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गणित व संख्याशास्त्र भाग 1 Mathematics and Statistics Part 1

११ वी (विज्ञान, कला) · ९ पाठ · ९ पाठांच्या नोट्स तयार

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पाठ

  1. 1Angle and its Measurementनोट्स
  2. 2Trigonometry - Iनोट्स
  3. 3Trigonometry - IIनोट्स
  4. 4Determinants and Matricesनोट्स
  5. 5Straight Lineनोट्स
  6. 6Circleनोट्स
  7. 7Conic Sectionsनोट्स
  8. 8Measures of Dispersionनोट्स
  9. 9Probabilityनोट्स

११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 1

Angle and its Measurement

थोडक्यात सारांश

This chapter explains directed angles, angles of any size (positive, negative, more than 360°), the two units of measuring angles – degree and radian – and how to find the length of an arc and the area of a sector of a circle.

महत्त्वाचे मुद्दे

  • A directed angle is formed when a ray rotates about its end point. Anticlockwise rotation gives a positive angle; clockwise rotation gives a negative angle.
  • The starting ray is the initial arm and the final position is the terminal arm.
  • An angle is in standard position when its vertex is at the origin and its initial arm is along the positive X-axis.
  • If the terminal arm lies on an axis, the angle is a quadrantal angle (0°, ±90°, ±180°, ±270°, …).
  • Co-terminal angles have the same initial and terminal arms; they differ by multiples of 360°.
  • Sexagesimal system: 1 complete rotation = 360°, 1° = 60′ (minutes), 1′ = 60″ (seconds).
  • Circular system: one radian is the angle subtended at the centre by an arc equal in length to the radius. The radian is a constant angle.
  • π radian = 180°. So 1 radian ≈ 57°17′45″ and 1° = π/180 radian ≈ 0.0175 radian.
  • Arc length and sector area use θ in radian only.
  • In a clock, the minute hand turns 6° per minute and the hour hand turns ½° per minute.
  • Each interior angle of a regular polygon of n sides = (n − 2)π/n radian.

महत्त्वाच्या संज्ञा

  • Directed angle – angle with a direction of rotation (positive or negative).
  • Radian – angle subtended at the centre by an arc of length equal to the radius.
  • Co-terminal angles – angles having the same terminal arm.
  • Sector – region bounded by two radii and an arc.

सूत्रे

  • π radian = 180°
  • Degree to radian: multiply by π/180 ; radian to degree: multiply by 180/π
  • Arc length: S = rθ (θ in radian)
  • Area of sector: A = ½ r²θ = ½ r S
  • Angle between clock hands at h hours m minutes: |30h − 5.5m| degrees
  • Interior angle of regular n-gon = (n − 2) × 180°/n

५ महत्त्वाचे प्रश्न

Q1. Convert 75° into radian and 5π/12 radian into degree.

75° = 75 × π/180 = 5π/12 radian. 5π/12 radian = (5π/12) × (180/π) = 75°.

Q2. Find the length of an arc of a circle of radius 10 cm which subtends an angle of 60° at the centre. Also find the area of the sector.

θ = 60° = π/3 radian. S = rθ = 10 × π/3 ≈ 10.47 cm. Area = ½ r²θ = ½ × 100 × π/3 = 50π/3 ≈ 52.36 cm².

Q3. Find the angle between the hour hand and minute hand at 3:40.

Minute hand from 12: 40 × 6° = 240°. Hour hand from 12: 3 × 30° + 40 × ½° = 90° + 20° = 110°. Angle = 240° − 110° = 130°.

Q4. Find the measure of each interior angle of a regular hexagon in degree and radian.

(n − 2) × 180°/n = 4 × 180°/6 = 120° = 2π/3 radian.

Q5. In which quadrant do the angles 405° and −135° lie?

405° = 360° + 45°, so it is co-terminal with 45° and lies in Quadrant I. −135° is a clockwise rotation of 135°; it is co-terminal with 225° and lies in Quadrant III.

परीक्षेसाठी टीप

Never use degrees in S = rθ or A = ½r²θ; convert to radian first.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 2

Trigonometry - I

थोडक्यात सारांश

This chapter defines trigonometric functions of any angle using the unit circle, finds their signs in the four quadrants, gives the fundamental identities, periodicity, and the domain, range and graphs of trigonometric functions.

महत्त्वाचे मुद्दे

  • If P(x, y) is a point on the unit circle (radius 1) on the terminal arm of θ, then cos θ = x and sin θ = y.
  • For a circle of radius r: sin θ = y/r, cos θ = x/r, tan θ = y/x, and cosec, sec, cot are their reciprocals.
  • Signs (ASTC rule, "All Silver Tea Cups"): Quadrant I – all positive; II – sin, cosec positive; III – tan, cot positive; IV – cos, sec positive.
  • −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. |sec θ| ≥ 1 and |cosec θ| ≥ 1. tan θ and cot θ can take any real value.
  • tan θ and sec θ are not defined when cos θ = 0 (θ = odd multiples of π/2). cot θ and cosec θ are not defined when sin θ = 0 (θ = multiples of π).
  • sin, cos, sec, cosec have period 2π; tan and cot have period π.
  • Allied angles: sin(−θ) = −sin θ, cos(−θ) = cos θ, tan(−θ) = −tan θ. So cos is an even function; sin and tan are odd functions.
  • Values of trigonometric functions of 0°, 30°, 45°, 60°, 90° must be memorised.

महत्त्वाच्या संज्ञा

  • Unit circle – circle of radius 1 with centre at the origin.
  • Periodic function – f(x + p) = f(x) for all x; the smallest such p > 0 is the period.
  • Domain – set of angles for which the function is defined.
  • Range – set of values taken by the function.

सूत्रे

  • sin²θ + cos²θ = 1
  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = cosec²θ
  • tan θ = sin θ / cos θ ; cot θ = cos θ / sin θ
  • sin(2nπ + θ) = sin θ ; tan(nπ + θ) = tan θ
  • Standard values: sin 30° = ½, sin 45° = 1/√2, sin 60° = √3/2 ; tan 30° = 1/√3, tan 45° = 1, tan 60° = √3

५ महत्त्वाचे प्रश्न

Q1. If sin θ = −3/5 and θ lies in the third quadrant, find the other trigonometric functions.

cos²θ = 1 − 9/25 = 16/25. In Quadrant III cos θ is negative, so cos θ = −4/5. tan θ = (−3/5)/(−4/5) = 3/4. cosec θ = −5/3, sec θ = −5/4, cot θ = 4/3.

Q2. Find the values of sin 690° and cos(−1110°).

sin 690° = sin(720° − 30°) = sin(−30°) = −½. cos(−1110°) = cos 1110° = cos(3 × 360° + 30°) = cos 30° = √3/2.

Q3. Prove that (1 − cos²θ)(1 + cot²θ) = 1.

LHS = sin²θ × cosec²θ = sin²θ × (1/sin²θ) = 1 = RHS.

Q4. State the signs of sin, cos and tan in each quadrant.

I: all positive. II: sin positive, cos and tan negative. III: tan positive, sin and cos negative. IV: cos positive, sin and tan negative.

Q5. If tan θ = 1/2, find (2 sin θ + 3 cos θ)/(4 cos θ + 3 sin θ).

Divide numerator and denominator by cos θ: (2 tan θ + 3)/(4 + 3 tan θ) = (1 + 3)/(4 + 1.5) = 4/5.5 = 8/11.

परीक्षेसाठी टीप

Always decide the sign from the quadrant first, then find the value — the sign is where marks are lost.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 3

Trigonometry - II

थोडक्यात सारांश

This chapter gives formulas for trigonometric functions of compound angles (A ± B), allied angles, multiple angles (2θ, 3θ), factorization formulas (sum ↔ product), and results for the angles of a triangle (A + B + C = π).

महत्त्वाचे मुद्दे

  • A compound angle is the sum or difference of two or more angles.
  • Allied angles are angles whose sum or difference with θ is a multiple of π/2, e.g. (π/2 − θ), (π + θ).
  • For allied angles with π or 2π the function stays the same; with π/2 or 3π/2 it changes (sin ↔ cos, tan ↔ cot, sec ↔ cosec). The sign comes from the quadrant.
  • Double and triple angle formulas come from putting B = A in compound angle formulas.
  • Factorization formulas convert sum/difference of sines or cosines into products, and products into sums.
  • In a triangle ABC, A + B + C = π, so sin(A + B) = sin C, cos(A + B) = −cos C, and sin((A + B)/2) = cos(C/2).

महत्त्वाच्या संज्ञा

  • Compound angle – algebraic sum of two or more angles.
  • Allied angles – angles related to θ by multiples of π/2.
  • Multiple angle – angle like 2θ, 3θ.

सूत्रे

  • sin(A ± B) = sin A cos B ± cos A sin B
  • cos(A ± B) = cos A cos B ∓ sin A sin B
  • tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
  • sin 2A = 2 sin A cos A = 2 tan A/(1 + tan²A)
  • cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A = (1 − tan²A)/(1 + tan²A)
  • tan 2A = 2 tan A/(1 − tan²A)
  • sin 3A = 3 sin A − 4 sin³A ; cos 3A = 4 cos³A − 3 cos A ; tan 3A = (3 tan A − tan³A)/(1 − 3 tan²A)
  • sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2)
  • sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2)
  • cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2)
  • cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2)
  • 2 sin A cos B = sin(A + B) + sin(A − B) ; 2 cos A cos B = cos(A + B) + cos(A − B) ; 2 sin A sin B = cos(A − B) − cos(A + B)

५ महत्त्वाचे प्रश्न

Q1. Find the value of sin 15° and cos 75°.

sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° = (1/√2)(√3/2) − (1/√2)(½) = (√3 − 1)/(2√2). cos 75° = cos(90° − 15°) = sin 15° = (√3 − 1)/(2√2).

Q2. Find tan 105°.

tan 105° = tan(60° + 45°) = (√3 + 1)/(1 − √3) = −(√3 + 1)/(√3 − 1) = −(2 + √3).

Q3. If sin A = 3/5 (A acute), find sin 2A and cos 2A.

cos A = 4/5. sin 2A = 2 × 3/5 × 4/5 = 24/25. cos 2A = 1 − 2 × 9/25 = 7/25.

Q4. Prove that (sin 3x + sin x)/(cos 3x + cos x) = tan 2x.

Numerator = 2 sin 2x cos x; denominator = 2 cos 2x cos x. Ratio = sin 2x/cos 2x = tan 2x.

Q5. In triangle ABC, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

sin 2A + sin 2B = 2 sin(A + B) cos(A − B) = 2 sin C cos(A − B). sin 2C = 2 sin C cos C = −2 sin C cos(A + B). LHS = 2 sin C [cos(A − B) − cos(A + B)] = 2 sin C × 2 sin A sin B = 4 sin A sin B sin C.

परीक्षेसाठी टीप

Write the factorization formulas on a card and revise daily; triangle identities are just these formulas plus A + B + C = π.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 4

Determinants and Matrices

थोडक्यात सारांश

This chapter covers determinants of order 2 and 3, minors and cofactors, properties of determinants, their uses (Cramer's rule, consistency of equations, area of a triangle, collinearity), and then matrices: types, operations and properties such as transpose.

महत्त्वाचे मुद्दे

  • A determinant is a number linked to a square arrangement of numbers. For order 2: |a b; c d| = ad − bc.
  • A 3 × 3 determinant can be expanded along any row or column using minors and cofactors.
  • Minor Mᵢⱼ = determinant left after deleting the i-th row and j-th column. Cofactor Cᵢⱼ = (−1)^(i+j) Mᵢⱼ.
  • Properties: value does not change if rows and columns are interchanged; interchanging two rows changes the sign; if two rows are identical the value is 0; a common factor of a row can be taken out; adding a multiple of one row to another does not change the value.
  • Cramer's rule solves linear equations: x = Dₓ/D, y = D_y/D, z = D_z/D (when D ≠ 0).
  • Three points are collinear if the area of the triangle formed by them is zero.
  • A matrix is a rectangular arrangement of numbers in rows and columns; order m × n. A matrix has no value, unlike a determinant.
  • Types: row, column, zero, square, diagonal, scalar, identity (unit), upper and lower triangular, symmetric (Aᵀ = A), skew-symmetric (Aᵀ = −A).
  • A square matrix is singular if |A| = 0 and non-singular if |A| ≠ 0.
  • Matrices of the same order can be added. AB is defined only when the number of columns of A = number of rows of B.
  • Matrix multiplication is associative and distributive but in general not commutative (AB ≠ BA).
  • (Aᵀ)ᵀ = A, (A + B)ᵀ = Aᵀ + Bᵀ, (AB)ᵀ = BᵀAᵀ.

महत्त्वाच्या संज्ञा

  • Order of a matrix – number of rows × number of columns.
  • Transpose (Aᵀ) – matrix obtained by interchanging rows and columns.
  • Identity matrix (I) – diagonal matrix with all diagonal elements 1.
  • Trace – sum of diagonal elements of a square matrix.

सूत्रे

  • 2 × 2: |a b; c d| = ad − bc
  • Cofactor: Cᵢⱼ = (−1)^(i+j) Mᵢⱼ
  • Expansion along R₁: Δ = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃
  • Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|
  • Cramer's rule: x = Dₓ/D, y = D_y/D, z = D_z/D
  • Every square matrix A = ½(A + Aᵀ) + ½(A − Aᵀ) (symmetric + skew-symmetric)

५ महत्त्वाचे प्रश्न

Q1. Evaluate |2 −1 3; 1 0 2; 3 1 1| by expanding along the first row.

= 2(0 × 1 − 2 × 1) − (−1)(1 × 1 − 2 × 3) + 3(1 × 1 − 0 × 3) = 2(−2) + 1(−5) + 3(1) = −4 − 5 + 3 = −6.

Q2. Solve by Cramer's rule: x + y = 5, 2x − y = 1.

D = |1 1; 2 −1| = −1 − 2 = −3. Dₓ = |5 1; 1 −1| = −5 − 1 = −6. D_y = |1 5; 2 1| = 1 − 10 = −9. x = −6/−3 = 2, y = −9/−3 = 3.

Q3. Find the area of the triangle with vertices A(1, 2), B(3, 5), C(4, 1).

Area = ½ |1(5 − 1) − 2(3 − 4) + 1(3 − 20)| = ½ |4 + 2 − 17| = ½ × 11 = 5.5 sq. units.

Q4. If A = [1 2; 3 4] and B = [0 1; 1 0], find AB and BA. Is AB = BA?

AB = [1×0+2×1, 1×1+2×0; 3×0+4×1, 3×1+4×0] = [2 1; 4 3]. BA = [3 4; 1 2]. AB ≠ BA, so matrix multiplication is not commutative.

Q5. Distinguish between a matrix and a determinant.

A determinant is always square and has a numerical value. A matrix can be of any order m × n and has no value; it is just an arrangement. Interchanging rows changes a determinant's sign but gives a different matrix.

परीक्षेसाठी टीप

Use properties (R₁ → R₁ + R₂ etc.) to create zeros before expanding a determinant; it saves time and reduces errors.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 5

Straight Line

थोडक्यात सारांश

This chapter covers locus, slope of a line, the different forms of the equation of a straight line, the general equation ax + by + c = 0, the angle between two lines, conditions for parallel and perpendicular lines, distance of a point from a line, and family of lines.

महत्त्वाचे मुद्दे

  • A locus is the set of points satisfying a given condition; its equation is satisfied by every such point and by no other point.
  • Inclination θ of a line is the angle it makes with the positive X-axis (0 ≤ θ < π). Slope m = tan θ.
  • Slope of a line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁).
  • Slope of X-axis is 0; slope of Y-axis is not defined.
  • Parallel lines have equal slopes (m₁ = m₂). Perpendicular lines satisfy m₁m₂ = −1.
  • Every first-degree equation in x and y, ax + by + c = 0 (a and b not both zero), represents a straight line with slope −a/b, x-intercept −c/a and y-intercept −c/b.
  • The family of lines through the intersection of L₁ = 0 and L₂ = 0 is L₁ + kL₂ = 0.

महत्त्वाच्या संज्ञा

  • Slope (gradient) – tangent of the inclination of a line.
  • Intercept – distance from the origin where a line cuts an axis.
  • Locus – path traced by a point moving under a given condition.
  • Normal form – equation of a line using the length p of the perpendicular from the origin and its angle α.

सूत्रे

  • 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
  • Angle between two lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
  • Distance of (x₁, y₁) from ax + by + c = 0: d = |ax₁ + by₁ + c|/√(a² + b²)
  • Distance between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0: d = |c₁ − c₂|/√(a² + b²)

५ महत्त्वाचे प्रश्न

Q1. Find the equation of the line passing through (2, 3) and (4, 7).

Slope m = (7 − 3)/(4 − 2) = 2. Equation: y − 3 = 2(x − 2) → y = 2x − 1, i.e. 2x − y − 1 = 0.

Q2. Find the equation of the line with x-intercept 3 and y-intercept −4.

x/3 + y/(−4) = 1 → 4x − 3y = 12, i.e. 4x − 3y − 12 = 0.

Q3. Find the acute angle between the lines y = 3x + 1 and y = −2x + 5.

m₁ = 3, m₂ = −2. tan θ = |(3 − (−2))/(1 + 3 × (−2))| = |5/(−5)| = 1, so θ = 45°.

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

d = |3(3) − 4(−2) − 2|/√(9 + 16) = |9 + 8 − 2|/5 = 15/5 = 3 units.

Q5. Find the equation of the line through (1, 2) perpendicular to 2x + 3y − 5 = 0.

Slope of given line = −2/3, so required slope = 3/2. y − 2 = (3/2)(x − 1) → 3x − 2y + 1 = 0.

परीक्षेसाठी टीप

For parallel/perpendicular line questions, write the slope of the given line first; then use slope-point form directly.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 6

Circle

थोडक्यात सारांश

This chapter studies the circle in coordinate geometry: its equation in centre-radius and general forms, diameter form, parametric form, the equation of a tangent, the condition for a line to touch a circle, and the director circle.

महत्त्वाचे मुद्दे

  • A circle is the locus of a point that stays at a constant distance (radius r) from a fixed point (centre).
  • The general equation x² + y² + 2gx + 2fy + c = 0 represents a circle with centre (−g, −f) and radius √(g² + f² − c).
  • If g² + f² − c > 0 the circle is real; if = 0 it is a point circle; if < 0 it is imaginary (no real circle).
  • A second-degree equation represents a circle when coefficients of x² and y² are equal and there is no xy term.
  • Parametric form: x = h + r cos θ, y = k + r sin θ.
  • A tangent is perpendicular to the radius at the point of contact.
  • A line touches a circle if the perpendicular distance of the centre from the line equals the radius.
  • The director circle is the locus of points from which the two tangents to the circle are perpendicular.

महत्त्वाच्या संज्ञा

  • Centre and radius – the fixed point and the constant distance of a circle.
  • Tangent – a line touching the circle at exactly one point.
  • Point circle – a circle with radius zero.
  • Director circle – locus of the point of intersection of perpendicular tangents.

सूत्रे

  • Centre-radius form: (x − h)² + (y − k)² = r² ; with centre at origin x² + y² = r²
  • General form: x² + y² + 2gx + 2fy + c = 0 ; centre (−g, −f), r = √(g² + f² − c)
  • Diameter form (ends (x₁, y₁), (x₂, y₂)): (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
  • Parametric form: x = h + r cos θ, y = k + r sin θ
  • Tangent to x² + y² = r² at (x₁, y₁): xx₁ + yy₁ = r²
  • Tangent to the general circle at (x₁, y₁): xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
  • Condition of tangency of y = mx + c to x² + y² = a²: c² = a²(1 + m²) ; tangents y = mx ± a√(1 + m²)
  • Director circle of x² + y² = a²: x² + y² = 2a²

५ महत्त्वाचे प्रश्न

Q1. Find the equation of the circle with centre (2, −3) and radius 5.

(x − 2)² + (y + 3)² = 25 → x² + y² − 4x + 6y + 4 + 9 − 25 = 0 → x² + y² − 4x + 6y − 12 = 0.

Q2. Find the centre and radius of x² + y² − 6x + 8y − 11 = 0.

2g = −6 → g = −3; 2f = 8 → f = 4; c = −11. Centre (3, −4). r = √(9 + 16 + 11) = √36 = 6.

Q3. Find the equation of the circle with ends of a diameter at (1, 2) and (3, 4).

(x − 1)(x − 3) + (y − 2)(y − 4) = 0 → x² − 4x + 3 + y² − 6y + 8 = 0 → x² + y² − 4x − 6y + 11 = 0.

Q4. Find the equation of the tangent to x² + y² = 25 at (3, 4).

xx₁ + yy₁ = r² → 3x + 4y = 25.

Q5. Show that the line 3x − 4y + 15 = 0 touches the circle x² + y² = 9.

Distance of centre (0, 0) from the line = |15|/√(9 + 16) = 15/5 = 3 = radius. Hence the line is a tangent.

परीक्षेसाठी टीप

Before using the general form, make the coefficients of x² and y² equal to 1 by dividing the whole equation.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 7

Conic Sections

थोडक्यात सारांश

Conic sections are curves obtained when a plane cuts a double cone: parabola, ellipse and hyperbola. This chapter gives their focus–directrix definition, standard equations, important features (focus, directrix, latus rectum, eccentricity), tangents and conditions of tangency.

महत्त्वाचे मुद्दे

  • A conic is the locus of a point P such that SP/PM = e (constant), where S is the focus, PM the distance from the directrix, and e the eccentricity.
  • e = 1: parabola; 0 < e < 1: ellipse; e > 1: hyperbola. A circle can be thought of as e = 0.
  • Parabola y² = 4ax: vertex (0, 0), focus (a, 0), directrix x = −a, axis along X-axis, latus rectum 4a.
  • Other parabolas: y² = −4ax, x² = 4by, x² = −4by.
  • Ellipse x²/a² + y²/b² = 1 (a > b): centre (0, 0), foci (±ae, 0), directrices x = ±a/e, major axis 2a, minor axis 2b, b² = a²(1 − e²).
  • For any point on an ellipse, sum of its distances from the two foci = 2a.
  • Hyperbola x²/a² − y²/b² = 1: foci (±ae, 0), directrices x = ±a/e, transverse axis 2a, conjugate axis 2b, b² = a²(e² − 1).
  • For any point on a hyperbola, difference of distances from the foci = 2a.
  • A rectangular hyperbola has a = b and e = √2.
  • Auxiliary circle of an ellipse: x² + y² = a². Director circle: x² + y² = a² + b².
  • Parametric forms: parabola (at², 2at); ellipse (a cos θ, b sin θ); hyperbola (a sec θ, b tan θ).

महत्त्वाच्या संज्ञा

  • Focus – fixed point used in defining the conic.
  • Directrix – fixed line used in defining the conic.
  • Eccentricity (e) – constant ratio SP/PM.
  • Latus rectum – chord through the focus perpendicular to the axis.
  • Focal distance – distance of a point on the conic from a focus.

सूत्रे

  • Parabola y² = 4ax: latus rectum = 4a ; focal distance of (x₁, y₁) = a + x₁ ; tangent at (x₁, y₁): yy₁ = 2a(x + x₁) ; tangent condition for y = mx + c: c = a/m
  • Ellipse: b² = a²(1 − e²) ; latus rectum = 2b²/a ; tangent at (x₁, y₁): xx₁/a² + yy₁/b² = 1 ; condition c² = a²m² + b²
  • Hyperbola: b² = a²(e² − 1) ; latus rectum = 2b²/a ; tangent at (x₁, y₁): xx₁/a² − yy₁/b² = 1 ; condition c² = a²m² − b²

५ महत्त्वाचे प्रश्न

Q1. Find the focus, directrix and length of latus rectum of y² = 12x.

Compare with y² = 4ax: 4a = 12, a = 3. Focus (3, 0), directrix x = −3, latus rectum = 12.

Q2. Find the eccentricity, foci and length of latus rectum of x²/25 + y²/9 = 1.

a² = 25, b² = 9 (a = 5, b = 3). b² = a²(1 − e²) → 9 = 25(1 − e²) → e² = 16/25, e = 4/5. Foci (±ae, 0) = (±4, 0). Latus rectum = 2b²/a = 18/5.

Q3. Find the eccentricity of the hyperbola x²/16 − y²/9 = 1.

a² = 16, b² = 9. 9 = 16(e² − 1) → e² = 25/16 → e = 5/4. Foci (±5, 0).

Q4. Find the equation of the tangent to y² = 8x at (2, 4).

4a = 8 → a = 2. Tangent: yy₁ = 2a(x + x₁) → 4y = 4(x + 2) → y = x + 2, i.e. x − y + 2 = 0.

Q5. Show that for any point on an ellipse, the sum of focal distances is 2a.

For P(x₁, y₁) on the ellipse, SP = e × PM = e(a/e − x₁) = a − ex₁, and S′P = e(x₁ + a/e) = a + ex₁. SP + S′P = 2a, a constant (length of the major axis).

परीक्षेसाठी टीप

Make a three-column table (parabola, ellipse, hyperbola) of focus, directrix, e, latus rectum and tangent — it covers most of this chapter's questions.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 8

Measures of Dispersion

थोडक्यात सारांश

Dispersion tells how much the observations of data are spread out around an average. This chapter covers range, variance and standard deviation for raw, ungrouped and grouped data, combined standard deviation and the coefficient of variation used to compare consistency.

महत्त्वाचे मुद्दे

  • Two data sets can have the same mean but very different spread; measures of dispersion show this spread.
  • Range = largest value − smallest value. It is easy but depends only on two extreme values.
  • Coefficient of range = (L − S)/(L + S).
  • Variance is the mean of the squares of deviations from the mean. Standard deviation (S.D., σ) is the positive square root of variance.
  • S.D. has the same unit as the data; variance has the square of that unit.
  • S.D. does not change if a constant is added to every observation (change of origin), but it gets multiplied if every observation is multiplied by a constant (change of scale).
  • Coefficient of variation (C.V.) compares variability of data sets with different units or means. The set with smaller C.V. is more consistent (stable); larger C.V. means more variable.

महत्त्वाच्या संज्ञा

  • Dispersion – extent of scatter of observations about an average.
  • Variance (σ²) – mean of squared deviations from the mean.
  • Standard deviation (σ) – square root of variance.
  • Coefficient of variation – S.D. expressed as a percentage of the mean.

सूत्रे

  • Range = L − S ; coefficient of range = (L − S)/(L + S)
  • Mean: x̄ = Σx/n (raw) ; x̄ = Σfx/N (frequency data)
  • Variance (raw data): σ² = Σ(x − x̄)²/n = Σx²/n − x̄²
  • Variance (frequency data): σ² = Σf(x − x̄)²/N = Σfx²/N − x̄²
  • Step deviation u = (x − A)/h: σₓ = h × σᵤ
  • Combined variance of two groups: σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂), where d₁ = x̄₁ − x̄, d₂ = x̄₂ − x̄
  • C.V. = (σ/x̄) × 100

५ महत्त्वाचे प्रश्न

Q1. Find the range and coefficient of range of 15, 22, 8, 30, 12.

L = 30, S = 8. Range = 22. Coefficient of range = 22/38 ≈ 0.579.

Q2. Find the variance and S.D. of 2, 4, 6, 8, 10.

Mean = 30/5 = 6. Deviations: −4, −2, 0, 2, 4; squares: 16, 4, 0, 4, 16; sum = 40. Variance = 40/5 = 8. S.D. = √8 ≈ 2.83.

Q3. For a data set the mean is 40 and S.D. is 8. Find the C.V.

C.V. = (8/40) × 100 = 20%.

Q4. Team A: mean runs 44, S.D. 5. Team B: mean 50, S.D. 8. Which team is more consistent?

C.V.(A) = 5/44 × 100 ≈ 11.4%. C.V.(B) = 8/50 × 100 = 16%. Team A has the smaller C.V., so it is more consistent.

Q5. What happens to the S.D. if each observation is increased by 5? If each is multiplied by 3?

Adding 5 shifts all values and the mean equally, so deviations and S.D. stay the same. Multiplying by 3 multiplies every deviation by 3, so S.D. becomes 3 times (variance 9 times).

परीक्षेसाठी टीप

Show a neat table with columns x, f, fx, fx² — even if the final answer slips, the table earns method marks.

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११ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 1 · पाठ 9

Probability

थोडक्यात सारांश

This chapter covers random experiments, sample space and events, the classical definition of probability, addition theorem, conditional probability, multiplication theorem, independent events, Bayes' theorem and odds.

महत्त्वाचे मुद्दे

  • A random experiment has more than one possible outcome and we cannot predict which will occur (tossing a coin, rolling a die).
  • Sample space (S) is the set of all possible outcomes. An event is a subset of S.
  • Types of events: sure (S), impossible (∅), elementary (one outcome), complementary (A′), mutually exclusive (A ∩ B = ∅), exhaustive (A ∪ B = S), equally likely.
  • Classical probability: P(A) = n(A)/n(S), when all outcomes are equally likely.
  • 0 ≤ P(A) ≤ 1 ; P(S) = 1 ; P(∅) = 0 ; P(A′) = 1 − P(A).
  • Addition theorem gives the probability of A or B.
  • Conditional probability P(A|B) is the probability of A when B has already happened.
  • Events are independent if the occurrence of one does not affect the other: P(A ∩ B) = P(A) P(B).
  • Bayes' theorem finds the probability of a cause (Eᵢ) when the result (A) is known – posterior probability.
  • Odds in favour of A = P(A) : P(A′); odds against A = P(A′) : P(A).

महत्त्वाच्या संज्ञा

  • Sample space – set of all possible outcomes of a random experiment.
  • Mutually exclusive events – events that cannot happen together.
  • Exhaustive events – events whose union is the whole sample space.
  • Conditional probability – probability of an event given another event has occurred.
  • Independent events – events where one does not affect the probability of the other.

सूत्रे

  • P(A) = n(A)/n(S) ; P(A′) = 1 − P(A)
  • Addition theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ; mutually exclusive: P(A ∪ B) = P(A) + P(B)
  • Three events: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) − P(A ∩ C) + P(A ∩ B ∩ C)
  • Conditional: P(A|B) = P(A ∩ B)/P(B), P(B) ≠ 0
  • Multiplication theorem: P(A ∩ B) = P(B) P(A|B) = P(A) P(B|A)
  • Independent: P(A ∩ B) = P(A) P(B)
  • Bayes' theorem: P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / Σ P(Eⱼ) P(A|Eⱼ)
  • Odds in favour a : b → P(A) = a/(a + b)

५ महत्त्वाचे प्रश्न

Q1. Two dice are thrown. Find the probability that the sum is 8.

n(S) = 36. Favourable: (2,6), (3,5), (4,4), (5,3), (6,2) → 5. P = 5/36.

Q2. A card is drawn from a pack of 52. Find the probability that it is a king or a heart.

P(K) = 4/52, P(H) = 13/52, P(K ∩ H) = 1/52. P(K ∪ H) = (4 + 13 − 1)/52 = 16/52 = 4/13.

Q3. If P(A) = 0.4, P(B) = 0.5 and A, B are independent, find P(A ∪ B).

P(A ∩ B) = 0.4 × 0.5 = 0.2. P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.

Q4. Bag I has 3 red and 2 black balls; Bag II has 2 red and 3 black. A bag is chosen at random and a red ball is drawn. Find the probability that it came from Bag I.

P(B₁) = P(B₂) = ½. P(R|B₁) = 3/5, P(R|B₂) = 2/5. P(B₁|R) = (½ × 3/5)/(½ × 3/5 + ½ × 2/5) = (3/10)/(5/10) = 3/5.

Q5. Find the odds in favour of getting a number greater than 4 on a die.

Favourable {5, 6} = 2, unfavourable = 4. Odds in favour = 2 : 4 = 1 : 2. Odds against = 2 : 1.

परीक्षेसाठी टीप

Write the sample space and events as sets before using formulas; in Bayes' questions, list P(Eᵢ) and P(A|Eᵢ) in a small table first.

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