module 9 of 10

18.10

Chord of Contact

T=0 gives the chord joining the two contact points of tangents from external point

📖 What is Chord of Contact?

From external point P(h,k), two tangents touch the parabola y²=4ax at points A and B. The line AB is the chord of contact. Its equation is T=0: yk = 2a(x+h). This is called the polar of P with respect to the parabola.

Chord of contact from (h,k)

T=0: yk = 2a(x+h)

Chord of contact equation = tangent equation (same formula T=0, different interpretation)
If P is inside parabola (S&sub1;<0), chord of contact is imaginary (no real tangents)
The chord of contact is the polar of P; P is the pole of the chord
Tangents at A and B to the parabola intersect at P (the external point)

✏️ Example 18.9 — Chord of contactStep-by-step

Find chord of contact from P(3,4) to y²=8x.

1

4a=8 → a=2. h=3, k=4.

2

T=0: yk = 2a(x+h) → 4y = 4(x+3) → y = x+3 → x−y+3=0

✔

Chord of contact: x − y + 3 = 0

Module 10 of 10 🎉

Chapter 19 · Data Analysis

Statistics

Statistics is the science of making sense of data. How do you find the “centre” of a data set? Which average is best for which situation? How do you measure how two variables move together? This chapter covers all of this — from drawing histograms to fitting regression lines — with real numerical examples at every step so you see exactly how every formula is applied.

10 Modules ⌛ ~60 min Class 11 & 12

19.1

Graphical Representation of Frequency Distributions

Histogram, frequency polygon, and cumulative frequency ogive — how to draw and read each

Before computing any statistic, we visualise data using graphs. Each graph type reveals something different about the data's shape and distribution.

📖 Three Key Graphs

Histogram: Bar chart where bars represent class intervals and heights represent frequencies. Bars touch (no gaps) because data is continuous.
Frequency Polygon: Connect midpoints of histogram tops with straight lines. Shows the shape of the distribution more clearly.
Ogive (Cumulative Frequency Curve): S-shaped curve plotting cumulative frequency vs upper class boundary. Used to read off median, quartiles, percentiles graphically.

Left: Histogram with frequency polygon (dashed). Right: Ogive — read median where cumulative frequency = N/2

Class width h

= Upper boundary − Lower boundary

Relative frequency

= f / N (f = class freq, N = total)

Cumulative frequency

= Running total of all frequencies up to that class

Histogram bars have no gaps (continuous data) — this distinguishes them from bar charts
The area of each bar = frequency (if class widths are equal)
From the ogive: read off median at cumulative frequency = N/2; Q1 at N/4; Q3 at 3N/4

Module 1 of 10

19.2

Measures of Central Tendency

Mean, median, mode — when to use which, and how they relate

📖 The Three Averages

Arithmetic Mean (AM): Sum divided by count. Affected by extreme values (outliers).
Median: Middle value when data is ordered. Not affected by outliers. Best for skewed data.
Mode: Most frequent value. Can be used for categorical data. A distribution can be bimodal (two modes).

Symmetric distribution

Mean = Median = Mode

Moderately skewed

Mode ≈ 3×Median − 2×Mean (Karl Pearson)

AM ≥ GM ≥ HM

Always true for positive values

💡 Which Average to Choose?

Use Mean when data has no extreme outliers (e.g. exam marks in a class).
Use Median when data is skewed or has outliers (e.g. income data where a few billionaires skew the mean).
Use Mode for categorical data or to find the most popular item (e.g. most popular shoe size).

Module 2 of 10

19.3

Arithmetic Mean

Simple, weighted, step-deviation (short-cut) methods for both ungrouped and grouped data

Simple (ungrouped)

x̅ = ∑xᵢ / n

Grouped data

x̅ = ∑fᵢxᵢ / ∑fᵢ (xᵢ = class midpoint)

Step-deviation (short-cut)

x̅ = A + h×(∑fᵢuᵢ/N) where uᵢ=(xᵢ−A)/h

Weighted Mean

x̅ᵂ = ∑wᵢxᵢ / ∑wᵢ

Combined Mean

x̅ = (n₁x̅₁ + n₂x̅₂) / (n₁+n₂)

💡 Step-Deviation Method — Why Use It?

When midpoints are large (e.g. 205, 215, 225…), multiplying by frequency gives huge numbers. The step-deviation trick: choose assumed mean A (usually the central class midpoint), and h = class width. Compute uᵢ = (xᵢ−A)/h (small integers!). This dramatically reduces calculation errors in exams.

✏️ Example 19.1 — Grouped mean by step-deviationStep-by-step

Find mean of: Classes 10–20, 20–30, 30–40, 40–50, 50–60 with frequencies 5, 8, 15, 9, 3.

1

Midpoints xᵢ: 15, 25, 35, 45, 55. N = 5+8+15+9+3 = 40. Choose A=35, h=10.

2

uᵢ = (xᵢ−35)/10: −2, −1, 0, 1, 2

3

fᵢuᵢ: 5(−2)+8(−1)+15(0)+9(1)+3(2) = −10−8+0+9+6 = −3

4

x̅ = 35 + 10×(−3/40) = 35 − 0.75 = 34.25

✔

Mean = 34.25

📖 Key Properties of AM

1. ∑(xᵢ−x̅) = 0 (deviations from mean always sum to zero)
2. If yᵢ = xᵢ+k, then ȳ = x̅+k (shift property)
3. If yᵢ = kxᵢ, then ȳ = kx̅ (scale property)
4. AM is the unique value that minimises ∑(xᵢ−a)² (least squares property)

Module 3 of 10

19.4

Geometric Mean

nth root of the product — best for ratios, growth rates, and index numbers

📖 When to Use GM

The GM is the right average when data represents rates, ratios, or multiplicative growth (e.g. percentage changes in stock prices, population growth rates, economic indices). The AM would overestimate in such cases.

GM (n values)

= (x₁×x₂×…×xₙ)^(1/n)

log(GM)

= (∑ log xᵢ) / n

Grouped data

log(GM) = ∑fᵢ log xᵢ / N

AM ≥ GM ≥ HM

Equality when all values equal

✏️ Example 19.2 — GM of growth ratesStep-by-step

A company grows at 10%, 20%, and 30% in three successive years. Find the average growth rate using GM.

1

Growth factors: 1.1, 1.2, 1.3. GM = (1.1×1.2×1.3)^(1/3) = (1.716)^(1/3)

2

= 1.197 approximately → average growth rate ≈ 19.7%

3

Compare: AM = (10+20+30)/3 = 20%. GM is slightly less due to compounding effect.

✔

Average growth ≈ 19.7% per year (by GM)

✏️ Example 19.3 — GM of simple valuesStep-by-step

Find GM of 4, 8, 16.

1

GM = (4×8×16)^(1/3) = (512)^(1/3) = 8

2

Verify using logs: log4+log8+log16 = 0.602+0.903+1.204 = 2.709. 2.709/3 = 0.903 = log8 → GM=8 ✔

✔

GM = 8

Module 4 of 10

19.5

Harmonic Mean

Reciprocal of AM of reciprocals — ideal for averaging rates and speeds

📖 When to Use HM

HM is the correct average when you are averaging rates. Classic example: if you travel equal distances at speeds v₁ and v₂, the average speed is the HM, not the AM. Why? Because time = distance/speed, and total time depends on the reciprocal of speed.

HM (n values)

= n / ∑(1/xᵢ)

Grouped data

= N / ∑(fᵢ/xᵢ)

Two values: AM×HM

= GM²

Speed application

Avg speed = 2v₁v₂ / (v₁+v₂)

✏️ Example 19.4 — Speed problem with HMStep-by-step

A car travels from A to B at 60 km/h and returns at 40 km/h. Find the average speed for the entire journey.

1

Equal distances → average speed = HM of 60 and 40.

2

HM = 2×60×40 / (60+40) = 4800/100 = 48 km/h

3

Verify: Let distance = 120 km each way. Time A→B = 2 h; Time B→A = 3 h. Total = 240 km in 5 h = 48 km/h ✔

✔

Average speed = 48 km/h (not 50, which is the AM!)

Module 5 of 10

19.6

Median

Middle value; interpolation formula for grouped data; quartiles

📖 Definition

The median is the value that divides an ordered data set into two equal halves. Half the values are below it, half above. It is the 50th percentile.

Odd n (ungrouped)

M = x₍(n+1)/2₎

Even n (ungrouped)

M = (x₍n/2₎ + x₍n/2+1₎) / 2

Grouped data

M = L + [(N/2 − cf) / f] × h

Q₁ (Lower Quartile)

= L + [(N/4 − cf) / f] × h

Q₃ (Upper Quartile)

= L + [(3N/4 − cf) / f] × h

📖 Notation for Grouped Median

L = lower class boundary of the median class
N = total frequency (∑f)
cf = cumulative frequency of classes before the median class
f = frequency of the median class
h = class width
Median class: the class containing the (N/2)th observation.

✏️ Example 19.5 — Grouped medianStep-by-step

Find median for: 0–10(f=5), 10–20(f=8), 20–30(f=15), 30–40(f=9), 40–50(f=3). N=40.

1

N/2 = 20. Cumulative frequencies: 5, 13, 28, 37, 40. The 20th obs falls in class 20–30 (cf before=13, f=15).

2

L=20, cf=13, f=15, h=10. M = 20 + [(20−13)/15]×10 = 20 + (7/15)×10 = 20 + 4.67 = 24.67

✔

Median ≈ 24.67

Module 6 of 10

19.7

Mode

Most frequent value; interpolation formula for grouped data

📖 Definition

The mode is the value (or class) that occurs most frequently. A data set can be unimodal (one mode), bimodal (two), or multimodal. For grouped data, the modal class has the highest frequency, and we interpolate within it.

Grouped data

Mo = L + [(f₁−f₀) / (2f₁−f₀−f₂)] × h

📖 Notation

L = lower boundary of modal class (class with highest f)
f₁ = frequency of modal class
f₀ = frequency of class just before modal class
f₂ = frequency of class just after modal class
h = class width

✏️ Example 19.6 — Mode for grouped dataStep-by-step

Classes 10–20(f=5), 20–30(f=8), 30–40(f=15), 40–50(f=9), 50–60(f=3). Find mode.

1

Modal class = 30–40 (highest frequency = 15). L=30, f₁=15, f₀=8, f₂=9, h=10.

2

Mo = 30 + [(15−8)/(2×15−8−9)]×10 = 30 + [7/(30−17)]×10 = 30 + (7/13)×10 ≈ 30 + 5.38 = 35.38

✔

Mode ≈ 35.38

Module 7 of 10

19.8

Covariance & Correlation

Measuring joint variability; Karl Pearson's correlation coefficient r

📖 Covariance — What It Means

Covariance measures whether two variables X and Y tend to move together. If large X is paired with large Y (both above their means), products (x−x̅)(y−ȳ) are positive → Cov > 0 (positive relationship). If large X is paired with small Y, Cov < 0. If no pattern, Cov ≈ 0.

Cov(X,Y)

= ∑(xᵢ−x̅)(yᵢ−ȳ) / n

Short-cut formula

= ∑xᵢyᵢ/n − x̅×ȳ

Pearson's r

= Cov(X,Y) / (σₓ×σᵧ)

Range of r

−1 ≤ r ≤ +1

💡 Interpreting r

r = +1: perfect positive linear relationship. r = −1: perfect negative. r = 0: no linear relationship (but could be non-linear!). |r| > 0.7: strong; 0.4–0.7: moderate; < 0.4: weak.

✏️ Example 19.7 — Computing rStep-by-step

X: 1,2,3,4,5. Y: 2,4,5,4,5. Find Cov(X,Y) and r.

1

x̅ = 3, ȳ = 4. ∑xᵢyᵢ = 1×2+2×4+3×5+4×4+5×5 = 2+8+15+16+25 = 66

2

Cov = 66/5 − 3×4 = 13.2 − 12 = 1.2

3

σₓ = √[∑xᵢ²/n − x̅²] = √[55/5−9] = √2 ≈ 1.414

4

σᵧ = √[∑yᵢ²/n − ȳ²] = √[(4+16+25+16+25)/5−16] = √[86/5−16] = √1.2 ≈ 1.095

5

r = 1.2 / (1.414×1.095) ≈ 1.2/1.548 ≈ 0.775

✔

Cov(X,Y) = 1.2; r ≈ 0.775 (strong positive correlation)

Module 8 of 10

19.9

Rank Correlation — Spearman's

Correlation between ranks when variables cannot be measured on an interval scale

📖 Why Spearman's?

When you cannot assign exact numerical values (e.g. judging beauty, ranking essays), you assign ranks instead. Spearman's rₛ measures how well the two sets of ranks agree. It uses the same concept as Pearson's r but applied to rank data.

Spearman's rₛ

= 1 − [6∑d² / n(n²−1)]

d

= difference in ranks for each pair

Ties: correction

Add (m³−m)/12 to ∑d² for each group of m tied ranks

💡 Interpreting rₛ

rₛ = +1: ranks perfectly agree. rₛ = −1: ranks perfectly disagree. rₛ = 0: no relationship between ranks. The interpretation is identical to Pearson's r.

✏️ Example 19.8 — Spearman's rank correlationStep-by-step

Two judges rank 5 students: Judge A: 1,2,3,4,5. Judge B: 2,1,4,3,5. Find rₛ.

1

d = A−B: −1, 1, −1, 1, 0

2

d²: 1, 1, 1, 1, 0. ∑d² = 4

3

rₛ = 1 − 6×4 / [5(25−1)] = 1 − 24/120 = 1 − 0.2 = 0.8

✔

rₛ = 0.8 (strong agreement between judges)

Module 9 of 10

19.10

Regression

Lines of best fit — predicting one variable from another using regression equations

📖 What is Regression?

If two variables X and Y are correlated, we can use X to predict Y (or Y to predict X). Regression gives us the best-fit straight line that minimises the sum of squared errors. There are TWO regression lines (Y on X, and X on Y) because each minimises errors in a different direction. They coincide only when r = ±1.

Y on X: y−ȳ = bᵧₓ(x−x̅)

bᵧₓ = r×σᵧ/σₓ = Cov(X,Y)/σₓ²

X on Y: x−x̅ = bₓᵧ(y−ȳ)

bₓᵧ = r×σₓ/σᵧ = Cov(X,Y)/σᵧ²

Relationship between r and b

r = √(bᵧₓ × bₓᵧ)

Both lines pass through

(x̅, ȳ) — the mean point

⚠️ Critical Properties

1. Both regression lines always pass through (x̅, ȳ).
2. r = +1 or −1: both lines coincide (perfect fit).
3. r = 0: lines are perpendicular (no linear relationship).
4. r² = bᵧₓ×bₓᵧ (product of regression coefficients = r²).
5. Both regression coefficients have the same sign (same as r).

✏️ Example 19.9 — Regression lines and predictionStep-by-step

x̅=5, ȳ=4, bᵧₓ=0.8, bₓᵧ=0.5. Find r, both regression lines, and predict y when x=8.

1

r = √(0.8×0.5) = √0.4 ≈ 0.632

2

Y on X: y−4 = 0.8(x−5) → y = 0.8x

3

X on Y: x−5 = 0.5(y−4) → x = 0.5y+3 → 2x−y=6

4

Predict y when x=8: y = 0.8×8 = 6.4

✔

r=0.632; Y on X: y=0.8x; X on Y: x=0.5y+3; y|x=8 = 6.4

Module 10 of 10 🎉

Master Maths withClarity & Confidence

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Step-by-Step Examples

Visual Diagrams

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