AQA · A-Level · Mathematics

AQA A-level Mathematics 7357, section A (Proof), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: algebraic manipulation from GCSE. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• The formula n² + n + 41 gives prime numbers for n = 1, 2, 3, all the way to 39. A student concludes it is always prime. How can one number destroy forty confirmations?
• Start with a prediction. State what kind of argument could settle a claim about EVERY case, and what could settle one about A SINGLE case.

Build the model

• A deductive proof 演绎证明 starts from stated definitions and follows logically: odd integers 2a+1 and 2b+1 sum to 2(a+b+1), which is even — proved for every pair.
• A counterexample 反例 refutes an all-values claim with one valid failing case: at n = 41, the value is 41 × 43 = 1763, not prime. A proof by contradiction 反证法 assumes the negation and derives an impossibility: if √2 = p/q in lowest terms, then p² = 2q² forces p even, and writing p = 2r forces q even too — contradicting lowest terms, so √2 is irrational.
• deduction: Proving a claim by logical steps from accepted facts; counterexample: A single valid case that disproves a universal claim; contradiction: Assuming the negation and deriving an impossibility.

Choose evidence that can test it

• Checking n = 0, 1, 2, …, 39 is strong evidence but not a proof: the claim concerns EVERY positive integer. A proof must cover all cases at once — by general algebra, by exhaustion of a complete finite list, or by contradiction.
• Before accepting a proof, check its structure: does it start from assumptions rather than from the conclusion? Does the contradiction attack the stated assumption? Checking examples verifies arithmetic but cannot replace the general argument.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: the candidate counterexample n = 41. Substitution: n² + n + 41 = 41² + 41 + 41 = 41(41 + 1 + 1) = 41 × 43 = 1763. The value has a factor of 41, so it is not prime — the universal claim fails at exactly one case.

Example:

Test whether the sum of two even numbers is always even using deduction: write the two numbers as 2a and 2b and show the structure. (Answer: 2a + 2b = 2(a + b), which is a multiple of 2, hence even for every pair — no examples needed.)

Check the conclusion and its limits

• Each method has a required shape: deduction needs general variables, exhaustion needs a genuinely complete case list, and contradiction needs the impossibility to follow from the negation itself. A proof about integers does not automatically cover real numbers, and an identity proved for all values differs from an equation true only at particular solutions.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Checking a claim for the first hundred integers proves it for every positive integer. This claim is false: a finite sample is evidence, never a proof — n² + n + 41 survives 40 checks and fails at the 41st, so only a general argument or a counterexample settles a universal claim.

Key:

Proof: deduction works from definitions with general variables; a counterexample kills an all-values claim with one case; contradiction assumes the negation and derives an impossibility. Examples verify and disprove — they never prove a universal.

AQA A-level Mathematics 7357, section C (Coordinate geometry), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: GCSE straight lines, Pythagoras and quadratic equations. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A satellite dish's cross-section and a wheel's rim are both circles, but only the rim can be written (x − a)² + (y − b)² = r² with a centre you can read at a glance. What does that equation buy you?
• Start with a prediction. State what the parameters a, b and r each control, and how a tangent direction could be found without differentiation.

Build the model

• A line through (x₁, y₁) with gradient 斜率 m is y − y₁ = m(x − x₁). Parallel lines share m; perpendicular finite gradients satisfy m₁ × m₂ = −1.
• A circle with centre (a, b) and radius r is (x − a)² + (y − b)² = r². The radius to a contact point is perpendicular to the tangent 切线 there, so the tangent's gradient is the negative reciprocal of the radius's. A diameter subtends a right angle at the circumference.
• gradient: The change in y per unit change in x; circle: The set of points at a fixed distance from a centre; tangent: A line touching the circle at one point, perpendicular to the radius there.

Choose evidence that can test it

• Check that a proposed point lies on the circle by substituting into (x − a)² + (y − b)² = r² before building a tangent at it. Complete the square to rescue a centre from the expanded form x² + y² + px + qy + s = 0 — and adjust the right side too.
• Intersect a line with the circle by substitution: a repeated root means tangency, two roots a chord, and a negative discriminant no meeting.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: the circle x² + y² = 25 and the point (3, 4) on it (3² + 4² = 25 ✓). The radius from (0, 0) to (3, 4) has gradient 4/3, so the tangent gradient is −3/4 by m₁ × m₂ = −1. The tangent is y − 4 = −(3/4)(x − 3), which rearranges to 3x + 4y = 25.

Example:

Find the centre and radius of x² + y² − 4x + 2y − 20 = 0 using the same sequence: group terms → complete each square → read the form. (Answer: (x − 2)² + (y + 1)² = 25, so the centre is (2, −1) and the radius is 5; note the bracket signs reverse in the centre.)

Check the conclusion and its limits

• The negative-reciprocal rule needs finite, nonzero gradients; a horizontal radius gives a vertical tangent, stated as x = constant. A chord bisected by a centre-perpendicular is a right-triangle shortcut: half-chord, radius and centre distance obey Pythagoras.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

The tangent at a circle point has the same gradient as the radius to that point. This claim is false: the radius and tangent are perpendicular at the contact point, so their gradients multiply to −1 — at (3, 4), the radius has 4/3 and the tangent has −3/4.

Key:

Coordinate geometry: lines by point-gradient, circles by (x − a)² + (y − b)² = r². Tangents use the radius-perpendicular relation; intersections use substitution and the discriminant.

AQA A-level Mathematics 7357, section D (Sequences and series), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: GCSE arithmetic sequences and simultaneous equations. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A saver deposits the same amount every year while an investment grows by the same percentage every year. After eight years, which balance can you total exactly in one line?
• Start with a prediction. State which of the two grows by addition and which by multiplication, and what each sum formula needs.

Build the model

• An arithmetic sequence 等差数列 adds a constant difference: uₙ = a + (n − 1)d. Its series sums by pairing ends: Sₙ = n(a + uₙ)/2. A geometric sequence 等比数列 multiplies by a constant ratio: uₙ = arⁿ⁻¹, with Sₙ = a(1 − rⁿ)/(1 − r) for r ≠ 1.
• An infinite geometric sum exists only when |r| < 1, because the terms shrink towards zero: S∞ = a/(1 − r). A recurrence relation 递推关系 defines each term from the previous one and needs an initial value; a periodic sequence 周期数列 repeats after a fixed number of steps.
• arithmetic sequence: A sequence with a constant difference between terms; geometric sequence: A sequence with a constant ratio between terms; recurrence relation: A rule giving each term from the preceding value; periodic sequence: A sequence whose terms repeat after a fixed number of steps.

Choose evidence that can test it

• Recover a and d from two stated terms by simultaneous equations before using a formula: u₃ = 10 and u₇ = 22 give 4d = 12, so d = 3 and a = 4. A term count must come out as a positive integer; a fraction or negative root is rejected, never rounded into place.
• Check a claimed sum against the first few terms directly, and verify an at-least target against the integers on BOTH sides of the crossing.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a = 5, d = 3, n = 8. The eighth term is u₈ = 5 + 7 × 3 = 26. The sum is S₈ = n(a + uₙ)/2 = 8(5 + 26)/2 = 124. Each of the four end-pairs totals 31, and 4 × 31 = 124 ✓.

Example:

For a = 12, r = 0.5, find the infinite sum and the first n whose partial sum exceeds 23.9, using the same sequence: known → formula → substitution → integer check. (Answer: S∞ = 12/(1 − 0.5) = 24. The partial sums are Sₙ = 24(1 − 0.5ⁿ): S₇ = 24(1 − 1/128) = 23.8125 < 23.9 and S₈ = 24(1 − 1/256) = 23.90625 ≥ 23.9, so n = 8.)

Check the conclusion and its limits

• The infinite-sum formula is invalid at r = 1 or r = −1, where the terms do not shrink; a negative ratio converges if its MAGNITUDE is below 1. A recurrence needs the operation order preserved exactly: B next = 1.1B + 50 differs from 1.1(B + 50).
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every geometric series has a finite sum to infinity. This claim is false: the sum exists only when |r| < 1. At r = 2 the terms grow without bound, and at r = −1 the partial sums oscillate forever — neither converges.

Key:

Sequences and series: arithmetic adds (pair the ends to sum), geometric multiplies (Sₙ = a(1−rⁿ)/(1−r), and S∞ = a/(1−r) only for |r| < 1). Recover parameters from stated terms, and demand integer answers for counts.

AQA A-level Mathematics 7357, section E (Trigonometry), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: GCSE right-triangle trigonometry and exact values. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• Two students are given A = 30°, a = 7 and b = 10. One reports one triangle; the other insists there are two, and both are right. What information separates the cases?
• Start with a prediction. State which rule pairs sides with opposite angles, and what extra check an inverse-sine answer needs.

Build the model

• On the unit circle 单位圆, the point at angle θ is (cos θ, sin θ), so signs follow the quadrant and exact values come from the 30-60-90 and 45-45-90 triangles. A radian 弧度 is the angle whose arc equals the radius: s = rθ and sector area = r²θ/2 require radians.
• The sine rule 正弦定理 pairs opposite sides and angles: a/sin A = b/sin B. The cosine rule 余弦定理 a² = b² + c² − 2bc cos A uses the angle opposite a. In the ambiguous case 二义情形, sin B = b sin A / a can give B or its supplement 180° − B — both must be tested against the remaining angle.
• unit circle: The circle of radius 1 on which (cos θ, sin θ) is the point at angle θ; radian: The angle subtended by an arc of length equal to the radius; sine rule: The rule a/sin A = b/sin B pairing opposite sides and angles; cosine rule: The rule a² = b² + c² − 2bc cos A; ambiguous case: The sine-rule situation where two triangles may fit the same data.

Choose evidence that can test it

• Choose the rule from the known information: a matching side-angle pair calls for the sine rule; two sides with the INCLUDED angle, or three sides, call for the cosine rule. For area, ab sin C / 2 needs the angle BETWEEN a and b.
• Test each candidate angle: the remaining angle C = 180° − A − B must be positive, and every side positive. A calculator's inverse sine returns only the acute principal value — the supplement is a separate candidate.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: sides 6 and 8 enclose 60°. Use the cosine rule: c² = 6² + 8² − 2 × 6 × 8 × cos 60° = 36 + 64 − 96 × 0.5 = 52. So c = √52 = 2√13 ≈ 7.21. The area is (1/2) × 6 × 8 × sin 60° = 24 × (√3/2) = 12√3.

Example:

For A = 30°, a = 7, b = 10, find the possible values of B using the same sequence: known → sine rule → both branches → validity check. (Answer: sin B = 10 × sin 30°/7 = 5/7, so B ≈ 45.6° or B ≈ 180° − 45.6° = 134.4°. Both leave a positive third angle, so TWO triangles fit.)

Check the conclusion and its limits

• The ambiguous case needs the count: with h = b sin A and acute A, a < h gives none, a = h one right triangle, h < a < b gives two, and a ≥ b gives one. Trig equations on an interval must list every solution: sin x = 1/2 on [0, 2π) has both π/6 and 5π/6.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

The calculator's inverse-sine value is the only possible angle in a sine-rule problem. This claim is false: sine is positive in two quadrants, so the supplement 180° − B is a second candidate and must be checked, which is exactly the ambiguous case.

Key:

Trigonometry: unit circle for values and signs, radians for arcs and sectors, sine rule with its ambiguous case, cosine rule for included angles, and ab sin C / 2 for area — always with a validity check on the remaining angle.

AQA A-level Mathematics 7357, section F (Exponentials and logarithms), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: index laws and scientific calculators from GCSE. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A sample halves every 3.5 hours, but the laboratory needs the level at 5 hours — no table gives it. How does one formula replace the missing table?
• Start with a prediction. State what the model y = Ae^(kt) needs as input, and how a log plot could recover k from data.

Build the model

• In y = Ae^(kt), A is the initial amount and k is the proportional rate: positive for growth, negative for decay. The half-life 半衰期 is ln 2/|k|; doubling time is ln 2/k.
• The logarithm is the inverse of the exponential: y = a^x means x = log_a y, defined only for positive y. The log laws follow from the index laws: log(uv) = log u + log v, log(u/v) = log u − log v, log(uᵏ) = k log u. Solving a^x = b gives x = ln b / ln a. Plotting ln y against t turns y = Ae^(kt) into the straight line ln y = ln A + kt.
• half-life: The time for a decaying quantity to fall to half its initial value; logarithm: The exponent to which a stated base must be raised to produce a given positive number.

Choose evidence that can test it

• A fitted exponential is a model: check its stated time unit and range before predicting, and compare a prediction with a fresh measurement (a residual) before trusting it. A negative k means decay; a negative starting amount usually contradicts the context.
• In log equations, record the original domain first: every argument of every log must be positive, so candidates produced after combining logs must be checked against it.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: y = 80e^(−0.2t), y in grams, t in hours. Half the amount means 40 = 80e^(−0.2t), so e^(−0.2t) = 1/2. Taking ln: −0.2t = −ln 2, so t = ln 2/0.2 ≈ 3.466 hours. The same k gives the half-life for any starting amount.

Example:

Solve 3^x = 20 using the same sequence: take logs → apply the power law → divide → interpret. (Answer: x ln 3 = ln 20, so x = ln 20/ln 3 ≈ 2.727. Check: 3^2.727 ≈ 20.)

Check the conclusion and its limits

• Log laws apply only to positive arguments; ln(x + y) has NO sum law (ln 2 + ln 3 = ln 6 ≠ ln 5). The exponential model needs its unit of time matched to k; an effective yearly rate of 10% means k = ln 1.1, not k = 0.1.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

For positive x and y, ln(x + y) = ln x + ln y. This claim is false: the product law is ln(xy) = ln x + ln y; there is no law for a sum inside a logarithm, as x = y = 1 shows (ln 2 ≈ 0.69 but ln 1 + ln 1 = 0).

Key:

Exponentials and logarithms: y = Ae^(kt) with half-life ln 2/|k|; logs invert exponentials, follow the product/quotient/power laws, and turn an exponential fit into a straight line. Check domains and units before solving.

AQA A-level Mathematics 7357, section G (Differentiation), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: algebraic manipulation and function notation from section B. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A car's distance-time curve keeps bending, so no single "speed" fits the whole journey — yet the speedometer shows one number now. What does that number mean geometrically?
• Start with a prediction. State how a tangent gradient differs from an average speed between two times.

Build the model

• The derivative 导数 is the limit of the difference quotient [f(x+h) − f(x)]/h as h → 0: the tangent gradient and instantaneous rate. For y = axⁿ, dy/dx = anxⁿ⁻¹; trigonometric and exponential forms follow: (sin x)′ = cos x, (cos x)′ = −sin x, (eˣ)′ = eˣ, (ln x)′ = 1/x, in radians.
• A stationary point 驻点 has dy/dx = 0; classify it by the sign change of the derivative or the sign of the second derivative 二阶导数 (positive: local minimum; negative: local maximum). The chain, product and quotient rules handle composite forms, with every inner-rate factor retained.
• derivative: The instantaneous rate of change, the tangent gradient; stationary point: A point where the derivative is zero; second derivative: The derivative of the derivative, measuring how the gradient changes.

Choose evidence that can test it

• A zero derivative is only a candidate: y = x³ is stationary at 0 yet keeps rising, so classification needs the sign change or f″. On a restricted interval, compare boundary values too — the global extreme can sit at an endpoint.
• After differentiating a product such as xeˣ, check at an easy input: the product rule gives eˣ + xeˣ, which at x = 0 is 1, not 0. Multiplying the separate derivatives would give the wrong value there.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: y = x³ − 3x, so dy/dx = 3x² − 3. Stationary where 3x² − 3 = 0: x = ±1. The second derivative is 6x: at x = 1 it is 6 > 0, so (1, −2) is a local minimum; at x = −1 it is −6 < 0, so (−1, 2) is a local maximum.

Example:

For y = x³ − 3x at x = 2, find the tangent and normal equations using the same sequence: point from f → gradient from f′ → line equations. (Answer: y = 2 at x = 2 and the gradient is 9, so the tangent is y − 2 = 9(x − 2) and the normal is y − 2 = −(1/9)(x − 2).)

Check the conclusion and its limits

• The derivative supplies the gradient, not the y-coordinate — the point always comes from the original curve. A horizontal tangent (gradient 0) has a vertical normal, stated as x = constant, since the negative reciprocal of zero does not exist.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every point with zero derivative is a local maximum or minimum. This claim is false: y = x³ has dy/dx = 0 at x = 0 but the derivative 3x² stays positive on both sides — a stationary point of inflection, with no extremum at all.

Key:

Differentiation: the derivative is the limit of the secant gradient, giving the tangent gradient and rate. Stationary candidates come from f′ = 0 and are classified by the sign change or f″; tangents and normals use the point from f and the gradient from f′.

AQA A-level Mathematics 7357, section H (Integration), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: differentiation from section G. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A runner's velocity graph dips below the time axis for a while, then climbs back. The "area under the curve" is nearly zero — did the runner go nowhere?
• Start with a prediction. State what a signed area means, and how total distance differs from displacement.

Build the model

• Integration reverses differentiation: an antiderivative 原函数 of axⁿ is axⁿ⁺¹/(n+1) + C for n ≠ −1, while 1/x integrates to ln|x| + C. A definite integral 定积分 from a to b is signed accumulation: F(b) − F(a).
• Signed area cancels where the curve crosses the axis, so total positive area needs the integral SPLIT at each crossing. The Fundamental Theorem 微积分基本定理 connects both directions: differentiating an accumulation function returns the integrand.
• antiderivative: A function whose derivative is the given integrand; definite integral: The signed accumulation of a function between two limits; Fundamental Theorem: The two-way link that differentiating the accumulation function gives back the integrand.

Choose evidence that can test it

• Verify every antiderivative by differentiating it back, and keep the constant C for indefinite integrals — a supplied point fixes it. The n = −1 case is separate: 1/x integrates to ln|x|, never the power rule with division by zero.
• Before computing total area, find the sign changes of the integrand on the interval and split there; a velocity with a sign change makes displacement and total distance different answers.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: v(t) = 3t² − 3 for 0 ≤ t ≤ 2. An antiderivative is t³ − 3t. Displacement = [t³ − 3t] from 0 to 2 = (8 − 6) − 0 = 2 m. But v = 0 at t = 1, so total distance = −[t³ − 3t] from 0 to 1 + [t³ − 3t] from 1 to 2 = 2 + 4 = 6 m.

Example:

Find the area between y = x² and the x-axis from 0 to 1, then the right-end rectangle estimate with 4 strips, using the same sequence. (Answer: exact ∫₀¹ x² dx = 1/3 ≈ 0.333; the estimate is (1/4)(1/16 + 1/4 + 9/16 + 1) = 15/32 ≈ 0.469, an overestimate for this increasing convex curve.)

Check the conclusion and its limits

• Integration is the limit of rectangle sums; a finite strip count only estimates it, and the error depends on curvature. Do not integrate across a discontinuity such as 1/x at x = 0, and remember signed area can be zero while total area is positive.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A definite integral always equals the total positive area between the curve and the axis. This claim is false: the integral is SIGNED — regions below the axis count negatively, so displacement can be small or zero while the total distance travelled is large.

Key:

Integration: antiderivatives verified by differentiating back, with C fixed by a supplied point; definite integrals are signed accumulation F(b) − F(a); total area needs a split at every sign change, and the Fundamental Theorem links accumulation to the integrand.

AQA A-level Mathematics 7357, section I (Numerical methods), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: function evaluation and sign analysis from sections B and G. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• The equation x³ − x − 1 = 0 has no tidy algebraic solution, yet an engineer needs the root to four decimal places. How can an approximate answer be PROVED correct to a stated accuracy?
• Start with a prediction. State what a sign change shows, and what could make a decimal answer trustworthy.

Build the model

• A continuous function with opposite signs at two endpoints has a root between them. Bisection 二分法 halves the interval, keeping the half whose endpoints still differ in sign.
• Iteration 迭代 x next = g(x) refines an estimate from a rearrangement; a limit must satisfy L = g(L) and be checked in the original equation. Newton–Raphson 牛顿迭代 uses x next = x − f(x)/f′(x) with a nonzero derivative. The trapezium rule 梯形法则 estimates a definite integral: (h/2)(y₀ + 2y₁ + … + 2yₙ₋₁ + yₙ).
• bisection: Halving a root bracket, keeping the half with a sign change; iteration: Repeating an update x next = g(x) from the previous estimate; Newton–Raphson: The root update x next = x − f(x)/f′(x); trapezium rule: Estimating an integral by adding trapezium strips.

Choose evidence that can test it

• A sign change proves a root only where the function is CONTINUOUS: 1/x changes sign across x = 0 with no root there. Report a rounded root with its bracket, such as [1.3245, 1.3255] for four decimal places, so the accuracy is provable.
• Iteration can diverge or cycle: check the iterates, the domain and the original equation. The trapezium rule over- or under-estimates depending on curvature, not on whether the function merely increases.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: f(x) = x² − 2 and x₀ = 1.5 for Newton–Raphson. f′(x) = 2x, so x₁ = x₀ − f(x₀)/f′(x₀) = 1.5 − (2.25 − 2)/3 = 1.5 − 0.0833… = 1.4166667. A second step: f(1.4166667) ≈ 0.0069444, f′ ≈ 2.8333333, so x₂ ≈ 1.4142157, closing on √2.

Example:

Estimate ∫₀² x² dx with two equal trapezium strips and compare with the exact value, using the same sequence. (Answer: h = 1, heights 0, 1, 4, so A = (1/2)(0 + 2 × 1 + 4) = 3. The exact value is 8/3 ≈ 2.667 — the convex curve makes the estimate an overestimate.)

Check the conclusion and its limits

• Each method has conditions: bisection needs a continuous sign change, Newton needs a nonzero derivative and can fail from a poor start, and a rearranged iteration converges only near a fixed point with |g′| < 1. More strips improve a trapezium estimate but never make it exact by assertion.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Every sign change proves a root, including one across a discontinuity. This claim is false: the intermediate-value reasoning needs continuity on the whole interval — across an asymptote the function can jump from negative to positive with no root at all.

Key:

Numerical methods: bracket a root with a continuous sign change and halve with bisection; iterate with checked convergence; Newton–Raphson uses the tangent; the trapezium rule estimates integrals with curvature deciding the error direction.

AQA A-level Mathematics 7357, section J (Vectors), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: coordinate geometry from section C and Pythagoras. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A drone flies from A(1, 2) to B(5, 5), but the mission software asks for the return leg too. Which single formula gives both the outward displacement and its exact reverse?
• Start with a prediction. State what a displacement vector needs beyond a distance, and how the return leg differs.

Build the model

• A vector carries magnitude and direction. Position vectors locate points from one origin; the displacement 位移向量 from A to B is AB = b − a, and BA = −AB reverses it. Add and subtract components; a scalar multiple keeps a vector parallel.
• The scalar product 数量积 a·b = a₁b₁ + a₂b₂ = |a||b| cos θ measures the angle: perpendicular vectors have a·b = 0, and a·a = |a|². Magnitude comes from Pythagoras: |(4, 3)| = 5.
• vector: A quantity with magnitude and direction; displacement: The vector from one point to another, b − a; scalar product: The sum of component products, equal to |a||b| cos θ.

Choose evidence that can test it

• Keep the order of subtraction: BA = −AB, so swapping endpoints reverses every component. A claimed parallel needs a scalar-multiple relation between the vectors; a sketch alone is not a proof.
• For collinearity of three points, show two displacements are parallel AND share a common point; parallel vectors at different locations prove nothing about one line.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: a = (4, 1) and b = (1, 3). The scalar product is a·b = 4 × 1 + 1 × 3 = 7. The magnitudes are |a| = √17 and |b| = √10, so cos θ = 7/(√17 × √10) ≈ 0.537, giving θ ≈ 57.5°. Since a·b ≠ 0, the vectors are not perpendicular.

Example:

With A(1, 2) and B(5, 5), find the displacement AB, its magnitude and the return displacement, using the same sequence. (Answer: AB = (5, 5) − (1, 2) = (4, 3); |AB| = √(4² + 3²) = 5; BA = (−4, −3), the same magnitude in the opposite direction.)

Check the conclusion and its limits

• A vector answers "how far AND which way"; a distance is only a scalar. The scalar product sees direction: it is largest when the vectors align, zero when perpendicular, and negative when they oppose.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

AB and BA have the same components. This claim is false: BA = −AB reverses every component — from B back to A is (−4, −3), not (4, 3). Only the magnitudes agree.

Key:

Vectors: displacement = end − start, reversed by the return leg; components add; scalar multiples stay parallel; the scalar product gives the angle, with perpendicularity exactly at a·b = 0.

AQA A-level Mathematics 7357, section K (Statistical sampling), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: percentages and proportional reasoning from GCSE. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• A school with 120 Year 12 students and 80 Year 13 students wants 30 opinions. A committee of 30 all comes from Year 12 "because they were easiest to find." What is wrong with the arithmetic AND the method?
• Start with a prediction. State what a representative sample requires, and whether a bigger convenience sample would fix it.

Build the model

• A population 总体 is the whole group the claim is about; a sample 样本 is the subset measured. A sampling frame 抽样框 is the list the sample is drawn from. Random, systematic, stratified, quota and opportunity sampling are different methods with different bias risks.
• A stratified sample 分层抽样 allocates in proportion to group size, then selects randomly within each group. Bias 偏倚 comes from the selection method, not the size: doubling a biased sample repeats the same flaw at scale. Missing data is not zero; check units, dates and variable definitions before comparing.
• population: The complete group about which a statistical claim is made; sample: The subset actually measured; sampling frame: The list or population definition the sample is selected from; stratified sample: A sample allocated in proportion to each group\u2019s size; bias: A systematic distortion from an unfair selection method.

Choose evidence that can test it

• Name the population, sampling unit and frame before collecting. Stratification is not "whoever is available from each group" — random selection is still needed within each stratum.
• A large biased sample remains biased: a voluntary online poll over-represents strong opinions however many respond. AQA large-data-set familiarity needs the actual supplied data and its metadata, not invented values.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: 120 students in group A and 80 in group B; a proportional stratified sample of 30. Total population = 200. Group A allocation = 30 × 120/200 = 18; group B allocation = 30 × 80/200 = 12. Each group is then sampled randomly at that size.

Example:

A sample includes 45 of 300 people. Find the sample percentage, and the allocation for a stratified sample of 40 from groups of 150 and 150, using the same sequence. (Answer: 45/300 × 100 = 15%. Equal groups split evenly: 20 and 20.)

Check the conclusion and its limits

• A proportion applied to a population gives an ESTIMATE with sampling variability, not an exact count. A census measures everyone; any sample, however large, inherits the method's selection effects.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

Doubling a biased sample automatically removes its selection bias. This claim is false: bias lives in HOW members are chosen, not how many — surveying twice as many volunteers still over-represents people with strong opinions; only changing the method helps.

Key:

Statistical sampling: define the population and frame; stratified sampling allocates proportionally with random selection inside groups; bias comes from the method, not the size, and estimates carry sampling variability.

AQA A-level Mathematics 7357, section L (Data presentation and interpretation), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.

Prerequisites: GCSE averages, frequency tables and box plots. Each lesson develops its own method before independent transfer.

These are original teaching examples, not official AQA material.

What would explain this observation?

• The tallest bar of a histogram contains FEWER observations than a shorter, wider one next to it. How can the picture be honest and the reading still be wrong?
• Start with a prediction. State what the AREA of a histogram bar means, and what height alone tells you.

Build the model

• A histogram uses AREA for frequency, so with unequal class widths the height is the frequency density 频率密度: frequency ÷ class width. Grouped means and medians are estimates — values inside a class are represented by its midpoint or interpolated position.
• A box plot 箱线图 shows minimum, quartiles, median and maximum. Compare the median for the typical value and the interquartile range 四分位距 (IQR = Q3 − Q1) for consistency. An outlier rule such as 1.5 × IQR beyond the quartiles is a stated convention, not a law.
• frequency density: Frequency divided by class width, used as histogram height; box plot: A display of the five-number summary for comparing distributions; interquartile range: The upper quartile minus the lower quartile, the spread of the middle half.

Choose evidence that can test it

• Before comparing displays, read the axis quantities, units and scale: a truncated axis exaggerates, and a grouped mean is an estimate because the true values inside each class are unknown.
• Compare both a centre and a spread in context: equal medians do not mean equal distributions, and the IQR describes only the middle half — a larger full range can still accompany a tighter box.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: classes 0–10, 10–20, 20–40 with frequencies 20, 30, 20. Widths are 10, 10, 20, so densities are 2, 3, 1. The last bar is shortest despite 20 observations because its width is double; the areas 20, 30, 20 recover the frequencies exactly.

Example:

For those grouped data, estimate the mean using midpoints, using the same sequence: midpoints → weighted sum → divide. (Answer: midpoints 5, 15, 30 give estimated sum 5×20 + 15×30 + 30×20 = 1,150; estimated mean = 1,150/70 ≈ 16.4. This is an estimate, since each class is represented by its midpoint.)

Check the conclusion and its limits

• The tallest bar is not automatically the most frequent: check widths first. A cumulative-frequency graph plots running totals at UPPER class boundaries, and quartiles read from it are interpolation estimates, not exact measurements.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A histogram bar's height always equals its frequency, even with unequal class widths. This claim is false: height is frequency DENSITY; only the area equals the frequency. The wide 20-frequency bar with density 1 holds exactly as much data as the narrow one with density 2.

Key:

Data presentation and interpretation: histogram area carries frequency, grouped statistics are midpoint estimates, box plots compare median and IQR in context, and every comparison starts from the axis units and scale.