SURYAMG

IIT/JEE Mathematics — Complete Context Masterclass (Part 1: Algebra & Calculus)

1. Sets, Relations & Functions

Key Formulas:
- Total Relations from A to B = 2^(m × n)
- Total Functions = n^m
- Total One-One Functions (Injective) = nPm (if n ≥ m)
- Fractional Part {x} = x - [x], Domain = R, Range = [0, 1)
Context 1: Functional Equations via Variable Substitution
Example 1: Find f(x) if 2f(x) + f(1/x) = x
  • Step 1: Original equation: 2f(x) + f(1/x) = x.
  • Step 2: Substitute x -> 1/x: 2f(1/x) + f(x) = 1/x.
  • Step 3: Multiply Eq 1 by 2: 4f(x) + 2f(1/x) = 2x.
  • Step 4: Subtract Eq 2: 3f(x) = 2x - 1/x.
  • Final Result: f(x) = (2x² - 1)/(3x)
Context 2: Finding Domain & Range of Inverse Trig and Radical Functions
Example 2: Find Domain of f(x) = sin⁻¹(2x - 3)
  • Step 1: Argument of sin⁻¹(u) must satisfy -1 ≤ u ≤ 1.
  • Step 2: Set inequality: -1 ≤ 2x - 3 ≤ 1.
  • Step 3: Add 3 to all parts: 2 ≤ 2x ≤ 4.
  • Step 4: Divide by 2: 1 ≤ x ≤ 2.
  • Final Result: Domain = [1, 2]
Context 3: Equivalence Relations (Reflexive, Symmetric, Transitive)
Example 3: Check if R = {(a, b) : a - b is divisible by 3} is Equivalence
  • Reflexive: a - a = 0 = 3(0) -> True.
  • Symmetric: If a - b = 3k, then b - a = -3k = 3(-k) -> True.
  • Transitive: If a - b = 3k and b - c = 3m, then (a - b) + (b - c) = a - c = 3(k + m) -> True.
  • Final Result: R is an Equivalence Relation

Homework Problems

1. Number of onto functions from A={1,2,3} to B={a,b}
Answer: 2³ - 2 = 6
2. Domain of 1 / sqrt(x² - 16)
Answer: (-∞, -4) ∪ (4, ∞)
3. Period of f(x) = |sin x| + |cos x|
Answer: π/2
4. Range of f(x) = x / (1 + x²)
Answer: [-1/2, 1/2]

2. Complex Numbers & Locus Problems

Key Formulas:
- Cube Roots of Unity: 1 + w + w² = 0, w³ = 1
- De Moivre's Theorem: (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)
- Locus |z - z1| / |z - z2| = k: Straight line if k = 1, Circle if k ≠ 1
Context 1: Cube Roots of Unity Simplifications
Example 1: Simplify (1 + w - w²)⁵ + (1 - w + w²)⁵
  • Step 1: Use 1 + w = -w² and 1 + w² = -w.
  • Step 2: First term: (-w² - w²)⁵ = (-2w²)⁵ = -32 w^10 = -32 w.
  • Step 3: Second term: (-w - w)⁵ = (-2w)⁵ = -32 w⁵ = -32 w².
  • Step 4: Combine: -32(w + w²) = -32(-1) = 32.
  • Final Result: 32
Context 2: Geometric Locus in Argand Plane
Example 2: Find Locus of |z - 3i| / |z + 3i| = 2
  • Step 1: Let z = x + iy. |x + i(y - 3)| = 2 |x + i(y + 3)|.
  • Step 2: Square both sides: x² + (y - 3)² = 4 [x² + (y + 3)²].
  • Step 3: Expand: x² + y² - 6y + 9 = 4x² + 4y² + 24y + 36.
  • Step 4: Rearrange: 3x² + 3y² + 30y + 27 = 0 -> x² + y² + 10y + 9 = 0.
  • Final Result: Circle with Center (0, -5) and Radius 4

Homework Problems

1. Value of (1 + i)^8
Answer: 16
2. Amplitude of -1 - i sqrt(3)
Answer: -2π/3
3. Value of w^2026 + w^2027
Answer: -1
4. Locus of |z - 1| = |z + 1|
Answer: Y-axis (x = 0)

3. Quadratic Equations & Newton's Method

Key Formulas:
- Newton's Theorem: If S_n = α^n + β^n, then a S_n + b S_(n-1) + c S_(n-2) = 0
- Location of Roots: Both roots > k -> D ≥ 0, -b/2a > k, a f(k) > 0
Context 1: Newton's Power Sum Recurrence
Example 1: Find (a10 - 2 a8) / (2 a9) if x² - 6x - 2 = 0
  • Step 1: Apply Newton's recurrence: a_n - 6 a_(n-1) - 2 a_(n-2) = 0.
  • Step 2: Set n = 10: a10 - 6 a9 - 2 a8 = 0.
  • Step 3: Rearrange: a10 - 2 a8 = 6 a9.
  • Step 4: Divide by 2 a9: (6 a9) / (2 a9) = 3.
  • Final Result: 3
Context 2: Location of Roots relative to Real Constants
Example 2: Find range of 'm' if both roots of x² - mx + 4 = 0 are real and > 1
  • Condition 1 (D ≥ 0): m² - 16 ≥ 0 -> m ≤ -4 or m ≥ 4.
  • Condition 2 (-b/2a > 1): m/2 > 1 -> m > 2.
  • Condition 3 (f(1) > 0): 1 - m + 4 > 0 -> m < 5.
  • Intersection: m ∈ [4, 5)
  • Final Result: [4, 5)

Homework Problems

1. Number of real roots of x² - 5|x| + 6 = 0
Answer: 4
2. Min value of 2x² - 8x + 11
Answer: 3
3. Condition for roots of ax² + bx + c = 0 to be reciprocal
Answer: a = c
4. Product of roots of x² + 3x - 10 = 0
Answer: -10

4. Matrices & Determinants

Key Formulas:
- |adj A| = |A|^(n-1)
- |adj(adj A)| = |A|^((n-1)²)
- Cramer's Rule: Infinite Solutions -> D = D1 = D2 = D3 = 0. No Solution -> D = 0 and at least one Di ≠ 0.
Context 1: Matrix Adjoint Determinant Properties
Example 1: If |A| = 4 for order 3, find |adj(3A)|
  • Step 1: Let B = 3A. Since order n = 3, |B| = |3A| = 3³ |A| = 27 × 4 = 108.
  • Step 2: Formula: |adj B| = |B|^(n-1) = |B|² = 108².
  • Step 3: 108² = 11,664.
  • Final Result: 11,664
Context 2: Consistency of Linear Equations System
Example 2: Find k for infinite solutions: x + y + z = 6, x + 2y + 3z = 10, x + 2y + kz = 10
  • Step 1: D = det|[1,1,1],[1,2,3],[1,2,k]| = 0.
  • Step 2: R3 -> R3 - R2: det|[1,1,1],[1,2,3],[0,0,k-3]| = 0.
  • Step 3: Expanding gives (k - 3)(2 - 1) = 0 -> k = 3.
  • Final Result: k = 3

Homework Problems

1. If A is orthogonal, |A| = ?
Answer: ±1
2. Trace of 3x3 Identity Matrix
Answer: 3
3. If A² = A, then A is called...
Answer: Idempotent Matrix
4. Determinant of skew-symmetric matrix of order 3
Answer: 0

5. Permutations & Combinations

Key Formulas:
- Derangement D_n = n! [1/2! - 1/3! + ... + (-1)^n / n!]
- Gap Method: Used when no two items can be together.
- String/Block Method: Used when specific items must always be together.
Context 1: Derangements (Complete Mismatch)
Example 1: Number of ways to put 4 letters into 4 wrong envelopes
  • Step 1: Apply D4 formula: 4! [1/2! - 1/3! + 1/4!].
  • Step 2: 24 × [1/2 - 1/6 + 1/24] = 12 - 4 + 1 = 9.
  • Final Result: 9
Context 2: Gap Method & Block Method Applications
Example 2: Arrange 4 boys and 4 girls such that no two girls are adjacent
  • Step 1: Arrange 4 boys first: 4! = 24.
  • Step 2: 4 boys create 5 gaps: _ B1 _ B2 _ B3 _ B4 _.
  • Step 3: Place 4 girls in 5 gaps: 5P4 = 120.
  • Step 4: Total = 24 × 120 = 2,880.
  • Final Result: 2,880

Homework Problems

1. Number of diagonals in an octagon (8 sides)
Answer: 8C2 - 8 = 20
2. Total handshakes among 12 people
Answer: 12C2 = 66
3. Derangements of 3 items (D3)
Answer: 2
4. Circular arrangements of 6 people
Answer: (6-1)! = 120

6. Binomial Theorem & Modular Remainder

Key Formulas:
- General Term T_(r+1) = nCr a^(n-r) b^r
- Remainder Trick: (a + 1)^n ÷ a gives Remainder = 1
Context 1: Modular Arithmetic Remainder Finding
Example 1: Find remainder when 7^103 is divided by 25
  • Step 1: Note that 7² = 49 = 50 - 1.
  • Step 2: Rewrite: 7^103 = 7 × (7²)^51 = 7 × (50 - 1)^51.
  • Step 3: Expand (50 - 1)^51 mod 25: 7 × (-1)^51 = -7.
  • Step 4: Positive remainder = 25 - 7 = 18.
  • Final Result: 18
Context 2: Term Independent of x in Expansion
Example 2: Find term independent of x in (x + 1/x)^6
  • Step 1: General term T_(r+1) = 6Cr · x^(6-r) · (x⁻¹)^r = 6Cr · x^(6-2r).
  • Step 2: Set exponent to 0: 6 - 2r = 0 -> r = 3.
  • Step 3: T4 = 6C3 = 20.
  • Final Result: 20

Homework Problems

1. Remainder when 8^n is divided by 7
Answer: 1
2. Number of terms in (x + y + z)^10
Answer: 66
3. Middle term coefficient of (1 + x)^8
Answer: 8C4 = 70
4. Sum of all coefficients in (2x - 1)^5
Answer: 1

7. Sequence & Series

Key Formulas:
- AM ≥ GM ≥ HM
- Infinite GP Sum = a / (1 - r) for |r| < 1
- Sum of squares Σ n² = n(n+1)(2n+1)/6
Context 1: AM-GM Inequality Optimization
Example 1: Find minimum value of 9^x + 9^(-x)
  • Step 1: AM = (9^x + 9^(-x)) / 2.
  • Step 2: GM = sqrt(9^x × 9^(-x)) = 1.
  • Step 3: Since AM ≥ GM -> (9^x + 9^(-x))/2 ≥ 1 -> 9^x + 9^(-x) ≥ 2.
  • Final Result: Minimum Value = 2
Context 2: Arithmetico-Geometric Progression (AGP)
Example 2: Find sum of S = 1 + 2/3 + 3/9 + 4/27 + ... to infinity
  • Step 1: Multiply S by 1/3: (1/3)S = 1/3 + 2/9 + 3/27 + ...
  • Step 2: Subtract: S - (1/3)S = 2/3 S = 1 + 1/3 + 1/9 + 1/27 + ...
  • Step 3: RHS is GP with a=1, r=1/3: 1 / (1 - 1/3) = 3/2.
  • Step 4: 2/3 S = 3/2 -> S = 9/4.
  • Final Result: 9/4 = 2.25

Homework Problems

1. Sum of 1 + 1/2 + 1/4 + 1/8...
Answer: 2
2. If AM = 10, GM = 8, find HM
Answer: GM² / AM = 64/10 = 6.4
3. Min value of x + 4/x for x > 0
Answer: 4
4. Sum of first 10 natural number cubes
Answer: (55)² = 3025

8. Limits, Continuity & Differentiability

Key Formulas:
- L'Hopital Rule for 0/0 or ∞/∞: lim f(x)/g(x) = lim f'(x)/g'(x)
- 1^∞ Form Shortcut: lim (f(x))^(g(x)) = e^(lim g(x)[f(x) - 1])
Context 1: L'Hopital Rule Application
Example 1: Evaluate lim (x->0) [tan x - x] / x³
  • Step 1: Form is 0/0. Differentiate: lim (sec² x - 1) / (3x²) = lim tan² x / (3x²).
  • Step 2: Use standard limit lim (tan x / x) = 1.
  • Step 3: Limit = (1/3) × 1² = 1/3.
  • Final Result: 1/3
Context 2: 1^∞ Indeterminate Form Exponential Shortcut
Example 2: Evaluate lim (x->0) (1 + 3x)^(1/x)
  • Step 1: Form is 1^∞. Apply e^(lim g(x)[f(x) - 1]).
  • Step 2: Exponent = lim (x->0) (1/x) × [1 + 3x - 1] = lim (3x/x) = 3.
  • Final Result:

Homework Problems

1. lim (x->0) [sin 5x / x]
Answer: 5
2. lim (x->0) (1 + x)^(2/x)
Answer: e²
3. Is f(x) = |x - 2| differentiable at x = 2?
Answer: No
4. lim (x->0) [(e^(4x) - 1) / x]
Answer: 4

9. Differential Calculus & AOD

Key Formulas:
- Tangent slope m = dy/dx. Normal slope = -1/m
- Rolle's Theorem: If f(a)=f(b), then f'(c) = 0 for some c ∈ (a, b)
Context 1: Monotonicity & Critical Points
Example 1: Find interval where f(x) = 2x³ - 9x² + 12x + 5 is strictly increasing
  • Step 1: f'(x) = 6x² - 18x + 12 = 6(x² - 3x + 2) = 6(x - 1)(x - 2).
  • Step 2: For increasing, f'(x) > 0 -> (x - 1)(x - 2) > 0.
  • Step 3: x < 1 or x > 2.
  • Final Result: (-∞, 1) ∪ (2, ∞)
Context 2: Tangent & Normal Equations
Example 2: Find equation of tangent to y = x² - 4x + 3 at (4, 3)
  • Step 1: dy/dx = 2x - 4. At x = 4, m = 2(4) - 4 = 4.
  • Step 2: Point-slope form: y - 3 = 4(x - 4) -> y - 3 = 4x - 16.
  • Final Result: 4x - y - 13 = 0

Homework Problems

1. Minimum value of x² - 8x + 15
Answer: -1 at x=4
2. Normal slope to y = sin x at x = 0
Answer: -1
3. d/dx [tan⁻¹ x + cot⁻¹ x]
Answer: 0
4. Point on y = x² with tangent parallel to y = 4x
Answer: (2, 4)