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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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'.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)

५ महत्त्वाचे प्रश्न

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).

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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).

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• ∫ 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).

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• ∫ 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

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• ∫[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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• ∫[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

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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).

परीक्षेसाठी टीप

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.