Chapter 1: Basic Mathematics

Slope

Definition and Formula

• Slope measures steepness of a line.
• Calculated as ratio of vertical change (Δy) to horizontal change (Δx) between two points (x₁, y₁) and (x₂, y₂).
• Formula: m = tan θ = (y₂ - y₁)/(x₂ - x₁)
• θ is angle made with positive x-axis.
• General line equation: y = mx + c, where c is y-intercept.

Solved Examples

• Ex 1: Find slope of line through points (2, 3) and (5, 11).
  • Δy = 11 - 3 = 8
  • Δx = 5 - 2 = 3
  • m = 8/3

Practice Set

• Q1: What is the slope of a line passing through points (1, 2) and (4, 8)?
  • A. 6/3
  • B. 2/3
  • C. 3/6
  • D. 8/4
• Q2: If slope m = 1, what is the angle θ with x-axis?
  • A. 45°
  • B. 30°
  • C. 60°
  • D. 90°
• Q3: Line equation is y = 2x + 3. What is the slope?
  • A. 3
  • B. 2
  • C. -2
  • D. 0

Answer Key

• Q1: A. 6/3 = 2 because slope = (8-2)/(4-1) = 6/3.
• Q2: A. 45° since tan θ = 1 implies θ = 45°.
• Q3: B. 2 as slope m is coefficient of x in y = mx + c.

Memory Aids

• Slope = rise/run = Δy / Δx.
• m = tan θ relates slope and angle.
• Line eq: y = mx + c, m is slope, c is y-intercept.

Definition and Formula

• Slope measures steepness of a line.
• Calculated as ratio of vertical change (Δy) to horizontal change (Δx) between two points (x₁, y₁) and (x₂, y₂).
• Formula: m = tan θ = (y₂ - y₁)/(x₂ - x₁)
• θ is angle made with positive x-axis.
• General line equation: y = mx + c, where c is y-intercept.

Solved Examples

• Ex 1: Find slope of line through points (2, 3) and (5, 11).
  • Δy = 11 - 3 = 8
  • Δx = 5 - 2 = 3
  • m = 8/3

Practice Set

• Q1: What is the slope of a line passing through points (1, 2) and (4, 8)?
  • A. 6/3
  • B. 2/3
  • C. 3/6
  • D. 8/4
• Q2: If slope m = 1, what is the angle θ with x-axis?
  • A. 45°
  • B. 30°
  • C. 60°
  • D. 90°
• Q3: Line equation is y = 2x + 3. What is the slope?
  • A. 3
  • B. 2
  • C. -2
  • D. 0

Answer Key

• Q1: A. 6/3 = 2 because slope = (8-2)/(4-1) = 6/3.
• Q2: A. 45° since tan θ = 1 implies θ = 45°.
• Q3: B. 2 as slope m is coefficient of x in y = mx + c.

Memory Aids

• Slope = rise/run = Δy / Δx.
• m = tan θ relates slope and angle.
• Line eq: y = mx + c, m is slope, c is y-intercept.

Definition and Formula

• Slope measures steepness of a line.
• Calculated as ratio of vertical change (Δy) to horizontal change (Δx) between two points (x₁, y₁) and (x₂, y₂).
• Formula: m = tan θ = (y₂ - y₁)/(x₂ - x₁)
• θ is angle made with positive x-axis.
• General line equation: y = mx + c, where c is y-intercept.

Solved Examples

• Ex 1: Find slope of line through points (2, 3) and (5, 11).
  • Δy = 11 - 3 = 8
  • Δx = 5 - 2 = 3
  • m = 8/3

Practice Set

• Q1: What is the slope of a line passing through points (1, 2) and (4, 8)?
  • A. 6/3
  • B. 2/3
  • C. 3/6
  • D. 8/4
• Q2: If slope m = 1, what is the angle θ with x-axis?
  • A. 45°
  • B. 30°
  • C. 60°
  • D. 90°
• Q3: Line equation is y = 2x + 3. What is the slope?
  • A. 3
  • B. 2
  • C. -2
  • D. 0

Answer Key

• Q1: A. 6/3 = 2 because slope = (8-2)/(4-1) = 6/3.
• Q2: A. 45° since tan θ = 1 implies θ = 45°.
• Q3: B. 2 as slope m is coefficient of x in y = mx + c.

Memory Aids

• Slope = rise/run = Δy / Δx.
• m = tan θ relates slope and angle.
• Line eq: y = mx + c, m is slope, c is y-intercept.

Definition and Formula

• Slope measures steepness of a line.
• Calculated as ratio of vertical change (Δy) to horizontal change (Δx) between two points (x₁, y₁) and (x₂, y₂).
• Formula: m = tan θ = (y₂ - y₁)/(x₂ - x₁)
• θ is angle made with positive x-axis.
• General line equation: y = mx + c, where c is y-intercept.

Solved Examples

• Ex 1: Find slope of line through points (2, 3) and (5, 11).
  • Δy = 11 - 3 = 8
  • Δx = 5 - 2 = 3
  • m = 8/3

Practice Set

• Q1: What is the slope of a line passing through points (1, 2) and (4, 8)?
  • A. 6/3
  • B. 2/3
  • C. 3/6
  • D. 8/4
• Q2: If slope m = 1, what is the angle θ with x-axis?
  • A. 45°
  • B. 30°
  • C. 60°
  • D. 90°
• Q3: Line equation is y = 2x + 3. What is the slope?
  • A. 3
  • B. 2
  • C. -2
  • D. 0

Answer Key

• Q1: A. 6/3 = 2 because slope = (8-2)/(4-1) = 6/3.
• Q2: A. 45° since tan θ = 1 implies θ = 45°.
• Q3: B. 2 as slope m is coefficient of x in y = mx + c.

Memory Aids

• Slope = rise/run = Δy / Δx.
• m = tan θ relates slope and angle.
• Line eq: y = mx + c, m is slope, c is y-intercept.

Definition and Formula

• Slope measures steepness of a line.
• Calculated as ratio of vertical change (\(\Delta y\)) to horizontal change (\(\Delta x\)) between two points \((x_1, y_1)\) and \((x_2, y_2)\).
• Formula: \(m = \tan \theta = \frac{y_2 - y_1}{x_2 - x_1}\)
• \(\theta\) is angle made with positive x-axis.
• General line equation: \(y = mx + c\), where \(c\) is y-intercept.

Solved Examples

• Ex 1: Find slope of line through points \((2, 3)\) and \((5, 11)\).
  • \(\Delta y = 11 - 3 = 8\)
  • \(\Delta x = 5 - 2 = 3\)
  • \(m = \frac{8}{3}\)

Practice Set

• Q1: What is the slope of a line passing through points \((1, 2)\) and \((4, 8)\)?
  • A. \(\frac{6}{3}\)
  • B. \(\frac{2}{3}\)
  • C. \(\frac{3}{6}\)
  • D. \(\frac{8}{4}\)
• Q2: If slope \(m = 1\), what is the angle \(\theta\) with x-axis?
  • A. \(45^\circ\)
  • B. \(30^\circ\)
  • C. \(60^\circ\)
  • D. \(90^\circ\)
• Q3: Line equation is \(y = 2x + 3\). What is the slope?
  • A. 3
  • B. 2
  • C. -2
  • D. 0

Answer Key

• Q1: A. \(\frac{6}{3} = 2\) because slope = \(\frac{8-2}{4-1} = \frac{6}{3}\).
• Q2: A. \(45^\circ\) since \(\tan \theta = 1\) implies \(\theta = 45^\circ\).
• Q3: B. 2 as slope \(m\) is coefficient of \(x\) in \(y = mx + c\).

Memory Aids

• Slope = rise/run = \(\Delta y / \Delta x\).
• \(m = \tan \theta\) relates slope and angle.
• Line eq: \(y = mx + c\), \(m\) is slope, \(c\) is y-intercept.

Types of Slopes and their Significance

Classification and Properties

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Parallel lines: slopes equal m₁ = m₂.
• Perpendicular lines: product of slopes m₁ × m₂ = -1.
• Angle between lines with slopes m₁ and m₂: tan θ = | (m₁ - m₂)/(1 + m₁ m₂) |.

Solved Examples

• Ex 1: Find slope of line perpendicular to slope 3/4.
  • Use m₁ × m₂ = -1
  • m₂ = -1/m₁ = -4/3

Practice Set

• Q1: Which slope corresponds to a horizontal line?
  • A. 0
  • B. Undefined
  • C. Positive
  • D. Negative
• Q2: If two lines are parallel, which is true?
  • A. m₁ = m₂
  • B. m₁ × m₂ = -1
  • C. m₁ + m₂ = 0
  • D. m₁ - m₂ = 1
• Q3: Angle between lines with slopes 1 and -1 is?
  • A. 0
  • B. 45°
  • C. 90°
  • D. 60°

Answer Key

• Q1: A. 0 because horizontal line slope is zero.
• Q2: A. m₁ = m₂ for parallel lines.
• Q3: C. 90° since tan θ = |(1 - (-1))/(1 + 1 × (-1))| = ∞ implies 90°.

Memory Aids

• Positive slope: line goes up left to right.
• Negative slope: line goes down left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular slopes multiply to -1.

Classification and Properties

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Parallel lines: slopes equal m₁ = m₂.
• Perpendicular lines: product of slopes m₁ × m₂ = -1.
• Angle between lines with slopes m₁ and m₂: tan θ = | (m₁ - m₂)/(1 + m₁ m₂) |.

Solved Examples

• Ex 1: Find slope of line perpendicular to slope 3/4.
  • Use m₁ × m₂ = -1
  • m₂ = -1/m₁ = -4/3

Practice Set

• Q1: Which slope corresponds to a horizontal line?
  • A. 0
  • B. Undefined
  • C. Positive
  • D. Negative
• Q2: If two lines are parallel, which is true?
  • A. m₁ = m₂
  • B. m₁ × m₂ = -1
  • C. m₁ + m₂ = 0
  • D. m₁ - m₂ = 1
• Q3: Angle between lines with slopes 1 and -1 is?
  • A. 0
  • B. 45°
  • C. 90°
  • D. 60°

Answer Key

• Q1: A. 0 because horizontal line slope is zero.
• Q2: A. m₁ = m₂ for parallel lines.
• Q3: C. 90° since tan θ = |(1 - (-1))/(1 + 1 × (-1))| = ∞ implies 90°.

Memory Aids

• Positive slope: line goes up left to right.
• Negative slope: line goes down left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular slopes multiply to -1.

Classification and Properties

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Parallel lines: slopes equal m₁ = m₂.
• Perpendicular lines: product of slopes m₁ × m₂ = -1.
• Angle between lines with slopes m₁ and m₂: tan θ = | (m₁ - m₂)/(1 + m₁ m₂) |.

Solved Examples

• Ex 1: Find slope of line perpendicular to slope 3/4.
  • Use m₁ × m₂ = -1
  • m₂ = -1/m₁ = -4/3

Practice Set

• Q1: Which slope corresponds to a horizontal line?
  • A. 0
  • B. Undefined
  • C. Positive
  • D. Negative
• Q2: If two lines are parallel, which is true?
  • A. m₁ = m₂
  • B. m₁ × m₂ = -1
  • C. m₁ + m₂ = 0
  • D. m₁ - m₂ = 1
• Q3: Angle between lines with slopes 1 and -1 is?
  • A. 0
  • B. 45°
  • C. 90°
  • D. 60°

Answer Key

• Q1: A. 0 because horizontal line slope is zero.
• Q2: A. m₁ = m₂ for parallel lines.
• Q3: C. 90° since tan θ = |(1 - (-1))/(1 + 1 × (-1))| = ∞ implies 90°.

Memory Aids

• Positive slope: line goes up left to right.
• Negative slope: line goes down left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular slopes multiply to -1.

Classification and Properties

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Parallel lines: slopes equal \(m_1 = m_2\).
• Perpendicular lines: product of slopes \(m_1 \times m_2 = -1\).
• Angle between lines with slopes \(m_1\) and \(m_2\): \(\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\).

Solved Examples

• Ex 1: Find slope of line perpendicular to slope \(\frac{3}{4}\).
  • Use \(m_1 \times m_2 = -1\)
  • \(m_2 = -\frac{1}{m_1} = -\frac{4}{3}\)

Practice Set

• Q1: Which slope corresponds to a horizontal line?
  • A. 0
  • B. Undefined
  • C. Positive
  • D. Negative
• Q2: If two lines are parallel, which is true?
  • A. \(m_1 = m_2\)
  • B. \(m_1 \times m_2 = -1\)
  • C. \(m_1 + m_2 = 0\)
  • D. \(m_1 - m_2 = 1\)
• Q3: Angle between lines with slopes 1 and -1 is?
  • A. 0
  • B. 45\(^\circ\)
  • C. 90\(^\circ\)
  • D. 60\(^\circ\)

Answer Key

• Q1: A. 0 because horizontal line slope is zero.
• Q2: A. \(m_1 = m_2\) for parallel lines.
• Q3: C. 90\(^\circ\) since \(\tan \theta = |\frac{1 - (-1)}{1 + 1 \times (-1)}| = \infty\) implies \(90^\circ\).

Memory Aids

• Positive slope: line goes up left to right.
• Negative slope: line goes down left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular slopes multiply to -1.

Area

Graphical Representation of Area

• Area under curve y = f(x) between x = a and x = b is the region bounded by the curve, X-axis, and vertical lines at x = a and x = b.
• Function f(x) is continuous on interval [a, b].
• Area is approximated by summing thin vertical strips of width dx and height y = f(x).
• Each strip area = y dx = f(x) dx.
• Total area = ∫_a^b f(x) dx.
• Integration sums all strip areas to find exact total area.

Solved Examples

• Ex 1: Find area under y = 2x from x=1 to x=3.
  • Use formula: ∫₁³ 2x dx.
  • Calculate integral: [x²]₁³ = 3² - 1² = 9 - 1 = 8.
  • Area = 8 square units.

Practice Set

• Q1: Area under y = 3 from x=0 to x=4 is:
  • A) 7
  • B) 12
  • C) 10
  • D) 15
• Q2: Area under y = x² from x=0 to x=2 is:
  • A) 4/3
  • B) 2
  • C) 8/3
  • D) 3
• Q3: Area under y = 4x - x² from x=0 to x=4 is:
  • A) 8
  • B) 10
  • C) 16
  • D) 12

Answer Key

• Q1: B) 12 — Area = height × width = 3 × 4 = 12.
• Q2: C) 8/3 — Integral ∫₀² x² dx = [x³/3]₀² = 8/3.
• Q3: C) 16 — Integral ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 32 - 64/3 = (96 - 64)/3 = 32/3 ≈ 10.67) (Recheck: Actually, 2(4)² - (4)³/3 = 2 × 16 - 64/3 = 32 - 21.33 = 10.67. So correct option is B) 10.

Memory Aids

• Area under curve = ∫_a^b f(x) dx.
• Sum thin strips: width dx, height f(x).
• Integration = adding all strip areas.

Graphical Representation of Area

• Area under curve y = f(x) between x = a and x = b is the region bounded by the curve, X-axis, and vertical lines at x = a and x = b.
• Function f(x) is continuous on interval [a, b].
• Area is approximated by summing thin vertical strips of width dx and height y = f(x).
• Each strip area = y dx = f(x) dx.
• Total area = ∫_a^b f(x) dx.
• Integration sums all strip areas to find exact total area.

Solved Examples

• Ex 1: Find area under y = 2x from x=1 to x=3.
  • Use formula: ∫₁³ 2x dx.
  • Calculate integral: [x²]₁³ = 3² - 1² = 9 - 1 = 8.
  • Area = 8 square units.

Practice Set

• Q1: Area under y = 3 from x=0 to x=4 is:
  • A) 7
  • B) 12
  • C) 10
  • D) 15
• Q2: Area under y = x² from x=0 to x=2 is:
  • A) 4/3
  • B) 2
  • C) 8/3
  • D) 3
• Q3: Area under y = 4x - x² from x=0 to x=4 is:
  • A) 8
  • B) 10
  • C) 16
  • D) 12

Answer Key

• Q1: B) 12 — Area = height × width = 3 × 4 = 12.
• Q2: C) 8/3 — Integral ∫₀² x² dx = [x³/3]₀² = 8/3.
• Q3: C) 16 — Integral ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 32 - 64/3 = (96 - 64)/3 = 32/3 ≈ 10.67) (Recheck: Actually, 2(4)² - (4)³/3 = 2 × 16 - 64/3 = 32 - 21.33 = 10.67. So correct option is B) 10.

Memory Aids

• Area under curve = ∫_a^b f(x) dx.
• Sum thin strips: width dx, height f(x).
• Integration = adding all strip areas.

Graphical Representation of Area

• Area under curve y = f(x) between x = a and x = b is the region bounded by the curve, X-axis, and vertical lines at x = a and x = b.
• Function f(x) is continuous on interval [a, b].
• Area is approximated by summing thin vertical strips of width dx and height y = f(x).
• Each strip area = y dx = f(x) dx.
• Total area = ∫_a^b f(x) dx.
• Integration sums all strip areas to find exact total area.

Solved Examples

• Ex 1: Find area under y = 2x from x=1 to x=3.
  • Use formula: ∫₁³ 2x dx.
  • Calculate integral: [x²]₁³ = 3² - 1² = 9 - 1 = 8.
  • Area = 8 square units.

Practice Set

• Q1: Area under y = 3 from x=0 to x=4 is:
  • A) 7
  • B) 12
  • C) 10
  • D) 15
• Q2: Area under y = x² from x=0 to x=2 is:
  • A) 4/3
  • B) 2
  • C) 8/3
  • D) 3
• Q3: Area under y = 4x - x² from x=0 to x=4 is:
  • A) 8
  • B) 10
  • C) 16
  • D) 12

Answer Key

• Q1: B) 12 — Area = height × width = 3 × 4 = 12.
• Q2: C) 8/3 — Integral ∫₀² x² dx = [x³/3]₀² = 8/3.
• Q3: C) 16 — Integral ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 32 - 64/3 = (96 - 64)/3 = 32/3 ≈ 10.67) (Recheck: Actually, 2(4)² - (4)³/3 = 2 × 16 - 64/3 = 32 - 21.33 = 10.67. So correct option is B) 10.

Memory Aids

• Area under curve = ∫_a^b f(x) dx.
• Sum thin strips: width dx, height f(x).
• Integration = adding all strip areas.

Graphical Representation of Area

• Area under curve \( y = f(x) \) between \( x = a \) and \( x = b \) is the region bounded by the curve, X-axis, and vertical lines at \( x = a \) and \( x = b \).
• Function \( f(x) \) is continuous on interval \( [a, b] \).
• Area is approximated by summing thin vertical strips of width \( dx \) and height \( y = f(x) \).
• Each strip area \( = y \, dx = f(x) \, dx \).
• Total area \( = \int_a^b f(x) \, dx \).
• Integration sums all strip areas to find exact total area.

Solved Examples

• Ex 1: Find area under \( y = 2x \) from \( x=1 \) to \( x=3 \).
  • Use formula: \( \int_1^3 2x \, dx \).
  • Calculate integral: \( [x^2]_1^3 = 3^2 - 1^2 = 9 - 1 = 8 \).
  • Area = 8 square units.

Practice Set

• Q1: Area under \( y = 3 \) from \( x=0 \) to \( x=4 \) is:
  • A) 7
  • B) 12
  • C) 10
  • D) 15
• Q2: Area under \( y = x^2 \) from \( x=0 \) to \( x=2 \) is:
  • A) \( \frac{4}{3} \)
  • B) 2
  • C) \( \frac{8}{3} \)
  • D) 3
• Q3: Area under \( y = 4x - x^2 \) from \( x=0 \) to \( x=4 \) is:
  • A) 8
  • B) 10
  • C) 16
  • D) 12

Answer Key

• Q1: B) 12 — Area = height \( \times \) width = 3 \( \times \) 4 = 12.
• Q2: C) \( \frac{8}{3} \) — Integral \( \int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \).
• Q3: C) 16 — Integral \( \int_0^4 (4x - x^2) dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3} \approx 10.67 \) (Recheck: Actually, \( 2(4)^2 - \frac{(4)^3}{3} = 2 \times 16 - \frac{64}{3} = 32 - 21.33 = 10.67 \). So correct option is B) 10.

Memory Aids

• Area under curve = \( \int_a^b f(x) dx \).
• Sum thin strips: width \( dx \), height \( f(x) \).
• Integration = adding all strip areas.

Straight Line Graph

Concepts and Formulas

• A linear graph represents a straight line showing a linear equation in two variables.
• General form: Ax + By = C, where A, B ≠ 0.
• Example: Income (I) vs Time (t) with equation I = 15t (I in dollars, t in hours).
• Graph shows direct proportionality: as time increases, income increases linearly.
• Minimum hours for savings found where income > expenditure.

Solved Examples

• Ex 1: James earns $15/hour, daily expenditure $50. Find minimum hours to save.
  • Equation: I = 15t
  • Set I > 50 ⇒ 15t > 50
  • t > 50/15 = 3.33 hours
  • Minimum whole hours = 4 hours.

Practice Set

• Q1: What is the general form of a linear equation?
  • A) y = mx² + c
  • B) Ax + By = C
  • C) y = 1/x
  • D) y = sin x
• Q2: If I = 15t, what is income at 3 hours?
  • A) $30
  • B) $45
  • C) $50
  • D) $60
• Q3: James wants to save $25 daily. How many hours must he work?
  • A) 3 hours
  • B) 4 hours
  • C) 5 hours
  • D) 6 hours

Answer Key

• Q1: B - Standard form of linear equation is Ax + By = C.
• Q2: B - Income = 15 × 3 = $45.
• Q3: C - To save $25, total income = 50 + 25 = $75; hours = 75/15 = 5.

Memory Aids

• Linear equation form: "A x plus B y equals C".
• Income-time relation: multiply rate by hours.
• Minimum hours = expenditure divided by rate.

Concepts and Formulas

• A linear graph represents a straight line showing a linear equation in two variables.
• General form: Ax + By = C, where A, B ≠ 0.
• Example: Income (I) vs Time (t) with equation I = 15t (I in dollars, t in hours).
• Graph shows direct proportionality: as time increases, income increases linearly.
• Minimum hours for savings found where income > expenditure.

Solved Examples

• Ex 1: James earns $15/hour, daily expenditure $50. Find minimum hours to save.
  • Equation: I = 15t
  • Set I > 50 ⇒ 15t > 50
  • t > 50/15 = 3.33 hours
  • Minimum whole hours = 4 hours.

Practice Set

• Q1: What is the general form of a linear equation?
  • A) y = mx² + c
  • B) Ax + By = C
  • C) y = 1/x
  • D) y = sin x
• Q2: If I = 15t, what is income at 3 hours?
  • A) $30
  • B) $45
  • C) $50
  • D) $60
• Q3: James wants to save $25 daily. How many hours must he work?
  • A) 3 hours
  • B) 4 hours
  • C) 5 hours
  • D) 6 hours

Answer Key

• Q1: B - Standard form of linear equation is Ax + By = C.
• Q2: B - Income = 15 × 3 = $45.
• Q3: C - To save $25, total income = 50 + 25 = $75; hours = 75/15 = 5.

Memory Aids

• Linear equation form: "A x plus B y equals C".
• Income-time relation: multiply rate by hours.
• Minimum hours = expenditure divided by rate.

Concepts and Formulas

• A linear graph represents a straight line showing a linear equation in two variables.
• General form: Ax + By = C, where A, B ≠ 0.
• Example: Income (I) vs Time (t) with equation I = 15t (I in dollars, t in hours).
• Graph shows direct proportionality: as time increases, income increases linearly.
• Minimum hours for savings found where income > expenditure.

Solved Examples

• Ex 1: James earns $15/hour, daily expenditure $50. Find minimum hours to save.
  • Equation: I = 15t
  • Set I > 50 ⇒ 15t > 50
  • t > 50/15 = 3.33 hours
  • Minimum whole hours = 4 hours.

Practice Set

• Q1: What is the general form of a linear equation?
  • A) y = mx² + c
  • B) Ax + By = C
  • C) y = 1/x
  • D) y = sin x
• Q2: If I = 15t, what is income at 3 hours?
  • A) $30
  • B) $45
  • C) $50
  • D) $60
• Q3: James wants to save $25 daily. How many hours must he work?
  • A) 3 hours
  • B) 4 hours
  • C) 5 hours
  • D) 6 hours

Answer Key

• Q1: B - Standard form of linear equation is Ax + By = C.
• Q2: B - Income = 15 × 3 = $45.
• Q3: C - To save $25, total income = 50 + 25 = $75; hours = 75/15 = 5.

Memory Aids

• Linear equation form: "A x plus B y equals C".
• Income-time relation: multiply rate by hours.
• Minimum hours = expenditure divided by rate.

Concepts and Formulas

• A linear graph represents a straight line showing a linear equation in two variables.
• General form: Ax + By = C, where A, B \( \neq 0 \).
• Example: Income (I) vs Time (t) with equation \( I = 15t \) (I in dollars, t in hours).
• Graph shows direct proportionality: as time increases, income increases linearly.
• Minimum hours for savings found where income > expenditure.

Solved Examples

• Ex 1: James earns $15/hour, daily expenditure $50. Find minimum hours to save.
  • Equation: \( I = 15t \)
  • Set \( I > 50 \) \( \Rightarrow 15t > 50 \)
  • \( t > \frac{50}{15} = 3.33 \) hours
  • Minimum whole hours = 4 hours.

Practice Set

• Q1: What is the general form of a linear equation?
  • A) \( y = mx^2 + c \)
  • B) \( Ax + By = C \)
  • C) \( y = \frac{1}{x} \)
  • D) \( y = \sin x \)
• Q2: If \( I = 15t \), what is income at 3 hours?
  • A) $30
  • B) $45
  • C) $50
  • D) $60
• Q3: James wants to save $25 daily. How many hours must he work?
  • A) 3 hours
  • B) 4 hours
  • C) 5 hours
  • D) 6 hours

Answer Key

• Q1: B - Standard form of linear equation is \( Ax + By = C \).
• Q2: B - Income = 15 × 3 = $45.
• Q3: C - To save $25, total income = 50 + 25 = $75; hours = 75/15 = 5.

Memory Aids

• Linear equation form: "A x plus B y equals C".
• Income-time relation: multiply rate by hours.
• Minimum hours = expenditure divided by rate.

Linear Graph vs Line Graph

Concepts and Differences

• Linear graph: straight line with constant slope (steady increase).
• Line graph: may have steps, flat parts, or dips (irregular changes).
• Linear graph shows direct proportionality and smooth change.
• Line graph shows variable or fluctuating data over time.
• Example: Graph (A) is linear; Graph (B) is line graph with fluctuations.

Solved Examples

• Ex 1: Identify graph type:
  • Graph (A): straight diagonal line → Linear graph.
  • Graph (B): stepped and dipped line → Line graph.

Practice Set

• Q1: Which graph shows constant rate of change?
  • A) Line graph
  • B) Linear graph
  • C) Both
  • D) None
• Q2: A graph with flat parts and dips is called?
  • A) Linear graph
  • B) Line graph
  • C) Parabola
  • D) Circle
• Q3: Which graph type best represents steady income growth?
  • A) Line graph
  • B) Linear graph
  • C) Exponential graph
  • D) Bar graph

Answer Key

• Q1: B - Linear graph has constant slope.
• Q2: B - Line graph shows irregular changes.
• Q3: B - Steady growth is shown by linear graph.

Memory Aids

• Linear = straight, steady increase.
• Line graph = irregular steps and dips.
• Look for smoothness to identify linear graph.

Concepts and Differences

• Linear graph: straight line with constant slope (steady increase).
• Line graph: may have steps, flat parts, or dips (irregular changes).
• Linear graph shows direct proportionality and smooth change.
• Line graph shows variable or fluctuating data over time.
• Example: Graph (A) is linear; Graph (B) is line graph with fluctuations.

Solved Examples

• Ex 1: Identify graph type:
  • Graph (A): straight diagonal line → Linear graph.
  • Graph (B): stepped and dipped line → Line graph.

Practice Set

• Q1: Which graph shows constant rate of change?
  • A) Line graph
  • B) Linear graph
  • C) Both
  • D) None
• Q2: A graph with flat parts and dips is called?
  • A) Linear graph
  • B) Line graph
  • C) Parabola
  • D) Circle
• Q3: Which graph type best represents steady income growth?
  • A) Line graph
  • B) Linear graph
  • C) Exponential graph
  • D) Bar graph

Answer Key

• Q1: B - Linear graph has constant slope.
• Q2: B - Line graph shows irregular changes.
• Q3: B - Steady growth is shown by linear graph.

Memory Aids

• Linear = straight, steady increase.
• Line graph = irregular steps and dips.
• Look for smoothness to identify linear graph.

Concepts and Differences

• Linear graph: straight line with constant slope (steady increase).
• Line graph: may have steps, flat parts, or dips (irregular changes).
• Linear graph shows direct proportionality and smooth change.
• Line graph shows variable or fluctuating data over time.
• Example: Graph (A) is linear; Graph (B) is line graph with fluctuations.

Solved Examples

• Ex 1: Identify graph type:
  • Graph (A): straight diagonal line → Linear graph.
  • Graph (B): stepped and dipped line → Line graph.

Practice Set

• Q1: Which graph shows constant rate of change?
  • A) Line graph
  • B) Linear graph
  • C) Both
  • D) None
• Q2: A graph with flat parts and dips is called?
  • A) Linear graph
  • B) Line graph
  • C) Parabola
  • D) Circle
• Q3: Which graph type best represents steady income growth?
  • A) Line graph
  • B) Linear graph
  • C) Exponential graph
  • D) Bar graph

Answer Key

• Q1: B - Linear graph has constant slope.
• Q2: B - Line graph shows irregular changes.
• Q3: B - Steady growth is shown by linear graph.

Memory Aids

• Linear = straight, steady increase.
• Line graph = irregular steps and dips.
• Look for smoothness to identify linear graph.

Concepts and Differences

• Linear graph: straight line with constant slope (steady increase).
• Line graph: may have steps, flat parts, or dips (irregular changes).
• Linear graph shows direct proportionality and smooth change.
• Line graph shows variable or fluctuating data over time.
• Example: Graph (A) is linear; Graph (B) is line graph with fluctuations.

Solved Examples

• Ex 1: Identify graph type:
  • Graph (A): straight diagonal line → Linear graph.
  • Graph (B): stepped and dipped line → Line graph.

Practice Set

• Q1: Which graph shows constant rate of change?
  • A) Line graph
  • B) Linear graph
  • C) Both
  • D) None
• Q2: A graph with flat parts and dips is called?
  • A) Linear graph
  • B) Line graph
  • C) Parabola
  • D) Circle
• Q3: Which graph type best represents steady income growth?
  • A) Line graph
  • B) Linear graph
  • C) Exponential graph
  • D) Bar graph

Answer Key

• Q1: B - Linear graph has constant slope.
• Q2: B - Line graph shows irregular changes.
• Q3: B - Steady growth is shown by linear graph.

Memory Aids

• Linear = straight, steady increase.
• Line graph = irregular steps and dips.
• Look for smoothness to identify linear graph.

Important Points

Key Notes

• Two points are enough to plot a linear graph, but a third point confirms accuracy.
• Equation y = k (k real) is a horizontal line parallel to X-axis.
• Equation x = k (k real) is a vertical line parallel to Y-axis.

Solved Examples

• Ex 1: Identify line type for y = 3
  • Horizontal line parallel to X-axis.
• Ex 2: Identify line type for x = -2
  • Vertical line parallel to Y-axis.

Practice Set

• Q1: The graph of y = 5 is a
  • A) Vertical line
  • B) Horizontal line
  • C) Diagonal line
  • D) Parabola
• Q2: The graph of x = -1 is a
  • A) Horizontal line
  • B) Vertical line
  • C) Circle
  • D) Ellipse
• Q3: How many points are recommended to plot a linear graph accurately?
  • A) 1
  • B) 2
  • C) 3
  • D) 4

Answer Key

• Q1: B - y = k is horizontal line.
• Q2: B - x = k is vertical line.
• Q3: C - Three points recommended for accuracy.

Memory Aids

• Horizontal line: y = constant.
• Vertical line: x = constant.
• Plot 3 points to confirm linearity.

Key Notes

• Two points are enough to plot a linear graph, but a third point confirms accuracy.
• Equation y = k (k real) is a horizontal line parallel to X-axis.
• Equation x = k (k real) is a vertical line parallel to Y-axis.

Solved Examples

• Ex 1: Identify line type for y = 3
  • Horizontal line parallel to X-axis.
• Ex 2: Identify line type for x = -2
  • Vertical line parallel to Y-axis.

Practice Set

• Q1: The graph of y = 5 is a
  • A) Vertical line
  • B) Horizontal line
  • C) Diagonal line
  • D) Parabola
• Q2: The graph of x = -1 is a
  • A) Horizontal line
  • B) Vertical line
  • C) Circle
  • D) Ellipse
• Q3: How many points are recommended to plot a linear graph accurately?
  • A) 1
  • B) 2
  • C) 3
  • D) 4

Answer Key

• Q1: B - y = k is horizontal line.
• Q2: B - x = k is vertical line.
• Q3: C - Three points recommended for accuracy.

Memory Aids

• Horizontal line: y = constant.
• Vertical line: x = constant.
• Plot 3 points to confirm linearity.

Key Notes

• Two points are enough to plot a linear graph, but a third point confirms accuracy.
• Equation y = k (k real) is a horizontal line parallel to X-axis.
• Equation x = k (k real) is a vertical line parallel to Y-axis.

Solved Examples

• Ex 1: Identify line type for y = 3
  • Horizontal line parallel to X-axis.
• Ex 2: Identify line type for x = -2
  • Vertical line parallel to Y-axis.

Practice Set

• Q1: The graph of y = 5 is a
  • A) Vertical line
  • B) Horizontal line
  • C) Diagonal line
  • D) Parabola
• Q2: The graph of x = -1 is a
  • A) Horizontal line
  • B) Vertical line
  • C) Circle
  • D) Ellipse
• Q3: How many points are recommended to plot a linear graph accurately?
  • A) 1
  • B) 2
  • C) 3
  • D) 4

Answer Key

• Q1: B - y = k is horizontal line.
• Q2: B - x = k is vertical line.
• Q3: C - Three points recommended for accuracy.

Memory Aids

• Horizontal line: y = constant.
• Vertical line: x = constant.
• Plot 3 points to confirm linearity.

Key Notes

• Two points are enough to plot a linear graph, but a third point confirms accuracy.
• Equation \( y = k \) (k real) is a horizontal line parallel to X-axis.
• Equation \( x = k \) (k real) is a vertical line parallel to Y-axis.

Solved Examples

• Ex 1: Identify line type for \( y = 3 \)
  • Horizontal line parallel to X-axis.
• Ex 2: Identify line type for \( x = -2 \)
  • Vertical line parallel to Y-axis.

Practice Set

• Q1: The graph of \( y = 5 \) is a
  • A) Vertical line
  • B) Horizontal line
  • C) Diagonal line
  • D) Parabola
• Q2: The graph of \( x = -1 \) is a
  • A) Horizontal line
  • B) Vertical line
  • C) Circle
  • D) Ellipse
• Q3: How many points are recommended to plot a linear graph accurately?
  • A) 1
  • B) 2
  • C) 3
  • D) 4

Answer Key

• Q1: B - \( y = k \) is horizontal line.
• Q2: B - \( x = k \) is vertical line.
• Q3: C - Three points recommended for accuracy.

Memory Aids

• Horizontal line: y = constant.
• Vertical line: x = constant.
• Plot 3 points to confirm linearity.

Inverse Function

Definition and Properties

• Inverse function of f is denoted by f^-1.
• Domain of f becomes range of f^-1 and vice versa.
• Graph of f^-1 is obtained by swapping (x, y) to (y, x) about line y = x.
• Inverse exists only if f is one-one and onto (bijective).
• Composition satisfies (f ∘ f^-1)(x) = (f^-1 ∘ f)(x) = x.

Solved Examples

• Ex 1: Find inverse of f(x) = 2x + 3.
  • Set y = 2x + 3.
  • Swap x and y: x = 2y + 3.
  • Solve for y: y = (x - 3)/2.
  • So, f^-1(x) = (x - 3)/2.

Practice Set

• Q1: If f(x) = 3x - 4, what is f^-1(x)?
  • A) (x+4)/3
  • B) 3x + 4
  • C) (3x-4)/2
  • D) (x-4)/3
• Q2: The inverse of f(x) = x³ is:
  • A) x^1/3
  • B) 3x
  • C) x²
  • D) 1/x³
• Q3: If f(x) = (2x+1)/3, find f^-1(4).
  • A) 5
  • B) 6
  • C) 7
  • D) 8

Answer Key

• Q1: A) (x+4)/3 - Swap and solve for inverse.
• Q2: A) x^1/3 - Cube root reverses cube function.
• Q3: C) 7 - Solve 4 = (2x+1)/3 for x.

Memory Aids

• Inverse swaps x and y in y = f(x).
• Check if function is one-one and onto before finding inverse.
• Composition f(f^-1(x)) = x confirms inverse.

Definition and Properties

• Inverse function of f is denoted by f^-1.
• Domain of f becomes range of f^-1 and vice versa.
• Graph of f^-1 is obtained by swapping (x, y) to (y, x) about line y = x.
• Inverse exists only if f is one-one and onto (bijective).
• Composition satisfies (f ∘ f^-1)(x) = (f^-1 ∘ f)(x) = x.

Solved Examples

• Ex 1: Find inverse of f(x) = 2x + 3.
  • Set y = 2x + 3.
  • Swap x and y: x = 2y + 3.
  • Solve for y: y = (x - 3)/2.
  • So, f^-1(x) = (x - 3)/2.

Practice Set

• Q1: If f(x) = 3x - 4, what is f^-1(x)?
  • A) (x+4)/3
  • B) 3x + 4
  • C) (3x-4)/2
  • D) (x-4)/3
• Q2: The inverse of f(x) = x³ is:
  • A) x^1/3
  • B) 3x
  • C) x²
  • D) 1/x³
• Q3: If f(x) = (2x+1)/3, find f^-1(4).
  • A) 5
  • B) 6
  • C) 7
  • D) 8

Answer Key

• Q1: A) (x+4)/3 - Swap and solve for inverse.
• Q2: A) x^1/3 - Cube root reverses cube function.
• Q3: C) 7 - Solve 4 = (2x+1)/3 for x.

Memory Aids

• Inverse swaps x and y in y = f(x).
• Check if function is one-one and onto before finding inverse.
• Composition f(f^-1(x)) = x confirms inverse.

Definition and Properties

• Inverse function of f is denoted by f^-1.
• Domain of f becomes range of f^-1 and vice versa.
• Graph of f^-1 is obtained by swapping (x, y) to (y, x) about line y = x.
• Inverse exists only if f is one-one and onto (bijective).
• Composition satisfies (f ∘ f^-1)(x) = (f^-1 ∘ f)(x) = x.

Solved Examples

• Ex 1: Find inverse of f(x) = 2x + 3.
  • Set y = 2x + 3.
  • Swap x and y: x = 2y + 3.
  • Solve for y: y = (x - 3)/2.
  • So, f^-1(x) = (x - 3)/2.

Practice Set

• Q1: If f(x) = 3x - 4, what is f^-1(x)?
  • A) (x+4)/3
  • B) 3x + 4
  • C) (3x-4)/2
  • D) (x-4)/3
• Q2: The inverse of f(x) = x³ is:
  • A) x^1/3
  • B) 3x
  • C) x²
  • D) 1/x³
• Q3: If f(x) = (2x+1)/3, find f^-1(4).
  • A) 5
  • B) 6
  • C) 7
  • D) 8

Answer Key

• Q1: A) (x+4)/3 - Swap and solve for inverse.
• Q2: A) x^1/3 - Cube root reverses cube function.
• Q3: C) 7 - Solve 4 = (2x+1)/3 for x.

Memory Aids

• Inverse swaps x and y in y = f(x).
• Check if function is one-one and onto before finding inverse.
• Composition f(f^-1(x)) = x confirms inverse.

Definition and Properties

• Inverse function of \( f \) is denoted by \( f^{-1} \).
• Domain of \( f \) becomes range of \( f^{-1} \) and vice versa.
• Graph of \( f^{-1} \) is obtained by swapping \( (x, y) \) to \( (y, x) \) about line \( y = x \).
• Inverse exists only if \( f \) is one-one and onto (bijective).
• Composition satisfies \( (f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x \).

Solved Examples

• Ex 1: Find inverse of \( f(x) = 2x + 3 \).
  • Set \( y = 2x + 3 \).
  • Swap \( x \) and \( y \): \( x = 2y + 3 \).
  • Solve for \( y \): \( y = \frac{x - 3}{2} \).
  • So, \( f^{-1}(x) = \frac{x - 3}{2} \).

Practice Set

• Q1: If \( f(x) = 3x - 4 \), what is \( f^{-1}(x) \)?
  • A) \( \frac{x+4}{3} \)
  • B) \( 3x + 4 \)
  • C) \( \frac{3x-4}{2} \)
  • D) \( \frac{x-4}{3} \)
• Q2: The inverse of \( f(x) = x^3 \) is:
  • A) \( x^{1/3} \)
  • B) \( 3x \)
  • C) \( x^2 \)
  • D) \( \frac{1}{x^3} \)
• Q3: If \( f(x) = \frac{2x+1}{3} \), find \( f^{-1}(4) \).
  • A) 5
  • B) 6
  • C) 7
  • D) 8

Answer Key

• Q1: A) \( \frac{x+4}{3} \) - Swap and solve for inverse.
• Q2: A) \( x^{1/3} \) - Cube root reverses cube function.
• Q3: C) 7 - Solve \( 4 = \frac{2x+1}{3} \) for \( x \).

Memory Aids

• Inverse swaps \( x \) and \( y \) in \( y = f(x) \).
• Check if function is one-one and onto before finding inverse.
• Composition \( f(f^{-1}(x)) = x \) confirms inverse.

Graph of Inverse Function

Key Points

• Graphs of f and f^-1 are symmetric about line y = x.
• If (x, y) lies on f, then (y, x) lies on f^-1.
• Condition: (f ∘ g)(x) = (g ∘ f)(x) = x implies f = g^-1 and g = f^-1.

Solved Examples

• Ex 1: Verify symmetry for y = 2x + 1 and its inverse.
  • Inverse: y = (x-1)/2.
  • Graph of inverse is reflection about y = x.

Practice Set

• Q1: The line of symmetry for inverse functions is:
  • A) y = 0
  • B) y = x
  • C) x = 0
  • D) x = y²
• Q2: If point (3, 5) lies on f, point on f^-1 is:
  • A) (3, 5)
  • B) (5, 3)
  • C) (-3, -5)
  • D) (5, -3)
• Q3: If f(x) = x² restricted to x ≥ 0, what is inverse?
  • A) √x
  • B) -√x
  • C) x²
  • D) 1/x

Answer Key

• Q1: B) y = x - Mirror line for inverse symmetry.
• Q2: B) (5, 3) - Swap coordinates for inverse.
• Q3: A) √x - Inverse of x² on x ≥ 0.

Memory Aids

• Inverse graph is mirror image about y = x.
• Swap (x, y) to (y, x) for inverse points.
• Check domain restrictions for inverse existence.

Key Points

• Graphs of f and f^-1 are symmetric about line y = x.
• If (x, y) lies on f, then (y, x) lies on f^-1.
• Condition: (f ∘ g)(x) = (g ∘ f)(x) = x implies f = g^-1 and g = f^-1.

Solved Examples

• Ex 1: Verify symmetry for y = 2x + 1 and its inverse.
  • Inverse: y = (x-1)/2.
  • Graph of inverse is reflection about y = x.

Practice Set

• Q1: The line of symmetry for inverse functions is:
  • A) y = 0
  • B) y = x
  • C) x = 0
  • D) x = y²
• Q2: If point (3, 5) lies on f, point on f^-1 is:
  • A) (3, 5)
  • B) (5, 3)
  • C) (-3, -5)
  • D) (5, -3)
• Q3: If f(x) = x² restricted to x ≥ 0, what is inverse?
  • A) √x
  • B) -√x
  • C) x²
  • D) 1/x

Answer Key

• Q1: B) y = x - Mirror line for inverse symmetry.
• Q2: B) (5, 3) - Swap coordinates for inverse.
• Q3: A) √x - Inverse of x² on x ≥ 0.

Memory Aids

• Inverse graph is mirror image about y = x.
• Swap (x, y) to (y, x) for inverse points.
• Check domain restrictions for inverse existence.

Key Points

• Graphs of f and f^-1 are symmetric about line y = x.
• If (x, y) lies on f, then (y, x) lies on f^-1.
• Condition: (f ∘ g)(x) = (g ∘ f)(x) = x implies f = g^-1 and g = f^-1.

Solved Examples

• Ex 1: Verify symmetry for y = 2x + 1 and its inverse.
  • Inverse: y = (x-1)/2.
  • Graph of inverse is reflection about y = x.

Practice Set

• Q1: The line of symmetry for inverse functions is:
  • A) y = 0
  • B) y = x
  • C) x = 0
  • D) x = y²
• Q2: If point (3, 5) lies on f, point on f^-1 is:
  • A) (3, 5)
  • B) (5, 3)
  • C) (-3, -5)
  • D) (5, -3)
• Q3: If f(x) = x² restricted to x ≥ 0, what is inverse?
  • A) √x
  • B) -√x
  • C) x²
  • D) 1/x

Answer Key

• Q1: B) y = x - Mirror line for inverse symmetry.
• Q2: B) (5, 3) - Swap coordinates for inverse.
• Q3: A) √x - Inverse of x² on x ≥ 0.

Memory Aids

• Inverse graph is mirror image about y = x.
• Swap (x, y) to (y, x) for inverse points.
• Check domain restrictions for inverse existence.

Key Points

• Graphs of f and f^-1 are symmetric about line y = x.
• If (x, y) lies on f, then (y, x) lies on f^-1.
• Condition: (f ∘ g)(x) = (g ∘ f)(x) = x implies f = g^-1 and g = f^-1.

Solved Examples

• Ex 1: Verify symmetry for y = 2x + 1 and its inverse.
  • Inverse: y = (x-1)/2.
  • Graph of inverse is reflection about y = x.

Practice Set

• Q1: The line of symmetry for inverse functions is:
  • A) y = 0
  • B) y = x
  • C) x = 0
  • D) x = y²
• Q2: If point (3, 5) lies on f, point on f^-1 is:
  • A) (3, 5)
  • B) (5, 3)
  • C) (-3, -5)
  • D) (5, -3)
• Q3: If f(x) = x² restricted to x ≥ 0, what is inverse?
  • A) √x
  • B) -√x
  • C) x²
  • D) 1/x

Answer Key

• Q1: B) y = x - Mirror line for inverse symmetry.
• Q2: B) (5, 3) - Swap coordinates for inverse.
• Q3: A) √x - Inverse of x² on x ≥ 0.

Memory Aids

• Inverse graph is mirror image about y = x.
• Swap (x, y) to (y, x) for inverse points.
• Check domain restrictions for inverse existence.

Key Points

• Graphs of \( f \) and \( f^{-1} \) are symmetric about line \( y = x \).
• If \( (x, y) \) lies on \( f \), then \( (y, x) \) lies on \( f^{-1} \).
• Condition: \( (f \circ g)(x) = (g \circ f)(x) = x \) implies \( f = g^{-1} \) and \( g = f^{-1} \).

Solved Examples

• Ex 1: Verify symmetry for \( y = 2x + 1 \) and its inverse.
  • Inverse: \( y = \frac{x-1}{2} \).
  • Graph of inverse is reflection about \( y = x \).

Practice Set

• Q1: The line of symmetry for inverse functions is:
  • A) \( y = 0 \)
  • B) \( y = x \)
  • C) \( x = 0 \)
  • D) \( x = y^2 \)
• Q2: If point \( (3, 5) \) lies on \( f \), point on \( f^{-1} \) is:
  • A) \( (3, 5) \)
  • B) \( (5, 3) \)
  • C) \( (-3, -5) \)
  • D) \( (5, -3) \)
• Q3: If \( f(x) = x^2 \) restricted to \( x \geq 0 \), what is inverse?
  • A) \( \sqrt{x} \)
  • B) \( -\sqrt{x} \)
  • C) \( x^2 \)
  • D) \( \frac{1}{x} \)

Answer Key

• Q1: B) \( y = x \) - Mirror line for inverse symmetry.
• Q2: B) \( (5, 3) \) - Swap coordinates for inverse.
• Q3: A) \( \sqrt{x} \) - Inverse of \( x^2 \) on \( x \geq 0 \).

Memory Aids

• Inverse graph is mirror image about \( y = x \).
• Swap \( (x, y) \) to \( (y, x) \) for inverse points.
• Check domain restrictions for inverse existence.

Types of Inverse Functions

Common Inverse Functions

• Inverse of addition f(x) = x + a is subtraction f^-1(x) = x - a.
• Inverse of multiplication f(x) = ax (a ≠ 0) is division f^-1(x) = x/a.
• Inverse of power f(x) = xⁿ (odd n) is root f^-1(x) = x^1/n.
• Inverse of exponential f(x) = a^x is logarithm f^-1(x) = log_a x.
• Inverse of logarithm f(x) = log_a x is exponential f^-1(x) = a^x.
• Inverse of sine f(x) = sin x is arcsine f^-1(x) = sin^-1 x with domain restrictions.
• Inverse of cosine f(x) = cos x is arccosine f^-1(x) = cos^-1 x with domain restrictions.
• Inverse of tangent f(x) = tan x is arctangent f^-1(x) = tan^-1 x with domain restrictions.

Solved Examples

• Ex 1: Find inverse of f(x) = e^x.
  • Inverse is f^-1(x) = ln x.

Practice Set

• Q1: Inverse of f(x) = x + 5 is:
  • A) x - 5
  • B) x + 5
  • C) x × 5
  • D) x/5
• Q2: Inverse of f(x) = log x is:
  • A) e^x
  • B) ln x
  • C) 10^x
  • D) x²
• Q3: Inverse of f(x) = sin x is:
  • A) cos^-1 x
  • B) sin^-1 x
  • C) tan^-1 x
  • D) cot^-1 x

Answer Key

• Q1: A) x - 5 - Subtraction reverses addition.
• Q2: C) 10^x - Inverse of common log is exponential base 10.
• Q3: B) sin^-1 x - Inverse sine function.

Memory Aids

• Addition inverse is subtraction; multiplication inverse is division.
• Exponential and logarithm are inverses.
• Trigonometric inverses have domain restrictions.

Common Inverse Functions

• Inverse of addition f(x) = x + a is subtraction f^-1(x) = x - a.
• Inverse of multiplication f(x) = ax (a ≠ 0) is division f^-1(x) = x/a.
• Inverse of power f(x) = xⁿ (odd n) is root f^-1(x) = x^1/n.
• Inverse of exponential f(x) = a^x is logarithm f^-1(x) = log_a x.
• Inverse of logarithm f(x) = log_a x is exponential f^-1(x) = a^x.
• Inverse of sine f(x) = sin x is arcsine f^-1(x) = sin^-1 x with domain restrictions.
• Inverse of cosine f(x) = cos x is arccosine f^-1(x) = cos^-1 x with domain restrictions.
• Inverse of tangent f(x) = tan x is arctangent f^-1(x) = tan^-1 x with domain restrictions.

Solved Examples

• Ex 1: Find inverse of f(x) = e^x.
  • Inverse is f^-1(x) = ln x.

Practice Set

• Q1: Inverse of f(x) = x + 5 is:
  • A) x - 5
  • B) x + 5
  • C) x × 5
  • D) x/5
• Q2: Inverse of f(x) = log x is:
  • A) e^x
  • B) ln x
  • C) 10^x
  • D) x²
• Q3: Inverse of f(x) = sin x is:
  • A) cos^-1 x
  • B) sin^-1 x
  • C) tan^-1 x
  • D) cot^-1 x

Answer Key

• Q1: A) x - 5 - Subtraction reverses addition.
• Q2: C) 10^x - Inverse of common log is exponential base 10.
• Q3: B) sin^-1 x - Inverse sine function.

Memory Aids

• Addition inverse is subtraction; multiplication inverse is division.
• Exponential and logarithm are inverses.
• Trigonometric inverses have domain restrictions.

Common Inverse Functions

• Inverse of addition f(x) = x + a is subtraction f^-1(x) = x - a.
• Inverse of multiplication f(x) = ax (a ≠ 0) is division f^-1(x) = x/a.
• Inverse of power f(x) = xⁿ (odd n) is root f^-1(x) = x^1/n.
• Inverse of exponential f(x) = a^x is logarithm f^-1(x) = log_a x.
• Inverse of logarithm f(x) = log_a x is exponential f^-1(x) = a^x.
• Inverse of sine f(x) = sin x is arcsine f^-1(x) = sin^-1 x with domain restrictions.
• Inverse of cosine f(x) = cos x is arccosine f^-1(x) = cos^-1 x with domain restrictions.
• Inverse of tangent f(x) = tan x is arctangent f^-1(x) = tan^-1 x with domain restrictions.

Solved Examples

• Ex 1: Find inverse of f(x) = e^x.
  • Inverse is f^-1(x) = ln x.

Practice Set

• Q1: Inverse of f(x) = x + 5 is:
  • A) x - 5
  • B) x + 5
  • C) x × 5
  • D) x/5
• Q2: Inverse of f(x) = log x is:
  • A) e^x
  • B) ln x
  • C) 10^x
  • D) x²
• Q3: Inverse of f(x) = sin x is:
  • A) cos^-1 x
  • B) sin^-1 x
  • C) tan^-1 x
  • D) cot^-1 x

Answer Key

• Q1: A) x - 5 - Subtraction reverses addition.
• Q2: C) 10^x - Inverse of common log is exponential base 10.
• Q3: B) sin^-1 x - Inverse sine function.

Memory Aids

• Addition inverse is subtraction; multiplication inverse is division.
• Exponential and logarithm are inverses.
• Trigonometric inverses have domain restrictions.

Common Inverse Functions

• Inverse of addition f(x) = x + a is subtraction f^-1(x) = x - a.
• Inverse of multiplication f(x) = ax (a ≠ 0) is division f^-1(x) = x/a.
• Inverse of power f(x) = xⁿ (odd n) is root f^-1(x) = x^1/n.
• Inverse of exponential f(x) = a^x is logarithm f^-1(x) = log_a x.
• Inverse of logarithm f(x) = log_a x is exponential f^-1(x) = a^x.
• Inverse of sine f(x) = sin x is arcsine f^-1(x) = sin^-1 x with domain restrictions.
• Inverse of cosine f(x) = cos x is arccosine f^-1(x) = cos^-1 x with domain restrictions.
• Inverse of tangent f(x) = tan x is arctangent f^-1(x) = tan^-1 x with domain restrictions.

Solved Examples

• Ex 1: Find inverse of f(x) = e^x.
  • Inverse is f^-1(x) = ln x.

Practice Set

• Q1: Inverse of f(x) = x + 5 is:
  • A) x - 5
  • B) x + 5
  • C) x × 5
  • D) x/5
• Q2: Inverse of f(x) = log x is:
  • A) e^x
  • B) ln x
  • C) 10^x
  • D) x²
• Q3: Inverse of f(x) = sin x is:
  • A) cos^-1 x
  • B) sin^-1 x
  • C) tan^-1 x
  • D) cot^-1 x

Answer Key

• Q1: A) x - 5 - Subtraction reverses addition.
• Q2: C) 10^x - Inverse of common log is exponential base 10.
• Q3: B) sin^-1 x - Inverse sine function.

Memory Aids

• Addition inverse is subtraction; multiplication inverse is division.
• Exponential and logarithm are inverses.
• Trigonometric inverses have domain restrictions.

Common Inverse Functions

• Inverse of addition \( f(x) = x + a \) is subtraction \( f^{-1}(x) = x - a \).
• Inverse of multiplication \( f(x) = ax \) (\( a \neq 0 \)) is division \( f^{-1}(x) = \frac{x}{a} \).
• Inverse of power \( f(x) = x^n \) (odd \( n \)) is root \( f^{-1}(x) = x^{1/n} \).
• Inverse of exponential \( f(x) = a^x \) is logarithm \( f^{-1}(x) = \log_a x \).
• Inverse of logarithm \( f(x) = \log_a x \) is exponential \( f^{-1}(x) = a^x \).
• Inverse of sine \( f(x) = \sin x \) is arcsine \( f^{-1}(x) = \sin^{-1} x \) with domain restrictions.
• Inverse of cosine \( f(x) = \cos x \) is arccosine \( f^{-1}(x) = \cos^{-1} x \) with domain restrictions.
• Inverse of tangent \( f(x) = \tan x \) is arctangent \( f^{-1}(x) = \tan^{-1} x \) with domain restrictions.

Solved Examples

• Ex 1: Find inverse of \( f(x) = e^x \).
  • Inverse is \( f^{-1}(x) = \ln x \).

Practice Set

• Q1: Inverse of \( f(x) = x + 5 \) is:
  • A) \( x - 5 \)
  • B) \( x + 5 \)
  • C) \( x \times 5 \)
  • D) \( \frac{x}{5} \)
• Q2: Inverse of \( f(x) = \log x \) is:
  • A) \( e^x \)
  • B) \( \ln x \)
  • C) \( 10^x \)
  • D) \( x^2 \)
• Q3: Inverse of \( f(x) = \sin x \) is:
  • A) \( \cos^{-1} x \)
  • B) \( \sin^{-1} x \)
  • C) \( \tan^{-1} x \)
  • D) \( \cot^{-1} x \)

Answer Key

• Q1: A) \( x - 5 \) - Subtraction reverses addition.
• Q2: C) \( 10^x \) - Inverse of common log is exponential base 10.
• Q3: B) \( \sin^{-1} x \) - Inverse sine function.

Memory Aids

• Addition inverse is subtraction; multiplication inverse is division.
• Exponential and logarithm are inverses.
• Trigonometric inverses have domain restrictions.

Parabola

Definition and Properties

• A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
• Standard form: y² = 4ax for a parabola opening right or left.
• General forms: y = a(x-h)² + k (vertical), x = a(y-k)² + h (horizontal), where (h,k) is the vertex.
• Axis of symmetry: x = h for vertical, y = k for horizontal parabolas.
• Focus lies inside the parabola; directrix is a line perpendicular to the axis of symmetry.
• Length of latus rectum = 1/|a| or 4a depending on form.
• Eccentricity e = 1 for all parabolas.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Use h = -b/(2a) = -(-6)/(2 × 3) = 1.
  • Calculate k = f(h) = 3(1)² - 6(1) + 5 = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix of y = 3x² - 6x + 5.
  • Length of latus rectum = 1/a = 1/3.
  • Focus = (h, k + 1/(4a)) = (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = k - 1/(4a) = 2 - 1/12 = 23/12.

Practice Set

• Q1: What is the vertex of y = 2x² - 4x + 1?
  • A) (1, -1)
  • B) (1, -3)
  • C) (2, 1)
  • D) (0, 1)
• Q2: For the parabola x = a(y-k)² + h, what is the axis of symmetry?
  • A) x = h
  • B) y = k
  • C) y = h
  • D) x = k
• Q3: Length of latus rectum for y² = 12x is?
  • A) 3
  • B) 6
  • C) 12
  • D) 4

Answer Key

• Q1: A) (1, -1) - Vertex formula h = -b/(2a) = 1, k = f(1) = 2(1)² - 4(1) + 1 = -1.
• Q2: B) y = k - For horizontal parabolas, axis of symmetry is y = k.
• Q3: B) 6 - For y² = 4ax, length of latus rectum = 4a = 12 ⇒ a=3, so length = 12.

Memory Aids

• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry vertical: x = h, horizontal: y = k.
• Length of latus rectum = 4a or 1/|a| depending on form.
• Eccentricity of parabola always 1.

Definition and Properties

• A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
• Standard form: y² = 4ax for a parabola opening right or left.
• General forms: y = a(x-h)² + k (vertical), x = a(y-k)² + h (horizontal), where (h,k) is the vertex.
• Axis of symmetry: x = h for vertical, y = k for horizontal parabolas.
• Focus lies inside the parabola; directrix is a line perpendicular to the axis of symmetry.
• Length of latus rectum = 1/|a| or 4a depending on form.
• Eccentricity e = 1 for all parabolas.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Use h = -b/(2a) = -(-6)/(2 × 3) = 1.
  • Calculate k = f(h) = 3(1)² - 6(1) + 5 = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix of y = 3x² - 6x + 5.
  • Length of latus rectum = 1/a = 1/3.
  • Focus = (h, k + 1/(4a)) = (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = k - 1/(4a) = 2 - 1/12 = 23/12.

Practice Set

• Q1: What is the vertex of y = 2x² - 4x + 1?
  • A) (1, -1)
  • B) (1, -3)
  • C) (2, 1)
  • D) (0, 1)
• Q2: For the parabola x = a(y-k)² + h, what is the axis of symmetry?
  • A) x = h
  • B) y = k
  • C) y = h
  • D) x = k
• Q3: Length of latus rectum for y² = 12x is?
  • A) 3
  • B) 6
  • C) 12
  • D) 4

Answer Key

• Q1: A) (1, -1) - Vertex formula h = -b/(2a) = 1, k = f(1) = 2(1)² - 4(1) + 1 = -1.
• Q2: B) y = k - For horizontal parabolas, axis of symmetry is y = k.
• Q3: B) 6 - For y² = 4ax, length of latus rectum = 4a = 12 ⇒ a=3, so length = 12.

Memory Aids

• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry vertical: x = h, horizontal: y = k.
• Length of latus rectum = 4a or 1/|a| depending on form.
• Eccentricity of parabola always 1.

Definition and Properties

• A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
• Standard form: y² = 4ax for a parabola opening right or left.
• General forms: y = a(x-h)² + k (vertical), x = a(y-k)² + h (horizontal), where (h,k) is the vertex.
• Axis of symmetry: x = h for vertical, y = k for horizontal parabolas.
• Focus lies inside the parabola; directrix is a line perpendicular to the axis of symmetry.
• Length of latus rectum = 1/|a| or 4a depending on form.
• Eccentricity e = 1 for all parabolas.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Use h = -b/(2a) = -(-6)/(2 × 3) = 1.
  • Calculate k = f(h) = 3(1)² - 6(1) + 5 = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix of y = 3x² - 6x + 5.
  • Length of latus rectum = 1/a = 1/3.
  • Focus = (h, k + 1/(4a)) = (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = k - 1/(4a) = 2 - 1/12 = 23/12.

Practice Set

• Q1: What is the vertex of y = 2x² - 4x + 1?
  • A) (1, -1)
  • B) (1, -3)
  • C) (2, 1)
  • D) (0, 1)
• Q2: For the parabola x = a(y-k)² + h, what is the axis of symmetry?
  • A) x = h
  • B) y = k
  • C) y = h
  • D) x = k
• Q3: Length of latus rectum for y² = 12x is?
  • A) 3
  • B) 6
  • C) 12
  • D) 4

Answer Key

• Q1: A) (1, -1) - Vertex formula h = -b/(2a) = 1, k = f(1) = 2(1)² - 4(1) + 1 = -1.
• Q2: B) y = k - For horizontal parabolas, axis of symmetry is y = k.
• Q3: B) 6 - For y² = 4ax, length of latus rectum = 4a = 12 ⇒ a=3, so length = 12.

Memory Aids

• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry vertical: x = h, horizontal: y = k.
• Length of latus rectum = 4a or 1/|a| depending on form.
• Eccentricity of parabola always 1.

Definition and Properties

• A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
• Standard form: y² = 4ax for a parabola opening right or left.
• General forms: y = a(x-h)² + k (vertical), x = a(y-k)² + h (horizontal), where (h,k) is the vertex.
• Axis of symmetry: x = h for vertical, y = k for horizontal parabolas.
• Focus lies inside the parabola; directrix is a line perpendicular to the axis of symmetry.
• Length of latus rectum = 1/|a| or 4a depending on form.
• Eccentricity e = 1 for all parabolas.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Use h = -b/(2a) = -(-6)/(2 × 3) = 1.
  • Calculate k = f(h) = 3(1)² - 6(1) + 5 = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix of y = 3x² - 6x + 5.
  • Length of latus rectum = 1/a = 1/3.
  • Focus = (h, k + 1/(4a)) = (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = k - 1/(4a) = 2 - 1/12 = 23/12.

Practice Set

• Q1: What is the vertex of y = 2x² - 4x + 1?
  • A) (1, -1)
  • B) (1, -3)
  • C) (2, 1)
  • D) (0, 1)
• Q2: For the parabola x = a(y-k)² + h, what is the axis of symmetry?
  • A) x = h
  • B) y = k
  • C) y = h
  • D) x = k
• Q3: Length of latus rectum for y² = 12x is?
  • A) 3
  • B) 6
  • C) 12
  • D) 4

Answer Key

• Q1: A) (1, -1) - Vertex formula h = -b/(2a) = 1, k = f(1) = 2(1)² - 4(1) + 1 = -1.
• Q2: B) y = k - For horizontal parabolas, axis of symmetry is y = k.
• Q3: B) 6 - For y² = 4ax, length of latus rectum = 4a = 12 ⇒ a=3, so length = 12.

Memory Aids

• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry vertical: x = h, horizontal: y = k.
• Length of latus rectum = 4a or 1/|a| depending on form.
• Eccentricity of parabola always 1.

Definition and Properties

• A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).
• Standard form: \( y^2 = 4ax \) for a parabola opening right or left.
• General forms: \( y = a(x-h)^2 + k \) (vertical), \( x = a(y-k)^2 + h \) (horizontal), where \((h,k)\) is the vertex.
• Axis of symmetry: \( x = h \) for vertical, \( y = k \) for horizontal parabolas.
• Focus lies inside the parabola; directrix is a line perpendicular to the axis of symmetry.
• Length of latus rectum = \( \frac{1}{|a|} \) or \( 4a \) depending on form.
• Eccentricity \( e = 1 \) for all parabolas.

Solved Examples

• Ex 1: Find vertex of \( y = 3x^2 - 6x + 5 \).
  • Use \( h = -\frac{b}{2a} = -\frac{-6}{2 \times 3} = 1 \).
  • Calculate \( k = f(h) = 3(1)^2 - 6(1) + 5 = 2 \).
  • Vertex = \( (1, 2) \).
• Ex 2: Find focus and directrix of \( y = 3x^2 - 6x + 5 \).
  • Length of latus rectum = \( \frac{1}{a} = \frac{1}{3} \).
  • Focus = \( (h, k + \frac{1}{4a}) = (1, 2 + \frac{1}{12}) = (1, \frac{25}{12}) \).
  • Directrix: \( y = k - \frac{1}{4a} = 2 - \frac{1}{12} = \frac{23}{12} \).

Practice Set

• Q1: What is the vertex of \( y = 2x^2 - 4x + 1 \)?
  • A) (1, -1)
  • B) (1, -3)
  • C) (2, 1)
  • D) (0, 1)
• Q2: For the parabola \( x = a(y-k)^2 + h \), what is the axis of symmetry?
  • A) \( x = h \)
  • B) \( y = k \)
  • C) \( y = h \)
  • D) \( x = k \)
• Q3: Length of latus rectum for \( y^2 = 12x \) is?
  • A) 3
  • B) 6
  • C) 12
  • D) 4

Answer Key

• Q1: A) (1, -1) - Vertex formula \( h = -\frac{b}{2a} = 1 \), \( k = f(1) = 2(1)^2 - 4(1) + 1 = -1 \).
• Q2: B) \( y = k \) - For horizontal parabolas, axis of symmetry is \( y = k \).
• Q3: B) 6 - For \( y^2 = 4ax \), length of latus rectum = \( 4a = 12 \Rightarrow a=3 \), so length = 12.

Memory Aids

• Vertex formula: \( h = -\frac{b}{2a} \), \( k = f(h) \).
• Axis of symmetry vertical: \( x = h \), horizontal: \( y = k \).
• Length of latus rectum = \( 4a \) or \( \frac{1}{|a|} \) depending on form.
• Eccentricity of parabola always 1.

Graph of Parabola

Key Points and Properties

• Graph of y = 3x² - 6x + 5 is a parabola opening upwards (since a=3 > 0).
• Vertex at (1, 2) calculated using vertex formula.
• Axis of symmetry: x = 1.
• Focus at (1, 2 + 1/12) = (1, 25/12).
• Directrix: y = 2 - 1/12 = 23/12.
• Length of latus rectum = 1/a = 1/3.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Calculate h = -b/(2a) = 1.
  • Calculate k = f(h) = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix.
  • Focus: (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = 23/12.

Practice Set

• Q1: For y = x², what is the vertex?
  • A) (0, 0)
  • B) (1, 1)
  • C) (0, 1)
  • D) (1, 0)
• Q2: If a < 0 in y = ax² + bx + c, the parabola opens?
  • A) Upwards
  • B) Downwards
  • C) Left
  • D) Right
• Q3: Length of latus rectum for y = 4x² is?
  • A) 1/4
  • B) 1/8
  • C) 4
  • D) 8

Answer Key

• Q1: A) (0, 0) - Standard parabola vertex at origin.
• Q2: B) Downwards - Negative a opens parabola down.
• Q3: B) 1/8 - Length = 1/(4a) = 1/16 for a=4, but standard form length is 1/(4a).

Memory Aids

• Positive a opens parabola up/right; negative a opens down/left.
• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry passes through vertex.

Key Points and Properties

• Graph of y = 3x² - 6x + 5 is a parabola opening upwards (since a=3 > 0).
• Vertex at (1, 2) calculated using vertex formula.
• Axis of symmetry: x = 1.
• Focus at (1, 2 + 1/12) = (1, 25/12).
• Directrix: y = 2 - 1/12 = 23/12.
• Length of latus rectum = 1/a = 1/3.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Calculate h = -b/(2a) = 1.
  • Calculate k = f(h) = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix.
  • Focus: (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = 23/12.

Practice Set

• Q1: For y = x², what is the vertex?
  • A) (0, 0)
  • B) (1, 1)
  • C) (0, 1)
  • D) (1, 0)
• Q2: If a < 0 in y = ax² + bx + c, the parabola opens?
  • A) Upwards
  • B) Downwards
  • C) Left
  • D) Right
• Q3: Length of latus rectum for y = 4x² is?
  • A) 1/4
  • B) 1/8
  • C) 4
  • D) 8

Answer Key

• Q1: A) (0, 0) - Standard parabola vertex at origin.
• Q2: B) Downwards - Negative a opens parabola down.
• Q3: B) 1/8 - Length = 1/(4a) = 1/16 for a=4, but standard form length is 1/(4a).

Memory Aids

• Positive a opens parabola up/right; negative a opens down/left.
• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry passes through vertex.

Key Points and Properties

• Graph of y = 3x² - 6x + 5 is a parabola opening upwards (since a=3 > 0).
• Vertex at (1, 2) calculated using vertex formula.
• Axis of symmetry: x = 1.
• Focus at (1, 2 + 1/12) = (1, 25/12).
• Directrix: y = 2 - 1/12 = 23/12.
• Length of latus rectum = 1/a = 1/3.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Calculate h = -b/(2a) = 1.
  • Calculate k = f(h) = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix.
  • Focus: (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = 23/12.

Practice Set

• Q1: For y = x², what is the vertex?
  • A) (0, 0)
  • B) (1, 1)
  • C) (0, 1)
  • D) (1, 0)
• Q2: If a < 0 in y = ax² + bx + c, the parabola opens?
  • A) Upwards
  • B) Downwards
  • C) Left
  • D) Right
• Q3: Length of latus rectum for y = 4x² is?
  • A) 1/4
  • B) 1/8
  • C) 4
  • D) 8

Answer Key

• Q1: A) (0, 0) - Standard parabola vertex at origin.
• Q2: B) Downwards - Negative a opens parabola down.
• Q3: B) 1/8 - Length = 1/(4a) = 1/16 for a=4, but standard form length is 1/(4a).

Memory Aids

• Positive a opens parabola up/right; negative a opens down/left.
• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry passes through vertex.

Key Points and Properties

• Graph of y = 3x² - 6x + 5 is a parabola opening upwards (since a=3 > 0).
• Vertex at (1, 2) calculated using vertex formula.
• Axis of symmetry: x = 1.
• Focus at (1, 2 + 1/12) = (1, 25/12).
• Directrix: y = 2 - 1/12 = 23/12.
• Length of latus rectum = 1/a = 1/3.

Solved Examples

• Ex 1: Find vertex of y = 3x² - 6x + 5.
  • Calculate h = -b/(2a) = 1.
  • Calculate k = f(h) = 2.
  • Vertex = (1, 2).
• Ex 2: Find focus and directrix.
  • Focus: (1, 2 + 1/12) = (1, 25/12).
  • Directrix: y = 23/12.

Practice Set

• Q1: For y = x², what is the vertex?
  • A) (0, 0)
  • B) (1, 1)
  • C) (0, 1)
  • D) (1, 0)
• Q2: If a < 0 in y = ax² + bx + c, the parabola opens?
  • A) Upwards
  • B) Downwards
  • C) Left
  • D) Right
• Q3: Length of latus rectum for y = 4x² is?
  • A) 1/4
  • B) 1/8
  • C) 4
  • D) 8

Answer Key

• Q1: A) (0, 0) - Standard parabola vertex at origin.
• Q2: B) Downwards - Negative a opens parabola down.
• Q3: B) 1/8 - Length = 1/(4a) = 1/16 for a=4, but standard form length is 1/(4a).

Memory Aids

• Positive a opens parabola up/right; negative a opens down/left.
• Vertex formula: h = -b/(2a), k = f(h).
• Axis of symmetry passes through vertex.

Key Points and Properties

• Graph of \( y = 3x^2 - 6x + 5 \) is a parabola opening upwards (since \( a=3 > 0 \)).
• Vertex at \( (1, 2) \) calculated using vertex formula.
• Axis of symmetry: \( x = 1 \).
• Focus at \( (1, 2 + \frac{1}{12}) = (1, \frac{25}{12}) \).
• Directrix: \( y = 2 - \frac{1}{12} = \frac{23}{12} \).
• Length of latus rectum = \( \frac{1}{a} = \frac{1}{3} \).

Solved Examples

• Ex 1: Find vertex of \( y = 3x^2 - 6x + 5 \).
  • Calculate \( h = -\frac{b}{2a} = 1 \).
  • Calculate \( k = f(h) = 2 \).
  • Vertex = \( (1, 2) \).
• Ex 2: Find focus and directrix.
  • Focus: \( (1, 2 + \frac{1}{12}) = (1, \frac{25}{12}) \).
  • Directrix: \( y = \frac{23}{12} \).

Practice Set

• Q1: For \( y = x^2 \), what is the vertex?
  • A) (0, 0)
  • B) (1, 1)
  • C) (0, 1)
  • D) (1, 0)
• Q2: If \( a < 0 \) in \( y = ax^2 + bx + c \), the parabola opens?
  • A) Upwards
  • B) Downwards
  • C) Left
  • D) Right
• Q3: Length of latus rectum for \( y = 4x^2 \) is?
  • A) 1/4
  • B) 1/8
  • C) 4
  • D) 8

Answer Key

• Q1: A) (0, 0) - Standard parabola vertex at origin.
• Q2: B) Downwards - Negative \( a \) opens parabola down.
• Q3: B) 1/8 - Length = \( \frac{1}{4a} = \frac{1}{16} \) for \( a=4 \), but standard form length is \( \frac{1}{4a} \).

Memory Aids

• Positive \( a \) opens parabola up/right; negative \( a \) opens down/left.
• Vertex formula: \( h = -\frac{b}{2a} \), \( k = f(h) \).
• Axis of symmetry passes through vertex.

Basic Formulae for Differentiation

Common Derivatives

• Power rule: d/dx xⁿ = n x^n-1.
• Derivative of constant c is 0.
• Derivative of c xⁿ is c n x^n-1.
• Trigonometric derivatives:
  • d/dx sin x = cos x
  • d/dx cos x = -sin x
  • d/dx tan x = sec² x
  • d/dx cot x = -csc² x
  • d/dx sec x = sec x tan x
  • d/dx csc x = -csc x cot x
• Logarithmic derivative: d/dx ln x = 1/x.
• Exponential derivative: d/dx e^x = e^x.

Chain Rule

• If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx.
• Used for composite functions.

Solved Examples

• Ex 1: Find derivative of y = x³.
  • Using power rule: dy/dx = 3x².
• Ex 2: Derivative of y = sin x.
  • dy/dx = cos x.

Practice Set

• Q1: Derivative of x⁴ is?
  • A) 4x³
  • B) x³
  • C) 3x⁴
  • D) 4x⁴
• Q2: Derivative of cos x is?
  • A) sin x
  • B) -sin x
  • C) cos x
  • D) -cos x
• Q3: Using chain rule, derivative of y = (3x² + 1)⁵ is?
  • A) 5(3x² + 1)⁴
  • B) 10x(3x² + 1)⁴
  • C) 30x(3x² + 1)⁴
  • D) (3x² + 1)⁵

Answer Key

• Q1: A) 4x³ - Power rule n x^n-1 applied.
• Q2: B) -sin x - Derivative of cosine is negative sine.
• Q3: C) 30x(3x² + 1)⁴ - Chain rule: outer derivative times inner derivative.

Memory Aids

• Power rule: multiply by exponent, reduce exponent by 1.
• Trig derivatives: sin → cos, cos → -sin, tan → sec².
• Chain rule: derivative of outer × derivative of inner.

Common Derivatives

• Power rule: d/dx xⁿ = n x^n-1.
• Derivative of constant c is 0.
• Derivative of c xⁿ is c n x^n-1.
• Trigonometric derivatives:
  • d/dx sin x = cos x
  • d/dx cos x = -sin x
  • d/dx tan x = sec² x
  • d/dx cot x = -csc² x
  • d/dx sec x = sec x tan x
  • d/dx csc x = -csc x cot x
• Logarithmic derivative: d/dx ln x = 1/x.
• Exponential derivative: d/dx e^x = e^x.

Chain Rule

• If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx.
• Used for composite functions.

Solved Examples

• Ex 1: Find derivative of y = x³.
  • Using power rule: dy/dx = 3x².
• Ex 2: Derivative of y = sin x.
  • dy/dx = cos x.

Practice Set

• Q1: Derivative of x⁴ is?
  • A) 4x³
  • B) x³
  • C) 3x⁴
  • D) 4x⁴
• Q2: Derivative of cos x is?
  • A) sin x
  • B) -sin x
  • C) cos x
  • D) -cos x
• Q3: Using chain rule, derivative of y = (3x² + 1)⁵ is?
  • A) 5(3x² + 1)⁴
  • B) 10x(3x² + 1)⁴
  • C) 30x(3x² + 1)⁴
  • D) (3x² + 1)⁵

Answer Key

• Q1: A) 4x³ - Power rule n x^n-1 applied.
• Q2: B) -sin x - Derivative of cosine is negative sine.
• Q3: C) 30x(3x² + 1)⁴ - Chain rule: outer derivative times inner derivative.

Memory Aids

• Power rule: multiply by exponent, reduce exponent by 1.
• Trig derivatives: sin → cos, cos → -sin, tan → sec².
• Chain rule: derivative of outer × derivative of inner.

Common Derivatives

• Power rule: d/dx xⁿ = n x^n-1.
• Derivative of constant c is 0.
• Derivative of c xⁿ is c n x^n-1.
• Trigonometric derivatives:
  • d/dx sin x = cos x
  • d/dx cos x = -sin x
  • d/dx tan x = sec² x
  • d/dx cot x = -csc² x
  • d/dx sec x = sec x tan x
  • d/dx csc x = -csc x cot x
• Logarithmic derivative: d/dx ln x = 1/x.
• Exponential derivative: d/dx e^x = e^x.

Chain Rule

• If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx.
• Used for composite functions.

Solved Examples

• Ex 1: Find derivative of y = x³.
  • Using power rule: dy/dx = 3x².
• Ex 2: Derivative of y = sin x.
  • dy/dx = cos x.

Practice Set

• Q1: Derivative of x⁴ is?
  • A) 4x³
  • B) x³
  • C) 3x⁴
  • D) 4x⁴
• Q2: Derivative of cos x is?
  • A) sin x
  • B) -sin x
  • C) cos x
  • D) -cos x
• Q3: Using chain rule, derivative of y = (3x² + 1)⁵ is?
  • A) 5(3x² + 1)⁴
  • B) 10x(3x² + 1)⁴
  • C) 30x(3x² + 1)⁴
  • D) (3x² + 1)⁵

Answer Key

• Q1: A) 4x³ - Power rule n x^n-1 applied.
• Q2: B) -sin x - Derivative of cosine is negative sine.
• Q3: C) 30x(3x² + 1)⁴ - Chain rule: outer derivative times inner derivative.

Memory Aids

• Power rule: multiply by exponent, reduce exponent by 1.
• Trig derivatives: sin → cos, cos → -sin, tan → sec².
• Chain rule: derivative of outer × derivative of inner.

Common Derivatives

• Power rule: d/dx xⁿ = n x^n-1.
• Derivative of constant c is 0.
• Derivative of c xⁿ is c n x^n-1.
• Trigonometric derivatives:
  • d/dx sin x = cos x
  • d/dx cos x = -sin x
  • d/dx tan x = sec² x
  • d/dx cot x = -csc² x
  • d/dx sec x = sec x tan x
  • d/dx csc x = -csc x cot x
• Logarithmic derivative: d/dx ln x = 1/x.
• Exponential derivative: d/dx e^x = e^x.

Chain Rule

• If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx.
• Used for composite functions.

Solved Examples

• Ex 1: Find derivative of y = x³.
  • Using power rule: dy/dx = 3x².
• Ex 2: Derivative of y = sin x.
  • dy/dx = cos x.

Practice Set

• Q1: Derivative of x⁴ is?
  • A) 4x³
  • B) x³
  • C) 3x⁴
  • D) 4x⁴
• Q2: Derivative of cos x is?
  • A) sin x
  • B) -sin x
  • C) cos x
  • D) -cos x
• Q3: Using chain rule, derivative of y = (3x² + 1)⁵ is?
  • A) 5(3x² + 1)⁴
  • B) 10x(3x² + 1)⁴
  • C) 30x(3x² + 1)⁴
  • D) (3x² + 1)⁵

Answer Key

• Q1: A) 4x³ - Power rule n x^n-1 applied.
• Q2: B) -sin x - Derivative of cosine is negative sine.
• Q3: C) 30x(3x² + 1)⁴ - Chain rule: outer derivative times inner derivative.

Memory Aids

• Power rule: multiply by exponent, reduce exponent by 1.
• Trig derivatives: sin → cos, cos → -sin, tan → sec².
• Chain rule: derivative of outer × derivative of inner.

Common Derivatives

• Power rule: d/dx xⁿ = n x^n-1.
• Derivative of constant c is 0.
• Derivative of c xⁿ is c n x^n-1.
• Trigonometric derivatives:
  • d/dx sin x = cos x
  • d/dx cos x = -sin x
  • d/dx tan x = sec² x
  • d/dx cot x = -csc² x
  • d/dx sec x = sec x tan x
  • d/dx csc x = -csc x cot x
• Logarithmic derivative: d/dx ln x = 1/x.
• Exponential derivative: d/dx e^x = e^x.

Chain Rule

• If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx.
• Used for composite functions.

Solved Examples

• Ex 1: Find derivative of y = x³.
  • Using power rule: dy/dx = 3x².
• Ex 2: Derivative of y = sin x.
  • dy/dx = cos x.

Practice Set

• Q1: Derivative of x⁴ is?
  • A) 4x³
  • B) x³
  • C) 3x⁴
  • D) 4x⁴
• Q2: Derivative of cos x is?
  • A) sin x
  • B) -sin x
  • C) cos x
  • D) -cos x
• Q3: Using chain rule, derivative of y = (3x² + 1)⁵ is?
  • A) 5(3x² + 1)⁴
  • B) 10x(3x² + 1)⁴
  • C) 30x(3x² + 1)⁴
  • D) (3x² + 1)⁵

Answer Key

• Q1: A) 4x³ - Power rule n x^n-1 applied.
• Q2: B) -sin x - Derivative of cosine is negative sine.
• Q3: C) 30x(3x² + 1)⁴ - Chain rule: outer derivative times inner derivative.

Memory Aids

• Power rule: multiply by exponent, reduce exponent by 1.
• Trig derivatives: sin → cos, cos → -sin, tan → sec².
• Chain rule: derivative of outer × derivative of inner.

Common Derivatives

• Power rule: \( \frac{d}{dx} x^n = n x^{n-1} \).
• Derivative of constant \( c \) is 0.
• Derivative of \( c x^n \) is \( c n x^{n-1} \).
• Trigonometric derivatives:
  • \( \frac{d}{dx} \sin x = \cos x \)
  • \( \frac{d}{dx} \cos x = -\sin x \)
  • \( \frac{d}{dx} \tan x = \sec^2 x \)
  • \( \frac{d}{dx} \cot x = -\csc^2 x \)
  • \( \frac{d}{dx} \sec x = \sec x \tan x \)
  • \( \frac{d}{dx} \csc x = -\csc x \cot x \)
• Logarithmic derivative: \( \frac{d}{dx} \ln x = \frac{1}{x} \).
• Exponential derivative: \( \frac{d}{dx} e^x = e^x \).

Chain Rule

• If \( y = f(u) \) and \( u = g(x) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
• Used for composite functions.

Solved Examples

• Ex 1: Find derivative of \( y = x^3 \).
  • Using power rule: \( \frac{dy}{dx} = 3x^2 \).
• Ex 2: Derivative of \( y = \sin x \).
  • \( \frac{dy}{dx} = \cos x \).

Practice Set

• Q1: Derivative of \( x^4 \) is?
  • A) \( 4x^3 \)
  • B) \( x^3 \)
  • C) \( 3x^4 \)
  • D) \( 4x^4 \)
• Q2: Derivative of \( \cos x \) is?
  • A) \( \sin x \)
  • B) \( -\sin x \)
  • C) \( \cos x \)
  • D) \( -\cos x \)
• Q3: Using chain rule, derivative of \( y = (3x^2 + 1)^5 \) is?
  • A) \( 5(3x^2 + 1)^4 \)
  • B) \( 10x(3x^2 + 1)^4 \)
  • C) \( 30x(3x^2 + 1)^4 \)
  • D) \( (3x^2 + 1)^5 \)

Answer Key

• Q1: A) \( 4x^3 \) - Power rule \( n x^{n-1} \) applied.
• Q2: B) \( -\sin x \) - Derivative of cosine is negative sine.
• Q3: C) \( 30x(3x^2 + 1)^4 \) - Chain rule: outer derivative times inner derivative.

Memory Aids

• Power rule: multiply by exponent, reduce exponent by 1.
• Trig derivatives: sin → cos, cos → -sin, tan → sec².
• Chain rule: derivative of outer × derivative of inner.

Differentiation as Rate of Change

Concept and Formula

• Derivative dy/dx measures rate of change of y with respect to x.
• Average rate of change: Δy/Δx over interval [x, x+Δx].
• Instantaneous rate of change: lim_{Δx → 0} Δy/Δx = dy/dx.
• At x = x₀, (dy/dx)_x=x0 gives rate of change at that point.

Rate of Change of Two Variables

• If x = φ(t) and y = ψ(t), then dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0.
• Rate of change of y with respect to x can be found via parameter t.

Solved Examples

• Ex 1: Given y = t², x = t, find dy/dx.
  • dy/dt = 2t, dx/dt = 1.
  • dy/dx = 2t/1 = 2t.

Practice Set

• Q1: If y = 3t³, x = t, find dy/dx.
  • A) 3t²
  • B) 9t²
  • C) t³
  • D) 3t³
• Q2: If y = sin t, x = cos t, find dy/dx.
  • A) -tan t
  • B) -cot t
  • C) -tan t
  • D) -cot t
• Q3: If y = e^t, x = t², find dy/dx at t=1.
  • A) e
  • B) e/2
  • C) 2e
  • D) 1/(2e)

Answer Key

• Q1: B) 9t² - dy/dt = 9t², dx/dt = 1, so dy/dx = 9t².
• Q2: A) -tan t - dy/dt = cos t, dx/dt = -sin t, so dy/dx = cos t/-sin t = -cot t.
• Q3: B) e/2 - dy/dt = e^t, dx/dt = 2t, at t=1, dy/dx = e/2.

Memory Aids

• Instantaneous rate = derivative (dy/dx).
• Rate of change of two variables via parameter: (dy/dx) = (dy/dt)/(dx/dt).

Concept and Formula

• Derivative dy/dx measures rate of change of y with respect to x.
• Average rate of change: Δy/Δx over interval [x, x+Δx].
• Instantaneous rate of change: lim_{Δx → 0} Δy/Δx = dy/dx.
• At x = x₀, (dy/dx)_x=x0 gives rate of change at that point.

Rate of Change of Two Variables

• If x = φ(t) and y = ψ(t), then dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0.
• Rate of change of y with respect to x can be found via parameter t.

Solved Examples

• Ex 1: Given y = t², x = t, find dy/dx.
  • dy/dt = 2t, dx/dt = 1.
  • dy/dx = 2t/1 = 2t.

Practice Set

• Q1: If y = 3t³, x = t, find dy/dx.
  • A) 3t²
  • B) 9t²
  • C) t³
  • D) 3t³
• Q2: If y = sin t, x = cos t, find dy/dx.
  • A) -tan t
  • B) -cot t
  • C) -tan t
  • D) -cot t
• Q3: If y = e^t, x = t², find dy/dx at t=1.
  • A) e
  • B) e/2
  • C) 2e
  • D) 1/(2e)

Answer Key

• Q1: B) 9t² - dy/dt = 9t², dx/dt = 1, so dy/dx = 9t².
• Q2: A) -tan t - dy/dt = cos t, dx/dt = -sin t, so dy/dx = cos t/-sin t = -cot t.
• Q3: B) e/2 - dy/dt = e^t, dx/dt = 2t, at t=1, dy/dx = e/2.

Memory Aids

• Instantaneous rate = derivative dy/dx.
• Rate of change of two variables via parameter: dy/dx = (dy/dt)/(dx/dt).

Concept and Formula

• Derivative dy/dx measures rate of change of y with respect to x.
• Average rate of change: Δy/Δx over interval [x, x+Δx].
• Instantaneous rate of change: lim_{Δx → 0} Δy/Δx = dy/dx.
• At x = x₀, (dy/dx)_x=x0 gives rate of change at that point.

Rate of Change of Two Variables

• If x = φ(t) and y = ψ(t), then dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0.
• Rate of change of y with respect to x can be found via parameter t.

Solved Examples

• Ex 1: Given y = t², x = t, find dy/dx.
  • dy/dt = 2t, dx/dt = 1.
  • dy/dx = 2t/1 = 2t.

Practice Set

• Q1: If y = 3t³, x = t, find dy/dx.
  • A) 3t²
  • B) 9t²
  • C) t³
  • D) 3t³
• Q2: If y = sin t, x = cos t, find dy/dx.
  • A) -tan t
  • B) -cot t
  • C) -tan t
  • D) -cot t
• Q3: If y = e^t, x = t², find dy/dx at t=1.
  • A) e
  • B) e/2
  • C) 2e
  • D) 1/(2e)

Answer Key

• Q1: B) 9t² - dy/dt = 9t², dx/dt = 1, so dy/dx = 9t².
• Q2: A) -tan t - dy/dt = cos t, dx/dt = -sin t, so dy/dx = cos t/-sin t = -cot t.
• Q3: B) e/2 - dy/dt = e^t, dx/dt = 2t, at t=1, dy/dx = e/2.

Memory Aids

• Instantaneous rate = derivative dy/dx.
• Rate of change of two variables via parameter: dy/dx = (dy/dt)/(dx/dt).

Concept and Formula

• Derivative (dy/dx) measures rate of change of y with respect to x.
• Average rate of change: (Δy/Δx) over interval [x, x+Δx].
• Instantaneous rate of change: lim_{Δx → 0} (Δy/Δx) = dy/dx.
• At x = x₀, ((dy/dx)_x=x0) gives rate of change at that point.

Rate of Change of Two Variables

• If x = φ(t) and y = ψ(t), then dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0.
• Rate of change of y with respect to x can be found via parameter t.

Solved Examples

• Ex 1: Given y = t², x = t, find dy/dx.
  • dy/dt = 2t, dx/dt = 1.
  • dy/dx = 2t/1 = 2t.

Practice Set

• Q1: If y = 3t³, x = t, find dy/dx.
  • A) 3t²
  • B) 9t²
  • C) t³
  • D) 3t³
• Q2: If y = sin t, x = cos t, find dy/dx.
  • A) -tan t
  • B) -cot t
  • C) -tan t
  • D) -cot t
• Q3: If y = e^t, x = t², find dy/dx at t=1.
  • A) e
  • B) e/2
  • C) 2e
  • D) 1/(2e)

Answer Key

• Q1: B) 9t² - dy/dt = 9t², dx/dt = 1, so dy/dx = 9t².
• Q2: A) -tan t - dy/dt = cos t, dx/dt = -sin t, so dy/dx = cos t/-sin t = -cot t.
• Q3: B) e/2 - dy/dt = e^t, dx/dt = 2t, at t=1, dy/dx = e/2.

Memory Aids

• Instantaneous rate = derivative (dy/dx).
• Rate of change of two variables via parameter: (dy/dx) = (dy/dt)/(dx/dt).

Concept and Formula

• Derivative dy/dx measures rate of change of y with respect to x.
• Average rate of change: Δy/Δx over interval [x, x+Δx].
• Instantaneous rate of change: lim_{Δx → 0} Δy/Δx = dy/dx.
• At x = x₀, (dy/dx)_x=x0 gives rate of change at that point.

Rate of Change of Two Variables

• If x = φ(t) and y = ψ(t), then dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0.
• Rate of change of y with respect to x can be found via parameter t.

Solved Examples

• Ex 1: Given y = t², x = t, find dy/dx.
  • dy/dt = 2t, dx/dt = 1.
  • dy/dx = 2t/1 = 2t.

Practice Set

• Q1: If y = 3t³, x = t, find dy/dx.
  • A) 3t²
  • B) 9t²
  • C) t³
  • D) 3t³
• Q2: If y = sin t, x = cos t, find dy/dx.
  • A) -tan t
  • B) -cot t
  • C) -tan t
  • D) -cot t
• Q3: If y = e^t, x = t², find dy/dx at t=1.
  • A) e
  • B) e/2
  • C) 2e
  • D) 1/(2e)

Answer Key

• Q1: B) 9t² - dy/dt = 9t², dx/dt = 1, so dy/dx = 9t².
• Q2: A) -tan t - dy/dt = cos t, dx/dt = -sin t, so dy/dx = cos t/-sin t = -cot t.
• Q3: B) e/2 - dy/dt = e^t, dx/dt = 2t, at t=1, dy/dx = e/2.

Memory Aids

• Instantaneous rate = derivative dy/dx.
• Rate of change of two variables via parameter: dy/dx = (dy/dt)/(dx/dt).

Concept and Formula

• Derivative \( \frac{dy}{dx} \) measures rate of change of \( y \) with respect to \( x \).
• Average rate of change: \( \frac{\Delta y}{\Delta x} \) over interval \( [x, x+\Delta x] \).
• Instantaneous rate of change: \( \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} \).
• At \( x = x_0 \), \( \left( \frac{dy}{dx} \right)_{x=x_0} \) gives rate of change at that point.

Rate of Change of Two Variables

• If \( x = \phi(t) \) and \( y = \psi(t) \), then \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) provided \( dx/dt \neq 0 \).
• Rate of change of \( y \) with respect to \( x \) can be found via parameter \( t \).

Solved Examples

• Ex 1: Given \( y = t^2 \), \( x = t \), find \( \frac{dy}{dx} \).
  • \( \frac{dy}{dt} = 2t \), \( \frac{dx}{dt} = 1 \).
  • \( \frac{dy}{dx} = \frac{2t}{1} = 2t \).

Practice Set

• Q1: If \( y = 3t^3 \), \( x = t \), find \( \frac{dy}{dx} \).
  • A) \( 3t^2 \)
  • B) \( 9t^2 \)
  • C) \( t^3 \)
  • D) \( 3t^3 \)
• Q2: If \( y = \sin t \), \( x = \cos t \), find \( \frac{dy}{dx} \).
  • A) \( -\tan t \)
  • B) \( -\cot t \)
  • C) \( -\tan t \)
  • D) \( -\cot t \)
• Q3: If \( y = e^t \), \( x = t^2 \), find \( \frac{dy}{dx} \) at \( t=1 \).
  • A) \( e \)
  • B) \( \frac{e}{2} \)
  • C) \( 2e \)
  • D) \( \frac{1}{2e} \)

Answer Key

• Q1: B) \( 9t^2 \) - \( \frac{dy}{dt} = 9t^2 \), \( \frac{dx}{dt} = 1 \), so \( \frac{dy}{dx} = 9t^2 \).
• Q2: A) \( -\tan t \) - \( \frac{dy}{dt} = \cos t \), \( \frac{dx}{dt} = -\sin t \), so \( \frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t \).
• Q3: B) \( \frac{e}{2} \) - \( \frac{dy}{dt} = e^t \), \( \frac{dx}{dt} = 2t \), at \( t=1 \), \( \frac{dy}{dx} = \frac{e}{2} \).

Memory Aids

• Instantaneous rate = derivative \( \frac{dy}{dx} \).
• Rate of change of two variables via parameter: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

Applications of Derivative

Slope of a Curve

• Function: y = f(x)
• Slope between two points A (x₁, y₁) and B (x₂, y₂): Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• As Δx → 0, points A and B coincide; slope becomes tangent slope.
• Tangent slope at a point: dy/dx = lim_{Δx → 0} Δy/Δx = tan θ

Solved Examples

• Ex 1: Find slope of line joining points A(1, 2) and B(3, 6).
  • Use Δy/Δx = (6 - 2)/(3 - 1) = 4/2 = 2
• Ex 2: Find slope of tangent to curve y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • At x=2, slope = 2 × 2 = 4

Practice Set

• Q1: Slope of line joining points (2, 3) and (5, 15) is?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: If y = 3x³, slope at x=1 is?
  • A) 3
  • B) 6
  • C) 9
  • D) 27
• Q3: Slope of tangent to y = √x at x=4 is?
  • A) 1/4
  • B) 1/2
  • C) 1
  • D) 2

Answer Key

• Q1: A) 3 — Slope = (15-3)/(5-2) = 12/3 = 4 (Correct option is B, corrected here)
• Q2: C) 9 — dy/dx = 9x², at x=1, slope = 9
• Q3: B) 1/2 — y = x^1/2, slope = 1/2 x^-1/2, at x=4, slope = 1/2 × 1/2 = 1/4 (Correct option is A, corrected here)

Memory Aids

• Slope between points: Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• Tangent slope: dy/dx = tan θ
• As Δx → 0, secant line becomes tangent

Slope of a Curve

• Function: y = f(x)
• Slope between two points A (x₁, y₁) and B (x₂, y₂): Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• As Δx → 0, points A and B coincide; slope becomes tangent slope.
• Tangent slope at a point: dy/dx = lim_{Δx → 0} Δy/Δx = tan θ

Solved Examples

• Ex 1: Find slope of line joining points A(1, 2) and B(3, 6).
  • Use Δy/Δx = (6 - 2)/(3 - 1) = 4/2 = 2
• Ex 2: Find slope of tangent to curve y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • At x=2, slope = 2 × 2 = 4

Practice Set

• Q1: Slope of line joining points (2, 3) and (5, 15) is?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: If y = 3x³, slope at x=1 is?
  • A) 3
  • B) 6
  • C) 9
  • D) 27
• Q3: Slope of tangent to y = √x at x=4 is?
  • A) 1/4
  • B) 1/2
  • C) 1
  • D) 2

Answer Key

• Q1: A) 3 — Slope = (15-3)/(5-2) = 12/3 = 4 (Correct option is B, corrected here)
• Q2: C) 9 — dy/dx = 9x², at x=1, slope = 9
• Q3: B) 1/2 — y = x^1/2, slope = 1/2 x^-1/2, at x=4, slope = 1/2 × 1/2 = 1/4 (Correct option is A, corrected here)

Memory Aids

• Slope between points: Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• Tangent slope: dy/dx = tan θ
• As Δx → 0, secant line becomes tangent

Slope of a Curve

• Function: y = f(x)
• Slope between two points A (x₁, y₁) and B (x₂, y₂): Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• As Δx → 0, points A and B coincide; slope becomes tangent slope.
• Tangent slope at a point: dy/dx = lim_{Δx → 0} Δy/Δx = tan θ

Solved Examples

• Ex 1: Find slope of line joining points A(1, 2) and B(3, 6).
  • Use (Δy/Δx) = (6 - 2)/(3 - 1) = (4/2) = 2
• Ex 2: Find slope of tangent to curve (y = x²) at (x = 2).
  • Derivative: (dy/dx) = 2x
  • At x=2, slope = 2 × 2 = 4

Practice Set

• Q1: Slope of line joining points (2, 3) and (5, 15) is?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: If (y = 3x³), slope at (x=1) is?
  • A) 3
  • B) 6
  • C) 9
  • D) 27
• Q3: Slope of tangent to (y = √x) at (x=4) is?
  • A) (1/4)
  • B) (1/2)
  • C) 1
  • D) 2

Answer Key

• Q1: A) 3 — Slope = ((15-3)/(5-2) = 12/3) = 4 (Correct option is B, corrected here)
• Q2: C) 9 — (dy/dx) = 9x², at (x=1), slope = 9
• Q3: B) (1/2) — (y = x^1/2), slope = (1/2 x^-1/2), at (x=4), slope = (1/2 × 1/2 = 1/4) (Correct option is A, corrected here)

Memory Aids

• Slope between points: (Δy/Δx) = (y₂ - y₁)/(x₂ - x₁)
• Tangent slope: (dy/dx) = tan θ
• As Δx → 0, secant line becomes tangent

Slope of a Curve

• Function: y = f(x)
• Slope between two points A (x₁, y₁) and B (x₂, y₂): Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• As Δx → 0, points A and B coincide; slope becomes tangent slope.
• Tangent slope at a point: dy/dx = lim_{Δx → 0} Δy/Δx = tan θ

Solved Examples

• Ex 1: Find slope of line joining points A(1, 2) and B(3, 6).
  • Use Δy/Δx = (6 - 2)/(3 - 1) = 4/2 = 2
• Ex 2: Find slope of tangent to curve y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • At x=2, slope = 2 × 2 = 4

Practice Set

• Q1: Slope of line joining points (2, 3) and (5, 15) is?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: If y = 3x³, slope at x=1 is?
  • A) 3
  • B) 6
  • C) 9
  • D) 27
• Q3: Slope of tangent to y = √x at x=4 is?
  • A) 1/4
  • B) 1/2
  • C) 1
  • D) 2

Answer Key

• Q1: A) 3 — Slope = (15-3)/(5-2) = 12/3 = 4 (Correct option is B, corrected here)
• Q2: C) 9 — dy/dx = 9x², at x=1, slope = 9
• Q3: B) 1/2 — y = x^1/2, slope = 1/2 x^-1/2, at x=4, slope = 1/2 × 1/2 = 1/4 (Correct option is A, corrected here)

Memory Aids

• Slope between points: Δy/Δx = (y₂ - y₁)/(x₂ - x₁)
• Tangent slope: dy/dx = tan θ
• As Δx → 0, secant line becomes tangent

Slope of a Curve

• Function: \( y = f(x) \)
• Slope between two points A \((x_1, y_1)\) and B \((x_2, y_2)\): \( \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \)
• As \( \Delta x \to 0 \), points A and B coincide; slope becomes tangent slope.
• Tangent slope at a point: \( \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \tan \theta \)

Solved Examples

• Ex 1: Find slope of line joining points A(1, 2) and B(3, 6).
  • Use \( \frac{\Delta y}{\Delta x} = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2 \)
• Ex 2: Find slope of tangent to curve \( y = x^2 \) at \( x = 2 \).
  • Derivative: \( \frac{dy}{dx} = 2x \)
  • At \( x=2 \), slope = \( 2 \times 2 = 4 \)

Practice Set

• Q1: Slope of line joining points (2, 3) and (5, 15) is?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: If \( y = 3x^3 \), slope at \( x=1 \) is?
  • A) 3
  • B) 6
  • C) 9
  • D) 27
• Q3: Slope of tangent to \( y = \sqrt{x} \) at \( x=4 \) is?
  • A) \( \frac{1}{4} \)
  • B) \( \frac{1}{2} \)
  • C) 1
  • D) 2

Answer Key

• Q1: A) 3 — Slope = \( \frac{15-3}{5-2} = \frac{12}{3} = 4 \) (Correct option is B, corrected here)
• Q2: C) 9 — \( \frac{dy}{dx} = 9x^2 \), at \( x=1 \), slope = 9
• Q3: B) \( \frac{1}{2} \) — \( y = x^{1/2} \), slope = \( \frac{1}{2} x^{-1/2} \), at \( x=4 \), slope = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \) (Correct option is A, corrected here)

Memory Aids

• Slope between points: \( \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \)
• Tangent slope: \( \frac{dy}{dx} = \tan \theta \)
• As \( \Delta x \to 0 \), secant line becomes tangent

Maxima and Minima

Critical Points and Derivatives

• At maxima or minima, slope of tangent dy/dx = 0
• Maxima: Tangent is horizontal, angle with x-axis = 0°
• Minima: Tangent is horizontal, angle with x-axis = 0°
• To distinguish maxima/minima, use second derivative test:
• d²y/dx² < 0 indicates maxima
• d²y/dx² > 0 indicates minima

Solved Examples

• Ex 1: Find maxima or minima of y = -x² + 4x + 1
  • dy/dx = -2x + 4
  • Set dy/dx = 0 gives x = 2
  • d²y/dx² = -2 < 0 so maxima at x=2
• Ex 2: Find minima of y = x² - 6x + 8
  • dy/dx = 2x - 6
  • Set dy/dx = 0 gives x = 3
  • d²y/dx² = 2 > 0 so minima at x=3

Practice Set

• Q1: At maxima or minima, dy/dx = ?
  • A) 1
  • B) 0
  • C) -1
  • D) Undefined
• Q2: If d²y/dx² < 0, the point is?
  • A) Minima
  • B) Maxima
  • C) Inflection point
  • D) None
• Q3: For y = x³ - 3x² + 4, find nature of critical point at x=2.
  • A) Maxima
  • B) Minima
  • C) Inflection point
  • D) None

Answer Key

• Q1: B) 0 — Slope at maxima/minima is zero
• Q2: B) Maxima — Negative second derivative means maxima
• Q3: A) Maxima — dy/dx = 3x² - 6x, zero at x=2; d²y/dx² = 6x - 6 = 6(2) - 6 = 6 > 0 (Correction: positive means minima, so correct answer is B)

Memory Aids

• Maxima: dy/dx = 0 and d²y/dx² < 0
• Minima: dy/dx = 0 and d²y/dx² > 0
• Second derivative test distinguishes maxima and minima

Critical Points and Derivatives

• At maxima or minima, slope of tangent dy/dx = 0
• Maxima: Tangent is horizontal, angle with x-axis = 0°
• Minima: Tangent is horizontal, angle with x-axis = 0°
• To distinguish maxima/minima, use second derivative test:
• d²y/dx² < 0 indicates maxima
• d²y/dx² > 0 indicates minima

Solved Examples

• Ex 1: Find maxima or minima of y = -x² + 4x + 1
  • dy/dx = -2x + 4
  • Set dy/dx = 0 gives x = 2
  • d²y/dx² = -2 < 0 so maxima at x=2
• Ex 2: Find minima of y = x² - 6x + 8
  • dy/dx = 2x - 6
  • Set dy/dx = 0 gives x = 3
  • d²y/dx² = 2 > 0 so minima at x=3

Practice Set

• Q1: At maxima or minima, dy/dx = ?
  • A) 1
  • B) 0
  • C) -1
  • D) Undefined
• Q2: If d²y/dx² < 0, the point is?
  • A) Minima
  • B) Maxima
  • C) Inflection point
  • D) None
• Q3: For y = x³ - 3x² + 4, find nature of critical point at x=2.
  • A) Maxima
  • B) Minima
  • C) Inflection point
  • D) None

Answer Key

• Q1: B) 0 — Slope at maxima/minima is zero
• Q2: B) Maxima — Negative second derivative means maxima
• Q3: A) Maxima — dy/dx = 3x² - 6x, zero at x=2; d²y/dx² = 6x - 6 = 6(2) - 6 = 6 > 0 (Correction: positive means minima, so correct answer is B)

Memory Aids

• Maxima: dy/dx = 0 and d²y/dx² < 0
• Minima: dy/dx = 0 and d²y/dx² > 0
• Second derivative test distinguishes maxima and minima

Critical Points and Derivatives

• At maxima or minima, slope of tangent (dy/dx) = 0
• Maxima: Tangent is horizontal, angle with x-axis = 0°
• Minima: Tangent is horizontal, angle with x-axis = 0°
• To distinguish maxima/minima, use second derivative test:
• (d² y/dx² < 0) indicates maxima
• (d² y/dx² > 0) indicates minima

Solved Examples

• Ex 1: Find maxima or minima of (y = -x² + 4x + 1)
  • (dy/dx) = -2x + 4
  • Set (dy/dx) = 0 gives (x = 2)
  • (d² y/dx² = -2 < 0) so maxima at (x=2)
• Ex 2: Find minima of (y = x² - 6x + 8)
  • (dy/dx) = 2x - 6
  • Set (dy/dx) = 0 gives (x = 3)
  • (d² y/dx² = 2 > 0) so minima at (x=3)

Practice Set

• Q1: At maxima or minima, (dy/dx = ?)
  • A) 1
  • B) 0
  • C) -1
  • D) Undefined
• Q2: If (d² y/dx² < 0), the point is?
  • A) Minima
  • B) Maxima
  • C) Inflection point
  • D) None
• Q3: For (y = x³ - 3x² + 4), find nature of critical point at (x=2).
  • A) Maxima
  • B) Minima
  • C) Inflection point
  • D) None

Answer Key

• Q1: B) 0 — Slope at maxima/minima is zero
• Q2: B) Maxima — Negative second derivative means maxima
• Q3: A) Maxima — (dy/dx = 3x² - 6x), zero at (x=2); (d² y/dx² = 6x - 6 = 6(2) - 6 = 6 > 0) (Correction: positive means minima, so correct answer is B)

Memory Aids

• Maxima: (dy/dx = 0) and (d² y/dx² < 0)
• Minima: (dy/dx = 0) and (d² y/dx² > 0)
• Second derivative test distinguishes maxima and minima

Critical Points and Derivatives

• At maxima or minima, slope of tangent dy/dx = 0
• Maxima: Tangent is horizontal, angle with x-axis = 0°
• Minima: Tangent is horizontal, angle with x-axis = 0°
• To distinguish maxima/minima, use second derivative test:
• d²y/dx² < 0 indicates maxima
• d²y/dx² > 0 indicates minima

Solved Examples

• Ex 1: Find maxima or minima of y = -x² + 4x + 1
  • dy/dx = -2x + 4
  • Set dy/dx = 0 gives x = 2
  • d²y/dx² = -2 < 0 so maxima at x=2
• Ex 2: Find minima of y = x² - 6x + 8
  • dy/dx = 2x - 6
  • Set dy/dx = 0 gives x = 3
  • d²y/dx² = 2 > 0 so minima at x=3

Practice Set

• Q1: At maxima or minima, dy/dx = ?
  • A) 1
  • B) 0
  • C) -1
  • D) Undefined
• Q2: If d²y/dx² < 0, the point is?
  • A) Minima
  • B) Maxima
  • C) Inflection point
  • D) None
• Q3: For y = x³ - 3x² + 4, find nature of critical point at x=2.
  • A) Maxima
  • B) Minima
  • C) Inflection point
  • D) None

Answer Key

• Q1: B) 0 — Slope at maxima/minima is zero
• Q2: B) Maxima — Negative second derivative means maxima
• Q3: A) Maxima — dy/dx = 3x² - 6x, zero at x=2; d²y/dx² = 6x - 6 = 6(2) - 6 = 6 > 0 (Correction: positive means minima, so correct answer is B)

Memory Aids

• Maxima: dy/dx = 0 and d²y/dx² < 0
• Minima: dy/dx = 0 and d²y/dx² > 0
• Second derivative test distinguishes maxima and minima

Critical Points and Derivatives

• At maxima or minima, slope of tangent \( \frac{dy}{dx} = 0 \)
• Maxima: Tangent is horizontal, angle with x-axis = 0°
• Minima: Tangent is horizontal, angle with x-axis = 0°
• To distinguish maxima/minima, use second derivative test:
• \( \frac{d^2 y}{dx^2} < 0 \) indicates maxima
• \( \frac{d^2 y}{dx^2} > 0 \) indicates minima

Solved Examples

• Ex 1: Find maxima or minima of \( y = -x^2 + 4x + 1 \)
  • \( \frac{dy}{dx} = -2x + 4 \)
  • Set \( \frac{dy}{dx} = 0 \) gives \( x = 2 \)
  • \( \frac{d^2 y}{dx^2} = -2 < 0 \) so maxima at \( x=2 \)
• Ex 2: Find minima of \( y = x^2 - 6x + 8 \)
  • \( \frac{dy}{dx} = 2x - 6 \)
  • Set \( \frac{dy}{dx} = 0 \) gives \( x = 3 \)
  • \( \frac{d^2 y}{dx^2} = 2 > 0 \) so minima at \( x=3 \)

Practice Set

• Q1: At maxima or minima, \( \frac{dy}{dx} = ? \)
  • A) 1
  • B) 0
  • C) -1
  • D) Undefined
• Q2: If \( \frac{d^2 y}{dx^2} < 0 \), the point is?
  • A) Minima
  • B) Maxima
  • C) Inflection point
  • D) None
• Q3: For \( y = x^3 - 3x^2 + 4 \), find nature of critical point at \( x=2 \).
  • A) Maxima
  • B) Minima
  • C) Inflection point
  • D) None

Answer Key

• Q1: B) 0 — Slope at maxima/minima is zero
• Q2: B) Maxima — Negative second derivative means maxima
• Q3: A) Maxima — \( \frac{dy}{dx} = 3x^2 - 6x \), zero at \( x=2 \); \( \frac{d^2 y}{dx^2} = 6x - 6 = 6(2) - 6 = 6 > 0 \) (Correction: positive means minima, so correct answer is B)

Memory Aids

• Maxima: \( \frac{dy}{dx} = 0 \) and \( \frac{d^2 y}{dx^2} < 0 \)
• Minima: \( \frac{dy}{dx} = 0 \) and \( \frac{d^2 y}{dx^2} > 0 \)
• Second derivative test distinguishes maxima and minima

Integration

Some Formulae for Integration

• Integration finds a function whose derivative is the given function.
• Indefinite integral: ∫ f(x) dx = F(x) + C, where C is constant.
• Common integrals include powers, trigonometric, exponential, and logarithmic functions.
• Example: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.

Solved Examples

• Ex 1: Find ∫ x² dx.
  • Use formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
  • Result: x³/3 + C.
• Ex 2: Find ∫ sin x dx.
  • Integral of sin x is -cos x + C.

Practice Set

• Q1: ∫ x³ dx = ?
  • A) x⁴/4 + C
  • B) x²/2 + C
  • C) 3x² + C
  • D) x³/3 + C
• Q2: ∫ e^x dx = ?
  • A) e^x + C
  • B) x e^x-1 + C
  • C) ln x + C
  • D) x + C
• Q3: ∫ 1/x dx = ?
  • A) ln |x| + C
  • B) 1/x² + C
  • C) x + C
  • D) e^x + C

Answer Key

• Q1: A - Use power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
• Q2: A - Integral of exponential e^x is itself plus constant.
• Q3: A - Integral of 1/x is natural log ln |x| + C.

Memory Aids

• Power rule: Increase power by 1, divide by new power.
• Integral of e^x is e^x + C.
• Integral of sin x is -cos x + C.
• Integral of 1/x is ln |x| + C.

Some Formulae for Integration

• Integration finds a function whose derivative is the given function.
• Indefinite integral: ∫ f(x) dx = F(x) + C, where C is constant.
• Common integrals include powers, trigonometric, exponential, and logarithmic functions.
• Example: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.

Solved Examples

• Ex 1: Find ∫ x² dx.
  • Use formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
  • Result: x³/3 + C.
• Ex 2: Find ∫ sin x dx.
  • Integral of sin x is -cos x + C.

Practice Set

• Q1: ∫ x³ dx = ?
  • A) x⁴/4 + C
  • B) x²/2 + C
  • C) 3x² + C
  • D) x³/3 + C
• Q2: ∫ e^x dx = ?
  • A) e^x + C
  • B) x e^x-1 + C
  • C) ln x + C
  • D) x + C
• Q3: ∫ 1/x dx = ?
  • A) ln |x| + C
  • B) 1/x² + C
  • C) x + C
  • D) e^x + C

Answer Key

• Q1: A - Use power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
• Q2: A - Integral of exponential e^x is itself plus constant.
• Q3: A - Integral of 1/x is natural log ln |x| + C.

Memory Aids

• Power rule: Increase power by 1, divide by new power.
• Integral of e^x is e^x + C.
• Integral of sin x is -cos x + C.
• Integral of 1/x is ln |x| + C.

Some Formulae for Integration

• Integration finds a function whose derivative is the given function.
• Indefinite integral: ∫ f(x) dx = F(x) + C, where C is constant.
• Common integrals include powers, trigonometric, exponential, and logarithmic functions.
• Example: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.

Solved Examples

• Ex 1: Find ∫ x² dx.
  • Use formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
  • Result: x³/3 + C.
• Ex 2: Find ∫ sin x dx.
  • Integral of sin x is -cos x + C.

Practice Set

• Q1: ∫ x³ dx = ?
  • A) x⁴/4 + C
  • B) x²/2 + C
  • C) 3x² + C
  • D) x³/3 + C
• Q2: ∫ e^x dx = ?
  • A) e^x + C
  • B) x e^x-1 + C
  • C) ln x + C
  • D) x + C
• Q3: ∫ 1/x dx = ?
  • A) ln |x| + C
  • B) 1/x² + C
  • C) x + C
  • D) e^x + C

Answer Key

• Q1: A - Use power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
• Q2: A - Integral of exponential e^x is itself plus constant.
• Q3: A - Integral of 1/x is natural log ln |x| + C.

Memory Aids

• Power rule: Increase power by 1, divide by new power.
• Integral of e^x is e^x + C.
• Integral of sin x is -cos x + C.
• Integral of 1/x is ln |x| + C.

Some Formulae for Integration

• Integration finds a function whose derivative is the given function.
• Indefinite integral: ∫ f(x) dx = F(x) + C, where C is constant.
• Common integrals include powers, trigonometric, exponential, and logarithmic functions.
• Example: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.

Solved Examples

• Ex 1: Find ∫ x² dx.
  • Use formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
  • Result: x³/3 + C.
• Ex 2: Find ∫ sin x dx.
  • Integral of sin x is -cos x + C.

Practice Set

• Q1: ∫ x³ dx = ?
  • A) x⁴/4 + C
  • B) x²/2 + C
  • C) 3x² + C
  • D) x³/3 + C
• Q2: ∫ e^x dx = ?
  • A) e^x + C
  • B) x e^x-1 + C
  • C) ln x + C
  • D) x + C
• Q3: ∫ 1/x dx = ?
  • A) ln |x| + C
  • B) 1/x² + C
  • C) x + C
  • D) e^x + C

Answer Key

• Q1: A - Use power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C.
• Q2: A - Integral of exponential e^x is itself plus constant.
• Q3: A - Integral of 1/x is natural log ln |x| + C.

Memory Aids

• Power rule: Increase power by 1, divide by new power.
• Integral of e^x is e^x + C.
• Integral of sin x is -cos x + C.
• Integral of 1/x is ln |x| + C.

Some Formulae for Integration

• Integration finds a function whose derivative is the given function.
• Indefinite integral: \( \int f(x) \, dx = F(x) + C \), where \( C \) is constant.
• Common integrals include powers, trigonometric, exponential, and logarithmic functions.
• Example: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, n \neq -1 \).

Solved Examples

• Ex 1: Find \( \int x^2 \, dx \).
  • Use formula: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
  • Result: \( \frac{x^3}{3} + C \).
• Ex 2: Find \( \int \sin x \, dx \).
  • Integral of \( \sin x \) is \( -\cos x + C \).

Practice Set

• Q1: \( \int x^3 \, dx = ? \)
  • A) \( \frac{x^4}{4} + C \)
  • B) \( \frac{x^2}{2} + C \)
  • C) \( 3x^2 + C \)
  • D) \( \frac{x^3}{3} + C \)
• Q2: \( \int e^x \, dx = ? \)
  • A) \( e^x + C \)
  • B) \( x e^{x-1} + C \)
  • C) \( \ln x + C \)
  • D) \( x + C \)
• Q3: \( \int \frac{1}{x} \, dx = ? \)
  • A) \( \ln |x| + C \)
  • B) \( \frac{1}{x^2} + C \)
  • C) \( x + C \)
  • D) \( e^x + C \)

Answer Key

• Q1: A - Use power rule \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \).
• Q2: A - Integral of exponential \( e^x \) is itself plus constant.
• Q3: A - Integral of \( \frac{1}{x} \) is natural log \( \ln |x| + C \).

Memory Aids

• Power rule: Increase power by 1, divide by new power.
• Integral of \( e^x \) is \( e^x + C \).
• Integral of \( \sin x \) is \( -\cos x + C \).
• Integral of \( \frac{1}{x} \) is \( \ln |x| + C \).

Indefinite Integration

Key Formula and Concept

• Formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.
• Integration reverses differentiation.
• Constant C added as integrals differ by constants.

Solved Examples

• Ex 1: ∫ x⁴ dx = x⁵/5 + C.
• Ex 2: ∫ x^-2 dx = -1/x + C.

Practice Set

• Q1: ∫ x⁵ dx = ?
  • A) x⁶/6 + C
  • B) 5x⁴ + C
  • C) x⁴/4 + C
  • D) 6x⁵ + C
• Q2: ∫ x^-3 dx = ?
  • A) -1/(2x²) + C
  • B) 1/(2x²) + C
  • C) 1/x² + C
  • D) -1/x² + C
• Q3: ∫ x^-1 dx = ?
  • A) ln |x| + C
  • B) x⁰/0 + C
  • C) x + C
  • D) x⁰/1 + C

Answer Key

• Q1: A - Power rule applies, add 1 to power and divide.
• Q2: A - ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, here n = -3.
• Q3: A - Special case n = -1, integral is natural log.

Memory Aids

• For n ≠ -1, add 1 to power, divide by new power.
• For n = -1, use ln |x| + C.

Key Formula and Concept

• Formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.
• Integration reverses differentiation.
• Constant C added as integrals differ by constants.

Solved Examples

• Ex 1: ∫ x⁴ dx = x⁵/5 + C.
• Ex 2: ∫ x^-2 dx = -1/x + C.

Practice Set

• Q1: ∫ x⁵ dx = ?
  • A) x⁶/6 + C
  • B) 5x⁴ + C
  • C) x⁴/4 + C
  • D) 6x⁵ + C
• Q2: ∫ x^-3 dx = ?
  • A) -1/(2x²) + C
  • B) 1/(2x²) + C
  • C) 1/x² + C
  • D) -1/x² + C
• Q3: ∫ x^-1 dx = ?
  • A) ln |x| + C
  • B) x⁰/0 + C
  • C) x + C
  • D) x⁰/1 + C

Answer Key

• Q1: A - Power rule applies, add 1 to power and divide.
• Q2: A - ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, here n = -3.
• Q3: A - Special case n = -1, integral is natural log.

Memory Aids

• For n ≠ -1, add 1 to power, divide by new power.
• For n = -1, use ln |x| + C.

Key Formula and Concept

• Formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.
• Integration reverses differentiation.
• Constant C added as integrals differ by constants.

Solved Examples

• Ex 1: ∫ x⁴ dx = x⁵/5 + C.
• Ex 2: ∫ x^-2 dx = -1/x + C.

Practice Set

• Q1: ∫ x⁵ dx = ?
  • A) x⁶/6 + C
  • B) 5x⁴ + C
  • C) x⁴/4 + C
  • D) 6x⁵ + C
• Q2: ∫ x^-3 dx = ?
  • A) -1/(2x²) + C
  • B) 1/(2x²) + C
  • C) 1/x² + C
  • D) -1/x² + C
• Q3: ∫ x^-1 dx = ?
  • A) ln |x| + C
  • B) x⁰/0 + C
  • C) x + C
  • D) x⁰/1 + C

Answer Key

• Q1: A - Power rule applies, add 1 to power and divide.
• Q2: A - ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, here n = -3.
• Q3: A - Special case n = -1, integral is natural log.

Memory Aids

• For n ≠ -1, add 1 to power, divide by new power.
• For n = -1, use ln |x| + C.

Key Formula and Concept

• Formula: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ -1.
• Integration reverses differentiation.
• Constant C added as integrals differ by constants.

Solved Examples

• Ex 1: ∫ x⁴ dx = x⁵/5 + C.
• Ex 2: ∫ x^-2 dx = -1/x + C.

Practice Set

• Q1: ∫ x⁵ dx = ?
  • A) x⁶/6 + C
  • B) 5x⁴ + C
  • C) x⁴/4 + C
  • D) 6x⁵ + C
• Q2: ∫ x^-3 dx = ?
  • A) -1/(2x²) + C
  • B) 1/(2x²) + C
  • C) 1/x² + C
  • D) -1/x² + C
• Q3: ∫ x^-1 dx = ?
  • A) ln |x| + C
  • B) x⁰/0 + C
  • C) x + C
  • D) x⁰/1 + C

Answer Key

• Q1: A - Power rule applies, add 1 to power and divide.
• Q2: A - ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, here n = -3.
• Q3: A - Special case n = -1, integral is natural log.

Memory Aids

• For n ≠ -1, add 1 to power, divide by new power.
• For n = -1, use ln |x| + C.

Key Formula and Concept

• Formula: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, n \neq -1 \).
• Integration reverses differentiation.
• Constant \( C \) added as integrals differ by constants.

Solved Examples

• Ex 1: \( \int x^4 \, dx = \frac{x^5}{5} + C \).
• Ex 2: \( \int x^{-2} \, dx = -\frac{1}{x} + C \).

Practice Set

• Q1: \( \int x^5 \, dx = ? \)
  • A) \( \frac{x^6}{6} + C \)
  • B) \( 5x^4 + C \)
  • C) \( \frac{x^4}{4} + C \)
  • D) \( 6x^5 + C \)
• Q2: \( \int x^{-3} \, dx = ? \)
  • A) \( -\frac{1}{2x^2} + C \)
  • B) \( \frac{1}{2x^2} + C \)
  • C) \( \frac{1}{x^2} + C \)
  • D) \( -\frac{1}{x^2} + C \)
• Q3: \( \int x^{-1} \, dx = ? \)
  • A) \( \ln |x| + C \)
  • B) \( \frac{x^0}{0} + C \)
  • C) \( x + C \)
  • D) \( \frac{x^0}{1} + C \)

Answer Key

• Q1: A - Power rule applies, add 1 to power and divide.
• Q2: A - \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), here \( n = -3 \).
• Q3: A - Special case \( n = -1 \), integral is natural log.

Memory Aids

• For \( n \neq -1 \), add 1 to power, divide by new power.
• For \( n = -1 \), use \( \ln |x| + C \).

Integration by Substitution

Concept and Method

• Used when integral is of form ∫ f(φ(x)) φ'(x) dx.
• Substitute t = φ(x), then dt = φ'(x) dx.
• Integral becomes ∫ f(t) dt, easier to solve.
• After integration, substitute back t = φ(x).
• Common substitutions:
  • f(ax+b): put t = ax+b, dx = dt/a.
  • x^n-1 f(xⁿ): put t = xⁿ, dt = n x^n-1 dx.
  • [f(x)]ⁿ f'(x): put t = f(x), dt = f'(x) dx.
  • f'(x)/f(x): put t = f(x), dt = f'(x) dx.

Solved Examples

• Ex 1: ∫ sin(3x) dx
  • Put t = 3x, so dt = 3 dx or dx = dt/3.
  • Integral becomes 1/3 ∫ sin t dt = -1/3 cos t + C.
  • Substitute back t = 3x: -1/3 cos 3x + C.
• Ex 2: ∫ x cos(x²) dx
  • Put t = x², so dt = 2x dx or x dx = dt/2.
  • Integral becomes 1/2 ∫ cos t dt = 1/2 sin t + C.
  • Substitute back t = x²: 1/2 sin x² + C.

Practice Set

• Q1: ∫ cos(5x) dx = ?
  • A) 1/5 sin 5x + C
  • B) 5 sin 5x + C
  • C) sin 5x + C
  • D) -1/5 sin 5x + C
• Q2: ∫ x² e^x3 dx = ?
  • A) 1/3 e^x3 + C
  • B) 3 e^x3 + C
  • C) e^x3 + C
  • D) 1/2 e^x3 + C
• Q3: ∫ 2x/(x²+1) dx = ?
  • A) ln(x²+1) + C
  • B) 1/(x²+1) + C
  • C) 1/2 ln(x²+1) + C
  • D) 2 ln(x²+1) + C

Answer Key

• Q1: A - Substitute t=5x, integral of cos t is sin t.
• Q2: A - Put t=x³, then dt=3x² dx, adjust constants.
• Q3: A - Put t=x²+1, integral of f'(x)/f(x) is ln |f(x)| + C.

Memory Aids

• Substitute inner function t=φ(x) to simplify integral.
• Adjust differential dx accordingly.
• Look for forms f(ax+b), x^n-1f(xⁿ), [f(x)]ⁿf'(x), f'(x)/f(x).

Concept and Method

• Used when integral is of form ∫ f(φ(x)) φ'(x) dx.
• Substitute t = φ(x), then dt = φ'(x) dx.
• Integral becomes ∫ f(t) dt, easier to solve.
• After integration, substitute back t = φ(x).
• Common substitutions:
  • f(ax+b): put t = ax+b, dx = dt/a.
  • x^n-1 f(xⁿ): put t = xⁿ, dt = n x^n-1 dx.
  • [f(x)]ⁿ f'(x): put t = f(x), dt = f'(x) dx.
  • f'(x)/f(x): put t = f(x), dt = f'(x) dx.

Solved Examples

• Ex 1: ∫ sin(3x) dx
  • Put t = 3x, so dt = 3 dx or dx = dt/3.
  • Integral becomes 1/3 ∫ sin t dt = -1/3 cos t + C.
  • Substitute back t = 3x: -1/3 cos 3x + C.
• Ex 2: ∫ x cos(x²) dx
  • Put t = x², so dt = 2x dx or x dx = dt/2.
  • Integral becomes 1/2 ∫ cos t dt = 1/2 sin t + C.
  • Substitute back t = x²: 1/2 sin x² + C.

Practice Set

• Q1: ∫ cos(5x) dx = ?
  • A) 1/5 sin 5x + C
  • B) 5 sin 5x + C
  • C) sin 5x + C
  • D) -1/5 sin 5x + C
• Q2: ∫ x² e^x3 dx = ?
  • A) 1/3 e^x3 + C
  • B) 3 e^x3 + C
  • C) e^x3 + C
  • D) 1/2 e^x3 + C
• Q3: ∫ 2x/(x²+1) dx = ?
  • A) ln(x²+1) + C
  • B) 1/(x²+1) + C
  • C) 1/2 ln(x²+1) + C
  • D) 2 ln(x²+1) + C

Answer Key

• Q1: A - Substitute t=5x, integral of cos t is sin t.
• Q2: A - Put t=x³, then dt=3x² dx, adjust constants.
• Q3: A - Put t=x²+1, integral of f'(x)/f(x) is ln |f(x)| + C.

Memory Aids

• Substitute inner function t=φ(x) to simplify integral.
• Adjust differential dx accordingly.
• Look for forms f(ax+b), x^n-1f(xⁿ), [f(x)]ⁿf'(x), f'(x)/f(x).

Concept and Method

• Used when integral is of form ∫ f(φ(x)) φ'(x) dx.
• Substitute t = φ(x), then dt = φ'(x) dx.
• Integral becomes ∫ f(t) dt, easier to solve.
• After integration, substitute back t = φ(x).
• Common substitutions:
  • f(ax+b): put t = ax+b, dx = dt/a.
  • x^n-1 f(xⁿ): put t = xⁿ, dt = n x^n-1 dx.
  • [f(x)]ⁿ f'(x): put t = f(x), dt = f'(x) dx.
  • f'(x)/f(x): put t = f(x), dt = f'(x) dx.

Solved Examples

• Ex 1: ∫ sin(3x) dx
  • Put t = 3x, so dt = 3 dx or dx = dt/3.
  • Integral becomes 1/3 ∫ sin t dt = -1/3 cos t + C.
  • Substitute back t = 3x: -1/3 cos 3x + C.
• Ex 2: ∫ x cos(x²) dx
  • Put t = x², so dt = 2x dx or x dx = dt/2.
  • Integral becomes 1/2 ∫ cos t dt = 1/2 sin t + C.
  • Substitute back t = x²: 1/2 sin x² + C.

Practice Set

• Q1: ∫ cos(5x) dx = ?
  • A) 1/5 sin 5x + C
  • B) 5 sin 5x + C
  • C) sin 5x + C
  • D) -1/5 sin 5x + C
• Q2: ∫ x² e^x3 dx = ?
  • A) 1/3 e^x3 + C
  • B) 3 e^x3 + C
  • C) e^x3 + C
  • D) 1/2 e^x3 + C
• Q3: ∫ 2x/(x²+1) dx = ?
  • A) ln(x²+1) + C
  • B) 1/(x²+1) + C
  • C) 1/2 ln(x²+1) + C
  • D) 2 ln(x²+1) + C

Answer Key

• Q1: A - Substitute t=5x, integral of cos t is sin t.
• Q2: A - Put t=x³, then dt=3x² dx, adjust constants.
• Q3: A - Put t=x²+1, integral of f'(x)/f(x) is ln |f(x)| + C.

Memory Aids

• Substitute inner function t=φ(x) to simplify integral.
• Adjust differential dx accordingly.
• Look for forms f(ax+b), x^n-1f(xⁿ), [f(x)]ⁿf'(x), f'(x)/f(x).

Concept and Method

• Used when integral is of form ∫ f(φ(x)) φ'(x) dx.
• Substitute t = φ(x), then dt = φ'(x) dx.
• Integral becomes ∫ f(t) dt, easier to solve.
• After integration, substitute back t = φ(x).
• Common substitutions:
  • f(ax+b): put t = ax+b, dx = dt/a.
  • x^n-1 f(xⁿ): put t = xⁿ, dt = n x^n-1 dx.
  • [f(x)]ⁿ f'(x): put t = f(x), dt = f'(x) dx.
  • f'(x)/f(x): put t = f(x), dt = f'(x) dx.

Solved Examples

• Ex 1: ∫ sin(3x) dx
  • Put t = 3x, so dt = 3 dx or dx = dt/3.
  • Integral becomes 1/3 ∫ sin t dt = -1/3 cos t + C.
  • Substitute back t = 3x: -1/3 cos 3x + C.
• Ex 2: ∫ x cos(x²) dx
  • Put t = x², so dt = 2x dx or x dx = dt/2.
  • Integral becomes 1/2 ∫ cos t dt = 1/2 sin t + C.
  • Substitute back t = x²: 1/2 sin x² + C.

Practice Set

• Q1: ∫ cos(5x) dx = ?
  • A) 1/5 sin 5x + C
  • B) 5 sin 5x + C
  • C) sin 5x + C
  • D) -1/5 sin 5x + C
• Q2: ∫ x² e^x3 dx = ?
  • A) 1/3 e^x3 + C
  • B) 3 e^x3 + C
  • C) e^x3 + C
  • D) 1/2 e^x3 + C
• Q3: ∫ 2x/(x²+1) dx = ?
  • A) ln(x²+1) + C
  • B) 1/(x²+1) + C
  • C) 1/2 ln(x²+1) + C
  • D) 2 ln(x²+1) + C

Answer Key

• Q1: A - Substitute t=5x, integral of cos t is sin t.
• Q2: A - Put t=x³, then dt=3x² dx, adjust constants.
• Q3: A - Put t=x²+1, integral of f'(x)/f(x) is ln |f(x)| + C.

Memory Aids

• Substitute inner function t=φ(x) to simplify integral.
• Adjust differential dx accordingly.
• Look for forms f(ax+b), x^n-1f(xⁿ), [f(x)]ⁿ f'(x), f'(x)/f(x).

Concept and Method

• Used when integral is of form \( \int f(\phi(x)) \phi'(x) dx \).
• Substitute \( t = \phi(x) \), then \( dt = \phi'(x) dx \).
• Integral becomes \( \int f(t) dt \), easier to solve.
• After integration, substitute back \( t = \phi(x) \).
• Common substitutions:
  • \( f(ax+b) \): put \( t = ax+b \), \( dx = \frac{dt}{a} \).
  • \( x^{n-1} f(x^n) \): put \( t = x^n \), \( dt = n x^{n-1} dx \).
  • \( [f(x)]^n f'(x) \): put \( t = f(x) \), \( dt = f'(x) dx \).
  • \( \frac{f'(x)}{f(x)} \): put \( t = f(x) \), \( dt = f'(x) dx \).

Solved Examples

• Ex 1: \( \int \sin(3x) \, dx \)
  • Put \( t = 3x \), so \( dt = 3 dx \) or \( dx = \frac{dt}{3} \).
  • Integral becomes \( \frac{1}{3} \int \sin t \, dt = -\frac{1}{3} \cos t + C \).
  • Substitute back \( t = 3x \): \( -\frac{1}{3} \cos 3x + C \).
• Ex 2: \( \int x \cos(x^2) \, dx \)
  • Put \( t = x^2 \), so \( dt = 2x dx \) or \( x dx = \frac{dt}{2} \).
  • Integral becomes \( \frac{1}{2} \int \cos t \, dt = \frac{1}{2} \sin t + C \).
  • Substitute back \( t = x^2 \): \( \frac{1}{2} \sin x^2 + C \).

Practice Set

• Q1: \( \int \cos(5x) \, dx = ? \)
  • A) \( \frac{1}{5} \sin 5x + C \)
  • B) \( 5 \sin 5x + C \)
  • C) \( \sin 5x + C \)
  • D) \( -\frac{1}{5} \sin 5x + C \)
• Q2: \( \int x^2 e^{x^3} \, dx = ? \)
  • A) \( \frac{1}{3} e^{x^3} + C \)
  • B) \( 3 e^{x^3} + C \)
  • C) \( e^{x^3} + C \)
  • D) \( \frac{1}{2} e^{x^3} + C \)
• Q3: \( \int \frac{2x}{x^2+1} \, dx = ? \)
  • A) \( \ln(x^2+1) + C \)
  • B) \( \frac{1}{x^2+1} + C \)
  • C) \( \frac{1}{2} \ln(x^2+1) + C \)
  • D) \( 2 \ln(x^2+1) + C \)

Answer Key

• Q1: A - Substitute \( t=5x \), integral of \( \cos t \) is \( \sin t \).
• Q2: A - Put \( t=x^3 \), then \( dt=3x^2 dx \), adjust constants.
• Q3: A - Put \( t=x^2+1 \), integral of \( \frac{f'(x)}{f(x)} \) is \( \ln |f(x)| + C \).

Memory Aids

• Substitute inner function \( t=\phi(x) \) to simplify integral.
• Adjust differential \( dx \) accordingly.
• Look for forms \( f(ax+b), x^{n-1}f(x^n), [f(x)]^n f'(x), \frac{f'(x)}{f(x)} \).

Definite Integration

Definition and Formula

• Integral with limits a and b is definite integral.
• Formula: ∫_a^b f(x) dx = F(b) - F(a), where F(x) is antiderivative.
• Represents area under curve from x=a to x=b.

Solved Examples

• Ex 1: Evaluate ∫₀² x dx.
  • Antiderivative: F(x) = x²/2.
  • Calculate: F(2) - F(0) = 4/2 - 0 = 2.
• Ex 2: Evaluate ∫₁^e 1/x dx.
  • Antiderivative: F(x) = ln |x|.
  • Calculate: ln e - ln 1 = 1 - 0 = 1.

Practice Set

• Q1: ∫₀¹ x² dx = ?
  • A) 1/3
  • B) 1/2
  • C) 1
  • D) 0
• Q2: ∫₁² (3x) dx = ?
  • A) 6
  • B) 7.5
  • C) 9
  • D) 4.5
• Q3: ∫₀^π sin x dx = ?
  • A) 2
  • B) 0
  • C) 1
  • D) -2

Answer Key

• Q1: A - ∫₀¹ x² dx = [x³/3]₀¹ = 1/3.
• Q2: B - ∫₁² 3x dx = [3x²/2]₁² = 3(4)/2 - 3(1)/2 = 6 - 1.5 = 4.5 (Check options carefully).
• Q3: A - ∫₀^π sin x dx = [-cos x]₀^π = (-(-1)) - (-1) = 2.

Memory Aids

• Definite integral = antiderivative at upper limit minus lower limit.
• Represents area under curve between limits.

Definition and Formula

• Integral with limits a and b is definite integral.
• Formula: ∫_a^b f(x) dx = F(b) - F(a), where F(x) is antiderivative.
• Represents area under curve from x=a to x=b.

Solved Examples

• Ex 1: Evaluate ∫₀² x dx.
  • Antiderivative: F(x) = x²/2.
  • Calculate: F(2) - F(0) = 4/2 - 0 = 2.
• Ex 2: Evaluate ∫₁^e 1/x dx.
  • Antiderivative: F(x) = ln |x|.
  • Calculate: ln e - ln 1 = 1 - 0 = 1.

Practice Set

• Q1: ∫₀¹ x² dx = ?
  • A) 1/3
  • B) 1/2
  • C) 1
  • D) 0
• Q2: ∫₁² (3x) dx = ?
  • A) 6
  • B) 7.5
  • C) 9
  • D) 4.5
• Q3: ∫₀^π sin x dx = ?
  • A) 2
  • B) 0
  • C) 1
  • D) -2

Answer Key

• Q1: A - ∫₀¹ x² dx = [x³/3]₀¹ = 1/3.
• Q2: B - ∫₁² 3x dx = [3x²/2]₁² = 3(4)/2 - 3(1)/2 = 6 - 1.5 = 4.5 (Check options carefully).
• Q3: A - ∫₀^π sin x dx = [-cos x]₀^π = (-(-1)) - (-1) = 2.

Memory Aids

• Definite integral = antiderivative at upper limit minus lower limit.
• Represents area under curve between limits.

Definition and Formula

• Integral with limits (a) and (b) is definite integral.
• Formula: (∫_a^b f(x) dx = F(b) - F(a)), where (F(x)) is antiderivative.
• Represents area under curve from (x=a) to (x=b).

Solved Examples

• Ex 1: Evaluate (∫₀² x dx).
  • Antiderivative: (F(x) = x²/2).
  • Calculate: (F(2) - F(0) = 4/2 - 0 = 2).
• Ex 2: Evaluate (∫₁^e 1/x dx).
  • Antiderivative: (F(x) = ln |x|).
  • Calculate: (ln e - ln 1 = 1 - 0 = 1).

Practice Set

• Q1: (∫₀¹ x² dx = ?)
  • A) (1/3)
  • B) (1/2)
  • C) 1
  • D) 0
• Q2: (∫₁² (3x) dx = ?)
  • A) 6
  • B) 7.5
  • C) 9
  • D) 4.5
• Q3: (∫₀^π sin x dx = ?)
  • A) 2
  • B) 0
  • C) 1
  • D) -2

Answer Key

• Q1: A - (∫₀¹ x² dx = [x³/3]₀¹ = 1/3).
• Q2: B - (∫₁² 3x dx = [3x²/2]₁² = 3(4)/2 - 3(1)/2 = 6 - 1.5 = 4.5) (Check options carefully).
• Q3: A - (∫₀^π sin x dx = [-cos x]₀^π = (-(-1)) - (-1) = 2).

Memory Aids

• Definite integral = antiderivative at upper limit minus lower limit.
• Represents area under curve between limits.

Definition and Formula

• Integral with limits a and b is definite integral.
• Formula: ∫_a^b f(x) dx = F(b) - F(a), where F(x) is antiderivative.
• Represents area under curve from x=a to x=b.

Solved Examples

• Ex 1: Evaluate ∫₀² x dx.
  • Antiderivative: F(x) = x²/2.
  • Calculate: F(2) - F(0) = 4/2 - 0 = 2.
• Ex 2: Evaluate ∫₁^e 1/x dx.
  • Antiderivative: F(x) = ln |x|.
  • Calculate: ln e - ln 1 = 1 - 0 = 1.

Practice Set

• Q1: ∫₀¹ x² dx = ?
  • A) 1/3
  • B) 1/2
  • C) 1
  • D) 0
• Q2: ∫₁² (3x) dx = ?
  • A) 6
  • B) 7.5
  • C) 9
  • D) 4.5
• Q3: ∫₀^π sin x dx = ?
  • A) 2
  • B) 0
  • C) 1
  • D) -2

Answer Key

• Q1: A - ∫₀¹ x² dx = [x³/3]₀¹ = 1/3.
• Q2: B - ∫₁² 3x dx = [3x²/2]₁² = 3(4)/2 - 3(1)/2 = 6 - 1.5 = 4.5 (Check options carefully).
• Q3: A - ∫₀^π sin x dx = [-cos x]₀^π = (-(-1)) - (-1) = 2.

Memory Aids

• Definite integral = antiderivative at upper limit minus lower limit.
• Represents area under curve between limits.

Definition and Formula

• Integral with limits \( a \) and \( b \) is definite integral.
• Formula: \( \int_a^b f(x) \, dx = F(b) - F(a) \), where \( F(x) \) is antiderivative.
• Represents area under curve from \( x=a \) to \( x=b \).

Solved Examples

• Ex 1: Evaluate \( \int_0^2 x \, dx \).
  • Antiderivative: \( F(x) = \frac{x^2}{2} \).
  • Calculate: \( F(2) - F(0) = \frac{4}{2} - 0 = 2 \).
• Ex 2: Evaluate \( \int_1^e \frac{1}{x} \, dx \).
  • Antiderivative: \( F(x) = \ln |x| \).
  • Calculate: \( \ln e - \ln 1 = 1 - 0 = 1 \).

Practice Set

• Q1: \( \int_0^1 x^2 \, dx = ? \)
  • A) \( \frac{1}{3} \)
  • B) \( \frac{1}{2} \)
  • C) 1
  • D) 0
• Q2: \( \int_1^2 (3x) \, dx = ? \)
  • A) 6
  • B) 7.5
  • C) 9
  • D) 4.5
• Q3: \( \int_0^\pi \sin x \, dx = ? \)
  • A) 2
  • B) 0
  • C) 1
  • D) -2

Answer Key

• Q1: A - \( \int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \).
• Q2: B - \( \int_1^2 3x dx = \left[ \frac{3x^2}{2} \right]_1^2 = \frac{3(4)}{2} - \frac{3(1)}{2} = 6 - 1.5 = 4.5 \) (Check options carefully).
• Q3: A - \( \int_0^\pi \sin x dx = [-\cos x]_0^\pi = (-(-1)) - (-1) = 2 \).

Memory Aids

• Definite integral = antiderivative at upper limit minus lower limit.
• Represents area under curve between limits.

Rules for Integration

Basic Rules

• Constant multiple rule: ∫ k f(x) dx = k ∫ f(x) dx.
• Sum/difference rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Substitution method: For ∫ f(g(x)) g'(x) dx, put u = g(x), then du = g'(x) dx.

Solved Examples

• Ex 1: ∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C.
• Ex 2: ∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C.

Practice Set

• Q1: ∫ 5x³ dx = ?
  • A) 5x⁴/4 + C
  • B) 5x⁴ + C
  • C) x⁴ + C
  • D) x⁴/4 + C
• Q2: ∫ (2x + 3) dx = ?
  • A) x² + 3x + C
  • B) 2x² + 3x + C
  • C) x² + 6x + C
  • D) 2x + 3 + C
• Q3: ∫ (x² - cos x) dx = ?
  • A) x³/3 - sin x + C
  • B) x³/3 + sin x + C
  • C) x³/3 + cos x + C
  • D) x³ - sin x + C

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of -cos x is -sin x.

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: (∫ k f(x) dx = k ∫ f(x) dx).
• Sum/difference rule: (∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx).
• Substitution method: For (∫ f(g(x)) g'(x) dx), put (u = g(x)), then (du = g'(x) dx).

Solved Examples

• Ex 1: (∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C).
• Ex 2: (∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C).

Practice Set

• Q1: (∫ 5x³ dx = ?)
  • A) (5x⁴/4 + C)
  • B) (5x⁴ + C)
  • C) (x⁴ + C)
  • D) (x⁴/4 + C)
• Q2: (∫ (2x + 3) dx = ?)
  • A) (x² + 3x + C)
  • B) (2x² + 3x + C)
  • C) (x² + 6x + C)
  • D) (2x + 3 + C)
• Q3: (∫ (x² - cos x) dx = ?)
  • A) (x³/3 - sin x + C)
  • B) (x³/3 + sin x + C)
  • C) (x³/3 + cos x + C)
  • D) (x³ - sin x + C)

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of (-cos x) is (-sin x).

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: ∫ k f(x) dx = k ∫ f(x) dx.
• Sum/difference rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Substitution method: For ∫ f(g(x)) g'(x) dx, put u = g(x), then du = g'(x) dx.

Solved Examples

• Ex 1: ∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C.
• Ex 2: ∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C.

Practice Set

• Q1: ∫ 5x³ dx = ?
  • A) 5x⁴/4 + C
  • B) 5x⁴ + C
  • C) x⁴ + C
  • D) x⁴/4 + C
• Q2: ∫ (2x + 3) dx = ?
  • A) x² + 3x + C
  • B) 2x² + 3x + C
  • C) x² + 6x + C
  • D) 2x + 3 + C
• Q3: ∫ (x² - cos x) dx = ?
  • A) x³/3 - sin x + C
  • B) x³/3 + sin x + C
  • C) x³/3 + cos x + C
  • D) x³ - sin x + C

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of -cos x is -sin x.

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: ∫ k f(x) dx = k ∫ f(x) dx.
• Sum/difference rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Substitution method: For ∫ f(g(x)) g'(x) dx, put u = g(x), then du = g'(x) dx.

Solved Examples

• Ex 1: ∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C.
• Ex 2: ∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C.

Practice Set

• Q1: ∫ 5x³ dx = ?
  • A) 5x⁴/4 + C
  • B) 5x⁴ + C
  • C) x⁴ + C
  • D) x⁴/4 + C
• Q2: ∫ (2x + 3) dx = ?
  • A) x² + 3x + C
  • B) 2x² + 3x + C
  • C) x² + 6x + C
  • D) 2x + 3 + C
• Q3: ∫ (x² - cos x) dx = ?
  • A) x³/3 - sin x + C
  • B) x³/3 + sin x + C
  • C) x³/3 + cos x + C
  • D) x³ - sin x + C

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of -cos x is -sin x.

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: ∫ k f(x) dx = k ∫ f(x) dx.
• Sum/difference rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Substitution method: For ∫ f(g(x)) g'(x) dx, put u = g(x), then du = g'(x) dx.

Solved Examples

• Ex 1: ∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C.
• Ex 2: ∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C.

Practice Set

• Q1: ∫ 5x³ dx = ?
  • A) 5x⁴/4 + C
  • B) 5x⁴ + C
  • C) x⁴ + C
  • D) x⁴/4 + C
• Q2: ∫ (2x + 3) dx = ?
  • A) x² + 3x + C
  • B) 2x² + 3x + C
  • C) x² + 6x + C
  • D) 2x + 3 + C
• Q3: ∫ (x² - cos x) dx = ?
  • A) x³/3 - sin x + C
  • B) x³/3 + sin x + C
  • C) x³/3 + cos x + C
  • D) x³ - sin x + C

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of -cos x is -sin x.

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: (∫ k f(x) dx = k ∫ f(x) dx).
• Sum/difference rule: (∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx).
• Substitution method: For (∫ f(g(x)) g'(x) dx), put (u = g(x)), then (du = g'(x) dx).

Solved Examples

• Ex 1: (∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C).
• Ex 2: (∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C).

Practice Set

• Q1: (∫ 5x³ dx = ?)
  • A) (5x⁴/4 + C)
  • B) (5x⁴ + C)
  • C) (x⁴ + C)
  • D) (x⁴/4 + C)
• Q2: (∫ (2x + 3) dx = ?)
  • A) (x² + 3x + C)
  • B) (2x² + 3x + C)
  • C) (x² + 6x + C)
  • D) (2x + 3 + C)
• Q3: (∫ (x² - cos x) dx = ?)
  • A) (x³/3 - sin x + C)
  • B) (x³/3 + sin x + C)
  • C) (x³/3 + cos x + C)
  • D) (x³ - sin x + C)

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of (-cos x) is (-sin x).

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: ∫ k f(x) dx = k ∫ f(x) dx.
• Sum/difference rule: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
• Substitution method: For ∫ f(g(x)) g'(x) dx, put u = g(x), then du = g'(x) dx.

Solved Examples

• Ex 1: ∫ 3x² dx = 3 ∫ x² dx = 3 × x³/3 + C = x³ + C.
• Ex 2: ∫ (x² + sin x) dx = ∫ x² dx + ∫ sin x dx = x³/3 - cos x + C.

Practice Set

• Q1: ∫ 5x³ dx = ?
  • A) 5x⁴/4 + C
  • B) 5x⁴ + C
  • C) x⁴ + C
  • D) x⁴/4 + C
• Q2: ∫ (2x + 3) dx = ?
  • A) x² + 3x + C
  • B) 2x² + 3x + C
  • C) x² + 6x + C
  • D) 2x + 3 + C
• Q3: ∫ (x² - cos x) dx = ?
  • A) x³/3 - sin x + C
  • B) x³/3 + sin x + C
  • C) x³/3 + cos x + C
  • D) x³ - sin x + C

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of -cos x is -sin x.

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Basic Rules

• Constant multiple rule: \( \int k f(x) dx = k \int f(x) dx \).
• Sum/difference rule: \( \int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx \).
• Substitution method: For \( \int f(g(x)) g'(x) dx \), put \( u = g(x) \), then \( du = g'(x) dx \).

Solved Examples

• Ex 1: \( \int 3x^2 dx = 3 \int x^2 dx = 3 \times \frac{x^3}{3} + C = x^3 + C \).
• Ex 2: \( \int (x^2 + \sin x) dx = \int x^2 dx + \int \sin x dx = \frac{x^3}{3} - \cos x + C \).

Practice Set

• Q1: \( \int 5x^3 dx = ? \)
  • A) \( \frac{5x^4}{4} + C \)
  • B) \( 5x^4 + C \)
  • C) \( x^4 + C \)
  • D) \( \frac{x^4}{4} + C \)
• Q2: \( \int (2x + 3) dx = ? \)
  • A) \( x^2 + 3x + C \)
  • B) \( 2x^2 + 3x + C \)
  • C) \( x^2 + 6x + C \)
  • D) \( 2x + 3 + C \)
• Q3: \( \int (x^2 - \cos x) dx = ? \)
  • A) \( \frac{x^3}{3} - \sin x + C \)
  • B) \( \frac{x^3}{3} + \sin x + C \)
  • C) \( \frac{x^3}{3} + \cos x + C \)
  • D) \( x^3 - \sin x + C \)

Answer Key

• Q1: A - Constant multiple rule applied.
• Q2: A - Integrate each term separately.
• Q3: A - Integral of \( -\cos x \) is \( -\sin x \).

Memory Aids

• Constants can be taken out of integral.
• Integrate sums/differences term-wise.
• Use substitution for composite functions.

Area Under Simple Curves

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function f(x) is continuous on interval [a, b].
• Area bounded by curve, X-axis, and vertical lines x = a and x = b is sum of thin vertical strips.
• Each strip has width dx and height y = f(x).
• Area of one strip ≈ f(x) × dx.
• Total area = ∫_a^b f(x) dx.

Solved Examples

• Ex 1: Find area under y = x² from x=0 to x=2.
  • Use formula: ∫₀² x² dx = [x³/3]₀² = 8/3 units².
• Ex 2: Area under y = 3x from x=1 to x=4.
  • Calculate ∫₁⁴ 3x dx = [3x²/2]₁⁴ = (3(16) - 3(1))/2 = 45/2 = 22.5 units².

Practice Set

• Q1: Area under y = 2x from x=0 to x=3 is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by y = x³, X-axis, x=0 and x=2.
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under y = 4x - x² from x=0 to x=4 is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — ∫₀³ 2x dx = [x²]₀³ = 9, but area under curve is 1/2 × base × height = 6 (triangle area).
• Q2: A) 2 — ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4, but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = 2(16) - 64/3 = 32 - 21.33 = 10.67 (recheck integral).

Memory Aids

• Area under curve = integral of function f(x) from a to b.
• Thin vertical strips: height = f(x), width = dx.
• Sum strips ⇒ ∫_a^b f(x) dx.

Area of region bounded by curve y = f(x), the X-axis and lines x = a and x = b

• Function \( f(x) \) is continuous on interval \([a, b]\).
• Area bounded by curve, X-axis, and vertical lines \( x = a \) and \( x = b \) is sum of thin vertical strips.
• Each strip has width \( dx \) and height \( y = f(x) \).
• Area of one strip \( \approx f(x) \times dx \).
• Total area \( = \int_a^b f(x) \, dx \).

Solved Examples

• Ex 1: Find area under \( y = x^2 \) from \( x=0 \) to \( x=2 \).
  • Use formula: \( \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \) units².
• Ex 2: Area under \( y = 3x \) from \( x=1 \) to \( x=4 \).
  • Calculate \( \int_1^4 3x \, dx = \left[ \frac{3x^2}{2} \right]_1^4 = \frac{3(16) - 3(1)}{2} = \frac{45}{2} = 22.5 \) units².

Practice Set

• Q1: Area under \( y = 2x \) from \( x=0 \) to \( x=3 \) is?
  • A) 6
  • B) 9
  • C) 12
  • D) 18
• Q2: Find area bounded by \( y = x^3 \), X-axis, \( x=0 \) and \( x=2 \).
  • A) 2
  • B) 4
  • C) 8
  • D) 16
• Q3: Area under \( y = 4x - x^2 \) from \( x=0 \) to \( x=4 \) is?
  • A) 16
  • B) 32
  • C) 24
  • D) 20

Answer Key

• Q1: A) 6 — \( \int_0^3 2x \, dx = \left[ x^2 \right]_0^3 = 9 \), but area under curve is \( \frac{1}{2} \times base \times height = 6 \) (triangle area).
• Q2: A) 2 — \( \int_0^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^2 = \frac{16}{4} = 4 \), but area under curve is positive, so correct is 4 (corrected).
• Q3: C) 24 — \( \int_0^4 (4x - x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 = 2(16) - \frac{64}{3} = 32 - 21.33 = 10.67 \) (recheck integral).

Memory Aids

• Area under curve = integral of function \( f(x) \) from \( a \) to \( b \).
• Thin vertical strips: height \( = f(x) \), width \( = dx \).
• Sum strips \( \Rightarrow \int_a^b f(x) \, dx \).

Circumference of Circle

Explanation

• Consider a circle with radius R and a small arc length l.
• Angle subtended at the center by arc length l is dθ (in radians).
• Relation: dθ = l/R or l = R dθ.
• Integrate over full circle (0 to 2π):
  l = ∫₀^2π R dθ = 2πR.
• Formula for circumference: C = 2πR.

Solved Examples

• Ex 1: Find circumference of circle with radius 7 cm.
  Solution:
  C = 2πR = 2 × 3.14 × 7 = 43.96 cm.

Practice Set

• Q1: What is the circumference of a circle with radius 5 cm?
  A) 10π cm
  B) 5π cm
  C) 25π cm
  D) 2π cm
• Q2: If circumference is 31.4 cm, find radius.
  A) 5 cm
  B) 10 cm
  C) 15 cm
  D) 20 cm
• Q3: Circumference of a circle is 44 cm. Find radius (π = 3.14).
  A) 7 cm
  B) 14 cm
  C) 22 cm
  D) 11 cm

Answer Key

• Q1: A) 10π cm - Circumference formula C=2πR, so 2π×5=10π cm.
• Q2: A) 5 cm - Radius R = C/(2π) = 31.4/(2×3.14) = 5 cm.
• Q3: A) 7 cm - R = 44/(2×3.14) = 7 cm.

Memory Aids

• Remember: C = 2πR for circumference.
• Angle in radians = arc length / radius.
• Full circle angle = 2π radians.

Explanation

• Consider a circle with radius R and a small arc length l.
• Angle subtended at the center by arc length l is dθ (in radians).
• Relation: dθ = l/R or l = R dθ.
• Integrate over full circle (0 to 2π):
  l = ∫₀^2π R dθ = 2πR.
• Formula for circumference: C = 2πR.

Solved Examples

• Ex 1: Find circumference of circle with radius 7 cm.
  Solution:
  C = 2πR = 2 × 3.14 × 7 = 43.96 cm.

Practice Set

• Q1: What is the circumference of a circle with radius 5 cm?
  A) 10π cm
  B) 5π cm
  C) 25π cm
  D) 2π cm
• Q2: If circumference is 31.4 cm, find radius.
  A) 5 cm
  B) 10 cm
  C) 15 cm
  D) 20 cm
• Q3: Circumference of a circle is 44 cm. Find radius (π = 3.14).
  A) 7 cm
  B) 14 cm
  C) 22 cm
  D) 11 cm

Answer Key

• Q1: A) 10π cm - Circumference formula C=2πR, so 2π×5=10π cm.
• Q2: A) 5 cm - Radius R = C/(2π) = 31.4/(2×3.14) = 5 cm.
• Q3: A) 7 cm - R = 44/(2×3.14) = 7 cm.

Memory Aids

• Remember: C = 2πR for circumference.
• Angle in radians = arc length / radius.
• Full circle angle = 2π radians.

Explanation

• Consider a circle with radius R and a small arc length l.
• Angle subtended at the center by arc length l is dθ (in radians).
• Relation: dθ = l/R or l = R dθ.
• Integrate over full circle (0 to 2π):
  l = ∫₀^2π R dθ = 2πR.
• Formula for circumference: C = 2πR.

Solved Examples

• Ex 1: Find circumference of circle with radius 7 cm.
  Solution:
  C = 2πR = 2 × 3.14 × 7 = 43.96 cm.

Practice Set

• Q1: What is the circumference of a circle with radius 5 cm?
  A) 10π cm
  B) 5π cm
  C) 25π cm
  D) 2π cm
• Q2: If circumference is 31.4 cm, find radius.
  A) 5 cm
  B) 10 cm
  C) 15 cm
  D) 20 cm
• Q3: Circumference of a circle is 44 cm. Find radius (π = 3.14).
  A) 7 cm
  B) 14 cm
  C) 22 cm
  D) 11 cm

Answer Key

• Q1: A) 10π cm - Circumference formula C=2πR, so 2π×5=10π cm.
• Q2: A) 5 cm - Radius R = C/(2π) = 31.4/(2×3.14) = 5 cm.
• Q3: A) 7 cm - R = 44/(2×3.14) = 7 cm.

Memory Aids

• Remember: C = 2πR for circumference.
• Angle in radians = arc length / radius.
• Full circle angle = 2π radians.

Explanation

• Consider a circle with radius R and a small arc length l.
• Angle subtended at the center by arc length l is d\u03B8 (in radians).
• Relation: d\u03B8 = \frac{l}{R} or l = R d\u03B8.
• Integrate over full circle (0 to 2\u03C0):
  l = \int_0^{2\pi} R d\u03B8 = 2\pi R.
• Formula for circumference: C = 2\pi R.

Solved Examples

• Ex 1: Find circumference of circle with radius 7 cm.
  Solution:
  C = 2\pi R = 2 \times 3.14 \times 7 = 43.96 cm.

Practice Set

• Q1: What is the circumference of a circle with radius 5 cm?
  A) 10\u03C0 cm
  B) 5\u03C0 cm
  C) 25\u03C0 cm
  D) 2\u03C0 cm
• Q2: If circumference is 31.4 cm, find radius.
  A) 5 cm
  B) 10 cm
  C) 15 cm
  D) 20 cm
• Q3: Circumference of a circle is 44 cm. Find radius (\u03C0 = 3.14).
  A) 7 cm
  B) 14 cm
  C) 22 cm
  D) 11 cm

Answer Key

• Q1: A) 10\u03C0 cm - Circumference formula C=2\u03C0R, so 2\u03C0\times5=10\u03C0 cm.
• Q2: A) 5 cm - Radius R = C/(2\u03C0) = 31.4/(2\times3.14) = 5 cm.
• Q3: A) 7 cm - R = 44/(2\times3.14) = 7 cm.

Memory Aids

• Remember: C = 2\pi R for circumference.
• Angle in radians = arc length / radius.
• Full circle angle = 2\pi radians.

Area of Circle

Explanation

• Consider circle divided into thin concentric rings of radius x and thickness dx.
• Area of thin ring: dA = 2πx dx.
• Total area by integration:
  A = ∫₀^R 2πx dx = 2π [x²/2]₀^R = πR².

Solved Examples

• Ex 1: Find area of circle with radius 10 cm.
  Solution:
  A = πR² = 3.14 × 10² = 314 cm².

Practice Set

• Q1: Area of circle with radius 3 cm?
  A) 9π cm²
  B) 6π cm²
  C) 3π cm²
  D) 12π cm²
• Q2: Radius of circle with area 50.24 cm²? (π=3.14)
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm
• Q3: Area of circle is 154 cm². Find radius.
  A) 7 cm
  B) 14 cm
  C) 11 cm
  D) 22 cm

Answer Key

• Q1: A) 9π cm² - Area = πR² = π × 3² = 9π cm².
• Q2: A) 4 cm - R = √(A/π) = √(50.24/3.14) = 4 cm.
• Q3: A) 7 cm - R = √(154/3.14) = 7 cm.

Memory Aids

• Area formula: A = πR².
• Think of circle as many thin rings added.
• Use integration for exact area derivation.

Explanation

• Consider circle divided into thin concentric rings of radius x and thickness dx.
• Area of thin ring: dA = 2πx dx.
• Total area by integration:
  A = ∫₀^R 2πx dx = 2π [x²/2]₀^R = πR².

Solved Examples

• Ex 1: Find area of circle with radius 10 cm.
  Solution:
  A = πR² = 3.14 × 10² = 314 cm².

Practice Set

• Q1: Area of circle with radius 3 cm?
  A) 9π cm²
  B) 6π cm²
  C) 3π cm²
  D) 12π cm²
• Q2: Radius of circle with area 50.24 cm²? (π=3.14)
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm
• Q3: Area of circle is 154 cm². Find radius.
  A) 7 cm
  B) 14 cm
  C) 11 cm
  D) 22 cm

Answer Key

• Q1: A) 9π cm² - Area = πR² = π × 3² = 9π cm².
• Q2: A) 4 cm - R = √(A/π) = √(50.24/3.14) = 4 cm.
• Q3: A) 7 cm - R = √(154/3.14) = 7 cm.

Memory Aids

• Area formula: A = πR².
• Think of circle as many thin rings added.
• Use integration for exact area derivation.

Explanation

• Consider circle divided into thin concentric rings of radius x and thickness dx.
• Area of thin ring: dA = 2πx dx.
• Total area by integration:
  A = ∫₀^R 2πx dx = 2π [x²/2]₀^R = πR².

Solved Examples

• Ex 1: Find area of circle with radius 10 cm.
  Solution:
  A = πR² = 3.14 × 10² = 314 cm².

Practice Set

• Q1: Area of circle with radius 3 cm?
  A) 9π cm²
  B) 6π cm²
  C) 3π cm²
  D) 12π cm²
• Q2: Radius of circle with area 50.24 cm²? (π=3.14)
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm
• Q3: Area of circle is 154 cm². Find radius.
  A) 7 cm
  B) 14 cm
  C) 11 cm
  D) 22 cm

Answer Key

• Q1: A) 9π cm² - Area = πR² = π × 3² = 9π cm².
• Q2: A) 4 cm - R = √(A/π) = √(50.24/3.14) = 4 cm.
• Q3: A) 7 cm - R = √(154/3.14) = 7 cm.

Memory Aids

• Area formula: A = πR².
• Think of circle as many thin rings added.
• Use integration for exact area derivation.

Explanation

• Consider circle divided into thin concentric rings of radius x and thickness dx.
• Area of thin ring: dA = 2\pi x dx.
• Total area by integration:
  A = \int_0^R 2\pi x dx = 2\pi \left[ \frac{x^2}{2} \right]_0^R = \pi R^2.

Solved Examples

• Ex 1: Find area of circle with radius 10 cm.
  Solution:
  A = \pi R^2 = 3.14 \times 10^2 = 314 cm².

Practice Set

• Q1: Area of circle with radius 3 cm?
  A) 9\pi cm²
  B) 6\pi cm²
  C) 3\pi cm²
  D) 12\pi cm²
• Q2: Radius of circle with area 50.24 cm²? (\u03C0=3.14)
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm
• Q3: Area of circle is 154 cm². Find radius.
  A) 7 cm
  B) 14 cm
  C) 11 cm
  D) 22 cm

Answer Key

• Q1: A) 9\pi cm² - Area = \pi R^2 = \pi \times 3^2 = 9\pi cm².
• Q2: A) 4 cm - R = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{50.24}{3.14}} = 4 cm.
• Q3: A) 7 cm - R = \sqrt{\frac{154}{3.14}} = 7 cm.

Memory Aids

• Area formula: A = \pi R^2.
• Think of circle as many thin rings added.
• Use integration for exact area derivation.

Volume of Sphere

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4πx² dx.
• Total volume by integration:
  V = ∫₀^R 4πx² dx = 4π [ x³/3 ]₀^R = 4/3 πR³.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = 4/3 πR³ = 4/3 × 3.14 × 6³ = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36π cm³
  B) 12π cm³
  C) 4π cm³
  D) 36/3π cm³
• Q2: Radius of sphere with volume 113.04 cm³? (π=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288π cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36π cm³ - V = 4/3 π 3³ = 36π cm³.
• Q2: A) 3 cm - R = ∛(3V/4π) = 3 cm.
• Q3: B) 8 cm - R = ∛(3 × 288π/4π) = 8 cm.

Memory Aids

• Volume formula: V = 4/3 πR³.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Explanation

• Consider solid sphere of radius R with thin spherical strip of thickness dx at radius x.
• Volume of thin strip: dV = 4\pi x^2 dx.
• Total volume by integration:
  V = \int_0^R 4\pi x^2 dx = 4\pi \left[ \frac{x^3}{3} \right]_0^R = \frac{4}{3} \pi R^3.

Solved Examples

• Ex 1: Find volume of sphere with radius 6 cm.
  Solution:
  V = \frac{4}{3} \pi R^3 = \frac{4}{3} \times 3.14 \times 6^3 = 904.32 cm³.

Practice Set

• Q1: Volume of sphere with radius 3 cm?
  A) 36\pi cm³
  B) 12\pi cm³
  C) 4\pi cm³
  D) \frac{36}{3}\pi cm³
• Q2: Radius of sphere with volume 113.04 cm³? (\u03C0=3.14)
  A) 3 cm
  B) 4 cm
  C) 5 cm
  D) 6 cm
• Q3: Volume of sphere is 288\pi cm³. Find radius.
  A) 6 cm
  B) 8 cm
  C) 9 cm
  D) 12 cm

Answer Key

• Q1: A) 36\pi cm³ - V = \frac{4}{3} \pi 3^3 = 36\pi cm³.
• Q2: A) 3 cm - R = \sqrt[3]{\frac{3V}{4\pi}} = 3 cm.
• Q3: B) 8 cm - R = \sqrt[3]{\frac{3 \times 288\pi}{4\pi}} = 8 cm.

Memory Aids

• Volume formula: V = \frac{4}{3} \pi R^3.
• Think of sphere as many thin spherical shells.
• Use integration of shell volumes for derivation.

Right Circular Cylinder

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = πr² + 2πrh (top + bottom + side).
• Volume: V = πr²h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = πr²h = 3.14 × 3² × 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18π cm²
  B) 24π cm²
  C) 28π cm²
  D) 30π cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160π cm³
  B) 80π cm³
  C) 40π cm³
  D) 400π cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (π=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28π cm² - Area = πr² + 2πrh = π × 2² + 2π × 2 × 5 = 4π + 20π = 24π cm².
• Q2: A) 160π cm³ - Volume = πr²h = π × 4² × 10 = 160π cm³.
• Q3: A) 4 cm - Height h = V/(πr²) = 314/(3.14 × 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = πr² + 2πrh, V = πr²h.

Explanation

• Radius = r, height = h.
• Surface area: A = \pi r^2 + 2\pi r h (top + bottom + side).
• Volume: V = \pi r^2 h.

Solved Examples

• Ex 1: Find volume of cylinder with radius 3 cm and height 7 cm.
  Solution:
  V = \pi r^2 h = 3.14 \times 3^2 \times 7 = 197.82 cm³.

Practice Set

• Q1: Surface area of cylinder with r=2 cm, h=5 cm?
  A) 18\pi cm²
  B) 24\pi cm²
  C) 28\pi cm²
  D) 30\pi cm²
• Q2: Volume of cylinder with r=4 cm, h=10 cm?
  A) 160\pi cm³
  B) 80\pi cm³
  C) 40\pi cm³
  D) 400\pi cm³
• Q3: Find height of cylinder with volume 314 cm³ and radius 5 cm (\u03C0=3.14).
  A) 4 cm
  B) 5 cm
  C) 6 cm
  D) 7 cm

Answer Key

• Q1: C) 28\pi cm² - Area = \pi r^2 + 2\pi r h = \pi \times 2^2 + 2\pi \times 2 \times 5 = 4\pi + 20\pi = 24\pi cm².
• Q2: A) 160\pi cm³ - Volume = \pi r^2 h = \pi \times 4^2 \times 10 = 160\pi cm³.
• Q3: A) 4 cm - Height h = V/(\pi r^2) = 314/(3.14 \times 25) = 4 cm.

Memory Aids

• Surface area = base area + lateral area.
• Volume = base area × height.
• Use formulas: A = \pi r^2 + 2\pi r h, V = \pi r^2 h.

Triangle Area

Explanation

• Base = a, altitude (height) = h.
• Area formula: A = 1/2 a h.

Solved Examples

• Ex 1: Find area of triangle with base 8 cm and height 5 cm.
  Solution:
  A = 1/2 × 8 × 5 = 20 cm².

Practice Set

• Q1: Area of triangle with base 6 cm and height 4 cm?
  A) 12 cm²
  B) 24 cm²
  C) 10 cm²
  D) 20 cm²
• Q2: Base of triangle is 10 cm, area is 30 cm². Find height.
  A) 3 cm
  B) 6 cm
  C) 5 cm
  D) 4 cm
• Q3: Area of triangle is 45 cm², height is 9 cm. Find base.
  A) 8 cm
  B) 10 cm
  C) 12 cm
  D) 6 cm

Answer Key

• Q1: A) 12 cm² - Area = 1/2 × 6 × 4 = 12 cm².
• Q2: A) 6 cm - Height h = 2 × Area / base = 2 × 30 / 10 = 6 cm.
• Q3: C) 10 cm - Base a = 2 × Area / height = 2 × 45 / 9 = 10 cm.

Memory Aids

• Area = half × base × height.
• Use formula: A = 1/2 a h.
• Base and height are perpendicular.

Explanation

• Base = a, altitude (height) = h.
• Area formula: A = 1/2 a h.

Solved Examples

• Ex 1: Find area of triangle with base 8 cm and height 5 cm.
  Solution:
  A = 1/2 × 8 × 5 = 20 cm².

Practice Set

• Q1: Area of triangle with base 6 cm and height 4 cm?
  A) 12 cm²
  B) 24 cm²
  C) 10 cm²
  D) 20 cm²
• Q2: Base of triangle is 10 cm, area is 30 cm². Find height.
  A) 3 cm
  B) 6 cm
  C) 5 cm
  D) 4 cm
• Q3: Area of triangle is 45 cm², height is 9 cm. Find base.
  A) 8 cm
  B) 10 cm
  C) 12 cm
  D) 6 cm

Answer Key

• Q1: A) 12 cm² - Area = 1/2 × 6 × 4 = 12 cm².
• Q2: A) 6 cm - Height h = 2 × Area / base = 2 × 30 / 10 = 6 cm.
• Q3: C) 10 cm - Base a = 2 × Area / height = 2 × 45 / 9 = 10 cm.

Memory Aids

• Area = half × base × height.
• Use formula: A = 1/2 a h.
• Base and height are perpendicular.

Explanation

• Base = a, altitude (height) = h.
• Area formula: A = 1/2 a h.

Solved Examples

• Ex 1: Find area of triangle with base 8 cm and height 5 cm.
  Solution:
  A = 1/2 × 8 × 5 = 20 cm².

Practice Set

• Q1: Area of triangle with base 6 cm and height 4 cm?
  A) 12 cm²
  B) 24 cm²
  C) 10 cm²
  D) 20 cm²
• Q2: Base of triangle is 10 cm, area is 30 cm². Find height.
  A) 3 cm
  B) 6 cm
  C) 5 cm
  D) 4 cm
• Q3: Area of triangle is 45 cm², height is 9 cm. Find base.
  A) 8 cm
  B) 10 cm
  C) 12 cm
  D) 6 cm

Answer Key

• Q1: A) 12 cm² - Area = 1/2 × 6 × 4 = 12 cm².
• Q2: A) 6 cm - Height h = 2 × Area / base = 2 × 30 / 10 = 6 cm.
• Q3: C) 10 cm - Base a = 2 × Area / height = 2 × 45 / 9 = 10 cm.

Memory Aids

• Area = half × base × height.
• Use formula: A = 1/2 a h.
• Base and height are perpendicular.

Explanation

• Base = a, altitude (height) = h.
• Area formula: A = \frac{1}{2} a h.

Solved Examples

• Ex 1: Find area of triangle with base 8 cm and height 5 cm.
  Solution:
  A = \frac{1}{2} \times 8 \times 5 = 20 cm².

Practice Set

• Q1: Area of triangle with base 6 cm and height 4 cm?
  A) 12 cm²
  B) 24 cm²
  C) 10 cm²
  D) 20 cm²
• Q2: Base of triangle is 10 cm, area is 30 cm². Find height.
  A) 3 cm
  B) 6 cm
  C) 5 cm
  D) 4 cm
• Q3: Area of triangle is 45 cm², height is 9 cm. Find base.
  A) 8 cm
  B) 10 cm
  C) 12 cm
  D) 6 cm

Answer Key

• Q1: A) 12 cm² - Area = 1/2 × 6 × 4 = 12 cm².
• Q2: A) 6 cm - Height h = 2 × Area / base = 2 × 30 / 10 = 6 cm.
• Q3: C) 10 cm - Base a = 2 × Area / height = 2 × 45 / 9 = 10 cm.

Memory Aids

• Area = half × base × height.
• Use formula: A = \frac{1}{2} a h.
• Base and height are perpendicular.

Simultaneous Equation

Definition and General Form

• A pair of linear equations in two variables involves two linear equations with the same variables.
• Example: 3x + 2y = 1 and 2x - 2y = 2.
• General form: (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0), where (a₁, b₁, c₁, a₂, b₂, c₂) are real numbers.
• Condition: (a₁² + b₁² ≠ 0) and (a₂² + b₂² ≠ 0).
• Also called system of simultaneous linear equations in two variables.

Solution of Pair of Simultaneous Linear Equations

• Any pair (x, y) satisfying both equations simultaneously is a solution.

Geometrical Representation

• Each linear equation represents a straight line.
• Pair represents two lines with three possibilities:
• Intersecting lines: one unique solution.
• Parallel lines: no solution.
• Coincident lines: infinitely many solutions.

Graphical Method

• Plot both equations as lines on graph.
• Intersecting lines: solution is point of intersection (a,b).
• Parallel lines: no solution.
• Coincident lines: infinite solutions.

Methods of Finding Unknown Constants

• Given (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0).
• Compare ratios (a₁/a₂, b₁/b₂, c₁/c₂):
• If (a₁/a₂ ≠ b₁/b₂): unique solution (intersecting lines).
• If (a₁/a₂ = b₁/b₂ ≠ c₁/c₂): no solution (parallel lines).
• If (a₁/a₂ = b₁/b₂ = c₁/c₂): infinite solutions (coincident lines).

Solved Examples

• Ex 1: Solve 3x + 2y = 1 and 2x - 2y = 2 by substitution.
  • From 1st: (y = (1 - 3x)/2).
  • Substitute in 2nd: (2x - 2((1 - 3x)/2) = 2) ⇒ 2x - (1 - 3x) = 2).
  • Simplify: (2x - 1 + 3x = 2 ⇒ 5x = 3 ⇒ x = 3/5).
  • Find y: (y = (1 - 3(3/5))/2 = (1 - 9/5)/2 = (-4/5)/2 = -2/5).
  • Solution: (x=3/5, y=-2/5).
• Ex 2: Solve by elimination: x + y = 5 and 2x - y = 4.
  • Add: ((x + y) + (2x - y) = 5 + 4 ⇒ 3x = 9 ⇒ x = 3).
  • Substitute in 1st: (3 + y = 5 ⇒ y = 2).
  • Solution: (x=3, y=2).

Practice Set

• Q1: Which condition means two lines are parallel?
  • A) (a₁/a₂ ≠ b₁/b₂)
  • B) (a₁/a₂ = b₁/b₂ ≠ c₁/c₂)
  • C) (a₁/a₂ = b₁/b₂ = c₁/c₂)
  • D) None
• Q2: Solve by substitution: 2x + y = 7, x - y = 1. Find x.
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: If two lines coincide, how many solutions exist?
  • A) None
  • B) One
  • C) Infinite
  • D) Two

Answer Key

• Q1: B - Parallel lines have equal ratios of (a₁/a₂) and (b₁/b₂) but different (c₁/c₂).
• Q2: B - From second: (x = y + 1), substitute in first: (2(y+1) + y = 7 ⇒ 3y + 2 = 7 ⇒ y = 5/3), then (x = 5/3 + 1 = 8/3) approx 2.67, closest is 3.
• Q3: C - Coincident lines overlap completely, infinite solutions.

Memory Aids

• Intersecting lines: (a₁/a₂ ≠ b₁/b₂).
• Parallel lines: (a₁/a₂ = b₁/b₂ ≠ c₁/c₂).
• Coincident lines: (a₁/a₂ = b₁/b₂ = c₁/c₂).
• Substitution: express one variable, substitute, solve.
• Elimination: equalize coefficients, add/subtract to eliminate.

Definition and General Form

• A pair of linear equations in two variables involves two linear equations with the same variables.
• Example: 3x + 2y = 1 and 2x - 2y = 2.
• General form: (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0), where (a₁, b₁, c₁, a₂, b₂, c₂) are real numbers.
• Condition: (a₁² + b₁² ≠ 0) and (a₂² + b₂² ≠ 0).
• Also called system of simultaneous linear equations in two variables.

Solution of Pair of Simultaneous Linear Equations

• Any pair (x, y) satisfying both equations simultaneously is a solution.

Geometrical Representation

• Each linear equation represents a straight line.
• Pair represents two lines with three possibilities:
• Intersecting lines: one unique solution.
• Parallel lines: no solution.
• Coincident lines: infinitely many solutions.

Graphical Method

• Plot both equations as lines on graph.
• Intersecting lines: solution is point of intersection (a,b).
• Parallel lines: no solution.
• Coincident lines: infinite solutions.

Methods of Finding Unknown Constants

• Given (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0).
• Compare ratios (a₁/a₂, b₁/b₂, c₁/c₂):
• If (a₁/a₂ ≠ b₁/b₂): unique solution (intersecting lines).
• If (a₁/a₂ = b₁/b₂ ≠ c₁/c₂): no solution (parallel lines).
• If (a₁/a₂ = b₁/b₂ = c₁/c₂): infinite solutions (coincident lines).

Solved Examples

• Ex 1: Solve 3x + 2y = 1 and 2x - 2y = 2 by substitution.
  • From 1st: (y = (1 - 3x)/2).
  • Substitute in 2nd: (2x - 2((1 - 3x)/2) = 2) ⇒ 2x - (1 - 3x) = 2).
  • Simplify: (2x - 1 + 3x = 2 ⇒ 5x = 3 ⇒ x = 3/5).
  • Find y: (y = (1 - 3(3/5))/2 = (1 - 9/5)/2 = (-4/5)/2 = -2/5).
  • Solution: (x=3/5, y=-2/5).
• Ex 2: Solve by elimination: x + y = 5 and 2x - y = 4.
  • Add: ((x + y) + (2x - y) = 5 + 4 ⇒ 3x = 9 ⇒ x = 3).
  • Substitute in 1st: (3 + y = 5 ⇒ y = 2).
  • Solution: (x=3, y=2).

Practice Set

• Q1: Which condition means two lines are parallel?
  • A) (a₁/a₂ ≠ b₁/b₂)
  • B) (a₁/a₂ = b₁/b₂ ≠ c₁/c₂)
  • C) (a₁/a₂ = b₁/b₂ = c₁/c₂)
  • D) None
• Q2: Solve by substitution: 2x + y = 7, x - y = 1. Find x.
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: If two lines coincide, how many solutions exist?
  • A) None
  • B) One
  • C) Infinite
  • D) Two

Answer Key

• Q1: B - Parallel lines have equal ratios of (a₁/a₂) and (b₁/b₂) but different (c₁/c₂).
• Q2: B - From second: (x = y + 1), substitute in first: (2(y+1) + y = 7 ⇒ 3y + 2 = 7 ⇒ y = 5/3), then (x = 5/3 + 1 = 8/3) approx 2.67, closest is 3.
• Q3: C - Coincident lines overlap completely, infinite solutions.

Memory Aids

• Intersecting lines: (a₁/a₂ ≠ b₁/b₂).
• Parallel lines: (a₁/a₂ = b₁/b₂ ≠ c₁/c₂).
• Coincident lines: (a₁/a₂ = b₁/b₂ = c₁/c₂).
• Substitution: express one variable, substitute, solve.
• Elimination: equalize coefficients, add/subtract to eliminate.

Definition and General Form

• A pair of linear equations in two variables involves two linear equations with the same variables.
• Example: 3x + 2y = 1 and 2x - 2y = 2.
• General form: \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\), where \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers.
• Condition: \(a_1^2 + b_1^2 \neq 0\) and \(a_2^2 + b_2^2 \neq 0\).
• Also called system of simultaneous linear equations in two variables.

Solution of Pair of Simultaneous Linear Equations

• Any pair \((x, y)\) satisfying both equations simultaneously is a solution.

Geometrical Representation

• Each linear equation represents a straight line.
• Pair represents two lines with three possibilities:
• Intersecting lines: one unique solution.
• Parallel lines: no solution.
• Coincident lines: infinitely many solutions.

Graphical Method

• Plot both equations as lines on graph.
• Intersecting lines: solution is point of intersection \((a,b)\).
• Parallel lines: no solution.
• Coincident lines: infinite solutions.

Methods of Finding Unknown Constants

• Given \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
• Compare ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}\):
• If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\): unique solution (intersecting lines).
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\): no solution (parallel lines).
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\): infinite solutions (coincident lines).

Solved Examples

• Ex 1: Solve 3x + 2y = 1 and 2x - 2y = 2 by substitution.
  • From 1st: \(y = \frac{1 - 3x}{2}\).
  • Substitute in 2nd: \(2x - 2(\frac{1 - 3x}{2}) = 2\) \Rightarrow 2x - (1 - 3x) = 2).
  • Simplify: \(2x - 1 + 3x = 2 \Rightarrow 5x = 3 \Rightarrow x = \frac{3}{5}\).
  • Find y: \(y = \frac{1 - 3(3/5)}{2} = \frac{1 - 9/5}{2} = \frac{-4/5}{2} = -\frac{2}{5}\).
  • Solution: \(x=\frac{3}{5}, y=-\frac{2}{5}\).
• Ex 2: Solve by elimination: x + y = 5 and 2x - y = 4.
  • Add: \((x + y) + (2x - y) = 5 + 4 \Rightarrow 3x = 9 \Rightarrow x = 3\).
  • Substitute in 1st: \(3 + y = 5 \Rightarrow y = 2\).
  • Solution: \(x=3, y=2\).

Practice Set

• Q1: Which condition means two lines are parallel?
  • A) \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
  • B) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
  • C) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
  • D) None
• Q2: Solve by substitution: 2x + y = 7, x - y = 1. Find x.
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: If two lines coincide, how many solutions exist?
  • A) None
  • B) One
  • C) Infinite
  • D) Two

Answer Key

• Q1: B - Parallel lines have equal ratios of \(a_1/a_2\) and \(b_1/b_2\) but different \(c_1/c_2\).
• Q2: B - From second: \(x = y + 1\), substitute in first: \(2(y+1) + y = 7 \Rightarrow 3y + 2 = 7 \Rightarrow y = 5/3\), then \(x = 5/3 + 1 = 8/3\) approx 2.67, closest is 3.
• Q3: C - Coincident lines overlap completely, infinite solutions.

Memory Aids

• Intersecting lines: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
• Parallel lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
• Coincident lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
• Substitution: express one variable, substitute, solve.
• Elimination: equalize coefficients, add/subtract to eliminate.

Definition and General Form

• A pair of linear equations in two variables involves two linear equations with the same variables.
• Example: 3x + 2y = 1 and 2x - 2y = 2.
• General form: (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0), where (a₁, b₁, c₁, a₂, b₂, c₂) are real numbers.
• Condition: (a₁² + b₁² ≠ 0) and (a₂² + b₂² ≠ 0).
• Also called system of simultaneous linear equations in two variables.

Solution of Pair of Simultaneous Linear Equations

• Any pair (x, y) satisfying both equations simultaneously is a solution.

Geometrical Representation

• Each linear equation represents a straight line.
• Pair represents two lines with three possibilities:
• Intersecting lines: one unique solution.
• Parallel lines: no solution.
• Coincident lines: infinitely many solutions.

Graphical Method

• Plot both equations as lines on graph.
• Intersecting lines: solution is point of intersection (a,b).
• Parallel lines: no solution.
• Coincident lines: infinite solutions.

Methods of Finding Unknown Constants

• Given (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0).
• Compare ratios (a₁/a₂, b₁/b₂, c₁/c₂):
• If (a₁/a₂ ≠ b₁/b₂): unique solution (intersecting lines).
• If (a₁/a₂ = b₁/b₂ ≠ c₁/c₂): no solution (parallel lines).
• If (a₁/a₂ = b₁/b₂ = c₁/c₂): infinite solutions (coincident lines).

Solved Examples

• Ex 1: Solve 3x + 2y = 1 and 2x - 2y = 2 by substitution.
  • From 1st: (y = (1 - 3x)/2).
  • Substitute in 2nd: (2x - 2((1 - 3x)/2) = 2) ⇒ 2x - (1 - 3x) = 2).
  • Simplify: (2x - 1 + 3x = 2 ⇒ 5x = 3 ⇒ x = 3/5).
  • Find y: (y = (1 - 3(3/5))/2 = (1 - 9/5)/2 = (-4/5)/2 = -2/5).
  • Solution: (x=3/5, y=-2/5).
• Ex 2: Solve by elimination: x + y = 5 and 2x - y = 4.
  • Add: ((x + y) + (2x - y) = 5 + 4 ⇒ 3x = 9 ⇒ x = 3).
  • Substitute in 1st: (3 + y = 5 ⇒ y = 2).
  • Solution: (x=3, y=2).

Practice Set

• Q1: Which condition means two lines are parallel?
  • A) (a₁/a₂ ≠ b₁/b₂)
  • B) (a₁/a₂ = b₁/b₂ ≠ c₁/c₂)
  • C) (a₁/a₂ = b₁/b₂ = c₁/c₂)
  • D) None
• Q2: Solve by substitution: 2x + y = 7, x - y = 1. Find x.
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: If two lines coincide, how many solutions exist?
  • A) None
  • B) One
  • C) Infinite
  • D) Two

Answer Key

• Q1: B - Parallel lines have equal ratios of (a₁/a₂) and (b₁/b₂) but different (c₁/c₂).
• Q2: B - From second: (x = y + 1), substitute in first: (2(y+1) + y = 7 ⇒ 3y + 2 = 7 ⇒ y = 5/3), then (x = 5/3 + 1 = 8/3) approx 2.67, closest is 3.
• Q3: C - Coincident lines overlap completely, infinite solutions.

Memory Aids

• Intersecting lines: (a₁/a₂ ≠ b₁/b₂).
• Parallel lines: (a₁/a₂ = b₁/b₂ ≠ c₁/c₂).
• Coincident lines: (a₁/a₂ = b₁/b₂ = c₁/c₂).
• Substitution: express one variable, substitute, solve.
• Elimination: equalize coefficients, add/subtract to eliminate.

Definition and General Form

• A pair of linear equations in two variables involves two linear equations with the same variables.
• Example: 3x + 2y = 1 and 2x - 2y = 2.
• General form: \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\), where \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers.
• Condition: \(a_1^2 + b_1^2 \neq 0\) and \(a_2^2 + b_2^2 \neq 0\).
• Also called system of simultaneous linear equations in two variables.

Solution of Pair of Simultaneous Linear Equations

• Any pair \((x, y)\) satisfying both equations simultaneously is a solution.

Geometrical Representation

• Each linear equation represents a straight line.
• Pair represents two lines with three possibilities:
• Intersecting lines: one unique solution.
• Parallel lines: no solution.
• Coincident lines: infinitely many solutions.

Graphical Method

• Plot both equations as lines on graph.
• Intersecting lines: solution is point of intersection \((a,b)\).
• Parallel lines: no solution.
• Coincident lines: infinite solutions.

Methods of Finding Unknown Constants

• Given \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
• Compare ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}\):
• If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\): unique solution (intersecting lines).
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\): no solution (parallel lines).
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\): infinite solutions (coincident lines).

Solved Examples

• Ex 1: Solve 3x + 2y = 1 and 2x - 2y = 2 by substitution.
  • From 1st: \(y = \frac{1 - 3x}{2}\).
  • Substitute in 2nd: \(2x - 2(\frac{1 - 3x}{2}) = 2\) \Rightarrow 2x - (1 - 3x) = 2\).
  • Simplify: \(2x - 1 + 3x = 2 \Rightarrow 5x = 3 \Rightarrow x = \frac{3}{5}\).
  • Find y: \(y = \frac{1 - 3(3/5)}{2} = \frac{1 - 9/5}{2} = \frac{-4/5}{2} = -\frac{2}{5}\).
  • Solution: \(x=\frac{3}{5}, y=-\frac{2}{5}\).
• Ex 2: Solve by elimination: x + y = 5 and 2x - y = 4.
  • Add: \((x + y) + (2x - y) = 5 + 4 \Rightarrow 3x = 9 \Rightarrow x = 3\).
  • Substitute in 1st: \(3 + y = 5 \Rightarrow y = 2\).
  • Solution: \(x=3, y=2\).

Practice Set

• Q1: Which condition means two lines are parallel?
  • A) \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
  • B) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
  • C) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
  • D) None
• Q2: Solve by substitution: 2x + y = 7, x - y = 1. Find x.
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: If two lines coincide, how many solutions exist?
  • A) None
  • B) One
  • C) Infinite
  • D) Two

Answer Key

• Q1: B - Parallel lines have equal ratios of \(a_1/a_2\) and \(b_1/b_2\) but different \(c_1/c_2\).
• Q2: B - From second: \(x = y + 1\), substitute in first: \(2(y+1) + y = 7 \Rightarrow 3y + 2 = 7 \Rightarrow y = 5/3\), then \(x = 5/3 + 1 = 8/3\) approx 2.67, closest is 3.
• Q3: C - Coincident lines overlap completely, infinite solutions.

Memory Aids

• Intersecting lines: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
• Parallel lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
• Coincident lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
• Substitution: express one variable, substitute, solve.
• Elimination: equalize coefficients, add/subtract to eliminate.

Algebraic Methods

Substitution Method

• Step 1: Express one variable in terms of the other from one equation.
• Step 2: Substitute in the other equation to get one variable.
• Step 3: Solve for the second variable by substitution.
• If no variable remains and statement is true: infinite solutions.
• If no variable remains and statement is false: no solution.

Elimination Method

• Given equations: (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0).
• Step 1: Multiply to equalize coefficients of one variable.
• Step 2: Add or subtract to eliminate one variable.
• Step 3: Solve resulting single variable equation.
• True statement with no variable: infinite solutions.
• False statement with no variable: no solution.
• More convenient than substitution in some cases.

Solved Examples

• Ex 1: Solve by substitution: x + 2y = 5, 2x - y = 1.
  • From 1st: (x = 5 - 2y).
  • Substitute in 2nd: (2(5 - 2y) - y = 1 ⇒ 10 - 4y - y = 1).
  • Simplify: (10 - 5y = 1 ⇒ 5y = 9 ⇒ y = 9/5).
  • Find x: (x = 5 - 2(9/5) = 5 - 18/5 = 7/5).
  • Solution: (x=7/5, y=9/5).
• Ex 2: Solve by elimination: 3x + 4y = 10, 6x - 4y = 2.
  • Add: (3x + 4y + 6x - 4y = 10 + 2 ⇒ 9x = 12 ⇒ x = 4/3).
  • Substitute in 1st: (3(4/3) + 4y = 10 ⇒ 4 + 4y = 10 ⇒ 4y = 6 ⇒ y = 3/2).
  • Solution: (x=4/3, y=3/2).

Practice Set

• Q1: In substitution method, what is first step?
  • A) Solve for both variables
  • B) Express one variable in terms of other
  • C) Add equations
  • D) Multiply equations
• Q2: In elimination method, what is done first?
  • A) Add equations directly
  • B) Multiply to equalize coefficients
  • C) Substitute variables
  • D) Graph equations
• Q3: If elimination gives a false statement, what is solution?
  • A) Unique solution
  • B) Infinite solutions
  • C) No solution
  • D) Two solutions

Answer Key

• Q1: B - First express one variable in terms of the other.
• Q2: B - Multiply to make coefficients equal before elimination.
• Q3: C - False statement means no solution.

Memory Aids

• Substitution: "Express & Substitute".
• Elimination: "Equalize & Eliminate".
• True/False statements indicate infinite/no solutions.

Substitution Method

• Step 1: Express one variable in terms of the other from one equation.
• Step 2: Substitute in the other equation to get one variable.
• Step 3: Solve for the second variable by substitution.
• If no variable remains and statement is true: infinite solutions.
• If no variable remains and statement is false: no solution.

Elimination Method

• Given equations: (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0).
• Step 1: Multiply to equalize coefficients of one variable.
• Step 2: Add or subtract to eliminate one variable.
• Step 3: Solve resulting single variable equation.
• True statement with no variable: infinite solutions.
• False statement with no variable: no solution.
• More convenient than substitution in some cases.

Solved Examples

• Ex 1: Solve by substitution: x + 2y = 5, 2x - y = 1.
  • From 1st: (x = 5 - 2y).
  • Substitute in 2nd: (2(5 - 2y) - y = 1 ⇒ 10 - 4y - y = 1).
  • Simplify: (10 - 5y = 1 ⇒ 5y = 9 ⇒ y = 9/5).
  • Find x: (x = 5 - 2(9/5) = 5 - 18/5 = 7/5).
  • Solution: (x=7/5, y=9/5).
• Ex 2: Solve by elimination: 3x + 4y = 10, 6x - 4y = 2.
  • Add: (3x + 4y + 6x - 4y = 10 + 2 ⇒ 9x = 12 ⇒ x = 4/3).
  • Substitute in 1st: (3(4/3) + 4y = 10 ⇒ 4 + 4y = 10 ⇒ 4y = 6 ⇒ y = 3/2).
  • Solution: (x=4/3, y=3/2).

Practice Set

• Q1: In substitution method, what is first step?
  • A) Solve for both variables
  • B) Express one variable in terms of other
  • C) Add equations
  • D) Multiply equations
• Q2: In elimination method, what is done first?
  • A) Add equations directly
  • B) Multiply to equalize coefficients
  • C) Substitute variables
  • D) Graph equations
• Q3: If elimination gives a false statement, what is solution?
  • A) Unique solution
  • B) Infinite solutions
  • C) No solution
  • D) Two solutions

Answer Key

• Q1: B - First express one variable in terms of the other.
• Q2: B - Multiply to make coefficients equal before elimination.
• Q3: C - False statement means no solution.

Memory Aids

• Substitution: "Express & Substitute".
• Elimination: "Equalize & Eliminate".
• True/False statements indicate infinite/no solutions.

Substitution Method

• Step 1: Express one variable in terms of the other from one equation.
• Step 2: Substitute in the other equation to get one variable.
• Step 3: Solve for the second variable by substitution.
• If no variable remains and statement is true: infinite solutions.
• If no variable remains and statement is false: no solution.

Elimination Method

• Given equations: \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
• Step 1: Multiply to equalize coefficients of one variable.
• Step 2: Add or subtract to eliminate one variable.
• Step 3: Solve resulting single variable equation.
• True statement with no variable: infinite solutions.
• False statement with no variable: no solution.
• More convenient than substitution in some cases.

Solved Examples

• Ex 1: Solve by substitution: x + 2y = 5, 2x - y = 1.
  • From 1st: \(x = 5 - 2y\).
  • Substitute in 2nd: \(2(5 - 2y) - y = 1 \Rightarrow 10 - 4y - y = 1\).
  • Simplify: \(10 - 5y = 1 \Rightarrow 5y = 9 \Rightarrow y = \frac{9}{5}\).
  • Find x: \(x = 5 - 2(9/5) = 5 - \frac{18}{5} = \frac{7}{5}\).
  • Solution: \(x=\frac{7}{5}, y=\frac{9}{5}\).
• Ex 2: Solve by elimination: 3x + 4y = 10, 6x - 4y = 2.
  • Add: \(3x + 4y + 6x - 4y = 10 + 2 \Rightarrow 9x = 12 \Rightarrow x = \frac{4}{3}\).
  • Substitute in 1st: \(3(4/3) + 4y = 10 \Rightarrow 4 + 4y = 10 \Rightarrow 4y = 6 \Rightarrow y = \frac{3}{2}\).
  • Solution: \(x=\frac{4}{3}, y=\frac{3}{2}\).

Practice Set

• Q1: In substitution method, what is first step?
  • A) Solve for both variables
  • B) Express one variable in terms of other
  • C) Add equations
  • D) Multiply equations
• Q2: In elimination method, what is done first?
  • A) Add equations directly
  • B) Multiply to equalize coefficients
  • C) Substitute variables
  • D) Graph equations
• Q3: If elimination gives a false statement, what is solution?
  • A) Unique solution
  • B) Infinite solutions
  • C) No solution
  • D) Two solutions

Answer Key

• Q1: B - First express one variable in terms of the other.
• Q2: B - Multiply to make coefficients equal before elimination.
• Q3: C - False statement means no solution.

Memory Aids

• Substitution: "Express & Substitute".
• Elimination: "Equalize & Eliminate".
• True/False statements indicate infinite/no solutions.

Substitution Method

• Step 1: Express one variable in terms of the other from one equation.
• Step 2: Substitute in the other equation to get one variable.
• Step 3: Solve for the second variable by substitution.
• If no variable remains and statement is true: infinite solutions.
• If no variable remains and statement is false: no solution.

Elimination Method

• Given equations: (a₁ x + b₁ y + c₁ = 0) and (a₂ x + b₂ y + c₂ = 0).
• Step 1: Multiply to equalize coefficients of one variable.
• Step 2: Add or subtract to eliminate one variable.
• Step 3: Solve resulting single variable equation.
• True statement with no variable: infinite solutions.
• False statement with no variable: no solution.
• More convenient than substitution in some cases.

Solved Examples

• Ex 1: Solve by substitution: x + 2y = 5, 2x - y = 1.
  • From 1st: (x = 5 - 2y).
  • Substitute in 2nd: (2(5 - 2y) - y = 1 ⇒ 10 - 4y - y = 1).
  • Simplify: (10 - 5y = 1 ⇒ 5y = 9 ⇒ y = 9/5).
  • Find x: (x = 5 - 2(9/5) = 5 - 18/5 = 7/5).
  • Solution: (x=7/5, y=9/5).
• Ex 2: Solve by elimination: 3x + 4y = 10, 6x - 4y = 2.
  • Add: (3x + 4y + 6x - 4y = 10 + 2 ⇒ 9x = 12 ⇒ x = 4/3).
  • Substitute in 1st: (3(4/3) + 4y = 10 ⇒ 4 + 4y = 10 ⇒ 4y = 6 ⇒ y = 3/2).
  • Solution: (x=4/3, y=3/2).

Practice Set

• Q1: In substitution method, what is first step?
  • A) Solve for both variables
  • B) Express one variable in terms of other
  • C) Add equations
  • D) Multiply equations
• Q2: In elimination method, what is done first?
  • A) Add equations directly
  • B) Multiply to equalize coefficients
  • C) Substitute variables
  • D) Graph equations
• Q3: If elimination gives a false statement, what is solution?
  • A) Unique solution
  • B) Infinite solutions
  • C) No solution
  • D) Two solutions

Answer Key

• Q1: B - First express one variable in terms of the other.
• Q2: B - Multiply to make coefficients equal before elimination.
• Q3: C - False statement means no solution.

Memory Aids

• Substitution: "Express & Substitute".
• Elimination: "Equalize & Eliminate".
• True/False statements indicate infinite/no solutions.

Substitution Method

• Step 1: Express one variable in terms of the other from one equation.
• Step 2: Substitute in the other equation to get one variable.
• Step 3: Solve for the second variable by substitution.
• If no variable remains and statement is true: infinite solutions.
• If no variable remains and statement is false: no solution.

Elimination Method

• Given equations: \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
• Step 1: Multiply to equalize coefficients of one variable.
• Step 2: Add or subtract to eliminate one variable.
• Step 3: Solve resulting single variable equation.
• True statement with no variable: infinite solutions.
• False statement with no variable: no solution.
• More convenient than substitution in some cases.

Solved Examples

• Ex 1: Solve by substitution: x + 2y = 5, 2x - y = 1.
  • From 1st: \(x = 5 - 2y\).
  • Substitute in 2nd: \(2(5 - 2y) - y = 1 \Rightarrow 10 - 4y - y = 1\).
  • Simplify: \(10 - 5y = 1 \Rightarrow 5y = 9 \Rightarrow y = \frac{9}{5}\).
  • Find x: \(x = 5 - 2(9/5) = 5 - \frac{18}{5} = \frac{7}{5}\).
  • Solution: \(x=\frac{7}{5}, y=\frac{9}{5}\).
• Ex 2: Solve by elimination: 3x + 4y = 10, 6x - 4y = 2.
  • Add: \(3x + 4y + 6x - 4y = 10 + 2 \Rightarrow 9x = 12 \Rightarrow x = \frac{4}{3}\).
  • Substitute in 1st: \(3(4/3) + 4y = 10 \Rightarrow 4 + 4y = 10 \Rightarrow 4y = 6 \Rightarrow y = \frac{3}{2}\).
  • Solution: \(x=\frac{4}{3}, y=\frac{3}{2}\).

Practice Set

• Q1: In substitution method, what is first step?
  • A) Solve for both variables
  • B) Express one variable in terms of other
  • C) Add equations
  • D) Multiply equations
• Q2: In elimination method, what is done first?
  • A) Add equations directly
  • B) Multiply to equalize coefficients
  • C) Substitute variables
  • D) Graph equations
• Q3: If elimination gives a false statement, what is solution?
  • A) Unique solution
  • B) Infinite solutions
  • C) No solution
  • D) Two solutions

Answer Key

• Q1: B - First express one variable in terms of the other.
• Q2: B - Multiply to make coefficients equal before elimination.
• Q3: C - False statement means no solution.

Memory Aids

• Substitution: "Express & Substitute".
• Elimination: "Equalize & Eliminate".
• True/False statements indicate infinite/no solutions.

Simultaneous Linear Equation in Three Variables

Definition and Procedure

• System contains at least three linear equations in three variables.
• Use elimination method in steps:
• (i) Eliminate one variable from two equations to get two-variable equation.
• (ii) Repeat for another pair to get second two-variable equation.
• (iii) Solve these two-variable equations by elimination.
• (iv) Substitute values to find third variable.

Solved Examples

• Ex 1: Solve system:
  • (x + y + z = 6)
  • (2x - y + 3z = 14)
  • (x - 2y + z = 2)
  • Eliminate z from (1) and (3): (x + y + z = 6) and (x - 2y + z = 2) subtract to get (3y = 4) ⇒ y = 4/3).
  • Eliminate z from (2) and (3): (2x - y + 3z = 14) and (x - 2y + z = 2) multiply (3) by 3 and subtract to eliminate z.
  • Solve resulting two-variable equations for x and y.
  • Substitute back to find z.

Practice Set

• Q1: How many equations needed for three variables?
  • A) 1
  • B) 2
  • C) 3
  • D) 4
• Q2: First step in elimination for three variables?
  • A) Eliminate one variable from two equations
  • B) Solve one equation directly
  • C) Substitute variables
  • D) Graph equations
• Q3: After eliminating one variable, what type of equations remain?
  • A) One variable
  • B) Two variables
  • C) Three variables
  • D) No variables

Answer Key

• Q1: C - Three equations needed for three variables.
• Q2: A - Eliminate one variable from two equations first.
• Q3: B - Two-variable equations remain after elimination.

Memory Aids

• 3 variables need 3 equations.
• Eliminate one variable at a time.
• Solve reduced two-variable system next.

Definition and Procedure

• System contains at least three linear equations in three variables.
• Use elimination method in steps:
• (i) Eliminate one variable from two equations to get two-variable equation.
• (ii) Repeat for another pair to get second two-variable equation.
• (iii) Solve these two-variable equations by elimination.
• (iv) Substitute values to find third variable.

Solved Examples

• Ex 1: Solve system:
  • (x + y + z = 6)
  • (2x - y + 3z = 14)
  • (x - 2y + z = 2)
  • Eliminate z from (1) and (3): (x + y + z = 6) and (x - 2y + z = 2) subtract to get (3y = 4) ⇒ y = 4/3).
  • Eliminate z from (2) and (3): (2x - y + 3z = 14) and (x - 2y + z = 2) multiply (3) by 3 and subtract to eliminate z.
  • Solve resulting two-variable equations for x and y.
  • Substitute back to find z.

Practice Set

• Q1: How many equations needed for three variables?
  • A) 1
  • B) 2
  • C) 3
  • D) 4
• Q2: First step in elimination for three variables?
  • A) Eliminate one variable from two equations
  • B) Solve one equation directly
  • C) Substitute variables
  • D) Graph equations
• Q3: After eliminating one variable, what type of equations remain?
  • A) One variable
  • B) Two variables
  • C) Three variables
  • D) No variables

Answer Key

• Q1: C - Three equations needed for three variables.
• Q2: A - Eliminate one variable from two equations first.
• Q3: B - Two-variable equations remain after elimination.

Memory Aids

• 3 variables need 3 equations.
• Eliminate one variable at a time.
• Solve reduced two-variable system next.

Definition and Procedure

• System contains at least three linear equations in three variables.
• Use elimination method in steps:
• (i) Eliminate one variable from two equations to get two-variable equation.
• (ii) Repeat for another pair to get second two-variable equation.
• (iii) Solve these two-variable equations by elimination.
• (iv) Substitute values to find third variable.

Solved Examples

• Ex 1: Solve system:
  • \(x + y + z = 6\)
  • \(2x - y + 3z = 14\)
  • \(x - 2y + z = 2\)
  • Eliminate z from (1) and (3): \(x + y + z = 6\) and \(x - 2y + z = 2\) subtract to get \(3y = 4\) \Rightarrow y = \frac{4}{3}\).
  • Eliminate z from (2) and (3): \(2x - y + 3z = 14\) and \(x - 2y + z = 2\) multiply (3) by 3 and subtract to eliminate z.
  • Solve resulting two-variable equations for x and y.
  • Substitute back to find z.

Practice Set

• Q1: How many equations needed for three variables?
  • A) 1
  • B) 2
  • C) 3
  • D) 4
• Q2: First step in elimination for three variables?
  • A) Eliminate one variable from two equations
  • B) Solve one equation directly
  • C) Substitute variables
  • D) Graph equations
• Q3: After eliminating one variable, what type of equations remain?
  • A) One variable
  • B) Two variables
  • C) Three variables
  • D) No variables

Answer Key

• Q1: C - Three equations needed for three variables.
• Q2: A - Eliminate one variable from two equations first.
• Q3: B - Two-variable equations remain after elimination.

Memory Aids

• 3 variables need 3 equations.
• Eliminate one variable at a time.
• Solve reduced two-variable system next.

Definition and Procedure

• System contains at least three linear equations in three variables.
• Use elimination method in steps:
• (i) Eliminate one variable from two equations to get two-variable equation.
• (ii) Repeat for another pair to get second two-variable equation.
• (iii) Solve these two-variable equations by elimination.
• (iv) Substitute values to find third variable.

Solved Examples

• Ex 1: Solve system:
  • (x + y + z = 6)
  • (2x - y + 3z = 14)
  • (x - 2y + z = 2)
  • Eliminate z from (1) and (3): (x + y + z = 6) and (x - 2y + z = 2) subtract to get (3y = 4) ⇒ y = 4/3).
  • Eliminate z from (2) and (3): (2x - y + 3z = 14) and (x - 2y + z = 2) multiply (3) by 3 and subtract to eliminate z.
  • Solve resulting two-variable equations for x and y.
  • Substitute back to find z.

Practice Set

• Q1: How many equations needed for three variables?
  • A) 1
  • B) 2
  • C) 3
  • D) 4
• Q2: First step in elimination for three variables?
  • A) Eliminate one variable from two equations
  • B) Solve one equation directly
  • C) Substitute variables
  • D) Graph equations
• Q3: After eliminating one variable, what type of equations remain?
  • A) One variable
  • B) Two variables
  • C) Three variables
  • D) No variables

Answer Key

• Q1: C - Three equations needed for three variables.
• Q2: A - Eliminate one variable from two equations first.
• Q3: B - Two-variable equations remain after elimination.

Memory Aids

• 3 variables need 3 equations.
• Eliminate one variable at a time.
• Solve reduced two-variable system next.

Definition and Procedure

• System contains at least three linear equations in three variables.
• Use elimination method in steps:
• (i) Eliminate one variable from two equations to get two-variable equation.
• (ii) Repeat for another pair to get second two-variable equation.
• (iii) Solve these two-variable equations by elimination.
• (iv) Substitute values to find third variable.

Solved Examples

• Ex 1: Solve system:
  • \(x + y + z = 6\)
  • \(2x - y + 3z = 14\)
  • \(x - 2y + z = 2\)
  • Eliminate z from (1) and (3): \(x + y + z = 6\) and \(x - 2y + z = 2\) subtract to get \(3y = 4\) \Rightarrow y = \frac{4}{3}\).
  • Eliminate z from (2) and (3): \(2x - y + 3z = 14\) and \(x - 2y + z = 2\) multiply (3) by 3 and subtract to eliminate z.
  • Solve resulting two-variable equations for x and y.
  • Substitute back to find z.

Practice Set

• Q1: How many equations needed for three variables?
  • A) 1
  • B) 2
  • C) 3
  • D) 4
• Q2: First step in elimination for three variables?
  • A) Eliminate one variable from two equations
  • B) Solve one equation directly
  • C) Substitute variables
  • D) Graph equations
• Q3: After eliminating one variable, what type of equations remain?
  • A) One variable
  • B) Two variables
  • C) Three variables
  • D) No variables

Answer Key

• Q1: C - Three equations needed for three variables.
• Q2: A - Eliminate one variable from two equations first.
• Q3: B - Two-variable equations remain after elimination.

Memory Aids

• 3 variables need 3 equations.
• Eliminate one variable at a time.
• Solve reduced two-variable system next.

Quadratic Equations

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form \(p(x) = 0\), where \(p(x)\) is a polynomial of degree 2.
• General form: \(ax^2 + bx + c = 0\), where \(a, b, c \in \mathbb{R}\) and \(a \neq 0\).

Method of Factorisation

• Express \(ax^2 + bx + c\) as product of two linear factors by splitting the middle term.
• Let factors be \((px + q)\) and \((rx + s)\), then:
  • \(a = pr\)
  • \(b = ps + qr\)
  • \(c = qs\)
  • Product of split terms \(ps \times qr = ac\)
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of \(ax^2 + bx + c = 0\) are given by:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  • Discriminant \(D = b^2 - 4ac\)
• Nature of roots based on \(D\):
  • \(D > 0\): Two distinct real roots \(\frac{-b + \sqrt{D}}{2a}\) and \(\frac{-b - \sqrt{D}}{2a}\)
  • \(D = 0\): Two equal real roots \(\frac{-b}{2a}\)
  • \(D < 0\): No real roots (complex roots)

Solved Examples

• Ex 1: Solve \(x^2 - 5x + 6 = 0\) by factorisation.
  • Split middle term: \(x^2 - 2x - 3x + 6 = 0\)
  • Factor: \(x(x - 2) - 3(x - 2) = 0\)
  • \((x - 3)(x - 2) = 0\)
  • Roots: \(x = 3, 2\)
• Ex 2: Find roots of \(2x^2 + 3x - 2 = 0\) using quadratic formula.
  • \(a=2, b=3, c=-2\)
  • \(D = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25\)
  • \(x = \frac{-3 \pm 5}{2 \times 2}\)
  • Roots: \(x = \frac{2}{4} = 0.5\), \(x = \frac{-8}{4} = -2\)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. \(x^3 + 2x + 1 = 0\)
  • B. \(x^2 - 4x + 4 = 0\)
  • C. \(2x + 3 = 0\)
  • D. \(x^4 - 1 = 0\)
• Q2: Find the roots of \(x^2 - 7x + 10 = 0\) by factorisation.
  • A. \(x=2, 5\)
  • B. \(x=3, 4\)
  • C. \(x=1, 10\)
  • D. \(x=5, 7\)
• Q3: For \(3x^2 + 6x + 3 = 0\), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. \(x^2 - 4x + 4 = 0\) is quadratic (degree 2).
• Q2: A. Factorisation: \(x^2 - 7x + 10 = (x-5)(x-2)\), roots \(2, 5\).
• Q3: B. Discriminant \(D = 6^2 - 4 \times 3 \times 3 = 36 - 36 = 0\), equal roots.

Memory Aids

• Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Discriminant \(D = b^2 - 4ac\) decides root nature.
• Factorisation: split middle term so product = \(ac\), sum = \(b\).

Definition and General Form

• Any equation of the form (p(x) = 0), where (p(x)) is a polynomial of degree 2.
• General form: (ax² + bx + c = 0), where (a, b, c ∈ ℝ) and (a ≠ 0).

Method of Factorisation

• Express (ax² + bx + c) as product of two linear factors by splitting the middle term.
• Let factors be ((px + q)) and ((rx + s)), then:
  • (a = pr)
  • (b = ps + qr)
  • (c = qs)
  • Product of split terms ps × qr = ac
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of (ax² + bx + c = 0) are given by:
  • (x = (-b ± √(b² - 4ac))/2a)
  • Discriminant D = b² - 4ac
• Nature of roots based on (D):
  • (D > 0): Two distinct real roots ((-b + √D)/2a) and ((-b - √D)/2a)
  • (D = 0): Two equal real roots ((-b)/2a)
  • (D < 0): No real roots (complex roots)

Solved Examples

• Ex 1: Solve (x² - 5x + 6 = 0) by factorisation.
  • Split middle term: (x² - 2x - 3x + 6 = 0)
  • Factor: x(x - 2) - 3(x - 2) = 0
  • ((x - 3)(x - 2) = 0)
  • Roots: (x = 3, 2)
• Ex 2: Find roots of (2x² + 3x - 2 = 0) using quadratic formula.
  • (a=2, b=3, c=-2)
  • (D = 3² - 4 × 2 × (-2) = 9 + 16 = 25)
  • (x = (-3 ± 5)/(2 × 2))
  • Roots: (x = 2/4 = 0.5), (x = -8/4 = -2)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. (x³ + 2x + 1 = 0)
  • B. (x² - 4x + 4 = 0)
  • C. (2x + 3 = 0)
  • D. (x⁴ - 1 = 0)
• Q2: Find the roots of (x² - 7x + 10 = 0) by factorisation.
  • A. (x=2, 5)
  • B. (x=3, 4)
  • C. (x=1, 10)
  • D. (x=5, 7)
• Q3: For (3x² + 6x + 3 = 0), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. (x² - 4x + 4 = 0) is quadratic (degree 2).
• Q2: A. Factorisation: (x² - 7x + 10 = (x-5)(x-2)), roots (2, 5).
• Q3: B. Discriminant D = 6² - 4 × 3 × 3 = 36 - 36 = 0, equal roots.

Memory Aids

• Quadratic formula: (x = (-b ± √(b² - 4ac))/2a)
• Discriminant D = b² - 4ac decides root nature.
• Factorisation: split middle term so product = (ac), sum = (b).

Definition and General Form

• Any equation of the form \(p(x) = 0\), where \(p(x)\) is a polynomial of degree 2.
• General form: \(ax^2 + bx + c = 0\), where \(a, b, c \in \mathbb{R}\) and \(a \neq 0\).

Method of Factorisation

• Express \(ax^2 + bx + c\) as product of two linear factors by splitting the middle term.
• Let factors be \((px + q)\) and \((rx + s)\), then:
  • \(a = pr\)
  • \(b = ps + qr\)
  • \(c = qs\)
  • Product of split terms \(ps \times qr = ac\)
• Set each factor to zero to find roots.

Quadratic Formula

• Roots of \(ax^2 + bx + c = 0\) are given by:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  • Discriminant \(D = b^2 - 4ac\)
• Nature of roots based on \(D\):
  • \(D > 0\): Two distinct real roots \(\frac{-b + \sqrt{D}}{2a}\) and \(\frac{-b - \sqrt{D}}{2a}\)
  • \(D = 0\): Two equal real roots \(\frac{-b}{2a}\)
  • \(D < 0\): No real roots (complex roots)

Solved Examples

• Ex 1: Solve \(x^2 - 5x + 6 = 0\) by factorisation.
  • Split middle term: \(x^2 - 2x - 3x + 6 = 0\)
  • Factor: \(x(x - 2) - 3(x - 2) = 0\)
  • \((x - 3)(x - 2) = 0\)
  • Roots: \(x = 3, 2\)
• Ex 2: Find roots of \(2x^2 + 3x - 2 = 0\) using quadratic formula.
  • \(a=2, b=3, c=-2\)
  • \(D = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25\)
  • \(x = \frac{-3 \pm 5}{2 \times 2}\)
  • Roots: \(x = \frac{2}{4} = 0.5\), \(x = \frac{-8}{4} = -2\)

Practice Set

• Q1: Which of the following is a quadratic equation?
  • A. \(x^3 + 2x + 1 = 0\)
  • B. \(x^2 - 4x + 4 = 0\)
  • C. \(2x + 3 = 0\)
  • D. \(x^4 - 1 = 0\)
• Q2: Find the roots of \(x^2 - 7x + 10 = 0\) by factorisation.
  • A. \(x=2, 5\)
  • B. \(x=3, 4\)
  • C. \(x=1, 10\)
  • D. \(x=5, 7\)
• Q3: For \(3x^2 + 6x + 3 = 0\), what is the nature of roots?
  • A. Two distinct real roots
  • B. Two equal real roots
  • C. No real roots
  • D. One real root

Answer Key

• Q1: B. \(x^2 - 4x + 4 = 0\) is quadratic (degree 2).
• Q2: A. Factorisation: \(x^2 - 7x + 10 = (x-5)(x-2)\), roots \(2, 5\).
• Q3: B. Discriminant \(D = 6^2 - 4 \times 3 \times 3 = 36 - 36 = 0\), equal roots.

Memory Aids

• Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Discriminant \(D = b^2 - 4ac\) decides root nature.
• Factorisation: split middle term so product = \(ac\), sum = \(b\).

Coordinate Geometry

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)²+(6-2)²) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)²+(y₂-y₁)²).

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)²+(6-2)²) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)²+(y₂-y₁)²).

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)^2+(6-2)^2) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)^2+(y₂-y₁)^2).

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)²+(6-2)²) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)²+(y₂-y₁)²).

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)^2+(6-2)^2) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)^2+(y₂-y₁)^2).

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)²+(6-2)²) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)²+(y₂-y₁)²).

Overview

• Describes points on a plane using ordered pairs \((x, y)\).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between \(P_1(2,3)\) and \(P_2(4,7)\).

• Midpoint formula: \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
• Calculate: \(\left(\frac{2+4}{2}, \frac{3+7}{2}\right) = (3, 5)\)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points \((1,2)\) and \((4,6)\) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = \(\sqrt{(4-1)^2+(6-2)^2} = 5\).

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).

Overview

• Describes points on a plane using ordered pairs (x, y).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between P₁(2,3) and P₂(4,7).

• Midpoint formula: ((x₁+x₂)/2, (y₁+y₂)/2)
• Calculate: ((2+4)/2, (3+7)/2) = (3, 5)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points (1,2) and (4,6) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = √((4-1)²+(6-2)²) = 5.

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: √((x₂-x₁)²+(y₂-y₁)²).

Overview

• Describes points on a plane using ordered pairs \((x, y)\).
• Key concepts: slope, tangent, secant.
• Used to analyze geometric shapes and their relations.

Solved Examples

Ex 1: Find coordinates of a point midway between \(P_1(2,3)\) and \(P_2(4,7)\).

• Midpoint formula: \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
• Calculate: \(\left(\frac{2+4}{2}, \frac{3+7}{2}\right) = (3, 5)\)

Practice Set

Q1: What are the coordinates of the origin?

• A) (1,0)
• B) (0,0)
• C) (0,1)
• D) (1,1)

Q2: Which coordinate pair lies on the x-axis?

• A) (0,5)
• B) (3,0)
• C) (2,2)
• D) (0,0)

Q3: The distance between points \((1,2)\) and \((4,6)\) is?

• A) 3 units
• B) 5 units
• C) 4 units
• D) 6 units

Answer Key

• Q1: B - Origin is at (0,0).
• Q2: B - Points on x-axis have y=0.
• Q3: B - Distance = \(\sqrt{(4-1)^2+(6-2)^2} = 5\).

Memory Aids

• Coordinates = (x, y) pairs.
• Origin = (0,0).
• Distance formula: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).

Slope

Definition and Formula

• Slope m = vertical change / horizontal change = (Δy/Δx) = (y₂ - y₁)/(x₂ - x₁).
• Represents steepness and direction of a line.
• Angle θ with x-axis: m = tan θ.
• Line equation: y = mx + c, where m is slope.
• Δy = change in y-axis, Δx = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: m₁ × m₂ = -1.
• Parallel lines: m₁ = m₂.

Solved Examples

Ex 1: Find slope of line through P₁(1,2) and P₂(4,8).

• Calculate Δy = 8 - 2 = 6, Δx = 4 - 1 = 3.
• Slope m = 6/3 = 2.

Practice Set

Q1: Slope of line through points (2,3) and (5,9) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are m₁ = 3, find m₂.

• A) -1/3
• B) 3
• C) 1
• D) -3

Answer Key

• Q1: A - m = (9-3)/(5-2) = 6/3 = 2.
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy m₁ × m₂ = -1, so m₂ = -1/3.

Memory Aids

• Slope formula: m = Δy/Δx.
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Definition and Formula

• Slope m = vertical change / horizontal change = (Δy/Δx) = (y₂ - y₁)/(x₂ - x₁).
• Represents steepness and direction of a line.
• Angle θ with x-axis: m = tan θ.
• Line equation: y = mx + c, where m is slope.
• Δy = change in y-axis, Δx = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: m₁ × m₂ = -1.
• Parallel lines: m₁ = m₂.

Solved Examples

Ex 1: Find slope of line through P₁(1,2) and P₂(4,8).

• Calculate Δy = 8 - 2 = 6, Δx = 4 - 1 = 3.
• Slope m = 6/3 = 2.

Practice Set

Q1: Slope of line through points (2,3) and (5,9) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are m₁ = 3, find m₂.

• A) -1/3
• B) 3
• C) 1
• D) -3

Answer Key

• Q1: A - m = (9-3)/(5-2) = 6/3 = 2.
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy m₁ × m₂ = -1, so m₂ = -1/3.

Memory Aids

• Slope formula: m = Δy/Δx.
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Definition and Formula

• Slope m = vertical change / horizontal change = (Δy/Δx) = (y₂ - y₁)/(x₂ - x₁).
• Represents steepness and direction of a line.
• Angle θ with x-axis: m = tan θ.
• Line equation: y = mx + c, where m is slope.
• Δy = change in y-axis, Δx = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: m₁ × m₂ = -1.
• Parallel lines: m₁ = m₂.

Solved Examples

Ex 1: Find slope of line through P₁(1,2) and P₂(4,8).

• Calculate Δy = 8 - 2 = 6, Δx = 4 - 1 = 3.
• Slope m = 6/3 = 2.

Practice Set

Q1: Slope of line through points (2,3) and (5,9) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are m₁ = 3, find m₂.

• A) -1/3
• B) 3
• C) 1
• D) -3

Answer Key

• Q1: A - m = (9-3)/(5-2) = 6/3 = 2.
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy m₁ × m₂ = -1, so m₂ = -1/3.

Memory Aids

• Slope formula: m = Δy/Δx.
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Definition and Formula

• Slope m = vertical change / horizontal change = (Δy/Δx) = (y₂ - y₁)/(x₂ - x₁).
• Represents steepness and direction of a line.
• Angle θ with x-axis: m = tan θ.
• Line equation: y = mx + c, where m is slope.
• Δy = change in y-axis, Δx = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: m₁ × m₂ = -1.
• Parallel lines: m₁ = m₂.

Solved Examples

Ex 1: Find slope of line through P₁(1,2) and P₂(4,8).

• Calculate Δy = 8 - 2 = 6, Δx = 4 - 1 = 3.
• Slope m = 6/3 = 2.

Practice Set

Q1: Slope of line through points (2,3) and (5,9) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are m₁ = 3, find m₂.

• A) -1/3
• B) 3
• C) 1
• D) -3

Answer Key

• Q1: A - m = (9-3)/(5-2) = 6/3 = 2.
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy m₁ × m₂ = -1, so m₂ = -1/3.

Memory Aids

• Slope formula: m = Δy/Δx.
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Definition and Formula

• Slope \(m\) = vertical change / horizontal change = \(\frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}\).
• Represents steepness and direction of a line.
• Angle \(\theta\) with x-axis: \(m = \tan \theta\).
• Line equation: \(y = mx + c\), where \(m\) is slope.
• \(\Delta y\) = change in y-axis, \(\Delta x\) = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: \(m_1 \times m_2 = -1\).
• Parallel lines: \(m_1 = m_2\).

Solved Examples

Ex 1: Find slope of line through \(P_1(1,2)\) and \(P_2(4,8)\).

• Calculate \(\Delta y = 8 - 2 = 6\), \(\Delta x = 4 - 1 = 3\).
• Slope \(m = \frac{6}{3} = 2\).

Practice Set

Q1: Slope of line through points \((2,3)\) and \((5,9)\) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are \(m_1 = 3\), find \(m_2\).

• A) \(-\frac{1}{3}\)
• B) 3
• C) 1
• D) \(-3\)

Answer Key

• Q1: A - \(m = \frac{9-3}{5-2} = \frac{6}{3} = 2\).
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy \(m_1 \times m_2 = -1\), so \(m_2 = -\frac{1}{3}\).

Memory Aids

• Slope formula: \(m = \frac{\Delta y}{\Delta x}\).
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Definition and Formula

• Slope m = vertical change / horizontal change = (Δy/Δx) = (y₂ - y₁)/(x₂ - x₁).
• Represents steepness and direction of a line.
• Angle θ with x-axis: m = tan θ.
• Line equation: y = mx + c, where m is slope.
• Δy = change in y-axis, Δx = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: m₁ × m₂ = -1.
• Parallel lines: m₁ = m₂.

Solved Examples

Ex 1: Find slope of line through P₁(1,2) and P₂(4,8).

• Calculate Δy = 8 - 2 = 6, Δx = 4 - 1 = 3.
• Slope m = 6/3 = 2.

Practice Set

Q1: Slope of line through points (2,3) and (5,9) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are m₁ = 3, find m₂.

• A) -1/3
• B) 3
• C) 1
• D) -3

Answer Key

• Q1: A - m = (9-3)/(5-2) = 6/3 = 2.
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy m₁ × m₂ = -1, so m₂ = -1/3.

Memory Aids

• Slope formula: m = Δy/Δx.
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Definition and Formula

• Slope \(m\) = vertical change / horizontal change = \(\frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}\).
• Represents steepness and direction of a line.
• Angle \(\theta\) with x-axis: \(m = \tan \theta\).
• Line equation: \(y = mx + c\), where \(m\) is slope.
• \(\Delta y\) = change in y-axis, \(\Delta x\) = change in x-axis.

Types of Slopes

• Positive slope: line rises left to right.
• Negative slope: line falls left to right.
• Zero slope: horizontal line.
• Undefined slope: vertical line.
• Perpendicular lines: \(m_1 \times m_2 = -1\).
• Parallel lines: \(m_1 = m_2\).

Solved Examples

Ex 1: Find slope of line through \(P_1(1,2)\) and \(P_2(4,8)\).

• Calculate \(\Delta y = 8 - 2 = 6\), \(\Delta x = 4 - 1 = 3\).
• Slope \(m = \frac{6}{3} = 2\).

Practice Set

Q1: Slope of line through points \((2,3)\) and \((5,9)\) is?

• A) 2
• B) 3
• C) 1
• D) 4

Q2: What is the slope of a horizontal line?

• A) 0
• B) Undefined
• C) 1
• D) -1

Q3: If slopes of two perpendicular lines are \(m_1 = 3\), find \(m_2\).

• A) \(-\frac{1}{3}\)
• B) 3
• C) 1
• D) \(-3\)

Answer Key

• Q1: A - \(m = \frac{9-3}{5-2} = \frac{6}{3} = 2\).
• Q2: A - Horizontal line slope = 0.
• Q3: A - Perpendicular slopes satisfy \(m_1 \times m_2 = -1\), so \(m_2 = -\frac{1}{3}\).

Memory Aids

• Slope formula: \(m = \frac{\Delta y}{\Delta x}\).
• Positive slope: line rises.
• Negative slope: line falls.
• Perpendicular slopes multiply to -1.

Tangent

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve \( y = f(x) \) at \( x = x_1 \) is the derivative \( \frac{dy}{dx} \) at that point.

Solved Examples

• Ex 1: Find slope of tangent to \( y = x^2 \) at \( x = 2 \).
  • Derivative: \( \frac{dy}{dx} = 2x \)
  • Slope at \( x=2 \): \( 2 \times 2 = 4 \)

Practice Set

• Q1: What is the slope of the tangent to \( y = 3x + 5 \) at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to \( y = x^3 \) at \( x = 1 \) is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of \( y=mx+c \) is constant \( m=3 \).
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative \( 3x^2 \) at \( x=1 \) is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve y = f(x) at x = x₁ is the derivative dy/dx at that point.

Solved Examples

• Ex 1: Find slope of tangent to y = x² at x = 2.
  • Derivative: dy/dx = 2x
  • Slope at x=2: 2 × 2 = 4

Practice Set

• Q1: What is the slope of the tangent to y = 3x + 5 at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to y = x³ at x = 1 is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of y=mx+c is constant m=3.
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative 3x² at x=1 is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Definition and Properties

• A tangent is a line that touches a curve at exactly one point without crossing it.
• For circles, the tangent is perpendicular to the radius at the point of contact.
• Slope of tangent to a curve \( y = f(x) \) at \( x = x_1 \) is the derivative \( \frac{dy}{dx} \) at that point.

Solved Examples

• Ex 1: Find slope of tangent to \( y = x^2 \) at \( x = 2 \).
  • Derivative: \( \frac{dy}{dx} = 2x \)
  • Slope at \( x=2 \): \( 2 \times 2 = 4 \)

Practice Set

• Q1: What is the slope of the tangent to \( y = 3x + 5 \) at any point?
  • A. 0
  • B. 3
  • C. 5
  • D. Variable
• Q2: The tangent to a circle is always:
  • A. Parallel to radius
  • B. Perpendicular to radius
  • C. Intersecting radius at two points
  • D. None of these
• Q3: Slope of tangent to \( y = x^3 \) at \( x = 1 \) is:
  • A. 1
  • B. 3
  • C. 0
  • D. 2

Answer Key

• Q1: B - Slope of \( y=mx+c \) is constant \( m=3 \).
• Q2: B - Tangent is perpendicular to radius at point of contact.
• Q3: B - Derivative \( 3x^2 \) at \( x=1 \) is 3.

Memory Aids

• Tangent touches curve at one point only.
• Slope of tangent = derivative at that point.
• Tangent ⟂ radius in circles.

Secant

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points \( P(x_0, f(x_0)) \) and \( Q(x_0+h, f(x_0+h)) \) is \( \frac{f(x_0+h) - f(x_0)}{h} \).
• As \( h \to 0 \), secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to \( y = x^2 \) between \( x=1 \) and \( x=3 \).
  • Slope = \( \frac{f(3) - f(1)}{3-1} = \frac{9 - 1}{2} = 4 \)

Practice Set

• Q1: Slope of secant line between \( x=2 \) and \( x=4 \) for \( y = x^2 \) is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between \( P(x_0, f(x_0)) \) and \( Q(x_0+h, f(x_0+h)) \) is:
  • A. \( \frac{f(x_0+h) + f(x_0)}{h} \)
  • B. \( \frac{f(x_0+h) - f(x_0)}{h} \)
  • C. \( \frac{f(x_0) - f(x_0+h)}{h} \)
  • D. \( \frac{f(x_0+h) - f(x_0)}{x_0} \)

Answer Key

• Q1: B - \( \frac{16 - 4}{2} = 6 \)
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is (f(x₀+h) - f(x₀))/h.
• As h → 0, secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to y = x² between x=1 and x=3.
  • Slope = (f(3) - f(1))/(3-1) = (9 - 1)/2 = 4

Practice Set

• Q1: Slope of secant line between x=2 and x=4 for y = x² is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between P(x₀, f(x₀)) and Q(x₀+h, f(x₀+h)) is:
  • A. (f(x₀+h) + f(x₀))/h
  • B. (f(x₀+h) - f(x₀))/h
  • C. (f(x₀) - f(x₀+h))/h
  • D. (f(x₀+h) - f(x₀))/x₀

Answer Key

• Q1: B - (16 - 4)/2 = 6
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Definition and Properties

• A secant is a line that intersects a curve at two or more points.
• For circles, a secant intersects at two distinct points.
• Slope of secant line between points \( P(x_0, f(x_0)) \) and \( Q(x_0+h, f(x_0+h)) \) is \( \frac{f(x_0+h) - f(x_0)}{h} \).
• As \( h \to 0 \), secant slope approaches tangent slope (derivative).

Solved Examples

• Ex 1: Find slope of secant to \( y = x^2 \) between \( x=1 \) and \( x=3 \).
  • Slope = \( \frac{f(3) - f(1)}{3-1} = \frac{9 - 1}{2} = 4 \)

Practice Set

• Q1: Slope of secant line between \( x=2 \) and \( x=4 \) for \( y = x^2 \) is:
  • A. 4
  • B. 6
  • C. 8
  • D. 10
• Q2: A secant line to a circle intersects it at:
  • A. One point
  • B. Two points
  • C. No points
  • D. Infinite points
• Q3: The slope of secant line between \( P(x_0, f(x_0)) \) and \( Q(x_0+h, f(x_0+h)) \) is:
  • A. \( \frac{f(x_0+h) + f(x_0)}{h} \)
  • B. \( \frac{f(x_0+h) - f(x_0)}{h} \)
  • C. \( \frac{f(x_0) - f(x_0+h)}{h} \)
  • D. \( \frac{f(x_0+h) - f(x_0)}{x_0} \)

Answer Key

• Q1: B - \( \frac{16 - 4}{2} = 6 \)
• Q2: B - Secant intersects circle at two points.
• Q3: B - Definition of slope of secant line.

Memory Aids

• Secant cuts curve at two or more points.
• Slope secant = average rate of change.
• As points merge, secant → tangent.

Comparing Tangent and Secant

Key Differences

• Tangent: Instantaneous rate of change at one point.
• Secant: Average rate of change between two points.

Solved Examples

• Ex 1: Identify if the line touching a curve at one point is tangent or secant.
  • Touches at one point → Tangent.
• Ex 2: Line intersecting curve at two points is:
  • Secant line.

Practice Set

• Q1: A line touching a curve at exactly one point is called:
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Radius
• Q2: The average rate of change between two points on a curve is given by:
  • A. Tangent slope
  • B. Secant slope
  • C. Radius length
  • D. None
• Q3: Which represents instantaneous rate of change?
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Diameter

Answer Key

• Q1: B - Tangent touches curve at one point only.
• Q2: B - Secant gives average rate of change.
• Q3: B - Tangent gives instantaneous rate of change.

Memory Aids

• Tangent = touch at one point.
• Secant = cut at two points.
• Instantaneous = tangent, Average = secant.

Key Differences

• Tangent: Instantaneous rate of change at one point.
• Secant: Average rate of change between two points.

Solved Examples

• Ex 1: Identify if the line touching a curve at one point is tangent or secant.
  • Touches at one point → Tangent.
• Ex 2: Line intersecting curve at two points is:
  • Secant line.

Practice Set

• Q1: A line touching a curve at exactly one point is called:
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Radius
• Q2: The average rate of change between two points on a curve is given by:
  • A. Tangent slope
  • B. Secant slope
  • C. Radius length
  • D. None
• Q3: Which represents instantaneous rate of change?
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Diameter

Answer Key

• Q1: B - Tangent touches curve at one point only.
• Q2: B - Secant gives average rate of change.
• Q3: B - Tangent gives instantaneous rate of change.

Memory Aids

• Tangent = touch at one point.
• Secant = cut at two points.
• Instantaneous = tangent, Average = secant.

Key Differences

• Tangent: Instantaneous rate of change at one point.
• Secant: Average rate of change between two points.

Solved Examples

• Ex 1: Identify if the line touching a curve at one point is tangent or secant.
  • Touches at one point → Tangent.
• Ex 2: Line intersecting curve at two points is:
  • Secant line.

Practice Set

• Q1: A line touching a curve at exactly one point is called:
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Radius
• Q2: The average rate of change between two points on a curve is given by:
  • A. Tangent slope
  • B. Secant slope
  • C. Radius length
  • D. None
• Q3: Which represents instantaneous rate of change?
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Diameter

Answer Key

• Q1: B - Tangent touches curve at one point only.
• Q2: B - Secant gives average rate of change.
• Q3: B - Tangent gives instantaneous rate of change.

Memory Aids

• Tangent = touch at one point.
• Secant = cut at two points.
• Instantaneous = tangent, Average = secant.

Key Differences

• Tangent: Instantaneous rate of change at one point.
• Secant: Average rate of change between two points.

Solved Examples

• Ex 1: Identify if the line touching a curve at one point is tangent or secant.
  • Touches at one point → Tangent.
• Ex 2: Line intersecting curve at two points is:
  • Secant line.

Practice Set

• Q1: A line touching a curve at exactly one point is called:
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Radius
• Q2: The average rate of change between two points on a curve is given by:
  • A. Tangent slope
  • B. Secant slope
  • C. Radius length
  • D. None
• Q3: Which represents instantaneous rate of change?
  • A. Secant
  • B. Tangent
  • C. Chord
  • D. Diameter

Answer Key

• Q1: B - Tangent touches curve at one point only.
• Q2: B - Secant gives average rate of change.
• Q3: B - Tangent gives instantaneous rate of change.

Memory Aids

• Tangent = touch at one point.
• Secant = cut at two points.
• Instantaneous = tangent, Average = secant.

Applications of Slope, Tangent, and Secant

Key Uses

• Analyzing geometric shapes and angles in coordinate geometry.
• Solving motion and optimization problems in calculus.
• Understanding properties of curves like parabolas, ellipses, and circles.

Solved Examples

• Ex 1: Use slope of tangent to find velocity at a point in motion.
• Ex 2: Use secant slope to find average speed between two times.

Practice Set

• Q1: Slope of tangent helps find:
  • A. Average speed
  • B. Instantaneous speed
  • C. Distance
  • D. Time
• Q2: Secant slope is used to calculate:
  • A. Instantaneous rate
  • B. Average rate
  • C. Acceleration
  • D. Displacement
• Q3: Which curve property can be studied using tangent and secant?
  • A. Shape only
  • B. Motion only
  • C. Curvature and rate of change
  • D. None

Answer Key

• Q1: B - Tangent slope = instantaneous speed.
• Q2: B - Secant slope = average rate of change.
• Q3: C - Tangent and secant help study curvature and rate of change.

Memory Aids

• Tangent = instantaneous, Secant = average.
• Used in motion, geometry, and calculus.
• Key for curve analysis.

Key Uses

• Analyzing geometric shapes and angles in coordinate geometry.
• Solving motion and optimization problems in calculus.
• Understanding properties of curves like parabolas, ellipses, and circles.

Solved Examples

• Ex 1: Use slope of tangent to find velocity at a point in motion.
• Ex 2: Use secant slope to find average speed between two times.

Practice Set

• Q1: Slope of tangent helps find:
  • A. Average speed
  • B. Instantaneous speed
  • C. Distance
  • D. Time
• Q2: Secant slope is used to calculate:
  • A. Instantaneous rate
  • B. Average rate
  • C. Acceleration
  • D. Displacement
• Q3: Which curve property can be studied using tangent and secant?
  • A. Shape only
  • B. Motion only
  • C. Curvature and rate of change
  • D. None

Answer Key

• Q1: B - Tangent slope = instantaneous speed.
• Q2: B - Secant slope = average rate of change.
• Q3: C - Tangent and secant help study curvature and rate of change.

Memory Aids

• Tangent = instantaneous, Secant = average.
• Used in motion, geometry, and calculus.
• Key for curve analysis.

Key Uses

• Analyzing geometric shapes and angles in coordinate geometry.
• Solving motion and optimization problems in calculus.
• Understanding properties of curves like parabolas, ellipses, and circles.

Solved Examples

• Ex 1: Use slope of tangent to find velocity at a point in motion.
• Ex 2: Use secant slope to find average speed between two times.

Practice Set

• Q1: Slope of tangent helps find:
  • A. Average speed
  • B. Instantaneous speed
  • C. Distance
  • D. Time
• Q2: Secant slope is used to calculate:
  • A. Instantaneous rate
  • B. Average rate
  • C. Acceleration
  • D. Displacement
• Q3: Which curve property can be studied using tangent and secant?
  • A. Shape only
  • B. Motion only
  • C. Curvature and rate of change
  • D. None

Answer Key

• Q1: B - Tangent slope = instantaneous speed.
• Q2: B - Secant slope = average rate of change.
• Q3: C - Tangent and secant help study curvature and rate of change.

Memory Aids

• Tangent = instantaneous, Secant = average.
• Used in motion, geometry, and calculus.
• Key for curve analysis.

Key Uses

• Analyzing geometric shapes and angles in coordinate geometry.
• Solving motion and optimization problems in calculus.
• Understanding properties of curves like parabolas, ellipses, and circles.

Solved Examples

• Ex 1: Use slope of tangent to find velocity at a point in motion.
• Ex 2: Use secant slope to find average speed between two times.

Practice Set

• Q1: Slope of tangent helps find:
  • A. Average speed
  • B. Instantaneous speed
  • C. Distance
  • D. Time
• Q2: Secant slope is used to calculate:
  • A. Instantaneous rate
  • B. Average rate
  • C. Acceleration
  • D. Displacement
• Q3: Which curve property can be studied using tangent and secant?
  • A. Shape only
  • B. Motion only
  • C. Curvature and rate of change
  • D. None

Answer Key

• Q1: B - Tangent slope = instantaneous speed.
• Q2: B - Secant slope = average rate of change.
• Q3: C - Tangent and secant help study curvature and rate of change.

Memory Aids

• Tangent = instantaneous, Secant = average.
• Used in motion, geometry, and calculus.
• Key for curve analysis.

Straight Line

Key Concepts

• General equation: y = mx - c
• m = tan θ is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Key Concepts

• General equation: y = mx - c
• m = tan θ is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Key Concepts

• General equation: y = mx - c
• m = tan θ is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Key Concepts

• General equation: y = mx - c
• m = tan θ is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Key Concepts

• General equation: y = mx - c
• m = tan θ is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Key Concepts

• General equation: y = mx - c
• m = tan θ is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Key Concepts

• General equation: y = mx - c
• m = \( \tan \theta \) is the slope of the line
• c is the y-intercept (point where line crosses Y-axis)

Solved Examples

• Ex 1: Find slope and intercept of line y = 2x - 3
  • Slope, m = 2
  • Intercept, c = 3

Practice Set

• Q1: What is the slope of the line y = 5x - 7?
  • A. 7
  • B. 5
  • C. -7
  • D. -5
• Q2: The line y = -3x + 4 crosses the y-axis at?
  • A. (0, -3)
  • B. (0, 4)
  • C. (4, 0)
  • D. (-3, 0)
• Q3: If slope m = 1 and y-intercept c = 2, what is the equation of the line?
  • A. y = x + 2
  • B. y = 2x + 1
  • C. y = x - 2
  • D. y = 2x - 1

Answer Key

• Q1: B - Slope is coefficient of x, which is 5.
• Q2: B - y-intercept is the constant term, here 4 at (0,4).
• Q3: A - Equation with m=1 and c=2 is y = x + 2.

Memory Aids

• Slope m = tan(θ), angle with x-axis
• Intercept c = y-value when x=0
• Equation form: y = mx - c

Circle

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x² + y² = a²
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = √16 = 4

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = √36 = 6.
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Key Concepts

• Equation: x^2 + y^2 = a^2
• Center at origin: (0, 0)
• Radius = a
• All points on circle satisfy distance from center = a

Solved Examples

• Ex 1: Find radius if circle equation is x² + y² = 16
  • Radius, a = \( \sqrt{16} = 4 \)

Practice Set

• Q1: What is the center of circle x² + y² = 25?
  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (5, 5)
• Q2: Radius of circle x² + y² = 36 is?
  • A. 6
  • B. 12
  • C. 18
  • D. 3
• Q3: Point (3, 4) lies on circle x² + y² = 25?
  • A. Yes
  • B. No
  • C. Only if radius > 5
  • D. Only if radius < 5

Answer Key

• Q1: A - Center is at origin (0,0) for given equation.
• Q2: A - Radius = \( \sqrt{36} = 6 \).
• Q3: A - Check: 3² + 4² = 9 + 16 = 25, lies on circle.

Memory Aids

• Circle equation: x² + y² = r²
• Center at (0,0) if no h,k terms
• Radius = distance from center to any point

Parabola

Key Concepts

• Equation: y² = kx
• Parabola passes through origin (0,0)
• Opens rightward along x-axis
• Shape is symmetric about x-axis

Solved Examples

• Ex 1: For y² = 4x, find y when x = 1
  • y² = 4(1) = 4
  • y = ±2

Practice Set

• Q1: Parabola y² = 9x opens in which direction?
  • A. Upward
  • B. Downward
  • C. Rightward
  • D. Leftward
• Q2: For y² = 16x, find y when x = 4
  • A. ±8
  • B. ±4
  • C. ±16
  • D. ±2
• Q3: Vertex of parabola y² = kx is at?
  • A. (0,0)
  • B. (k,0)
  • C. (0,k)
  • D. (1,1)

Answer Key

• Q1: C - Parabola opens rightward along x-axis.
• Q2: A - y² = 16*4 = 64, y = ±8.
• Q3: A - Vertex is at origin (0,0).

Memory Aids

• Parabola y² = kx opens right
• Vertex at origin
• Symmetric about x-axis

Key Concepts

• Equation: y² = kx
• Parabola passes through origin (0,0)
• Opens rightward along x-axis
• Shape is symmetric about x-axis

Solved Examples

• Ex 1: For y² = 4x, find y when x = 1
  • y² = 4(1) = 4
  • y = ±2

Practice Set

• Q1: Parabola y² = 9x opens in which direction?
  • A. Upward
  • B. Downward
  • C. Rightward
  • D. Leftward
• Q2: For y² = 16x, find y when x = 4
  • A. ±8
  • B. ±4
  • C. ±16
  • D. ±2
• Q3: Vertex of parabola y² = kx is at?
  • A. (0,0)
  • B. (k,0)
  • C. (0,k)
  • D. (1,1)

Answer Key

• Q1: C - Parabola opens rightward along x-axis.
• Q2: A - y² = 16*4 = 64, y = ±8.
• Q3: A - Vertex is at origin (0,0).

Memory Aids

• Parabola y² = kx opens right
• Vertex at origin
• Symmetric about x-axis

Key Concepts

• Equation: y² = kx
• Parabola passes through origin (0,0)
• Opens rightward along x-axis
• Shape is symmetric about x-axis

Solved Examples

• Ex 1: For y² = 4x, find y when x = 1
  • y² = 4(1) = 4
  • y = ±2

Practice Set

• Q1: Parabola y² = 9x opens in which direction?
  • A. Upward
  • B. Downward
  • C. Rightward
  • D. Leftward
• Q2: For y² = 16x, find y when x = 4
  • A. ±8
  • B. ±4
  • C. ±16
  • D. ±2
• Q3: Vertex of parabola y² = kx is at?
  • A. (0,0)
  • B. (k,0)
  • C. (0,k)
  • D. (1,1)

Answer Key

• Q1: C - Parabola opens rightward along x-axis.
• Q2: A - y² = 16*4 = 64, y = ±8.
• Q3: A - Vertex is at origin (0,0).

Memory Aids

• Parabola y² = kx opens right
• Vertex at origin
• Symmetric about x-axis

Key Concepts

• Equation: y² = kx
• Parabola passes through origin (0,0)
• Opens rightward along x-axis
• Shape is symmetric about x-axis

Solved Examples

• Ex 1: For y² = 4x, find y when x = 1
  • y² = 4(1) = 4
  • y = ±2

Practice Set

• Q1: Parabola y² = 9x opens in which direction?
  • A. Upward
  • B. Downward
  • C. Rightward
  • D. Leftward
• Q2: For y² = 16x, find y when x = 4
  • A. ±8
  • B. ±4
  • C. ±16
  • D. ±2
• Q3: Vertex of parabola y² = kx is at?
  • A. (0,0)
  • B. (k,0)
  • C. (0,k)
  • D. (1,1)

Answer Key

• Q1: C - Parabola opens rightward along x-axis.
• Q2: A - y² = 16*4 = 64, y = ±8.
• Q3: A - Vertex is at origin (0,0).

Memory Aids

• Parabola y² = kx opens right
• Vertex at origin
• Symmetric about x-axis

Key Concepts

• Equation: y² = kx
• Parabola passes through origin (0,0)
• Opens rightward along x-axis
• Shape is symmetric about x-axis

Solved Examples

• Ex 1: For y² = 4x, find y when x = 1
  • y² = 4(1) = 4
  • y = ±2

Practice Set

• Q1: Parabola y² = 9x opens in which direction?
  • A. Upward
  • B. Downward
  • C. Rightward
  • D. Leftward
• Q2: For y² = 16x, find y when x = 4
  • A. ±8
  • B. ±4
  • C. ±16
  • D. ±2
• Q3: Vertex of parabola y² = kx is at?
  • A. (0,0)
  • B. (k,0)
  • C. (0,k)
  • D. (1,1)

Answer Key

• Q1: C - Parabola opens rightward along x-axis.
• Q2: A - y² = 16*4 = 64, y = ±8.
• Q3: A - Vertex is at origin (0,0).

Memory Aids

• Parabola y² = kx opens right
• Vertex at origin
• Symmetric about x-axis

Ellipse

Key Concepts

• Equation: (x²/a² + y²/b² = 1)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = √(1 - b²/a²), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = πab

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = √(1 - 9/25) = √(16/25) = 0.8

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Key Concepts

• Equation: (x²/a² + y²/b² = 1)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = √(1 - b²/a²), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = πab

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = √(1 - 9/25) = √(16/25) = 0.8

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Key Concepts

• Equation: (x²/a² + y²/b² = 1)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = √(1 - b²/a²), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = πab

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = √(1 - 9/25) = √(16/25) = 0.8

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Key Concepts

• Equation: (x²/a² + y²/b² = 1)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = √(1 - b²/a²), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = πab

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = √(1 - 9/25) = √(16/25) = 0.8

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Key Concepts

• Equation: (x²/a² + y²/b² = 1)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = √(1 - b²/a²), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = πab

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = √(1 - 9/25) = √(16/25) = 0.8

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Key Concepts

• Equation: (x²/a² + y²/b² = 1)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = √(1 - b²/a²), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = πab

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = √(1 - 9/25) = √(16/25) = 0.8

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Key Concepts

• Equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
• a = semi-major axis, b = semi-minor axis
• Coefficients of x² and y² are different
• Eccentricity: e = \sqrt{1 - \frac{b^2}{a^2}} \), e < 1
• Foci: S = (ae, 0), S' = (-ae, 0)
• Area = \( \pi ab \)

Solved Examples

• Ex 1: Find eccentricity if a = 5, b = 3
  • e = \( \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = 0.8 \)

Practice Set

• Q1: For ellipse, eccentricity e is?
  • A. > 1
  • B. = 1
  • C. < 1
  • D. = 0
• Q2: Area of ellipse with a=4, b=3 is?
  • A. 12π
  • B. 7π
  • C. 25π
  • D. 1
• Q3: Coordinates of foci for a=6, e=0.5 are?
  • A. (3,0) and (-3,0)
  • B. (6,0) and (-6,0)
  • C. (0,3) and (0,-3)
  • D. (0,6) and (0,-6)

Answer Key

• Q1: C - For ellipse, e < 1 always.
• Q2: A - Area = πab = π*4*3 = 12π.
• Q3: A - Foci at (ae,0) = (6*0.5,0) = (3,0) and (-3,0).

Memory Aids

• Ellipse: x²/a² + y²/b² = 1
• Eccentricity < 1
• Area = πab

Hyperbola

Key Concepts

• Equation: (x²/a² - y²/b² = 1)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola (x²/9 - y²/4 = 1)
  • Calculate: (3²/9 - 2²/4 = 1 - 1 = 0) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. (x²/a² + y²/b² = 1)
  • B. (x²/a² - y²/b² = 1)
  • C. xy = c
  • D. y² = kx
• Q3: For hyperbola (x²/16 - y²/9 = 1), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - (4²/16 - 3²/9 = 1 - 1 = 0) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Key Concepts

• Equation: (x²/a² - y²/b² = 1)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola (x²/9 - y²/4 = 1)
  • Calculate: (3²/9 - 2²/4 = 1 - 1 = 0) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. (x²/a² + y²/b² = 1)
  • B. (x²/a² - y²/b² = 1)
  • C. xy = c
  • D. y² = kx
• Q3: For hyperbola (x²/16 - y²/9 = 1), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - (4²/16 - 3²/9 = 1 - 1 = 0) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Key Concepts

• Equation: (x²/a² - y²/b² = 1)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola (x²/9 - y²/4 = 1)
  • Calculate: (3²/9 - 2²/4 = 1 - 1 = 0) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. (x²/a² + y²/b² = 1)
  • B. (x²/a² - y²/b² = 1)
  • C. xy = c
  • D. y² = kx
• Q3: For hyperbola (x²/16 - y²/9 = 1), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - (4²/16 - 3²/9 = 1 - 1 = 0) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Key Concepts

• Equation: (x²/a² - y²/b² = 1)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola (x²/9 - y²/4 = 1)
  • Calculate: (3²/9 - 2²/4 = 1 - 1 = 0) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. (x²/a² + y²/b² = 1)
  • B. (x²/a² - y²/b² = 1)
  • C. xy = c
  • D. y² = kx
• Q3: For hyperbola (x²/16 - y²/9 = 1), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - (4²/16 - 3²/9 = 1 - 1 = 0) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Key Concepts

• Equation: (x²/a² - y²/b² = 1)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola (x²/9 - y²/4 = 1)
  • Calculate: (3²/9 - 2²/4 = 1 - 1 = 0) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. (x²/a² + y²/b² = 1)
  • B. (x²/a² - y²/b² = 1)
  • C. xy = c
  • D. y² = kx
• Q3: For hyperbola (x²/16 - y²/9 = 1), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - (4²/16 - 3²/9 = 1 - 1 = 0) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Key Concepts

• Equation: (x²/a² - y²/b² = 1)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola (x²/9 - y²/4 = 1)
  • Calculate: (3²/9 - 2²/4 = 1 - 1 = 0) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. (x²/a² + y²/b² = 1)
  • B. (x²/a² - y²/b² = 1)
  • C. xy = c
  • D. y² = kx
• Q3: For hyperbola (x²/16 - y²/9 = 1), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - (4²/16 - 3²/9 = 1 - 1 = 0) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Key Concepts

• Equation: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• Center at origin (0,0)
• Two branches opening left and right
• Symmetric about x-axis and y-axis

Solved Examples

• Ex 1: Check if point (3,2) lies on hyperbola \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \)
  • Calculate: \( \frac{3^2}{9} - \frac{2^2}{4} = 1 - 1 = 0 \) (No)

Practice Set

• Q1: Hyperbola opens along?
  • A. y-axis
  • B. x-axis
  • C. both axes
  • D. none
• Q2: Equation of hyperbola is?
  • A. \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
  • B. \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
  • C. \( xy = c \)
  • D. \( y^2 = kx \)
• Q3: For hyperbola \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \), find if point (4,3) lies on it
  • A. Yes
  • B. No
  • C. Only if c=1
  • D. Only if c=0

Answer Key

• Q1: B - Hyperbola opens left and right along x-axis.
• Q2: B - Standard hyperbola equation.
• Q3: A - \( \frac{4^2}{16} - \frac{3^2}{9} = 1 - 1 = 0 \) (Check carefully, actually 0, so no. Correct answer is B)

Memory Aids

• Hyperbola: x²/a² - y²/b² = 1
• Two branches open left and right
• Symmetric about both axes

Rectangular Hyperbola

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Key Concepts

• If xy = constant, curve is rectangular hyperbola
• Equation form: xy = k
• Curve decreases steeply then flattens near axes
• Asymptotes are x-axis and y-axis

Solved Examples

• Ex 1: If xy = 4, find y when x = 2
  • y = 4/2 = 2

Practice Set

• Q1: Rectangular hyperbola equation is?
  • A. x + y = k
  • B. xy = k
  • C. x² - y² = k
  • D. y² = kx
• Q2: If xy = 9, find y when x = 3
  • A. 3
  • B. 6
  • C. 9
  • D. 27
• Q3: As x increases in xy = k, y?
  • A. Increases
  • B. Decreases
  • C. Remains constant
  • D. Becomes zero

Answer Key

• Q1: B - Rectangular hyperbola equation is xy = k.
• Q2: A - y = 9/3 = 3.
• Q3: B - y decreases as x increases to keep product constant.

Memory Aids

• Rectangular hyperbola: xy = constant
• As x increases, y decreases
• Asymptotes: x and y axes

Exponential Graph

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^{-x} (exponential decay)
  • y = 1 - e^{-x} (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e^1 = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^{-x} represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^{-x} approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e^0 = 1.
• Q2: B - y = e^{-x} is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^{-x}: decay curve
• y = 1 - e^{-x}: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^{-x} (exponential decay)
  • y = 1 - e^{-x} (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e^1 = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^{-x} represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^{-x} approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e^0 = 1.
• Q2: B - y = e^{-x} is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^{-x}: decay curve
• y = 1 - e^{-x}: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^-x (exponential decay)
  • y = 1 - e^-x (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e¹ = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^-x represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^-x approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e⁰ = 1.
• Q2: B - y = e^-x is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^-x: decay curve
• y = 1 - e^-x: growth to 1

Key Concepts

• Common forms:
  • y = e^x (exponential growth)
  • y = e^{-x} (exponential decay)
  • y = 1 - e^{-x} (growth approaching 1)
• At x=0, y=1 for first two forms
• Graphs show rapid increase or decrease

Solved Examples

• Ex 1: Find y for y = e^x at x=1
  • y = e^1 = e ≈ 2.718

Practice Set

• Q1: For y = e^x, what is y when x=0?
  • A. 0
  • B. 1
  • C. e
  • D. -1
• Q2: y = e^{-x} represents?
  • A. Exponential growth
  • B. Exponential decay
  • C. Linear growth
  • D. Constant function
• Q3: y = 1 - e^{-x} approaches which value as x → ∞?
  • A. 0
  • B. 1
  • C. e
  • D. ∞

Answer Key

• Q1: B - At x=0, y = e^0 = 1.
• Q2: B - y = e^{-x} is exponential decay.
• Q3: B - y approaches 1 as x → ∞.

Memory Aids

• y = e^x: growth curve
• y = e^{-x}: decay curve
• y = 1 - e^{-x}: growth to 1

Trigonometric Identities

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• 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)
• Double angle formulas:
  • sin 2A = 2 sin A cos A
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
  • tan 2A = (2 tan A)/(1 - tan² A)
• Product to sum formulas:
  • sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
  • sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
  • cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
  • cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2
• Half angle formulas:
  • sin A/2 = ±√((1 - cos A)/2)
  • cos A/2 = ±√((1 + cos A)/2)
  • tan A/2 = ±√((1 - cos A)/(1 + cos A)) = sin A/(1 + cos A) = (1 - cos A)/sin A

Solved Examples

• Ex 1: Find sin 75° using sum formula.
  • Use sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
  • Calculate: √2/2 × √3/2 + √2/2 × 1/2 = (√6 + √2)/4
• Ex 2: Calculate tan 75° using tangent sum formula.
  • tan(45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = 2 + √3

Practice Set

• Q1: sin 60° + sin 30° = ?
  • A) √3
  • B) 1
  • C) 3/2
  • D) 1/2
• Q2: tan 45° - tan 15° = ?
  • A) √3
  • B) 1
  • C) 2 - √3
  • D) 1/√3
• Q3: Using double angle formula, cos 2θ = 2 cos² θ - 1, find cos 60°.
  • A) 1/2
  • B) 1
  • C) 0
  • D) -1/2

Answer Key

• Q1: A) √3 - Using sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2, value is √3.
• Q2: C) 2 - √3 - Using tangent difference formula (tan A - tan B)/(1 + tan A tan B).
• Q3: A) 1/2 - cos 60° = 2 cos² 30° - 1 = 1/2.

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Key Formulas

• \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \)
• \( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \)
• \( \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \)
• Double angle formulas:
  • \( \sin 2A = 2 \sin A \cos A \)
  • \( \cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A \)
  • \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \)
• Product to sum formulas:
  • \( \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \)
  • \( \sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \)
  • \( \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \)
  • \( \cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} \)
• Half angle formulas:
  • \( \sin \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{2}} \)
  • \( \cos \frac{A}{2} = \pm \sqrt{\frac{1 + \cos A}{2}} \)
  • \( \tan \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{\sin A} \)

Solved Examples

• Ex 1: Find \( \sin 75^\circ \) using sum formula.
  • Use \( \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
  • Calculate: \( \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \)
• Ex 2: Calculate \( \tan 75^\circ \) using tangent sum formula.
  • \( \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = 2 + \sqrt{3} \)

Practice Set

• Q1: \( \sin 60^\circ + \sin 30^\circ = ? \)
  • A) \( \sqrt{3} \)
  • B) 1
  • C) \( \frac{3}{2} \)
  • D) \( \frac{1}{2} \)
• Q2: \( \tan 45^\circ - \tan 15^\circ = ? \)
  • A) \( \sqrt{3} \)
  • B) 1
  • C) \( 2 - \sqrt{3} \)
  • D) \( \frac{1}{\sqrt{3}} \)
• Q3: Using double angle formula, \( \cos 2\theta = 2 \cos^2 \theta - 1 \), find \( \cos 60^\circ \).
  • A) \( \frac{1}{2} \)
  • B) 1
  • C) 0
  • D) \( -\frac{1}{2} \)

Answer Key

• Q1: A) \( \sqrt{3} \) - Using \( \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \), value is \( \sqrt{3} \).
• Q2: C) \( 2 - \sqrt{3} \) - Using tangent difference formula \( \frac{\tan A - \tan B}{1 + \tan A \tan B} \).
• Q3: A) \( \frac{1}{2} \) - \( \cos 60^\circ = 2 \cos^2 30^\circ - 1 = \frac{1}{2} \).

Memory Aids

• Sum formulas: "Sine sum uses sine-cos + cos-sin"
• Double angle: "Double angle sine = 2 sin cos"
• Half angle: "Half angle sine uses sqrt((1-cos)/2)"

Trigonometric Values

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• tan 90° is undefined
• sin 0° = 0, cos 0° = 1, cos 180° = -1

Solved Examples

• Ex 1: Find sin 45°.
  • Value from table: √2/2
• Ex 2: Calculate tan 60°.
  • Value from table: √3

Practice Set

• Q1: sin 90° = ?
  • A) 0
  • B) 1
  • C) 1/2
  • D) √3
• Q2: cos 180° = ?
  • A) 1
  • B) 0
  • C) -1
  • D) 1/2
• Q3: tan 45° = ?
  • A) 0
  • B) 1
  • C) √3
  • D) Undefined

Answer Key

• Q1: B) 1 - sin 90° = 1
• Q2: C) -1 - cos 180° = -1
• Q3: B) 1 - tan 45° = 1

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• sin 30° = 1/2, cos 60° = 1/2
• tan 45° = 1

Key Points

• Standard angles: 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°
• Values of sin, cos, tan for these angles are exact and useful for problems
• Example: \( \sin 30^\circ = \frac{1}{2} \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)
• \( \tan 90^\circ \) is undefined
• \( \sin 0^\circ = 0 \), \( \cos 0^\circ = 1 \), \( \cos 180^\circ = -1 \)

Solved Examples

• Ex 1: Find \( \sin 45^\circ \).
  • Value from table: \( \frac{\sqrt{2}}{2} \)
• Ex 2: Calculate \( \tan 60^\circ \).
  • Value from table: \( \sqrt{3} \)

Practice Set

• Q1: \( \sin 90^\circ = ? \)
  • A) 0
  • B) 1
  • C) \( \frac{1}{2} \)
  • D) \( \sqrt{3} \)
• Q2: \( \cos 180^\circ = ? \)
  • A) 1
  • B) 0
  • C) -1
  • D) \( \frac{1}{2} \)
• Q3: \( \tan 45^\circ = ? \)
  • A) 0
  • B) 1
  • C) \( \sqrt{3} \)
  • D) Undefined

Answer Key

• Q1: B) 1 - \( \sin 90^\circ = 1 \)
• Q2: C) -1 - \( \cos 180^\circ = -1 \)
• Q3: B) 1 - \( \tan 45^\circ = 1 \)

Memory Aids

• Remember key angles: 0°, 30°, 45°, 60°, 90°
• \( \sin 30^\circ = 1/2 \), \( \cos 60^\circ = 1/2 \)
• \( \tan 45^\circ = 1 \)

Trigonometric Inverse Formulae

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• tan^-1(-x) = -tan^-1 x, x ∈ ℝ
• cot^-1(-x) = π - cot^-1 x, x ∈ ℝ
• csc^-1(-x) = -csc^-1 x, |x| > 1
• sec^-1(-x) = π - sec^-1 x, |x| > 1
• sin(sin^-1 x) = x or sin^-1(sin x) = x
• sin^-1 1/x = csc^-1 x, x ≥ 1 or x ≤ -1
• cos^-1 1/x = sec^-1 x, x ≥ 1 or x ≤ -1
• tan^-1 1/x = cot^-1 x, x > 0
• sin^-1(-x) = -sin^-1 x, x ∈ [-1,1]
• cos^-1(-x) = π - cos^-1 x, x ∈ [-1,1]
• tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)), xy < 1
• tan^-1 x + tan^-1 y = tan^-1((x-y)/(1+xy)), xy > -1

Solved Examples

• Ex 1: Find tan^-1(-1).
  • Use formula: tan^-1(-x) = -tan^-1 x
  • Result: -π/4
• Ex 2: Calculate sin^-1 1/2.
  • Use sin^-1 x = θ ⇒ sin θ = x
  • Result: π/6

Practice Set

• Q1: tan^-1(-1) = ?
  • A) π/4
  • B) -π/4
  • C) 0
  • D) π
• Q2: sin^-1(-1/2) = ?
  • A) -π/6
  • B) π/6
  • C) 0
  • D) π/2
• Q3: tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy)) holds if?
  • A) xy > 1
  • B) xy < 1
  • C) xy = 1
  • D) xy = 0

Answer Key

• Q1: B) -π/4 - Negative argument flips sign of inverse tangent.
• Q2: A) -π/6 - sin^-1(-x) = -sin^-1 x.
• Q3: B) xy < 1 - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from π
• Sum of inverse tan: valid if product less than 1

Key Formulas

• \( \tan^{-1}(-x) = -\tan^{-1} x, x \in \mathbb{R} \)
• \( \cot^{-1}(-x) = \pi - \cot^{-1} x, x \in \mathbb{R} \)
• \( \csc^{-1}(-x) = -\csc^{-1} x, |x| > 1 \)
• \( \sec^{-1}(-x) = \pi - \sec^{-1} x, |x| > 1 \)
• \( \sin(\sin^{-1} x) = x \) or \( \sin^{-1}(\sin x) = x \)
• \( \sin^{-1} \frac{1}{x} = \csc^{-1} x, x \geq 1 \text{ or } x \leq -1 \)
• \( \cos^{-1} \frac{1}{x} = \sec^{-1} x, x \geq 1 \text{ or } x \leq -1 \)
• \( \tan^{-1} \frac{1}{x} = \cot^{-1} x, x > 0 \)
• \( \sin^{-1}(-x) = -\sin^{-1} x, x \in [-1,1] \)
• \( \cos^{-1}(-x) = \pi - \cos^{-1} x, x \in [-1,1] \)
• \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right), xy < 1 \)
• \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right), xy > -1 \)

Solved Examples

• Ex 1: Find \( \tan^{-1}(-1) \).
  • Use formula: \( \tan^{-1}(-x) = -\tan^{-1} x \)
  • Result: \( -\frac{\pi}{4} \)
• Ex 2: Calculate \( \sin^{-1} \frac{1}{2} \).
  • Use \( \sin^{-1} x = \theta \Rightarrow \sin \theta = x \)
  • Result: \( \frac{\pi}{6} \)

Practice Set

• Q1: \( \tan^{-1}(-1) = ? \)
  • A) \( \frac{\pi}{4} \)
  • B) \( -\frac{\pi}{4} \)
  • C) 0
  • D) \( \pi \)
• Q2: \( \sin^{-1}(-\frac{1}{2}) = ? \)
  • A) \( -\frac{\pi}{6} \)
  • B) \( \frac{\pi}{6} \)
  • C) 0
  • D) \( \frac{\pi}{2} \)
• Q3: \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \) holds if?
  • A) \( xy > 1 \)
  • B) \( xy < 1 \)
  • C) \( xy = 1 \)
  • D) \( xy = 0 \)

Answer Key

• Q1: B) \( -\frac{\pi}{4} \) - Negative argument flips sign of inverse tangent.
• Q2: A) \( -\frac{\pi}{6} \) - \( \sin^{-1}(-x) = -\sin^{-1} x \).
• Q3: B) \( xy < 1 \) - Condition for sum formula of inverse tangent.

Memory Aids

• Inverse tan negative: flip sign
• Inverse cot negative: subtract from \( \pi \)
• Sum of inverse tan: valid if product less than 1

Trigonometric Approximation

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when theta 0 (in radians):
  • sin theta approx theta
  • cos theta approx 1
  • tan theta approx theta
• Trigonometric expansions (Taylor series):
  • sin theta = theta - theta³/3! + theta⁵/5! - ...
  • cos theta = 1 - theta²/2! + theta⁴/4! - ...
  • tan theta = theta - theta³/3 + 2theta⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 approx 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 approx 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small theta, sin theta approx ?
  • A) theta²
  • B) theta
  • C) 1
  • D) 0
• Q2: Using expansion, cos theta approx ? for small theta
  • A) 1
  • B) 1 - theta²/2
  • C) theta
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) theta - Small angle approximation for sine.
• Q2: B) 1 - theta²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when θ → 0 (in radians):
  • sin θ ≈ θ
  • cos θ ≈ 1
  • tan θ ≈ θ
• Trigonometric expansions (Taylor series):
  • sin θ = θ - θ³/3! + θ⁵/5! - ...
  • cos θ = 1 - θ²/2! + θ⁴/4! - ...
  • tan θ = θ - θ³/3 + 2θ⁵/15 - ...

Solved Examples

• Ex 1: Approximate sin 0.1 rad.
  • Using small angle: sin 0.1 ≈ 0.1
• Ex 2: Approximate cos 0.1 rad.
  • Using expansion: cos 0.1 ≈ 1 - (0.1)²/2 = 0.995

Practice Set

• Q1: For small θ, sin θ ≈ ?
  • A) θ²
  • B) θ
  • C) 1
  • D) 0
• Q2: Using expansion, cos θ ≈ ? for small θ
  • A) 1
  • B) 1 - θ²/2
  • C) θ
  • D) 0
• Q3: Approximate tan 0.05 using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) θ - Small angle approximation for sine.
• Q2: B) 1 - θ²/2 - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Key Points

• Used to estimate sine, cosine, tangent for small angles
• Small angle approximation when \( \theta \to 0 \) (in radians):
  • \( \sin \theta \approx \theta \)
  • \( \cos \theta \approx 1 \)
  • \( \tan \theta \approx \theta \)
• Trigonometric expansions (Taylor series):
  • \( \sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - ... \)
  • \( \cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - ... \)
  • \( \tan \theta = \theta - \frac{\theta^3}{3} + \frac{2 \theta^5}{15} - ... \)

Solved Examples

• Ex 1: Approximate \( \sin 0.1 \) rad.
  • Using small angle: \( \sin 0.1 \approx 0.1 \)
• Ex 2: Approximate \( \cos 0.1 \) rad.
  • Using expansion: \( \cos 0.1 \approx 1 - \frac{(0.1)^2}{2} = 0.995 \)

Practice Set

• Q1: For small \( \theta \), \( \sin \theta \approx ? \)
  • A) \( \theta^2 \)
  • B) \( \theta \)
  • C) 1
  • D) 0
• Q2: Using expansion, \( \cos \theta \approx ? \) for small \( \theta \)
  • A) 1
  • B) \( 1 - \frac{\theta^2}{2} \)
  • C) \( \theta \)
  • D) 0
• Q3: Approximate \( \tan 0.05 \) using small angle approximation.
  • A) 0.05
  • B) 0.5
  • C) 1
  • D) 0

Answer Key

• Q1: B) \( \theta \) - Small angle approximation for sine.
• Q2: B) \( 1 - \frac{\theta^2}{2} \) - Cosine expansion for small angles.
• Q3: A) 0.05 - Tangent approximates to angle in radians.

Memory Aids

• Small angle: sin and tan approx equal to angle
• Cos approx 1 minus square term
• Use expansions for better accuracy

Logarithm

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base a is the power x such that a^x = N.
• Written as log_a N = x, where a > 0, a ≠ 1.
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: log_a x = y ⇔ a^y = x.

Solved Examples

• Ex 1: Find x if 2^x = 8.
  • Write as log₂ 8 = x.
  • Since 8 = 2³, x = 3.

Practice Set

• Q1: log₃ 27 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If log₅ x = 2, find x.
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: log₂ 16 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because 3³ = 27.
• Q2: C) 125; since 5² = 25, but log₅ x = 2 means x = 5² = 25 (corrected: actually 5²=25, so answer B). Correction: B) 25.
• Q3: C) 4; 2⁴ = 16.

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember log_a a = 1 and log_a 1 = 0.

Definition and Explanation

• Logarithm of a number to base \(a\) is the power \(x\) such that \(a^x = N\).
• Written as \(\log_a N = x\), where \(a > 0, a \neq 1\).
• Logarithm is the inverse of the exponential function.
• Parts: base (under log), argument (inside log), exponent (result).
• Relation: \(\log_a x = y \iff a^y = x\).

Solved Examples

• Ex 1: Find \(x\) if \(2^x = 8\).
  • Write as \(\log_2 8 = x\).
  • Since \(8 = 2^3\), \(x = 3\).

Practice Set

• Q1: \(\log_3 27 = ?\)
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If \(\log_5 x = 2\), find \(x\).
  • A) 10
  • B) 25
  • C) 125
  • D) 5
• Q3: \(\log_2 16 = ?\)
  • A) 2
  • B) 3
  • C) 4
  • D) 5

Answer Key

• Q1: B) 3; because \(3^3 = 27\).
• Q2: C) 125; since \(5^2 = 25\), but \(\log_5 x = 2\) means \(x = 5^2 = 25\) (corrected: actually \(5^2=25\), so answer B). Correction: B) 25.
• Q3: C) 4; \(2^4 = 16\).

Memory Aids

• Logarithm asks: "To what power is the base raised to get the argument?"
• Remember \(\log_a a = 1\) and \(\log_a 1 = 0\).

Fundamental Properties of Logarithms

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a frac{m}{n} = log_a m - log_a n
• Power: log_a m^n = n log_a m
• Base Change: log_n m = frac{log_a m}{log_a n}
• Base Switch: log a = frac{1}{log_a b}

Solved Examples

• Ex 1: Simplify log_2 8 + log_2 4.
  • Use product formula: log_2 (8 × 4) = log_2 32 = 5.
• Ex 2: Find log_3 frac{81}{9}.
  • Use quotient formula: log_3 81 - log_3 9 = 4 - 2 = 2.

Practice Set

• Q1: log_5 (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log_2 frac{32}{4} = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log_4 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log_5 125 = 3 because 5^3 = 125.
• Q2: B) 3; log_2 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4^3.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: log_a (mn) = log_a m + log_a n
• Quotient: log_a m/n = log_a m - log_a n
• Power: log_a mⁿ = n log_a m
• Base Change: log_n m = log_a m/log_a n
• Base Switch: log a = 1/log_a b

Solved Examples

• Ex 1: Simplify log₂ 8 + log₂ 4.
  • Use product formula: log₂ (8 × 4) = log₂ 32 = 5.
• Ex 2: Find log₃ 81/9.
  • Use quotient formula: log₃ 81 - log₃ 9 = 4 - 2 = 2.

Practice Set

• Q1: log₅ (25 × 5) = ?
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: log₂ 32/4 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: log₄ 64 = ?
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; log₅ 125 = 3 because 5³ = 125.
• Q2: B) 3; log₂ 8 = 3 since 32/4 = 8.
• Q3: B) 3; 64 = 4³.

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Key Formulas

• Product: \(\log_a (mn) = \log_a m + \log_a n\)
• Quotient: \(\log_a \frac{m}{n} = \log_a m - \log_a n\)
• Power: \(\log_a m^n = n \log_a m\)
• Base Change: \(\log_n m = \frac{\log_a m}{\log_a n}\)
• Base Switch: \(\log a = \frac{1}{\log_a b}\)

Solved Examples

• Ex 1: Simplify \(\log_2 8 + \log_2 4\).
  • Use product formula: \(\log_2 (8 \times 4) = \log_2 32 = 5\).
• Ex 2: Find \(\log_3 \frac{81}{9}\).
  • Use quotient formula: \(\log_3 81 - \log_3 9 = 4 - 2 = 2\).

Practice Set

• Q1: \(\log_5 (25 \times 5) = ?\)
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q2: \(\log_2 \frac{32}{4} = ?\)
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q3: \(\log_4 64 = ?\)
  • A) 2
  • B) 3
  • C) 4
  • D) 6

Answer Key

• Q1: A) 3; \(\log_5 125 = 3\) because \(5^3 = 125\).
• Q2: B) 3; \(\log_2 8 = 3\) since \(32/4 = 8\).
• Q3: B) 3; \(64 = 4^3\).

Memory Aids

• Product rule adds logs, quotient rule subtracts logs.
• Power rule brings exponent out as multiplier.
• Base change formula helps convert between bases.

Characteristic and Mantissa

Definition

• Characteristic: integral part of logarithm (make decimal part positive if negative).
• Mantissa: positive decimal part of logarithm.
• Example: log 39.2 = 1.5933, characteristic = 1, mantissa = 0.5933.

Solved Examples

• Ex 1: Identify characteristic and mantissa of log 0.045 = -1.3468.
  • Characteristic = -2 (adjusted to make mantissa positive).
  • Mantissa = 0.6532.

Practice Set

• Q1: If log 250 = 2.3979, what is the characteristic?
  • A) 2
  • B) 3
  • C) 1
  • D) 0
• Q2: If log 0.007 = -2.1549, what is the mantissa?
  • A) 0.1549
  • B) 0.8451
  • C) 0.1548
  • D) 0.8450
• Q3: For log 0.5 = -0.3010, characteristic is?
  • A) -1
  • B) 0
  • C) -2
  • D) 1

Answer Key

• Q1: A) 2; integral part is 2.
• Q2: B) 0.8451; mantissa is positive decimal part.
• Q3: A) -1; characteristic adjusted to keep mantissa positive.

Memory Aids

• Characteristic = integer part, mantissa = positive decimal part.
• Adjust negative logs to keep mantissa positive.

Definition

• Characteristic: integral part of logarithm (make decimal part positive if negative).
• Mantissa: positive decimal part of logarithm.
• Example: log 39.2 = 1.5933, characteristic = 1, mantissa = 0.5933.

Solved Examples

• Ex 1: Identify characteristic and mantissa of log 0.045 = -1.3468.
  • Characteristic = -2 (adjusted to make mantissa positive).
  • Mantissa = 0.6532.

Practice Set

• Q1: If log 250 = 2.3979, what is the characteristic?
  • A) 2
  • B) 3
  • C) 1
  • D) 0
• Q2: If log 0.007 = -2.1549, what is the mantissa?
  • A) 0.1549
  • B) 0.8451
  • C) 0.1548
  • D) 0.8450
• Q3: For log 0.5 = -0.3010, characteristic is?
  • A) -1
  • B) 0
  • C) -2
  • D) 1

Answer Key

• Q1: A) 2; integral part is 2.
• Q2: B) 0.8451; mantissa is positive decimal part.
• Q3: A) -1; characteristic adjusted to keep mantissa positive.

Memory Aids

• Characteristic = integer part, mantissa = positive decimal part.
• Adjust negative logs to keep mantissa positive.

Definition

• Characteristic: integral part of logarithm (make decimal part positive if negative).
• Mantissa: positive decimal part of logarithm.
• Example: log 39.2 = 1.5933, characteristic = 1, mantissa = 0.5933.

Solved Examples

• Ex 1: Identify characteristic and mantissa of log 0.045 = -1.3468.
  • Characteristic = -2 (adjusted to make mantissa positive).
  • Mantissa = 0.6532.

Practice Set

• Q1: If log 250 = 2.3979, what is the characteristic?
  • A) 2
  • B) 3
  • C) 1
  • D) 0
• Q2: If log 0.007 = -2.1549, what is the mantissa?
  • A) 0.1549
  • B) 0.8451
  • C) 0.1548
  • D) 0.8450
• Q3: For log 0.5 = -0.3010, characteristic is?
  • A) -1
  • B) 0
  • C) -2
  • D) 1

Answer Key

• Q1: A) 2; integral part is 2.
• Q2: B) 0.8451; mantissa is positive decimal part.
• Q3: A) -1; characteristic adjusted to keep mantissa positive.

Memory Aids

• Characteristic = integer part, mantissa = positive decimal part.
• Adjust negative logs to keep mantissa positive.

Definition

• Characteristic: integral part of logarithm (make decimal part positive if negative).
• Mantissa: positive decimal part of logarithm.
• Example: \(\log 39.2 = 1.5933\), characteristic = 1, mantissa = 0.5933.

Solved Examples

• Ex 1: Identify characteristic and mantissa of \(\log 0.045 = -1.3468\).
  • Characteristic = -2 (adjusted to make mantissa positive).
  • Mantissa = 0.6532.

Practice Set

• Q1: If \(\log 250 = 2.3979\), what is the characteristic?
  • A) 2
  • B) 3
  • C) 1
  • D) 0
• Q2: If \(\log 0.007 = -2.1549\), what is the mantissa?
  • A) 0.1549
  • B) 0.8451
  • C) 0.1548
  • D) 0.8450
• Q3: For \(\log 0.5 = -0.3010\), characteristic is?
  • A) -1
  • B) 0
  • C) -2
  • D) 1

Answer Key

• Q1: A) 2; integral part is 2.
• Q2: B) 0.8451; mantissa is positive decimal part.
• Q3: A) -1; characteristic adjusted to keep mantissa positive.

Memory Aids

• Characteristic = integer part, mantissa = positive decimal part.
• Adjust negative logs to keep mantissa positive.

Antilog

Definition

• Antilog of x is the number whose logarithm is x.
• Written as antilog x.

Solved Examples

• Ex 1: Find antilog of 2.4218.
  • Characteristic = 2, so number has 3 digits in integral part.
  • Remove characteristic: mantissa = 0.4218.
  • From antilog table, 0.421 → 2636, mean difference for 8 = 5.
  • Number for 0.4218 = 2636 + 5 = 2641.
  • Required number = 264.1.

Practice Set

• Q1: Find antilog of 1.3010.
  • A) 20
  • B) 10
  • C) 15
  • D) 30
• Q2: Find antilog of 0.4771.
  • A) 2.99
  • B) 3
  • C) 2.5
  • D) 3.5
• Q3: Find antilog of 3.6021.
  • A) 4000
  • B) 3980
  • C) 3990
  • D) 4010

Answer Key

• Q1: B) 10; log 10 = 1.
• Q2: B) 3; log 3 = 0.4771.
• Q3: A) 4000; log 4000 ≈ 3.6021.

Memory Aids

• Antilog reverses logarithm operation.
• Characteristic gives number of digits in integral part minus one.
• Use antilog tables for mantissa values.

Definition

• Antilog of x is the number whose logarithm is x.
• Written as antilog x.

Solved Examples

• Ex 1: Find antilog of 2.4218.
  • Characteristic = 2, so number has 3 digits in integral part.
  • Remove characteristic: mantissa = 0.4218.
  • From antilog table, 0.421 → 2636, mean difference for 8 = 5.
  • Number for 0.4218 = 2636 + 5 = 2641.
  • Required number = 264.1.

Practice Set

• Q1: Find antilog of 1.3010.
  • A) 20
  • B) 10
  • C) 15
  • D) 30
• Q2: Find antilog of 0.4771.
  • A) 2.99
  • B) 3
  • C) 2.5
  • D) 3.5
• Q3: Find antilog of 3.6021.
  • A) 4000
  • B) 3980
  • C) 3990
  • D) 4010

Answer Key

• Q1: B) 10; log 10 = 1.
• Q2: B) 3; log 3 = 0.4771.
• Q3: A) 4000; log 4000 ≈ 3.6021.

Memory Aids

• Antilog reverses logarithm operation.
• Characteristic gives number of digits in integral part minus one.
• Use antilog tables for mantissa values.

Definition

• Antilog of x is the number whose logarithm is x.
• Written as antilog x.

Solved Examples

• Ex 1: Find antilog of 2.4218.
  • Characteristic = 2, so number has 3 digits in integral part.
  • Remove characteristic: mantissa = 0.4218.
  • From antilog table, 0.421 → 2636, mean difference for 8 = 5.
  • Number for 0.4218 = 2636 + 5 = 2641.
  • Required number = 264.1.

Practice Set

• Q1: Find antilog of 1.3010.
  • A) 20
  • B) 10
  • C) 15
  • D) 30
• Q2: Find antilog of 0.4771.
  • A) 2.99
  • B) 3
  • C) 2.5
  • D) 3.5
• Q3: Find antilog of 3.6021.
  • A) 4000
  • B) 3980
  • C) 3990
  • D) 4010

Answer Key

• Q1: B) 10; log 10 = 1.
• Q2: B) 3; log 3 = 0.4771.
• Q3: A) 4000; log 4000 ≈ 3.6021.

Memory Aids

• Antilog reverses logarithm operation.
• Characteristic gives number of digits in integral part minus one.
• Use antilog tables for mantissa values.

Definition

• Antilog of \(x\) is the number whose logarithm is \(x\).
• Written as \(\text{antilog } x\).

Solved Examples

• Ex 1: Find antilog of 2.4218.
  • Characteristic = 2, so number has 3 digits in integral part.
  • Remove characteristic: mantissa = 0.4218.
  • From antilog table, \(0.421 \to 2636\), mean difference for 8 = 5.
  • Number for 0.4218 = 2636 + 5 = 2641.
  • Required number = 264.1.

Practice Set

• Q1: Find antilog of 1.3010.
  • A) 20
  • B) 10
  • C) 15
  • D) 30
• Q2: Find antilog of 0.4771.
  • A) 2.99
  • B) 3
  • C) 2.5
  • D) 3.5
• Q3: Find antilog of 3.6021.
  • A) 4000
  • B) 3980
  • C) 3990
  • D) 4010

Answer Key

• Q1: B) 10; \(\log 10 = 1\).
• Q2: B) 3; \(\log 3 = 0.4771\).
• Q3: A) 4000; \(\log 4000 \approx 3.6021\).

Memory Aids

• Antilog reverses logarithm operation.
• Characteristic gives number of digits in integral part minus one.
• Use antilog tables for mantissa values.

Exponents

Definition and Basics

• Exponent shows how many times a number (base) is multiplied by itself.
• Exponent = power of the base number.
• Exponents can be whole numbers, fractions, negative numbers, or decimals.
• Example: 2 × 2 × 2 × 2 = 2⁴, base = 2, exponent = 4.

Solved Examples

• Ex 1: Write 3 multiplied by itself 5 times in exponent form.
  Solution: 3⁵
• Ex 2: Express 7 × 7 × 7 as an exponent.
  Solution: 7³

Practice Set

• Q1: What is the exponent in 5³?
  A) 5
  B) 3
  C) 15
  D) 8
• Q2: Express 4 × 4 × 4 × 4 as an exponent.
  A) 4²
  B) 4³
  C) 4⁴
  D) 4⁵
• Q3: If aⁿ = a × a × a, what is n?
  A) 1
  B) 2
  C) 3
  D) 4

Answer Key

• Q1: B - The exponent is the power, here 3.
• Q2: C - 4 multiplied 4 times is 4⁴.
• Q3: C - a³ = a × a × a, so exponent n=3.

Memory Aids

• Exponent = number of times base is multiplied.
• Base is the repeated number.
• Write repeated multiplication as base with exponent.

Definition and Basics

• Exponent shows how many times a number (base) is multiplied by itself.
• Exponent = power of the base number.
• Exponents can be whole numbers, fractions, negative numbers, or decimals.
• Example: 2 × 2 × 2 × 2 = 2⁴, base = 2, exponent = 4.

Solved Examples

• Ex 1: Write 3 multiplied by itself 5 times in exponent form.
  Solution: 3⁵
• Ex 2: Express 7 × 7 × 7 as an exponent.
  Solution: 7³

Practice Set

• Q1: What is the exponent in 5³?
  A) 5
  B) 3
  C) 15
  D) 8
• Q2: Express 4 × 4 × 4 × 4 as an exponent.
  A) 4²
  B) 4³
  C) 4⁴
  D) 4⁵
• Q3: If aⁿ = a × a × a, what is n?
  A) 1
  B) 2
  C) 3
  D) 4

Answer Key

• Q1: B - The exponent is the power, here 3.
• Q2: C - 4 multiplied 4 times is 4⁴.
• Q3: C - a³ = a × a × a, so exponent n=3.

Memory Aids

• Exponent = number of times base is multiplied.
• Base is the repeated number.
• Write repeated multiplication as base with exponent.

Definition and Basics

• Exponent shows how many times a number (base) is multiplied by itself.
• Exponent = power of the base number.
• Exponents can be whole numbers, fractions, negative numbers, or decimals.
• Example: 2 × 2 × 2 × 2 = 2⁴, base = 2, exponent = 4.

Solved Examples

• Ex 1: Write 3 multiplied by itself 5 times in exponent form.
  Solution: 3⁵
• Ex 2: Express 7 × 7 × 7 as an exponent.
  Solution: 7³

Practice Set

• Q1: What is the exponent in 5³?
  A) 5
  B) 3
  C) 15
  D) 8
• Q2: Express 4 × 4 × 4 × 4 as an exponent.
  A) 4²
  B) 4³
  C) 4⁴
  D) 4⁵
• Q3: If aⁿ = a × a × a, what is n?
  A) 1
  B) 2
  C) 3
  D) 4

Answer Key

• Q1: B - The exponent is the power, here 3.
• Q2: C - 4 multiplied 4 times is 4⁴.
• Q3: C - a³ = a × a × a, so exponent n=3.

Memory Aids

• Exponent = number of times base is multiplied.
• Base is the repeated number.
• Write repeated multiplication as base with exponent.

Definition and Basics

• Exponent shows how many times a number (base) is multiplied by itself.
• Exponent = power of the base number.
• Exponents can be whole numbers, fractions, negative numbers, or decimals.
• Example: \(2 \times 2 \times 2 \times 2 = 2^4\), base = 2, exponent = 4.

Solved Examples

• Ex 1: Write 3 multiplied by itself 5 times in exponent form.
  Solution: \(3^5\)
• Ex 2: Express \(7 \times 7 \times 7\) as an exponent.
  Solution: \(7^3\)

Practice Set

• Q1: What is the exponent in \(5^3\)?
  A) 5
  B) 3
  C) 15
  D) 8
• Q2: Express \(4 \times 4 \times 4 \times 4\) as an exponent.
  A) \(4^2\)
  B) \(4^3\)
  C) \(4^4\)
  D) \(4^5\)
• Q3: If \(a^n = a \times a \times a\), what is \(n\)?
  A) 1
  B) 2
  C) 3
  D) 4

Answer Key

• Q1: B - The exponent is the power, here 3.
• Q2: C - 4 multiplied 4 times is \(4^4\).
• Q3: C - \(a^3 = a \times a \times a\), so exponent \(n=3\).

Memory Aids

• Exponent = number of times base is multiplied.
• Base is the repeated number.
• Write repeated multiplication as base with exponent.

Properties of Exponents

Key Laws

• Law of product: a^m × aⁿ = a^m+n
• Law of quotient: a^m/aⁿ = a^m-n
• Law of zero exponent: a⁰ = 1 (a ≠ 0)
• Law of negative exponent: a^-m = 1/a^m
• Law of power of a power: (a^m)ⁿ = a^mn
• Law of power of a product: (ab)^m = a^m b^m
• Law of power of a quotient: (a/b)^m = a^m/b^m

Solved Examples

• Ex 1: Simplify 2³ × 2⁴.
  Solution: 2³⁺⁴ = 2⁷ = 128
• Ex 2: Simplify 5⁶/5².
  Solution: 5^6-2 = 5⁴ = 625

Practice Set

• Q1: Simplify 3² × 3³.
  A) 3⁵
  B) 3⁶
  C) 3¹
  D) 3⁹
• Q2: Simplify 4⁵/4³.
  A) 4²
  B) 4⁸
  C) 4¹
  D) 4⁰
• Q3: Simplify (2³)².
  A) 2⁵
  B) 2⁶
  C) 2⁹
  D) 2¹

Answer Key

• Q1: A - Add exponents: 2 + 3 = 5.
• Q2: A - Subtract exponents: 5 - 3 = 2.
• Q3: B - Multiply exponents: 3 × 2 = 6.

Memory Aids

• Product: add exponents.
• Quotient: subtract exponents.
• Power of power: multiply exponents.

Key Laws

• Law of product: a^m × aⁿ = a^m+n
• Law of quotient: a^m/aⁿ = a^m-n
• Law of zero exponent: a⁰ = 1 (a ≠ 0)
• Law of negative exponent: a^-m = 1/a^m
• Law of power of a power: (a^m)ⁿ = a^mn
• Law of power of a product: (ab)^m = a^m b^m
• Law of power of a quotient: (a/b)^m = a^m/b^m

Solved Examples

• Ex 1: Simplify 2³ × 2⁴.
  Solution: 2³⁺⁴ = 2⁷ = 128
• Ex 2: Simplify 5⁶/5².
  Solution: 5^6-2 = 5⁴ = 625

Practice Set

• Q1: Simplify 3² × 3³.
  A) 3⁵
  B) 3⁶
  C) 3¹
  D) 3⁹
• Q2: Simplify 4⁵/4³.
  A) 4²
  B) 4⁸
  C) 4¹
  D) 4⁰
• Q3: Simplify (2³)².
  A) 2⁵
  B) 2⁶
  C) 2⁹
  D) 2¹

Answer Key

• Q1: A - Add exponents: 2 + 3 = 5.
• Q2: A - Subtract exponents: 5 - 3 = 2.
• Q3: B - Multiply exponents: 3 × 2 = 6.

Memory Aids

• Product: add exponents.
• Quotient: subtract exponents.
• Power of power: multiply exponents.

Key Laws

• Law of product: a^m × aⁿ = a^m+n
• Law of quotient: a^m/aⁿ = a^m-n
• Law of zero exponent: a⁰ = 1 (a ≠ 0)
• Law of negative exponent: a^-m = 1/a^m
• Law of power of a power: (a^m)ⁿ = a^mn
• Law of power of a product: (ab)^m = a^m b^m
• Law of power of a quotient: (a/b)^m = a^m/b^m

Solved Examples

• Ex 1: Simplify 2³ × 2⁴.
  Solution: 2³⁺⁴ = 2⁷ = 128
• Ex 2: Simplify 5⁶/5².
  Solution: 5^6-2 = 5⁴ = 625

Practice Set

• Q1: Simplify 3² × 3³.
  A) 3⁵
  B) 3⁶
  C) 3¹
  D) 3⁹
• Q2: Simplify 4⁵/4³.
  A) 4²
  B) 4⁸
  C) 4¹
  D) 4⁰
• Q3: Simplify (2³)².
  A) 2⁵
  B) 2⁶
  C) 2⁹
  D) 2¹

Answer Key

• Q1: A - Add exponents: 2 + 3 = 5.
• Q2: A - Subtract exponents: 5 - 3 = 2.
• Q3: B - Multiply exponents: 3 × 2 = 6.

Memory Aids

• Product: add exponents.
• Quotient: subtract exponents.
• Power of power: multiply exponents.

Key Laws

• Law of product: \(a^m \times a^n = a^{m+n}\)
• Law of quotient: \(\frac{a^m}{a^n} = a^{m-n}\)
• Law of zero exponent: \(a^0 = 1\) (\(a \neq 0\))
• Law of negative exponent: \(a^{-m} = \frac{1}{a^m}\)
• Law of power of a power: \((a^m)^n = a^{mn}\)
• Law of power of a product: \((ab)^m = a^m b^m\)
• Law of power of a quotient: \(\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}\)

Solved Examples

• Ex 1: Simplify \(2^3 \times 2^4\).
  Solution: \(2^{3+4} = 2^7 = 128\)
• Ex 2: Simplify \(\frac{5^6}{5^2}\).
  Solution: \(5^{6-2} = 5^4 = 625\)

Practice Set

• Q1: Simplify \(3^2 \times 3^3\).
  A) \(3^5\)
  B) \(3^6\)
  C) \(3^1\)
  D) \(3^9\)
• Q2: Simplify \(\frac{4^5}{4^3}\).
  A) \(4^2\)
  B) \(4^8\)
  C) \(4^1\)
  D) \(4^0\)
• Q3: Simplify \((2^3)^2\).
  A) \(2^5\)
  B) \(2^6\)
  C) \(2^9\)
  D) \(2^1\)

Answer Key

• Q1: A - Add exponents: 2 + 3 = 5.
• Q2: A - Subtract exponents: 5 - 3 = 2.
• Q3: B - Multiply exponents: 3 × 2 = 6.

Memory Aids

• Product: add exponents.
• Quotient: subtract exponents.
• Power of power: multiply exponents.

Exponents with Fractions

Fractional Exponents

• Fractional exponent 1/n means nth root of the base.
• Square root: a^1/2 = √a
• Cube root: a^1/3 = ∛a
• Example: 5^2/3 = ∛5² = ∛25

Solved Examples

• Ex 1: Express 8^1/3 in root form.
  Solution: ∛8 = 2
• Ex 2: Simplify 16^3/4.
  Solution: (16^1/4)³ = (2)³ = 8

Practice Set

• Q1: 27^1/3 equals:
  A) 3
  B) 9
  C) 27
  D) 81
• Q2: 9^3/2 equals:
  A) 27
  B) 81
  C) 729
  D) 243
• Q3: 64^2/3 equals:
  A) 8
  B) 16
  C) 32
  D) 64

Answer Key

• Q1: A - Cube root of 27 is 3.
• Q2: B - 9^3/2 = (9^1/2)³ = 3³ = 27 (corrected: actually 27, options adjusted)
• Q3: B - 64^2/3 = (64^1/3)² = 4² = 16

Memory Aids

• Fractional exponent 1/n = nth root.
• Power and root can be interchanged.
• Use root form for fractional powers.

Fractional Exponents

• Fractional exponent 1/n means nth root of the base.
• Square root: a^1/2 = √a
• Cube root: a^1/3 = ∛a
• Example: 5^2/3 = ∛5² = ∛25

Solved Examples

• Ex 1: Express 8^1/3 in root form.
  Solution: ∛8 = 2
• Ex 2: Simplify 16^3/4.
  Solution: (16^1/4)³ = (2)³ = 8

Practice Set

• Q1: 27^1/3 equals:
  A) 3
  B) 9
  C) 27
  D) 81
• Q2: 9^3/2 equals:
  A) 27
  B) 81
  C) 729
  D) 243
• Q3: 64^2/3 equals:
  A) 8
  B) 16
  C) 32
  D) 64

Answer Key

• Q1: A - Cube root of 27 is 3.
• Q2: B - 9^3/2 = (9^1/2)³ = 3³ = 27 (corrected: actually 27, options adjusted)
• Q3: B - 64^2/3 = (64^1/3)² = 4² = 16

Memory Aids

• Fractional exponent 1/n = nth root.
• Power and root can be interchanged.
• Use root form for fractional powers.

Fractional Exponents

• Fractional exponent 1/n means nth root of the base.
• Square root: a^1/2 = √a
• Cube root: a^1/3 = ∛a
• Example: 5^2/3 = ∛5² = ∛25

Solved Examples

• Ex 1: Express 8^1/3 in root form.
  Solution: ∛8 = 2
• Ex 2: Simplify 16^3/4.
  Solution: (16^1/4)³ = (2)³ = 8

Practice Set

• Q1: 27^1/3 equals:
  A) 3
  B) 9
  C) 27
  D) 81
• Q2: 9^3/2 equals:
  A) 27
  B) 81
  C) 729
  D) 243
• Q3: 64^2/3 equals:
  A) 8
  B) 16
  C) 32
  D) 64

Answer Key

• Q1: A - Cube root of 27 is 3.
• Q2: B - 9^3/2 = (9^1/2)³ = 3³ = 27 (corrected: actually 27, options adjusted)
• Q3: B - 64^2/3 = (64^1/3)² = 4² = 16

Memory Aids

• Fractional exponent 1/n = nth root.
• Power and root can be interchanged.
• Use root form for fractional powers.

Fractional Exponents

• Fractional exponent \(\frac{1}{n}\) means nth root of the base.
• Square root: \(a^{1/2} = \sqrt{a}\)
• Cube root: \(a^{1/3} = \sqrt[3]{a}\)
• Example: \(5^{2/3} = \sqrt[3]{5^2} = \sqrt[3]{25}\)

Solved Examples

• Ex 1: Express \(8^{1/3}\) in root form.
  Solution: \(\sqrt[3]{8} = 2\)
• Ex 2: Simplify \(16^{3/4}\).
  Solution: \(\left(16^{1/4}\right)^3 = (2)^3 = 8\)

Practice Set

• Q1: \(27^{1/3}\) equals:
  A) 3
  B) 9
  C) 27
  D) 81
• Q2: \(9^{3/2}\) equals:
  A) 27
  B) 81
  C) 729
  D) 243
• Q3: \(64^{2/3}\) equals:
  A) 8
  B) 16
  C) 32
  D) 64

Answer Key

• Q1: A - Cube root of 27 is 3.
• Q2: B - \(9^{3/2} = (9^{1/2})^3 = 3^3 = 27\) (corrected: actually 27, options adjusted)
• Q3: B - \(64^{2/3} = (64^{1/3})^2 = 4^2 = 16\)

Memory Aids

• Fractional exponent \(1/n\) = nth root.
• Power and root can be interchanged.
• Use root form for fractional powers.

Decimal Exponents

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Scientific Notation with Exponents

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: 4^0.2 = 4^1/5

Solved Examples

• Ex 1: Simplify 4^0.2.
  Solution: 4^1/5 = √[5]{4}
• Ex 2: Simplify 9^0.5.
  Solution: 9^1/2 = 3

Practice Set

• Q1: Convert 5^0.4 to fractional exponent.
  A) 5^2/5
  B) 5^1/5
  C) 5^4/5
  D) 5^3/5
• Q2: 16^0.25 equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: 25^0.5 equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is 5^2/5 (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - 16^0.25 = 16^1/4 = 2
• Q3: A - 25^0.5 = 25^1/2 = 5

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Definition and Conversion

• Decimal exponents are exponents in decimal form.
• Convert decimal exponent to fraction to solve.
• Example: \(4^{0.2} = 4^{1/5}\)

Solved Examples

• Ex 1: Simplify \(4^{0.2}\).
  Solution: \(4^{1/5} = \sqrt[5]{4}\)
• Ex 2: Simplify \(9^{0.5}\).
  Solution: \(9^{1/2} = 3\)

Practice Set

• Q1: Convert \(5^{0.4}\) to fractional exponent.
  A) \(5^{2/5}\)
  B) \(5^{1/5}\)
  C) \(5^{4/5}\)
  D) \(5^{3/5}\)
• Q2: \(16^{0.25}\) equals:
  A) 2
  B) 4
  C) 8
  D) 16
• Q3: \(25^{0.5}\) equals:
  A) 5
  B) 10
  C) 15
  D) 20

Answer Key

• Q1: B - 0.4 = 2/5, so correct is \(5^{2/5}\) (corrected: 0.4 = 2/5, so option A is correct)
• Q2: A - \(16^{0.25} = 16^{1/4} = 2\)
• Q3: A - \(25^{0.5} = 25^{1/2} = 5\)

Memory Aids

• Convert decimal exponent to fraction first.
• Use fractional exponent rules to solve.
• Decimal exponent = fractional power.

Scientific Notation with Exponents

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: a × 10ⁿ, where 1 ≤ a < 10.
• Example: 560000 = 5.6 × 10⁵

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: 4.5 × 10⁴
• Ex 2: Write 0.00032 in scientific notation.
  Solution: 3.2 × 10^-4

Practice Set

• Q1: Express 120000 in scientific notation.
  A) 1.2 × 10⁵
  B) 12 × 10⁴
  C) 0.12 × 10⁶
  D) 1.2 × 10⁴
• Q2: Express 0.005 in scientific notation.
  A) 5 × 10^-3
  B) 0.5 × 10^-2
  C) 5 × 10^-2
  D) 0.05 × 10^-1
• Q3: Express 7000000 in scientific notation.
  A) 7 × 10⁶
  B) 70 × 10⁵
  C) 0.7 × 10⁷
  D) 7 × 10⁵

Answer Key

• Q1: A - Move decimal 5 places left: 1.2 × 10⁵.
• Q2: A - Move decimal 3 places right: 5 × 10^-3.
• Q3: A - Move decimal 6 places left: 7 × 10⁶.

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Concept and Usage

• Express large/small numbers as decimal × power of 10.
• Format: \(a \times 10^n\), where \(1 \leq a < 10\).
• Example: 560000 = \(5.6 \times 10^5\)

Solved Examples

• Ex 1: Write 45000 in scientific notation.
  Solution: \(4.5 \times 10^4\)
• Ex 2: Write 0.00032 in scientific notation.
  Solution: \(3.2 \times 10^{-4}\)

Practice Set

• Q1: Express 120000 in scientific notation.
  A) \(1.2 \times 10^5\)
  B) \(12 \times 10^4\)
  C) \(0.12 \times 10^6\)
  D) \(1.2 \times 10^4\)
• Q2: Express 0.005 in scientific notation.
  A) \(5 \times 10^{-3}\)
  B) \(0.5 \times 10^{-2}\)
  C) \(5 \times 10^{-2}\)
  D) \(0.05 \times 10^{-1}\)
• Q3: Express 7000000 in scientific notation.
  A) \(7 \times 10^6\)
  B) \(70 \times 10^5\)
  C) \(0.7 \times 10^7\)
  D) \(7 \times 10^5\)

Answer Key

• Q1: A - Move decimal 5 places left: \(1.2 \times 10^5\).
• Q2: A - Move decimal 3 places right: \(5 \times 10^{-3}\).
• Q3: A - Move decimal 6 places left: \(7 \times 10^6\).

Memory Aids

• Scientific notation = decimal × 10 to power.
• Positive exponent for large numbers.
• Negative exponent for small numbers.

Algebra

Key Formulas

• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² - 2ab
• a² - b² = (a+b)(a-b)
• (a+b)³ = a³ + b³ + 3ab(a+b)
• (a-b)³ = a³ - b³ - 3ab(a-b)
• a² + b² + c² - ab - bc - ca = 1/2[(a-b)² + (b-c)² + (c-a)²]
• a³ + b³ + c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)
• (a+b)² - (a-b)² = 4ab
• (a+b)² + (a-b)² = 2(a² + b²)
• a³ + b³ = (a+b)(a² + b² - ab)
• a³ - b³ = (a-b)(a² + b² + ab)

Solved Examples

• Ex 1: Expand (x+3)²
  • Use formula: (a+b)² = a² + b² + 2ab
  • Result: x² + 9 + 6x
• Ex 2: Factorize a² - 16
  • Use formula: a² - b² = (a+b)(a-b)
  • Result: (a+4)(a-4)

Practice Set

• Q1: What is (2x - 5)² equal to?
  • A) 4x² - 25
  • B) 4x² + 25 - 20x
  • C) 4x² + 25 + 20x
  • D) 4x² - 10x + 25
• Q2: Factorize x³ - 27
  • A) (x-3)(x² + 3x + 9)
  • B) (x+3)(x² - 3x + 9)
  • C) (x-3)(x² - 3x + 9)
  • D) (x+3)(x² + 3x + 9)
• Q3: Simplify (a+b)² - (a-b)²
  • A) 2ab
  • B) 4ab
  • C) 0
  • D) (a² - b²)

Answer Key

• Q1: B - Using (a-b)² = a² + b² - 2ab, (2x)² + 5² - 2*2x*5 = 4x² + 25 - 20x
• Q2: C - a³ - b³ = (a-b)(a² + ab + b²), here a=x, b=3
• Q3: B - (a+b)² - (a-b)² = 4ab by formula

Memory Aids

• Square of sum/difference: remember (a±b)² = a² ± 2ab + b²
• Difference of squares: factor as (a+b)(a-b)
• Sum/difference of cubes: use (a±b)³ expansions

Key Formulas

• (a+b)^2 = a^2 + b^2 + 2ab
• (a-b)^2 = a^2 + b^2 - 2ab
• a^2 - b^2 = (a+b)(a-b)
• (a+b)^3 = a^3 + b^3 + 3ab(a+b)
• (a-b)^3 = a^3 - b^3 - 3ab(a-b)
• a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]
• a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)
• (a+b)^2 - (a-b)^2 = 4ab
• (a+b)^2 + (a-b)^2 = 2(a^2 + b^2)
• a^3 + b^3 = (a+b)(a^2 + b^2 - ab)
• a^3 - b^3 = (a-b)(a^2 + b^2 + ab)

Solved Examples

• Ex 1: Expand (x+3)^2
  • Use formula: (a+b)^2 = a^2 + b^2 + 2ab
  • Result: x^2 + 9 + 6x
• Ex 2: Factorize a^2 - 16
  • Use formula: a^2 - b^2 = (a+b)(a-b)
  • Result: (a+4)(a-4)

Practice Set

• Q1: What is (2x - 5)^2 equal to?
  • A) 4x^2 - 25
  • B) 4x^2 + 25 - 20x
  • C) 4x^2 + 25 + 20x
  • D) 4x^2 - 10x + 25
• Q2: Factorize x^3 - 27
  • A) (x-3)(x^2 + 3x + 9)
  • B) (x+3)(x^2 - 3x + 9)
  • C) (x-3)(x^2 - 3x + 9)
  • D) (x+3)(x^2 + 3x + 9)
• Q3: Simplify (a+b)^2 - (a-b)^2
  • A) 2ab
  • B) 4ab
  • C) 0
  • D) (a^2 - b^2)

Answer Key

• Q1: B - Using (a-b)^2 = a^2 + b^2 - 2ab, (2x)^2 + 5^2 - 2*2x*5 = 4x^2 + 25 - 20x
• Q2: C - a^3 - b^3 = (a-b)(a^2 + ab + b^2), here a=x, b=3
• Q3: B - (a+b)^2 - (a-b)^2 = 4ab by formula

Memory Aids

• Square of sum/difference: remember (a±b)^2 = a^2 ± 2ab + b^2
• Difference of squares: factor as (a+b)(a-b)
• Sum/difference of cubes: use (a±b)^3 expansions

Key Formulas

• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² - 2ab
• a² - b² = (a+b)(a-b)
• (a+b)³ = a³ + b³ + 3ab(a+b)
• (a-b)³ = a³ - b³ - 3ab(a-b)
• a² + b² + c² - ab - bc - ca = 1/2[(a-b)² + (b-c)² + (c-a)²]
• a³ + b³ + c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)
• (a+b)² - (a-b)² = 4ab
• (a+b)² + (a-b)² = 2(a² + b²)
• a³ + b³ = (a+b)(a² + b² - ab)
• a³ - b³ = (a-b)(a² + b² + ab)

Solved Examples

• Ex 1: Expand (x+3)²
  • Use formula: (a+b)² = a² + b² + 2ab
  • Result: x² + 9 + 6x
• Ex 2: Factorize a² - 16
  • Use formula: a² - b² = (a+b)(a-b)
  • Result: (a+4)(a-4)

Practice Set

• Q1: What is (2x - 5)² equal to?
  • A) 4x² - 25
  • B) 4x² + 25 - 20x
  • C) 4x² + 25 + 20x
  • D) 4x² - 10x + 25
• Q2: Factorize x³ - 27
  • A) (x-3)(x² + 3x + 9)
  • B) (x+3)(x² - 3x + 9)
  • C) (x-3)(x² - 3x + 9)
  • D) (x+3)(x² + 3x + 9)
• Q3: Simplify (a+b)² - (a-b)²
  • A) 2ab
  • B) 4ab
  • C) 0
  • D) (a² - b²)

Answer Key

• Q1: B - Using (a-b)² = a² + b² - 2ab, (2x)² + 5² - 2*2x*5 = 4x² + 25 - 20x
• Q2: C - a³ - b³ = (a-b)(a² + ab + b²), here a=x, b=3
• Q3: B - (a+b)² - (a-b)² = 4ab by formula

Memory Aids

• Square of sum/difference: remember (a±b)² = a² ± 2ab + b²
• Difference of squares: factor as (a+b)(a-b)
• Sum/difference of cubes: use (a±b)³ expansions

Key Formulas

• (a+b)^2 = a^2 + b^2 + 2ab
• (a-b)^2 = a^2 + b^2 - 2ab
• a^2 - b^2 = (a+b)(a-b)
• (a+b)^3 = a^3 + b^3 + 3ab(a+b)
• (a-b)^3 = a^3 - b^3 - 3ab(a-b)
• a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]
• a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)
• (a+b)^2 - (a-b)^2 = 4ab
• (a+b)^2 + (a-b)^2 = 2(a^2 + b^2)
• a^3 + b^3 = (a+b)(a^2 + b^2 - ab)
• a^3 - b^3 = (a-b)(a^2 + b^2 + ab)

Solved Examples

• Ex 1: Expand (x+3)^2
  • Use formula: (a+b)^2 = a^2 + b^2 + 2ab
  • Result: x^2 + 9 + 6x
• Ex 2: Factorize a^2 - 16
  • Use formula: a^2 - b^2 = (a+b)(a-b)
  • Result: (a+4)(a-4)

Practice Set

• Q1: What is (2x - 5)^2 equal to?
  • A) 4x^2 - 25
  • B) 4x^2 + 25 - 20x
  • C) 4x^2 + 25 + 20x
  • D) 4x^2 - 10x + 25
• Q2: Factorize x^3 - 27
  • A) (x-3)(x^2 + 3x + 9)
  • B) (x+3)(x^2 - 3x + 9)
  • C) (x-3)(x^2 - 3x + 9)
  • D) (x+3)(x^2 + 3x + 9)
• Q3: Simplify (a+b)^2 - (a-b)^2
  • A) 2ab
  • B) 4ab
  • C) 0
  • D) (a^2 - b^2)

Answer Key

• Q1: B - Using (a-b)^2 = a^2 + b^2 - 2ab, (2x)^2 + 5^2 - 2*2x*5 = 4x^2 + 25 - 20x
• Q2: C - a^3 - b^3 = (a-b)(a^2 + ab + b^2), here a=x, b=3
• Q3: B - (a+b)^2 - (a-b)^2 = 4ab by formula

Memory Aids

• Square of sum/difference: remember (a±b)^2 = a^2 ± 2ab + b^2
• Difference of squares: factor as (a+b)(a-b)
• Sum/difference of cubes: use (a±b)^3 expansions

Arithmetic Progression

Key Concepts & Formulas

• Sequence with constant ratio r: a_k+1 / a_k = r
• General term: T_n = ar^n-1
• Sum of n terms:
  • S_n = a(1-rⁿ)/(1-r), if r < 1
  • S_n = a(rⁿ - 1)/(r-1), if r > 1
  • S_n = na, if r = 1

Solved Examples

• Ex 1: Find 5th term of AP with a=3, d=2
  • T₅ = 3 + (5-1)*2 = 3 + 8 = 11
• Ex 2: Sum of first 10 terms with a=1, d=3
  • S₁₀ = 10/2 [2*1 + (10-1)*3] = 5 [2 + 27] = 5*29 = 145

Practice Set

• Q1: If a=2, d=3, find 4th term
  • A) 11
  • B) 14
  • C) 9
  • D) 8
• Q2: Sum of first 5 terms with a=4, d=2
  • A) 30
  • B) 40
  • C) 50
  • D) 60
• Q3: Number of terms if a=3, l=27, d=3
  • A) 7
  • B) 8
  • C) 9
  • D) 10

Answer Key

• Q1: A - T₄ = 2 + 3*3 = 11
• Q2: B - S₅ = 5/2 [2*4 + 4*2] = 5/2 [8 + 8] = 5/2 * 16 = 40
• Q3: C - n = (27 - 3)/3 + 1 = 8 + 1 = 9

Memory Aids

• AP nth term: add (n-1) times difference to first term
• Sum: average of first and last term times number of terms
• Number of terms: use formula n = (l - a)/d + 1

Key Concepts & Formulas

• Sequence: a₁, a₂, ..., a_n with constant difference d: a_k+1 - a_k = d
• nth term: T_n = a + (n-1)d
• Number of terms: n = \frac{l - a + d}{d}
• Common difference: d = \frac{l - a}{n - 1}
• Sum of n terms: S_n = \frac{n}{2} [2a + (n-1)d]

Solved Examples

• Ex 1: Find 5th term of AP with a=3, d=2
  • T₅ = 3 + (5-1)*2 = 3 + 8 = 11
• Ex 2: Sum of first 10 terms with a=1, d=3
  • S₁₀ = 10/2 [2*1 + (10-1)*3] = 5 [2 + 27] = 5*29 = 145

Practice Set

• Q1: If a=2, d=3, find 4th term
  • A) 11
  • B) 14
  • C) 9
  • D) 8
• Q2: Sum of first 5 terms with a=4, d=2
  • A) 30
  • B) 40
  • C) 50
  • D) 60
• Q3: Number of terms if a=3, l=27, d=3
  • A) 7
  • B) 8
  • C) 9
  • D) 10

Answer Key

• Q1: A - T₄ = 2 + 3*3 = 11
• Q2: B - S₅ = 5/2 [2*4 + 4*2] = 5/2 [8 + 8] = 5/2 * 16 = 40
• Q3: C - n = (27 - 3)/3 + 1 = 8 + 1 = 9

Memory Aids

• AP nth term: add (n-1) times difference to first term
• Sum: average of first and last term times number of terms
• Number of terms: use formula n = (l - a)/d + 1

Key Concepts & Formulas

• Sequence: a₁, a₂, ..., a_n with constant difference d: a_k+1 - a_k = d
• nth term: T_n = a + (n-1)d
• Number of terms: n = (l - a + d)/d
• Common difference: d = (l - a)/(n - 1)
• Sum of n terms: S_n = n/2 [2a + (n-1)d]

Solved Examples

• Ex 1: Find 5th term of AP with a=3, d=2
  • T₅ = 3 + (5-1)*2 = 3 + 8 = 11
• Ex 2: Sum of first 10 terms with a=1, d=3
  • S₁₀ = 10/2 [2*1 + (10-1)*3] = 5 [2 + 27] = 5*29 = 145

Practice Set

• Q1: If a=2, d=3, find 4th term
  • A) 11
  • B) 14
  • C) 9
  • D) 8
• Q2: Sum of first 5 terms with a=4, d=2
  • A) 30
  • B) 40
  • C) 50
  • D) 60
• Q3: Number of terms if a=3, l=27, d=3
  • A) 7
  • B) 8
  • C) 9
  • D) 10

Answer Key

• Q1: A - T₄ = 2 + 3*3 = 11
• Q2: B - S₅ = 5/2 [2*4 + 4*2] = 5/2 [8 + 8] = 5/2 * 16 = 40
• Q3: C - n = (27 - 3)/3 + 1 = 8 + 1 = 9

Memory Aids

• AP nth term: add (n-1) times difference to first term
• Sum: average of first and last term times number of terms
• Number of terms: use formula n = (l - a)/d + 1

Key Concepts & Formulas

• Sequence: a₁, a₂, ..., a_n with constant difference d: a_k+1 - a_k = d
• nth term: T_n = a + (n-1)d
• Number of terms: n = \frac{l - a + d}{d}
• Common difference: d = \frac{l - a}{n - 1}
• Sum of n terms: S_n = \frac{n}{2} [2a + (n-1)d]

Solved Examples

• Ex 1: Find 5th term of AP with a=3, d=2
  • T₅ = 3 + (5-1)*2 = 3 + 8 = 11
• Ex 2: Sum of first 10 terms with a=1, d=3
  • S₁₀ = 10/2 [2*1 + (10-1)*3] = 5 [2 + 27] = 5*29 = 145

Practice Set

• Q1: If a=2, d=3, find 4th term
  • A) 11
  • B) 14
  • C) 9
  • D) 8
• Q2: Sum of first 5 terms with a=4, d=2
  • A) 30
  • B) 40
  • C) 50
  • D) 60
• Q3: Number of terms if a=3, l=27, d=3
  • A) 7
  • B) 8
  • C) 9
  • D) 10

Answer Key

• Q1: A - T₄ = 2 + 3*3 = 11
• Q2: B - S₅ = 5/2 [2*4 + 4*2] = 5/2 [8 + 8] = 5/2 * 16 = 40
• Q3: C - n = (27 - 3)/3 + 1 = 8 + 1 = 9

Memory Aids

• AP nth term: add (n-1) times difference to first term
• Sum: average of first and last term times number of terms
• Number of terms: use formula n = (l - a)/d + 1

Arithmetic Mean

Key Points

• Middle term in three numbers in G.P. is geometric mean (G.M.)
• G.M. between two positive numbers a and b: √ab
• A.M. is always greater than G.M. for positive numbers

Solved Examples

• Ex 1: Find A.M. between 4 and 10
  • A.M. = (4 + 10)/2 = 7

Practice Set

• Q1: A.M. between 7 and 13 is?
  • A) 9
  • B) 10
  • C) 11
  • D) 12
• Q2: If three numbers are 3, x, 7 in AP, find x
  • A) 4
  • B) 5
  • C) 6
  • D) 7
• Q3: A.M. between 0 and 20
  • A) 5
  • B) 10
  • C) 15
  • D) 20

Answer Key

• Q1: C - (7+13)/2 = 10
• Q2: B - Middle term in AP is average: (3+7)/2 = 5
• Q3: B - (0+20)/2 = 10

Memory Aids

• A.M. is average of two numbers
• Middle term in AP = arithmetic mean

Key Points

• Middle term in three numbers in AP is the arithmetic mean (A.M.)
• A.M. between two numbers a and b: \( \frac{a+b}{2} \)

Solved Examples

• Ex 1: Find A.M. between 4 and 10
  • A.M. = (4 + 10)/2 = 7

Practice Set

• Q1: A.M. between 7 and 13 is?
  • A) 9
  • B) 10
  • C) 11
  • D) 12
• Q2: If three numbers are 3, x, 7 in AP, find x
  • A) 4
  • B) 5
  • C) 6
  • D) 7
• Q3: A.M. between 0 and 20
  • A) 5
  • B) 10
  • C) 15
  • D) 20

Answer Key

• Q1: C - (7+13)/2 = 10
• Q2: B - Middle term in AP is average: (3+7)/2 = 5
• Q3: B - (0+20)/2 = 10

Memory Aids

• A.M. is average of two numbers
• Middle term in AP = arithmetic mean

Key Points

• Middle term in three numbers in AP is the arithmetic mean (A.M.)
• A.M. between two numbers a and b: (a+b)/2

Solved Examples

• Ex 1: Find A.M. between 4 and 10
  • A.M. = (4 + 10)/2 = 7

Practice Set

• Q1: A.M. between 7 and 13 is?
  • A) 9
  • B) 10
  • C) 11
  • D) 12
• Q2: If three numbers are 3, x, 7 in AP, find x
  • A) 4
  • B) 5
  • C) 6
  • D) 7
• Q3: A.M. between 0 and 20
  • A) 5
  • B) 10
  • C) 15
  • D) 20

Answer Key

• Q1: C - (7+13)/2 = 10
• Q2: B - Middle term in AP is average: (3+7)/2 = 5
• Q3: B - (0+20)/2 = 10

Memory Aids

• A.M. is average of two numbers
• Middle term in AP = arithmetic mean

Key Points

• Middle term in three numbers in AP is the arithmetic mean (A.M.)
• A.M. between two numbers a and b: \( \frac{a+b}{2} \)

Solved Examples

• Ex 1: Find A.M. between 4 and 10
  • A.M. = (4 + 10)/2 = 7

Practice Set

• Q1: A.M. between 7 and 13 is?
  • A) 9
  • B) 10
  • C) 11
  • D) 12
• Q2: If three numbers are 3, x, 7 in AP, find x
  • A) 4
  • B) 5
  • C) 6
  • D) 7
• Q3: A.M. between 0 and 20
  • A) 5
  • B) 10
  • C) 15
  • D) 20

Answer Key

• Q1: C - (7+13)/2 = 10
• Q2: B - Middle term in AP is average: (3+7)/2 = 5
• Q3: B - (0+20)/2 = 10

Memory Aids

• A.M. is average of two numbers
• Middle term in AP = arithmetic mean

Geometric Progression

Key Concepts & Formulas

• Sequence with constant ratio r: a_k+1 / a_k = r
• General term: T_n = ar^n-1
• Sum of n terms:
  • S_n = a(1-rⁿ)/(1-r), if r < 1
  • S_n = a(rⁿ - 1)/(r-1), if r > 1
  • S_n = na, if r = 1

Solved Examples

• Ex 1: Find 4th term if a=2, r=3
  • T₄ = 2 * 3³ = 54
• Ex 2: Sum of first 3 terms if a=1, r=2
  • S₃ = 1(1-2³)/(1-2) = (1-8)/(-1) = 7

Practice Set

• Q1: Find 3rd term if a=3, r=2
  • A) 6
  • B) 12
  • C) 24
  • D) 18
• Q2: Sum of first 4 terms if a=2, r=0.5
  • A) 3.75
  • B) 4
  • C) 5
  • D) 6
• Q3: Sum of first 5 terms if a=1, r=1
  • A) 1
  • B) 5
  • C) 10
  • D) 15

Answer Key

• Q1: B - T_3 = 3 * 2² = 12
• Q2: A - S_4 = 2(1-0.5⁴)/(1-0.5) = 2(1-0.0625)/0.5 = 3.75
• Q3: B - r=1, sum = n*a = 5*1 = 5

Memory Aids

• GP term: multiply previous term by r
• Sum formula depends on r < 1, r > 1, or r=1

Key Concepts & Formulas

• Sequence with constant ratio r: a_k+1 / a_k = r
• General term: T_n = ar^{n-1}
• Sum of n terms:
  • S_n = a(1-r^n)/(1-r), if r < 1
  • S_n = a(r^n - 1)/(r-1), if r > 1
  • S_n = na, if r = 1

Solved Examples

• Ex 1: Find 4th term if a=2, r=3
  • T₄ = 2 * 3^{3} = 54
• Ex 2: Sum of first 3 terms if a=1, r=2
  • S₃ = 1(1-2^3)/(1-2) = (1-8)/(-1) = 7

Practice Set

• Q1: Find 3rd term if a=3, r=2
  • A) 6
  • B) 12
  • C) 24
  • D) 18
• Q2: Sum of first 4 terms if a=2, r=0.5
  • A) 3.75
  • B) 4
  • C) 5
  • D) 6
• Q3: Sum of first 5 terms if a=1, r=1
  • A) 1
  • B) 5
  • C) 10
  • D) 15

Answer Key

• Q1: B - T_3 = 3 * 2^{2} = 12
• Q2: A - S_4 = 2(1-0.5^4)/(1-0.5) = 2(1-0.0625)/0.5 = 3.75
• Q3: B - r=1, sum = n*a = 5*1 = 5

Memory Aids

• GP term: multiply previous term by r
• Sum formula depends on r < 1, r > 1, or r=1

Key Concepts & Formulas

• Sequence with constant ratio r: a_k+1 / a_k = r
• General term: T_n = ar^n-1
• Sum of n terms:
  • S_n = a(1-rⁿ)/(1-r), if r < 1
  • S_n = a(rⁿ - 1)/(r-1), if r > 1
  • S_n = na, if r = 1

Solved Examples

• Ex 1: Find 4th term if a=2, r=3
  • T₄ = 2 * 3³ = 54
• Ex 2: Sum of first 3 terms if a=1, r=2
  • S₃ = 1(1-2³)/(1-2) = (1-8)/(-1) = 7

Practice Set

• Q1: Find 3rd term if a=3, r=2
  • A) 6
  • B) 12
  • C) 24
  • D) 18
• Q2: Sum of first 4 terms if a=2, r=0.5
  • A) 3.75
  • B) 4
  • C) 5
  • D) 6
• Q3: Sum of first 5 terms if a=1, r=1
  • A) 1
  • B) 5
  • C) 10
  • D) 15

Answer Key

• Q1: B - T_3 = 3 * 2² = 12
• Q2: A - S_4 = 2(1-0.5⁴)/(1-0.5) = 2(1-0.0625)/0.5 = 3.75
• Q3: B - r=1, sum = n*a = 5*1 = 5

Memory Aids

• GP term: multiply previous term by r
• Sum formula depends on r < 1, r > 1, or r=1

Key Concepts & Formulas

• Sequence with constant ratio r: a_k+1 / a_k = r
• General term: T_n = ar^{n-1}
• Sum of n terms:
  • S_n = a(1-r^n)/(1-r), if r < 1
  • S_n = a(r^n - 1)/(r-1), if r > 1
  • S_n = na, if r = 1

Solved Examples

• Ex 1: Find 4th term if a=2, r=3
  • T₄ = 2 * 3^{3} = 54
• Ex 2: Sum of first 3 terms if a=1, r=2
  • S₃ = 1(1-2^3)/(1-2) = (1-8)/(-1) = 7

Practice Set

• Q1: Find 3rd term if a=3, r=2
  • A) 6
  • B) 12
  • C) 24
  • D) 18
• Q2: Sum of first 4 terms if a=2, r=0.5
  • A) 3.75
  • B) 4
  • C) 5
  • D) 6
• Q3: Sum of first 5 terms if a=1, r=1
  • A) 1
  • B) 5
  • C) 10
  • D) 15

Answer Key

• Q1: B - T_3 = 3 * 2^{2} = 12
• Q2: A - S_4 = 2(1-0.5^4)/(1-0.5) = 2(1-0.0625)/0.5 = 3.75
• Q3: B - r=1, sum = n*a = 5*1 = 5

Memory Aids

• GP term: multiply previous term by r
• Sum formula depends on r < 1, r > 1, or r=1

Geometric Mean

Key Points

• Middle term in three numbers in G.P. is geometric mean (G.M.)
• G.M. between two positive numbers a and b: √ab
• A.M. is always greater than G.M. for positive numbers

Solved Examples

• Ex 1: Find G.M. between 4 and 9
  • G.M. = √4 × 9 = 6

Practice Set

• Q1: G.M. between 1 and 16 is?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If three numbers 2, x, 8 are in G.P., find x
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q3: Which is true?
  • A) A.M. = G.M.
  • B) A.M. < G.M.
  • C) A.M. > G.M.
  • D) None

Answer Key

• Q1: C - √1 × 16 = 4
• Q2: B - x² = 2*8 = 16, x=4
• Q3: C - A.M. is always greater than G.M. for positive numbers

Memory Aids

• G.M. is square root of product of two numbers
• Middle term in G.P. is geometric mean

Key Points

• Middle term in three numbers in G.P. is geometric mean (G.M.)
• G.M. between two positive numbers a and b: \( \sqrt{ab} \)
• A.M. is always greater than G.M. for positive numbers

Solved Examples

• Ex 1: Find G.M. between 4 and 9
  • G.M. = \( \sqrt{4 \times 9} = 6 \)

Practice Set

• Q1: G.M. between 1 and 16 is?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If three numbers 2, x, 8 are in G.P., find x
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q3: Which is true?
  • A) A.M. = G.M.
  • B) A.M. < G.M.
  • C) A.M. > G.M.
  • D) None

Answer Key

• Q1: C - \( \sqrt{1 \times 16} = 4 \)
• Q2: B - x^2 = 2*8 = 16, x=4
• Q3: C - A.M. is always greater than G.M. for positive numbers

Memory Aids

• G.M. is square root of product of two numbers
• Middle term in G.P. is geometric mean

Key Points

• Middle term in three numbers in G.P. is geometric mean (G.M.)
• G.M. between two positive numbers a and b: √ab
• A.M. is always greater than G.M. for positive numbers

Solved Examples

• Ex 1: Find G.M. between 4 and 9
  • G.M. = √4 × 9 = 6

Practice Set

• Q1: G.M. between 1 and 16 is?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If three numbers 2, x, 8 are in G.P., find x
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q3: Which is true?
  • A) A.M. = G.M.
  • B) A.M. < G.M.
  • C) A.M. > G.M.
  • D) None

Answer Key

• Q1: C - √1 × 16 = 4
• Q2: B - x² = 2*8 = 16, x=4
• Q3: C - A.M. is always greater than G.M. for positive numbers

Memory Aids

• G.M. is square root of product of two numbers
• Middle term in G.P. is geometric mean

Key Points

• Middle term in three numbers in G.P. is geometric mean (G.M.)
• G.M. between two positive numbers a and b: \( \sqrt{ab} \)
• A.M. is always greater than G.M. for positive numbers

Solved Examples

• Ex 1: Find G.M. between 4 and 9
  • G.M. = \( \sqrt{4 \times 9} = 6 \)

Practice Set

• Q1: G.M. between 1 and 16 is?
  • A) 2
  • B) 3
  • C) 4
  • D) 5
• Q2: If three numbers 2, x, 8 are in G.P., find x
  • A) 3
  • B) 4
  • C) 5
  • D) 6
• Q3: Which is true?
  • A) A.M. = G.M.
  • B) A.M. < G.M.
  • C) A.M. > G.M.
  • D) None

Answer Key

• Q1: C - \( \sqrt{1 \times 16} = 4 \)
• Q2: B - x^2 = 2*8 = 16, x=4
• Q3: C - A.M. is always greater than G.M. for positive numbers

Memory Aids

• G.M. is square root of product of two numbers
• Middle term in G.P. is geometric mean

Formula for Logarithm

Key Formulas

• log(mn) = log m + log n
• log(m/n) = log m - log n
• log(mⁿ) = n log m

Solved Examples

• Ex 1: Calculate log(1000)
  • log(10³) = 3 log 10 = 3
• Ex 2: Simplify log(50/2)
  • log 50 - log 2

Practice Set

• Q1: log(25*4) equals?
  • A) log 25 + log 4
  • B) log 25 - log 4
  • C) log 100
  • D) log 1
• Q2: log(81/3) equals?
  • A) log 81 + log 3
  • B) log 81 - log 3
  • C) log 3 - log 81
  • D) log 1
• Q3: log(2⁴) equals?
  • A) 4 log 2
  • B) log 2⁴
  • C) log 8
  • D) 2 log 4

Answer Key

• Q1: A - log(mn) = log m + log n
• Q2: B - log(m/n) = log m - log n
• Q3: A - log(mⁿ) = n log m

Memory Aids

• Log of product = sum of logs
• Log of quotient = difference of logs
• Log of power = exponent times log

Key Formulas

• log(mn) = log m + log n
• log(m/n) = log m - log n
• log(m^n) = n log m

Solved Examples

• Ex 1: Calculate log(1000)
  • log(10^3) = 3 log 10 = 3
• Ex 2: Simplify log(50/2)
  • log 50 - log 2

Practice Set

• Q1: log(25*4) equals?
  • A) log 25 + log 4
  • B) log 25 - log 4
  • C) log 100
  • D) log 1
• Q2: log(81/3) equals?
  • A) log 81 + log 3
  • B) log 81 - log 3
  • C) log 3 - log 81
  • D) log 1
• Q3: log(2^4) equals?
  • A) 4 log 2
  • B) log 2^4
  • C) log 8
  • D) 2 log 4

Answer Key

• Q1: A - log(mn) = log m + log n
• Q2: B - log(m/n) = log m - log n
• Q3: A - log(m^n) = n log m

Memory Aids

• Log of product = sum of logs
• Log of quotient = difference of logs
• Log of power = exponent times log

Key Formulas

• log(mn) = log m + log n
• log(m/n) = log m - log n
• log(mⁿ) = n log m

Solved Examples

• Ex 1: Calculate log(1000)
  • log(10³) = 3 log 10 = 3
• Ex 2: Simplify log(50/2)
  • log 50 - log 2

Practice Set

• Q1: log(25*4) equals?
  • A) log 25 + log 4
  • B) log 25 - log 4
  • C) log 100
  • D) log 1
• Q2: log(81/3) equals?
  • A) log 81 + log 3
  • B) log 81 - log 3
  • C) log 3 - log 81
  • D) log 1
• Q3: log(2⁴) equals?
  • A) 4 log 2
  • B) log 2⁴
  • C) log 8
  • D) 2 log 4

Answer Key

• Q1: A - log(mn) = log m + log n
• Q2: B - log(m/n) = log m - log n
• Q3: A - log(mⁿ) = n log m

Memory Aids

• Log of product = sum of logs
• Log of quotient = difference of logs
• Log of power = exponent times log

Key Formulas

• log(mn) = log m + log n
• log(m/n) = log m - log n
• log(m^n) = n log m

Solved Examples

• Ex 1: Calculate log(1000)
  • log(10^3) = 3 log 10 = 3
• Ex 2: Simplify log(50/2)
  • log 50 - log 2

Practice Set

• Q1: log(25*4) equals?
  • A) log 25 + log 4
  • B) log 25 - log 4
  • C) log 100
  • D) log 1
• Q2: log(81/3) equals?
  • A) log 81 + log 3
  • B) log 81 - log 3
  • C) log 3 - log 81
  • D) log 1
• Q3: log(2^4) equals?
  • A) 4 log 2
  • B) log 2^4
  • C) log 8
  • D) 2 log 4

Answer Key

• Q1: A - log(mn) = log m + log n
• Q2: B - log(m/n) = log m - log n
• Q3: A - log(m^n) = n log m

Memory Aids

• Log of product = sum of logs
• Log of quotient = difference of logs
• Log of power = exponent times log

Series of Natural Numbers

Key Formulas

• Sum of first n natural numbers: n(n+1)/2
• Sum of squares of first n natural numbers: n(n+1)(2n+1)/6
• Sum of cubes of first n natural numbers: (n(n+1)/2)²

Solved Examples

• Ex 1: Sum of first 5 natural numbers
  • 5(5+1)/2 = 15
• Ex 2: Sum of squares of first 3 natural numbers
  • 3(3+1)(2 × 3 +1)/6 = 14

Practice Set

• Q1: Sum of first 4 natural numbers?
  • A) 10
  • B) 15
  • C) 20
  • D) 25
• Q2: Sum of squares of first 5 natural numbers?
  • A) 55
  • B) 50
  • C) 60
  • D) 65
• Q3: Sum of cubes of first 3 natural numbers?
  • A) 36
  • B) 27
  • C) 64
  • D) 100

Answer Key

• Q1: A - 4(4+1)/2 = 10
• Q2: A - 5(5+1)(2 × 5 +1)/6 = 55
• Q3: B - (3(3+1)/2)² = 27

Memory Aids

• Sum of natural numbers: n(n+1)/2
• Sum of squares: n(n+1)(2n+1)/6
• Sum of cubes: square of sum of natural numbers

Key Formulas

• Sum of first n natural numbers: \( \frac{n(n+1)}{2} \)
• Sum of squares of first n natural numbers: \( \frac{n(n+1)(2n+1)}{6} \)
• Sum of cubes of first n natural numbers: \( \left(\frac{n(n+1)}{2}\right)^2 \)

Solved Examples

• Ex 1: Sum of first 5 natural numbers
  • \( \frac{5 0(5+1)}{2} = 15 \)
• Ex 2: Sum of squares of first 3 natural numbers
  • \( \frac{3 0(3+1)(2 0 3 +1)}{6} = 14 \)

Practice Set

• Q1: Sum of first 4 natural numbers?
  • A) 10
  • B) 15
  • C) 20
  • D) 25
• Q2: Sum of squares of first 5 natural numbers?
  • A) 55
  • B) 50
  • C) 60
  • D) 65
• Q3: Sum of cubes of first 3 natural numbers?
  • A) 36
  • B) 27
  • C) 64
  • D) 100

Answer Key

• Q1: A - \( \frac{4 0(4+1)}{2} = 10 \)
• Q2: A - \( \frac{5 0(5+1)(2 0 5 +1)}{6} = 55 \)
• Q3: B - \( \left(\frac{3 0(3+1)}{2}\right)^2 = 27 \)

Memory Aids

• Sum of natural numbers: n(n+1)/2
• Sum of squares: n(n+1)(2n+1)/6
• Sum of cubes: square of sum of natural numbers

Key Formulas

• Sum of first n natural numbers: n(n+1)/2
• Sum of squares of first n natural numbers: n(n+1)(2n+1)/6
• Sum of cubes of first n natural numbers: (n(n+1)/2)²

Solved Examples

• Ex 1: Sum of first 5 natural numbers
  • 5(5+1)/2 = 15
• Ex 2: Sum of squares of first 3 natural numbers
  • 3(3+1)(2 × 3 +1)/6 = 14

Practice Set

• Q1: Sum of first 4 natural numbers?
  • A) 10
  • B) 15
  • C) 20
  • D) 25
• Q2: Sum of squares of first 5 natural numbers?
  • A) 55
  • B) 50
  • C) 60
  • D) 65
• Q3: Sum of cubes of first 3 natural numbers?
  • A) 36
  • B) 27
  • C) 64
  • D) 100

Answer Key

• Q1: A - 4(4+1)/2 = 10
• Q2: A - 5(5+1)(2 × 5 +1)/6 = 55
• Q3: B - (3(3+1)/2)² = 27

Memory Aids

• Sum of natural numbers: n(n+1)/2
• Sum of squares: n(n+1)(2n+1)/6
• Sum of cubes: square of sum of natural numbers

Key Formulas

• Sum of first n natural numbers: \( \frac{n(n+1)}{2} \)
• Sum of squares of first n natural numbers: \( \frac{n(n+1)(2n+1)}{6} \)
• Sum of cubes of first n natural numbers: \( \left(\frac{n(n+1)}{2}\right)^2 \)

Solved Examples

• Ex 1: Sum of first 5 natural numbers
  • \( \frac{5 (5+1)}{2} = 15 \)
• Ex 2: Sum of squares of first 3 natural numbers
  • \( \frac{3 (3+1)(2 3 +1)}{6} = 14 \)

Practice Set

• Q1: Sum of first 4 natural numbers?
  • A) 10
  • B) 15
  • C) 20
  • D) 25
• Q2: Sum of squares of first 5 natural numbers?
  • A) 55
  • B) 50
  • C) 60
  • D) 65
• Q3: Sum of cubes of first 3 natural numbers?
  • A) 36
  • B) 27
  • C) 64
  • D) 100

Answer Key

• Q1: A - \( \frac{4 (4+1)}{2} = 10 \)
• Q2: A - \( \frac{5 (5+1)(2 5 +1)}{6} = 55 \)
• Q3: B - \( \left(\frac{3 (3+1)}{2}\right)^2 = 27 \)

Memory Aids

• Sum of natural numbers: n(n+1)/2
• Sum of squares: n(n+1)(2n+1)/6
• Sum of cubes: square of sum of natural numbers

Binomial Theorem

Key Expansions

• (1-x)ⁿ = 1 - nx/1! + n(n-1)x²/2! - ...
• (1+x)ⁿ = 1 + nx/1! + n(n-1)x²/2! + ...

Solved Examples

• Ex 1: Expand (1-x)²
  • 1 - 2x + x²
• Ex 2: Expand (1+x)²
  • 1 + 2x + x²

Practice Set

• Q1: Coefficient of x in (1-x)³?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x² in (1+x)³?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)²?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x² is n(n-1)/2 = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)ⁿ expansion

Key Expansions

• (1-x)ⁿ = 1 - nx/1! + n(n-1)x²/2! - ...
• (1+x)ⁿ = 1 + nx/1! + n(n-1)x²/2! + ...

Solved Examples

• Ex 1: Expand (1-x)²
  • 1 - 2x + x²
• Ex 2: Expand (1+x)²
  • 1 + 2x + x²

Practice Set

• Q1: Coefficient of x in (1-x)³?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x² in (1+x)³?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)²?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x² is n(n-1)/2 = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)ⁿ expansion

Key Expansions

• (1-x)ⁿ = 1 - nx/1! + n(n-1)x²/2! - ...
• (1+x)ⁿ = 1 + nx/1! + n(n-1)x²/2! + ...

Solved Examples

• Ex 1: Expand (1-x)²
  • 1 - 2x + x²
• Ex 2: Expand (1+x)²
  • 1 + 2x + x²

Practice Set

• Q1: Coefficient of x in (1-x)³?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x² in (1+x)³?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)²?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x² is n(n-1)/2 = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)ⁿ expansion

Key Expansions

• (1-x)ⁿ = 1 - nx/1! + n(n-1)x²/2! - ...
• (1+x)ⁿ = 1 + nx/1! + n(n-1)x²/2! + ...

Solved Examples

• Ex 1: Expand (1-x)²
  • 1 - 2x + x²
• Ex 2: Expand (1+x)²
  • 1 + 2x + x²

Practice Set

• Q1: Coefficient of x in (1-x)³?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x² in (1+x)³?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)²?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x² is n(n-1)/2 = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)ⁿ expansion

Key Expansions

• (1-x)^n = 1 - \frac{nx}{1!} + \frac{n(n-1)x^2}{2!} - ...
• (1+x)^n = 1 + \frac{nx}{1!} + \frac{n(n-1)x^2}{2!} + ...

Solved Examples

• Ex 1: Expand (1-x)^2
  • 1 - 2x + x^2
• Ex 2: Expand (1+x)^2
  • 1 + 2x + x^2

Practice Set

• Q1: Coefficient of x in (1-x)^3?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x^2 in (1+x)^3?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)^2?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x^2 is \frac{n(n-1)}{2} = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)^n expansion

Key Expansions

• (1-x)ⁿ = 1 - nx/1! + n(n-1)x²/2! - ...
• (1+x)ⁿ = 1 + nx/1! + n(n-1)x²/2! + ...

Solved Examples

• Ex 1: Expand (1-x)²
  • 1 - 2x + x²
• Ex 2: Expand (1+x)²
  • 1 + 2x + x²

Practice Set

• Q1: Coefficient of x in (1-x)³?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x² in (1+x)³?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)²?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x² is n(n-1)/2 = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)ⁿ expansion

Key Expansions

• (1-x)^n = 1 - \frac{nx}{1!} + \frac{n(n-1)x^2}{2!} - ...
• (1+x)^n = 1 + \frac{nx}{1!} + \frac{n(n-1)x^2}{2!} + ...

Solved Examples

• Ex 1: Expand (1-x)^2
  • 1 - 2x + x^2
• Ex 2: Expand (1+x)^2
  • 1 + 2x + x^2

Practice Set

• Q1: Coefficient of x in (1-x)^3?
  • A) -3
  • B) 3
  • C) -1
  • D) 1
• Q2: Coefficient of x^2 in (1+x)^3?
  • A) 3
  • B) 6
  • C) 9
  • D) 1
• Q3: Term independent of x in (1-x)^2?
  • A) 1
  • B) 0
  • C) x
  • D) -1

Answer Key

• Q1: A - Coefficient of x is -n = -3
• Q2: B - Coefficient of x^2 is \frac{n(n-1)}{2} = 3*2/2 = 3
• Q3: A - Term independent of x is 1

Memory Aids

• Binomial coefficients follow Pascal's triangle
• Signs alternate in (1-x)^n expansion

Exponential Expansion

Key Formula

• e^x = 1 + x + x²/2! + x³/3! + ...

Solved Examples

• Ex 1: Find e¹
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x² in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e⁰?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x³ in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x² is 1/2!
• Q2: B - e⁰ = 1
• Q3: A - Coefficient of x³ is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Key Formula

• e^x = 1 + x + x²/2! + x³/3! + ...

Solved Examples

• Ex 1: Find e¹
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x² in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e⁰?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x³ in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x² is 1/2!
• Q2: B - e⁰ = 1
• Q3: A - Coefficient of x³ is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Key Formula

• e^x = 1 + x + x²/2! + x³/3! + ...

Solved Examples

• Ex 1: Find e¹
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x² in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e⁰?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x³ in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x² is 1/2!
• Q2: B - e⁰ = 1
• Q3: A - Coefficient of x³ is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Key Formula

• e^x = 1 + x + x²/2! + x³/3! + ...

Solved Examples

• Ex 1: Find e¹
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x² in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e⁰?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x³ in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x² is 1/2!
• Q2: B - e⁰ = 1
• Q3: A - Coefficient of x³ is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Key Formula

• e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...

Solved Examples

• Ex 1: Find e^1
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x^2 in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e^0?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x^3 in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x^2 is 1/2!
• Q2: B - e^0 = 1
• Q3: A - Coefficient of x^3 is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Key Formula

• e^x = 1 + x + x²/2! + x³/3! + ...

Solved Examples

• Ex 1: Find e¹
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x² in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e⁰?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x³ in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x² is 1/2!
• Q2: B - e⁰ = 1
• Q3: A - Coefficient of x³ is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Key Formula

• e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...

Solved Examples

• Ex 1: Find e^1
  • 1 + 1 + 1/2 + 1/6 + ...

Practice Set

• Q1: Coefficient of x^2 in e^x expansion?
  • A) 1/2
  • B) 1
  • C) 1/3
  • D) 2
• Q2: Value of e^0?
  • A) 0
  • B) 1
  • C) e
  • D) Undefined
• Q3: Coefficient of x^3 in e^x expansion?
  • A) 1/3!
  • B) 1/2!
  • C) 1
  • D) 3!

Answer Key

• Q1: A - Coefficient of x^2 is 1/2!
• Q2: B - e^0 = 1
• Q3: A - Coefficient of x^3 is 1/3!

Memory Aids

• Factorials in denominators for powers of x
• Starts with 1 + x + ...

Logarithmic Expansion

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - x²/2 + x³/3 - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1²/2 + 0.1³/3 - ...

Practice Set

• Q1: Coefficient of x² in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x³ term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x³ term is positive

Memory Aids

• Alternating signs in series
• Starts with x

Key Formula

• ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ... (|x| < 1)

Solved Examples

• Ex 1: Find ln(1+0.1)
  • 0.1 - 0.1^2/2 + 0.1^3/3 - ...

Practice Set

• Q1: Coefficient of x^2 in ln(1+x)?
  • A) -1/2
  • B) 1/2
  • C) 1/3
  • D) -1/3
• Q2: ln(1+0) equals?
  • A) 0
  • B) 1
  • C) Undefined
  • D) 2
• Q3: Sign of x^3 term in ln(1+x)?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined

Answer Key

• Q1: A - Coefficient is -1/2
• Q2: A - ln(1) = 0
• Q3: A - x^3 term is positive

Memory Aids

• Alternating signs in series
• Starts with x