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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
सूत्रे
∫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)
५ महत्त्वाचे प्रश्न
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.
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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).
महत्त्वाच्या संज्ञा
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 ′.
सूत्रे
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
५ महत्त्वाचे प्रश्न
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.
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
सूत्रे
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
५ महत्त्वाचे प्रश्न
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.
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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).
महत्त्वाच्या संज्ञा
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.
सूत्रे
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
५ महत्त्वाचे प्रश्न
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.
परीक्षेसाठी टीप
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.
महत्त्वाच्या संज्ञा
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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
सूत्रे
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
५ महत्त्वाचे प्रश्न
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).
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
सूत्रे
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
५ महत्त्वाचे प्रश्न
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.
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
सूत्रे
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%.
परीक्षेसाठी टीप
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.
महत्त्वाचे मुद्दे
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.
महत्त्वाच्या संज्ञा
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.
सूत्रे
r = Σdxdy/√(Σdx² × Σdy²)
bᵧₓ = Σdxdy/Σdx², bₓᵧ = Σdxdy/Σdy², r = ±√(bᵧₓbₓᵧ)
ρ = 1 − 6Σd²/(n(n² − 1))
५ महत्त्वाचे प्रश्न
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).
परीक्षेसाठी टीप
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.