Module 1 · NDA Mathematics (Paper 1)

Algebra

Sets, complex numbers, quadratic equations, sequences, permutations, binomial theorem.
Sets · Complex Numbers · Quadratics · Sequences · Binomial
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Learning Objectives

  • Master set theory, complex numbers, quadratic equations, and sequences for NDA Paper 1
  • Apply shortcut techniques to solve algebra MCQs in under 45 seconds
  • Recognise NDA-specific question patterns and elimination strategies
  • Build speed through practice with 15+ topic-wise solved examples and 12 practice MCQs
🎯 NDA Exam Strategy Algebra
NDA Maths (Paper 1) has 120 questions for 300 marks in 2.5 hours. Algebra typically contributes 2530 questions. The key to scoring is formula recall speed and elimination. Never spend more than 60 seconds on a single MCQ. Use option substitution, dimensional analysis, and boundary-value testing to skip lengthy derivations. Mark and skip if a question takes more than 90 seconds come back to it if time permits.

1. Sets & Relations

Sets form the foundational language of mathematics. In NDA exams, set theory is tested through questions on set operations, Venn diagrams, cardinality of unions and intersections, and De Morgan’s laws. Relations and functions appear as interconnected topics with moderate weightage.

Set Theory Basics

Definitions & Terminology

A set is a well-defined collection of distinct objects called elements (or members). Sets are denoted by capital letters (A, B, C, …) and elements by lowercase letters. The notation aA means a belongs to set A. Sets can be described in roster form (listing elements within curly braces) or set-builder form (stating the property that defines membership).

Types of Sets Complete Classification

Type Definition Example NDA Note
Empty / Null Set Set with no elements φ or { } φ ≠ {0}
Singleton Set Exactly one element {5} Cardinality = 1
Finite Set Countable number of elements {1, 2, 3, 4} |A| = n
Infinite Set Uncountable elements ℕ = {1, 2, 3, …} Natural numbers
Equal Sets Exactly the same elements {1,2} = {2,1} Order irrelevant
Equivalent Sets Same cardinality {a,b} ~ {1,2} n(A) = n(B)
Subset Every element of A is in B {1} ⊆ {1,2} ∅ ⊆ every set
Proper Subset A ⊆ B and A ≠ B {1} ⊂ {1,2} Strict subset
Power Set Set of all subsets of A P({a}) = {φ, {a}} |P(A)| = 2n
Universal Set All elements under discussion U Context-dependent

Subsets, Power Set & Cardinality Formulas

If |A| = n, then:
• Number of subsets = 2n
• Number of proper subsets = 2n − 1
• Number of non-empty subsets = 2n − 1
• Number of non-empty proper subsets = 2n − 2

Venn Diagrams & Set Operations

Venn diagrams use overlapping circles to visually represent relationships between sets. They are particularly useful for solving NDA word problems involving surveys, preferences, and group memberships.

Operation Symbol Set-Builder Venn Region
Union AB {x | xA or xB} Entire shaded area
Intersection AB {x | xA and xB} Overlapping region
Difference AB {x | xA and xB} A only
Complement A′ or Ac {x | xU and xA} Outside circle A
Symmetric Diff. A Δ B (AB) ∪ (BA) Non-overlapping parts

Key Cardinality Formulas

Two sets: n(AB) = n(A) + n(B) − n(AB)

Three sets: n(ABC) = n(A) + n(B) + n(C) − n(AB) − n(BC) − n(CA) + n(ABC)

De Morgan’s Laws

  • (AB)′ = A′ ∩ B
  • (AB)′ = A′ ∪ B
Example 1.1 NDA 2020 Pattern
In a class of 100 students, 55 like Mathematics, 60 like Physics, and 35 like both. How many like neither subject?

a) 10    b) 15    c) 20    d) 25
Solution: c) 20. n(M ∪ P) = 55 + 60 − 35 = 80. Neither = 100 − 80 = 20. NDA Approach: Draw a 2-circle Venn. Fill the intersection (35) first, then the remaining parts of each circle, and finally the outside region.
Example 1.2 De Morgan’s Law
If U = {1,2,3,4,5,6}, A = {1,2,3}, B = {3,4,5}, verify (AB)′ = A′ ∩ B′.
Solution: AB = {1,2,3,4,5}, so (AB)′ = {6}. A′ = {4,5,6}, B′ = {1,2,6}. A′ ∩ B′ = {6}. Verified. This is a common NDA verification question.
⚡ NDA Shortcut Set Cardinality
When asked “how many like neither”, always compute n(A ∪ B) first then subtract from total. Drawing a quick 2-circle Venn in the margin takes 10 seconds and prevents mistakes. For “exactly one” questions, use: n(only A) + n(only B) = n(A) + n(B) − 2·n(A ∩ B).

Relations & Functions NDA Essentials

A relation from set A to set B is a subset of A × B (Cartesian product). A function f: AB is a special relation where each element of A has exactly one image in B. Key function types tested in NDA:

  • One-one (injective): f(x1) = f(x2) ⇒ x1 = x2
  • Onto (surjective): For every yB, there exists xA with f(x) = y
  • Bijective: Both one-one and onto
  • Even function: f(−x) = f(x) symmetric about y-axis
  • Odd function: f(−x) = −f(x) symmetric about origin
⚡ NDA Shortcut Even/Odd Functions
To test if a function is even, odd, or neither, substitute −x and simplify. If f(−x) = f(x) ? even; if f(−x) = −f(x) ? odd; otherwise ? neither. Memorise: cos x, x are even; sin x, x are odd.

2. Complex Numbers

Complex numbers extend the real number system to include solutions of equations like x2 + 1 = 0. A complex number is written as z = a + ib, where i = √(−1). The real part is Re(z) = a and the imaginary part is Im(z) = b. NDA consistently tests modulus, argument, polar form, De Moivre’s theorem, and cube roots of unity.

Argand Plane & Modulus

Geometric Representation

Every complex number z = x + iy corresponds uniquely to a point (x, y) in the Argand plane (also called the complex plane). The x-axis is the real axis and the y-axis is the imaginary axis.

Modulus, Argument & Conjugate

  • Modulus: |z| = √(x2 + y2) distance from origin
  • Argument: arg(z) = tan−1(y/x) angle with positive real axis, range (−π, π]
  • Conjugate: z̅ = xiy reflection across the real axis

Three Representations of Complex Numbers

Form Expression Best For
Cartesian (Rectangular) z = x + iy Addition, subtraction, conjugate
Polar (Trigonometric) z = r(cos θ + i sin θ) Multiplication, division, powers
Exponential (Euler) z = reiθ De Moivre, roots of unity

Properties of Modulus and Conjugate

  • |z1 z2| = |z1| · |z2|
  • |z1 / z2| = |z1| / |z2| (z2 ≠ 0)
  • |z1 + z2| ≤ |z1| + |z2| (Triangle inequality)
  • |z1z2| ≥ ||z1| − |z2||
  • zz̅ = |z|2
  • z1 + z2 = z̅1 + z̅2
  • z1 z2 = z̅1 · z̅2
Example 2.1 NDA 2019
If |z − 1| = |z + 1|, then z lies on which of the following?

a) The real axis    b) The imaginary axis    c) A circle    d) The line y = x
Solution: b) The imaginary axis. Let z = x + iy. Then |x − 1 + iy| = |x + 1 + iy| ⇒ (x − 1)2 + y2 = (x + 1)2 + y2 ⇒ 4x = 0 ⇒ x = 0. So z lies on the imaginary axis (y-axis).
Example 2.2 Modulus Calculation
Find |(3 + 4i)(1 − 2i)|.

a) 5√5    b) 25    c) 5    d) √5
Solution: a) 5√5. |3+4i| = √(9+16) = 5, |1−2i| = √(1+4) = √5. By product property, |(3+4i)(1−2i)| = 5 × √5 = 5√5. NDA Approach: Use |z1z2| = |z1|·|z2| instead of expanding saves 30 seconds.

Polar Form & De Moivre’s Theorem

To convert to polar form: z = r(cos θ + i sin θ), where r = |z| and θ = arg(z).

De Moivre’s Theorem: (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) for integer n.

To find nth roots: z1/n = r1/n[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], for k = 0, 1, …, n−1.

⚡ NDA Shortcut Modulus of Sum/Difference
For NDA, when you see |z1 ± z2|, square both sides and use |z|2 = zz̅. Avoid expanding into x + iy unless necessary. Using properties instead of brute expansion saves 30+ seconds per question critical for NDA pace.

Cube Roots of Unity

The cube roots of unity are the three solutions of z3 = 1: 1, ω, ω2, where ω = (−1 + i√3)/2 and ω2 = (−1 − i√3)/2.

Essential Properties (memorise for NDA):

  • 1 + ω + ω2 = 0
  • ω3 = 1 and (ω2)3 = 1
  • ω3k = 1, ω3k+1 = ω, ω3k+2 = ω2
  • ω and ω2 are complex conjugates
Example 2.3 NDA 2021
Evaluate (1 + ω + ω2)100, where ω is a cube root of unity.

a) 1    b) 0    c) ω    d) ω2
Solution: b) 0. Since 1 + ω + ω2 = 0, we have 0100 = 0. This is a direct application of the fundamental property.

3. Quadratic Equations

A quadratic equation in x is of the form ax2 + bx + c = 0, where a, b, c are real and a ≠ 0. NDA tests nature of roots, sum and product relations, formation of equations, and common root conditions.

Standard Form & Quadratic Formula

The roots are given by: x = [−b ± √(b2 − 4ac)] / 2a. The discriminant D = b2 − 4ac determines the nature.

Nature of Roots Complete Classification

Discriminant Nature of Roots Example
D > 0, perfect square Real, distinct, rational x2 − 5x + 6 = 0
D > 0, not perfect square Real, distinct, irrational x2 − 4x + 2 = 0
D = 0 Real and equal (repeated) x2 − 4x + 4 = 0
D < 0 Complex conjugate pair x2 + x + 1 = 0

Sum & Product of Roots

If α and β are roots, then:
Sum: α + β = −b/a
Product: αβ = c/a
Form equation: x2 − (α+β)x + αβ = 0

Useful Expressions

  • α2 + β2 = (α + β)2 − 2αβ
  • α3 + β3 = (α + β)3 − 3αβ(α + β)
  • |α − β| = √D / |a|
Example 3.1 NDA 2018
If α and β are roots of x2 − 3x + 2 = 0, find α2β + αβ2.

a) 4    b) 6    c) 8    d) 10
Solution: b) 6. α+β = 3, αβ = 2. Then α2β + αβ2 = αβ(α+β) = 2×3 = 6. NDA Approach: Use sum and product directly no need to find actual roots.
Example 3.2 Equal Roots Condition
For what k does kx2 − 6x + 1 = 0 have equal roots?

a) 9    b) 3    c) 6    d) 0
Solution: a) 9. Set D = 0: 36 − 4k(1) = 0 ⇒ k = 9.

Common Roots

One common root: Condition: (c1a2c2a1)2 = (a1b2a2b1)(b1c2b2c1). Both roots common: a1/a2 = b1/b2 = c1/c2.

⚡ NDA Shortcut Common Root Substitution
For “common root” questions, let α be the common root and substitute into both equations. Eliminate α2 by subtraction. This avoids memorising the complicated condition and is faster in an exam.

Graphical Interpretation

The parabola y = ax2 + bx + c opens upward if a > 0, downward if a < 0. Vertex at x = −b/2a. If a > 0 and D < 0, expression is always positive; if a < 0 and D < 0, always negative.

4. Sequences & Series

A sequence is an ordered list of numbers following a pattern (e.g., 2, 4, 6, 8, …). A series is the sum of terms. NDA tests AP, GP, HP, and standard summation formulas.

Arithmetic Progression (AP)

General form: a, a+d, a+2d, … Common difference = d.

  • nth term: Tn = a + (n − 1)d
  • Sum: Sn = n/2 [2a + (n − 1)d] = n/2 (a + l)
  • Arithmetic Mean: AM of a, b = (a + b)/2
Example 4.1 NDA 2020
The 5th term of an AP is 17 and the 10th term is 32. Find d.

a) 2    b) 3    c) 4    d) 5
Solution: b) 3. T5 = a+4d = 17, T10 = a+9d = 32. Subtract: 5d = 15 ⇒ d = 3.

Geometric Progression (GP)

General form: a, ar, ar2, … Common ratio = r.

  • nth term: Tn = arn−1
  • Sum (n terms): Sn = a(rn − 1)/(r − 1), r ≠ 1
  • Sum (∞): S = a/(1 − r), |r| < 1
  • Geometric Mean: GM of a, b = √(ab)
⚡ NDA Shortcut Infinite GP & Recurring Decimals
For recurring decimals: 0.3 = 3/10 + 3/100 + … = (3/10)/(1−1/10) = 1/3. Same principle for any recurring decimal split into a GP and use S.
Example 4.2 NDA 2019
Find sum of infinite GP: 1 + 1/3 + 1/9 + 1/27 + …

a) 4/3    b) 3/2    c) 2    d) 3
Solution: b) 3/2. a = 1, r = 1/3. S = 1/(1 − 1/3) = 1/(2/3) = 3/2.

Harmonic Progression (HP)

Reciprocals of terms form an AP. HM of a, b = 2ab/(a+b). Key relationship: AM ≥ GM ≥ HM (equality when numbers are equal).

Sum of Special Series

Series Sum of First n Terms Memory Aid
1 + 2 + 3 + … + n n(n + 1)/2 “n n plus one by two”
12 + 22 + … + n2 n(n+1)(2n+1)/6 “n n+1 2n+1 by six”
13 + 23 + … + n3 [n(n+1)/2]2 “Square of the sum”
1 + 3 + 5 + … + (2n−1) n2 “Sum of odds = n2
⚡ NDA Shortcut Series with General Term
Given Tn, split into standard forms: Σ(n2+3n+1) = Σn2 + 3Σn + Σ1. Apply formulas individually. Do NOT expand fully.

5. Permutations & Combinations

Permutations count arrangements (order matters), combinations count selections (order does not). High-weightage topic in NDA with 4–6 questions on word arrangements, committee formation, and geometric counting.

Fundamental Principle of Counting

Multiplication: If event 1 in m ways and event 2 in n ways, both occur together in m × n ways.
Addition: If events are mutually exclusive, total = m + n.

nPr & nCr Formulas

Concept Formula Example
Permutation nPr = n!/(nr)! 5P3 = 60
Combination nCr = n!/[r!(nr)!] 5C3 = 10
All distinct n! 5! = 120
With repetitions n!/(p!q!r!…) MISSISSIPPI: 11!/(4!4!2!)
Combination identity nCr = nCnr 7C4 = 7C3 = 35
⚡ NDA Shortcut Combination Identity
Always check nCr = nCnr. Example: 100C98 = 100C2 = 4950. Also memorise: nC0 = nCn = 1, nC1 = nCn−1 = n.
Example 5.1 NDA 2019
How many 3-digit numbers using digits 1–5 without repetition?

a) 60    b) 120    c) 20    d) 125
Solution: a) 60. 5P3 = 5!/2! = 60. NDA Approach: 5 choices × 4 × 3 = 60. Direct multiplication is often faster than formula.

Circular Permutations

• Distinct objects around circle: (n − 1)!
• Necklace/garland (clockwise = anticlockwise): (n − 1)! / 2

Example 5.2 Circular
Ways to seat 5 people around a circular table?

a) 120    b) 24    c) 60    d) 12
Solution: b) 24. (5 − 1)! = 4! = 24.

6. Binomial Theorem

The binomial theorem expands (x + y)n for positive integer n. NDA tests general term, middle term, term independent of x, and coefficient properties.

Statement & Expansion

(x + y)n = Σ nCr xnr yr, where r = 0 to n. The expansion has (n+1) terms.

General Term & Middle Term

Term Type Formula Condition
General term (r+1)th Tr+1 = nCr xnr yr 0 ≤ rn
Middle term (n even) Tn/2+1 One middle term
Middle terms (n odd) T(n+1)/2 and T(n+3)/2 Two middle terms
Independent of x Set exponent of x = 0 Solve for r
Example 6.1 NDA 2021
Coefficient of x3 in (1+x)7?

a) 21    b) 35    c) 28    d) 15
Solution: b) 35. Tr+1 = 7Cr xr. For x3, r = 3. Coefficient = 7C3 = 35.

Binomial Coefficients Properties

  • nC0 + nC1 + … + nCn = 2n
  • nC0nC1 + … + (−1)n nCn = 0
  • Sum of even-indexed coefficients = Sum of odd-indexed coefficients = 2n−1
  • nCr + nCr−1 = n+1Cr (Pascal’s rule)
⚡ NDA Shortcut Substitution Method
For Σ nCr = 2n, substitute x = 1 in (1+x)n. For alternating sum, substitute x = −1. This substitution trick handles all NDA binomial summation problems in seconds.

7. Matrices & Determinants

A matrix is a rectangular array of numbers (rows × columns). Determinants are scalars for square matrices. NDA tests operations, determinant properties, and Cramer’s rule.

Matrix Operations

Operation Rule Note
Addition Element-wise (same order) Commutative
Scalar Multiplication Each element × scalar Distributive
Multiplication Row × Column (dot product) NOT commutative
Transpose Swap rows & columns (AB)T = BTAT
Inverse (2×2) A−1 = (1/det A) adj(A) det A ≠ 0

Matrix Properties

  • ABBA (non-commutative)
  • (AB)C = A(BC) (associative)
  • A(B+C) = AB + AC (distributive)
  • (AB)−1 = B−1A−1

Determinant Properties

  • det(kA) = kn det(A) for n×n matrix
  • det(AB) = det(A) · det(B)
  • det(AT) = det(A)
  • Two identical rows/columns ⇒ det = 0
  • Row interchange changes sign of det
  • det(A−1) = 1 / det(A)
  • Triangular matrix: det = product of diagonal entries
Example 7.1 NDA 2020
Find det of [[1,2],[3,4]].

a) −2    b) 2    c) −1    d) 4
Solution: a) −2. det = (1×4) − (2×3) = 4 − 6 = −2.
Example 7.2 Singular Matrix
For what k is [[k,2],[4,8]] singular?

a) 1    b) 2    c) 0    d) −1
Solution: a) 1. det = 8k − 8 = 0 ⇒ k = 1.

Cramer’s Rule

For a1x + b1y = c1, a2x + b2y = c2:
D = det[[a1,b1],[a2,b2]], Dx = det[[c1,b1],[c2,b2]], Dy = det[[a1,c1],[a2,c2]]
x = Dx/D, y = Dy/D (provided D ≠ 0).

⚡ NDA Shortcut Singular Matrix
A matrix with determinant = 0 is singular (no inverse). NDA often gives a matrix with a variable and asks for the value making it singular. Just set det = 0 and solve. This is a 10-second question.

8. Linear Inequalities

Linear inequalities involve >, ≥, <, ≤. NDA tests solving, interval representation, and sign reversal rules.

Interval Notation

  • (a, b) → a < x < b
  • [a, b] → axb
  • (a, ∞) → x > a
  • (−∞, b] → xb

Solving Rules

  • Adding/subtracting preserves inequality
  • Multiplying/dividing by positive preserves inequality
  • Multiplying/dividing by negative reverses the inequality
  • Never multiply/divide by a variable with unknown sign
Example 8.1 NDA 2019
Solve 3x − 7 > 2x + 1.

a) x > 8    b) x > 6    c) x < 8    d) x < 6
Solution: a) x > 8. 3x − 2x > 1+7 ⇒ x > 8.
Example 8.2 Negative Coefficient
Solve −2x + 5 ≤ 9.

a) x ≥ −2    b) x ≤ −2    c) x ≥ 2    d) x ≤ 2
Solution: a) x ≥ −2. −2x ≤ 4 ⇒ x ≥ −2 (sign reversed).
⚡ NDA Shortcut Sign Reversal Trap
Whenever you multiply/divide by a negative, flip the inequality sign. NDA sets traps here. Example: −3x < 9 ⇒ x > −3, NOT x < −3.

9. Probability

Probability ranges from 0 to 1. NDA tests classical probability, conditional probability, and Bayes’ theorem at a basic level.

Basic Terminology

  • Experiment: Process with well-defined outcomes (rolling a die)
  • Sample Space (S): All possible outcomes; n(S) = total count
  • Event (E): Subset of S; n(E) = favourable count
  • Probability: P(E) = n(E)/n(S), 0 ≤ P(E) ≤ 1

Key Formulas

Concept Formula
Union P(AB) = P(A) + P(B) − P(AB)
Complement P(A′) = 1 − P(A)
Conditional P(A|B) = P(AB)/P(B)
Independent P(AB) = P(AP(B)
Mutually Exclusive P(AB) = 0
Example 9.1 NDA 2021
Two dice rolled. Probability of sum 7?

a) 1/6    b) 1/12    c) 5/36    d) 1/9
Solution: a) 1/6. Total = 36. Favourable: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6. P = 6/36 = 1/6.
⚡ NDA Shortcut Dice Distribution
For two dice, sum frequencies: 2?1, 3?2, 4?3, 5?4, 6?5, 7?6, 8?5, 9?4, 10?3, 11?2, 12?1. Sum 7 has max probability (1/6). This triangular distribution helps eliminate options instantly.

10. Statistics (basic)

Statistics covers collection, organisation, and analysis of data. NDA tests mean, median, mode, variance, and standard deviation.

Measures of Central Tendency

  • Mean (𝑥): For ungrouped: (Σxi)/n. For grouped: (Σfixi)/(Σfi)
  • Median: Middle value when sorted. n odd: (n+1)/2th term. n even: average of n/2th and (n/2+1)th terms
  • Mode: Most frequent value

Measures of Dispersion

Measure Definition Formula
Range Max − Min
Variance (σ2) Avg of squared deviations Σ(xi−𝑥)2/n
Std Dev (σ) √(variance) √[Σ(xi−𝑥)2/n]
Coeff of Variation Relative dispersion (σ/𝑥) × 100%

Computational Formula for Variance

σ2 = [Σxi2/n] − (𝑥)2. This avoids individual deviations and is much faster for NDA.

Example 10.1 NDA 2018
Find the mean of 4, 8, 12, 16, 20.

a) 10    b) 12    c) 14    d) 16
Solution: b) 12. Sum = 60, Mean = 60/5 = 12.
Example 10.2 Standard Deviation
Find SD of 2, 4, 6, 8.

a) √5    b) √6    c) 5    d) 2√2
Solution: a) √5. Mean = 5. Σ(xi−5)2 = 9+1+1+9 = 20. Variance = 20/4 = 5. SD = √5.
⚡ NDA Shortcut Variance by Computation
Use σ2 = (Σxi2/n) − (𝑥)2. For 2,4,6,8: Σxi2 = 120, and = 30, 𝑥 = 5, 𝑥2 = 25, so σ2 = 30−25 = 5. This is 3x faster than computing deviations.

Practice Questions

Calculus →