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Mathematics flashcards

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Topic 1: Pure Mathematics 1

ax² + bx + c = a(x + p)² + q: what are the vertex and the line of symmetry?: Vertex (−p, q); line of symmetry x = −p. If a > 0, q is the minimum value; if a < 0, q is the maximum value.
Write 2x² − 12x + 5 in completed-square form and state its minimum point.: 2(x² − 6x) + 5 = 2[(x − 3)² − 9] + 5 = 2(x − 3)² − 13. Minimum point (3, −13). The classic slip is not multiplying the −9 by 2.
The quadratic formula for ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a) It always works, even when the quadratic does not factorise.
What does the discriminant b² − 4ac tell you about the roots?: > 0: two distinct real roots. = 0: one repeated root (the curve touches the x-axis). < 0: no real roots.
A line y = mx + k and a curve: how do you find k for which the line is a tangent?: 1. Substitute the line into the curve to get one quadratic. 2. Set its discriminant b² − 4ac = 0 and solve for k. (Two points of intersection: > 0. No intersection: < 0.)
Which two conditions make ax² + bx + c positive for all x?: a > 0 and b² − 4ac < 0. Both are needed; giving only the discriminant loses a mark.
Solve (x − 2)(x − 5) < 0 and (x − 2)(x − 5) > 0.: < 0: 2 < x < 5 (below the axis, between the roots). > 0: x < 2 or x > 5 (above the axis, outside the roots). Always sketch the parabola first.
Solve x − 7√x + 12 = 0.: Let u = √x: u² − 7u + 12 = 0, so u = 3 or u = 4 (both positive, so both valid). x = 9 or x = 16. Always substitute back, and reject u < 0 when u = √x.
One-one function, and when does f⁻¹ exist?: A one-one function gives different outputs for different inputs. f⁻¹ exists only when f is one-one on its domain (test: a horizontal line meets the graph at most once).
When can the composite gf be formed, and what does gf(x) mean?: Only when the range of f lies within the domain of g. gf(x) = g(f(x)): apply f first, then g.
How are the domain and range of f⁻¹ related to f?: Domain of f⁻¹ = range of f. Range of f⁻¹ = domain of f.
How do you find f⁻¹(x)?: 1. Check f is one-one. 2. Write y = f(x) and make x the subject (choose the square-root sign that matches the domain of f). 3. Swap letters to give f⁻¹(x). 4. State its domain: the range of f.
How is the graph of y = f⁻¹(x) related to y = f(x)?: It is the reflection of y = f(x) in the line y = x. A point (a, b) on f becomes (b, a) on f⁻¹. Draw and label y = x on the sketch.
Describe the transformations y = f(x) + a, y = f(x + a), y = af(x) and y = f(ax).: f(x) + a: translation by the vector (0, a). f(x + a): translation by the vector (−a, 0). af(x): stretch parallel to the y-axis, scale factor a. f(ax): stretch parallel to the x-axis, scale factor 1/a.
(2, 5) lies on y = f(x). Where does it move to on y = 3f(2x) − 1?: Halve the x-coordinate; multiply the y-coordinate by 3, then subtract 1. The image is (1, 14).
f(x) = 2x² − 12x + 7 for x ≥ 4. Find the range of f.: f(x) = 2(x − 3)² − 11. The vertex x = 3 is outside the domain and f is increasing for x ≥ 4, so the least value is f(4) = −9. Range: f(x) ≥ −9.
Gradient, midpoint and length of the line joining (x₁, y₁) and (x₂, y₂): Gradient m = (y₂ − y₁)/(x₂ − x₁). Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2). Length √((x₂ − x₁)² + (y₂ − y₁)²).
Gradients of parallel and perpendicular lines: Parallel: m₁ = m₂. Perpendicular: m₁m₂ = −1, so m₂ = −1/m₁ (the negative reciprocal, not just −m₁).
Equation of a circle with centre (a, b) and radius r, and the centre and radius of x² + y² + 2gx + 2fy + c = 0: (x − a)² + (y − b)² = r². Expanded form: centre (−g, −f), radius √(g² + f² − c). If g² + f² − c ≤ 0 it is not a real circle.
Find the centre and radius of x² + y² − 10x + 4y + 13 = 0.: (x − 5)² − 25 + (y + 2)² − 4 + 13 = 0, so (x − 5)² + (y + 2)² = 16. Centre (5, −2), radius 4.
How do you find the tangent to a circle at a point P on it?: 1. Find the centre C. 2. Find the gradient of the radius CP. 3. Tangent gradient = −1/(gradient of CP), because a tangent is perpendicular to the radius. 4. Use y − y₁ = m(x − x₁) through P.
How do you find the perpendicular bisector of AB?: 1. Find the midpoint M of AB. 2. Find the gradient of AB, then the perpendicular gradient −1/m. 3. Write the line through M with that gradient.
A line and a curve give a quadratic after substitution. What do "two distinct points", "meets" and "touches" each need?: Two distinct points: b² − 4ac > 0. Meets (at least once): b² − 4ac ≥ 0. Touches (tangent): b² − 4ac = 0.
Radian: The angle at the centre of a circle subtended by an arc equal in length to the radius. π radians = 180°, so 1 radian ≈ 57.3°.
How do you convert between degrees and radians?: Degrees to radians: × π/180 (for example 72° = 2π/5). Radians to degrees: × 180/π (for example 1.5 rad ≈ 85.9°).
Arc length and sector area for radius r and angle θ: Arc length s = rθ. Sector area A = ½r²θ. θ must be in radians.
Area of a segment cut off by a chord that subtends θ at the centre: Segment = sector − triangle = ½r²θ − ½r² sin θ = ½r²(θ − sin θ), with θ in radians (and the calculator in radian mode).
Length of a chord that subtends θ at the centre of a circle of radius r: 2r sin(θ/2), or from the cosine rule: AB² = r² + r² − 2r² cos θ.
A sector has radius 12 cm and angle 0.75 rad. Find its arc length, area and perimeter.: Arc = 12 × 0.75 = 9 cm. Area = ½ × 144 × 0.75 = 54 cm². Perimeter = 2 × 12 + 9 = 33 cm (include both radii).
Exact values of sin, cos and tan for π/6, π/4 and π/3: sin: 1/2, √2/2, √3/2. cos: √3/2, √2/2, 1/2. tan: 1/√3 (= √3/3), 1, √3.
Principal value ranges of sin⁻¹x, cos⁻¹x and tan⁻¹x: sin⁻¹x: −π/2 ≤ y ≤ π/2. cos⁻¹x: 0 ≤ y ≤ π. tan⁻¹x: −π/2 < y < π/2.
After the calculator value x₁, how do you find the second solution of sin x = k, cos x = k and tan x = k?: sin x = k: 180° − x₁ (π − x₁). cos x = k: 360° − x₁ (2π − x₁). tan x = k: x₁ + 180° (x₁ + π). Then add or subtract 360° (2π) to find every value in the interval.
The two identities in Pure Mathematics 1 trigonometry: tan θ ≡ sin θ / cos θ sin²θ + cos²θ ≡ 1
Greatest value, least value and period of y = a sin bx + c (a, b > 0): Greatest c + a; least c − a; period 360°/b (2π/b). For tan bx the period is 180°/b.
Why must you not cancel sin x in 7 sin x cos x = 2 sin x?: Cancelling loses the solutions of sin x = 0. Factorise instead: sin x(7 cos x − 2) = 0. For 0° ≤ x ≤ 360° this gives x = 0°, 180°, 360°, 73.4° and 286.6°: five solutions.
Binomial coefficient ⁿCᵣ, and the general term of (a + b)ⁿ: ⁿCᵣ = n!/(r!(n − r)!). The general term ⁿCᵣ aⁿ⁻ʳ bʳ is the (r + 1)th term, so the third term uses ⁿC₂.
Find the coefficient of x³ in (1 + 2x)⁶.: ⁶C₃ × 2³ = 20 × 8 = 160. Raise the whole of 2x to the power, not just x.
nth term and sum of an arithmetic progression (first term a, common difference d): uₙ = a + (n − 1)d Sₙ = ½n(a + l) = ½n{2a + (n − 1)d}, where l is the last term.
nth term, sum and sum to infinity of a geometric progression (first term a, ratio r): uₙ = arⁿ⁻¹ Sₙ = a(1 − rⁿ)/(1 − r), r ≠ 1 S∞ = a/(1 − r), only when |r| < 1.
Three-term tests: when are a, b, c in AP, and when in GP?: AP: 2b = a + c. GP: b² = ac.
Show that the GP 12, 8, 16/3, … converges and find its sum to infinity.: r = 8/12 = 2/3 and |2/3| < 1, so it converges. S∞ = 12/(1 − 2/3) = 36.
d/dx (xⁿ), and what must you do first with terms like 6/x⁴ or ∛x?: d/dx (xⁿ) = nxⁿ⁻¹ for rational n. Rewrite every term as a single power first: 6/x⁴ = 6x⁻⁴, ∛x = x^(1/3).
Chain rule, with an example: dy/dx = (dy/du) × (du/dx). d/dx (x² + 1)⁴ = 4(x² + 1)³ × 2x = 8x(x² + 1)³.
Gradient of the tangent and of the normal at x = a: Tangent: m = dy/dx at x = a. Normal: −1/m. Then use y − y₁ = m(x − x₁).
How do you find and classify stationary points?: 1. Solve dy/dx = 0 and find each y. 2. Find d²y/dx² at each point: negative → maximum, positive → minimum. 3. If d²y/dx² = 0, test the sign of dy/dx either side.
Connected rates of change: the rule, and the side of a square, x cm, grows at 0.2 cm s⁻¹. Rate of change of area at x = 6?: dA/dt = (dA/dx) × (dx/dt). A = x², dA/dx = 2x = 12, so dA/dt = 12 × 0.2 = 2.4 cm² s⁻¹. Differentiate before substituting the value.
Increasing and decreasing functions: f is increasing where f′(x) > 0 and decreasing where f′(x) < 0. To show f is increasing, prove f′(x) > 0 for all x in the domain (often by completing the square) and say so.
∫ xⁿ dx and ∫ (ax + b)ⁿ dx: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, n ≠ −1. ∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/(a(n + 1)) + c, n ≠ −1: divide by a as well as by the new power.
How do you find a curve from its gradient function and a point on it?: 1. Integrate dy/dx, including + c. 2. Substitute the given point to find c. 3. Write y = … in full.
Area between two graphs: ∫ (upper − lower) dx between the x-coordinates of the intersection points. Sketch first to see which graph is on top.
Volume of revolution about the x-axis and about the y-axis: About the x-axis: V = π ∫ y² dx (x-limits). About the y-axis: V = π ∫ x² dy (y-limits). With a hole: π ∫ (outer² − inner²), never π ∫ (outer − inner)².
Evaluate ∫₁^∞ 2/x² dx.: [−2/x]₁^∞. As x → ∞, −2/x → 0. Value = 0 − (−2) = 2. State the limit in words; do not substitute ∞ as a number.
Why is ∫ y dx not always the area between a curve and the x-axis?: Parts of the curve below the x-axis give negative values. Split the integral at each root, find each part, and add their sizes. An area is never negative.

Topic 2: Pure Mathematics 2

Rewrite |x − a| < b and |x − a| > b without the modulus (b > 0).: |x − a| < b ⇔ a − b < x < a + b. |x − a| > b ⇔ x < a − b or x > a + b.
How do you solve |a| = |b| when both sides are moduli?: Square both sides: |a| = |b| ⇔ a² = b². When one side is not a modulus, solve the two cases separately and check each answer instead.
Remainder theorem: The remainder when p(x) is divided by (x − a) is p(a). When divided by (ax + b) it is p(−b/a).
Factor theorem: (ax + b) is a factor of p(x) ⇔ p(−b/a) = 0. For (3x + 1), substitute x = −1/3.
What form does the remainder take when you divide by a linear polynomial, and by a quadratic?: Linear divisor: a constant. Quadratic divisor: Ax + B (A or B may be zero). Identity: p(x) ≡ divisor × quotient + remainder.
Find the remainder when 9x³ − 3x + 5 is divided by (3x + 1).: p(−1/3) = 9(−1/27) − 3(−1/3) + 5 = −1/3 + 1 + 5 = 17/3.
Definition of a logarithm: logₐ b = x ⇔ aˣ = b, with a > 0, a ≠ 1 and b > 0. lg x = log₁₀ x; ln x = logₑ x.
The three laws of logarithms: log(xy) = log x + log y log(x/y) = log x − log y log(xⁿ) = n log x log(x + y) cannot be split.
eˣ and ln x as inverses: the key facts and how their graphs are related: e^(ln x) = x for x > 0; ln(eˣ) = x for all x. y = eˣ passes through (0, 1), asymptote y = 0; y = ln x passes through (1, 0), asymptote x = 0. They are reflections in y = x.
Solve 6ˣ = 11.: Take logs: x ln 6 = ln 11, so x = ln 11/ln 6 = 1.34 (3 s.f.).
Which graph is a straight line for y = kxⁿ, and which for y = k(aˣ)?: y = kxⁿ: ln y = ln k + n ln x, so plot ln y against ln x (gradient n). y = k(aˣ): ln y = ln k + x ln a, so plot ln y against x (gradient ln a). In both, the intercept is ln k, so k = e^(intercept).
Solving an inequality such as 0.6ⁿ < 0.01 with logs: what is the trap?: n ln 0.6 < ln 0.01. ln 0.6 is negative, so dividing by it reverses the inequality: n > ln 0.01/ln 0.6 = 9.02, so the least integer is n = 10.
Define sec θ, cosec θ and cot θ.: sec θ = 1/cos θ cosec θ = 1/sin θ cot θ = 1/tan θ = cos θ/sin θ Each has the same sign as its parent function.
The two Pythagorean identities involving sec, cosec, tan and cot: 1 + tan²θ ≡ sec²θ 1 + cot²θ ≡ cosec²θ
Compound angle formulae for sin(A ± B) and cos(A ± B): sin(A ± B) ≡ sin A cos B ± cos A sin B cos(A ± B) ≡ cos A cos B ∓ sin A sin B (the sign changes).
sin 2A and the three forms of cos 2A: sin 2A ≡ 2 sin A cos A cos 2A ≡ cos²A − sin²A ≡ 2cos²A − 1 ≡ 1 − 2sin²A
How do you write a sin θ + b cos θ as R sin(θ + α)?: 1. Expand: R sin θ cos α + R cos θ sin α. 2. Match: R cos α = a, R sin α = b. 3. R = √(a² + b²), tan α = b/a. Greatest value R, least value −R.
Express 5 sin θ + 2 cos θ in the form R sin(θ + α), 0° < α < 90°.: R = √(25 + 4) = √29; tan α = 2/5, α = 21.80°. 5 sin θ + 2 cos θ = √29 sin(θ + 21.80°).
Derivatives of eˣ, ln x, sin x, cos x and tan x: eˣ → eˣ ln x → 1/x sin x → cos x cos x → −sin x tan x → sec²x (x in radians)
Product rule, with an example: d/dx (uv) = u dv/dx + v du/dx. d/dx (x e⁻ˣ) = −x e⁻ˣ + e⁻ˣ = e⁻ˣ(1 − x).
Quotient rule: d/dx (u/v) = (v du/dx − u dv/dx)/v². The minus sign makes the order matter: v du/dx comes first.
Gradient of a parametric curve x = f(t), y = g(t): dy/dx = (dy/dt) ÷ (dx/dt), usually left in terms of t. Horizontal tangent where dy/dt = 0; vertical tangent where dx/dt = 0.
Implicit differentiation: how do you differentiate y² and xy with respect to x?: d/dx (y²) = 2y dy/dx. d/dx (xy) = x dy/dx + y (product rule). Then collect the dy/dx terms on one side and factorise.
Differentiate e^(kx), ln(ax + b) and sin(kx).: e^(kx) → k e^(kx) ln(ax + b) → a/(ax + b) sin(kx) → k cos(kx)
∫ e^(ax + b) dx and ∫ 1/(ax + b) dx: ∫ e^(ax + b) dx = (1/a) e^(ax + b) + c ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + c
∫ sin(ax + b) dx, ∫ cos(ax + b) dx and ∫ sec²(ax + b) dx: −(1/a) cos(ax + b) + c (1/a) sin(ax + b) + c (1/a) tan(ax + b) + c (radians)
How do you integrate sin²x and cos²x?: Use the double-angle formula: sin²x = ½(1 − cos 2x), integral ½x − ¼ sin 2x + c. cos²x = ½(1 + cos 2x), integral ½x + ¼ sin 2x + c.
The trapezium rule: ∫ₐᵇ y dx ≈ ½h{y₀ + 2(y₁ + … + yₙ₋₁) + yₙ}, where h = (b − a)/n. n strips need n + 1 ordinates. It is not printed in MF19.
When does the trapezium rule over-estimate, and when does it under-estimate?: Curve bends upward (chords above the curve, e.g. eˣ): over-estimate. Curve bends downward (chords below the curve, e.g. ln x, √x): under-estimate. Give the reason from a sketch.
Find the exact value of ∫₀² 3/(4 − x) dx.: [−3 ln|4 − x|]₀² = −3 ln 2 + 3 ln 4 = 3 ln 2. Here a = −1, so the minus sign comes out in front.
How do you show that f(x) = 0 has a root between a and b?: Work out f(a) and f(b) and show they have opposite signs. State: "sign change and f is continuous, so a root lies between a and b".
When can the sign-change method mislead you?: A repeated (touching) root, or two roots close together, gives no sign change. A discontinuity, such as an asymptote of 1/x or tan x, can give a sign change with no root.
An iteration xₙ₊₁ = F(xₙ) converges to α. What equation does α satisfy?: α = F(α), so α is a root of x = F(x). To find α exactly, replace xₙ₊₁ and xₙ by α and solve.
When do you stop iterating, and how can you confirm a root is 2.66 to 2 d.p.?: Stop when two successive iterates round to the same value (use the stored calculator value each time). Confirm by a sign change of f on [2.655, 2.665].
xₙ₊₁ = 6/(xₙ + 1), x₁ = 1, converges to α. Find α exactly.: α = 6/(α + 1) gives α² + α − 6 = 0, so (α + 3)(α − 2) = 0. All iterates are positive, so α = 2.

Topic 3: Pure Mathematics 3

Partial fractions: what form do you use for a linear factor, a repeated factor (cx + d)² and a quadratic factor cx² + d?: Linear (ax + b): A/(ax + b). Repeated: B/(cx + d) + C/(cx + d)². Quadratic: (Bx + C)/(cx² + d). The numerator must have lower degree than the denominator.
Express 1/((x + 1)(x + 2)) in partial fractions.: 1/(x + 1) − 1/(x + 2). Check: ((x + 2) − (x + 1))/((x + 1)(x + 2)) = 1/((x + 1)(x + 2)).
Binomial expansion of (1 + x)ⁿ for rational n, and when is it valid?: (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + … Valid for |x| < 1.
Expand √(1 + x) up to the term in x².: (1 + x)^(1/2) = 1 + ½x + (½)(−½)x²/2 + … = 1 + ½x − ⅛x² + …, for |x| < 1.
How do you expand (a + bx)ⁿ when n is not a positive integer?: Take out aⁿ: (a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ, then expand. Valid for |bx/a| < 1, that is |x| < |a/b|.
Solve 3^(2x) − 5 × 3ˣ + 6 = 0.: Let u = 3ˣ: u² − 5u + 6 = 0, so u = 2 or u = 3. 3ˣ = 2 gives x = ln 2/ln 3 = 0.631 (3 s.f.); 3ˣ = 3 gives x = 1.
How do you solve an equation with the unknown in an index, such as 5^(2x − 1) = 3^(x + 2)?: 1. Take ln of both sides. 2. Bring the indices down: (2x − 1) ln 5 = (x + 2) ln 3. 3. Expand, collect the x terms, factorise out x and divide.
A graph of ln y against x is a line with gradient 0.2 and intercept 1.5, for y = k(aˣ). Find a and k.: ln y = ln k + x ln a, so ln a = 0.2 and ln k = 1.5. a = e^0.2 = 1.22, k = e^1.5 = 4.48 (3 s.f.).
In a hidden quadratic in eˣ, why must you reject a root such as eˣ = −3?: eˣ > 0 for every real x (and aˣ > 0 for a > 0), so eˣ = −3 has no solution. Say why you reject it.
Sketch y = e^(kx) for k > 0 and for k < 0.: Both pass through (0, 1) with asymptote y = 0 and lie above the x-axis. k > 0: growth (steeper for larger k). k < 0: decay towards the x-axis.
sec x vs cos⁻¹ x: sec x = 1/cos x; cos⁻¹ x is the inverse function.
Find the exact value of tan 75°.: tan(45° + 30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1) = 2 + √3.
tan(A ± B) and tan 2A: tan(A ± B) ≡ (tan A ± tan B)/(1 ∓ tan A tan B) tan 2A ≡ 2 tan A/(1 − tan²A)
cos²A and sin²A in terms of cos 2A: cos²A ≡ ½(1 + cos 2A) sin²A ≡ ½(1 − cos 2A)
Key features of y = sec θ, y = cosec θ and y = cot θ: sec and cosec: |y| ≥ 1, period 360°; asymptotes where cos θ = 0 (sec) or sin θ = 0 (cosec). cot: period 180°, asymptotes where sin θ = 0.
To solve R sin(θ + α) = c for 0° ≤ θ ≤ 360°, what must you do before finding the angles?: Shift the interval: 0° + α ≤ θ + α ≤ 360° + α. Find every value of θ + α in that range, then subtract α.
d/dx tan⁻¹x, and d/dx tan⁻¹(2x): d/dx tan⁻¹x = 1/(1 + x²). d/dx tan⁻¹(2x) = 2/(1 + 4x²): do not forget the inner derivative 2.
d/dx ln(f(x)) and d/dx e^(f(x)): d/dx ln(f(x)) = f′(x)/f(x) d/dx e^(f(x)) = f′(x) e^(f(x))
For y = eˣ/(x − 2), find the stationary point.: Quotient rule: dy/dx = ((x − 2)eˣ − eˣ)/(x − 2)² = eˣ(x − 3)/(x − 2)². eˣ > 0, so dy/dx = 0 only at x = 3. Stationary point (3, e³).
x = t² + 1, y = t³ − 3t. Where is the tangent horizontal, and where vertical?: dx/dt = 2t, dy/dt = 3t² − 3. Horizontal: t = ±1, at (2, −2) and (2, 2). Vertical: t = 0, at (1, 0).
What must you substitute to find the gradient of an implicit curve at a point?: Both the x- and the y-coordinate of the point, after making dy/dx the subject.
ln|x| vs ln x: The modulus is needed when x may be negative.
∫ f′(x)/f(x) dx: ln|f(x)| + c. Use it when the numerator is a constant multiple of the derivative of the denominator, e.g. ∫ 2x/(x² + 1) dx = ln(x² + 1) + c.
Integration by parts: the formula and how to choose u: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. Choose u as the part that becomes simpler when differentiated (a power of x, or ln x).
Find ∫ x eˣ dx and ∫ ln x dx.: ∫ x eˣ dx = x eˣ − eˣ + c (u = x). ∫ ln x dx = x ln x − x + c (u = ln x, dv/dx = 1).
∫ 1/(x² + a²) dx: (1/a) tan⁻¹(x/a) + c
Integration by substitution: what must change besides the integrand?: Replace dx using du/dx, and change the limits to values of the new variable (or substitute back before using x-limits).
Show that 3 sin x = x has a root between 2 and 3.: f(x) = 3 sin x − x (radians). f(2) = 0.728 > 0, f(3) = −2.58 < 0. Sign change and f is continuous, so a root lies between 2 and 3.
Iterating to a given accuracy: three rules for the working: Use the full stored value (Ans) each time, not a rounded one. Write each iterate to at least one more decimal place than the answer needs. Stop when two successive iterates round to the same value.
How do you show the limit α of xₙ₊₁ = F(xₙ) is a root of a given equation?: Put xₙ₊₁ = xₙ = α, so α = F(α), then rearrange this to the given equation, showing every step.
How can you tell from the numbers that an iteration is failing?: The iterates move away from the root, or oscillate with growing size, instead of settling down. A different rearrangement of the same equation may converge.
Parallel vs skew: Parallel lines have direction vectors that are multiples of each other; skew lines are not parallel and do not meet.
Scalar product and the angle between two vectors: a.b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ, so cos θ = a.b/(|a||b|). a.b = 0 means the vectors are perpendicular.
Vector equation of a line: r = a + tb, where a is the position vector of a point on the line, b is a direction vector and t is a scalar parameter.
How do you find the angle between two lines?: Use the scalar product of their direction vectors only, not the position vectors.
How do you decide whether two lines in 3D intersect?: 1. Equate the general points: a + sb = c + td. 2. Solve two of the three component equations for s and t. 3. Check the third. If it holds, they meet; if not (and they are not parallel), they are skew.
Magnitude and unit vector of a = xi + yj + zk: |a| = √(x² + y² + z²). Unit vector â = a/|a|.
How do you find the foot of the perpendicular from a point C to a line?: 1. Write a general point P on the line in terms of t. 2. Form CP. 3. Solve CP . (direction vector) = 0 for t. 4. Substitute t to get P; |CP| is the shortest distance.
How do you solve a first-order differential equation by separating the variables?: 1. Rearrange so all y terms are with dy and all x terms with dx. 2. Integrate both sides, with one constant c. 3. Use the given condition to find c. 4. Rearrange if asked, and interpret in context.
Translate "the rate of decrease of x is proportional to x" into a differential equation.: dx/dt = −kx, where k is a positive constant.
Solve dy/dx = ky.: ∫ (1/y) dy = ∫ k dx, so ln|y| = kx + c, giving y = Ae^(kx) (A a constant).
Solve dy/dx = 2xy given y = 1 when x = 0.: ∫ (1/y) dy = ∫ 2x dx, so ln y = x² + c. At x = 0, y = 1: c = 0. y = e^(x²).
When must the constant of integration be added?: At the integration step, before you take exponentials or rearrange. Adding it afterwards gives the wrong general solution.
|z − a| = k vs |z − a| = |z − b|: A circle vs a perpendicular bisector.
arg(z − a) = α: A half-line from a, not a full line.
Complex conjugate of z = x + iy, and zz*: z* = x − iy. zz* = x² + y² = |z|², which is real.
Express (3 + 4i)/(1 + 2i) in the form x + iy.: Multiply top and bottom by 1 − 2i: (3 − 6i + 4i + 8)/5 = (11 − 2i)/5 = 2.2 − 0.4i.
Modulus-argument (polar) form, and the range of the principal argument: z = r(cos θ + i sin θ) = re^(iθ), where r = |z| and θ = arg z. Principal argument: −π < θ ≤ π. Place z in the correct quadrant; do not just use tan⁻¹(y/x).
Multiplying and dividing in polar form: |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg z₁ + arg z₂. |z₁/z₂| = |z₁|/|z₂| and arg(z₁/z₂) = arg z₁ − arg z₂. Adjust the argument into −π < θ ≤ π if needed.
How do you find the square roots of a complex number z?: Set (a + ib)² = z. Compare real parts (a² − b² = Re z) and imaginary parts (2ab = Im z), then solve for real a and b.
Non-real roots of a polynomial equation with real coefficients: They occur in conjugate pairs: if a + ib is a root, so is a − ib.

Topic 4: Mechanics

Limiting vs not limiting: F = μR only at the point of slipping or while sliding.
Components of the weight of a particle on a plane inclined at α: mg sin α down the slope (along the plane). mg cos α into the slope (perpendicular to the plane).
Condition for a particle to be in equilibrium: The resultant force is zero: the sum of the components in every direction is zero (resolve in two perpendicular directions).
Friction: the inequality, and which way it acts: F ≤ μR, with F = μR when limiting or sliding. Friction opposes the motion that would happen (or is happening). A smooth surface has no friction.
A particle is about to slip down a rough plane inclined at α, with no other forces. Find μ.: Perpendicular: R = mg cos α. Along: mg sin α = μR = μmg cos α. So μ = tan α.
Newton’s third law: When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction.
Distance vs displacement: Distance adds all motion; displacement is the net change in position.
Speed vs velocity: Velocity has a sign; speed does not.
The constant-acceleration (suvat) equations: v = u + at s = ½(u + v)t s = ut + ½at² v² = u² + 2as Only for constant acceleration.
What do the gradient and area of a displacement–time and a velocity–time graph give?: s–t graph: gradient = velocity. v–t graph: gradient = acceleration; area under the graph = displacement.
Variable acceleration (a function of t): how do you move between s, v and a?: Differentiate: v = ds/dt, a = dv/dt. Integrate: v = ∫ a dt, s = ∫ v dt (with a constant from the initial conditions). Maximum velocity: a = 0. Instantaneous rest: v = 0.
P starts from rest at O with acceleration 2 m s⁻². At the same moment Q passes O at a constant 12 m s⁻¹ in the same direction. When does P catch Q?: P: s = ½ × 2 × t² = t². Q: s = 12t. t² = 12t gives t = 12 s, at 144 m from O.
Momentum vs kinetic energy: Momentum is conserved in impacts; kinetic energy is generally not.
Momentum: definition and unit: Momentum = mass × velocity = mv. Unit: kg m s⁻¹ (equivalently N s). It is a vector, so its sign shows direction.
Principle of conservation of momentum (direct impact in one dimension): Total momentum before = total momentum after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Choose a positive direction first.
Two particles coalesce. What does that mean for the equation?: They join and move with one common velocity v: m₁u₁ + m₂u₂ = (m₁ + m₂)v.
A 0.4 kg ball at 5 m s⁻¹ hits a stationary 0.6 kg ball and rebounds at 1 m s⁻¹. Find the speed of the 0.6 kg ball.: 0.4 × 5 = 0.4 × (−1) + 0.6v, so 2 = −0.4 + 0.6v and v = 4 m s⁻¹. The rebound velocity is −1, not +1.
Mass vs weight: Mass in kg; weight mg in newtons.
Tension vs thrust: A string can only pull; a rod or tow-bar can pull (tension) or push (thrust).
Newton’s second law: Resultant force = mass × acceleration (F = ma), applied in the direction of the acceleration.
How do you solve a connected-particles problem?: 1. Draw a separate force diagram for each particle. 2. Write F = ma for each in its own direction of motion; both have the same acceleration. 3. Add the equations to eliminate the tension. 4. Find the tension from one equation.
Particles of 3 kg and 2 kg hang either side of a smooth pulley (g = 10). Find the acceleration and the tension.: System: 30 − 20 = 5a, so a = 2 m s⁻². 2 kg particle: T − 20 = 2 × 2, so T = 24 N. Check: 30 − 24 = 3 × 2.
Acceleration of a particle sliding down a smooth plane inclined at α: a = g sin α (5 m s⁻² at 30° with g = 10). On a rough plane, the acceleration going up differs from coming down, because friction reverses.
Driving force vs resultant force: P = Fv uses the driving force, not the resultant.
Work done by a constant force F moving through distance d at angle θ to the force: Work done = Fd cos θ, in joules (J).
Kinetic energy and gravitational potential energy: KE = ½mv² GPE = mgh (h is the height gained)
The work–energy principle: Initial energy + work done by driving forces = final energy + work done against resistances (resistance × distance).
Power: the definitions, and what happens at maximum (steady) speed: Power = work done ÷ time, and P = Fv where F is the driving force. At maximum speed a = 0, so the driving force equals the total resistance.
A 50 kg load is raised 6 m at constant speed in 4 s (g = 10). Find the average power.: Work = 50 × 10 × 6 = 3000 J. Power = 3000 ÷ 4 = 750 W.

Topic 5: Probability & Statistics 1

Histogram: what does the height of each bar show?: Frequency density = frequency ÷ class width. The area of each bar is proportional to the frequency.
Mean and standard deviation of raw data: x̄ = Σx/n σ = √(Σx²/n − x̄²) Grouped: use mid-points, x̄ = Σxf/Σf and σ = √(Σx²f/Σf − x̄²).
Coded data: if y = x − a, how are the mean and standard deviation of x found from y?: x̄ = a + ȳ. The standard deviation of x equals the standard deviation of y (subtracting a constant does not change the spread).
Cumulative frequency graph: where do you plot points, and how do you read the median and quartiles for n values?: Plot at upper class boundaries. Median at n/2, Q₁ at n/4, Q₃ at 3n/4 on the cumulative frequency axis.
How do you compare two data sets?: One comment on centre (median or mean) and one on spread (IQR or standard deviation), both in the context of the data.
How do you find the mean and standard deviation of two data sets combined?: Add n, Σx and Σx² across the two sets, then use the formulas once. Do not average the two means unless the sets are the same size.
Permutation vs combination: Permutation: order matters, ⁿPᵣ = n!/(n − r)!. Combination: order does not matter, ⁿCᵣ = n!/(r!(n − r)!).
How many arrangements are there of the letters of LETTER?: 6 letters with E twice and T twice: 6!/(2! × 2!) = 720/4 = 180.
Arrangements where certain items must be together: Glue the items into one block, arrange the blocks, then multiply by the number of arrangements inside the block.
7 people stand in a line. In how many ways can A and B not be next to each other?: Total − together = 7! − 2 × 6! = 5040 − 1440 = 3600.
How do you handle "at least one" or "at most" in a selection?: List the exact cases (for example exactly 1, exactly 2, …), count each with combinations, and add. A single product usually double-counts.
Mutually exclusive events: They cannot happen together: P(A ∩ B) = 0, so P(A or B) = P(A) + P(B).
Independent events: definition and test: One does not affect the probability of the other. Test: P(A ∩ B) = P(A) × P(B). Compare the numbers to justify it.
Conditional probability formula: P(A | B) = P(A ∩ B)/P(B).
Can two events with non-zero probabilities be both mutually exclusive and independent?: No. Exclusive gives P(A ∩ B) = 0, but independence needs P(A ∩ B) = P(A)P(B) > 0.
How do you find P(A | B) from a tree diagram?: Numerator: the probability of the branches where both A and B happen. Denominator: the total probability of all branches where B happens. Divide.
Discrete random variable: the probability rule, E(X) and Var(X): Σp = 1. E(X) = Σxp. Var(X) = Σx²p − {E(X)}².
Conditions for a binomial distribution B(n, p): A fixed number n of independent trials, each with two outcomes (success or failure), and a constant probability p of success. X is the number of successes.
Binomial B(n, p): P(X = r), mean and variance: P(X = r) = ⁿCᵣ pʳ(1 − p)ⁿ⁻ʳ. Mean np; variance np(1 − p).
Geometric Geo(p): P(X = r), mean, and P(X > r): P(X = r) = p(1 − p)ʳ⁻¹, r = 1, 2, 3, … Mean 1/p. P(X > r) = (1 − p)ʳ (the first r trials all fail).
Binomial or geometric?: Binomial: fixed number of trials; X counts the successes (0 to n). Geometric: trials continue until the first success; X is the trial on which it occurs (1, 2, 3, …).
X takes the values 1, 2, 3, 4 with probabilities 0.1, a, b, 0.3, and E(X) = 2.7. Find a and b.: Σp = 1: a + b = 0.6. E(X): 0.1 + 2a + 3b + 1.2 = 2.7, so 2a + 3b = 1.4. b = 0.2, a = 0.4.
Standardising X ~ N(μ, σ²): Z = (X − μ)/σ, where Z ~ N(0, 1). Then P(X < x) = Φ((x − μ)/σ).
How do you find Φ(−z), and P(Z > z)?: Φ(−z) = 1 − Φ(z). P(Z > z) = 1 − Φ(z). Between two values: Φ(z₂) − Φ(z₁).
Normal approximation to B(n, p): conditions and parameters: Use when np > 5 and nq > 5 (q = 1 − p). Approximate by N(np, npq), with a continuity correction.
Continuity corrections for P(X ≤ 15), P(X < 15), P(X ≥ 15) and P(X = 15): P(X ≤ 15) → P(Y < 15.5) P(X < 15) → P(Y < 14.5) P(X ≥ 15) → P(Y > 14.5) P(X = 15) → P(14.5 < Y < 15.5)
What z-value marks the top 10% of a normal distribution?: z = +1.282 (the area to its left is 0.9). The bottom 10% is z = −1.282.
P(X < 40) = 0.05 and P(X > 70) = 0.1 for X ~ N(μ, σ²). Find μ and σ.: (40 − μ)/σ = −1.645 and (70 − μ)/σ = 1.282. Subtract: 30 = 2.927σ, so σ = 10.2 and μ = 40 + 1.645σ = 56.9 (3 s.f.). Keep σ unrounded when finding μ.

Topic 6: Probability & Statistics 2

Poisson Po(λ): P(X = r), mean and variance: P(X = r) = e^(−λ) λʳ/r!, r = 0, 1, 2, … Mean = variance = λ.
Conditions for a Poisson model: Events occur singly, independently, at random and at a constant mean rate.
Faults occur at a mean rate of 0.8 per metre. What is the distribution for 3 metres?: Scale λ with the interval: λ = 3 × 0.8 = 2.4, so X ~ Po(2.4).
When can B(n, p) be approximated by a Poisson distribution?: When n is large and np is small: the syllabus guide is n > 50 and np < 5. Use Po(np), with no continuity correction.
When can Po(λ) be approximated by a normal distribution?: When λ > 15 (approximately). Use N(λ, λ) with a continuity correction.
X ~ Po(λ₁) and Y ~ Po(λ₂) are independent. What is the distribution of X + Y?: X + Y ~ Po(λ₁ + λ₂). This holds only for sums: 2X is not Poisson.
E(aX + b) and Var(aX + b): E(aX + b) = aE(X) + b Var(aX + b) = a²Var(X) Adding a constant does not change the variance.
E(aX ± bY) and Var(aX ± bY) for independent X and Y: E(aX ± bY) = aE(X) ± bE(Y) Var(aX ± bY) = a²Var(X) + b²Var(Y) Variances always add, even for a difference.
X₁ + X₂ + X₃ (three separate items) vs 3X (one item tripled): Both have mean 3μ. X₁ + X₂ + X₃ has variance 3σ². 3X has variance 9σ².
Var(X) = 3. Find Var(5 − 2X).: (−2)² × 3 = 12. The constant 5 does not change the variance.
What is the distribution of a linear combination of independent normal variables?: It is also normal, with mean and variance found from the rules for linear combinations.
Properties of a probability density function f(x): f(x) ≥ 0 for all x, and the total area under f is 1: ∫ f(x) dx = 1 over the whole range. P(a < X < b) = ∫ₐᵇ f(x) dx.
E(X) and Var(X) for a continuous random variable: E(X) = ∫ x f(x) dx Var(X) = ∫ x² f(x) dx − {E(X)}²
How do you find the median of a continuous random variable?: Solve ∫ f(x) dx = ½, integrating from the lower end of the range up to m. Quartiles use ¼ and ¾.
f(x) = (3/8)x² for 0 ≤ x ≤ 2. Find E(X) and Var(X).: E(X) = (3/8)∫₀² x³ dx = (3/8)(4) = 1.5. E(X²) = (3/8)∫₀² x⁴ dx = (3/8)(32/5) = 2.4. Var(X) = 2.4 − 1.5² = 0.15.
For a continuous random variable, what is P(X = a)?: 0, because a single value has no area. So P(X < a) = P(X ≤ a).
Unbiased estimate of the population variance: s² = (1/(n − 1))(Σx² − (Σx)²/n). Dividing by n would, on average, underestimate the population variance.
Why must a sample be random?: A random sample gives every member of the population an equal chance of selection, so the results are not biased. A method that leaves out part of the population (one time, one place, volunteers only) is unsatisfactory.
Distribution of the sample mean X̄: Normal population: X̄ ~ N(μ, σ²/n) exactly, for any n. Other or unknown population, large n: X̄ ~ N(μ, σ²/n) approximately, by the Central Limit Theorem.
Confidence interval for a population mean: x̄ ± z × σ/√n (or s/√n for a large sample). Interpretation: about 95% of intervals built this way contain the true mean (for a 95% interval).
z-values for 90%, 95%, 98% and 99% confidence intervals: 90%: 1.645 95%: 1.960 98%: 2.326 99%: 2.576
Approximate confidence interval for a population proportion: p̂ ± z √(p̂(1 − p̂)/n), where p̂ is the sample proportion.
σ = 5. What is the smallest n for which a 95% confidence interval for μ has width less than 2?: 2 × 1.96 × 5/√n < 2, so √n > 9.8 and n > 96.04. Smallest n = 97.
Steps of a hypothesis test: 1. State H₀ and H₁ in terms of the parameter (p, λ or μ). 2. State the distribution under H₀. 3. Find the tail probability or test statistic. 4. Compare with the significance level. 5. Conclude in context, without overstating.
Type I and Type II errors: Type I: rejecting H₀ when it is true. Type II: accepting H₀ when it is false.
P(Type I error) for a normal test and for a binomial or Poisson test: Normal test: equals the significance level. Binomial or Poisson test: the actual probability of the rejection region under H₀, usually less than the significance level.
How do you compare in a two-tailed test?: Compare each tail probability with half the significance level (for example 2.5% in each tail at the 5% level).
H₀: p = 0.4, H₁: p > 0.4, n = 15, 5% level. P(X ≥ 9) = 0.0950 and P(X ≥ 10) = 0.0338. Find the rejection region.: The first upper tail below 0.05 starts at 10, so the rejection region is X ≥ 10. P(Type I error) = 0.0338.
How do you find P(Type II error)?: Find the probability of landing in the acceptance region, using the stated alternative value of the parameter (not the H₀ value).
Test statistic for a population mean with σ known (normal population): z = (x̄ − μ)/(σ/√n), using the H₀ value of μ. For a large sample with σ unknown, use s in place of σ.

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Written for Cambridge International AS & A Level Mathematics (9709), syllabus 9709 (2026-2027).

Cards by Marlbridge Academic Team; 12 restate definitions from the 9709 revision notes. This deck was reviewed by Sajawal Zahid on 2026-11-05. Report an error.

All 9709 notes, practice and topics