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

११ वी वाणिज्य · ९ पाठ · ९ पाठांच्या नोट्स तयार

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पाठ

  1. 1Sets and Relationsनोट्स
  2. 2Functionsनोट्स
  3. 3Complex Numbersनोट्स
  4. 4Sequences and Seriesनोट्स
  5. 5Locus and Straight Lineनोट्स
  6. 6Determinantsनोट्स
  7. 7Limitsनोट्स
  8. 8Continuityनोट्स
  9. 9Differentiationनोट्स

११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 1

Sets and Relations

थोडक्यात सारांश

A set is a well-defined collection of objects. This chapter covers how to represent sets, intervals, the types of sets, operations on sets (union, intersection, difference, complement) and Venn diagrams, ordered pairs, the Cartesian product and relations.

महत्त्वाचे मुद्दे

  • Representation: roster form lists the elements, e.g. A = {1, 2, 3}. Set-builder form states a rule, e.g. A = {x | x ∈ N, x < 4}.
  • Intervals: open (a, b) = {x | a < x < b}; closed [a, b] = {x | a ≤ x ≤ b}; semi-open [a, b) and (a, b]; infinite intervals such as (a, ∞) and (−∞, b].
  • Types of sets: empty (null) set ∅, singleton, finite, infinite, equal sets (same elements), equivalent sets (same number of elements), subset (A ⊆ B), proper subset, universal set U, power set P(A).
  • A set with n elements has 2ⁿ subsets, so P(A) has 2ⁿ elements.
  • Operations: A ∪ B = {x | x ∈ A or x ∈ B}; A ∩ B = {x | x ∈ A and x ∈ B}; A − B = {x | x ∈ A, x ∉ B}; A′ = U − A. Disjoint sets have A ∩ B = ∅.
  • De Morgan's laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.
  • Counting: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). For three sets: n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C).
  • Ordered pair (a, b): the order matters, and (a, b) = (c, d) only if a = c and b = d.
  • Cartesian product: A × B = {(a, b) | a ∈ A, b ∈ B}, and n(A × B) = n(A) · n(B).
  • Relation: any subset of A × B. Its domain is the set of first elements and its range is the set of second elements.
  • Types of relations on a set A: reflexive ((a, a) ∈ R for all a), symmetric ((a, b) ∈ R ⇒ (b, a) ∈ R), transitive ((a, b), (b, c) ∈ R ⇒ (a, c) ∈ R). A relation that is all three is an equivalence relation.

महत्त्वाच्या संज्ञा

  • Well-defined – it is clear whether or not an object belongs to the set.
  • Power set – the set of all subsets.
  • Universal set – the set containing all elements under discussion.
  • Equivalence relation – reflexive, symmetric and transitive.

नमुने व सूत्रे

FormulaMeaning
n(P(A)) = 2ⁿNumber of subsets
n(A∪B) = n(A)+n(B)−n(A∩B)Union count
n(A×B) = n(A)·n(B)Cartesian product count
(A∪B)′ = A′∩B′De Morgan's law

५ महत्त्वाचे प्रश्न

Q1. Write A = {x | x ∈ N, x² < 20} in roster form.

x² < 20 gives x = 1, 2, 3, 4, so A = {1, 2, 3, 4}.

Q2. If A = {1, 2}, write P(A).

P(A) = {∅, {1}, {2}, {1, 2}}, which has 2² = 4 elements.

Q3. In a group of 50 people, 35 speak Marathi, 25 speak Hindi and all speak at least one. How many speak both?

n(M∪H) = 50 = 35 + 25 − n(M∩H), so n(M∩H) = 10.

Q4. If A = {1, 2} and B = {a, b, c}, find n(A × B).

n(A × B) = 2 × 3 = 6.

Q5. Is R = {(a, b) | a ≤ b} on N an equivalence relation?

It is reflexive and transitive but not symmetric (2 ≤ 3 but 3 ≰ 2), so it is not an equivalence relation.

परीक्षेसाठी टीप

Draw a Venn diagram for every counting problem. It prevents double counting.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 2

Functions

थोडक्यात सारांश

A function is a special relation in which every element of the first set is associated with exactly one element of the second set. This chapter covers the domain, co-domain and range, one-one and onto functions, ways of representing a function, evaluating functions, the fundamental types of functions and special functions such as modulus, signum and greatest integer.

महत्त्वाचे मुद्दे

  • Function f : A → B: every element of A has one and only one image in B.
  • Domain = A; co-domain = B; range = the set of images actually obtained (range ⊆ co-domain).
  • One-one (injective): different elements have different images, i.e. f(x₁) = f(x₂) ⇒ x₁ = x₂.
  • Onto (surjective): range = co-domain. A function that is both one-one and onto is bijective.
  • Representation: by an arrow diagram, ordered pairs, a table, a formula or a graph.
  • Fundamental types: constant function f(x) = k; identity function f(x) = x; linear function f(x) = ax + b; quadratic function; polynomial function; rational function p(x)/q(x) with q(x) ≠ 0; exponential function f(x) = aˣ (a > 0, a ≠ 1); logarithmic function f(x) = logₐx (x > 0).
  • Composite function: (fog)(x) = f(g(x)). In general fog ≠ gof.
  • Inverse function f⁻¹ exists only when f is bijective. To find it, write y = f(x) and solve for x.
  • Special functions:
  • Modulus: |x| = x if x ≥ 0, and −x if x < 0.
  • Signum: f(x) = 1 if x > 0, 0 if x = 0, −1 if x < 0.
  • Greatest integer: [x] is the greatest integer ≤ x, e.g. [2.7] = 2 and [−1.2] = −2.
  • Even function: f(−x) = f(x). Odd function: f(−x) = −f(x).

महत्त्वाच्या संज्ञा

  • Image – the element f(x) in B associated with x.
  • Pre-image – the element x whose image is f(x).
  • Bijective – both one-one and onto.
  • Composite function – a function of a function.

नमुने व सूत्रे

FunctionRule
Modulusf(x) = x if x ≥ 0; −x if x < 0
Signumf(x) = 1, 0 or −1 for x > 0, x = 0, x < 0
Greatest integerf(x) = [x]
Logarithm lawslog(mn) = log m + log n; log(m/n) = log m − log n; log mⁿ = n log m

५ महत्त्वाचे प्रश्न

Q1. If f(x) = x² + 3x + 1, find f(2) and f(−1).

f(2) = 4 + 6 + 1 = 11 and f(−1) = 1 − 3 + 1 = −1.

Q2. Show that f(x) = 2x + 3 on R is one-one.

f(x₁) = f(x₂) ⇒ 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂, so f is one-one.

Q3. Find the inverse of f(x) = (x + 4)/3.

y = (x + 4)/3 ⇒ x = 3y − 4, so f⁻¹(x) = 3x − 4.

Q4. If f(x) = x² and g(x) = x + 1, find fog and gof.

fog(x) = (x + 1)² and gof(x) = x² + 1. They are not equal.

Q5. Evaluate [3.8] + [−2.5] + |−4|.

3 + (−3) + 4 = 4.

परीक्षेसाठी टीप

For the greatest integer of a negative number, move to the left on the number line: [−2.5] = −3, not −2.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 3

Complex Numbers

थोडक्यात सारांश

The square of a real number is never negative, so we need the imaginary unit i = √−1. This chapter covers imaginary and complex numbers, the algebra of complex numbers (addition, subtraction, multiplication, division and conjugate), the square root of a complex number, solving quadratic equations with complex roots, and the cube roots of unity.

महत्त्वाचे मुद्दे

  • Imaginary number: ki, where k ∈ R, k ≠ 0, and i = √−1.
  • Powers of i: i² = −1, i³ = −i, i⁴ = 1. In general i⁴ⁿ = 1, i⁴ⁿ⁺¹ = i, i⁴ⁿ⁺² = −1, i⁴ⁿ⁺³ = −i.
  • Complex number z = a + ib: a = Re(z) is the real part and b = Im(z) is the imaginary part.
  • Equality: a + ib = c + id ⇔ a = c and b = d.
  • Conjugate: z̄ = a − ib. Then z + z̄ = 2a and z · z̄ = a² + b².
  • Addition: (a + ib) + (c + id) = (a + c) + i(b + d).
  • Multiplication: (a + ib)(c + id) = (ac − bd) + i(ad + bc).
  • Division: multiply the numerator and the denominator by the conjugate of the denominator.
  • Square root of a + ib: let √(a + ib) = x + iy, square both sides, compare the real and imaginary parts, and solve.
  • Quadratic equation ax² + bx + c = 0 with Δ = b² − 4ac < 0 has complex roots (−b ± i√(4ac − b²))/2a, which are conjugates of each other.
  • Cube roots of unity: 1, ω and ω², where ω = (−1 + i√3)/2 and ω² = (−1 − i√3)/2.
  • Properties: 1 + ω + ω² = 0 and ω³ = 1.

महत्त्वाच्या संज्ञा

  • Purely imaginary – a complex number with a = 0.
  • Purely real – a complex number with b = 0.
  • Conjugate – a − ib for the number a + ib.
  • ω (omega) – a complex cube root of unity.

नमुने व सूत्रे

ResultValue
i²−1
i⁴1
z · z̄a² + b²
1 + ω + ω²0
ω³1

५ महत्त्वाचे प्रश्न

Q1. Simplify i³⁵.

35 = 4 × 8 + 3, so i³⁵ = i³ = −i.

Q2. Express (2 + 3i)/(1 − i) in the form a + ib.

Multiply by (1 + i)/(1 + i): (2 + 2i + 3i + 3i²)/2 = (−1 + 5i)/2 = −1/2 + (5/2)i.

Q3. Find the square root of 3 + 4i.

(x + iy)² = 3 + 4i gives x² − y² = 3 and xy = 2, so x = 2 and y = 1. The answer is ±(2 + i).

Q4. Solve x² + x + 1 = 0.

x = (−1 ± √(1 − 4))/2 = (−1 ± i√3)/2, which are ω and ω².

Q5. If ω is a complex cube root of unity, find (1 + ω²)³.

1 + ω² = −ω, so (−ω)³ = −ω³ = −1.

परीक्षेसाठी टीप

In ω questions, first replace 1 + ω by −ω² or 1 + ω² by −ω. This shortens the work.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 4

Sequences and Series

थोडक्यात सारांश

A sequence is an ordered list of numbers that follow a rule. This chapter covers geometric progressions (G.P.): the nth term, the sum of n terms and the sum to infinity. It also covers writing recurring decimals as fractions, harmonic progressions (H.P.), the arithmetic, geometric and harmonic means (A.M., G.M., H.M.) and sigma (Σ) notation for special series.

महत्त्वाचे मुद्दे

  • A.P.: t(n+1) − t(n) = d, a constant. G.P.: t(n+1)/t(n) = r, a constant called the common ratio.
  • G.P. nth term: tₙ = a·rⁿ⁻¹.
  • Sum of n terms of a G.P.: Sₙ = a(rⁿ − 1)/(r − 1) for r > 1, or a(1 − rⁿ)/(1 − r) for r < 1. If r = 1, Sₙ = na.
  • Sum to infinity (only when |r| < 1): S∞ = a/(1 − r).
  • A recurring decimal is an infinite G.P. For example, 0.333… = 3/10 + 3/100 + … = (3/10)/(1 − 1/10) = 1/3.
  • H.P.: a sequence whose reciprocals are in A.P. To find the nth term, find the nth term of the A.P. of reciprocals and then take its reciprocal.
  • Means of two positive numbers a and b: A = (a + b)/2, G = √(ab), H = 2ab/(a + b).
  • Relations: G² = A·H and A ≥ G ≥ H.
  • Sigma notation: Σ₁ⁿ r = n(n + 1)/2; Σ r² = n(n + 1)(2n + 1)/6; Σ r³ = [n(n + 1)/2]²; Σ k = nk.
  • Properties: Σ(aᵣ + bᵣ) = Σaᵣ + Σbᵣ, and Σ k·aᵣ = k·Σaᵣ.

महत्त्वाच्या संज्ञा

  • Common ratio (r) – the ratio of any term to the term before it in a G.P.
  • Harmonic progression – the reciprocals of an A.P.
  • Geometric mean – √(ab) for two positive numbers a and b.
  • Sigma (Σ) – the symbol for summation.

नमुने व सूत्रे

ItemFormula
G.P. nth terma·rⁿ⁻¹
G.P. sum (r ≠ 1)a(rⁿ − 1)/(r − 1)
Infinite G.P.a/(1 − r), for −1 < r < 1
A.M., G.M., H.M.(a+b)/2, √(ab), 2ab/(a+b)
Σ r, Σ r², Σ r³n(n+1)/2, n(n+1)(2n+1)/6, [n(n+1)/2]²

५ महत्त्वाचे प्रश्न

Q1. Find the 8th term of the G.P. 3, 6, 12, …

a = 3 and r = 2, so t₈ = 3 × 2⁷ = 384.

Q2. Find the sum of the first 6 terms of 2, 6, 18, …

a = 2 and r = 3, so S₆ = 2(3⁶ − 1)/(3 − 1) = 728.

Q3. Express 0.4545… as a rational number.

It equals 45/100 + 45/10000 + … = (45/100)/(1 − 1/100) = 45/99 = 5/11.

Q4. Find the A.M., G.M. and H.M. of 4 and 16, and verify G² = AH.

A = 10, G = 8 and H = 128/20 = 6.4. Then A·H = 64 = G², which verifies the result.

Q5. Find Σ₁¹⁰ (2r + 1).

2 × (10 × 11/2) + 10 = 110 + 10 = 120.

परीक्षेसाठी टीप

Before you use S∞ = a/(1 − r), check that −1 < r < 1. Otherwise the infinite sum does not exist.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 5

Locus and Straight Line

थोडक्यात सारांश

A locus is the path of a point that moves according to a given condition. This chapter covers the equation of a locus, the slope of a line, the conditions for parallel and perpendicular lines, the angle between two lines, the equations of a line in different forms, the distance of a point from a line and the distance between two parallel lines.

महत्त्वाचे मुद्दे

  • Locus: the set of points in a plane that satisfy a given geometric condition. To find its equation, take a general point P(x, y), write the condition in terms of x and y, and simplify.
  • Slope of a line: m = tan θ (where θ is the angle the line makes with the positive X-axis). Through two points: m = (y₂ − y₁)/(x₂ − x₁). For the line ax + by + c = 0, m = −a/b.
  • Parallel lines: m₁ = m₂. Perpendicular lines: m₁·m₂ = −1.
  • Angle between two lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.
  • Forms of the equation of a line:
  • Slope-point form: y − y₁ = m(x − x₁).
  • Slope-intercept form: y = mx + c.
  • Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁).
  • Double-intercept form: x/a + y/b = 1.
  • Normal form: x cos α + y sin α = p.
  • General form: ax + by + c = 0.
  • The X-axis is y = 0 and the Y-axis is x = 0. A line parallel to the X-axis is y = k, and a line parallel to the Y-axis is x = h.
  • Distance of the point (x₁, y₁) from ax + by + c = 0: |ax₁ + by₁ + c|/√(a² + b²).
  • Distance between the parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0: |c₁ − c₂|/√(a² + b²).

महत्त्वाच्या संज्ञा

  • Slope (gradient) – the steepness of a line, tan θ.
  • Intercepts – the points where a line cuts the X-axis and the Y-axis.
  • Normal – the perpendicular from the origin to the line.
  • Locus – the path traced by a moving point.

नमुने व सूत्रे

FormEquation
Slope-pointy − y₁ = m(x − x₁)
Slope-intercepty = mx + c
Double-interceptx/a + y/b = 1
Distance from a point(absolute value of ax₁ + by₁ + c) ÷ √(a² + b²)

५ महत्त्वाचे प्रश्न

Q1. Find the slope of the line through (2, 3) and (6, 11).

m = (11 − 3)/(6 − 2) = 2.

Q2. Find the equation of the line through (1, 2) with slope 3.

y − 2 = 3(x − 1), so 3x − y − 1 = 0.

Q3. Find the equation of the line with intercepts 3 and 4 on the axes.

x/3 + y/4 = 1, so 4x + 3y − 12 = 0.

Q4. Find the distance of (1, 2) from 3x + 4y − 1 = 0.

The absolute value of 3 + 8 − 1 is 10, and √(9 + 16) = 5, so the distance is 2 units.

Q5. Find the locus of a point that is equidistant from A(1, 0) and B(0, 1).

PA² = PB² ⇒ (x − 1)² + y² = x² + (y − 1)² ⇒ x − y = 0.

परीक्षेसाठी टीप

Always convert the final answer to the general form ax + by + c = 0 unless the question asks for another form.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 6

Determinants

थोडक्यात सारांश

A determinant is a number linked to a square arrangement of numbers. This chapter covers how to evaluate 2×2 and 3×3 determinants, the properties of determinants, and their applications: Cramer's rule for solving linear equations, the consistency of three equations, the area of a triangle and the collinearity of three points.

महत्त्वाचे मुद्दे

  • 2×2 determinant: the value of det[a b; c d] is ad − bc.
  • 3×3 determinant: expand along the first row, using the signs +, −, +: a₁(b₂c₃ − b₃c₂) − b₁(a₂c₃ − a₃c₂) + c₁(a₂b₃ − a₃b₂).
  • Properties:
  • The value is unchanged when rows and columns are interchanged.
  • Interchanging two rows (or columns) changes the sign.
  • If two rows (or columns) are identical, the value is 0.
  • If every element of a row is multiplied by k, the value is multiplied by k.
  • If each element of a row is a sum of two terms, the determinant splits into the sum of two determinants.
  • Adding a multiple of one row to another row (Rᵢ → Rᵢ + kRⱼ) does not change the value.
  • If any row (or column) is all zeros, the value is 0.
  • Cramer's rule for three equations in three unknowns: x = Dx/D, y = Dy/D, z = Dz/D, provided D ≠ 0.
  • Consistency: the three equations a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are consistent if the determinant of their coefficients and constants equals 0.
  • Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) = (1/2) × det[x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1]. Take the absolute value of the result.
  • Collinearity: three points are collinear if this determinant equals 0, i.e. the area is zero.

महत्त्वाच्या संज्ञा

  • Determinant – a number calculated from a square array.
  • Collinear points – points that lie on one straight line.
  • Cramer's rule – a method of solving linear equations using determinants.
  • Consistent equations – equations that have a common solution.

नमुने व सूत्रे

ItemFormula
2×2 valuead − bc
Sign pattern for expansion+ − +
Cramer's rulex = Dx/D, y = Dy/D, z = Dz/D
Area of triangle(1/2) × det[x y 1 for each vertex]

५ महत्त्वाचे प्रश्न

Q1. Evaluate det[3 −2; 4 5].

3 × 5 − (−2) × 4 = 15 + 8 = 23.

Q2. Solve using Cramer's rule: x + y = 5, x − y = 1.

D = −2, Dx = −6 and Dy = −4, so x = 3 and y = 2.

Q3. Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).

(1/2) × det[0 0 1; 4 0 1; 0 3 1] = (1/2) × 12 = 6 sq. units.

Q4. Show that (1, 1), (2, 2) and (3, 3) are collinear.

det[1 1 1; 2 2 1; 3 3 1] = 0 because column 1 equals column 2. So the points are collinear.

Q5. Without expanding, find the value of det[1 2 3; 2 4 6; 5 7 9].

Row 2 = 2 × Row 1, so the value is 0.

परीक्षेसाठी टीप

Look for proportional rows or a property you can use before expanding. It saves time and avoids arithmetic slips.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 7

Limits

थोडक्यात सारांश

The limit of a function describes the value f(x) approaches as x gets closer and closer to a number a, without being equal to a. This chapter covers the definition of a limit, the algebra of limits, ways of evaluating limits (direct substitution, factorisation, rationalisation, the standard result for xⁿ − aⁿ) and the limits of exponential and logarithmic functions.

महत्त्वाचे मुद्दे

  • x → a means that x comes very close to a but x ≠ a, so (x − a) ≠ 0.
  • The limit exists only when the left-hand limit equals the right-hand limit.
  • Algebra of limits: the limit of a sum, difference, product or quotient is the sum, difference, product or quotient of the limits (for a quotient, the limit of the denominator must not be 0). Also lim k·f(x) = k·lim f(x).
  • Direct method: substitute x = a if the result is not of the form 0/0.
  • Factorisation: if you get 0/0, factorise the expression, cancel (x − a) and then substitute.
  • Rationalisation: if the expression has square roots, multiply by the conjugate.
  • Standard result: lim (xⁿ − aⁿ)/(x − a) as x → a equals n·aⁿ⁻¹.
  • Exponential and logarithmic limits (as x → 0):
  • lim (aˣ − 1)/x = log a (a > 0).
  • lim (eˣ − 1)/x = 1.
  • lim (1 + x)^(1/x) = e.
  • lim log(1 + x)/x = 1.
  • Useful extension: lim (1 + kx)^(1/x) = eᵏ as x → 0.

महत्त्वाच्या संज्ञा

  • Limit – the value a function approaches.
  • Indeterminate form – 0/0, which needs simplifying first.
  • Left-hand limit – the limit as x approaches a from values less than a.
  • e – the base of natural logarithms (about 2.718).

नमुने व सूत्रे

Limit (x → 0 unless stated)Value
(xⁿ − aⁿ)/(x − a), as x → an·aⁿ⁻¹
(aˣ − 1)/xlog a
(eˣ − 1)/x1
(1 + x)^(1/x)e
log(1 + x)/x1

५ महत्त्वाचे प्रश्न

Q1. Evaluate lim (x² − 9)/(x − 3) as x → 3.

(x − 3)(x + 3)/(x − 3) = x + 3, so the limit is 6.

Q2. Evaluate lim (x⁵ − 32)/(x − 2) as x → 2.

Using n·aⁿ⁻¹: 5 × 2⁴ = 80.

Q3. Evaluate lim (5ˣ − 3ˣ)/x as x → 0.

[(5ˣ − 1) − (3ˣ − 1)]/x gives log 5 − log 3 = log(5/3).

Q4. Evaluate lim log(1 + 4x)/x as x → 0.

4 × log(1 + 4x)/(4x) → 4 × 1 = 4.

Q5. Evaluate lim (√(x + 4) − 2)/x as x → 0.

Multiply by the conjugate: x/[x(√(x + 4) + 2)] = 1/(√(x + 4) + 2), so the limit is 1/4.

परीक्षेसाठी टीप

If substitution gives 0/0, simplify first by factorising, rationalising or using a standard result. Never write 0/0 as the answer.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 8

Continuity

थोडक्यात सारांश

A function is continuous when its graph has no break, jump or hole. This chapter covers the definition of continuity at a point, left and right continuity, discontinuous functions, continuity over an interval, the continuity of standard functions, and how to find unknown constants that make a function continuous.

महत्त्वाचे मुद्दे

  • Everyday examples of continuity: an unbroken road, the flow of river water, railway tracks, and the temperature changing during a day.
  • f(x) is continuous at x = a if all three conditions hold:
  • (i) f is defined at a (on an open interval containing a);
  • (ii) lim f(x) as x → a exists;
  • (iii) lim f(x) as x → a = f(a).
  • Right continuity: lim f(x) as x → a⁺ = f(a). Left continuity: lim f(x) as x → a⁻ = f(a).
  • f is continuous at a if and only if LHL = RHL = f(a).
  • Discontinuity: if any of the three conditions fails, f is discontinuous at a. This happens, for example, when f(a) is not defined, when LHL ≠ RHL (a jump), or when the limit exists but is not equal to f(a) (a hole).
  • Continuity over an interval: f is continuous on the open interval (a, b) if it is continuous at every point of it. On the closed interval [a, b], it must also be right continuous at a and left continuous at b.
  • Standard continuous functions: constant, polynomial, exponential and logarithmic functions (for x > 0) are continuous on their domains. A rational function p(x)/q(x) is continuous wherever q(x) ≠ 0.
  • Sums, differences, products and quotients (with a non-zero denominator) of continuous functions are continuous.
  • Finding constants: for a function defined in parts, set LHL = RHL = f(a) and solve for the unknown constant.

महत्त्वाच्या संज्ञा

  • Continuous function – a function whose graph has no break.
  • Discontinuous function – a function with a break, jump or hole at some point.
  • LHL / RHL – the left-hand and right-hand limits.
  • Piecewise function – a function defined by different formulas on different intervals.

नमुने व सूत्रे

ConditionRequirement
Continuity at aLHL = RHL = f(a)
Right continuitylim f(x) as x → a⁺ = f(a)
Left continuitylim f(x) as x → a⁻ = f(a)
Rational functionContinuous where q(x) ≠ 0

५ महत्त्वाचे प्रश्न

Q1. State the conditions for continuity of f(x) at x = a.

f(a) is defined, lim f(x) as x → a exists, and that limit equals f(a).

Q2. Discuss the continuity of f(x) = x² at x = 2.

f(2) = 4 and lim x² as x → 2 is 4, so f is continuous at x = 2.

Q3. Discuss the continuity of f(x) = (x² − 4)/(x − 2) for x ≠ 2, with f(2) = 5, at x = 2.

lim (x + 2) as x → 2 is 4, which is not equal to f(2) = 5, so f is discontinuous at x = 2.

Q4. Find k if f(x) = kx + 1 for x ≤ 3 and f(x) = 2x − 1 for x > 3 is continuous at x = 3.

LHL = 3k + 1 and RHL = 5. Setting 3k + 1 = 5 gives k = 4/3.

Q5. Where is f(x) = (x + 1)/(x² − 9) discontinuous?

Where x² − 9 = 0, i.e. at x = 3 and x = −3.

परीक्षेसाठी टीप

For a piecewise function, always find the LHL, the RHL and f(a) separately, then compare the three.

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११ वी वाणिज्य · गणित व संख्याशास्त्र (वाणिज्य) भाग 1 · पाठ 9

Differentiation

थोडक्यात सारांश

The derivative of a function at a point is the rate of change of the function with respect to its variable at that point. Finding a derivative is called differentiation. This chapter covers the derivative from first principles, the derivatives of standard functions, the rules of differentiation (sum, product and quotient) and applications in business and economics: marginal cost, marginal revenue, marginal demand and marginal supply.

महत्त्वाचे मुद्दे

  • First principle: f′(x) = lim [f(x + h) − f(x)]/h as h → 0. Finding a derivative from this definition is called the derivative from first principles.
  • Standard derivatives:
  • d/dx (k) = 0, where k is a constant.
  • d/dx (xⁿ) = n·xⁿ⁻¹.
  • d/dx (√x) = 1/(2√x).
  • d/dx (eˣ) = eˣ.
  • d/dx (aˣ) = aˣ·log a.
  • d/dx (log x) = 1/x.
  • Rules:
  • d/dx [k·f(x)] = k·f′(x).
  • (u ± v)′ = u′ ± v′.
  • Product rule: (uv)′ = u·v′ + v·u′.
  • Quotient rule: (u/v)′ = (v·u′ − u·v′)/v², with v ≠ 0.
  • Applications (as given in the textbook):
  • Demand D = f(p). Marginal demand MD = dD/dp.
  • Supply S = g(p). Marginal supply MS = dS/dp.
  • Total cost C = f(x), where x is the output. Marginal cost MC = dC/dx. Average cost AC = C/x.
  • Total revenue R = p·D (price × quantity demanded). Average revenue AR = R/D = p. Marginal revenue MR = dR/dD.
  • Profit = Revenue − Cost (P = R − C).

महत्त्वाच्या संज्ञा

  • Derivative – the rate of change of a function.
  • Marginal cost – the rate of change of total cost with respect to output.
  • Marginal revenue – the rate of change of total revenue with respect to quantity.
  • Average cost – the total cost divided by the output.

नमुने व सूत्रे

FunctionDerivative
xⁿn·xⁿ⁻¹
eˣeˣ
aˣaˣ·log a
log x1/x
uvu·v′ + v·u′
u/v(v·u′ − u·v′)/v²

५ महत्त्वाचे प्रश्न

Q1. Find the derivative of x² from first principles.

[(x + h)² − x²]/h = 2x + h, which tends to 2x as h → 0.

Q2. Differentiate y = x³ + 5x² − 3x + 7.

dy/dx = 3x² + 10x − 3.

Q3. Differentiate y = x·log x.

x·(1/x) + log x·1 = 1 + log x.

Q4. Differentiate y = (x² + 1)/(x − 1).

Using the quotient rule: [(x − 1)(2x) − (x² + 1)]/(x − 1)² = (x² − 2x − 1)/(x − 1)².

Q5. The cost is C = 2x³ + 20x² − 30x + 45. Find the marginal cost when x = 2.

MC = 6x² + 40x − 30. At x = 2: 24 + 80 − 30 = 74.

परीक्षेसाठी टीप

In application questions, write the formula (MC = dC/dx) first, differentiate, and only then substitute the value. Show every step.

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