Chapter 1: Sets - JEE Main & Advanced

1. Introduction to Sets

A set is a well-defined collection of distinct objects. The objects in a set are called its elements or members.

Notation

Sets are usually denoted by capital letters: A, B, C,...
Elements are denoted by small letters: a, b, c,...
a ∈ A means 'a is an element of set A'
a ∉ A means 'a is not an element of set A'

Methods of Representing Sets

Roster/Tabular Form: Listing all elements {1, 2, 3}
Set-builder Form: {x | x has property P}

2. Types of Sets

Type	Definition	Example
Empty/Null Set	Set with no elements	∅ or {}
Singleton Set	Set with one element	{5}
Finite Set	Set with countable number of elements	{1, 2, 3, 4}
Infinite Set	Set with unlimited elements	Set of natural numbers
Equal Sets	Sets with exactly same elements	A = {1, 2}, B = {2, 1}
Equivalent Sets	Sets with same number of elements	A = {1, 2}, B = {a, b}

JEE Tip: Questions often test understanding of empty set vs {∅} - the first has no elements, the second has one element which is the empty set.

3. Subsets and Power Sets

Subset: A is a subset of B (A ⊆ B) if every element of A is in B

Proper Subset: A ⊂ B if A ⊆ B and A ≠ B

Important Properties

Every set is a subset of itself (A ⊆ A)
Empty set is subset of every set (∅ ⊆ A)
Number of subsets of a set with n elements = 2ⁿ
Number of proper subsets = 2ⁿ - 1

Example

Find all subsets of A = {1, 2}

Solution: ∅, {1}, {2}, {1, 2} (Total 4 = 2² subsets)

4. Operations on Sets

Operation	Notation	Definition
Union	A ∪ B	{x \| x ∈ A or x ∈ B}
Intersection	A ∩ B	{x \| x ∈ A and x ∈ B}
Difference	A - B	{x \| x ∈ A and x ∉ B}
Complement	A' or A^c	{x \| x ∉ A in universal set}

De Morgan's Laws

(A ∪ B)' = A' ∩ B'

(A ∩ B)' = A' ∪ B'

5. Venn Diagrams

Visual representation of sets and their relationships using overlapping circles.

Venn Diagram Representation:

[Three overlapping circles representing sets A, B, and C]

Shaded regions show intersections and unions

JEE Tip: Venn diagram problems often appear in JEE Main. Practice problems with 3 sets and their various intersections.

6. Important Formulas

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

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

n(A - B) = n(A) - n(A ∩ B)

7. Practice Problems for JEE/Unit Test

If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, find A ∪ B, A ∩ B, A - B
Prove that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
In a class of 50 students, 30 take math, 25 take physics, and 10 take both. How many take neither?
If n(A) = 5, n(B) = 7, and n(A ∩ B) = 3, find n(A ∪ B)
Find the power set of {a, b, c}

Common Mistakes to Avoid

Confusing ∈ and ⊆
Forgetting the empty set in power sets
Miscounting elements in union/intersection problems
Applying De Morgan's laws incorrectly

8. Previous Year JEE Questions

JEE Main 2022: Let A and B be two sets containing 3 and 4 elements respectively. The number of subsets of A × B having at least 5 elements is?

JEE Advanced 2021: For sets A and B, let f: A → B and g: B → A be functions such that f(g(x)) = x for each x ∈ B. Which of the following must be true?

Chapter 2: Relations & Functions - JEE Main & Advanced

1. Cartesian Product of Sets

Definition: The Cartesian product A × B of two non-empty sets A and B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

A × B = {(a, b) : a ∈ A, b ∈ B}

Number of elements: If n(A) = p and n(B) = q, then n(A × B) = p × q

Example

If A = {1, 2} and B = {3, 4, 5}, find A × B and B × A.

Solution:

A × B = {(1,3), (1,4), (1,5), (2,3), (2,4), (2,5)}

B × A = {(3,1), (3,2), (4,1), (4,2), (5,1), (5,2)}

Important Properties

A × B ≠ B × A (unless A = B)
A × (B ∪ C) = (A × B) ∪ (A × C)
A × (B ∩ C) = (A × B) ∩ (A × C)
If A ⊆ B, then A × C ⊆ B × C for any set C

2. Relations

Definition: A relation R from set A to set B is a subset of A × B.

If (a, b) ∈ R, we say 'a is related to b' and write a R b.

Types of Relations

Relation Type	Definition	Example
Empty Relation	R = ∅ ⊆ A × A	No elements are related
Universal Relation	R = A × A	All elements are related
Identity Relation	I_A = {(a, a) : a ∈ A}	Each element related to itself
Reflexive	(a, a) ∈ R ∀ a ∈ A	Every element related to itself
Symmetric	(a, b) ∈ R ⇒ (b, a) ∈ R	If a~b then b~a
Transitive	(a, b), (b, c) ∈ R ⇒ (a, c) ∈ R	If a~b and b~c then a~c
Equivalence	Reflexive, Symmetric and Transitive	Congruence modulo n

JEE Tip: Equivalence relations partition the set into disjoint equivalence classes. Questions often ask to verify if a given relation is equivalence.

3. Functions

Definition: A function f: A → B is a relation that associates each element of A to exactly one element of B.

Domain: Set A (input values)

Codomain: Set B (possible output values)

Range: Actual outputs {f(x) : x ∈ A} ⊆ B

Types of Functions

Function Type	Definition	Example
One-one (Injective)	f(x₁) = f(x₂) ⇒ x₁ = x₂	f(x) = 2x + 3
Onto (Surjective)	Range = Codomain	f: R → R, f(x) = x³
Bijective	Both one-one and onto	f(x) = x + 1
Polynomial	f(x) = a_nxⁿ + ... + a₀	f(x) = x² - 3x + 2
Rational	Ratio of two polynomials	f(x) = (x+1)/(x-2)
Modulus	f(x) = \|x\|	f(x) = \|x - 3\|
Signum	sgn(x) = -1, 0, or 1	sgn(x) = x/\|x\| for x≠0

Important Function Graphs:

[Graphs would be inserted here in actual implementation]

1. Linear: f(x) = ax + b

2. Quadratic: f(x) = ax² + bx + c

3. Modulus: f(x) = |x|

4. Greatest Integer: f(x) = [x]

4. Composition of Functions

For f: A → B and g: B → C, the composition g∘f: A → C is defined as:

(g∘f)(x) = g(f(x))

Properties

Associative: h∘(g∘f) = (h∘g)∘f
Not commutative: g∘f ≠ f∘g in general
If f and g are injective, then g∘f is injective
If f and g are surjective, then g∘f is surjective

5. Inverse of a Function

Definition: For a bijective function f: A → B, the inverse f^-1: B → A is defined by:

f^-1(y) = x ⇔ f(x) = y

f∘f^-1 = I_B and f^-1∘f = I_A

(g∘f)^-1 = f^-1∘g^-1

JEE Tip: To find if a function is invertible, check if it's bijective. Graphically, a function is invertible if any horizontal line intersects the graph at most once (Horizontal Line Test).

6. Binary Operations

Definition: A binary operation * on a set A is a function *: A × A → A.

Properties of Binary Operations

Property	Definition	Example
Commutative	ab = ba ∀ a,b ∈ A	Addition on R
Associative	(ab)c = a(bc)	Matrix multiplication
Identity	∃ e ∈ A such that ae = ea = a	0 for addition, 1 for multiplication
Inverse	For each a ∈ A, ∃ b ∈ A such that ab = ba = e	-a is additive inverse of a

7. Practice Problems for JEE

Let A = {1, 2, 3} and R = {(1,2), (2,3)}. Find the smallest relation containing R that is reflexive and transitive.
Show that the function f: R → R defined by f(x) = x/(x²+1) is neither one-one nor onto.
If f(x) = (3x+2)/(5x-3), prove that f∘f is an identity function.
Let * be a binary operation on Q defined by a*b = a + b - ab. Find the identity element and inverses.
Determine whether the relation R in R defined by R = {(a,b): a ≤ b³} is reflexive, symmetric or transitive.
Let f: N → N be defined by f(n) = n + 1 if n is odd, f(n) = n - 1 if n is even. Show that f is bijective.
Find the domain of the function f(x) = √(x - [x]), where [x] is greatest integer function.
If f: R → R is defined by f(x) = x² - 3x + 2, find f(f(x)).

Previous Year JEE Questions

JEE Main 2022: Let R be a relation on the set N of natural numbers defined by R = {(x,y): y = x + 5, x < 4}. Which of the following is false?
JEE Advanced 2021: Let f: R → R be defined by f(x) = x/(1 + x²). Then the range of f is?
JEE Main 2020: Let A = {1, 2, 3, 4, 5}. The number of onto functions from A to A such that f(x) ≠ x for all x ∈ A is?

Exam Strategy:

For function questions, always check domain restrictions first
Practice drawing graphs of standard functions quickly
Remember that composition of functions is associative but not commutative
For binary operations, verify properties step by step

Chapter 3: Trigonometric Functions - JEE Main & Advanced

1. Trigonometric Ratios

Trigonometric ratios are defined based on a right-angled triangle, where:

Sine: sin(θ) = Opposite / Hypotenuse
Cosine: cos(θ) = Adjacent / Hypotenuse
Tangent: tan(θ) = Opposite / Adjacent
Secant: sec(θ) = 1 / cos(θ)
Cosecant: csc(θ) = 1 / sin(θ)
Cotangent: cot(θ) = 1 / tan(θ)

Right Triangle Diagram:

[Right triangle with sides labeled: Opposite, Adjacent, Hypotenuse]

Angle θ at one vertex

2. Unit Circle Approach

The unit circle plays a vital role in defining trigonometric functions for all angles.

Unit Circle Diagram:

[Circle with radius 1 centered at origin]

Point P(x,y) on circle, angle θ from positive x-axis

sin(θ) = y, cos(θ) = x, tan(θ) = y/x

Key Points on Unit Circle

Angle (θ)	Coordinates (cosθ, sinθ)
0° or 0	(1, 0)
30° or π/6	(√3/2, 1/2)
45° or π/4	(√2/2, √2/2)
60° or π/3	(1/2, √3/2)
90° or π/2	(0, 1)

3. Trigonometric Identities

Pythagorean Identity: sin²(θ) + cos²(θ) = 1
Sum and Difference Formulas:
- sin(A ± B) = sinA cosB ± cosA sinB
- cos(A ± B) = cosA cosB ∓ sinA sinB
- tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB)
Double Angle Formulas:
- sin(2θ) = 2sin(θ)cos(θ)
- cos(2θ) = cos²(θ) - sin²(θ) = 2cos²(θ) - 1 = 1 - 2sin²(θ)
- tan(2θ) = 2tan(θ)/(1 - tan²(θ))
Half Angle Formulas:
- sin(θ/2) = ±√[(1 - cosθ)/2]
- cos(θ/2) = ±√[(1 + cosθ)/2]
- tan(θ/2) = ±√[(1 - cosθ)/(1 + cosθ)] = sinθ/(1 + cosθ) = (1 - cosθ)/sinθ

Chapter 4: Mathematical Induction - JEE Main & Advanced

1. Introduction to Mathematical Induction

What is Mathematical Induction?

Mathematical Induction is a powerful mathematical proof technique used to prove that a statement P(n) is true for all natural numbers n (n ∈ N).

1.1 The Principle of Mathematical Induction

The principle consists of two main steps:

Step 1 (Base Case): Prove that P(1) is true

Step 2 (Inductive Step): Assume P(k) is true for some k ∈ N, then prove that P(k+1) is true

Conclusion: If both steps hold, then P(n) is true for all n ∈ N

Induction Process Flow:

P(1) true → P(k) true ⇒ P(k+1) true → P(n) true ∀ n ∈ N

The process creates a "domino effect" where proving one case implies the next

2. First Principle of Mathematical Induction

2.1 Formal Statement

Let P(n) be a statement involving natural number n such that:

(i) P(1) is true, and

(ii) P(k) is true ⇒ P(k+1) is true for all k ∈ N

Then P(n) is true for all n ∈ N

2.2 Step-by-Step Procedure

Detailed Steps:

Base Case: Verify P(1) or P(n₀) for the smallest natural number
Inductive Hypothesis: Assume P(k) is true for some arbitrary k ∈ N
Inductive Step: Using the assumption, prove that P(k+1) is true
Conclusion: By principle of mathematical induction, P(n) is true ∀ n ∈ N

3. Proving Summation Formulas

3.1 Sum of First n Natural Numbers

Theorem:

1 + 2 + 3 + ... + n = n(n + 1)/2 for all n ∈ N

Proof:

Step 1: Base Case (n = 1)

LHS = 1

RHS = 1(1 + 1)/2 = 1

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

1 + 2 + 3 + ... + k = k(k + 1)/2

Step 3: Inductive Step (Prove P(k+1))

Consider 1 + 2 + 3 + ... + k + (k + 1)

= [1 + 2 + 3 + ... + k] + (k + 1)

= k(k + 1)/2 + (k + 1) [Using inductive hypothesis]

= (k + 1)[k/2 + 1]

= (k + 1)(k + 2)/2

= (k + 1)[(k + 1) + 1]/2

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, the formula is true for all n ∈ N

3.2 Sum of Squares of First n Natural Numbers

Theorem:

1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6 for all n ∈ N

Proof:

Step 1: Base Case (n = 1)

LHS = 1² = 1

RHS = 1(1 + 1)(2×1 + 1)/6 = (1×2×3)/6 = 1

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

1² + 2² + 3² + ... + k² = k(k + 1)(2k + 1)/6

Step 3: Inductive Step (Prove P(k+1))

Consider 1² + 2² + 3² + ... + k² + (k + 1)²

= [1² + 2² + 3² + ... + k²] + (k + 1)²

= k(k + 1)(2k + 1)/6 + (k + 1)² [Using inductive hypothesis]

= (k + 1)[k(2k + 1)/6 + (k + 1)]

= (k + 1)[(2k² + k) + 6(k + 1)]/6

= (k + 1)(2k² + 7k + 6)/6

= (k + 1)(k + 2)(2k + 3)/6

= (k + 1)[(k + 1) + 1][2(k + 1) + 1]/6

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, the formula is true for all n ∈ N

3.3 Sum of Cubes of First n Natural Numbers

Theorem:

1³ + 2³ + 3³ + ... + n³ = [n(n + 1)/2]² for all n ∈ N

Proof:

Step 1: Base Case (n = 1)

LHS = 1³ = 1

RHS = [1(1 + 1)/2]² = (2/2)² = 1

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

1³ + 2³ + 3³ + ... + k³ = [k(k + 1)/2]²

Step 3: Inductive Step (Prove P(k+1))

Consider 1³ + 2³ + 3³ + ... + k³ + (k + 1)³

= [1³ + 2³ + 3³ + ... + k³] + (k + 1)³

= [k(k + 1)/2]² + (k + 1)³ [Using inductive hypothesis]

= (k + 1)²[k²/4 + (k + 1)]

= (k + 1)²[(k² + 4k + 4)/4]

= (k + 1)²(k + 2)²/4

= [(k + 1)(k + 2)/2]²

= [((k + 1)((k + 1) + 1))/2]²

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, the formula is true for all n ∈ N

4. Divisibility Problems

4.1 Proving Divisibility by a Number

Theorem:

3²ⁿ - 1 is divisible by 8 for all n ∈ N

Proof:

Step 1: Base Case (n = 1)

3²⁽¹⁾ - 1 = 9 - 1 = 8, which is divisible by 8

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

3²ᵏ - 1 = 8m, where m ∈ Z

Step 3: Inductive Step (Prove P(k+1))

Consider 3²⁽ᵏ⁺¹⁾ - 1 = 3²ᵏ⁺² - 1

= 9·3²ᵏ - 1

= 9(3²ᵏ - 1) + 9 - 1

= 9(8m) + 8 [Using inductive hypothesis]

= 8(9m + 1)

Which is divisible by 8

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, 3²ⁿ - 1 is divisible by 8 for all n ∈ N

4.2 Another Divisibility Example

Theorem:

n³ + 2n is divisible by 3 for all n ∈ N

Proof:

Step 1: Base Case (n = 1)

1³ + 2(1) = 3, which is divisible by 3

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

k³ + 2k = 3m, where m ∈ Z

Step 3: Inductive Step (Prove P(k+1))

Consider (k + 1)³ + 2(k + 1)

= k³ + 3k² + 3k + 1 + 2k + 2

= (k³ + 2k) + 3k² + 3k + 3

= 3m + 3(k² + k + 1) [Using inductive hypothesis]

= 3(m + k² + k + 1)

Which is divisible by 3

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, n³ + 2n is divisible by 3 for all n ∈ N

5. Inequality Problems

5.1 Bernoulli's Inequality

Theorem:

(1 + x)ⁿ ≥ 1 + nx for all n ∈ N and x > -1

Proof:

Step 1: Base Case (n = 1)

LHS = 1 + x, RHS = 1 + x

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

(1 + x)ᵏ ≥ 1 + kx

Step 3: Inductive Step (Prove P(k+1))

Consider (1 + x)ᵏ⁺¹ = (1 + x)ᵏ(1 + x)

≥ (1 + kx)(1 + x) [Using inductive hypothesis and 1 + x > 0]

= 1 + (k + 1)x + kx²

≥ 1 + (k + 1)x [Since kx² ≥ 0]

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, (1 + x)ⁿ ≥ 1 + nx for all n ∈ N and x > -1

5.2 Another Inequality Example

Theorem:

2ⁿ > n for all n ∈ N

Proof:

Step 1: Base Case (n = 1)

2¹ = 2 > 1

∴ P(1) is true

Step 2: Inductive Hypothesis

Assume P(k) is true for some k ∈ N

2ᵏ > k

Step 3: Inductive Step (Prove P(k+1))

Consider 2ᵏ⁺¹ = 2·2ᵏ

> 2k [Using inductive hypothesis]

= k + k ≥ k + 1 [Since k ≥ 1]

∴ 2ᵏ⁺¹ > k + 1

∴ P(k+1) is true

Step 4: Conclusion

By mathematical induction, 2ⁿ > n for all n ∈ N

6. Second Principle of Mathematical Induction (Strong Induction)

6.1 Formal Statement

Let P(n) be a statement involving natural number n such that:

(i) P(1) is true, and

(ii) P(1), P(2), ..., P(k) are true ⇒ P(k+1) is true for all k ∈ N

Then P(n) is true for all n ∈ N

Key Difference:

In strong induction, we assume ALL previous cases (P(1) through P(k)) are true, not just P(k)

6.2 Application: Fundamental Theorem of Arithmetic

Theorem:

Every integer n > 1 can be expressed as a product of prime numbers

Proof Sketch (Using Strong Induction):

Base Case: n = 2 is prime, so it's a product of primes (itself)

Inductive Hypothesis: Assume every integer from 2 to k can be expressed as a product of primes

Inductive Step: Consider k+1

If k+1 is prime, it's already a product of primes
If k+1 is composite, then k+1 = ab where 1 < a, b ≤ k
By inductive hypothesis, both a and b are products of primes
Therefore, k+1 is also a product of primes

7. Common Mistakes and Pitfalls

Common Errors to Avoid:

Skipping Base Case: Always verify the base case
Wrong Base Case: Ensure you're starting with the correct initial value
Circular Reasoning: Don't assume what you're trying to prove
Incomplete Inductive Step: Make sure you actually use the inductive hypothesis
Assuming P(k+1) in Proof: You can only assume P(k), not P(k+1)

JEE Tip: In JEE problems, always clearly label:

Base Case
Inductive Hypothesis
Inductive Step
Conclusion

This makes your proof structured and easy to follow for examiners.

8. Practice Problems for JEE

Basic Level:

Prove that 1 + 3 + 5 + ... + (2n - 1) = n² for all n ∈ N
Prove that 1·2 + 2·3 + 3·4 + ... + n(n + 1) = n(n + 1)(n + 2)/3
Prove that 7ⁿ - 1 is divisible by 6 for all n ∈ N
Prove that n² > n + 1 for all n ≥ 2

Intermediate Level:

Prove that 1² + 3² + 5² + ... + (2n - 1)² = n(2n - 1)(2n + 1)/3
Prove that 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133 for all n ∈ N
Prove that n! > 2ⁿ for all n ≥ 4
Prove that 1/(1·2) + 1/(2·3) + ... + 1/[n(n + 1)] = n/(n + 1)

Advanced Level:

Prove that (1 + 1/1)(1 + 1/2)(1 + 1/3)...(1 + 1/n) = n + 1
Prove that cosθ·cos2θ·cos4θ...cos2ⁿ⁻¹θ = sin2ⁿθ/(2ⁿsinθ) for all n ∈ N
Prove that (1 + x)ⁿ ≥ 1 + nx + [n(n - 1)/2]x² for x ≥ 0 and n ∈ N
Prove that the number of subsets of a set with n elements is 2ⁿ

JEE Main Previous Year Question:

Prove by mathematical induction that for all n ∈ N:

1/(1·2·3) + 1/(2·3·4) + ... + 1/[n(n + 1)(n + 2)] = n(n + 3)/[4(n + 1)(n + 2)]

JEE Advanced Previous Year Question:

Let P(n) be the statement: "3²ⁿ⁺¹ + 2ⁿ⁺² is divisible by 7." Prove that P(n) is true for all n ∈ N.

9. Key Takeaways and Summary

Essential Points:

Mathematical induction is used to prove statements for all natural numbers
Always verify the base case (it may not be n = 1)
The inductive hypothesis must be used in the inductive step
Strong induction assumes all previous cases, not just the immediate predecessor
Practice various types: summation formulas, divisibility, inequalities

Exam Strategy:

Clearly label each step of your proof
Show all algebraic manipulations clearly
For divisibility problems, explicitly show the factor
For inequalities, ensure each step is justified
Always write a clear conclusion statement

Chapter 5: Complex Numbers - JEE Main & Advanced

1. Introduction to Complex Numbers

What are Complex Numbers?

Complex numbers are numbers of the form a + ib, where a and b are real numbers and i = √(-1) is the imaginary unit.

z = a + ib

Where: a = Real part (Re(z)), b = Imaginary part (Im(z))

1.1 Need for Complex Numbers

Complex numbers were introduced to solve equations that have no real solutions, particularly equations like x² + 1 = 0.

Historical Context:

The equation x² + 1 = 0 has no real solution because no real number squared gives -1.

By defining i = √(-1), we can write the solutions as x = ±i

2. Algebra of Complex Numbers

2.1 Equality of Complex Numbers

Two complex numbers z₁ = a + ib and z₂ = c + id are equal if and only if:

a = c and b = d

2.2 Operations on Complex Numbers

Addition:

(a + ib) + (c + id) = (a + c) + i(b + d)

Subtraction:

(a + ib) - (c + id) = (a - c) + i(b - d)

Multiplication:

(a + ib)(c + id) = (ac - bd) + i(ad + bc)

Division:

(a + ib)/(c + id) = [(ac + bd) + i(bc - ad)]/(c² + d²)

Example: Complex Arithmetic

Let z₁ = 3 + 4i, z₂ = 1 - 2i

z₁ + z₂ = (3 + 1) + i(4 - 2) = 4 + 2i

z₁ × z₂ = (3×1 - 4×(-2)) + i(3×(-2) + 4×1) = (3 + 8) + i(-6 + 4) = 11 - 2i

z₁/z₂ = [(3×1 + 4×(-2)) + i(4×1 - 3×(-2))]/(1² + (-2)²) = [(3 - 8) + i(4 + 6)]/5 = (-5 + 10i)/5 = -1 + 2i

3. Conjugate and Modulus of Complex Numbers

3.1 Complex Conjugate

If z = a + ib, then its conjugate is defined as: z̄ = a - ib

Properties of Conjugate:

z₁ + z₂ = z̄₁ + z̄₂
z₁ - z₂ = z̄₁ - z̄₂
z₁ × z₂ = z̄₁ × z̄₂
(z₁/z₂) = z̄₁/z̄₂ (z₂ ≠ 0)
(z̄)̄ = z
z + z̄ = 2Re(z)
z - z̄ = 2iIm(z)
z × z̄ = |z|²

3.2 Modulus of Complex Number

If z = a + ib, then its modulus is defined as: |z| = √(a² + b²)

Properties of Modulus:

|z| ≥ 0 and |z| = 0 ⇔ z = 0
|z₁ × z₂| = |z₁| × |z₂|
|z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0)
|z₁ + z₂| ≤ |z₁| + |z₂| (Triangle Inequality)
|z₁ - z₂| ≥ ||z₁| - |z₂||
|z̄| = |z|

Example: Conjugate and Modulus

Let z = 3 + 4i

Conjugate: z̄ = 3 - 4i

Modulus: |z| = √(3² + 4²) = √(9 + 16) = √25 = 5

z × z̄ = (3 + 4i)(3 - 4i) = 9 + 16 = 25 = |z|²

4. Argand Plane and Polar Representation

4.1 Argand Plane

The complex number z = x + iy can be represented as a point (x, y) in the Cartesian plane, called the Argand plane.

Argand Plane Diagram:

[Cartesian plane with x-axis as real axis, y-axis as imaginary axis]

Point P(x, y) represents complex number z = x + iy

Distance OP = |z| = √(x² + y²)

Angle θ = arg(z) = tan⁻¹(y/x)

4.2 Polar Form of Complex Number

z = r(cosθ + isinθ) = re^(iθ)

Where: r = |z| = √(x² + y²), θ = arg(z) = tan⁻¹(y/x)

Euler's Formula:

e^(iθ) = cosθ + isinθ

4.3 Argument of Complex Number

The argument of z (arg(z)) is the angle between the positive real axis and the line joining origin to z.

Principal Value of Argument:

The principal value of argument lies in the interval (-π, π]

For z = x + iy:

If x > 0: arg(z) = tan⁻¹(y/x)
If x < 0, y > 0: arg(z) = π + tan⁻¹(y/x)
If x < 0, y < 0: arg(z) = -π + tan⁻¹(y/x)
If x = 0, y > 0: arg(z) = π/2
If x = 0, y < 0: arg(z) = -π/2

5. De Moivre's Theorem and Applications

5.1 De Moivre's Theorem

(cosθ + isinθ)ⁿ = cos(nθ) + isin(nθ) for all n ∈ Z

Proof:

Using Euler's formula:

(cosθ + isinθ)ⁿ = (e^(iθ))ⁿ = e^(inθ) = cos(nθ) + isin(nθ)

5.2 Applications of De Moivre's Theorem

Finding nth Roots:

The nth roots of a complex number z = r(cosθ + isinθ) are given by:

z^(1/n) = r^(1/n)[cos((θ + 2kπ)/n) + isin((θ + 2kπ)/n)]

where k = 0, 1, 2, ..., n-1

Example: Cube Roots of Unity

Find the cube roots of 1:

1 = cos0 + isin0

Cube roots: cos(2kπ/3) + isin(2kπ/3) for k = 0, 1, 2

ω = cos(2π/3) + isin(2π/3) = -1/2 + i√3/2

ω² = cos(4π/3) + isin(4π/3) = -1/2 - i√3/2

Properties: 1 + ω + ω² = 0, ω³ = 1

5.3 Cube Roots of Unity Properties

1 + ω + ω² = 0
ω³ = 1
ω³ⁿ = 1, ω³ⁿ⁺¹ = ω, ω³ⁿ⁺² = ω²
|ω| = |ω²| = 1
ω̄ = ω²

6. Geometry in Complex Plane

6.1 Locus Problems

Common Loci:

|z - z₀| = r → Circle with center z₀ and radius r
|z - z₁| = |z - z₂| → Perpendicular bisector of segment joining z₁ and z₂
|z - z₁| + |z - z₂| = 2a (a > |z₁ - z₂|/2) → Ellipse with foci z₁, z₂
|z - z₁| - |z - z₂| = 2a (a < |z₁ - z₂|/2) → Hyperbola with foci z₁, z₂
arg((z - z₁)/(z - z₂)) = α → Arc of circle

6.2 Rotation Theorem

If z₁ and z₂ are two complex numbers, then:

z₂ - z₀ = e^(iθ)(z₁ - z₀)

represents rotation of z₁ about z₀ by angle θ to get z₂

Example: Rotation

If z₂ = iz₁, then z₂ is obtained by rotating z₁ by 90° counterclockwise about origin.

Since i = e^(iπ/2) = cos(π/2) + isin(π/2)

7. Practice Problems - Complex Numbers

Basic Level:

Simplify: (1 + i)/(1 - i)
Find modulus and argument of: √3 + i
Solve: z² + 2z + 5 = 0
If z = (1 + i)/(1 - i), find z⁴

Intermediate Level:

If |z - 3i| = 3, find the maximum value of |z + 2|
Find all roots of z⁴ + z³ + z² + z + 1 = 0
If z₁, z₂, z₃ are vertices of equilateral triangle, prove: z₁² + z₂² + z₃² = z₁z₂ + z₂z₃ + z₃z₁
If (1 + i)(1 + 2i)(1 + 3i)...(1 + ni) = x + iy, show that: 2·5·10·...·(1 + n²) = x² + y²

Advanced Level:

If z₁ and z₂ are two complex numbers such that |z₁| = |z₂| and arg(z₁) + arg(z₂) = π, then show that z₁ = -z̄₂
Find the locus of z if Re(z²) = 0
If |z| = 1, prove that |(z - 1)/(z + 1)| = |Im(z)|
If ω is cube root of unity, find value of (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)... to 2n factors

Chapter 6: Linear Inequalities - JEE Main & Advanced

1. Introduction to Inequalities

What are Inequalities?

Inequalities are mathematical statements that compare two expressions using inequality symbols:

> : Greater than
< : Less than
≥ : Greater than or equal to
≤ : Less than or equal to

1.1 Types of Inequalities

Type	Definition	Example
Linear Inequality	Inequality of degree 1	ax + b > 0
Quadratic Inequality	Inequality of degree 2	ax² + bx + c > 0
Rational Inequality	Inequality with rational expressions	(x + 1)/(x - 2) > 0
Absolute Value Inequality	Inequality with absolute values	\|x - 3\| < 5

2. Properties of Inequalities

Fundamental Properties:

Transitive Property: If a > b and b > c, then a > c
Addition Property: If a > b, then a + c > b + c
Subtraction Property: If a > b, then a - c > b - c
Multiplication Property:
- If a > b and c > 0, then ac > bc
- If a > b and c < 0, then ac < bc
Division Property:
- If a > b and c > 0, then a/c > b/c
- If a > b and c < 0, then a/c < b/c

Example: Properties Application

Given: 3x + 2 > 8

Subtract 2 from both sides: 3x > 6

Divide by 3 (positive): x > 2

Solution: x ∈ (2, ∞)

2.1 Important Results

a² ≥ 0 for all real a
a² + b² ≥ 2ab (AM-GM inequality)
(a + b)² ≥ 4ab
a/b + b/a ≥ 2 for a, b > 0

3. Solving Linear Inequalities

3.1 Single Variable Linear Inequalities

Example 1: Basic Linear Inequality

Solve: 2x - 3 < 7

2x - 3 < 7

2x < 10

x < 5

Solution: x ∈ (-∞, 5)

Example 2: Inequality with Negative Coefficient

Solve: -3x + 5 ≥ 11

-3x + 5 ≥ 11

-3x ≥ 6

x ≤ -2 (Inequality reverses when dividing by negative)

Solution: x ∈ (-∞, -2]

3.2 Double Inequalities

Example: Double Inequality

Solve: -4 < 2x - 3 ≤ 5

-4 < 2x - 3 ≤ 5

-1 < 2x ≤ 8 (Add 3 to all parts)

-1/2 < x ≤ 4 (Divide all parts by 2)

Solution: x ∈ (-1/2, 4]

4. Graphical Method for Linear Inequalities

4.1 Graphing Linear Inequalities in Two Variables

Steps for Graphing:

Graph the corresponding equation (replace inequality with equality)
Use dashed line for < or >, solid line for ≤ or ≥
Test a point not on the line (usually (0,0))
Shade the region that satisfies the inequality

Graphical Representation:

[Coordinate plane showing line and shaded region]

For 2x + 3y ≤ 6:

Line: 2x + 3y = 6

Test (0,0): 0 ≤ 6 ✓ (shade region containing origin)

4.2 System of Linear Inequalities

Example: System of Inequalities

Solve graphically:

x + y ≤ 4

x - y ≥ 1

x ≥ 0, y ≥ 0

Solution:

Graph each inequality and find the intersection of all shaded regions.

The solution is the feasible region (usually a polygon).

5. Quadratic Inequalities

5.1 Solving Quadratic Inequalities

Method:

Solve the corresponding quadratic equation ax² + bx + c = 0
Plot the roots on the number line
Determine the sign of the quadratic in each interval
Select intervals that satisfy the inequality

Example 1: x² - 5x + 6 > 0

Solve: x² - 5x + 6 = 0 → (x - 2)(x - 3) = 0 → x = 2, 3

Number line: ---(+)---2---(-)---3---(+)---

For > 0, we need positive regions: x ∈ (-∞, 2) ∪ (3, ∞)

Example 2: x² - 4x + 4 ≤ 0

Solve: x² - 4x + 4 = 0 → (x - 2)² = 0 → x = 2 (repeated root)

Since (x - 2)² is always ≥ 0, and equals 0 only at x = 2

Solution: x = 2

5.2 Wavy Curve Method (Sign Scheme Method)

Steps for Wavy Curve Method:

Factorize the expression completely
Mark critical points (where expression = 0 or undefined) on number line
Start from rightmost region with positive sign
Change sign when passing through odd multiplicity roots
Keep same sign when passing through even multiplicity roots

6. Rational Inequalities

6.1 Solving Rational Inequalities

Example: (x + 1)/(x - 2) > 0

Critical points: x = -1 (numerator zero), x = 2 (denominator zero, undefined)

Number line: ---(+)---(-1)---(-)---(2)---(+)---

For > 0, we need positive regions: x ∈ (-∞, -1) ∪ (2, ∞)

Example: (x² - 4)/(x - 1) ≤ 0

Factorize: [(x - 2)(x + 2)]/(x - 1) ≤ 0

Critical points: x = -2, 1, 2

Number line: ---(+)---(-2)---(-)---(1)---(+)---(2)---(-)---

For ≤ 0, we need negative or zero: x ∈ [-2, 1) ∪ [2, ∞)

Note: x = 1 is excluded (undefined)

7. Absolute Value Inequalities

7.1 Basic Absolute Value Inequalities

Important Results:

|x| < a ⇔ -a < x < a (for a > 0)
|x| > a ⇔ x < -a or x > a (for a > 0)
|x| ≤ a ⇔ -a ≤ x ≤ a (for a > 0)
|x| ≥ a ⇔ x ≤ -a or x ≥ a (for a > 0)

Example 1: |2x - 3| < 5

-5 < 2x - 3 < 5

-2 < 2x < 8

-1 < x < 4

Solution: x ∈ (-1, 4)

Example 2: |3x + 2| ≥ 4

3x + 2 ≤ -4 or 3x + 2 ≥ 4

3x ≤ -6 or 3x ≥ 2

x ≤ -2 or x ≥ 2/3

Solution: x ∈ (-∞, -2] ∪ [2/3, ∞)

7.2 Inequalities with Multiple Absolute Values

Example: |x - 1| + |x - 2| < 5

Critical points: x = 1, 2

Case 1: x < 1: -(x - 1) - (x - 2) < 5 → -2x + 3 < 5 → x > -1

Case 2: 1 ≤ x < 2: (x - 1) - (x - 2) < 5 → 1 < 5 (always true)

Case 3: x ≥ 2: (x - 1) + (x - 2) < 5 → 2x - 3 < 5 → x < 4

Solution: x ∈ (-1, 4)

8. AM-GM Inequality and Applications

8.1 Arithmetic Mean - Geometric Mean Inequality

For positive real numbers a₁, a₂, ..., aₙ:

(a₁ + a₂ + ... + aₙ)/n ≥ (a₁a₂...aₙ)^(1/n)

Equality holds when a₁ = a₂ = ... = aₙ

Example: Two Variable Case

For a, b > 0: (a + b)/2 ≥ √(ab)

This implies: a + b ≥ 2√(ab)

And: ab ≤ [(a + b)/2]²

8.2 Applications of AM-GM

Example 1: Maximum Product

If x + y = 10, find maximum value of xy

Using AM ≥ GM: (x + y)/2 ≥ √(xy)

10/2 ≥ √(xy) → 5 ≥ √(xy) → xy ≤ 25

Maximum product = 25 when x = y = 5

Example 2: Minimum Sum

If xy = 16, find minimum value of x + y

Using AM ≥ GM: (x + y)/2 ≥ √(xy) = √16 = 4

x + y ≥ 8

Minimum sum = 8 when x = y = 4

9. Practice Problems - Linear Inequalities

Basic Level:

Solve: 3x - 7 > 5x + 1
Solve: -2 ≤ (3x - 4)/2 < 5
Solve: x² - 5x + 6 < 0
Solve: |2x - 3| ≤ 7

Intermediate Level:

Solve: (x - 1)/(x + 2) > 0
Solve: |x - 1| + |x - 2| > 4
Solve graphically: x + 2y ≤ 6, 3x + 4y ≥ 12, x ≥ 0, y ≥ 0
If x is real, find the range of (x² + x + 1)/(x² - x + 1)

Advanced Level:

Solve: (x² - 4)/(x² - 2x - 3) ≥ 0
Find all real x satisfying: |x - 1| - |x - 2| + |x - 3| < 4
Using AM-GM, prove that for positive reals a, b, c: (a + b + c)(1/a + 1/b + 1/c) ≥ 9
Find the minimum value of 4x + 9y when xy = 36 and x, y > 0

JEE Strategy:

For inequality problems, always check critical points and test intervals
Remember to reverse inequality when multiplying/dividing by negative numbers
For absolute value inequalities, consider all cases
AM-GM is frequently used in optimization problems
Practice graphical method for linear programming problems

Chapter 7: Permutations and Combinations - JEE Main & Advanced

1. Fundamental Principle of Counting

1.1 Multiplication Principle

If an operation can be performed in m ways and following which another operation can be performed in n ways, then both operations can be performed in m × n ways.

Example:

If there are 3 shirts and 4 pants, then total outfits = 3 × 4 = 12

1.2 Addition Principle

If an operation can be performed in m ways and another operation can be performed in n ways, and both operations cannot be performed together, then either of the operations can be performed in m + n ways.

Example:

If you can go to Delhi by 3 trains or 2 flights, then total ways = 3 + 2 = 5

2. Factorial Notation

n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1

0! = 1 (by definition)

Properties of Factorials:

n! = n × (n-1)!
n! = n × (n-1) × (n-2)!
(2n)! = 2ⁿ × n! × 1 × 3 × 5 × ... × (2n-1)
n! ends with zeros = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...

Example:

5! = 5 × 4 × 3 × 2 × 1 = 120

Number of zeros in 100! = ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24

3. Permutations

3.1 Definition

A permutation is an arrangement of objects in a specific order.

ⁿPᵣ = n!/(n-r)!

Where: n = total objects, r = objects to arrange

3.2 Properties of ⁿPᵣ

ⁿP₀ = 1
ⁿP₁ = n
ⁿPₙ = n!
ⁿPᵣ = n × ⁿ⁻¹Pᵣ₋₁
ⁿPᵣ = ⁿ⁻¹Pᵣ + r × ⁿ⁻¹Pᵣ₋₁

Example:

⁵P₂ = 5!/(5-2)! = 120/6 = 20

Number of ways to arrange 3 books out of 7 = ⁷P₃ = 7 × 6 × 5 = 210

3.3 Permutations with Repetition

Number of permutations of n objects with repetition = nʳ

Example:

Number of 3-digit numbers = 9 × 10 × 10 = 900

4. Circular Permutations

4.1 Circular Arrangements

Number of circular permutations of n distinct objects = (n-1)!

Example:

Number of ways to seat 5 people around a circular table = (5-1)! = 4! = 24

4.2 Clockwise and Anti-clockwise

If clockwise and anti-clockwise arrangements are considered same, then:

Number of circular permutations = (n-1)!/2

Example:

Number of necklaces with 6 different beads = (6-1)!/2 = 5!/2 = 60

5. Combinations

5.1 Definition

A combination is a selection of objects where order doesn't matter.

ⁿCᵣ = n!/[r!(n-r)!]

5.2 Properties of ⁿCᵣ

ⁿC₀ = 1
ⁿC₁ = n
ⁿCₙ = 1
ⁿCᵣ = ⁿCₙ₋ᵣ
ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's Rule)
ⁿCᵣ = (n/r) × ⁿ⁻¹Cᵣ₋₁

Example:

⁵C₂ = 5!/(2!3!) = 120/(2×6) = 10

Number of ways to select 2 students from 5 = ⁵C₂ = 10

5.3 Important Results

Number of diagonals in n-sided polygon = ⁿC₂ - n
Number of handshakes among n people = ⁿC₂
ⁿCᵣ is maximum when:
- If n is even: r = n/2
- If n is odd: r = (n-1)/2 or (n+1)/2

6. Restricted Permutations and Combinations

6.1 Permutations with Restrictions

Example 1: Specific objects together

Number of ways to arrange 5 people with 2 specific persons together = 4! × 2! = 48

Example 2: Specific objects not together

Number of ways to arrange 5 people with 2 specific persons not together = Total - Together = 5! - 4! × 2! = 120 - 48 = 72

6.2 Combinations with Restrictions

Example 1: At least one

Number of ways to select at least one from n objects = 2ⁿ - 1

Example 2: At most one

Number of ways to select at most r from n objects = ⁿC₀ + ⁿC₁ + ... + ⁿCᵣ

7. Division and Distribution

7.1 Division into Groups

Case 1: Groups of different sizes (distinct groups)

Number of ways = n!/(n₁!n₂!...nₖ!)

Case 2: Groups of equal sizes (identical groups)

Number of ways = n!/[(n₁!n₂!...nₖ!) × k!]

Example:

Divide 10 students into groups of 4, 3, 3:

Number of ways = 10!/(4!3!3!) = 4200

7.2 Distribution Problems

Case 1: Distinct objects to distinct persons

Number of ways = mⁿ (if no restriction)

Case 2: Identical objects to distinct persons

Number of ways = ⁿ⁺ᵐ⁻¹Cₙ₋₁

8. Practice Problems - Permutations & Combinations

Basic Level:

Evaluate: ⁸P₃ and ⁸C₃
How many 3-digit numbers can be formed using 1,2,3,4,5 without repetition?
In how many ways can 5 books be arranged on a shelf?
How many triangles can be formed from 8 points, no three collinear?

Intermediate Level:

Find the number of ways to arrange letters of the word "ARRANGE"
In how many ways can 10 people be seated around a circular table?
How many words can be formed from "MATHEMATICS" with all vowels together?
Find the number of ways to distribute 10 identical balls in 4 distinct boxes

Advanced Level:

Find the number of ways to select a cricket team of 11 from 15 players if a particular player is always included and another is always excluded
How many 6-digit numbers contain exactly 3 even digits?
Find the number of ways to arrange 5 red, 4 blue, and 3 green balls in a row so that balls of same color are together
If ⁿC₈ = ⁿC₂, find ⁿC₂

Chapter 8: Binomial Theorem - JEE Main & Advanced

1. Binomial Theorem for Positive Integral Index

1.1 Binomial Theorem

(x + y)ⁿ = ⁿC₀xⁿy⁰ + ⁿC₁xⁿ⁻¹y¹ + ⁿC₂xⁿ⁻²y² + ... + ⁿCᵣxⁿ⁻ʳyʳ + ... + ⁿCₙx⁰yⁿ

Where n is a positive integer

Example: (a + b)³

(a + b)³ = ³C₀a³ + ³C₁a²b + ³C₂ab² + ³C₃b³ = a³ + 3a²b + 3ab² + b³

1.2 Observations

Total number of terms = n + 1
Sum of indices of x and y in each term = n
Coefficients are symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ
Coefficients follow Pascal's Triangle

2. General Term and Middle Term

2.1 General Term

Tᵣ₊₁ = ⁿCᵣxⁿ⁻ʳyʳ

This is the (r+1)th term in the expansion

Example:

Find 5th term in expansion of (2x - 3y)⁷

T₅ = T₄₊₁ = ⁷C₄(2x)³(-3y)⁴ = 35 × 8x³ × 81y⁴ = 22680x³y⁴

2.2 Middle Term

Case 1: When n is even

Middle term = T₍ₙ/₂₎₊₁

Case 2: When n is odd

Middle terms = T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂

Example:

Middle term of (x + y)⁶: n=6 (even), so middle term = T₄ = ⁶C₃x³y³ = 20x³y³

Middle terms of (x + y)⁷: n=7 (odd), so middle terms = T₄ and T₅ = ⁷C₃x⁴y³ and ⁷C₄x³y⁴ = 35x⁴y³ and 35x³y⁴

3. Independent Term and Greatest Coefficient

3.1 Term Independent of x

The term independent of x is the term where power of x is zero.

Example:

Find term independent of x in (x² + 1/x)⁶

Tᵣ₊₁ = ⁶Cᵣ(x²)⁶⁻ʳ(1/x)ʳ = ⁶Cᵣx¹²⁻³ʳ

For independent term: 12 - 3r = 0 ⇒ r = 4

T₅ = ⁶C₄ = 15

3.2 Greatest Coefficient

For (x + y)ⁿ, the greatest binomial coefficient is:

If n is even: ⁿCₙ/₂
If n is odd: ⁿC₍ₙ₋₁₎/₂ and ⁿC₍ₙ₊₁₎/₂

Example:

Greatest coefficient in (1 + x)¹⁰: n=10 (even), so greatest coefficient = ¹⁰C₅ = 252

3.3 Greatest Term

The numerically greatest term in (x + y)ⁿ can be found using:

If (n+1)|y|/(|x|+|y|) is an integer = m, then Tₘ and Tₘ₊₁ are greatest terms

Otherwise, T₍ₘ₊₁₎ is the greatest term, where m = ⌊(n+1)|y|/(|x|+|y|)⌋

4. Properties of Binomial Coefficients

4.1 Sum of Binomial Coefficients

ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ
ⁿC₀ + ⁿC₂ + ⁿC₄ + ... = 2ⁿ⁻¹
ⁿC₁ + ⁿC₃ + ⁿC₅ + ... = 2ⁿ⁻¹

Example:

⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = 2⁵ = 32

⁵C₀ + ⁵C₂ + ⁵C₄ = 1 + 10 + 5 = 16 = 2⁴

4.2 Other Important Results

ⁿC₀² + ⁿC₁² + ⁿC₂² + ... + ⁿCₙ² = ²ⁿCₙ
ⁿC₁ + 2·ⁿC₂ + 3·ⁿC₃ + ... + n·ⁿCₙ = n·2ⁿ⁻¹
ⁿC₀ - ⁿC₁ + ⁿC₂ - ⁿC₃ + ... + (-1)ⁿ·ⁿCₙ = 0
ⁿC₀ + 2·ⁿC₁ + 3·ⁿC₂ + ... + (n+1)·ⁿCₙ = (n+2)·2ⁿ⁻¹

5. Multinomial Theorem

5.1 Multinomial Expansion

(x₁ + x₂ + ... + xₖ)ⁿ = Σ [n!/(n₁!n₂!...nₖ!)] × x₁ⁿ¹x₂ⁿ²...xₖⁿᵏ

Where n₁ + n₂ + ... + nₖ = n

Example:

Find coefficient of x²y³z² in (x + y + z)⁷

Coefficient = 7!/(2!3!2!) = 210

5.2 Number of Terms

Number of terms in (x₁ + x₂ + ... + xₖ)ⁿ = ⁿ⁺ᵏ⁻¹Cₖ₋₁

Example:

Number of terms in (x + y + z)⁵ = ⁵⁺³⁻¹C₃₋₁ = ⁷C₂ = 21

6. Binomial Series for Rational Index

6.1 Binomial Theorem for Any Index

(1 + x)ⁿ = 1 + nx + [n(n-1)/2!]x² + [n(n-1)(n-2)/3!]x³ + ...

Valid when |x| < 1 and n is any rational number

6.2 Important Expansions

(1 + x)⁻¹ = 1 - x + x² - x³ + x⁴ - ...
(1 - x)⁻¹ = 1 + x + x² + x³ + x⁴ + ...
(1 + x)⁻² = 1 - 2x + 3x² - 4x³ + 5x⁴ - ...
(1 - x)⁻² = 1 + 2x + 3x² + 4x³ + 5x⁴ + ...

6.3 General Term

Tᵣ₊₁ = [n(n-1)(n-2)...(n-r+1)/r!] × xʳ

Example:

Find general term in expansion of (1 + x)⁻²

Tᵣ₊₁ = [(-2)(-3)(-4)...(-2-r+1)/r!] × xʳ = [(-1)ʳ(r+1)!/r!] × xʳ = (-1)ʳ(r+1)xʳ

7. Applications of Binomial Theorem

7.1 Approximations

Binomial theorem is used for numerical approximations.

Example:

Find approximate value of (1.01)⁵

(1.01)⁵ = (1 + 0.01)⁵ ≈ 1 + 5(0.01) + 10(0.01)² = 1 + 0.05 + 0.001 = 1.051

7.2 Divisibility Problems

Example:

Prove that 6ⁿ - 5n always leaves remainder 1 when divided by 25

6ⁿ = (1 + 5)ⁿ = 1 + 5n + 25k (where k is some integer)

6ⁿ - 5n = 1 + 25k, which leaves remainder 1 when divided by 25

7.3 Finding Remainders

Example:

Find remainder when 7¹⁰⁰ is divided by 9

7¹⁰⁰ = (9 - 2)¹⁰⁰ = 9k + 2¹⁰⁰

2¹⁰⁰ = (2³)³³ × 2 = 8³³ × 2 = (9 - 1)³³ × 2 = (9m - 1) × 2 = 18m - 2

7¹⁰⁰ = 9k + 18m - 2 = 9(k + 2m) - 2 = 9N + 7

Remainder = 7

8. Practice Problems - Binomial Theorem

Basic Level:

Expand: (2x - 3y)⁴
Find the middle term in expansion of (x + 1/x)¹⁰
Find the coefficient of x⁵ in (1 + 2x)⁸
Find the term independent of x in (x² - 1/x)⁶

Intermediate Level:

If the coefficients of x⁷ and x⁸ in (2 + x/3)ⁿ are equal, find n
Find the sum of coefficients in expansion of (1 - 3x + 2x²)⁵
Find the greatest term in expansion of (2 + 3x)⁹ when x = 3/2
Prove that: ⁵⁰C₀² + ⁵⁰C₁² + ... + ⁵⁰C₅₀² = ¹⁰⁰C₅₀

Advanced Level:

If (1 + x)ⁿ = C₀ + C₁x + C₂x² + ... + Cₙxⁿ, prove that: C₁ - 2C₂ + 3C₃ - ... + (-1)ⁿ⁻¹nCₙ = 0
Find the coefficient of x⁵ in expansion of (1 + x²)⁵(1 + x)⁴
If the three consecutive coefficients in (1 + x)ⁿ are 28, 56, 70, find n
Find approximate value of (0.99)¹⁵ using binomial theorem

JEE Strategy:

Remember the general term formula Tᵣ₊₁ = ⁿCᵣxⁿ⁻ʳyʳ
For term independent of x, set power of x to zero
Use properties of binomial coefficients for summation problems
Practice numerical approximations using binomial series
Learn to handle divisibility problems using binomial theorem

Chapter 9: Sequences and Series - JEE Main & Advanced

1. Introduction to Sequences and Series

1.1 Sequence

A sequence is an ordered list of numbers arranged in a specific pattern.

a₁, a₂, a₃, ..., aₙ, ...

1.2 Series

A series is the sum of terms of a sequence.

Sₙ = a₁ + a₂ + a₃ + ... + aₙ

1.3 Types of Sequences

Type	Definition	Example
Finite Sequence	Has limited number of terms	1, 3, 5, 7, 9
Infinite Sequence	Has unlimited number of terms	1, 1/2, 1/3, 1/4, ...
Arithmetic Sequence	Constant difference between terms	2, 5, 8, 11, ...
Geometric Sequence	Constant ratio between terms	3, 6, 12, 24, ...

2. Arithmetic Progression (AP)

2.1 Definition

A sequence where the difference between consecutive terms is constant.

a, a + d, a + 2d, a + 3d, ...

Where: a = first term, d = common difference

2.2 General Term of AP

nth term: aₙ = a + (n - 1)d

Derivation:

a₁ = a

a₂ = a + d

a₃ = a + 2d

...

aₙ = a + (n - 1)d

2.3 Sum of n Terms of AP

Sₙ = n/2 × [2a + (n - 1)d]

or Sₙ = n/2 × (first term + last term)

Derivation:

Sₙ = a + (a + d) + (a + 2d) + ... + [a + (n - 1)d]

Also, Sₙ = [a + (n - 1)d] + [a + (n - 2)d] + ... + a

Adding both: 2Sₙ = n[2a + (n - 1)d]

∴ Sₙ = n/2 × [2a + (n - 1)d]

2.4 Properties of AP

If a, b, c are in AP, then 2b = a + c
The sum of terms equidistant from beginning and end is constant
aₙ = Sₙ - Sₙ₋₁ for n ≥ 2
If each term of AP is increased/decreased/multiplied/divided by same number, the resulting sequence is also AP

Example:

Find 10th term and sum of first 10 terms of AP: 3, 7, 11, 15, ...

a = 3, d = 4

a₁₀ = 3 + (10 - 1)×4 = 3 + 36 = 39

S₁₀ = 10/2 × [2×3 + (10-1)×4] = 5 × [6 + 36] = 5 × 42 = 210

3. Geometric Progression (GP)

3.1 Definition

A sequence where the ratio between consecutive terms is constant.

a, ar, ar², ar³, ...

Where: a = first term, r = common ratio

3.2 General Term of GP

nth term: aₙ = arⁿ⁻¹

Derivation:

a₁ = a

a₂ = ar

a₃ = ar²

...

aₙ = arⁿ⁻¹

3.3 Sum of n Terms of GP

When r ≠ 1: Sₙ = a(1 - rⁿ)/(1 - r)

When r = 1: Sₙ = na

Derivation:

Sₙ = a + ar + ar² + ... + arⁿ⁻¹

rSₙ = ar + ar² + ... + arⁿ⁻¹ + arⁿ

Subtracting: Sₙ - rSₙ = a - arⁿ

Sₙ(1 - r) = a(1 - rⁿ)

∴ Sₙ = a(1 - rⁿ)/(1 - r)

3.4 Sum of Infinite GP

When |r| < 1: S∞ = a/(1 - r)

Derivation:

As n → ∞, rⁿ → 0 when |r| < 1

∴ S∞ = lim(n→∞) a(1 - rⁿ)/(1 - r) = a/(1 - r)

3.5 Properties of GP

If a, b, c are in GP, then b² = ac
The product of terms equidistant from beginning and end is constant
If each term of GP is multiplied/divided by same number, the resulting sequence is also GP

Example:

Find 6th term and sum of first 6 terms of GP: 2, 6, 18, 54, ...

a = 2, r = 3

a₆ = 2 × 3⁵ = 2 × 243 = 486

S₆ = 2(1 - 3⁶)/(1 - 3) = 2(1 - 729)/(-2) = 728

4. Harmonic Progression (HP)

4.1 Definition

A sequence where the reciprocals of terms form an AP.

1/a, 1/(a + d), 1/(a + 2d), ...

4.2 General Term of HP

If 1/a, 1/b, 1/c are in HP, then a, b, c are in AP

nth term of HP = 1/[a + (n - 1)d]

4.3 Harmonic Mean

Harmonic mean of two numbers a and b is:

HM = 2ab/(a + b)

Example:

Find 5th term of HP: 1/2, 1/5, 1/8, 1/11, ...

Reciprocals: 2, 5, 8, 11, ... (AP with a=2, d=3)

5th term of AP = 2 + (5-1)×3 = 14

5th term of HP = 1/14

5. Arithmetic, Geometric and Harmonic Means

5.1 Definitions

Arithmetic Mean (AM): (a + b)/2

Geometric Mean (GM): √(ab)

Harmonic Mean (HM): 2ab/(a + b)

5.2 Relationship between AM, GM, HM

AM ≥ GM ≥ HM

GM² = AM × HM

Proof:

For two positive numbers a and b:

AM = (a + b)/2, GM = √(ab), HM = 2ab/(a + b)

AM - GM = (a + b)/2 - √(ab) = (√a - √b)²/2 ≥ 0

∴ AM ≥ GM

GM² = ab, AM × HM = [(a + b)/2] × [2ab/(a + b)] = ab

∴ GM² = AM × HM

5.3 Insertion of Means

Arithmetic Means:

Between a and b, insert n AMs: d = (b - a)/(n + 1)

The AMs are: a + d, a + 2d, ..., a + nd

Geometric Means:

Between a and b, insert n GMs: r = (b/a)^(1/(n+1))

The GMs are: ar, ar², ..., arⁿ

Example:

Insert 3 AMs between 3 and 11

d = (11 - 3)/(3 + 1) = 8/4 = 2

AMs: 3+2=5, 3+4=7, 3+6=9

Sequence: 3, 5, 7, 9, 11

6. Special Series

6.1 Sum of First n Natural Numbers

1 + 2 + 3 + ... + n = n(n + 1)/2

6.2 Sum of Squares of First n Natural Numbers

1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6

Derivation (Method of Differences):

(n + 1)³ - n³ = 3n² + 3n + 1

n³ - (n - 1)³ = 3(n - 1)² + 3(n - 1) + 1

...

1³ - 0³ = 3(0)² + 3(0) + 1

Adding: (n + 1)³ = 3Σn² + 3Σn + (n + 1)

3Σn² = (n + 1)³ - 3[n(n + 1)/2] - (n + 1)

Simplifying: Σn² = n(n + 1)(2n + 1)/6

6.3 Sum of Cubes of First n Natural Numbers

1³ + 2³ + 3³ + ... + n³ = [n(n + 1)/2]²

6.4 Sum of Arithmetic-Geometric Series

S = a + (a + d)r + (a + 2d)r² + ... + [a + (n - 1)d]rⁿ⁻¹

S = a/(1 - r) + dr(1 - rⁿ⁻¹)/(1 - r)² - [a + (n - 1)d]rⁿ/(1 - r)

7. Practice Problems - Sequences and Series

Basic Level:

Find 15th term of AP: 7, 12, 17, 22, ...
Find sum of first 20 terms of AP: 3, 7, 11, 15, ...
Find 8th term of GP: 2, 6, 18, 54, ...
Find sum of infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...

Intermediate Level:

If the sum of n terms of AP is 3n² + 5n, find its 10th term
Insert 4 arithmetic means between 3 and 18
Find the sum: 1² + 3² + 5² + ... to n terms
If a, b, c are in GP, prove that a² + b², ab + bc, b² + c² are in GP

Advanced Level:

Find the sum: 1×2 + 2×3 + 3×4 + ... + n(n + 1)
If S₁, S₂, S₃ are sums of n, 2n, 3n terms of an AP, prove that 3(S₂ - S₁) = S₃
Find the sum: 1/1×2 + 1/2×3 + 1/3×4 + ... + 1/n(n + 1)
If a, b, c are in AP; b, c, d are in GP; c, d, e are in HP, prove that a, c, e are in GP

Chapter 10: Straight Lines - JEE Main & Advanced

1. Coordinate Geometry Basics

1.1 Cartesian Coordinate System

A system that uses coordinates to represent points in a plane.

Cartesian Plane:

[Coordinate plane with x-axis, y-axis, and four quadrants]

Point P(x, y) where x = abscissa, y = ordinate

1.2 Distance Formula

Distance between P(x₁, y₁) and Q(x₂, y₂):

PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]

Derivation:

Using Pythagoras theorem:

PQ² = (x₂ - x₁)² + (y₂ - y₁)²

∴ PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]

1.3 Section Formula

Point dividing P(x₁, y₁) and Q(x₂, y₂) in ratio m:n:

Internally: [(mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)]

Externally: [(mx₂ - nx₁)/(m - n), (my₂ - ny₁)/(m - n)]

Example:

Find point dividing (2, 3) and (4, 7) in ratio 2:1 internally

x = (2×4 + 1×2)/(2 + 1) = (8 + 2)/3 = 10/3

y = (2×7 + 1×3)/(2 + 1) = (14 + 3)/3 = 17/3

Point: (10/3, 17/3)

1.4 Midpoint Formula

Midpoint of P(x₁, y₁) and Q(x₂, y₂):

[(x₁ + x₂)/2, (y₁ + y₂)/2]

1.5 Area of Triangle

Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):

Δ = 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

2. Slope of a Line

2.1 Definition

Slope (m) of a line is the tangent of the angle it makes with positive x-axis.

m = tanθ

2.2 Slope Formula

Slope through points (x₁, y₁) and (x₂, y₂):

m = (y₂ - y₁)/(x₂ - x₁)

2.3 Angle between Two Lines

If slopes are m₁ and m₂, then angle between lines:

tanθ = |(m₁ - m₂)/(1 + m₁m₂)|

Derivation:

Let lines make angles θ₁ and θ₂ with x-axis

m₁ = tanθ₁, m₂ = tanθ₂

θ = θ₁ - θ₂

tanθ = tan(θ₁ - θ₂) = (tanθ₁ - tanθ₂)/(1 + tanθ₁tanθ₂)

∴ tanθ = (m₁ - m₂)/(1 + m₁m₂)

2.4 Conditions for Parallel and Perpendicular Lines

Parallel lines: m₁ = m₂
Perpendicular lines: m₁ × m₂ = -1

Example:

Find slope of line perpendicular to line joining (1, 2) and (3, 4)

Slope of given line = (4 - 2)/(3 - 1) = 2/2 = 1

Slope of perpendicular line = -1/1 = -1

3. Various Forms of Straight Line Equations

3.1 Slope-Intercept Form

y = mx + c

Where: m = slope, c = y-intercept

Slope-Intercept Form:

[Line with slope m and y-intercept c]

3.2 Point-Slope Form

y - y₁ = m(x - x₁)

Where: (x₁, y₁) is a point on line, m = slope

3.3 Two-Point Form

(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)

Where: (x₁, y₁) and (x₂, y₂) are two points on line

3.4 Intercept Form

x/a + y/b = 1

Where: a = x-intercept, b = y-intercept

Intercept Form:

[Line cutting x-axis at (a, 0) and y-axis at (0, b)]

3.5 Normal Form

x cosα + y sinα = p

Where: p = perpendicular distance from origin, α = angle with x-axis

Example:

Find equation of line with slope 2 and y-intercept 3

y = 2x + 3 (Slope-intercept form)

Find equation of line through (1, 2) with slope 3

y - 2 = 3(x - 1) ⇒ y = 3x - 1

4. General Equation of Straight Line

4.1 Standard Form

Ax + By + C = 0

Where A, B, C are constants and A, B are not both zero

4.2 Conversion to Other Forms

Slope-Intercept Form:

y = (-A/B)x - C/B (when B ≠ 0)

Slope m = -A/B, y-intercept c = -C/B

Intercept Form:

x/(-C/A) + y/(-C/B) = 1 (when C ≠ 0)

x-intercept = -C/A, y-intercept = -C/B

Normal Form:

x cosα + y sinα = p

Where: cosα = A/√(A² + B²), sinα = B/√(A² + B²), p = -C/√(A² + B²)

Example:

Convert 3x + 4y - 12 = 0 to intercept form

3x + 4y = 12

x/4 + y/3 = 1

x-intercept = 4, y-intercept = 3

5. Distance Formulas

5.1 Distance from a Point to a Line

Distance from point (x₁, y₁) to line Ax + By + C = 0:

d = |Ax₁ + By₁ + C|/√(A² + B²)

Derivation:

Let the perpendicular from (x₁, y₁) to line meet at (x₂, y₂)

Slope of line = -A/B, slope of perpendicular = B/A

Using point-slope form and solving gives the formula

5.2 Distance between Two Parallel Lines

Distance between lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0:

d = |C₁ - C₂|/√(A² + B²)

Example:

Find distance from (1, 2) to line 3x + 4y - 5 = 0

d = |3(1) + 4(2) - 5|/√(3² + 4²) = |3 + 8 - 5|/5 = 6/5

Find distance between 3x + 4y - 5 = 0 and 3x + 4y + 10 = 0

d = |-5 - 10|/√(3² + 4²) = 15/5 = 3

6. Concurrent Lines and Family of Lines

6.1 Concurrent Lines

Three or more lines are concurrent if they pass through a common point.

Condition for Concurrency:

Lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 are concurrent if:

|a₁ b₁ c₁; a₂ b₂ c₂; a₃ b₃ c₃| = 0

Example:

Check if lines x + 2y - 3 = 0, 2x + 3y - 4 = 0, 3x + 4y - 5 = 0 are concurrent

|1 2 -3; 2 3 -4; 3 4 -5| = 1(3×(-5) - 4×(-4)) - 2(2×(-5) - 3×(-4)) + (-3)(2×4 - 3×3)

= 1(-15 + 16) - 2(-10 + 12) - 3(8 - 9) = 1 - 4 + 3 = 0

∴ Lines are concurrent

6.2 Family of Lines

Lines through Intersection of Two Lines:

(a₁x + b₁y + c₁) + λ(a₂x + b₂y + c₂) = 0

This represents all lines passing through the intersection of given lines

Example:

Find equation of line through intersection of x + y - 1 = 0 and 2x - y + 3 = 0 and parallel to 3x + 4y = 0

Family: (x + y - 1) + λ(2x - y + 3) = 0

Slope = -(1 + 2λ)/(1 - λ) = -3/4 (parallel to given line)

Solving: λ = -7/2

Equation: 3x + 4y - 13 = 0

7. Angle Bisectors

7.1 Equation of Angle Bisectors

For lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the angle bisectors are:

(a₁x + b₁y + c₁)/√(a₁² + b₁²) = ± (a₂x + b₂y + c₂)/√(a₂² + b₂²)

Angle Bisectors:

[Two lines with their acute and obtuse angle bisectors]

7.2 Identifying Acute and Obtuse Angle Bisectors

Method:

If a₁a₂ + b₁b₂ > 0, then:

Positive sign gives obtuse angle bisector
Negative sign gives acute angle bisector

If a₁a₂ + b₁b₂ < 0, then:

Positive sign gives acute angle bisector
Negative sign gives obtuse angle bisector

Example:

Find the angle bisectors of lines 3x + 4y - 7 = 0 and 12x - 5y + 8 = 0

√(3² + 4²) = 5, √(12² + (-5)²) = 13

(3x + 4y - 7)/5 = ± (12x - 5y + 8)/13

Positive sign: 39x + 52y - 91 = 60x - 25y + 40 ⇒ 21x - 77y + 131 = 0

Negative sign: 39x + 52y - 91 = -60x + 25y - 40 ⇒ 99x + 27y - 51 = 0

8. Pair of Straight Lines

8.1 General Equation

ax² + 2hxy + by² + 2gx + 2fy + c = 0

represents a pair of straight lines if:

abc + 2fgh - af² - bg² - ch² = 0

8.2 Homogeneous Equation

ax² + 2hxy + by² = 0 represents two lines through origin

Slopes: m₁, m₂ = [-h ± √(h² - ab)]/b

8.3 Angle between Lines

Angle between lines represented by ax² + 2hxy + by² = 0:

tanθ = |2√(h² - ab)/(a + b)|

Example:

Find the angle between lines represented by 6x² - 5xy - 6y² = 0

a = 6, h = -2.5, b = -6

tanθ = |2√(6.25 + 36)/(6 - 6)| = ∞

∴ θ = 90° (lines are perpendicular)

8.4 Condition for Parallel and Perpendicular Lines

Parallel lines: h² = ab
Perpendicular lines: a + b = 0

9. Practice Problems - Straight Lines

Basic Level:

Find distance between points (3, 4) and (7, 1)
Find equation of line through (2, 3) with slope 4
Find slope of line perpendicular to 2x + 3y = 7
Convert 4x - 3y + 12 = 0 to intercept form

Intermediate Level:

Find distance from (2, -1) to line 3x - 4y + 5 = 0
Find equation of line through intersection of x + 2y = 3 and 3x - y = 4, parallel to x-axis
Find the acute angle between lines 2x + y = 3 and x - 3y = 5
Find the angle bisectors of lines x + y - 1 = 0 and 7x - y + 5 = 0

Advanced Level:

Prove that lines 3x + 4y = 10, 4x - 3y = 5, and x = 1 are concurrent
Find the equation of line through (1, 2) making equal intercepts on axes
A line through (2, 3) makes angle 45° with x-axis. Find its equation and distance from (4, 1)
Find the pair of lines through origin perpendicular to x² - 5xy + 4y² = 0

Chapter 11: Conic Sections - JEE Main & Advanced

1. Introduction to Conic Sections

1.1 Definition and Formation

Conic sections are curves obtained by intersecting a right circular cone with a plane at different angles.

Types: Circle, Parabola, Ellipse, and Hyperbola

Formation of Conic Sections:

[Diagram showing cone intersected by planes at different angles]

Circle: Plane perpendicular to axis
Parabola: Plane parallel to generator
Ellipse: Plane oblique to axis (not parallel to generator)
Hyperbola: Plane parallel to axis

1.2 General Second Degree Equation

General equation: ax² + 2hxy + by² + 2gx + 2fy + c = 0

Discriminant (Δ) = abc + 2fgh - af² - bg² - ch²

Nature of Conic:

Δ = 0: Pair of straight lines
Δ ≠ 0 and h² = ab: Parabola
Δ ≠ 0 and h² < ab: Ellipse
Δ ≠ 0 and h² > ab: Hyperbola
Δ ≠ 0, h² < ab, and a = b: Circle

2. Circle

2.1 Definition

A circle is the locus of a point that moves in a plane such that its distance from a fixed point (center) is constant (radius).

2.2 Standard Equations

Standard form: (x - h)² + (y - k)² = r²

Where: Center = (h, k), Radius = r

General form: x² + y² + 2gx + 2fy + c = 0

Where: Center = (-g, -f), Radius = √(g² + f² - c)

2.3 Parametric Form

x = h + r cosθ, y = k + r sinθ

Where θ is the parameter (0 ≤ θ < 2π)

2.4 Important Results

Diameter form: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
Tangent at (x₁, y₁): xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
Length of tangent: √(x₁² + y₁² + 2gx₁ + 2fy₁ + c)
Power of point: PT² = (x₁² + y₁² + 2gx₁ + 2fy₁ + c)

Example:

Find center and radius of circle: x² + y² - 6x + 8y - 11 = 0

Comparing with x² + y² + 2gx + 2fy + c = 0

2g = -6 ⇒ g = -3, 2f = 8 ⇒ f = 4, c = -11

Center = (-g, -f) = (3, -4)

Radius = √(g² + f² - c) = √(9 + 16 + 11) = √36 = 6

3. Parabola

3.1 Definition

A parabola is the locus of a point that moves in a plane such that its distance from a fixed point (focus) equals its distance from a fixed line (directrix).

3.2 Standard Parabolas

y² = 4ax (Right opening)

Focus: (a, 0)
Directrix: x = -a
Axis: y = 0
Vertex: (0, 0)
Latus Rectum: 4a

x² = 4ay (Upward opening)

Focus: (0, a)
Directrix: y = -a
Axis: x = 0
Vertex: (0, 0)
Latus Rectum: 4a

3.3 Parametric Form

For y² = 4ax: x = at², y = 2at

Parameter: t (any real number)

3.4 Tangent and Normal

Tangent at (x₁, y₁): yy₁ = 2a(x + x₁)
Tangent at parameter t: ty = x + at²
Normal at (x₁, y₁): y - y₁ = -y₁/2a (x - x₁)
Normal at parameter t: y + tx = 2at + at³

Example:

Find focus, directrix and latus rectum of parabola y² = 12x

Comparing with y² = 4ax: 4a = 12 ⇒ a = 3

Focus: (a, 0) = (3, 0)

Directrix: x = -a = -3

Latus Rectum = 4a = 12 units

4. Ellipse

4.1 Definition

An ellipse is the locus of a point that moves in a plane such that the sum of its distances from two fixed points (foci) is constant and greater than the distance between foci.

4.2 Standard Equation

x²/a² + y²/b² = 1 (a > b)

Center: (0, 0)
Vertices: (±a, 0)
Foci: (±c, 0) where c² = a² - b²
Directrices: x = ±a/e
Eccentricity: e = c/a (0 < e < 1)
Length of Latus Rectum: 2b²/a

x²/b² + y²/a² = 1 (a > b)

Center: (0, 0)
Vertices: (0, ±a)
Foci: (0, ±c) where c² = a² - b²
Directrices: y = ±a/e
Eccentricity: e = c/a (0 < e < 1)
Length of Latus Rectum: 2b²/a

4.3 Parametric Form

x = a cosθ, y = b sinθ

Where θ is the parameter (0 ≤ θ < 2π)

4.4 Important Properties

Sum of focal distances = 2a (constant)
Area of ellipse = πab
Perimeter ≈ π[3(a + b) - √{(3a + b)(a + 3b)}] (Ramanujan's approximation)
Tangent at (x₁, y₁): xx₁/a² + yy₁/b² = 1
Tangent at parameter θ: (x cosθ)/a + (y sinθ)/b = 1

Example:

Find eccentricity and foci of ellipse: 9x² + 25y² = 225

Dividing by 225: x²/25 + y²/9 = 1

a² = 25 ⇒ a = 5, b² = 9 ⇒ b = 3

c² = a² - b² = 25 - 9 = 16 ⇒ c = 4

Eccentricity e = c/a = 4/5 = 0.8

Foci: (±c, 0) = (±4, 0)

5. Hyperbola

5.1 Definition

A hyperbola is the locus of a point that moves in a plane such that the difference of its distances from two fixed points (foci) is constant.

5.2 Standard Equation

x²/a² - y²/b² = 1

Center: (0, 0)
Vertices: (±a, 0)
Foci: (±c, 0) where c² = a² + b²
Directrices: x = ±a/e
Eccentricity: e = c/a (e > 1)
Length of Latus Rectum: 2b²/a
Asymptotes: y = ±(b/a)x

y²/a² - x²/b² = 1

Center: (0, 0)
Vertices: (0, ±a)
Foci: (0, ±c) where c² = a² + b²
Directrices: y = ±a/e
Eccentricity: e = c/a (e > 1)
Length of Latus Rectum: 2b²/a
Asymptotes: y = ±(a/b)x

5.3 Parametric Form

x = a secθ, y = b tanθ

or x = a coshθ, y = b sinhθ

Where θ is the parameter

5.4 Rectangular Hyperbola

xy = c²

Vertices: (c, c) and (-c, -c)
Foci: (√2c, √2c) and (-√2c, -√2c)
Eccentricity: e = √2
Asymptotes: x = 0, y = 0 (coordinate axes)

Example:

Find eccentricity and asymptotes of hyperbola: 16x² - 9y² = 144

Dividing by 144: x²/9 - y²/16 = 1

a² = 9 ⇒ a = 3, b² = 16 ⇒ b = 4

c² = a² + b² = 9 + 16 = 25 ⇒ c = 5

Eccentricity e = c/a = 5/3

Asymptotes: y = ±(b/a)x = ±(4/3)x

6. Practice Problems - Conic Sections

Basic Level:

Find center and radius of circle: x² + y² - 4x + 6y - 3 = 0
Find focus and directrix of parabola: y² = 16x
Find eccentricity of ellipse: 4x² + 9y² = 36
Find asymptotes of hyperbola: x²/16 - y²/9 = 1

Intermediate Level:

Find equation of circle with center (2, -3) and passing through (5, 1)
Find equation of tangent to parabola y² = 8x at point (2, 4)
Find foci and vertices of ellipse: 9x² + 16y² = 144
Find equation of hyperbola with foci (±5, 0) and vertices (±3, 0)

Advanced Level:

Find locus of point of intersection of perpendicular tangents to parabola y² = 4ax
Prove that the chord of contact of tangents from (h, k) to ellipse x²/a² + y²/b² = 1 is hx/a² + ky/b² = 1
Find angle between asymptotes of hyperbola x²/a² - y²/b² = 1
A variable chord of circle x² + y² = 4 subtends right angle at vertex of parabola y² = 4x. Find locus of midpoint of chord.

Chapter 12: Three Dimensional Geometry - JEE Main & Advanced

1. Coordinates in Three Dimensions

1.1 3D Coordinate System

A system with three mutually perpendicular axes: x-axis, y-axis, and z-axis.

Point P(x, y, z) where x = abscissa, y = ordinate, z = applicate

3D Coordinate System:

[Diagram showing octants with coordinates]

Octants: I: (+, +, +), II: (-, +, +), III: (-, -, +), IV: (+, -, +), V: (+, +, -), VI: (-, +, -), VII: (-, -, -), VIII: (+, -, -)

1.2 Distance Formula

Distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):

PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

1.3 Section Formula

Point dividing P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in ratio m:n:

Internally: [(mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n)]

Externally: [(mx₂ - nx₁)/(m - n), (my₂ - ny₁)/(m - n), (mz₂ - nz₁)/(m - n)]

Example:

Find distance between points (1, 2, 3) and (4, 6, 8)

Distance = √[(4 - 1)² + (6 - 2)² + (8 - 3)²] = √[9 + 16 + 25] = √50 = 5√2

2. Direction Cosines and Ratios

2.1 Direction Cosines

If a line makes angles α, β, γ with x, y, z axes respectively, then:

l = cosα, m = cosβ, n = cosγ

l² + m² + n² = 1

2.2 Direction Ratios

If a, b, c are proportional to direction cosines, then:

l = a/√(a² + b² + c²), m = b/√(a² + b² + c²), n = c/√(a² + b² + c²)

2.3 Angle between Two Lines

If direction ratios are (a₁, b₁, c₁) and (a₂, b₂, c₂), then:

cosθ = (a₁a₂ + b₁b₂ + c₁c₂)/[√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)]

Conditions:

Parallel lines: a₁/a₂ = b₁/b₂ = c₁/c₂
Perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Example:

Find direction cosines of line with direction ratios 2, -1, 2

√(2² + (-1)² + 2²) = √(4 + 1 + 4) = √9 = 3

l = 2/3, m = -1/3, n = 2/3

3. Equation of a Line in 3D

3.1 Various Forms

Vector form:

r = a + λb

Where: a = position vector of point, b = direction vector, λ = parameter

Cartesian form:

(x - x₁)/a = (y - y₁)/b = (z - z₁)/c

Where: (x₁, y₁, z₁) is a point, (a, b, c) are direction ratios

Two-point form:

(x - x₁)/(x₂ - x₁) = (y - y₁)/(y₂ - y₁) = (z - z₁)/(z₂ - z₁)

Where: (x₁, y₁, z₁) and (x₂, y₂, z₂) are two points

3.2 Angle between Two Lines

If lines are: (x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁ and (x - x₂)/a₂ = (y - y₂)/b₂ = (z - z₂)/c₂

Then: cosθ = (a₁a₂ + b₁b₂ + c₁c₂)/[√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)]

3.3 Shortest Distance between Skew Lines

For lines: r = a₁ + λb₁ and r = a₂ + μb₂

Shortest distance = |(a₂ - a₁) · (b₁ × b₂)|/|b₁ × b₂|

Example:

Find equation of line through (1, 2, 3) with direction ratios 2, -1, 4

(x - 1)/2 = (y - 2)/(-1) = (z - 3)/4

4. Equation of a Plane

4.1 Various Forms

General form:

ax + by + cz + d = 0

Where: (a, b, c) are direction ratios of normal to plane

Normal form:

lx + my + nz = p

Where: (l, m, n) are direction cosines of normal, p = perpendicular distance from origin

Intercept form:

x/a + y/b + z/c = 1

Where: a, b, c are intercepts on x, y, z axes respectively

Three-point form:

|x y z 1; x₁ y₁ z₁ 1; x₂ y₂ z₂ 1; x₃ y₃ z₃ 1| = 0

4.2 Angle between Two Planes

If planes are: a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0

Then: cosθ = (a₁a₂ + b₁b₂ + c₁c₂)/[√(a₁² + b₁² + c₁²) √(a₂² + b₂² + c₂²)]

4.3 Angle between Line and Plane

If line: (x - x₁)/a = (y - y₁)/b = (z - z₁)/c and plane: ax + by + cz + d = 0

Then: sinφ = (aa + bb + cc)/[√(a² + b² + c²) √(a² + b² + c²)]

Example:

Find equation of plane with intercepts 2, 3, 4 on x, y, z axes

x/2 + y/3 + z/4 = 1

6x + 4y + 3z = 12 (multiplying by 12)

5. Coplanarity of Lines

5.1 Condition for Coplanarity

Two lines: r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar if:

(a₂ - a₁) · (b₁ × b₂) = 0

5.2 Equation of Plane containing Two Lines

If lines are coplanar, equation of plane containing them is:

|x - x₁ y - y₁ z - z₁; a₁ b₁ c₁; a₂ b₂ c₂| = 0

Example:

Check if lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5 are coplanar

Points: (1, 2, 3) and (2, 3, 4), Direction ratios: (2, 3, 4) and (3, 4, 5)

(a₂ - a₁) = (1, 1, 1), b₁ × b₂ = |i j k; 2 3 4; 3 4 5| = (-1, 2, -1)

(a₂ - a₁) · (b₁ × b₂) = (1)(-1) + (1)(2) + (1)(-1) = 0

∴ Lines are coplanar

6. Distance Formulas in 3D

6.1 Distance from Point to Plane

Distance from point (x₁, y₁, z₁) to plane ax + by + cz + d = 0:

d = |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²)

6.2 Distance between Parallel Planes

Distance between planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0:

d = |d₁ - d₂|/√(a² + b² + c²)

6.3 Distance between Point and Line

Distance from point P to line through A with direction vector b:

d = |AP × b|/|b|

6.4 Distance between Two Skew Lines

For lines: r = a₁ + λb₁ and r = a₂ + μb₂

Shortest distance = |(a₂ - a₁) · (b₁ × b₂)|/|b₁ × b₂|

Example:

Find distance from point (1, 2, 3) to plane 2x - y + 2z - 4 = 0

d = |2(1) - 1(2) + 2(3) - 4|/√(4 + 1 + 4) = |2 - 2 + 6 - 4|/3 = |2|/3 = 2/3

7. Sphere

7.1 Definition

A sphere is the locus of a point that moves in space such that its distance from a fixed point (center) is constant (radius).

7.2 Standard Equations

Standard form: (x - h)² + (y - k)² + (z - l)² = r²

Where: Center = (h, k, l), Radius = r

General form: x² + y² + z² + 2ux + 2vy + 2wz + d = 0

Where: Center = (-u, -v, -w), Radius = √(u² + v² + w² - d)

7.3 Equation of Tangent Plane

Tangent plane at (x₁, y₁, z₁) to sphere x² + y² + z² + 2ux + 2vy + 2wz + d = 0:

xx₁ + yy₁ + zz₁ + u(x + x₁) + v(y + y₁) + w(z + z₁) + d = 0

Example:

Find center and radius of sphere: x² + y² + z² - 4x + 6y - 8z + 4 = 0

Comparing with x² + y² + z² + 2ux + 2vy + 2wz + d = 0

2u = -4 ⇒ u = -2, 2v = 6 ⇒ v = 3, 2w = -8 ⇒ w = -4, d = 4

Center = (-u, -v, -w) = (2, -3, 4)

Radius = √(u² + v² + w² - d) = √(4 + 9 + 16 - 4) = √25 = 5

8. Practice Problems - 3D Geometry

Basic Level:

Find distance between points (1, -2, 3) and (4, 1, -2)
Find direction cosines of line with direction ratios 1, -2, 2
Find equation of plane with intercepts 3, 4, 5 on coordinate axes
Find center and radius of sphere: x² + y² + z² - 2x + 4y - 6z + 8 = 0

Intermediate Level:

Find angle between lines: (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5
Find distance from point (2, -1, 3) to plane 3x - 4y + 12z - 5 = 0
Find equation of line through (1, 2, 3) parallel to line (x-2)/3 = (y+1)/4 = (z-5)/2
Check if points (1, 2, 3), (2, 3, 4), (3, 4, 5) are collinear

Advanced Level:

Find shortest distance between lines: r = i + j + λ(2i - j + k) and r = 2i + j - k + μ(3i - 5j + 2k)
Find equation of plane through intersection of planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0 and perpendicular to plane 5x + 3y + 6z + 8 = 0
Find locus of point equidistant from points (1, 2, 3) and (3, 2, -1)
A variable plane is at constant distance p from origin and meets axes at A, B, C. Show that locus of centroid of triangle ABC is 1/x² + 1/y² + 1/z² = 9/p²

Chapter 13: Limits and Derivatives - JEE Main & Advanced

1. Introduction to Limits

1.1 Concept of Limit

The limit of a function f(x) as x approaches a is the value that f(x) approaches as x gets closer to a.

lim_x→a f(x) = L

1.2 Left Hand and Right Hand Limits

Left Hand Limit (LHL): lim_x→a⁻ f(x)

Right Hand Limit (RHL): lim_x→a⁺ f(x)

Limit exists if: LHL = RHL = lim_x→a f(x)

1.3 Basic Limits

lim_x→a c = c (where c is constant)
lim_x→a x = a
lim_x→a xⁿ = aⁿ
lim_x→a [f(x) ± g(x)] = lim_x→a f(x) ± lim_x→a g(x)
lim_x→a [f(x) · g(x)] = lim_x→a f(x) · lim_x→a g(x)
lim_x→a [f(x)/g(x)] = lim_x→a f(x)/lim_x→a g(x), if lim_x→a g(x) ≠ 0

Example:

Evaluate lim_x→2 (x² + 3x - 1)

= (2)² + 3(2) - 1 = 4 + 6 - 1 = 9

2. Evaluation of Limits

2.1 Factorization Method

For 0/0 form: Factorize numerator and denominator, cancel common factors

2.2 Rationalization Method

For forms involving radicals: Multiply numerator and denominator by conjugate

2.3 Standard Limits

lim_x→0 sinx/x = 1
lim_x→0 tanx/x = 1
lim_x→0 (eˣ - 1)/x = 1
lim_x→0 (aˣ - 1)/x = logₑa
lim_x→0 (1 + x)¹/ˣ = e
lim_x→∞ (1 + 1/x)ˣ = e
lim_x→0 log(1 + x)/x = 1

Example:

Evaluate lim_x→0 (sin3x)/x

= lim_x→0 3(sin3x)/(3x) = 3 × 1 = 3

3. Continuity

3.1 Definition

A function f(x) is continuous at x = a if:

f(a) exists
lim_x→a f(x) exists
lim_x→a f(x) = f(a)

3.2 Types of Discontinuity

Removable discontinuity: Limit exists but f(a) ≠ limit or f(a) not defined
Jump discontinuity: LHL and RHL exist but are unequal
Infinite discontinuity: Function approaches ±∞
Oscillatory discontinuity: Function oscillates between values

3.3 Continuous Functions

All polynomial functions are continuous everywhere
Rational functions are continuous except where denominator = 0
Trigonometric functions are continuous in their domains
Exponential functions are continuous everywhere
Logarithmic functions are continuous in their domains

Example:

Check continuity of f(x) = (x² - 4)/(x - 2) at x = 2

f(2) is not defined ⇒ Discontinuous at x = 2

But lim_x→2 f(x) = lim_x→2 (x + 2) = 4 exists

∴ Removable discontinuity

4. Introduction to Derivatives

4.1 Definition

The derivative of f(x) at x = a is:

f'(a) = lim_h→0 [f(a + h) - f(a)]/h

It represents the instantaneous rate of change of f(x) at x = a

4.2 Geometrical Interpretation

Derivative as Slope:

[Graph showing tangent line to curve at point P]

f'(a) = slope of tangent to y = f(x) at point (a, f(a))

4.3 First Principle of Derivatives

f'(x) = lim_h→0 [f(x + h) - f(x)]/h

Example:

Find derivative of f(x) = x² from first principles

f'(x) = lim_h→0 [(x + h)² - x²]/h

= lim_h→0 [x² + 2xh + h² - x²]/h

= lim_h→0 (2x + h) = 2x

5. Rules of Differentiation

5.1 Standard Derivatives

d/dx (c) = 0 (c is constant)
d/dx (xⁿ) = nxⁿ⁻¹ (Power Rule)
d/dx (sinx) = cosx
d/dx (cosx) = -sinx
d/dx (tanx) = sec²x
d/dx (cotx) = -cosec²x
d/dx (secx) = secx tanx
d/dx (cosecx) = -cosecx cotx
d/dx (eˣ) = eˣ
d/dx (aˣ) = aˣ logₑa
d/dx (logₑx) = 1/x
d/dx (logₐx) = 1/(x logₑa)

5.2 Algebra of Derivatives

Sum Rule: d/dx [u(x) ± v(x)] = u'(x) ± v'(x)

Product Rule: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)

Quotient Rule: d/dx [u(x)/v(x)] = [u'(x)v(x) - u(x)v'(x)]/[v(x)]²

5.3 Chain Rule

If y = f(u) and u = g(x), then:

dy/dx = dy/du · du/dx

Example:

Differentiate y = (2x³ + 5x)⁴

Let u = 2x³ + 5x, then y = u⁴

dy/dx = 4u³ · (6x² + 5) = 4(2x³ + 5x)³(6x² + 5)

6. Implicit and Parametric Differentiation

6.1 Implicit Differentiation

For equations where y is not explicitly expressed in terms of x:

Differentiate both sides w.r.t. x, treating y as function of x

6.2 Parametric Differentiation

If x = f(t) and y = g(t), then:

dy/dx = (dy/dt)/(dx/dt)

6.3 Logarithmic Differentiation

For functions of type [f(x)]ᵍ⁽ˣ⁾ or products/quotients:

Take log on both sides, then differentiate

Example:

Find dy/dx if x² + y² = 25

Differentiating: 2x + 2y(dy/dx) = 0

∴ dy/dx = -x/y

7. Higher Order Derivatives

7.1 Second Order Derivative

d²y/dx² = d/dx (dy/dx)

Represents rate of change of slope

7.2 nth Order Derivative

dⁿy/dxⁿ = d/dx (dⁿ⁻¹y/dxⁿ⁻¹)

7.3 Standard Higher Derivatives

dⁿ/dxⁿ (xᵐ) = m(m-1)...(m-n+1)xᵐ⁻ⁿ
dⁿ/dxⁿ (eᵃˣ) = aⁿeᵃˣ
dⁿ/dxⁿ (sin(ax+b)) = aⁿ sin(ax+b + nπ/2)
dⁿ/dxⁿ (cos(ax+b)) = aⁿ cos(ax+b + nπ/2)
dⁿ/dxⁿ (1/(ax+b)) = (-1)ⁿ n! aⁿ/(ax+b)ⁿ⁺¹

Example:

Find second derivative of y = x³ + 2x² - 5x + 1

dy/dx = 3x² + 4x - 5

d²y/dx² = 6x + 4

8. Practice Problems - Limits and Derivatives

Basic Level:

Evaluate lim_x→2 (x² - 4)/(x - 2)
Find derivative of f(x) = 3x⁴ - 2x³ + 5x - 1
Evaluate lim_x→0 (sin5x)/x
Differentiate y = √(x² + 1)

Intermediate Level:

Evaluate lim_x→0 (eˣ - 1 - x)/x²
Find dy/dx if x³ + y³ = 3axy
Check continuity of f(x) = |x| at x = 0
Differentiate y = xˣ using logarithmic differentiation

Advanced Level:

Evaluate lim_x→0 (tanx - sinx)/x³
If y = sin(m sin⁻¹x), prove that (1 - x²)y₂ - xy₁ + m²y = 0
Find nth derivative of y = 1/(2x + 3)
Evaluate lim_x→∞ [√(x² + x + 1) - √(x² + 1)]

Chapter 14: Mathematical Reasoning - JEE Main & Advanced

1. Statements

1.1 Definition

A statement is a declarative sentence which is either true or false, but not both simultaneously.

1.2 Types of Statements

Simple statement: Cannot be broken into smaller statements
Compound statement: Formed by combining two or more simple statements using connectives
Atomic statement: Simple statement with no connectives
Molecular statement: Compound statement with connectives

1.3 Truth Value

Every statement has a truth value:

True (T or 1)
False (F or 0)

Example:

Identify which are statements:

"New Delhi is capital of India" - Statement (True)
"Close the door" - Not a statement (command)
"x + 5 = 10" - Not a statement (open sentence)
"The sun rises in the east" - Statement (True)

2. Logical Connectives

2.1 Negation (NOT)

If p is a statement, then negation of p is denoted by ~p or ¬p

Truth table:

p	~p
T	F
F	T

2.2 Conjunction (AND)

p ∧ q is true only when both p and q are true

Truth table:

p	q	p ∧ q
T	T	T
T	F	F
F	T	F
F	F	F

2.3 Disjunction (OR)

p ∨ q is false only when both p and q are false

Truth table:

p	q	p ∨ q
T	T	T
T	F	T
F	T	T
F	F	F

2.4 Conditional (IMPLIES)

p → q is false only when p is true and q is false

Truth table:

p	q	p → q
T	T	T
T	F	F
F	T	T
F	F	T

2.5 Biconditional (IF AND ONLY IF)

p ↔ q is true when both p and q have same truth values

Truth table:

p	q	p ↔ q
T	T	T
T	F	F
F	T	F
F	F	T

Example:

Let p: "It is raining", q: "I carry an umbrella"

p → q: "If it is raining, then I carry an umbrella"

~p: "It is not raining"

p ∧ q: "It is raining and I carry an umbrella"

3. Tautology and Contradiction

3.1 Tautology

A compound statement that is always true, regardless of the truth values of its components.

3.2 Contradiction

A compound statement that is always false, regardless of the truth values of its components.

3.3 Contingency

A compound statement that is neither a tautology nor a contradiction.

Example:

Verify if p ∨ ~p is a tautology

p	~p	p ∨ ~p
T	F	T
F	T	T

Always true ⇒ Tautology

4. Quantifiers

4.1 Universal Quantifier (∀)

∀ means "for all" or "for every"

Example: ∀x ∈ R, x² ≥ 0 (For all real numbers x, x² is non-negative)

4.2 Existential Quantifier (∃)

∃ means "there exists" or "for some"

Example: ∃x ∈ R such that x² = 4 (There exists a real number x such that x² = 4)

4.3 Negation of Quantified Statements

~[∀x, p(x)] ≡ ∃x such that ~p(x)
~[∃x such that p(x)] ≡ ∀x, ~p(x)

Example:

Write negation of: "All students passed the exam"

Negation: "There exists at least one student who did not pass the exam"

5. Implications and Related Concepts

5.1 Converse, Inverse, and Contrapositive

For conditional statement p → q:

Converse: q → p
Inverse: ~p → ~q
Contrapositive: ~q → ~p

5.2 Equivalence of Statements

Conditional and its contrapositive are logically equivalent
Converse and inverse are logically equivalent
p → q ≡ ~p ∨ q
~(p → q) ≡ p ∧ ~q

Example:

Let p → q: "If it rains, then the ground is wet"

Converse: "If the ground is wet, then it rains"

Inverse: "If it does not rain, then the ground is not wet"

Contrapositive: "If the ground is not wet, then it does not rain"

6. Validity of Statements

6.1 Valid Statement

A statement is valid if its conclusion logically follows from its premises.

6.2 Methods to Check Validity

Truth table method: Construct truth table and check
Direct proof: Assume premises are true and show conclusion must be true
Contrapositive proof: Prove contrapositive
Contradiction method: Assume conclusion is false and derive contradiction

6.3 Common Valid Argument Forms

Modus Ponens: [(p → q) ∧ p] → q
Modus Tollens: [(p → q) ∧ ~q] → ~p
Hypothetical Syllogism: [(p → q) ∧ (q → r)] → (p → r)
Disjunctive Syllogism: [(p ∨ q) ∧ ~p] → q

Example:

Check validity: "If it rains, I carry umbrella. It is raining. Therefore, I carry umbrella."

This is Modus Ponens - Valid argument

7. Practice Problems - Mathematical Reasoning

Basic Level:

Identify which are statements: "x + 3 = 8", "Close the window", "2 + 2 = 4"
Write negation of "All prime numbers are odd"
Construct truth table for (p ∧ q) → p
Write converse of "If a number is divisible by 6, then it is divisible by 3"

Intermediate Level:

Show that (p → q) ↔ (~q → ~p) is a tautology
Write contrapositive of "If a triangle is equilateral, then it is isosceles"
Check validity: "If I study, I will pass. I passed. Therefore, I studied."
Write negation of "∃x ∈ R such that x² < 0"

Advanced Level:

Prove that [(p → q) ∧ (q → r)] → (p → r) is a tautology
Show that (p → q) ∧ (p → r) is equivalent to p → (q ∧ r)
Check validity using truth table: (p → q) ∧ (r → s) ∧ (p ∨ r) → (q ∨ s)
Write the dual of (p ∧ q) ∨ (~p ∧ r)

Chapter 15: Statistics - JEE Main & Advanced

1. Introduction to Statistics

1.1 Basic Concepts

Statistics: Science of collecting, organizing, analyzing, interpreting, and presenting data.

Data: Collection of facts and figures.

1.2 Types of Data

Ungrouped Data: Raw data without any organization
Grouped Data: Data organized into classes or intervals
Discrete Data: Countable values (e.g., number of students)
Continuous Data: Measurable values (e.g., height, weight)

1.3 Frequency Distribution

Frequency: Number of times a value occurs

Cumulative Frequency: Sum of all frequencies up to that class

Example:

Create frequency distribution for: 5, 8, 6, 5, 7, 8, 9, 5, 6, 7

Value	Frequency
5	3
6	2
7	2
8	2
9	1

2. Measures of Central Tendency

2.1 Mean (Arithmetic Mean)

Ungrouped Data: Mean = (∑xᵢ)/n

Grouped Data: Mean = (∑fᵢxᵢ)/(∑fᵢ)

Assumed Mean Method: Mean = A + (∑fᵢdᵢ)/(∑fᵢ), where dᵢ = xᵢ - A

Step Deviation Method: Mean = A + h(∑fᵢuᵢ)/(∑fᵢ), where uᵢ = (xᵢ - A)/h

2.2 Median

Ungrouped Data: Middle value when data is arranged in order

For n observations: Median = [(n+1)/2]th term (if n is odd)

Median = [n/2th term + (n/2+1)th term]/2 (if n is even)

Grouped Data: Median = l + [(n/2 - cf)/f] × h

Where: l = lower limit of median class, n = total frequency, cf = cumulative frequency of class before median class, f = frequency of median class, h = class width

2.3 Mode

Ungrouped Data: Value that occurs most frequently

Grouped Data: Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h

Where: l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class preceding modal class, f₂ = frequency of class succeeding modal class, h = class width

Example:

Find mean, median, mode of: 2, 3, 4, 5, 5, 6, 7

Mean = (2+3+4+5+5+6+7)/7 = 32/7 = 4.57

Median = 5 (4th term)

Mode = 5 (occurs twice)

3. Measures of Dispersion

3.1 Range

Range = Maximum value - Minimum value

3.2 Mean Deviation

Ungrouped Data: MD = (∑|xᵢ - mean|)/n

Grouped Data: MD = (∑fᵢ|xᵢ - mean|)/(∑fᵢ)

MD about median = (∑fᵢ|xᵢ - median|)/(∑fᵢ)

3.3 Variance and Standard Deviation

Variance (σ²): Average of squared deviations from mean

Ungrouped Data: σ² = (∑(xᵢ - mean)²)/n

Grouped Data: σ² = (∑fᵢ(xᵢ - mean)²)/(∑fᵢ)

Standard Deviation (σ): √Variance

3.4 Coefficient of Variation

CV = (σ/mean) × 100%

Used to compare variability of different datasets

Example:

Find variance of: 2, 4, 6, 8

Mean = (2+4+6+8)/4 = 5

Variance = [(2-5)² + (4-5)² + (6-5)² + (8-5)²]/4 = (9+1+1+9)/4 = 20/4 = 5

Standard Deviation = √5 ≈ 2.236

4. Correlation and Regression

4.1 Correlation

Measures the strength and direction of relationship between two variables

Karl Pearson's Coefficient: r = (∑(xᵢ - x̄)(yᵢ - ȳ))/√[∑(xᵢ - x̄)² ∑(yᵢ - ȳ)²]

Simplified: r = [n∑xy - (∑x)(∑y)]/√[n∑x² - (∑x)²][n∑y² - (∑y)²]

4.2 Regression

Regression line of y on x: y - ȳ = bᵧₓ(x - x̄)

Where: bᵧₓ = r(σᵧ/σₓ) = [n∑xy - (∑x)(∑y)]/[n∑x² - (∑x)²]

Regression line of x on y: x - x̄ = bₓᵧ(y - ȳ)

Where: bₓᵧ = r(σₓ/σᵧ) = [n∑xy - (∑x)(∑y)]/[n∑y² - (∑y)²]

4.3 Properties

-1 ≤ r ≤ 1
r = 1: Perfect positive correlation
r = -1: Perfect negative correlation
r = 0: No correlation
r² = bᵧₓ × bₓᵧ (Coefficient of determination)

Example:

Find correlation coefficient for:

x: 1, 2, 3, 4, 5 and y: 2, 4, 6, 8, 10

∑x = 15, ∑y = 30, ∑xy = 110, ∑x² = 55, ∑y² = 220, n = 5

r = [5×110 - 15×30]/√[(5×55 - 225)(5×220 - 900)] = [550-450]/√[(275-225)(1100-900)] = 100/√[50×200] = 100/100 = 1

Perfect positive correlation

5. Probability Basics

5.1 Basic Concepts

Random Experiment: Experiment with uncertain outcome

Sample Space (S): Set of all possible outcomes

Event (E): Subset of sample space

5.2 Probability Definition

Classical Definition: P(E) = n(E)/n(S)

Axiomatic Definition:

0 ≤ P(E) ≤ 1
P(S) = 1
For mutually exclusive events, P(E₁ ∪ E₂) = P(E₁) + P(E₂)

5.3 Basic Theorems

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

P(A') = 1 - P(A)

P(A ∩ B') = P(A) - P(A ∩ B)

Example:

Find probability of getting even number when die is thrown

S = {1,2,3,4,5,6}, E = {2,4,6}

P(E) = n(E)/n(S) = 3/6 = 1/2

6. Conditional Probability

6.1 Definition

Probability of event A given that event B has occurred:

P(A|B) = P(A ∩ B)/P(B), where P(B) ≠ 0

6.2 Multiplication Theorem

P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)

6.3 Independent Events

Events A and B are independent if:

P(A ∩ B) = P(A) × P(B)

or P(A|B) = P(A)

or P(B|A) = P(B)

6.4 Bayes' Theorem

If E₁, E₂, ..., Eₙ are mutually exclusive and exhaustive events, then:

P(Eᵢ|A) = [P(Eᵢ) × P(A|Eᵢ)] / [∑P(Eⱼ) × P(A|Eⱼ)]

Example:

In a class, 60% are boys, 40% are girls. 30% of boys and 50% of girls wear glasses. If a student wears glasses, find probability it's a girl.

P(Girl|Glasses) = [P(Girl)×P(Glasses|Girl)] / [P(Boy)×P(Glasses|Boy) + P(Girl)×P(Glasses|Girl)]

= [0.4×0.5] / [0.6×0.3 + 0.4×0.5] = 0.2 / [0.18 + 0.2] = 0.2/0.38 ≈ 0.526

7. Random Variables and Probability Distributions

7.1 Random Variable

A function that assigns a real number to each outcome of a random experiment.

Discrete Random Variable: Takes countable values

Continuous Random Variable: Takes values in an interval

7.2 Probability Distribution

For discrete random variable X with values x₁, x₂, ..., xₙ:

∑P(X = xᵢ) = 1

7.3 Mean and Variance

Mean (Expected Value): E(X) = ∑xᵢP(X = xᵢ)

Variance: Var(X) = E(X²) - [E(X)]² = ∑xᵢ²P(X = xᵢ) - [∑xᵢP(X = xᵢ)]²

Standard Deviation: σ = √Var(X)

7.4 Binomial Distribution

P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, where q = 1 - p

Mean = np, Variance = npq

Example:

Find probability of getting exactly 3 heads in 5 tosses of fair coin

n = 5, r = 3, p = 0.5, q = 0.5

P(X = 3) = ⁵C₃ (0.5)³ (0.5)² = 10 × 0.125 × 0.25 = 0.3125

8. Practice Problems - Statistics

Basic Level:

Find mean, median, mode of: 4, 6, 8, 10, 12, 12, 15
Calculate variance and standard deviation of: 2, 4, 6, 8, 10
Find probability of drawing a red card from deck of 52 cards
In a binomial distribution with n=6, p=0.4, find P(X=3)

Intermediate Level:

Find correlation coefficient for: x: 10, 20, 30, 40 and y: 15, 25, 35, 45
For grouped data with classes 0-10, 10-20, 20-30 with frequencies 5, 8, 7, find mean and median
If P(A)=0.4, P(B)=0.5, P(A∩B)=0.2, find P(A|B) and check if A and B are independent
Find regression line of y on x for: x: 1,2,3,4 and y: 2,4,5,8

Advanced Level:

Prove that Var(aX + b) = a²Var(X)
Using Bayes' theorem, solve: In a factory, machines A, B, C produce 50%, 30%, 20% items. Their defect rates are 2%, 3%, 4%. If an item is defective, find probability it came from machine A.
For normal distribution with mean 50 and SD 10, find P(40 ≤ X ≤ 60)
Show that for two variables, byx × bxy = r²

Chapter 16: Probability - JEE Main & Advanced

1. Advanced Probability Concepts

1.1 Sample Space for Various Experiments

Coin Toss: S = {H, T} for one coin
Die Throw: S = {1,2,3,4,5,6} for one die
Cards: 52 cards in deck (13 each of ♠♥♦♣)
Birthday Problem: S = {Jan 1, Jan 2, ..., Dec 31}

1.2 Algebra of Events

Complement: A' = S - A

Union: A ∪ B (A or B or both)

Intersection: A ∩ B (Both A and B)

Difference: A - B = A ∩ B'

Symmetric Difference: A Δ B = (A - B) ∪ (B - A)

1.3 De Morgan's Laws

(A ∪ B)' = A' ∩ B'

(A ∩ B)' = A' ∪ B'

Example:

In throwing two dice, find probability of getting sum 7 or 11

Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes

Sum 11: (5,6),(6,5) → 2 outcomes

Total favorable = 8, Total outcomes = 36

P = 8/36 = 2/9

2. Addition Theorems

2.1 For Two Events

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

2.2 For Three Events

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

2.3 Mutually Exclusive Events

If A ∩ B = φ, then P(A ∪ B) = P(A) + P(B)

2.4 Exhaustive Events

If A₁ ∪ A₂ ∪ ... ∪ Aₙ = S, then ∑P(Aᵢ) = 1

Example:

If P(A)=0.5, P(B)=0.4, P(A∩B)=0.2, find P(A∪B)

P(A∪B) = 0.5 + 0.4 - 0.2 = 0.7

3. Conditional Probability (Advanced)

3.1 Properties

0 ≤ P(A|B) ≤ 1
P(S|B) = 1
P(A₁ ∪ A₂|B) = P(A₁|B) + P(A₂|B) - P(A₁ ∩ A₂|B)
P(A'|B) = 1 - P(A|B)

3.2 Multiplication Rule for Multiple Events

P(A ∩ B ∩ C) = P(A) × P(B|A) × P(C|A ∩ B)

3.3 Total Probability Theorem

If B₁, B₂, ..., Bₙ form a partition of S, then for any event A:

P(A) = ∑P(Bᵢ) × P(A|Bᵢ)

3.4 Bayes' Theorem (General Form)

P(Bᵢ|A) = [P(Bᵢ) × P(A|Bᵢ)] / [∑P(Bⱼ) × P(A|Bⱼ)]

Example:

Three urns contain: Urn1: 2W,3B; Urn2: 4W,1B; Urn3: 3W,4B. An urn is chosen at random and a ball is drawn. If it's white, find probability it came from Urn1.

P(Urn1|White) = [(1/3)×(2/5)] / [(1/3)×(2/5) + (1/3)×(4/5) + (1/3)×(3/7)]

= (2/15) / (2/15 + 4/15 + 3/21) = (2/15) / (6/15 + 1/7) = (2/15) / (42/105 + 15/105) = (2/15)/(57/105) = 14/57

4. Independent Events

4.1 Pairwise Independence

Events A, B, C are pairwise independent if:

P(A ∩ B) = P(A)P(B), P(B ∩ C) = P(B)P(C), P(C ∩ A) = P(C)P(A)

4.2 Mutual Independence

Events A, B, C are mutually independent if:

P(A ∩ B) = P(A)P(B), P(B ∩ C) = P(B)P(C), P(C ∩ A) = P(C)P(A), and P(A ∩ B ∩ C) = P(A)P(B)P(C)

4.3 Properties

If A and B are independent, then A' and B are independent
If A and B are independent, then A' and B' are independent
Mutual independence implies pairwise independence, but converse is not true

Example:

Three coins are tossed. Let A: Head on first, B: Head on second, C: Exactly two heads. Check independence.

P(A)=1/2, P(B)=1/2, P(C)=3/8

P(A∩B)=1/4 = P(A)P(B) ⇒ A,B independent

P(A∩C)=1/4 ≠ P(A)P(C)=3/16 ⇒ Not mutually independent

5. Probability Distributions

5.1 Bernoulli Distribution

P(X = 1) = p, P(X = 0) = 1 - p

Mean = p, Variance = p(1-p)

5.2 Binomial Distribution

X ~ B(n, p): P(X = r) = ⁿCᵣ pʳ (1-p)ⁿ⁻ʳ

Mean = np, Variance = np(1-p)

Mode: Integer between (n+1)p - 1 and (n+1)p

5.3 Poisson Distribution

P(X = r) = (e⁻λ λʳ)/r!, r = 0,1,2,...

Mean = λ, Variance = λ

Used for rare events

5.4 Normal Distribution

f(x) = (1/σ√2π) e^{-(x-μ)²/2σ²}

Mean = μ, Variance = σ²

Standard Normal: Z = (X-μ)/σ ~ N(0,1)

Example:

If X ~ B(10, 0.4), find P(X ≥ 3)

P(X ≥ 3) = 1 - P(X ≤ 2) = 1 - [P(X=0)+P(X=1)+P(X=2)]

= 1 - [⁰C₀(0.6)¹⁰ + ¹⁰C₁(0.4)(0.6)⁹ + ¹⁰C₂(0.4)²(0.6)⁸]

= 1 - [0.0060 + 0.0403 + 0.1209] = 1 - 0.1672 = 0.8328

6. Random Variables (Advanced)

6.1 Probability Mass Function (PMF)

For discrete random variable X:

p(xᵢ) = P(X = xᵢ) such that ∑p(xᵢ) = 1

6.2 Probability Density Function (PDF)

For continuous random variable X:

f(x) ≥ 0 and ∫f(x)dx = 1

P(a ≤ X ≤ b) = ∫_a^b f(x)dx

6.3 Cumulative Distribution Function (CDF)

F(x) = P(X ≤ x)

For discrete: F(x) = ∑_xᵢ≤x p(xᵢ)

For continuous: F(x) = ∫_-∞^x f(t)dt

6.4 Expectation and Variance Properties

E(aX + b) = aE(X) + b
Var(aX + b) = a²Var(X)
E(X + Y) = E(X) + E(Y)
If X,Y independent: E(XY) = E(X)E(Y), Var(X+Y) = Var(X)+Var(Y)

Example:

If E(X)=3, Var(X)=4, find E(2X+1) and Var(2X+1)

E(2X+1) = 2E(X)+1 = 2×3+1 = 7

Var(2X+1) = 4Var(X) = 4×4 = 16

7. Advanced Probability Problems

7.1 Problems on Cards

Total cards: 52 (13 each suit)
Face cards: Jack, Queen, King (12 total)
Honor cards: Ace, King, Queen, Jack (16 total)

7.2 Problems on Dice

One die: 6 outcomes
Two dice: 36 outcomes
Three dice: 216 outcomes

7.3 Problems on Balls and Urns

With/without replacement problems

7.4 Birthday Problems

Probability that at least two people share birthday

Example:

From deck of 52 cards, 3 cards are drawn. Find probability of getting exactly 2 kings.

P = [⁴C₂ × ⁴⁸C₁] / ⁵²C₃ = [6 × 48] / 22100 = 288/22100 ≈ 0.013

8. Practice Problems - Probability

Basic Level:

Find probability of getting sum 9 when two dice are thrown
Three coins are tossed. Find probability of getting at least 2 heads
From deck of 52 cards, one card is drawn. Find probability it's a heart or a king
If P(A)=0.6, P(B)=0.5, P(A∩B)=0.3, find P(A|B) and check independence

Intermediate Level:

A problem is given to three students whose chances of solving are 1/2, 1/3, 1/4. Find probability problem is solved
In a binomial distribution with n=8, p=0.5, find P(X≥6)
A speaks truth 3/4 times, B 4/5 times. They agree on a statement. Find probability statement is true
Find probability that in a room of 10 people, at least two share birthday

Advanced Level:

If X ~ N(50, 100), find P(40 ≤ X ≤ 60)
A man takes steps forward with probability 2/3 and backward with probability 1/3. Find probability that after 5 steps, he is one step away from starting point
For events A,B,C which are pairwise independent with P(A)=P(B)=P(C)=p and P(A∩B∩C)=0, find maximum value of p
A random variable X has PDF f(x)=kx² for 0≤x≤1, 0 otherwise. Find k, mean, variance