A vector space is a set whose elements (vectors) can be added and multiplied by scalars, following ten simple rules. Rⁿ, sets of matrices, polynomials and functions are all vector spaces. This topic covers the definition, subspaces, linear combinations and span, linear dependence and independence, basis and dimension.
Key points
A vector space V over a field F (usually R): addition is closed, commutative and associative; a zero vector exists; every v has −v; scalar multiplication is closed; and α(u + v) = αu + αv, (α + β)v = αv + βv, (αβ)v = α(βv), 1·v = v.
Examples: Rⁿ; the set Mₘₓₙ of m × n matrices; Pₙ, polynomials of degree ≤ n; the set of continuous functions on [a, b].
Subspace: a non-empty subset W of V that is itself a vector space under the same operations.
Subspace test: W is a subspace if for all u, v ∈ W and scalars α, β, αu + βv ∈ W.
Every subspace contains the zero vector. So a set not containing 0 (like the line x + y = 1) is not a subspace.
Intersection of subspaces is a subspace; union usually is not. W₁ + W₂ is a subspace.
Linear combination: α₁v₁ + α₂v₂ + … + αₙvₙ. Span L(S): set of all linear combinations of vectors of S; it is the smallest subspace containing S.
Linearly independent: α₁v₁ + … + αₙvₙ = 0 only when all αᵢ = 0. Otherwise dependent.
Any set containing the zero vector is dependent. In Rⁿ, n vectors are independent iff the determinant of their components is non-zero.
Basis: a linearly independent set that spans V. Dimension = number of vectors in any basis (all bases have the same size).
dim Rⁿ = n; dim Pₙ = n + 1; dim Mₘₓₙ = mn.
In an n-dimensional space, any n independent vectors form a basis; any n + 1 vectors are dependent.
dim(W₁ + W₂) = dim W₁ + dim W₂ − dim(W₁ ∩ W₂).
Important terms
Scalar – element of the field F (a number).
Standard basis of R³ – e₁ = (1, 0, 0), e₂ = (0, 1, 0), e₃ = (0, 0, 1).
Direct sum – V = W₁ ⊕ W₂ when V = W₁ + W₂ and W₁ ∩ W₂ = {0}.
Vector space (सदिश अवकाश) – set with vector addition and scalar multiplication obeying the axioms.
Formulas
Subspace test: αu + βv ∈ W
Independence test in Rⁿ: det[v₁ v₂ … vₙ] ≠ 0
dim(W₁ + W₂) = dim W₁ + dim W₂ − dim(W₁ ∩ W₂)
5 important questions
Q1. Define a subspace and give an example of a subset of R² that is not a subspace.
A non-empty subset W of V is a subspace if it is a vector space under the same operations (equivalently αu + βv ∈ W). The line {(x, y) : x + y = 1} is not a subspace because it does not contain (0, 0).
Q2. Show that any set containing the zero vector is linearly dependent.
If S = {0, v₂, …, vₙ}, then 1·0 + 0·v₂ + … + 0·vₙ = 0 with a non-zero coefficient (1). So S is linearly dependent.
Q3. Show that (1, 2, 1), (2, 1, 0), (1, −1, 2) form a basis of R³.
Three vectors in R³ (dimension 3): enough to check independence.
Determinant with these as rows: 1(1·2 − 0·(−1)) − 2(2·2 − 0·1) + 1(2·(−1) − 1·1).
= 1(2) − 2(4) + 1(−3) = 2 − 8 − 3 = −9.
Determinant ≠ 0, so the vectors are linearly independent.
Three independent vectors in R³ form a basis.
Q4. Show that W = {(x, y, z) : x + y + z = 0} is a subspace of R³ and find its dimension.
(0, 0, 0) ∈ W, so W is non-empty.
Let u = (x₁, y₁, z₁), v = (x₂, y₂, z₂) ∈ W. Then αu + βv has coordinate sum α(x₁ + y₁ + z₁) + β(x₂ + y₂ + z₂) = 0.
So αu + βv ∈ W: W is a subspace.
Any element: z = −x − y, so (x, y, −x − y) = x(1, 0, −1) + y(0, 1, −1).
(1, 0, −1) and (0, 1, −1) are independent and span W, so dim W = 2 (a plane through the origin).
Q5. If W₁ and W₂ are subspaces of V, prove that W₁ ∩ W₂ is a subspace.
0 ∈ W₁ and 0 ∈ W₂, so 0 ∈ W₁ ∩ W₂ (non-empty).
Let u, v ∈ W₁ ∩ W₂ and α, β scalars.
Since W₁ is a subspace, αu + βv ∈ W₁; since W₂ is a subspace, αu + βv ∈ W₂.
So αu + βv ∈ W₁ ∩ W₂.
Hence W₁ ∩ W₂ is a subspace.
Exam tip
For subspace questions, first check that the zero vector is in the set — if not, you are done in one line. For basis questions in Rⁿ, the determinant test is the fastest proof of independence.
A linear transformation is a function between vector spaces that preserves addition and scalar multiplication. Every linear transformation between finite-dimensional spaces can be represented by a matrix. This topic covers the definition and examples, kernel and range, the rank–nullity theorem, one-one and onto maps, isomorphism and the matrix of a transformation.
Key points
T: U → V is linear if T(u + v) = T(u) + T(v) and T(αu) = αT(u); equivalently T(αu + βv) = αT(u) + βT(v).
Every linear T maps 0 to 0. So T(x, y) = (x + 1, y) is not linear.
Examples: rotation and reflection in R², differentiation on polynomials, multiplication by a fixed matrix.
Kernel (null space): ker T = {u ∈ U : T(u) = 0}, a subspace of U. Its dimension is the nullity.
Range (image): R(T) = {T(u) : u ∈ U}, a subspace of V. Its dimension is the rank.
Rank–nullity theorem: rank T + nullity T = dim U.
T is one-one (injective) iff ker T = {0}.
T is onto (surjective) iff R(T) = V.
An isomorphism is a linear map that is one-one and onto. Two finite-dimensional spaces over the same field are isomorphic iff they have the same dimension.
A linear map is completely fixed by its values on a basis.
Matrix of T: write T of each basis vector of U as a combination of the basis of V; the coefficients form the columns of the matrix.
Composition of linear maps corresponds to multiplication of their matrices.
Eigenvalue of an operator: T(v) = λv for some non-zero v.
Important terms
Linear operator – linear map from a space to itself.
Linear transformation (रेखीय रूपांतरण) – map preserving vector operations.
Formulas
T(αu + βv) = αT(u) + βT(v)
dim U = rank(T) + nullity(T)
[T∘S] = [T][S] (matrices of compositions)
T one-one ⇔ ker T = {0}
5 important questions
Q1. Show that T(x, y) = (x + 1, y) is not a linear transformation.
T(0, 0) = (1, 0) ≠ (0, 0). A linear transformation must map the zero vector to the zero vector, so T is not linear.
Q2. State the rank–nullity theorem.
If T: U → V is a linear transformation and U is finite-dimensional, then rank(T) + nullity(T) = dim U, where rank is the dimension of the range and nullity the dimension of the kernel.
Q3. Show that T: R³ → R², T(x, y, z) = (x + y, y + z) is linear.
Let u = (x₁, y₁, z₁), v = (x₂, y₂, z₂), scalars α, β.
To prove linearity, use the single condition T(αu + βv) = αT(u) + βT(v). For the matrix, remember that images of basis vectors go into columns, not rows.
An inner product space is a vector space with an extra operation, the inner product, that generalises the dot product. It lets us measure length (norm), distance and angle, and talk about orthogonal vectors. The Cauchy–Schwarz inequality and the Gram–Schmidt process are the two most examined results.
Key points
An inner product on a real vector space V assigns a number ⟨u, v⟩ to each pair, such that: (i) ⟨u, u⟩ ≥ 0, and = 0 only for u = 0; (ii) ⟨u, v⟩ = ⟨v, u⟩; (iii) ⟨au + bv, w⟩ = a⟨u, w⟩ + b⟨v, w⟩. (Over C, (ii) becomes ⟨u, v⟩ = conjugate of ⟨v, u⟩.)
Standard examples: the dot product on Rⁿ; ⟨f, g⟩ = ∫ₐᵇ f(x)g(x) dx on continuous functions.
Norm: ‖u‖ = √⟨u, u⟩. Distance: d(u, v) = ‖u − v‖.
Cauchy–Schwarz inequality: |⟨u, v⟩| ≤ ‖u‖ ‖v‖, with equality iff u, v are linearly dependent.
Triangle inequality: ‖u + v‖ ≤ ‖u‖ + ‖v‖.
Angle: cos θ = ⟨u, v⟩/(‖u‖ ‖v‖). u and v are orthogonal if ⟨u, v⟩ = 0.
Pythagoras: if u ⊥ v then ‖u + v‖² = ‖u‖² + ‖v‖². Parallelogram law: ‖u + v‖² + ‖u − v‖² = 2‖u‖² + 2‖v‖².
An orthonormal set has unit vectors, each pair orthogonal. Orthogonal non-zero vectors are linearly independent.
Gram–Schmidt process turns any basis into an orthonormal basis.
Orthogonal complement W⊥ = {v : ⟨v, w⟩ = 0 for all w ∈ W}; for finite-dimensional V, V = W ⊕ W⊥.
Important terms
Norm (मानांक) – length of a vector.
Orthonormal basis – a basis of mutually perpendicular unit vectors.
For any vectors u, v in an inner product space, |⟨u, v⟩| ≤ ‖u‖ ‖v‖. Equality holds if and only if u and v are linearly dependent.
Q2. Prove that orthogonal non-zero vectors are linearly independent.
Let a₁v₁ + … + aₙvₙ = 0 with ⟨vᵢ, vⱼ⟩ = 0 for i ≠ j. Take the inner product with vₖ: aₖ⟨vₖ, vₖ⟩ = 0. Since vₖ ≠ 0, ⟨vₖ, vₖ⟩ > 0, so aₖ = 0 for every k.
Q3. Verify the Cauchy–Schwarz inequality for u = (1, 2, 2) and v = (3, 0, 4) in R³ and find the angle.
Q5. With ⟨f, g⟩ = ∫₀¹ f(x)g(x) dx, find ⟨x, x²⟩, ‖x‖ and check Cauchy–Schwarz.
⟨x, x²⟩ = ∫₀¹ x³ dx = ¼.
‖x‖² = ∫₀¹ x² dx = ⅓, so ‖x‖ = 1/√3.
‖x²‖² = ∫₀¹ x⁴ dx = ⅕, so ‖x²‖ = 1/√5.
‖x‖ ‖x²‖ = 1/√15 ≈ 0.258 ≥ 0.25 ✓.
Exam tip
In Gram–Schmidt problems, check orthogonality after each step before normalising — one arithmetic slip spoils every later vector. Keep fractions, not decimals.
A complex function f(z) = u(x, y) + iv(x, y) is analytic in a region if it has a derivative at every point there. The Cauchy–Riemann equations give the test for differentiability. This topic covers limits and derivatives of complex functions, the Cauchy–Riemann equations in Cartesian and polar forms, harmonic functions, harmonic conjugates and the Milne-Thomson method.
Key points
z = x + iy; a complex function w = f(z) = u(x, y) + iv(x, y), where u and v are real.
Derivative: f'(z) = lim (Δz → 0) [f(z + Δz) − f(z)]/Δz; the limit must be the same along every direction of approach.
Analytic (holomorphic) function: differentiable at every point of a neighbourhood. A function analytic in the whole plane is entire (polynomials, eᶻ, sin z).
Cauchy–Riemann (C–R) equations: uₓ = vᵧ and uᵧ = −vₓ. They are necessary for differentiability.
Sufficient condition: if uₓ, uᵧ, vₓ, vᵧ are continuous and satisfy the C–R equations, f is analytic.
f'(z) = uₓ + ivₓ = vᵧ − iuᵧ.
Polar form: uᵣ = (1/r)v_θ and vᵣ = −(1/r)u_θ.
f(z) = z̄ = x − iy is continuous everywhere but differentiable nowhere (uₓ = 1, vᵧ = −1).
If f is analytic, u and v are harmonic: ∇²u = uₓₓ + uᵧᵧ = 0 and ∇²v = 0.
v is a harmonic conjugate of u if f = u + iv is analytic.
The curves u = constant and v = constant cut orthogonally.
Milne-Thomson method: find f(z) from u alone: f'(z) = uₓ(z, 0) − iuᵧ(z, 0), then integrate.
Important terms
Singular point – point where f fails to be analytic.
Harmonic function (संनादी फलन) – real function satisfying Laplace's equation.
Entire function – analytic at every point of the complex plane.
Orthogonal trajectories – families of curves cutting at right angles.
Formulas
C–R (Cartesian): uₓ = vᵧ, uᵧ = −vₓ
C–R (polar): uᵣ = (1/r)v_θ, vᵣ = −(1/r)u_θ
f'(z) = uₓ + ivₓ
Laplace: uₓₓ + uᵧᵧ = 0
Milne-Thomson: f'(z) = uₓ(z, 0) − iuᵧ(z, 0)
Conjugate: dv = vₓ dx + vᵧ dy = −uᵧ dx + uₓ dy
5 important questions
Q1. State the Cauchy–Riemann equations.
If f(z) = u + iv is differentiable at z = x + iy, then uₓ = vᵧ and uᵧ = −vₓ at that point. In polar form: uᵣ = (1/r)v_θ, vᵣ = −(1/r)u_θ.
Q2. Show that f(z) = z̄ is not analytic anywhere.
u = x, v = −y. uₓ = 1, vᵧ = −1. Since uₓ ≠ vᵧ at every point, the C–R equations fail everywhere, so z̄ is nowhere differentiable and hence not analytic.
Q3. Verify the C–R equations for f(z) = z² and find f'(z).
z² = (x + iy)² = x² − y² + 2ixy; u = x² − y², v = 2xy.
uₓ = 2x, vᵧ = 2x → uₓ = vᵧ ✓.
uᵧ = −2y, vₓ = 2y → uᵧ = −vₓ ✓.
Partial derivatives are continuous, so f is analytic everywhere.
f'(z) = uₓ + ivₓ = 2x + 2iy = 2z.
Q4. Show that u = x³ − 3xy² is harmonic, find its harmonic conjugate and f(z).
Q5. Using the Milne-Thomson method, find the analytic function whose real part is u = x² − y² + 2x.
uₓ = 2x + 2, uᵧ = −2y.
Put x = z, y = 0: uₓ(z, 0) = 2z + 2, uᵧ(z, 0) = 0.
f'(z) = uₓ(z, 0) − iuᵧ(z, 0) = 2z + 2.
Integrate: f(z) = z² + 2z + ic.
Check: real part of z² + 2z is x² − y² + 2x ✓.
Exam tip
Always check both C–R equations and mention continuity of the partial derivatives to claim analyticity. In harmonic conjugate problems, verify your answer by checking that f(z) comes out as a function of z alone.
A complex function w = f(z) maps points of the z-plane to the w-plane. When f is analytic with f′(z) ≠ 0, the mapping is conformal: it preserves angles in size and sense. Bilinear (Möbius) transformations w = (az + b)/(cz + d) are the most important conformal maps; they send circles and lines to circles and lines.
Key points
A mapping is conformal at z₀ if the angle between any two curves through z₀ equals the angle between their images, both in size and direction.
If f is analytic and f′(z₀) ≠ 0, then w = f(z) is conformal at z₀. Points where f′(z) = 0 are critical points (angles are multiplied there).
Basic maps: translation w = z + c; rotation w = e^(iα)z; magnification w = kz (k > 0); inversion w = 1/z.
Bilinear transformation: w = (az + b)/(cz + d) with ad − bc ≠ 0. It is a combination of translation, rotation, magnification and inversion.
Its inverse is also bilinear: z = (−dw + b)/(cw − a).
A bilinear map sends circles and straight lines (a line = circle through ∞) to circles or straight lines.
Cross-ratio (z₁, z₂, z₃, z₄) = (z₁ − z₂)(z₃ − z₄)/((z₁ − z₄)(z₃ − z₂)) is unchanged by a bilinear map.
There is exactly one bilinear map sending three given points z₁, z₂, z₃ to three given points w₁, w₂, w₃.
Fixed (invariant) points: solve z = (az + b)/(cz + d), i.e. cz² + (d − a)z − b = 0. One repeated fixed point → parabolic transformation.
Important terms
Conformal mapping (संरूपी प्रतिचित्रण) – an angle-preserving map.
Möbius transformation – another name for a bilinear transformation.
Critical point – a point where f′(z) = 0.
Formulas
w = (az + b)/(cz + d), ad − bc ≠ 0
(w − w₁)(w₂ − w₃)/((w − w₃)(w₂ − w₁)) = (z − z₁)(z₂ − z₃)/((z − z₃)(z₂ − z₁))
Fixed points: cz² + (d − a)z − b = 0
5 important questions
Q1. When is w = f(z) conformal?
At every point where f is analytic and f′(z) ≠ 0. For example, w = z² is conformal everywhere except at z = 0, where angles are doubled.
Q2. Find the image of the circle |z| = 2 under w = 1/z.
|w| = 1/|z| = ½. So the image is the circle |w| = ½.
Q3. Find the bilinear map that sends z = 0, 1, ∞ to w = −1, 0, 1.
Let w = (az + b)/(cz + d).
z = ∞ → w = a/c = 1, so a = c.
z = 0 → w = b/d = −1, so b = −d.
z = 1 → w = (a + b)/(c + d) = 0, so b = −a, hence d = a.
With a = c = d = 1, b = −1: w = (z − 1)/(z + 1).
Check: 0 → −1 ✓, 1 → 0 ✓, ∞ → 1 ✓.
Q4. Find the fixed points of w = (3z − 4)/(z − 1). What type is it?
Put w = z: z(z − 1) = 3z − 4, so z² − 4z + 4 = 0.
(z − 2)² = 0, so z = 2 is the only (repeated) fixed point.
One repeated fixed point, so the transformation is parabolic.
Q5. Under w = 1/z, find the image of the line x = 1.
z = 1/w. With w = u + iv: z = (u − iv)/(u² + v²).
So x = u/(u² + v²).
x = 1 gives u² + v² = u, i.e. (u − ½)² + v² = ¼.
The image is the circle with centre (½, 0) and radius ½ (it passes through w = 0, the image of z = ∞).
Exam tip
For "find the bilinear map" questions, either use the cross-ratio formula or solve for a, b, c, d as above — and always check all three given points at the end.
Complex integration is integration of a complex function along a curve (contour) in the complex plane. For analytic functions it gives powerful results: Cauchy's theorem, Cauchy's integral formula, Taylor and Laurent series, and the residue theorem, which makes many difficult integrals easy.
Key points
Contour integral: ∫_C f(z) dz along a curve C; if z = z(t), a ≤ t ≤ b, then ∫_C f(z) dz = ∫ₐᵇ f(z(t)) z'(t) dt.
Cauchy's (Cauchy–Goursat) theorem: if f is analytic inside and on a simple closed curve C, then ∮_C f(z) dz = 0.
Consequence: for analytic f in a simply connected region, the integral between two points does not depend on the path.
Cauchy's integral formula: if f is analytic inside and on C and a is inside C, f(a) = (1/2πi)∮_C f(z)/(z − a) dz.
Derivatives: f⁽ⁿ⁾(a) = (n!/2πi)∮_C f(z)/(z − a)ⁿ⁺¹ dz. So an analytic function has derivatives of all orders.
Basic result: ∮_C dz/(z − a) = 2πi if a is inside C, and 0 if outside.
Taylor series: an analytic function can be expanded about a point inside its disc of analyticity.
Laurent series: about a singular point, f(z) = Σ aₙ(z − a)ⁿ + Σ bₙ/(z − a)ⁿ in an annulus.
Singularities: removable (no negative powers), pole of order m (finitely many negative powers, highest (z − a)⁻ᵐ), essential (infinitely many, e.g. e^(1/z) at 0).
Residue at a: the coefficient b₁ of 1/(z − a) in the Laurent series.
Residue theorem: ∮_C f(z) dz = 2πi × (sum of residues at poles inside C).
Uses: evaluating real integrals such as ∫₀^2π dθ/(a + b cos θ) and ∫₋∞^∞ dx/(1 + x²).
Important terms
Simple closed curve – a closed curve that does not cross itself.
Pole (ध्रुव) – isolated singularity where |f| → ∞ in a definite way.
Residue – coefficient of 1/(z − a) in the Laurent expansion.
Positive direction – anticlockwise around the contour.
Formulas
f(a) = (1/2πi)∮ f(z)/(z − a) dz
f⁽ⁿ⁾(a) = (n!/2πi)∮ f(z)/(z − a)ⁿ⁺¹ dz
Residue at a simple pole: Res = lim (z → a) (z − a)f(z); for f = φ/ψ with ψ(a) = 0: Res = φ(a)/ψ'(a)
Residue at a pole of order m: Res = (1/(m − 1)!) lim d^(m−1)/dz^(m−1) [(z − a)ᵐ f(z)]
Residue theorem: ∮_C f(z) dz = 2πi Σ Res
5 important questions
Q1. State Cauchy's integral theorem.
If f(z) is analytic inside and on a simple closed contour C (with f' continuous, in the classical form), then ∮_C f(z) dz = 0.
Q2. Classify the singularity of f(z) = sin z/z at z = 0.
sin z/z = (z − z³/3! + …)/z = 1 − z²/3! + z⁴/5! − … There are no negative powers of z, so z = 0 is a removable singularity.
Q3. Evaluate ∮_C eᶻ/(z − 1) dz, where C is the circle |z| = 2.
f(z) = eᶻ is analytic everywhere; a = 1 lies inside |z| = 2.
Cauchy's integral formula: ∮ f(z)/(z − a) dz = 2πi f(a).
= 2πi e¹ = 2πie.
Q4. Evaluate ∮_C z/((z − 1)(z − 2)) dz, where C is |z| = 3, using the residue theorem.
Simple poles at z = 1 and z = 2, both inside |z| = 3.
Res at 1: lim (z → 1) z/(z − 2) = 1/(1 − 2) = −1.
Res at 2: lim (z → 2) z/(z − 1) = 2/1 = 2.
Sum of residues = −1 + 2 = 1.
∮ = 2πi × 1 = 2πi.
Q5. Evaluate ∮_C sin z/(z − π/2)³ dz, where C is |z| = 2.
a = π/2 ≈ 1.57 lies inside |z| = 2; f(z) = sin z is analytic.
Use the derivative formula with n = 2: ∮ f(z)/(z − a)³ dz = (2πi/2!) f''(a).
f''(z) = −sin z, so f''(π/2) = −1.
∮ = πi × (−1) = −πi.
Exam tip
Before using any formula, check which singular points lie inside the contour — mark them on a rough sketch. Points outside contribute nothing.
Cauchy's residue theorem gives a powerful way to evaluate real definite integrals that are hard by ordinary calculus. We change the real integral into a contour integral, find the poles inside the contour, add their residues and multiply by 2πi. Three standard types are asked in exams.
Key points
Residue theorem: if f is analytic inside and on a closed curve C except at poles a₁, …, aₙ inside C, then ∮_C f(z) dz = 2πi × (sum of residues at a₁, …, aₙ).
Residue at a simple pole a: lim (z − a)f(z); for f = P/Q with Q(a) = 0, Q′(a) ≠ 0, the residue is P(a)/Q′(a).
Residue at a pole of order m: (1/(m − 1)!) × lim d^(m−1)/dz^(m−1) [(z − a)ᵐ f(z)].
Type 1: ∫₀^(2π) F(cos θ, sin θ) dθ. Put z = e^(iθ): cos θ = (z + 1/z)/2, sin θ = (z − 1/z)/(2i), dθ = dz/(iz). The contour is |z| = 1.
Type 2: ∫₋∞^∞ P(x)/Q(x) dx where deg Q ≥ deg P + 2 and Q has no real zeros. Integrate over a large semicircle in the upper half plane; the arc integral tends to 0. Answer = 2πi × (sum of residues in the upper half plane).
Type 3: ∫₋∞^∞ (P/Q) cos mx dx or sin mx dx (m > 0). Use f(z) = (P/Q)e^(imz); by Jordan's lemma the arc part → 0. Take the real part for cos and the imaginary part for sin.
Only poles inside the contour (inside |z| = 1, or in the upper half plane) are counted.
Important terms
Residue (अवशेष) – the coefficient of 1/(z − a) in the Laurent series about a.
Contour – a closed piecewise-smooth path.
Jordan's lemma – shows the arc integral of e^(imz)g(z) vanishes when g → 0.
Formulas
∫₀^(2π) dθ/(a + b cos θ) = 2π/√(a² − b²), a > |b|
∫₋∞^∞ dx/(x² + a²) = π/a
Residue of P/Q at a simple pole a = P(a)/Q′(a)
5 important questions
Q1. State Cauchy's residue theorem.
If f(z) is analytic inside and on a simple closed curve C except at a finite number of poles inside C, then ∮_C f(z) dz = 2πi times the sum of the residues of f at those poles.
Q2. Evaluate ∫₋∞^∞ dx/(1 + x²) using residues.
f(z) = 1/(1 + z²) has poles ±i; only z = i is in the upper half plane. Residue = 1/(2i). Integral = 2πi × 1/(2i) = π.
Q3. Evaluate ∫₀^(2π) dθ/(2 + cos θ).
Put z = e^(iθ): integral = ∮ [1/(2 + (z + 1/z)/2)] dz/(iz) = (2/i)∮ dz/(z² + 4z + 1).
Poles: z = −2 ± √3. Only α = −2 + √3 ≈ −0.27 lies inside |z| = 1.
Residue at α = 1/(2α + 4) = 1/(2√3).
Integral = (2/i) × 2πi × 1/(2√3) = 2π/√3.
Check with the formula 2π/√(a² − b²) = 2π/√(4 − 1) = 2π/√3 ✓.
Q4. Evaluate ∫₋∞^∞ dx/((x² + 1)(x² + 4)).
Poles in the upper half plane: z = i and z = 2i (both simple).
Residue at i: 1/(2i × (−1 + 4)) = 1/(6i).
Residue at 2i: 1/((−4 + 1) × 4i) = −1/(12i).
Sum = 2/(12i) − 1/(12i) = 1/(12i).
Integral = 2πi × 1/(12i) = π/6.
Q5. Evaluate ∫₋∞^∞ cos x/(x² + 1) dx.
Take f(z) = e^(iz)/(z² + 1). Upper-half-plane pole: z = i.
Real part: ∫ cos x/(x² + 1) dx = π/e. (Imaginary part gives ∫ sin x/(x² + 1) dx = 0, as expected for an odd integrand.)
Exam tip
Always list all poles first, then clearly mark which lie inside the contour — including an outside pole is the most common mistake. Check a Type 1 answer with the a + b cos θ formula when it applies.
This topic compares the "sizes" of infinite sets and studies the basic topology of the real line. A set is countable if its elements can be listed as a sequence. N, Z and Q are countable, but R is not. We also study neighbourhoods, interior and limit points, open and closed sets, and the Bolzano–Weierstrass and Heine–Borel theorems on R.
Key points
A set A is countable if it is finite or there is a one-one onto map f : N → A (then A is countably infinite, or denumerable). Otherwise A is uncountable.
Every subset of a countable set is countable. A countable union of countable sets is countable. N × N is countable.
Z and Q are countable; R, the interval (0, 1) and the set of irrational numbers are uncountable (Cantor's diagonal argument).
The set of algebraic numbers is countable, so "most" real numbers are transcendental.
Interior point: x is interior to A if some interval (x − ε, x + ε) ⊆ A. A is open if every point of A is interior.
Limit (accumulation) point: every neighbourhood of x contains a point of A other than x. The set of limit points is the derived set A′.
A is closed if it contains all its limit points, i.e. A′ ⊆ A. Equivalently, its complement is open. Closure Ā = A ∪ A′.
Any union of open sets is open; a finite intersection of open sets is open. ∩(−1/n, 1/n) = {0} shows an infinite intersection may not be open.
Bolzano–Weierstrass theorem: every bounded infinite subset of R has at least one limit point.
Heine–Borel theorem: a subset of R is compact (every open cover has a finite subcover) iff it is closed and bounded.
Important terms
Countable set (गणनीय संच) – a set that can be listed as a₁, a₂, a₃, …
Neighbourhood – a set containing an open interval around the point.
Isolated point – a point of A that is not a limit point of A.
Formulas
Open: every x ∈ A has (x − ε, x + ε) ⊆ A
Closed: A′ ⊆ A; Ā = A ∪ A′
Q̄ = R, Q′ = R
5 important questions
Q1. Define a countable set and give one countable and one uncountable example.
A set is countable if it is finite or in one-one correspondence with N. Example: Z is countable; the interval (0, 1) is uncountable.
Q2. Find the limit points of A = {1/n : n ∈ N}. Is A open or closed?
The only limit point is 0. Since 0 ∉ A, A is not closed. A is not open either, because no interval around 1 lies inside A. So A is neither open nor closed; Ā = A ∪ {0}.
Q3. Show that Z is countable.
Define f : N → Z by f(n) = n/2 if n is even and f(n) = −(n − 1)/2 if n is odd.
This lists Z as 0, 1, −1, 2, −2, 3, … (f(1) = 0, f(2) = 1, f(3) = −1, …).
Every integer appears exactly once, so f is one-one and onto.
Hence Z is countable.
Q4. Outline Cantor's proof that (0, 1) is uncountable.
Suppose (0, 1) is countable: x₁, x₂, x₃, … with decimal expansions xₙ = 0.aₙ₁aₙ₂aₙ₃…
Build y = 0.b₁b₂b₃… where bₙ = 5 if aₙₙ ≠ 5, and bₙ = 4 if aₙₙ = 5.
Then y differs from each xₙ in the n-th digit, so y is not in the list.
But y ∈ (0, 1), a contradiction. Hence (0, 1) is uncountable, and so is R.
Q5. Classify as open, closed, both or neither: (a, b), [a, b], [a, b), R, Q.
(a, b): open (each point has a small interval inside); not closed (a is a limit point outside it).
[a, b]: closed (complement (−∞, a) ∪ (b, ∞) is open); not open (no interval around a lies inside).
[a, b): neither.
R: both open and closed (so is ∅).
Q: neither — every interval contains irrationals, and every real number is a limit point of Q.
Exam tip
For "open/closed" questions, test the end points: an end point inside the set spoils openness, an end point outside spoils closedness. Write the explicit listing map when proving a set countable.
A metric space is a set with a distance function (metric) that behaves like ordinary distance. It generalises ideas from real analysis — open and closed sets, convergence, completeness, continuity and compactness — to many settings. This topic covers the definition, standard examples, open balls and open/closed sets, sequences and completeness, and continuity.
Key points
A metric on X is a function d: X × X → R such that: (1) d(x, y) ≥ 0; (2) d(x, y) = 0 iff x = y; (3) d(x, y) = d(y, x); (4) d(x, z) ≤ d(x, y) + d(y, z) (triangle inequality).
Examples: usual metric on R, d(x, y) = |x − y|; Euclidean metric on R², √((x₁ − y₁)² + (x₂ − y₂)²); taxicab metric |x₁ − y₁| + |x₂ − y₂|; max metric max(|x₁ − y₁|, |x₂ − y₂|); discrete metric (0 if x = y, 1 otherwise).
A set G is open if every point of G has an open ball around it lying inside G. Every open ball is open.
A set F is closed if its complement is open; equivalently F contains all its limit points.
In the discrete metric every set is both open and closed.
Arbitrary unions and finite intersections of open sets are open.
Convergence: xₙ → x if d(xₙ, x) → 0. Limits are unique.
Cauchy sequence: d(xₘ, xₙ) → 0 as m, n → ∞. Every convergent sequence is Cauchy.
Complete metric space: every Cauchy sequence converges in it. R and Rⁿ are complete; Q is not (a sequence of rationals can converge to √2).
Continuity: f is continuous at a if for each ε > 0 there is δ > 0 with d(x, a) < δ ⇒ d(f(x), f(a)) < ε. Equivalently, the inverse image of every open set is open.
Compactness: every open cover has a finite subcover. In Rⁿ, a set is compact iff closed and bounded (Heine–Borel).
Banach's contraction principle: a contraction on a complete metric space has a unique fixed point.
Important terms
Limit point – every ball around it contains a point of the set other than itself.
Closure Ā – A together with its limit points.
Bounded set – contained in some ball.
Metric (मापक/अंतर फलन) – distance function obeying the four axioms.
Q2. Is Q (rationals) with the usual metric complete? Why?
No. The sequence 1, 1.4, 1.41, 1.414, … (decimal approximations of √2) is a Cauchy sequence of rationals, but its limit √2 is not rational. So it does not converge in Q.
Q3. Prove that the discrete metric is a metric.
d(x, y) = 0 if x = y, 1 if x ≠ y. Clearly d ≥ 0 and d = 0 iff x = y.
d(x, y) = d(y, x), since x = y iff y = x.
Triangle inequality: if x = z, d(x, z) = 0 ≤ anything.
If x ≠ z, then y differs from at least one of x, z, so d(x, y) + d(y, z) ≥ 1 = d(x, z).
All four axioms hold.
Q4. Prove that every convergent sequence in a metric space is a Cauchy sequence.
Let xₙ → x. Given ε > 0, there is N with d(xₙ, x) < ε/2 for all n ≥ N.
For m, n ≥ N, by the triangle inequality: d(xₘ, xₙ) ≤ d(xₘ, x) + d(x, xₙ).
< ε/2 + ε/2 = ε.
So (xₙ) is Cauchy. (The converse fails in incomplete spaces like Q.)
Q5. Show that d'(x, y) = |x − y|/(1 + |x − y|) is a metric on R.
d' ≥ 0, d' = 0 iff x = y, and d' is symmetric — clear from |x − y|.
For the triangle inequality, f(t) = t/(1 + t) is increasing for t ≥ 0 (f'(t) = 1/(1 + t)² > 0).
Let a = |x − y|, b = |y − z|; then |x − z| ≤ a + b, so d'(x, z) ≤ f(a + b).
f(a + b) = a/(1 + a + b) + b/(1 + a + b) ≤ a/(1 + a) + b/(1 + b).
So d'(x, z) ≤ d'(x, y) + d'(y, z). Hence d' is a metric (bounded by 1).
Exam tip
Proving something is a metric means checking all four axioms in order — the triangle inequality is where the marks are. Learn the standard examples (Q incomplete, discrete metric) as ready counterexamples.
Continuity, uniform continuity and connectedness extend familiar ideas from R to any metric space. Continuous functions preserve compactness and connectedness, which gives general versions of the extreme value and intermediate value theorems. This topic builds on open sets, compactness and completeness studied earlier.
Key points
f : (X, d) → (Y, ρ) is continuous at a if for every ε > 0 there is δ > 0 with d(x, a) < δ ⇒ ρ(f(x), f(a)) < ε.
Sequential criterion: f is continuous at a iff xₙ → a implies f(xₙ) → f(a).
Open-set criterion: f is continuous on X iff f⁻¹(G) is open in X for every open G in Y (equivalently, inverse images of closed sets are closed).
f is uniformly continuous if one δ works for all points: d(x, y) < δ ⇒ ρ(f(x), f(y)) < ε. Uniform continuity implies continuity, not conversely.
A continuous function on a compact metric space is uniformly continuous, is bounded, and attains its maximum and minimum.
The continuous image of a compact set is compact.
X is connected if it cannot be written as A ∪ B with A, B non-empty, disjoint and open. Otherwise it is disconnected.
A subset of R is connected iff it is an interval.
The continuous image of a connected set is connected. This gives the intermediate value theorem.
A discrete metric space with more than one point is not connected.
Important terms
Uniform continuity (एकसमान सातत्य) – continuity with δ independent of the point.
Separation – a split of X into two disjoint non-empty open sets.
Homeomorphism – a continuous bijection with continuous inverse.
f : X → Y is uniformly continuous if for every ε > 0 there is a δ > 0 (depending only on ε) such that d(x, y) < δ implies ρ(f(x), f(y)) < ε for all x, y ∈ X.
Q2. Show that Q (with the usual metric) is not connected.
Take A = Q ∩ (−∞, √2) and B = Q ∩ (√2, ∞). Both are open in Q, non-empty, disjoint, and A ∪ B = Q because √2 ∉ Q. So Q is disconnected.
Q3. Prove that f(x) = x² is uniformly continuous on [0, 1].
For x, y ∈ [0, 1]: |x² − y²| = |x + y| |x − y| ≤ 2|x − y|.
Given ε > 0, choose δ = ε/2.
If |x − y| < δ then |x² − y²| < 2δ = ε.
δ does not depend on x or y, so f is uniformly continuous on [0, 1].
Q4. Show that f(x) = 1/x is not uniformly continuous on (0, 1).
Take xₙ = 1/n and yₙ = 1/(n + 1), both in (0, 1) for n ≥ 2.
|xₙ − yₙ| = 1/(n(n + 1)) → 0.
But |f(xₙ) − f(yₙ)| = |n − (n + 1)| = 1 for every n.
So for ε = 1/2 no δ can work. Hence f is not uniformly continuous (though it is continuous on (0, 1)).
Q5. Prove that the continuous image of a connected space is connected.
Let f : X → Y be continuous and onto, X connected. Suppose Y = A ∪ B is a separation.
Then X = f⁻¹(A) ∪ f⁻¹(B). Both sets are open (continuity), disjoint, and non-empty (f is onto).
This is a separation of X, contradicting connectedness.
Hence Y = f(X) is connected. On R this gives the intermediate value theorem.
Exam tip
To disprove uniform continuity, use two sequences that get closer while their images stay a fixed distance apart. To prove it, try to get a bound |f(x) − f(y)| ≤ K|x − y|.
Riemann integration makes the idea of "area under a curve" exact using upper and lower sums over partitions. A bounded function is Riemann integrable when its upper and lower integrals are equal. This topic covers partitions, Darboux sums, the criterion of integrability, classes of integrable functions, properties, mean value theorems and the fundamental theorem of calculus.
Key points
A partition P of [a, b] is a finite set a = x₀ < x₁ < … < xₙ = b. Let Δxᵢ = xᵢ − xᵢ₋₁.
On each sub-interval, Mᵢ = sup f and mᵢ = inf f (f bounded).
Upper sum U(P, f) = ΣMᵢΔxᵢ; lower sum L(P, f) = ΣmᵢΔxᵢ. Always L(P, f) ≤ U(P, f).
A refinement P* ⊇ P makes L larger (or equal) and U smaller (or equal).
Upper integral = inf of all U(P, f); lower integral = sup of all L(P, f).
f is Riemann integrable on [a, b] if upper integral = lower integral; the common value is ∫ₐᵇ f dx.
Riemann's criterion: f is integrable iff for every ε > 0 there is a partition P with U(P, f) − L(P, f) < ε.
Every continuous function on [a, b] is integrable. Every monotonic function on [a, b] is integrable. A bounded function with finitely many discontinuities is integrable.
Dirichlet's function (1 on rationals, 0 on irrationals) is bounded but not Riemann integrable.
Properties: linearity; ∫ₐᵇ f = ∫ₐᶜ f + ∫ᶜᵇ f; if f ≤ g then ∫f ≤ ∫g; |∫f| ≤ ∫|f|.
First mean value theorem: for continuous f, ∫ₐᵇ f dx = f(c)(b − a) for some c in [a, b].
Fundamental theorem of calculus: if f is integrable and F' = f, then ∫ₐᵇ f dx = F(b) − F(a). If f is continuous, F(x) = ∫ₐˣ f(t) dt is differentiable with F'(x) = f(x).
Riemann sums give limits of series as integrals: lim (1/n) Σ f(r/n) = ∫₀¹ f(x) dx.
Important terms
Norm of partition – length of its largest sub-interval.
Darboux sums – the upper and lower sums U and L.
Riemann sum – Σ f(tᵢ)Δxᵢ with tᵢ any point of the i-th sub-interval.
Integrable (समाकलनीय) – upper and lower integrals agree.
A bounded function f on [a, b] is Riemann integrable if and only if for every ε > 0 there exists a partition P of [a, b] such that U(P, f) − L(P, f) < ε.
Q2. Show that Dirichlet's function is not Riemann integrable on [0, 1].
Every sub-interval contains both rationals and irrationals, so Mᵢ = 1 and mᵢ = 0 for every partition. Hence U(P, f) = 1 and L(P, f) = 0 always; upper integral 1 ≠ lower integral 0. So f is not integrable.
Q3. Show that f(x) = x is integrable on [0, 1] and find ∫₀¹ x dx using upper and lower sums.
Q4. Using the integral as a limit of a sum, evaluate lim (n → ∞) Σ (r = 1 to n) 1/(n + r).
Write 1/(n + r) = (1/n) × 1/(1 + r/n).
This is (1/n) Σ f(r/n) with f(x) = 1/(1 + x).
Limit = ∫₀¹ dx/(1 + x) = [log(1 + x)]₀¹.
= log 2 − log 1 = log 2.
Q5. State and prove the first mean value theorem of integral calculus.
Statement: if f is continuous on [a, b], there is c in [a, b] with ∫ₐᵇ f dx = f(c)(b − a).
f is continuous, so it attains a minimum m and a maximum M on [a, b].
m ≤ f(x) ≤ M ⇒ m(b − a) ≤ ∫ₐᵇ f dx ≤ M(b − a).
So μ = (1/(b − a))∫ₐᵇ f dx lies between m and M.
By the intermediate value theorem, f(c) = μ for some c in [a, b]. Hence proved.
Exam tip
In upper/lower sum questions, use equal partitions and the formulas for Σr and Σr². Mention monotonicity to say where sup and inf are attained on each sub-interval.
An integral is improper when its range is infinite or the integrand becomes unbounded inside the range. Such an integral is defined as a limit of ordinary (Riemann) integrals and may converge or diverge. We learn the two kinds of improper integrals, the standard p-test, comparison tests and absolute convergence.
Key points
First kind: infinite limit, e.g. ∫ₐ^∞ f(x) dx = lim(t→∞) ∫ₐᵗ f(x) dx. It converges if this limit is finite.
Second kind: f unbounded at a point of [a, b], e.g. at a: ∫ₐᵇ f dx = lim(ε→0⁺) ∫ₐ₊ε^b f dx.
∫₋∞^∞ f dx converges only if both ∫₋∞^c and ∫_c^∞ converge separately.
p-test (first kind): ∫₁^∞ dx/xᵖ converges iff p > 1.
p-test (second kind): ∫₀¹ dx/xᵖ converges iff p < 1.
Comparison test (0 ≤ f ≤ g): if ∫g converges, ∫f converges; if ∫f diverges, ∫g diverges.
Limit comparison test: if f, g > 0 and f/g → l, with 0 < l < ∞, then ∫f and ∫g converge or diverge together.
Absolute convergence: if ∫|f| converges, ∫f converges. ∫₀^∞ (sin x/x) dx converges (to π/2) but not absolutely — it is conditionally convergent.
Gamma function Γ(n) = ∫₀^∞ e^(−x)x^(n−1) dx is an important improper integral, convergent for n > 0.
Important terms
Convergent integral (अभिसारी संकल) – the defining limit exists and is finite.
Cauchy principal value – lim(t→∞) ∫₋ₜᵗ f dx, which may exist even when the integral diverges.
Conditional convergence – convergent, but not absolutely.
Formulas
∫₁^∞ x⁻ᵖ dx = 1/(p − 1) for p > 1
∫₀¹ x⁻ᵖ dx = 1/(1 − p) for p < 1
∫₀^∞ e^(−ax) dx = 1/a (a > 0)
5 important questions
Q1. Distinguish between improper integrals of the first and second kind.
First kind: the range of integration is infinite (e.g. ∫₁^∞ dx/x²). Second kind: the range is finite but the integrand is unbounded at some point (e.g. ∫₀¹ dx/√x).
Q2. Test the convergence of ∫₁^∞ dx/x.
∫₁ᵗ dx/x = log t → ∞ as t → ∞. So the integral diverges (p = 1 in the p-test).
Q3. Evaluate ∫₀^∞ dx/(1 + x²) and ∫₀¹ dx/√x.
∫₀ᵗ dx/(1 + x²) = tan⁻¹t → π/2 as t → ∞. Converges to π/2.
∫ε¹ x^(−1/2) dx = [2√x] from ε to 1 = 2 − 2√ε → 2 as ε → 0⁺. Converges to 2 (p = ½ < 1).
Q4. Test ∫₁^∞ x/(x³ + 1) dx for convergence.
For large x, x/(x³ + 1) behaves like 1/x². Take g = 1/x².
f/g = x³/(x³ + 1) → 1, a finite non-zero limit.
∫₁^∞ dx/x² converges (p = 2 > 1).
By the limit comparison test, the given integral converges.
Q5. Show that ∫₁^∞ (sin x)/x² dx converges absolutely.
|sin x/x²| ≤ 1/x² for x ≥ 1.
∫₁^∞ dx/x² = 1 converges.
By comparison, ∫₁^∞ |sin x|/x² dx converges.
So the integral converges absolutely, hence converges.
Exam tip
Always find the "dominant term" of the integrand at the trouble point (∞ or the singular point) and compare with 1/xᵖ — this one idea solves most convergence questions.
A lattice is a partially ordered set in which every pair of elements has a least upper bound and a greatest lower bound. A Boolean algebra is a lattice that is distributive and complemented. Boolean algebra is the mathematics behind logic, set operations and switching (digital) circuits.
Key points
A partially ordered set (poset) is a set with a relation ≤ that is reflexive, antisymmetric and transitive. It is drawn as a Hasse diagram.
A lattice is a poset in which every pair a, b has a join a ∨ b = lub{a, b} and a meet a ∧ b = glb{a, b}.
Lattice laws: idempotent (a ∨ a = a), commutative, associative and absorption: a ∨ (a ∧ b) = a, a ∧ (a ∨ b) = a.
Examples: (P(S), ⊆) with ∨ = ∪ and ∧ = ∩; (Dₙ, |), the divisors of n, with ∨ = lcm and ∧ = gcd.
A lattice is bounded if it has a least element 0 and a greatest element 1.
Distributive lattice: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) (and its dual).
Complement of a: an element a′ with a ∨ a′ = 1 and a ∧ a′ = 0. In a distributive lattice a complement, if it exists, is unique.
Boolean algebra: a bounded, distributive, complemented lattice. Usually written with + (join), · (meet) and ′.
De Morgan's laws: (a + b)′ = a′b′, (ab)′ = a′ + b′. Duality: swapping + with · and 0 with 1 in a valid law gives another valid law.
Dₙ is a Boolean algebra iff n is square-free (a product of distinct primes), e.g. D₃₀.
In circuits, + is a parallel connection (OR) and · is a series connection (AND).
Important terms
Hasse diagram – a picture of a finite poset with upward lines for "covers".
Join / meet (संयोग / छेद) – lub and glb of two elements.
Boolean function – a function of Boolean variables built from +, · and ′.
Formulas
x + x′ = 1, x·x′ = 0, x + 1 = 1, x·0 = 0
x + xy = x; x + x′y = x + y
In Dₙ: a ∨ b = lcm(a, b), a ∧ b = gcd(a, b), 0 = 1, 1 = n
5 important questions
Q1. Define a Boolean algebra.
A Boolean algebra is a lattice with least element 0 and greatest element 1 which is distributive and in which every element has a complement.
Q2. In D₃₀, find 6 ∨ 10, 6 ∧ 10 and the complement of 6.
6 ∨ 10 = lcm(6, 10) = 30; 6 ∧ 10 = gcd(6, 10) = 2. Complement of 6 is 5, since lcm(6, 5) = 30 (the top) and gcd(6, 5) = 1 (the bottom).
Q3. Show that D₁₂ is not a Boolean algebra.
D₁₂ = {1, 2, 3, 4, 6, 12}, with 0 = 1 and 1 = 12.
A complement x of 2 needs gcd(2, x) = 1, so x must be odd: x = 1 or 3.
lcm(2, 1) = 2 and lcm(2, 3) = 6; neither is 12.
So 2 has no complement, and D₁₂ is not a Boolean algebra (12 = 2² × 3 is not square-free).
Q4. Simplify f = xy + xy′ + x′y.
xy + xy′ = x(y + y′) = x·1 = x.
So f = x + x′y.
x + x′y = (x + x′)(x + y) = 1·(x + y) = x + y.
f = x + y (a single OR gate).
Q5. Write the truth table of f(x, y) = x′y + xy′ and name the function.
x
y
x′y
xy′
f
0
0
0
0
0
0
1
1
0
1
1
0
0
1
1
1
1
0
0
0
f is 1 exactly when x and y differ: it is the exclusive OR (XOR) function.
Exam tip
For any Dₙ question, list all divisors first and remember: join = lcm, meet = gcd. Show each simplification step with the law used (absorption, complement, distributive).
A graph is a set of points (vertices) joined by lines (edges). Graph theory models networks such as roads, circuits, computer networks and social links. This topic covers basic terms, degree and the handshaking lemma, special graphs, paths and connectivity, Euler and Hamiltonian graphs, trees and spanning trees, and planar graphs.
Key points
A graph G = (V, E): V is a set of vertices, E a set of edges joining pairs of vertices.
A loop joins a vertex to itself; parallel (multiple) edges join the same pair. A simple graph has neither.
Degree of a vertex = number of edges at it (a loop counts 2).
Handshaking lemma: the sum of degrees of all vertices = 2 × number of edges. Hence the number of odd-degree vertices is always even.
Complete graph Kₙ: every pair of vertices joined; it has n(n − 1)/2 edges.
Bipartite graph: vertices split into two sets with edges only between the sets; K_{m,n} is complete bipartite with mn edges.
Walk, trail, path, cycle: walk = sequence of vertices and edges; trail = no repeated edge; path = no repeated vertex; cycle = closed path.
Connected graph: a path exists between every pair of vertices.
Euler circuit: a closed trail using every edge exactly once. A connected graph has one iff every vertex has even degree. It has an open Euler trail iff exactly two vertices have odd degree. (Euler, Königsberg bridges problem, 1736.)
Hamiltonian cycle: a cycle passing through every vertex exactly once. No simple necessary-and-sufficient test is known.
Tree: a connected graph with no cycles. A tree with n vertices has n − 1 edges, and there is exactly one path between any two vertices.
Spanning tree: a tree containing all vertices of a connected graph. Minimum spanning tree (least total weight) is found by Kruskal's or Prim's algorithm.
Planar graph: can be drawn with no edges crossing. Euler's formula for a connected planar graph: v − e + f = 2. K₅ and K₃,₃ are not planar.
Graphs can be stored as an adjacency matrix: entry aᵢⱼ = number of edges between vᵢ and vⱼ.
Important terms
Isolated vertex – degree 0; pendant vertex – degree 1.
Regular graph – all vertices have the same degree.
Weighted graph – each edge carries a number (cost, distance).
Graph (आलेख) – vertices and edges.
Formulas
Σ deg(v) = 2e
Edges of Kₙ = n(n − 1)/2; edges of K_{m,n} = mn
Tree: e = n − 1
Euler's formula (planar): v − e + f = 2
Simple connected planar graph (v ≥ 3): e ≤ 3v − 6
5 important questions
Q1. State the handshaking lemma and one consequence.
In any graph, the sum of the degrees of all vertices equals twice the number of edges, because each edge adds 2 to the total degree. Consequence: the number of vertices of odd degree is even.
Q2. How many edges does K₆ have? How many edges does a tree with 12 vertices have?
K₆ has 6 × 5/2 = 15 edges. A tree with 12 vertices has 12 − 1 = 11 edges.
Q3. A connected planar graph has 8 vertices and 12 edges. How many faces does it have? Can a simple graph with 5 vertices have degrees 3, 3, 3, 3, 3?
Euler's formula: v − e + f = 2 → 8 − 12 + f = 2 → f = 6.
Degrees 3, 3, 3, 3, 3 have sum 15, which is odd.
The sum of degrees must be even (= 2e), so no such graph exists.
Q4. Find a minimum spanning tree by Kruskal's algorithm for the graph with edges AB = 1, BC = 2, AC = 3, CD = 4, BD = 5.
Arrange edges by weight: AB 1, BC 2, AC 3, CD 4, BD 5.
Choose AB (1). Choose BC (2).
AC (3) would form cycle A–B–C — reject.
Choose CD (4). Now 3 edges for 4 vertices — the tree is complete.
Q5. Prove that a tree with n vertices has n − 1 edges.
Use induction on n. For n = 1: no edges, 0 = 1 − 1 ✓.
Assume true for all trees with fewer than n vertices.
In a tree with n ≥ 2 vertices, take a longest path; its end vertex has degree 1 (a pendant vertex), else there would be a cycle or a longer path.
Remove that vertex and its edge: what remains is a tree with n − 1 vertices, having n − 2 edges by hypothesis.
Adding back gives (n − 2) + 1 = n − 1 edges. Proved.
Exam tip
Draw small, clear figures for every definition you write (path, cycle, tree, bipartite). In Kruskal/Prim questions, list the edges chosen and rejected step by step with reasons.
Linear programming (LP) finds the best (maximum or minimum) value of a linear objective function subject to linear constraints. It is used for planning production, diets, transport and resource allocation. This topic covers formulation, the graphical method, the simplex method and duality.
Key points
LP problem: optimise Z = c₁x₁ + c₂x₂ + … subject to linear inequalities (constraints) and x₁, x₂, … ≥ 0 (non-negativity).
Decision variables are the unknown quantities to be decided.
Feasible solution: satisfies all constraints. The set of all such points is the feasible region, a convex set.
Corner point (fundamental) theorem: if an optimal solution exists, it occurs at a corner (vertex) of the feasible region.
Graphical method (two variables): draw each constraint as a line, shade the feasible region, find corner points, evaluate Z at each.
Special cases: unbounded solution, infeasible problem (no common region), multiple optimal solutions (objective line parallel to a boundary edge).
Standard form for simplex: constraints as equations with non-negative right sides. Add a slack variable for ≤, subtract a surplus variable (and add an artificial variable) for ≥.
Simplex method (Dantzig): start from a basic feasible solution; at each step bring in the variable that most improves Z (entering variable) and remove one by the minimum ratio test (leaving variable); stop when no improvement is possible.
Big-M and two-phase methods handle artificial variables.
Duality: every LP (primal) has a dual. Max ↔ min, constraint constants ↔ objective coefficients, the coefficient matrix is transposed, ≤ ↔ ≥.
Duality theorem: if both have optimal solutions, max Z (primal) = min W (dual).
Important terms
Objective function (उद्दिष्ट फलन) – the linear function to be optimised.
Basic feasible solution – a corner-point solution in simplex language.
Optimal solution – feasible solution giving the best value of Z.
Slack variable – added to a ≤ constraint to make it an equation.
Formulas
Primal: max Z = cᵀx, Ax ≤ b, x ≥ 0
Dual: min W = bᵀy, Aᵀy ≥ c, y ≥ 0
Minimum ratio test: choose the row with least (bᵢ/aᵢⱼ) for aᵢⱼ > 0
At optimum: max Z = min W
5 important questions
Q1. State the fundamental theorem of linear programming.
If an LP problem has an optimal solution, then at least one optimal solution occurs at a corner point (vertex) of the feasible region. If the region is bounded, both maximum and minimum exist and occur at corner points.
Q2. What are slack and surplus variables?
A slack variable is added to a ≤ constraint to turn it into an equation (it measures unused resource). A surplus variable is subtracted from a ≥ constraint to make it an equation (it measures the excess over the requirement).
Q3. Solve graphically: maximise Z = 5x + 4y subject to 6x + 4y ≤ 24, x + 2y ≤ 6, x, y ≥ 0.
Line 6x + 4y = 24 meets axes at (4, 0) and (0, 6); line x + 2y = 6 at (6, 0) and (0, 3).
Feasible region (towards origin) has corners O(0, 0), A(4, 0), B, C(0, 3).
B: solve 6x + 4y = 24 and x + 2y = 6. From the second, x = 6 − 2y; 36 − 12y + 4y = 24 → y = 1.5, x = 3.
Z at O = 0; A = 20; B = 15 + 6 = 21; C = 12.
Maximum Z = 21 at x = 3, y = 1.5.
Q4. Write the dual of the LP in Q3 and verify the duality theorem.
Dual: minimise W = 24u + 6v subject to 6u + v ≥ 5, 4u + 2v ≥ 4, u, v ≥ 0.
Both primal variables are positive at the optimum, so both dual constraints hold as equations.
Solve 6u + v = 5 and 4u + 2v = 4: v = 5 − 6u → 4u + 10 − 12u = 4 → u = 0.75, v = 0.5.
W = 24(0.75) + 6(0.5) = 18 + 3 = 21.
min W = 21 = max Z ✓.
Q5. Explain the steps of the simplex method for a maximisation problem.
Write the problem in standard form by adding slack variables; slacks form the initial basis.
Build the initial simplex table and compute the net evaluations (Zⱼ − Cⱼ).
If all Zⱼ − Cⱼ ≥ 0, the current solution is optimal.
Otherwise, choose the most negative Zⱼ − Cⱼ: that column is the entering variable.
Use the minimum ratio test to find the leaving variable; the intersection is the pivot element.
Make the pivot 1 and other entries in its column 0 by row operations; repeat from step 3.
Exam tip
In graphical problems, label every line and corner point on the graph and tabulate Z at each corner. Check that your optimum point satisfies all constraints before writing the answer.
Transportation and assignment problems are special linear programming problems. A transportation problem finds the cheapest way to send goods from several sources to several destinations. An assignment problem gives n jobs to n persons, one each, at minimum total cost. Both have quick special methods instead of the simplex method.
Balanced problem: total supply = total demand. If not, add a dummy row or column with zero costs.
A basic feasible solution has m + n − 1 allocations. Fewer means the solution is degenerate (add a tiny ε in an empty cell).
Initial solution methods: North-West Corner Rule (start top-left, ignore costs), Least Cost Method (allocate to the cheapest cell first), Vogel's Approximation Method (use penalties = difference of two lowest costs; usually the best start).
Optimality test (MODI / u-v method): for occupied cells uᵢ + vⱼ = cᵢⱼ (start with one u = 0). For empty cells find Δᵢⱼ = cᵢⱼ − (uᵢ + vⱼ). If every Δ ≥ 0, the solution is optimal; otherwise improve along a closed loop.
Assignment problem: an n × n cost matrix; each row and column gets exactly one assignment.
Hungarian method: (1) subtract each row's minimum from the row; (2) subtract each column's minimum; (3) cover all zeros with the minimum number of lines; (4) if lines = n, make assignments on zeros; otherwise subtract the smallest uncovered number from uncovered cells, add it at line crossings, and repeat.
For maximisation, subtract all entries from the largest entry first. An unbalanced (non-square) matrix gets a dummy row or column of zeros.
Important terms
Transportation problem (वाहतूक समस्या) – minimum-cost shipping from sources to destinations.
Degeneracy – fewer than m + n − 1 occupied cells.
Opportunity cost – Δᵢⱼ, the change in cost from using an empty cell.
Statics studies bodies at rest under forces. We find the resultant of forces acting at a point, use Lami's theorem and the triangle of forces for three forces in equilibrium, combine parallel forces, and study moments and couples. The conditions of equilibrium of a rigid body follow from these ideas.
Key points
Parallelogram law: two forces P and Q at angle α have resultant R with R² = P² + Q² + 2PQ cos α, making angle θ with P where tan θ = Q sin α/(P + Q cos α).
Resolution: a force F at angle θ to the x-axis has components F cos θ and F sin θ. For many forces, X = ΣF cos θ, Y = ΣF sin θ, R = √(X² + Y²), tan φ = Y/X.
Triangle of forces: if three forces at a point can be drawn as the sides of a triangle taken in order, they are in equilibrium.
Lami's theorem: if three forces P, Q, R at a point are in equilibrium, each is proportional to the sine of the angle between the other two: P/sin α = Q/sin β = R/sin γ.
Like parallel forces P and Q at A and B: resultant P + Q acts at C on AB with P·AC = Q·CB. Unlike forces: resultant P − Q, outside AB.
Moment of a force about a point = force × perpendicular distance. Varignon's theorem: the moment of the resultant equals the sum of the moments of the forces.
A couple is two equal, unlike, parallel forces; its moment (force × arm) is the same about every point. A couple cannot be balanced by a single force.
Equilibrium of a rigid body under coplanar forces: ΣX = 0, ΣY = 0 and Σ(moments about any point) = 0.
Friction: limiting friction F = μR; μ = tan λ, where λ is the angle of friction.
Important terms
Resultant (परिणामी बल) – the single force with the same effect as several forces.
Equilibrium – the state in which the resultant force and the total moment are zero.
Couple – a pair of equal, opposite, parallel forces with different lines of action.
Formulas
R² = P² + Q² + 2PQ cos α; tan θ = Q sin α/(P + Q cos α)
Lami: P/sin α = Q/sin β = R/sin γ
Like parallel forces: P·AC = Q·CB
5 important questions
Q1. State Lami's theorem.
If three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces.
Q2. Can a couple be balanced by a single force? Why?
No. A couple has zero resultant force but a non-zero moment. A single force would give a non-zero resultant, so equilibrium is impossible; a couple can only be balanced by an equal and opposite couple.
Q3. Forces of 3 N and 5 N act at a point at 60° to each other. Find the resultant.
R² = 3² + 5² + 2(3)(5) cos 60° = 9 + 25 + 15 = 49, so R = 7 N.
tan θ = 5 sin 60°/(3 + 5 cos 60°) = (5 × 0.866)/(3 + 2.5) = 4.33/5.5 ≈ 0.787.
θ ≈ 38.2° with the 3 N force.
Q4. A weight of 10 N hangs from two strings making 30° and 60° with the vertical. Find the tensions.
The strings are at 90° to each other. Angle between T₂ (the 60° string) and the weight = 180° − 60° = 120°; between T₁ and the weight = 180° − 30° = 150°.
Lami: T₁/sin 120° = T₂/sin 150° = 10/sin 90°.
T₁ = 10 × 0.866 = 8.66 N (5√3 N); T₂ = 10 × 0.5 = 5 N.
Check horizontal: 8.66 sin 30° = 4.33 and 5 sin 60° = 4.33 ✓. Vertical: 8.66 cos 30° + 5 cos 60° = 7.5 + 2.5 = 10 ✓.
Q5. Like parallel forces of 4 N and 6 N act at A and B, 1 m apart. Find the resultant and its position.
Resultant R = 4 + 6 = 10 N, in the same direction.
Its point C on AB satisfies 4 × AC = 6 × CB, with AC + CB = 1.
4 AC = 6(1 − AC) gives 10 AC = 6, so AC = 0.6 m from A (nearer the larger force).
Exam tip
Draw a clear force diagram and mark every angle before using Lami's theorem — the angles must be between the OTHER two forces. Always check equilibrium by resolving horizontally and vertically.
Dynamics studies how bodies move under forces. For motion in a plane we express velocity and acceleration in Cartesian, radial–transverse and tangential–normal components. We then apply these to central orbits, simple harmonic motion and projectiles — the most common exam applications.
Key points
Cartesian: v = (ẋ, ẏ), a = (ẍ, ÿ), where a dot means d/dt.
Tangential and normal components: a_t = dv/dt = v dv/ds, a_n = v²/ρ (ρ = radius of curvature). For uniform circular motion, a_n = v²/r towards the centre.
Central orbit: the force is always directed to a fixed point. Then a_θ = 0, so r²θ̇ = h (constant), and the areal velocity h/2 is constant — Kepler's second law.
Under an inverse-square attraction the orbit is a conic (ellipse, parabola or hyperbola).
Simple harmonic motion (SHM): ẍ = −μx. Solution x = a cos(√μ t + ε); period T = 2π/√μ; v² = μ(a² − x²); maximum speed √μ·a at the centre.
Projectile (speed u at angle α, no air resistance): time of flight 2u sin α/g, greatest height u² sin²α/(2g), horizontal range u² sin 2α/g (maximum u²/g at α = 45°). The path is a parabola.
Important terms
Radial component (त्रिज्य घटक) – along the line from the origin to the particle.
Transverse component – perpendicular to the radius, in the direction of increasing θ.
Amplitude – the greatest distance from the centre in SHM.
Formulas
a_r = r̈ − rθ̇², a_θ = rθ̈ + 2ṙθ̇
r²θ̇ = h (central orbit)
SHM: T = 2π/√μ, v² = μ(a² − x²)
Projectile: R = u² sin 2α/g, H = u² sin²α/(2g), T = 2u sin α/g
5 important questions
Q1. Write the radial and transverse components of acceleration.
Q2. Show that for a central orbit the areal velocity is constant.
For a central force the transverse acceleration is zero: (1/r) d/dt(r²θ̇) = 0, so r²θ̇ = h, a constant. The area swept per unit time is ½r²θ̇ = h/2, which is constant.
Q3. A particle moves on the curve r = aθ with constant angular speed ω. Find its radial and transverse acceleration.
θ̇ = ω, θ̈ = 0. ṙ = aθ̇ = aω, r̈ = 0.
a_r = r̈ − rθ̇² = 0 − aθω² = −aθω².
a_θ = rθ̈ + 2ṙθ̇ = 0 + 2(aω)(ω) = 2aω².
Q4. A particle moves in SHM with amplitude 5 cm and period 4 s. Find the maximum speed and the speed at 3 cm from the centre.
T = 2π/√μ = 4, so √μ = π/2.
Maximum speed = √μ × a = (π/2) × 5 = 2.5π ≈ 7.85 cm/s.
At x = 3: v = √μ √(a² − x²) = (π/2)√(25 − 9) = (π/2)(4) = 2π ≈ 6.28 cm/s.
Q5. A ball is projected at 20 m/s at 30° to the horizontal (g = 9.8 m/s²). Find the time of flight, greatest height and range.
Time of flight = 2 × 20 × sin 30°/9.8 = 20/9.8 ≈ 2.04 s.
Range = 20² × sin 60°/9.8 = 400 × 0.866/9.8 ≈ 35.35 m.
Exam tip
Write ṙ, r̈, θ̇ and θ̈ separately before substituting into the component formulas. In SHM problems, find √μ from the period first; everything else follows.
A probability distribution tells how probabilities are spread over the values of a random variable. This topic covers random variables, probability mass and density functions, expectation and variance, and the three most important distributions: binomial, Poisson and normal.
Key points
A random variable X assigns a real number to each outcome of a random experiment. It is discrete (countable values) or continuous (any value in an interval).
Probability mass function (pmf) p(x) for discrete X: p(x) ≥ 0 and Σp(x) = 1.
Probability density function (pdf) f(x) for continuous X: f(x) ≥ 0 and ∫f(x) dx = 1; P(a < X < b) = ∫ₐᵇ f(x) dx.
Expectation (mean) E(X) and variance V(X) = E(X²) − [E(X)]². Standard deviation σ = √V(X).
E(aX + b) = aE(X) + b; V(aX + b) = a²V(X).
Binomial distribution B(n, p): number of successes in n independent trials, each with success probability p (q = 1 − p). Mean np, variance npq (variance < mean).
Poisson distribution with parameter m: for rare events (number of accidents, misprints). It is the limit of binomial when n is large, p small and np = m. Mean = variance = m.
Normal distribution N(μ, σ²): continuous, bell-shaped, symmetric about the mean μ; mean = median = mode; total area 1.
Standard normal variable Z = (X − μ)/σ has mean 0 and SD 1; probabilities are read from the normal table.
Areas: about 68.27% of values lie within μ ± σ, 95.45% within μ ± 2σ, 99.73% within μ ± 3σ.
The moment generating function M(t) = E(e^(tX)) gives moments by differentiation.
Binomial and Poisson tend to normal for large n or large m.
Important terms
Trial (प्रयोग) – one performance of a random experiment.
Bernoulli trial – a trial with only two outcomes, success or failure.
Parameter – constant that fixes a distribution (n, p; m; μ, σ).
Standard normal table – gives area under N(0, 1) from 0 to z.
Formulas
Binomial: P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, r = 0, 1, …, n; mean np; variance npq
Poisson: P(X = r) = e⁻ᵐ mʳ/r!, r = 0, 1, 2, …; mean = variance = m
Normal pdf: f(x) = (1/(σ√(2π))) e^(−(x − μ)²/(2σ²))
Q1. For a binomial distribution, mean = 6 and variance = 4.2. Find n and p.
np = 6 and npq = 4.2, so q = 4.2/6 = 0.7 and p = 0.3. Then n = 6/0.3 = 20.
Q2. State two properties of the normal distribution.
The curve is bell-shaped and symmetric about x = μ, so mean, median and mode are equal. The total area under it is 1, and about 68.27% of the area lies within one standard deviation of the mean.
Q3. Ten coins are tossed. Find the probability of getting exactly 4 heads and at least one head.
n = 10, p = q = ½.
P(X = 4) = ¹⁰C₄ (½)¹⁰ = 210/1024 ≈ 0.2051.
P(X ≥ 1) = 1 − P(X = 0) = 1 − (½)¹⁰ = 1 − 1/1024.
= 1023/1024 ≈ 0.999.
Q4. The number of accidents per day on a road follows a Poisson distribution with mean 2. Find the probability of no accident and of at most one accident on a day (e⁻² = 0.1353).
P(X = 0) = e⁻² × 2⁰/0! = 0.1353.
P(X = 1) = e⁻² × 2¹/1! = 2 × 0.1353 = 0.2706.
P(X ≤ 1) = 0.1353 + 0.2706 = 0.4059.
Q5. Marks of students are normally distributed with mean 60 and SD 10. Find the percentage of students scoring above 80 and between 50 and 70 (area from 0 to z = 1 is 0.3413, to z = 2 is 0.4772).
For X = 80: Z = (80 − 60)/10 = 2.
P(X > 80) = P(Z > 2) = 0.5 − 0.4772 = 0.0228, i.e. about 2.28%.
For 50 and 70: Z = −1 and +1.
P(50 < X < 70) = P(−1 < Z < 1) = 2 × 0.3413 = 0.6826, i.e. about 68.26%.
Exam tip
Write the distribution, its parameters and the formula before substituting numbers. In normal-distribution questions, draw a small bell curve and shade the required area — it prevents adding when you should subtract.
Correlation measures how strongly two variables move together, and regression gives the best straight line to predict one variable from the other. Karl Pearson's coefficient r and Spearman's rank coefficient are the main measures. The two regression lines, y on x and x on y, meet at the means and are linked to r through the regression coefficients.
Key points
Positive correlation: both variables increase together; negative: one increases as the other decreases. A scatter diagram shows this visually.
Karl Pearson's coefficient: r = cov(x, y)/(σₓσᵧ) = Σdxdy/√(Σdx² Σdy²), where dx = x − x̄, dy = y − ȳ.
−1 ≤ r ≤ 1. r = ±1 means perfect linear relation; r = 0 means no linear relation (not necessarily independence).
r does not change if we shift the origin or change the scale (with positive factors).
Spearman's rank correlation: ρ = 1 − 6Σd²/(n(n² − 1)), where d = difference of ranks.
Regression line of y on x: y − ȳ = bᵧₓ(x − x̄), with bᵧₓ = rσᵧ/σₓ = Σdxdy/Σdx². Used to estimate y for a given x.
Regression line of x on y: x − x̄ = bₓᵧ(y − ȳ), with bₓᵧ = rσₓ/σᵧ = Σdxdy/Σdy².
r² = bᵧₓ × bₓᵧ; r has the same sign as both regression coefficients; both cannot exceed 1 in absolute value together (their product is at most 1).
The two regression lines intersect at (x̄, ȳ). If r = ±1 the lines coincide; if r = 0 they are perpendicular.
Important terms
Covariance – cov(x, y) = (1/n)Σ(x − x̄)(y − ȳ).
Correlation (सहसंबंध) – the degree of linear relationship between two variables.
Regression coefficient – the slope of a regression line.
Formulas
r = Σdxdy/√(Σdx² × Σdy²)
bᵧₓ = Σdxdy/Σdx², bₓᵧ = Σdxdy/Σdy², r = ±√(bᵧₓbₓᵧ)
ρ = 1 − 6Σd²/(n(n² − 1))
5 important questions
Q1. State any two properties of the correlation coefficient.
(1) −1 ≤ r ≤ 1. (2) r is independent of change of origin and scale. Also, r is the geometric mean of the two regression coefficients.
Q2. Find Spearman's rank correlation when the ranks are x: 1, 2, 3, 4, 5 and y: 2, 1, 4, 3, 5.
Q4. Find both regression lines for the data in Q3 and estimate y when x = 6.
bᵧₓ = 6/10 = 0.6. Line y on x: y − 4 = 0.6(x − 3), i.e. y = 0.6x + 2.2.
bₓᵧ = 6/6 = 1. Line x on y: x − 3 = 1(y − 4), i.e. x = y − 1.
Check: bᵧₓ × bₓᵧ = 0.6 = r², so r = √0.6 ≈ 0.775 ✓.
For x = 6: y = 0.6 × 6 + 2.2 = 5.8.
Q5. The regression lines are 8x − 10y + 66 = 0 and 40x − 18y = 214. Find x̄, ȳ and r.
Solve together: 5 × (first) gives 40x − 50y = −330. Subtract from the second: 32y = 544, so ȳ = 17. Then 8x = 170 − 66 = 104, so x̄ = 13.
Take the first as y on x: y = 0.8x + 6.6, so bᵧₓ = 0.8. Take the second as x on y: x = (214 + 18y)/40, so bₓᵧ = 0.45.
r² = 0.8 × 0.45 = 0.36 ≤ 1 (so the choice is correct); r = +0.6 (both coefficients positive).
Exam tip
Use the deviation table — it keeps calculations tidy and earns step marks. In "given two regression lines" problems, check that the product of the chosen coefficients is at most 1.