Mathematics and Statistics Part 1 गणित व संख्याशास्त्र भाग 1

Quick summary

Mathematical logic studies statements and how they combine. You learn statements and truth values, logical connectives, truth tables, statement patterns, logical equivalence, tautology/contradiction/contingency, quantifiers, duality, negation, algebra of statements and application to switching circuits.

Key points

• A statement is a declarative sentence that is either true (T) or false (F), not both. Questions, commands, exclamations and open sentences are not statements.
• Connectives: conjunction p ∧ q (and), disjunction p ∨ q (or), negation ~p (not), conditional p → q (if…then), biconditional p ↔ q (if and only if).
• p ∧ q is true only when both are true. p ∨ q is false only when both are false.
• p → q is false only when p is true and q is false.
• p ↔ q is true when p and q have the same truth value.
• For p → q: converse q → p, inverse ~p → ~q, contrapositive ~q → ~p. A conditional is equivalent to its contrapositive.
• A tautology is always true; a contradiction is always false; a contingency is neither.
• Quantifiers: "for all" (∀) and "there exists" (∃). Negation swaps them.
• Dual: interchange ∧ and ∨, and T and F.
• Switching circuits: switches in series ↔ ∧, in parallel ↔ ∨.

Important terms

• Truth value – T or F of a statement.
• Statement pattern – a compound expression built from simple statements and connectives.
• Logical equivalence (≡) – two patterns with identical truth tables.
• Open sentence – sentence with a variable whose truth depends on the value, e.g. "x + 2 = 5".

Formulas

5 important questions

Q1. Using a truth table, show that p → q ≡ ~q → ~p.

The last two columns are identical, so the statements are equivalent.

Q2. Examine whether (p ∧ q) → p is a tautology.

p ∧ q is true only when both p, q are true; then p is true, so the conditional is true. In all other rows p ∧ q is false, so the conditional is true. All values are T, so it is a tautology.

Q3. Write the negation of: "If it rains, then the match is cancelled."

~(p → q) ≡ p ∧ ~q: "It rains and the match is not cancelled."

Q4. Write the converse, inverse and contrapositive of: "If x is a prime number, then x is odd."

Converse: If x is odd, then x is prime. Inverse: If x is not prime, then x is not odd. Contrapositive: If x is not odd, then x is not prime.

Q5. Write the negation of "∀ n ∈ N, n + 3 > 2" and write the dual of p ∨ (q ∧ r).

Negation: ∃ n ∈ N such that n + 3 ≤ 2. Dual: p ∧ (q ∨ r).

Exam tip

Truth tables carry easy marks; draw them neatly with every column. In circuit questions, first write the statement form, then simplify using laws.

Quick summary

This chapter builds on Class 11 matrices. It covers elementary row and column transformations, the inverse of a matrix by the adjoint method and by elementary transformations, and solving a system of linear equations by the inversion method and by reduction method.

Key points

• A square matrix A is non-singular if |A| ≠ 0. Only non-singular matrices have an inverse.
• Elementary transformations: interchanging two rows (Rᵢ ↔ Rⱼ), multiplying a row by a non-zero number (k Rᵢ), adding a multiple of one row to another (Rᵢ + k Rⱼ). Same for columns.
• Inverse by adjoint method: A⁻¹ = (1/|A|) adj A.
• adj A is the transpose of the matrix of cofactors. Cofactor Aᵢⱼ = (−1)^(i+j) Mᵢⱼ.
• Inverse by row transformations: start with A = I A, apply row operations only until left side becomes I; then the right side gives A⁻¹.
• Properties: (AB)⁻¹ = B⁻¹A⁻¹, (A⁻¹)ᵀ = (Aᵀ)⁻¹, A · adj A = |A| I.
• For a system AX = B with |A| ≠ 0, the unique solution is X = A⁻¹B.
• Method of reduction: write AX = B, use row operations to reduce A to upper triangular form, then back-substitute.

Important terms

• Minor (Mᵢⱼ) – determinant left after deleting the ith row and jth column.
• Cofactor – minor with sign (−1)^(i+j).
• Adjoint – transpose of the cofactor matrix.
• Singular matrix – |A| = 0.
• Equivalent matrices – one obtained from the other by elementary transformations.

Formulas

5 important questions

Q1. Find the inverse of A = [2 3; 1 4] by the adjoint method.

|A| = 8 − 3 = 5 ≠ 0. adj A = [4 −3; −1 2]. A⁻¹ = (1/5)[4 −3; −1 2].
Check: A A⁻¹ = (1/5)[8 − 3, −6 + 6; 4 − 4, −3 + 8] = I.

Q2. Solve by the inversion method: x + y = 3, 2x − y = 0.

A = [1 1; 2 −1], B = [3; 0]. |A| = −1 − 2 = −3. adj A = [−1 −1; −2 1]. A⁻¹ = (−1/3)[−1 −1; −2 1].
X = A⁻¹B = (−1/3)[−3; −6] = [1; 2]. So x = 1, y = 2.

Q3. Find the inverse of A = [1 2; 3 7] using elementary row transformations.

Write [1 2; 3 7] = I · A. R₂ − 3R₁: [1 2; 0 1] = [1 0; −3 1] A. R₁ − 2R₂: [1 0; 0 1] = [7 −2; −3 1] A. So A⁻¹ = [7 −2; −3 1].

Q4. If A is a non-singular matrix, show that (Aᵀ)⁻¹ = (A⁻¹)ᵀ.

A A⁻¹ = I. Taking transpose: (A⁻¹)ᵀ Aᵀ = Iᵀ = I. So (A⁻¹)ᵀ is the inverse of Aᵀ, i.e. (Aᵀ)⁻¹ = (A⁻¹)ᵀ.

Q5. Solve by the method of reduction: x + y + z = 6, y + 3z = 11, x − 2y + z = 0.

R₃ − R₁ on the system: −3y = −6, so y = 2. Then 2 + 3z = 11 gives z = 3. Then x = 6 − 2 − 3 = 1. Solution: x = 1, y = 2, z = 3.

Exam tip

Always check |A| ≠ 0 first, and verify the inverse by multiplying; in row transformation questions use only row operations throughout.

Quick summary

This chapter covers trigonometric equations and their general solutions, solutions of a triangle using the sine rule, cosine rule and projection rule, and inverse trigonometric functions with their principal values and properties.

Key points

• Principal solutions lie in 0 ≤ x < 2π. General solutions cover all angles.
• General solutions: sin θ = sin α ⇒ θ = nπ + (−1)ⁿ α; cos θ = cos α ⇒ θ = 2nπ ± α; tan θ = tan α ⇒ θ = nπ + α (n ∈ Z).
• sin θ = 0 ⇒ θ = nπ; cos θ = 0 ⇒ θ = (2n + 1)π/2; tan θ = 0 ⇒ θ = nπ.
• sin²θ = sin²α (or cos², tan²) ⇒ θ = nπ ± α.
• In triangle ABC with sides a, b, c opposite angles A, B, C: sine rule, cosine rule and projection rule connect sides and angles.
• Half-angle formulas use semi-perimeter s = (a + b + c)/2. Area = ½ bc sin A = √(s(s − a)(s − b)(s − c)) (Heron's formula).
• Napier's analogy: tan((B − C)/2) = ((b − c)/(b + c)) cot(A/2).
• Inverse functions are defined on restricted domains. Principal value ranges: sin⁻¹: [−π/2, π/2]; cos⁻¹: [0, π]; tan⁻¹: (−π/2, π/2).

Important terms

• Trigonometric equation – an equation involving trigonometric functions of an unknown angle.
• General solution – expression giving all solutions using an integer n.
• Solving a triangle – finding unknown sides and angles from given ones.
• Principal value – the value of an inverse trigonometric function within its principal range.

Formulas

5 important questions

Q1. Find the general solution of sin θ = 1/2.

sin θ = 1/2 = sin(π/6). So θ = nπ + (−1)ⁿ π/6, n ∈ Z.

Q2. Find the principal solutions of tan x = −√3.

tan(π/3) = √3. tan is negative in the 2nd and 4th quadrants: x = π − π/3 = 2π/3 and x = 2π − π/3 = 5π/3.

Q3. In triangle ABC, if a = 3, b = 5, c = 7, find angle C.

cos C = (a² + b² − c²)/(2ab) = (9 + 25 − 49)/30 = −15/30 = −1/2. So C = 2π/3 (120°).

Q4. Prove that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4.

Using the sum formula: tan⁻¹((1/2 + 1/3)/(1 − 1/6)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹(1) = π/4.

Q5. Prove the sine rule: a/sin A = b/sin B = c/sin C.

Draw altitude AD from A to BC. In triangle ABD, AD = c sin B; in triangle ACD, AD = b sin C. So c sin B = b sin C, i.e. b/sin B = c/sin C. Similarly, drawing an altitude from B gives a/sin A = c/sin C.

Exam tip

Write the general solution formula first, then substitute. For inverse functions, check that the answer lies in the principal range.

Quick summary

Two straight lines through the origin can be represented by one second-degree homogeneous equation. This chapter covers the combined equation of two lines, homogeneous equation ax² + 2hxy + by² = 0, the angle between the lines, conditions for perpendicular and coincident lines, and the general second-degree equation representing a pair of lines.

Key points

• If u₁ = 0 and u₂ = 0 are two lines, their combined equation is u₁ · u₂ = 0.
• Every homogeneous equation of degree two, ax² + 2hxy + by² = 0, represents a pair of lines through the origin if h² − ab ≥ 0.
• If the lines are y = m₁x and y = m₂x, then m₁ + m₂ = −2h/b and m₁ m₂ = a/b (b ≠ 0).
• Lines are real and distinct if h² > ab, coincident if h² = ab, imaginary if h² < ab.
• Lines are perpendicular if a + b = 0 (coefficient of x² + coefficient of y² = 0).
• The general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines if the determinant |a h g; h b f; g f c| = 0, i.e. abc + 2fgh − af² − bg² − ch² = 0, and h² − ab ≥ 0.
• The pair through the origin parallel to the lines given by the general equation is ax² + 2hxy + by² = 0 (same angle between them).

Important terms

• Homogeneous equation – each term has the same total degree in x and y.
• Combined (joint) equation – one equation representing both lines.
• Acute angle between lines – the angle θ with 0 < θ ≤ π/2.

Formulas

5 important questions

Q1. Find the combined equation of the lines x + 2y = 0 and 3x − y = 0.

(x + 2y)(3x − y) = 0 ⇒ 3x² − xy + 6xy − 2y² = 0 ⇒ 3x² + 5xy − 2y² = 0.

Q2. Find the separate equations of the lines given by x² − 5xy + 6y² = 0.

Factorise: (x − 2y)(x − 3y) = 0. So the lines are x − 2y = 0 and x − 3y = 0.

Q3. Find the acute angle between the lines represented by x² + 4xy + y² = 0.

a = 1, h = 2, b = 1. tan θ = |2√(4 − 1)/(1 + 1)| = √3. So θ = 60° (π/3).

Q4. Show that the lines 3x² − 4xy − 3y² = 0 are perpendicular.

Coefficient of x² + coefficient of y² = 3 + (−3) = 0, so the lines are perpendicular.

Q5. Show that 2x² + xy − y² + x + 4y − 3 = 0 represents a pair of lines.

Compare: a = 2, h = 1/2, b = −1, g = 1/2, f = 2, c = −3.
abc + 2fgh − af² − bg² − ch² = 6 + 1 − 8 + 1/4 + 3/4 = 0, and h² − ab = 1/4 + 2 > 0.
So it represents a pair of lines. In fact it factorises as (2x − y + 3)(x + y − 1) = 0.

Exam tip

Learn to derive m₁ + m₂ and m₁m₂ by dividing ax² + 2hxy + by² = 0 by x² and treating it as a quadratic in m = y/x.

Quick summary

This chapter covers vectors in three dimensions: representation, position vectors, collinearity and coplanarity, section formula, scalar (dot) product, vector (cross) product, scalar triple product, vector triple product, and applications to geometry such as medians and altitudes of a triangle.

Key points

• A vector has magnitude and direction. In 3D, a = a₁ i + a₂ j + a₃ k and |a| = √(a₁² + a₂² + a₃²).
• Unit vector along a: â = a/|a|.
• Direction cosines l, m, n satisfy l² + m² + n² = 1.
• Two vectors are collinear if one is a scalar multiple of the other. Three vectors are coplanar if their scalar triple product is zero.
• Section formula: point dividing AB internally in ratio m : n has position vector r = (m b + n a)/(m + n).
• Dot product a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃. a · b = 0 means a ⊥ b.
• Cross product a × b = |a||b| sin θ n̂; it is perpendicular to both a and b. a × b = 0 means a ∥ b. a × b = −(b × a).
• |a × b| = area of the parallelogram with adjacent sides a and b. Area of triangle = ½|a × b|.
• Scalar triple product [a b c] = a · (b × c) = volume of the parallelepiped.
• Projection of a on b = (a · b)/|b|.

Important terms

• Position vector – vector from the origin to a point.
• Collinear vectors – vectors along the same or parallel lines.
• Coplanar vectors – vectors lying in the same plane.
• Centroid – point where medians meet: (a + b + c)/3.

Formulas

5 important questions

Q1. Find the angle between a = i + j + 2k and b = 2i − j + k.

a · b = 2 − 1 + 2 = 3. |a| = √6, |b| = √6. cos θ = 3/6 = 1/2, so θ = π/3.

Q2. Find a unit vector perpendicular to both a = i + j and b = j + k.

a × b = det[i j k; 1 1 0; 0 1 1] = i(1 − 0) − j(1 − 0) + k(1 − 0) = i − j + k. Magnitude √3. Unit vector = (i − j + k)/√3.

Q3. Find the area of the triangle with vertices A(1, 0, 0), B(0, 1, 0), C(0, 0, 1).

AB = −i + j, AC = −i + k. AB × AC = det[i j k; −1 1 0; −1 0 1] = i(1) − j(−1) + k(1) = i + j + k. Area = ½ √3 sq units.

Q4. Show that the vectors a = i + 2j + 3k, b = 2i + j + 3k, c = 3i + 3j + 6k are coplanar.

[a b c] = 1(6 − 9) − 2(12 − 9) + 3(6 − 3) = −3 − 6 + 9 = 0. So they are coplanar (indeed c = a + b).

Q5. Prove by the vector method that the medians of a triangle are concurrent.

Let A, B, C have position vectors a, b, c. Midpoint D of BC is (b + c)/2. The point dividing AD in ratio 2 : 1 is (2(b + c)/2 + a)/3 = (a + b + c)/3. By symmetry the same point lies on the other two medians, so the medians meet at G = (a + b + c)/3.

Exam tip

Expand determinants carefully with the alternating signs (+, −, +). Vector geometry proofs are often asked as 4-mark questions.

Quick summary

This chapter gives the vector and Cartesian equations of a line and a plane in space. It covers lines through a point parallel to a vector, lines through two points, angle between lines, shortest distance between skew and parallel lines, distance of a point from a line, equations of a plane in different forms, angle between planes, angle between a line and a plane, coplanarity of lines and distance of a point from a plane.

Key points

• Line through point A(a) parallel to b: r = a + λ b. Cartesian: (x − x₁)/b₁ = (y − y₁)/b₂ = (z − z₁)/b₃.
• Line through two points A(a) and B(b): r = a + λ(b − a).
• Angle between two lines = angle between their direction vectors.
• Skew lines are neither parallel nor intersecting.
• Plane through a point A(a) perpendicular to n: (r − a) · n = 0, i.e. r · n = a · n.
• Normal form: r · n̂ = p, where p is the distance of the plane from the origin.
• Cartesian form: a x + b y + c z + d = 0, where (a, b, c) are direction ratios of the normal.
• Plane through three non-collinear points A, B, C: [r − a, b − a, c − a] = 0.
• Angle between planes = angle between their normals.
• Lines r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar if (a₂ − a₁) · (b₁ × b₂) = 0.

Important terms

• Direction ratios – numbers proportional to the direction cosines of a line.
• Normal to a plane – a vector perpendicular to every line in the plane.
• Shortest distance – length of the common perpendicular between two lines.
• Intercept form – x/a + y/b + z/c = 1.

Formulas

5 important questions

Q1. Find the vector and Cartesian equations of the line through (1, 2, 3) parallel to 2i − j + 4k.

Vector: r = (i + 2j + 3k) + λ(2i − j + 4k). Cartesian: (x − 1)/2 = (y − 2)/(−1) = (z − 3)/4.

Q2. Find the distance of the point (2, 1, −1) from the plane x − 2y + 4z − 9 = 0.

d = |2 − 2 − 4 − 9|/√(1 + 4 + 16) = 13/√21 units.

Q3. Find the angle between the planes 2x − y + z = 6 and x + y + 2z = 7.

n₁ = (2, −1, 1), n₂ = (1, 1, 2). n₁ · n₂ = 2 − 1 + 2 = 3. |n₁| = |n₂| = √6. cos θ = 3/6 = 1/2, so θ = π/3.

Q4. Find the shortest distance between r = (i + 2j + k) + λ(i − j + k) and r = (2i − j − k) + μ(2i + j + 2k).

a₂ − a₁ = i − 3j − 2k. b₁ × b₂ = det[i j k; 1 −1 1; 2 1 2] = i(−2 − 1) − j(2 − 2) + k(1 + 2) = −3i + 3k. |b₁ × b₂| = 3√2. (a₂ − a₁) · (b₁ × b₂) = −3 + 0 − 6 = −9. d = 9/(3√2) = 3/√2 units.

Q5. Find the equation of the plane through (1, 1, 1) perpendicular to the vector 2i + 3j − k.

2(x − 1) + 3(y − 1) − 1(z − 1) = 0 ⇒ 2x + 3y − z = 4.

Exam tip

For angle between a line and a plane use sin, not cos. Write both vector and Cartesian forms when the question asks for "equation".

Quick summary

Linear programming finds the best (maximum or minimum) value of a linear function subject to linear inequalities. This chapter covers graphing linear inequations, formulating a linear programming problem (LPP) from a word problem, and solving an LPP by the graphical (corner point) method.

Key points

• A linear inequation like a x + b y ≤ c represents a half-plane. Draw the line a x + b y = c, then test a point (usually the origin) to choose the side.
• Objective function Z = a x + b y is the function to maximise or minimise (profit, cost).
• Constraints are the linear inequalities the variables must satisfy. Non-negativity constraints: x ≥ 0, y ≥ 0.
• The feasible region is the common region satisfying all constraints. Every point in it is a feasible solution.
• Corner point theorem: if an optimum exists, it occurs at a corner (vertex) of the feasible region.
• If the feasible region is bounded, both maximum and minimum exist.
• If it is unbounded, an optimum may not exist; check carefully.
• If two corners give the same optimum value, every point on the segment joining them is also optimal (multiple optimal solutions).

Important terms

• Decision variables – the unknowns x, y to be decided.
• Feasible region – set of all points satisfying all constraints.
• Optimal solution – feasible point giving the best value of Z.
• Infeasible problem – no point satisfies all constraints.
• Bounded region – region that can be enclosed in a circle.

Formulas

Steps of the graphical method:

1. Read the problem and define x and y.
2. Write the objective function Z.
3. Write all constraints as inequalities, including x ≥ 0, y ≥ 0.
4. Draw the lines, shade the feasible region.
5. Find coordinates of all corner points (solve pairs of line equations).
6. Find Z at each corner and pick the maximum or minimum.

5 important questions

Q1. Maximise Z = 3x + 2y subject to x + y ≤ 4, x ≤ 3, x ≥ 0, y ≥ 0.

Corner points: O(0, 0), A(3, 0), B(3, 1), C(0, 4).
Z(O) = 0, Z(A) = 9, Z(B) = 11, Z(C) = 8.
Maximum Z = 11 at x = 3, y = 1.

Q2. Minimise Z = 2x + 3y subject to x + y ≥ 5, x ≥ 1, y ≥ 1.

The region is unbounded. Corners: A(4, 1) and B(1, 4). Z(A) = 8 + 3 = 11, Z(B) = 2 + 12 = 14. Testing 2x + 3y < 11 shows no common point with the region, so minimum Z = 11 at (4, 1).

Q3. Formulate: A company makes chairs (profit ₹40) and tables (profit ₹60). A chair needs 2 hours of work and a table 3 hours; 120 hours are available. At most 30 tables can be made. Maximise profit.

Let x = chairs, y = tables. Maximise Z = 40x + 60y subject to 2x + 3y ≤ 120, y ≤ 30, x ≥ 0, y ≥ 0.

Q4. Solve Q3 graphically.

Corners: O(0, 0), A(60, 0), B(15, 30), C(0, 30). Z: 0, 2400, 600 + 1800 = 2400, 1800. Max Z = 2400 at both A and B, so every point on segment AB is optimal (multiple optimal solutions).

Q5. Define the feasible region and state the corner point theorem.

Feasible region: the common region of all constraints including non-negativity. Corner point theorem: the optimum value of a linear objective function over a bounded feasible region occurs at one of its corner points.

Exam tip

Draw the graph neatly on a proper scale, mark the feasible region by shading, and show a table of Z values at all corners.