Mathematics and Statistics (Commerce) Part 1 गणित व संख्याशास्त्र (वाणिज्य) भाग 1

Quick summary

Mathematical logic studies statements that are either true or false and how they combine using logical connectives. This chapter covers statements and truth values, logical connectives and truth tables, quantifiers, statement patterns, tautology/contradiction/contingency, duality, negation, algebra of statements and Venn diagrams.

Key points

• A statement is a declarative sentence that is either true (T) or false (F), not both. Questions, orders, exclamations and opinions 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 T and q is F; p ↔ q is true when both have the same truth value.
• Quantifiers: "for all" (∀, universal) and "there exists" (∃, existential).
• Tautology: always true (e.g., p ∨ ~p). Contradiction: always false (e.g., p ∧ ~p). Contingency: neither.
• Dual: interchange ∧ and ∨, and T and F.
• Converse of p → q is q → p; inverse is ~p → ~q; contrapositive is ~q → ~p (logically equivalent to p → q).
• Laws: idempotent, commutative, associative, distributive, identity, complement, De Morgan's: ~(p ∧ q) ≡ ~p ∨ ~q and ~(p ∨ q) ≡ ~p ∧ ~q.
• Negations: ~(p → q) ≡ p ∧ ~q; ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p); ~(∀x, P(x)) ≡ ∃x, ~P(x).
• p → q ≡ ~p ∨ q. Venn diagrams can represent statements using sets.

Important terms

• Truth value – T or F of a statement.
• Compound statement – statement made of simple statements and connectives.
• Logically equivalent – two statement patterns with the same truth table (≡).
• Open sentence – sentence with a variable whose truth depends on its value.

Formulas

5 important questions

Q1. Which of the following are statements? (i) 2 is a prime number. (ii) Close the door.

(i) is a statement (true). (ii) is an order, not a statement.

Q2. If p is true and q is false, find the truth value of (p ∨ q) → (p ∧ q).

p ∨ q = T, p ∧ q = F, so T → F = F. The statement is false.

Q3. Show that p ∨ ~p is a tautology.

All values are T, so it is a tautology.

Q4. Write the negation of "If it rains, then the match is cancelled."

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

Q5. Write the converse, inverse and contrapositive of "If x is even, then x² is even."

Converse: If x² is even then x is even. Inverse: If x is not even then x² is not even. Contrapositive: If x² is not even then x is not even.

Exam tip

Always build a full truth table column by column — examiners give marks for each correct column.

Quick summary

A matrix is a rectangular arrangement of numbers in rows and columns. This chapter covers types of matrices, algebra of matrices (addition, scalar multiplication, multiplication), transpose, properties, elementary transformations, inverse of a matrix (adjoint method and elementary transformations) and solving linear equations using matrices.

Key points

• Theory of matrices was developed by Arthur Cayley. An m × n matrix has m rows and n columns.
• Types: row, column, square, zero (null), diagonal, scalar, identity (unit), upper/lower triangular, symmetric (Aᵀ = A), skew-symmetric (Aᵀ = −A), singular (|A| = 0), non-singular (|A| ≠ 0).
• Addition is possible only for matrices of the same order. AB is defined if columns of A = rows of B.
• Matrix multiplication is generally not commutative (AB ≠ BA), but is associative and distributive.
• Transpose properties: (Aᵀ)ᵀ = A, (A + B)ᵀ = Aᵀ + Bᵀ, (AB)ᵀ = BᵀAᵀ.
• Elementary transformations: interchange of rows/columns, multiplying a row/column by a non-zero number, adding a multiple of one row/column to another.
• Inverse exists only for a non-singular square matrix: A⁻¹ = adj(A)/|A|.
• Solving AX = B: X = A⁻¹B (method of inversion) or by reduction.

Important terms

• Order – number of rows × number of columns.
• Cofactor Aij = (−1)^(i+j) × Mij, where Mij is the minor.
• Adjoint – transpose of the cofactor matrix.
• Identity matrix I – diagonal entries 1, others 0; AI = IA = A.

Formulas

• A⁻¹ = (1/|A|) × adj(A), |A| ≠ 0
• For A = [a b; c d]: A⁻¹ = (1/(ad − bc)) × [d −b; −c a]
• AA⁻¹ = A⁻¹A = I; (AB)⁻¹ = B⁻¹A⁻¹
• Every square matrix = symmetric part + skew-symmetric part: A = ½(A + Aᵀ) + ½(A − Aᵀ)

5 important questions

Q1. If A = [1 2; 3 4] and B = [2 0; 1 3], find AB.

AB = [1×2 + 2×1, 1×0 + 2×3; 3×2 + 4×1, 3×0 + 4×3] = [4 6; 10 12].

Q2. Find the inverse of A = [2 3; 1 4].

|A| = 8 − 3 = 5 ≠ 0. adj A = [4 −3; −1 2]. A⁻¹ = (1/5)[4 −3; −1 2].

Q3. Solve using matrices: 2x + 3y = 8, x + 4y = 9.

A = [2 3; 1 4], X = [x; y], B = [8; 9]. X = A⁻¹B = (1/5)[4×8 − 3×9; −1×8 + 2×9] = (1/5)[5; 10] = [1; 2]. So x = 1, y = 2.

Q4. When is a matrix symmetric? Give an example.

When Aᵀ = A, i.e., aij = aji. Example: [1 5; 5 2].

Q5. Show that matrix multiplication is not commutative using A = [1 2; 3 4], B = [2 0; 1 3].

AB = [4 6; 10 12]; BA = [2×1 + 0×3, 2×2 + 0×4; 1×1 + 3×3, 1×2 + 3×4] = [2 4; 10 14]. AB ≠ BA.

Exam tip

Check |A| ≠ 0 before finding the inverse, and verify your answer by putting x and y back into the equations.

Quick summary

This chapter extends differentiation learnt in Class 11 to new types of functions: composite functions (chain rule), inverse functions, logarithmic functions (logarithmic differentiation), implicit functions, parametric functions, and second-order derivatives.

Key points

• Chain rule (composite function): if y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx).
• Inverse function: if y = f(x) and x = f⁻¹(y), then dx/dy = 1 ÷ (dy/dx), provided dy/dx ≠ 0.
• Logarithmic differentiation: take log on both sides before differentiating — useful for y = [f(x)]^g(x) and for products/quotients of many factors.
• Implicit function: x and y are mixed, e.g., x² + y² = 25. Differentiate each term w.r.t. x, treating y as a function of x, and solve for dy/dx.
• Parametric function: x = f(t), y = g(t). Then dy/dx = (dy/dt) ÷ (dx/dt).
• Second-order derivative: d²y/dx² = d/dx (dy/dx).
• Rules recalled: (u ± v)' = u' ± v'; (uv)' = uv' + vu'; (u/v)' = (vu' − uv')/v²; (ku)' = k u'.

Important terms

• Composite function – function of a function, f(g(x)).
• Implicit function – y not expressed directly in terms of x.
• Parameter – third variable (t) through which x and y are expressed.

Formulas

• d/dx (xⁿ) = n xⁿ⁻¹; d/dx (eˣ) = eˣ; d/dx (aˣ) = aˣ log a
• d/dx (log x) = 1/x; d/dx (constant) = 0
• d/dx [f(x)]ⁿ = n[f(x)]ⁿ⁻¹ × f'(x)
• d/dx log f(x) = f'(x)/f(x); d/dx e^f(x) = e^f(x) × f'(x)

5 important questions

Q1. Differentiate y = (x² + 5)⁴ w.r.t. x.

dy/dx = 4(x² + 5)³ × 2x = 8x(x² + 5)³.

Q2. Find dy/dx if y = log(x² + 1).

dy/dx = 2x ÷ (x² + 1).

Q3. Find dy/dx if x² + y² = 25.

2x + 2y (dy/dx) = 0, so dy/dx = −x/y.

Q4. Find dy/dx if x = t², y = 2t.

dx/dt = 2t, dy/dt = 2. dy/dx = 2 ÷ 2t = 1/t.

Q5. Differentiate y = xˣ.

log y = x log x. (1/y)(dy/dx) = x × (1/x) + log x × 1 = 1 + log x. So dy/dx = xˣ(1 + log x).

Exam tip

For implicit and logarithmic differentiation, write every step; forgetting the dy/dx factor on y-terms is the most common mistake.

Quick summary

Derivatives measure the rate of change. This chapter uses them to find the slope of tangents, to decide whether a function is increasing or decreasing, to find maximum and minimum values, and to solve economics problems involving cost, revenue, profit and elasticity of demand.

Key points

• f'(a) = slope of the tangent to the curve y = f(x) at x = a.
• f is increasing where f'(x) > 0 and decreasing where f'(x) < 0.
• Maxima/minima: solve f'(x) = 0 to get critical points. If f''(x) < 0 there → maximum; if f''(x) > 0 → minimum (second derivative test).
• Economics: total cost C, average cost CA = C/x, marginal cost CM = dC/dx; revenue R = P × D, average revenue RA = R/D = P, marginal revenue RM = dR/dD; profit π = R − C.
• Profit is maximum where dπ/dx = 0 and d²π/dx² < 0 (i.e., MR = MC).
• Elasticity of demand: η = −(P/D) × (dD/dP). If η = 0 demand is perfectly inelastic; η < 1 inelastic; η = 1 unitary; η > 1 elastic.
• Relation: RM = RA (1 − 1/η).

Important terms

• Critical point – where f'(x) = 0.
• Marginal cost – extra cost of producing one more unit.
• Marginal revenue – extra revenue from selling one more unit.
• Elasticity of demand – responsiveness of demand to price.

Formulas

• Slope of tangent = f'(a)
• CA = C/x; CM = dC/dx
• R = P × D; RA = P; RM = dR/dD
• π = R − C
• η = −(P/D)(dD/dP)
• RM = RA(1 − 1/η)

5 important questions

Q1. Find the intervals in which f(x) = x² − 4x + 3 is increasing or decreasing.

f'(x) = 2x − 4. f'(x) > 0 when x > 2 (increasing); f'(x) < 0 when x < 2 (decreasing).

Q2. Find the maximum and minimum values of f(x) = 2x³ − 21x² + 36x − 20.

f'(x) = 6x² − 42x + 36 = 6(x − 1)(x − 6) = 0 gives x = 1, 6. f''(x) = 12x − 42. At x = 1, f'' = −30 < 0 → maximum f(1) = 2 − 21 + 36 − 20 = −3. At x = 6, f'' = 30 > 0 → minimum f(6) = 432 − 756 + 216 − 20 = −128.

Q3. Divide 20 into two parts such that their product is maximum.

Let parts be x and 20 − x. P = 20x − x². dP/dx = 20 − 2x = 0 → x = 10; d²P/dx² = −2 < 0. Parts are 10 and 10.

Q4. Cost function C = x² + 10x + 50. Find average cost and marginal cost at x = 5.

CA = x + 10 + 50/x = 5 + 10 + 10 = 25. CM = 2x + 10 = 20.

Q5. Demand D = 50 − 3P. Find elasticity at P = 5.

D = 35, dD/dP = −3. η = −(5/35)(−3) = 15/35 = 3/7 < 1, so demand is inelastic.

Exam tip

In maxima–minima problems, always state the second-derivative sign before concluding 'maximum' or 'minimum'.

Quick summary

Integration is the reverse process of differentiation. If d/dx [f(x)] = g(x), then f(x) is an integral (antiderivative) of g(x). This chapter covers standard integrals, integration by substitution, some special integrals, integration by parts and integration by partial fractions.

Key points

• ∫ g(x) dx = f(x) + c, where c is the constant of integration.
• ∫ k f(x) dx = k ∫ f(x) dx; ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Substitution: put u = g(x) so that du = g'(x) dx; useful when the integrand contains a function and its derivative.
• Useful result: ∫ f'(x)/f(x) dx = log|f(x)| + c; ∫ eˣ[f(x) + f'(x)] dx = eˣ f(x) + c.
• Integration by parts: ∫ u v dx = u ∫ v dx − ∫ [u' ∫ v dx] dx. Choose u by the LIATE order (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential).
• Partial fractions: break a proper rational function into simpler fractions, e.g., 1/((x − a)(x − b)) = A/(x − a) + B/(x − b).

Important terms

• Integrand – the function being integrated.
• Antiderivative / primitive – a function whose derivative is the integrand.
• Proper rational function – degree of numerator less than degree of denominator.

Formulas

• ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, n ≠ −1
• ∫ (1/x) dx = log|x| + c
• ∫ eˣ dx = eˣ + c; ∫ aˣ dx = aˣ/log a + c
• ∫ 1/(x² − a²) dx = (1/2a) log|(x − a)/(x + a)| + c
• ∫ 1/(a² − x²) dx = (1/2a) log|(a + x)/(a − x)| + c
• ∫ 1/√(x² + a²) dx = log|x + √(x² + a²)| + c
• ∫ 1/√(x² − a²) dx = log|x + √(x² − a²)| + c

5 important questions

Q1. Evaluate ∫ (3x² + 4x + 5) dx.

= x³ + 2x² + 5x + c.

Q2. Evaluate ∫ 2x/(x² + 1) dx.

Numerator is the derivative of the denominator, so the answer is log|x² + 1| + c.

Q3. Evaluate ∫ x eˣ dx.

By parts with u = x, v = eˣ: x eˣ − ∫ 1 × eˣ dx = x eˣ − eˣ + c = eˣ(x − 1) + c.

Q4. Evaluate ∫ 1/(x² − 9) dx.

Here a = 3: (1/6) log|(x − 3)/(x + 3)| + c.

Q5. Evaluate ∫ 1/((x − 1)(x − 2)) dx.

1/((x − 1)(x − 2)) = −1/(x − 1) + 1/(x − 2). Integral = −log|x − 1| + log|x − 2| + c = log|(x − 2)/(x − 1)| + c.

Exam tip

Never forget "+ c" in indefinite integrals — it costs marks in every answer.

Quick summary

A definite integral gives a number, not a function. If φ'(x) = f(x), then the definite integral of f(x) from a to b is φ(b) − φ(a). This chapter explains how to evaluate definite integrals, the fundamental theorem of integral calculus and the properties of definite integrals that make evaluation easier.

Key points

• ∫[a to b] f(x) dx = [φ(x)] from a to b = φ(b) − φ(a). No constant c is needed.
• a = lower limit, b = upper limit.
• With substitution, change the limits as well.
• Properties:
• P1: ∫[a to b] f(x) dx = ∫[a to b] f(t) dt (variable does not matter).
• P2: ∫[a to b] f(x) dx = −∫[b to a] f(x) dx; ∫[a to a] f(x) dx = 0.
• P3: ∫[a to b] f(x) dx = ∫[a to c] f(x) dx + ∫[c to b] f(x) dx, where a < c < b.
• P4: ∫[a to b] f(x) dx = ∫[a to b] f(a + b − x) dx.
• P5: ∫[0 to a] f(x) dx = ∫[0 to a] f(a − x) dx.
• P6: ∫[−a to a] f(x) dx = 2∫[0 to a] f(x) dx if f is even; = 0 if f is odd.
• P4/P5 are very useful when adding the original and new integrals gives a simple integrand.

Important terms

• Even function – f(−x) = f(x), e.g., x².
• Odd function – f(−x) = −f(x), e.g., x³.
• Limits of integration – the values a and b.

Formulas

• ∫[a to b] xⁿ dx = (bⁿ⁺¹ − aⁿ⁺¹)/(n + 1)
• ∫[a to b] (1/x) dx = log b − log a = log(b/a)
• ∫[a to b] eˣ dx = eᵇ − eᵃ

5 important questions

Q1. Evaluate ∫[1 to 3] x² dx.

[x³/3] from 1 to 3 = 27/3 − 1/3 = 26/3.

Q2. Evaluate ∫[0 to 1] (2x + 3) dx.

[x² + 3x] from 0 to 1 = 1 + 3 − 0 = 4.

Q3. Evaluate ∫[−2 to 2] x³ dx.

x³ is odd, so the integral = 0.

Q4. Evaluate ∫[1 to e] (1/x) dx.

[log x] from 1 to e = log e − log 1 = 1 − 0 = 1.

Q5. Evaluate ∫[0 to 3] √x/(√x + √(3 − x)) dx.

Let I be the integral. By P5, I = ∫[0 to 3] √(3 − x)/(√(3 − x) + √x) dx. Adding, 2I = ∫[0 to 3] 1 dx = 3, so I = 3/2.

Exam tip

Learn property P5 well — 'add the two forms' questions are asked almost every year.

Quick summary

Definite integration can be used to find the area of a region bounded by curves and lines. This chapter shows how to find the area under a curve between two ordinates (x = a, x = b), the area bounded by a curve and the y-axis, and areas related to standard curves like the parabola and the ellipse.

Key points

• Area bounded by y = f(x), the X-axis and the lines x = a, x = b is A = ∫[a to b] y dx (when the curve is above the X-axis).
• If the curve is below the X-axis, the integral is negative; take its absolute value for area.
• Area bounded by x = g(y), the Y-axis and the lines y = c, y = d is A = ∫[c to d] x dy.
• Area between two curves y = f(x) and y = g(x) from x = a to x = b, with f(x) ≥ g(x): A = ∫[a to b] [f(x) − g(x)] dx.
• Always draw a rough sketch first to see which part is above and where the curves meet (solve the equations together to get limits).
• Standard curves: parabola y² = 4ax (opens right), x² = 4by (opens up), x² = −4by (opens down); ellipse x²/a² + y²/b² = 1; circle x² + y² = r².
• Use symmetry: for a circle or ellipse, find area in the first quadrant and multiply by 4.
• Area is always positive and is in square units.

Important terms

• Ordinate – a vertical line x = a.
• Bounded region – area enclosed fully by curves and lines.
• Point of intersection – where two curves meet; gives the limits.

Formulas

• A = ∫[a to b] y dx
• A = ∫[c to d] x dy
• Area of circle x² + y² = r² is πr²
• Area of ellipse x²/a² + y²/b² = 1 is πab

5 important questions

Q1. Find the area bounded by y = x², the X-axis and x = 1 to x = 3.

A = ∫[1 to 3] x² dx = [x³/3] from 1 to 3 = (27 − 1)/3 = 26/3 sq. units.

Q2. Find the area bounded by y = 2x, the X-axis and x = 0 to x = 4.

A = ∫[0 to 4] 2x dx = [x²] from 0 to 4 = 16 sq. units.

Q3. Find the area bounded by y = x³, the X-axis and x = −1 to x = 0.

∫[−1 to 0] x³ dx = [x⁴/4] from −1 to 0 = 0 − 1/4 = −1/4. Area = 1/4 sq. unit.

Q4. Find the area bounded by y² = 4x, the X-axis and the line x = 1 (in the first quadrant).

y = 2√x. A = ∫[0 to 1] 2x^(1/2) dx = 2 × (2/3)[x^(3/2)] from 0 to 1 = 4/3 sq. units.

Q5. Find the area of the region between y = x and y = x².

They meet at x = 0 and x = 1, and x ≥ x² there. A = ∫[0 to 1] (x − x²) dx = 1/2 − 1/3 = 1/6 sq. unit.

Exam tip

Write 'sq. units' with every area answer and draw a quick sketch — it helps you choose correct limits.

Quick summary

A differential equation is an equation that contains derivatives of a dependent variable. This chapter covers order and degree, solution and formation of differential equations, methods of solving (variables separable, homogeneous and linear equations), and applications such as population growth and growth of bacteria.

Key points

• Ordinary differential equation: dependent variable depends on only one independent variable.
• Order: order of the highest derivative present. Degree: power of the highest-order derivative after removing radicals and fractions.
• General solution has as many arbitrary constants as the order; a particular solution is obtained by giving values to the constants.
• Formation: differentiate the given equation as many times as the number of arbitrary constants and eliminate them.
• Variables separable: write as f(y) dy = g(x) dx and integrate both sides.
• Homogeneous equation: dy/dx = f(y/x); put y = vx, so dy/dx = v + x dv/dx, then separate variables.
• Linear equation: dy/dx + Py = Q (P, Q functions of x). Integrating factor I.F. = e^(∫P dx). Solution: y × (I.F.) = ∫ Q × (I.F.) dx + c.
• Applications: if a quantity grows (or decays) at a rate proportional to its present amount, dx/dt = kx, which gives x = a e^(kt). Used for population growth, bacteria growth, compound interest.

Important terms

• Arbitrary constant – constant that can take any value in the general solution.
• Integrating factor – factor that makes a linear equation exactly integrable.
• Constant of proportionality (k) – growth rate constant.

Formulas

• I.F. = e^(∫P dx)
• y(I.F.) = ∫ Q(I.F.) dx + c
• dP/dt = kP ⇒ P = a e^(kt), where a = initial value (at t = 0)
• e^(log x) = x

5 important questions

Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.

Highest derivative is d²y/dx², so order = 2; its power is 3, so degree = 3.

Q2. Solve dy/dx = x/y.

y dy = x dx. Integrating: y²/2 = x²/2 + c, i.e., y² − x² = 2c (or y² − x² = C).

Q3. Solve dy/dx + y/x = x².

P = 1/x, Q = x². I.F. = e^(log x) = x. Solution: xy = ∫ x³ dx = x⁴/4 + c.

Q4. Form the differential equation of y = A eˣ.

dy/dx = A eˣ = y. So the equation is dy/dx − y = 0.

Q5. The population of a town grows at a rate proportional to the population. It was 20,000 and doubles in 50 years. Write the population law.

dP/dt = kP ⇒ P = 20000 e^(kt). Doubling: 40000 = 20000 e^(50k) ⇒ k = (log 2)/50. So P = 20000 e^((t log 2)/50).

Exam tip

For linear equations, first write the equation in the form dy/dx + Py = Q and identify P and Q clearly before finding the I.F.