Technique Masterclasses

Pure Mathematics 4

Flashcard Bank

Provenance: written 2026-09-29. This is a reformatting exercise, not new research: every flashcard below repackages a formula, definition, method step, worked micro-example, or documented real-exam finding that is already written and sourced in this paper's own codex-formula-and-method-bank.md and the twelve codex-deep-dive-*.md files, plus codex-complete-worked-paper.md — all read in full before writing this document. No new fact, worked example, likelihood, grade-boundary figure, or exam-board claim is introduced here. Every card's back is traceable to a specific source file and section, cited inline directly under the card using the short tags below. Where a source file itself flags a piece of content as course-original mathematical reasoning rather than a Pearson-sourced fact (e.g. the foot-of-perpendicular closed form, the Maclaurin derivation of the binomial coefficient, the four-way line-classification proof), that same flag is carried onto the card rather than upgraded into an exam-board claim. Any prose fragment quoted directly from a mark scheme or examiner's report is reproduced under fifteen words with its own series/question attribution, exactly as the source document already quotes it — nothing below extends a quotation beyond what the source already carries. Not affiliated with or endorsed by Pearson Edexcel.

Source tags used below — each is one of the twelve existing WMA14 technique-hub files in this same directory:

WMA14 · CODEX — Flashcard Bank

How to use this deck

148 cards, organized by the spec's own topic order (matching FORMULA's own Part 1 numbering). Each card is front, then back, with a source tag underneath the back so any card can be checked against its original document. Cards mix four real card-types, each still traceable to already-verified content:

No card states a likelihood, a grade boundary, or a "confirmed in every series" claim that its source document does not itself state — where a source flags a mechanism as evidenced only once, or as course-original reasoning rather than exam-board fact, this deck's card says so too.

1. Proof by Contradiction (spec 1.1)

Card 1 Front: What is the four-part logical shape every real P4 proof-by-contradiction answer needs? Back: Assumption → algebraic derivation → identified impossibility → concluding sentence. Source: FORMULA §1 (Part 1, "Proof (spec 1.1)").

Card 2 Front: State the five-step generic method skeleton for a proof-by-contradiction answer. Back: 1. State "Assume [negation of the claim]" as an explicit sentence. 2. Set up a genuine algebraic consequence of the assumption. 3. Derive forward, naming each intermediate result in words. 4. Identify the specific impossibility — don't stop at "this has no solutions." 5. Close with an explicit sentence tying the contradiction back to the original claim. Source: FORMULA §2.1.

Card 3 Front: What real mark-scheme generosity applies to the opening "assume" statement, and what does the final A1* mark actually require? Back: "no need for explicit statement of assumption... accept if just a suitable equation is set up" (Jun 2022, Q9 MS) — but the final A1* mark needs "a clear assumption at the start that is contradicted." Source: FORMULA §2.1; PROOF §1.1.

Card 4 Front: Name the three real contradiction mechanisms confirmed across independent WMA14 series. Back: Parity/divisibility contradiction; forcing a real square negative; exhaustion nested inside a contradiction. Source: FORMULA §2.1; PROOF §2.

Card 5 Front: Why is the parity/divisibility mechanism's "even or odd" case split exhaustive? Back: The division algorithm guarantees every integer n can be written as n=2k or n=2k+1 for a unique integer k — no third possibility, no overlap. Source: PROOF §2.1.

Card 6 Front: Prove by contradiction that there is no integer n for which n²+n is odd. What's the key factorisation and how does the case split close it? Back: n²+n = n(n+1), the product of two consecutive integers. If n is even, n(n+1) is even immediately; if n is odd, n+1 is even, so the product is still even. Either branch of the exhaustive even/odd split forces n(n+1) even, contradicting the assumption it is odd — so no such integer exists. Source: PROOF §2.1 (fresh micro-example).

Card 7 Front: Why does the "forcing a real square negative" mechanism need no case split at all? Back: It rests on one always-true fact — x²≥0 for every real x, with no exceptions to enumerate. A negative real number can never equal a real square, so the real mark schemes close it with one short sentence ("no real x satisfies this") rather than a branching argument. Source: PROOF §2.2.

Card 8 Front: Prove by contradiction that the curves y=x²+4 and y=−x²−1 do not intersect. Back: Assume they intersect at some real x. Then x²+4=−x²−1 ⟹ 2x²=−5 ⟹ x²=−5/2. But x²≥0 for every real x, so no real x satisfies this — a direct contradiction. So the curves do not intersect. Source: PROOF §2.2 (fresh micro-example).

Card 9 Front: In the "exhaustion nested in contradiction" mechanism, why is a divisor-pair list provably complete? Back: A fixed positive integer N has only finitely many positive divisors, and therefore only finitely many ways to be written as a product of two positive integers — the list is complete because it is determined entirely by N's own factor structure, not because anyone chose to stop searching. Source: PROOF §2.3.

Card 10 Front: Prove by contradiction that there are no positive integers a>b>0 with a²−b²=34. Back: Assume such a,b exist. Then (a−b)(a+b)=34, and since (a+b)+(a−b)=2a and (a+b)−(a−b)=2b are both even, a+b and a−b must share the same parity. 34's only positive factor pairs, 1×34 and 2×17, are both mixed parity — so neither pair can be (a−b,a+b), and no such a,b exist. (Faster route: 34≡2 mod 4, and a²−b² can only ever be 0, 1 or 3 mod 4, never 2 — so 34 was never reachable as a difference of two squares.) Source: PROOF §2.3 (fresh micro-example).

Card 11 Front: What is the real, quoted example of a one-line assertion that reads like a derivation step but is simply false? Back: "n is an integer so n² + 1 is odd" — grammatically a derivation step, but untrue (Jan 2021, Q3). Source: FORMULA §2.0.

Card 12 Front: What is proof by contrapositive, and how does it differ from proof by contradiction? Back: Proof by contrapositive proves P⇒Q by proving the logically equivalent ¬Q⇒¬P directly — a direct proof of a different (equivalent) implication; no absurdity is ever reached. Proof by contradiction proves a statement P by assuming ¬P and deriving an actual impossibility. Source: PROOF §1.2.

Card 13 Front: In the standard √2-irrationality proof, which stage is actually a proof by contrapositive nested inside the outer contradiction? Back: Stage 3, establishing "p² even ⟹ p even," which is proved by showing the contrapositive "p odd ⟹ p² odd" directly (p=2k+1 ⟹ p²=2(2k²+2k)+1, a plain odd number) and invoking the logical equivalence of an implication and its contrapositive. Source: PROOF §1.2.

Card 14 Front: On the "exhaustion nested" mechanism (4p²−q²=46, Jan 2026 Q8), what does the real mark scheme's dependent mark require? Back: "Attempts to solve both valid pairs," not just one — the scheme accepts three separate valid routes to the same conclusion (factor-pair exhaustion, an odd/even parity split, or a direct substitution q=2m) on exactly equal footing. Source: FORMULA §2.1.

Card 15 Front: WMA14 June 2022 Q9 asks to prove n²−2 is never divisible by 4 for integer n. What is the mark scheme's default route (distinct from the curriculum's own "Alt 3" case-split route)? Back: Assume n²−2=4k for some integer k. Then n²=4k+2=2(2k+1), so n² is even, so n is even; write n=2m, so n²=4m². Then n²−2=4m²−2=2(2m²−1), and since 2m²−1 is odd, this is 2×(an odd number) — never a multiple of 4. This contradicts the assumption, so n²−2 is never divisible by 4. Source: PAPER, Question 9.

2. Partial Fractions — Decomposition and Integration (spec 2.1, 6.3)

Card 16 Front: State the two partial-fraction decomposition forms spec 2.1 requires. Back: Three distinct linear factors: N(x)/[(ax+b)(cx+d)(ex+f)] = A/(ax+b)+B/(cx+d)+C/(ex+f). One repeated linear factor: N(x)/[(ax+b)(cx+d)²] = A/(ax+b)+B/(cx+d)+C/(cx+d)². Source: FORMULA §1 (Part 1, "Algebra and functions — partial fractions").

Card 17 Front: What real condition (verbatim from spec guidance) means a top-heavy fraction needs extra work before decomposition? Back: "The degree of the numerator may equal or exceed the degree of the denominator" — this requires polynomial long division first, to reach a proper fraction plus a polynomial remainder, before either decomposition form applies. Source: FORMULA §1.

Card 18 Front: What denominator shape is genuinely excluded from WMA14 partial fractions, verbatim? Back: "Quadratic factors in the denominator such as (x² + a), a > 0, are not required." Source: FORMULA §1.

Card 19 Front: Why is an irreducible quadratic factor x²+a (a>0) genuinely excluded, not arbitrarily excluded? Back: Cover-up substitution needs a real root of the factor being isolated. A linear factor always has exactly one; x²+a with a>0 has discriminant −4a<0, so no real root exists — there's no real x that isolates a constant over it the way cover-up isolates every other constant in this topic. Source: PARTFRAC §1.4.

Card 20 Front: What single fact licenses both cover-up substitution and coefficient comparison as valid solving methods for partial fractions? Back: A partial-fraction decomposition is asserted with ≡, not =: a claim that two expressions are identical for every value of x, not merely equal for some. Source: PARTFRAC §1.1.

Card 21 Front: Why is a partial-fraction decomposition, once it exists, unique? Back: Suppose two decompositions of the same proper fraction over k distinct linear factors both held. Subtracting and clearing denominators gives a polynomial identity true for every x. Evaluating at x=r₁, every term except the i=1 term vanishes (it contains the factor x−r₁); the surviving term is (A₁−A₁′)×∏(r₁−rⱼ), which is nonzero because the roots are distinct — forcing A₁=A₁′. The same substitution at every other root forces every remaining constant to match. Source: PARTFRAC §1.3.

Card 22 Front: When both substitution (cover-up) and full simultaneous equations are viable, which is preferred, and why? Back: Substitution — plug in the x-value that zeroes one factor. Real-mark-scheme-confirmed as less error-prone: simultaneous equations were "more prone to error than attempts via substitution" (Jan 2024, Q2a). Source: FORMULA §2.2.

Card 23 Front: What is the single most-repeated confirmed recurring slip in the entire WMA14 facts bank, related to partial-fraction integration? Back: Forgetting to attach the 1/a scaling factor to every ∫1/(ax+b)dx term the moment it's written down — confirmed independently across two series a year apart (Oct 2022, Q2; Jan 2024, Q2b). Source: FORMULA §2.2; PARTFRAC §4.

Card 24 Front: On the long-division route into a partial-fraction identity, what specific sign-tracking trap is confirmed? Back: Setting a numerator equal to −6 but then re-substituting +6 (Oct 2021, Q3a). Source: FORMULA §2.2.

Card 25 Front: Decompose (2x²+5x−1)/[(x−1)(x+1)(x+2)] into partial fractions. Back: 1/(x−1) + 2/(x+1) − 1/(x+2). Source: PARTFRAC §2.1.

Card 26 Front: Decompose (3x²−6x+9)/[(x+1)(x−2)²] into partial fractions. Back: 2/(x+1) + 1/(x−2) + 3/(x−2)². Source: PARTFRAC §2.2.

Card 27 Front: Decompose (x²+x−1)/[(x−1)(x+2)] (numerator degree equals denominator degree). Back: Long division gives quotient 1, remainder 1: 1 + 1/[(x−1)(x+2)] = 1 + 1/[3(x−1)] − 1/[3(x+2)]. Source: PARTFRAC §2.3.

Card 28 Front: Once a rational function is decomposed into partial fractions, what three applications does spec 2.1's own guidance name for it? Back: Integration, differentiation, series expansion. Source: PARTFRAC §3 intro; FORMULA §1.

Card 29 Front: Differentiate the decomposed form f(x) ≡ 1/(x−1) + 2/(x+1) − 1/(x+2) term by term. Back: f'(x) = −1/(x−1)² − 2/(x+1)² + 1/(x+2)². Source: PARTFRAC §3.3.

Card 30 Front: When expanding a partial-fraction sum as a binomial series, how is the combined validity range found? Back: The combined range is the intersection (the most restrictive) of the individual ranges, not their average or union — e.g. for 1/(1−x)+1/(1+2x), individual ranges |x|<1 and |x|<1/2 combine to |x|<1/2. Source: BINOM §3.3; PARTFRAC §3.2.

3. Parametric Equations — Conversion & Domain/Range (spec 3.1)

Card 31 Front: What is the default elimination route for converting parametric equations to Cartesian form, and what does the real Jan 2021 Q4 mark scheme credit first? Back: Direct substitution — rearrange whichever equation is simpler to solve for t, substitute into the other at every occurrence of t. The Jan 2021 Q4 scheme's main method line credits: "Attempts to find t in terms of x and substitutes into y." Source: PARAM §2.

Card 32 Front: What real, documented failure occurs when a candidate only "partially rearranges" a parametric elimination? Back: The resulting "cartesian" equation still contains t in one term while it's been substituted out of another — "ended up with y in terms of both x and t" (Oct 2022, Q1). Source: PARAM §2.

Card 33 Front: Convert x=2t+3, y=t²−4t to Cartesian form. Back: t=(x−3)/2; substituting and simplifying gives 4y = x²−14x+33, equivalently y=(x−3)(x−11)/4. Source: PARAM §2.

Card 34 Front: What is the "assume the target's general form and match coefficients" elimination route, and when is it used? Back: Used when direct substitution is intractable (e.g. a reciprocal-in-t pair): assume the Cartesian relationship has a general algebraic form (e.g. (ax+b)/(cx+d)), substitute x(t) into it, set equal to y(t) as a genuine identity in t, and match coefficients across the whole identity — real-credited as "Alternative 1" on Jan 2021 Q4. Source: PARAM §4.

Card 35 Front: What is the real documented trap on the "assume general form" method, and how does it differ from the credited version? Back: Some candidates "assumed the general form of the answer... but this strategy was often unsuccessful" — because they guessed and checked against a couple of points rather than carrying the coefficient-matching through as a full identity. The credited version completes the matching as a genuine identity, not eyeballed against sample points. Source: PARAM §4; FORMULA §2.0.

Card 36 Front: For a trig-parameter pair like x=2+5cosθ, y=−1+2sinθ, what is the elimination method, and what must happen before squaring? Back: Isolate cosθ and sinθ separately, each with coefficient exactly 1 on its own side, before squaring anything; only once both are isolated, square each and add, using sin²θ+cos²θ≡1. Here: cosθ=(x−2)/5, sinθ=(y+1)/2, giving (x−2)²/25+(y+1)²/4=1. Source: PARAM §5.

Card 37 Front: Domain — for a closed, bounded parameter domain t∈[p,q] with x=f(t) monotonic, how is the x-domain found? Back: Substitute both literal endpoint values of t; the direction of the mapping depends on whether f is increasing or decreasing. E.g. x=5−2t, t∈[0,4]: dx/dt=−2<0 so x is decreasing — t=0 gives the larger x-value (x=5), t=4 gives the smaller (x=−3); domain −3≤x≤5. Source: PARAM §6.1.

Card 38 Front: Domain — for an open or unbounded parameter domain (e.g. t>0), how must the endpoints be found? Back: Each excluded or infinite end must be reasoned about as a limit, not substituted directly. The real Jan 2021 Q4 scheme states "k = 2 or x > 2" for x=1/t+2, t>0, reached only via the limits t→0⁺ and t→∞. Source: PARAM §6.2.

Card 39 Front: Range — why is checking only the two endpoints of a domain not enough? Back: y=g(t) is very often not monotonic over the domain; if its one turning point sits interior to the domain, the values g takes near that point aren't bounded between the two endpoint values at all. Real finding: candidates who substituted only the domain's edges had "the minimum value missing" (Oct 2021, Q5c). Source: PARAM §7.

Card 40 Front: Find the range of y=x²−2x for x∈[−1,3]. Back: Completing the square, y=(x−1)²−1: minimum −1 at x=1, interior to [−1,3]. Both endpoints happen to give the same value (y(−1)=3, y(3)=3) — the true range is −1≤y≤3, not just {3}. Source: PARAM §7.1.

Card 41 Front: Find the range of g(x)=(2x−1)/(x+2) for x>1 (strictly increasing, no turning point). Back: g'(x)=5/(x+2)²>0 throughout. g(1)=1/3 (the excluded-endpoint boundary); as x→∞, g(x)→2 (the leading-coefficient boundary). Range: 1/3<g(x)<2. Source: PARAM §7.2.

Card 42 Front: What is the single most bluntly attested real finding for the whole parametric conversion/domain/ range topic? Back: Candidates "confused range with domain and gave answers in terms of x rather than y" (Jan 2021, Q4). Domain is about inputs (x-values), fixed by how t is restricted; range is about outputs (y-values), depending on the full behaviour of g across the domain. Source: PARAM §8.

Card 43 Front: What real, documented "wrong tool" trap occurs on a pure parametric-elimination question? Back: "A few candidates thought it appropriate to use calculus and scored no marks" (Oct 2022, Q1) — forming dy/dx answers the curve's gradient, not the direct relationship between x and y that the question asked for. Source: PARAM §10; FORMULA §1.

Card 44 Front: Given x=8cos³θ, y=6sin²θ, find the coordinates at θ=π/3. Back: cos(π/3)=1/2 ⟹ x=8×1/8=1; sin(π/3)=√3/2 ⟹ y=6×3/4=4.5. Point: (1, 4.5). Source: PARAM §9.

4. Binomial Expansion for Rational n (spec 4.1)

Card 45 Front: State the general binomial series for (1+x)ⁿ, and its range of validity. Back: (1+x)ⁿ = 1 + nx + [n(n−1)/2!]x² + ... + [n(n−1)···(n−r+1)/r!]xʳ + ..., valid for |x|<1, n∈ℝ. Source: FORMULA §1 (Part 1, "Binomial expansion for any rational n").

Card 46 Front: To expand (ax+b)ⁿ, what factoring step is required, and what is the resulting validity condition? Back: (ax+b)ⁿ = bⁿ(1+(a/b)x)ⁿ, valid for |x| < |b/a| — re-deriving the validity condition from this question's actual a and b (not copying it from a different question) is the single most consequential step in this topic. Source: FORMULA §1.

Card 47 Front: Where does the general binomial coefficient n(n−1)···(n−r+1)/r! actually come from? Back: Build f(x)=(1+x)ⁿ's own Maclaurin series by repeated differentiation: f(0)=1, f'(0)=n, f''(0)=n(n−1), and each further differentiation brings down one more factor — so f⁽ʳ⁾(0)=n(n−1)···(n−r+1). Maclaurin's theorem says the coefficient of xʳ is f⁽ʳ⁾(0)/r!, exactly the booklet's formula. Source: BINOM §2.2.

Card 48 Front: Why does the series become infinite unless n is a non-negative integer? Back: The coefficient's running product n(n−1)···(n−r+1) contains a factor equal to exactly zero once r climbs past n, but only when n is a non-negative integer the count-down actually lands on. For a genuine fraction or negative n, the count-down never touches zero, so no term is ever exactly zero. Source: BINOM §2.3.

Card 49 Front: Expand (4+2x)^(−1/2) to the x² term, and state the validity condition. Back: (4+2x)^(−1/2) = 4^(−1/2)(1+x/2)^(−1/2) = 1/2 − x/8 + (3/64)x² − ..., valid for |x/2|<1, i.e. |x|<2. Source: BINOM §3.1.

Card 50 Front: Expand (1+2x²)^(−1/2) to the x² term, and state the validity condition. What error does this bracket-shape invite? Back: (1+2x²)^(−1/2) = 1 − x² + ..., valid for |2x²|<1, i.e. |x|<1/√2. The documented failure mode is treating the power inside the bracket as though it weren't there — collapsing 4x² to 4x (Oct 2021, Q4) — a confusion that measurably improved on a later series (Oct 2022, Q4). Source: BINOM §3.2.

Card 51 Front: What is the "notation hazard" this topic carries across three equivalent presentations of the same formula? Back: Spec 4.1's own guidance uses (ax+b)ⁿ with range "|x| < b/a"; this course's lessons use (a+bx)ⁿ with range |x|<|a/b|; the formula booklet's base form is (1+x)ⁿ with range |x|<1. All three describe the identical fact; only which letter plays which role changes. Source: BINOM §4.

Card 52 Front: WMA14 Oct 2024 Q1: f(x)=(8−3x)^(−1/3), expand to the x³ term. State the four-term expansion and the mark-code chain. Back: 1/2 + x/16 + x²/64 + 7x³/1536. Mark-code chain: B1 (8^(−1/3)=1/2 evaluated) → M1 (general expansion attempted) → A1 (any two of three non-constant terms correct) → A1 (the complete four-term expansion). Source: BINOM §5.

Card 53 Front: In the same Oct 2024 Q1, part (b) uses x=2/3 to estimate a cube root. What must a candidate do beyond substituting x, and what is the final answer? Back: Attempt the reciprocal of the resulting fraction — the reciprocal step is its own separately-earnable method mark, not automatic — reaching 5184/2851 (equivalently 1 2333/2851), since the series approximates 6^(−1/3), the reciprocal of the target ∛6. Source: BINOM §5; FORMULA §2.4.

Card 54 Front: What does the "isw" (ignore subsequent working) marking convention mean, illustrated on this topic's real anchor? Back: Once a correct final answer is written down (the four-term expansion, or 5184/2851), further wrong simplification underneath it does not retract the mark. Source: BINOM §5.

Card 55 Front: On the same anchor, what happens to the marks if a candidate misreads the index's sign (+1/3 instead of −1/3)? Back: A maximum of B1M1A0A0 for part (a) and M0A0 for part (b) — the method-shaped marks survive a sign misread; only the marks depending on the misread quantity's value are lost. Source: BINOM §5.

Card 56 Front: What real, documented error concerns submitting a decimal instead of an exact answer on a binomial-approximation question? Back: A minority of candidates "gave a decimal approximation for their answer" where an exact fraction or surd was required (Jan 2021, Q1). Source: BINOM §6.

Card 57 Front: What is the single most severe documented outcome for this topic, and on what kind of question? Back: "Scores of 0 marks were very common" (Oct 2021, Q4) — on an approximation question answered from the series alone, never reconnected to the original expression. Source: BINOM §6.

Card 58 Front: WMA14 June 2022 Q1: for (3+kx)^(−2), valid for |kx|<3, expanding as A+Bx+Cx²+Dx³+..., find A, and the equation relating C and B (given C=3B). Back: A = 3^(−2) = 1/9. Setting C=3B leads to k²=−6k, i.e. k²+6k=0. Solving (k non-zero): k=−6, and D=32/9. Source: PAPER, Question 1.

5. Implicit & Parametric Differentiation, Tangents/Normals (spec 5.1)

Card 59 Front: State the parametric differentiation formula relating dy/dx to dy/dt and dx/dt. Back: dy/dx = (dy/dt)/(dx/dt), for dx/dt ≠ 0 — differentiate x(t) and y(t) separately, then divide. Source: FORMULA §1 (Part 1, "Differentiation").

Card 60 Front: State the tangent and normal line rules at a point, using gradient m. Back: Tangent: y−y₁=m(x−x₁). Normal: gradient = −1/m (the perpendicular-gradient relation). Source: FORMULA §1.

Card 61 Front: For a bare power of y, yⁿ, what is the chain-rule differentiation result? Back: d/dx(yⁿ) = n yⁿ⁻¹ (dy/dx). Source: FORMULA §2.5; IMPPAR §2.1.

Card 62 Front: Differentiate 3x²y with respect to x (a mixed x-power times y-power term). Back: Product rule with u=3x², v=y: d/dx[3x²y] = 3x²(dy/dx) + 6xy — the chain rule fires only on the y-factor, never on the x²-factor. Source: IMPPAR §2.2.

Card 63 Front: What is the confirmed single hardest mechanical step in implicit differentiation across the whole reviewed record? Back: A mixed term needing product rule then chain rule together in the same term (e.g. y²x) — Jan 2024, Q3(a): candidates "obtained one term as a result of the product rule" (dropping the other factor) or "obtained an incorrect term" on the other. Source: IMPPAR §2.2.

Card 64 Front: Why does an x-only term never pick up a dy/dx factor, and what real trap over-applies the opposite rule? Back: Because d/dx(x)=1, the "invisible" multiplier is exactly 1 — no chain-rule factor attached. Oct 2021, Q1's report documents candidates "surprisingly" writing 4x²+8 → 8x+8(dy/dx), reflexively applying the "y-terms pick up dy/dx" pattern to a term with no y in it at all. Source: IMPPAR §2.3.

Card 65 Front: Differentiate 6ye^(−2x) with respect to x (a mixed term where the x-factor is itself composite). Back: d/dx[6ye^(−2x)] = 6e^(−2x)(dy/dx) − 12ye^(−2x) — product rule across the y-factor and the e^(−2x)-factor, plus a separate chain-rule application on the x-side factor itself. Matches the real mark scheme's A1 criterion for this line (Jan 2021, Q6). Source: IMPPAR §2.4.

Card 66 Front: For (x+y)³, name the two methods the real Oct 2022 Q11(a) mark scheme credits on exactly equal footing. Back: Route A: direct chain rule, d/dx[(x+y)³] = 3(x+y)²(1+dy/dx). Route B: expand first to x³+3x²y+3xy²+y³, then differentiate term by term. Source: IMPPAR §2.5.

Card 67 Front: What is the one real documented failure mode for parametric differentiation itself (not the setup)? Back: A simplification error occurring after the correct differentiation and division — Oct 2022, Q6: candidates simplifying 1/(3sin²t) incorrectly to 3cos²t, rather than the correct (1/3)csc²t or equivalent. Source: IMPPAR §3.

Card 68 Front: What is the dominant, most consistently cited failure on the tangent/normal construction step, once dy/dx is known? Back: Which point gets used, not the mechanism. Jan 2021 Q6's report records "a surprisingly large number of candidates" who took the named point P to be the origin outright, and a second depth of the same failure — finding P's correct coordinates but then evaluating the gradient at (0,0) anyway. Source: IMPPAR §4.

6. Connected Rates of Change (spec 5.2)

Card 69 Front: State the general chain-rule composition used to set up a connected-rates-of-change equation. Back: dV/dt = dV/dr × dr/dt (or the equivalent for whichever two quantities the modelling context connects). Source: FORMULA §1 (Part 1, "Connected rates of change").

Card 70 Front: What real, documented setup error uses the wrong solid's volume formula? Back: October 2021, Q9's report documents candidates "using the volume of the cylinder as V = (4/3)πr²h" — a sphere's volume formula, misapplied to a cylinder and even mistyped with an h the sphere formula never has. Source: SEPDE §3; FORMULA §1.

Card 71 Front: What real, documented error treats a genuinely varying geometric quantity as a constant? Back: January 2024, Q4's report records "treating l in the given S formula as a constant," where the slant height l was actually a function of the radius via Pythagoras and needed substituting in before differentiating. The report calls the question "not well understood by most." Source: SEPDE §3.

Card 72 Front: What real, documented error drops half of a two-term rate? Back: Oct 2021 Q9's report documents candidates "using either dV/dt = 0.6π or dV/dt = −0.15πh" — each exactly half of a genuine two-term rate (an inflow and an outflow), the other term silently discarded. Source: SEPDE §3.

Card 73 Front: What real, documented final-answer error concerns units on a rates-of-change question? Back: A numerically correct answer of 208 lost marks for "giving the units for the answer 208 as seconds rather than minutes" — every quantity in that question had been defined in minutes from its first line (Oct 2021, Q9). Source: SEPDE §3.

Card 74 Front: WMA14 June 2022 Q3: a tablet is a cylinder of radius x mm, length 3x mm; cross-sectional area decreases at constant 0.5 mm² s⁻¹. Find dx/dt when x=7. Back: A=πx² ⟹ dA/dx=2πx. dx/dt = (dA/dt)/(dA/dx) = −0.5/(2πx). At x=7: dx/dt = −0.5/(14π) = −1/(28π) ≈ −0.0114 mm s⁻¹. Source: PAPER, Question 3.

7. Volume of Revolution (spec 6.1)

Card 75 Front: State the volume of revolution formula, and what axis is it restricted to on this spec? Back: V = π∫y² dx, about the x-axis only — π∫x² dy is explicitly NOT required by this spec. Source: FORMULA §1 (Part 1, "Volume of revolution").

Card 76 Front: State the parametric form of the volume-of-revolution formula. Back: V = π∫y² (dx/dt) dt. Source: FORMULA §1.

Card 77 Front: What real, confirmed mark-scheme distinction applies to using 2π∫y²dx instead of π∫y²dx? Back: It is scored as a wrong constant (partial credit possible), distinct from and better than setting up an entirely wrong integral (Jan 2021, Q9). Source: FORMULA §1.

Card 78 Front: Which of two trigonometric identities is needed for a sum/difference y versus a bare sin²θ or cos²θ term when squaring y for volume of revolution? Back: A sum/difference y (e.g. y=sinθ+cosθ) squares to include a cross term — resolved via sin2θ=2sinθcosθ. A bare sin²θ or cos²θ term needs the power-reduction form instead: sin²θ=(1−cos2θ)/2, cos²θ=(1+cos2θ)/2. Source: FORMULA §1.

Card 79 Front: State the six-step method skeleton for volume of revolution. Back: 1. Quote V=π∫y²dx first. 2. Square y fully, including any cross term. 3. Identify which identity (power-reduction or double-angle) the squared expression needs. 4. Check for a recognisable reverse-chain- rule form before defaulting to parts/substitution out of habit. 5. If parametric, convert dx into (dx/dt)dt and convert x-limits into t-limits. 6. Use π, not 2π, unless the solid genuinely requires the doubled constant. Source: FORMULA §2.6.

Card 80 Front: Find the volume of revolution for y=x²−1/x on [1,2], about the x-axis. Back: y² = x⁴ − 2x + 1/x² (the cross term 2·x²·(1/x)=2x is the term most likely to be dropped). V = π[x⁵/5 − x² − 1/x] from 1 to 2 = 37π/10. Source: VOLREV §3, Case A.

Card 81 Front: What real, documented failure occurs on a Cartesian volume-of-revolution question with a compound y? Back: Jan 2024 Q7(b): candidates who "made no further progress, usually due to being unable to see how to square y." Source: VOLREV §3, §4.

Card 82 Front: What real, documented failure occurs when the resulting integral needs technique recognition rather than routine integration? Back: Oct 2022 Q5: candidates "wasted much time pursuing incorrect methods" including integration by parts, when the efficient route was recognising the integrand "as being the result of a chain rule differentiation." Source: VOLREV §3, §4.

Card 83 Front: What real, documented finding concerns the double-angle step on a parametric volume-of-revolution question? Back: Jan 2021 Q9's general finding: "there were a lot of slips in working with y²... failing to show that they had used sin2θ=2sinθcosθ." Source: VOLREV §3, Case C.

Card 84 Front: WMA14 June 2022 Q8: y=10xe^(−x/2) on [0,10] rotated about the x-axis. Show V=k∫x²e^(−x)dx and state k. Back: V = π∫y²dx = π∫[10xe^(−x/2)]²dx = π∫100x²e^(−x)dx = 100π∫x²e^(−x)dx, so k=100π (since [e^(−x/2)]²=e^(−x)). Source: PAPER, Question 8.

Card 85 Front: What plausible (not documented) origin does the deep-dive suggest for the confirmed real "2π" slip in volume-of-revolution answers? Back: A stray 2π is exactly the constant that genuinely belongs to two related but different formulas a student may be pattern-matching against: the surface area of revolution, and Pappus's centroid theorem for a region rotated about an external axis — offered as reasoning, not as something any examiner report attributes it to. Source: VOLREV §2.3.

Card 86 Front: What general theorem is the disk method V=π∫y²dx a special case of? Back: If a solid's cross-sectional area at position x is a known function A(x), then V=∫A(x)dx from a to b — the disk method is this theorem with A(x)=πy², because rotating a boundary curve about the x-axis makes every cross-section a circle of radius y. Source: VOLREV §2.1.

8. Integration by Parts & Substitution (spec 6.2)

Card 87 Front: State the integration-by-parts formula (booklet-given, P4 section). Back: ∫u (dv/dx) dx = uv − ∫v (du/dx) dx. Source: FORMULA §1; PARTS Part 1.

Card 88 Front: Why can ln x never be chosen as dv/dx in an integration-by-parts application? Back: No elementary antiderivative of ln x exists independently of the "ln x = ln x × 1" trick — ln x is always u. Source: PARTS §2.1.

Card 89 Front: Find ∫x sin2x dx by parts. Back: u=x, dv/dx=sin2x ⟹ du/dx=1, v=−(1/2)cos2x. ∫x sin2x dx = −(x/2)cos2x + (1/4)sin2x + c. Source: PARTS §2.2.

Card 90 Front: State the general reduction formula for ∫P(x)e^(kx)dx (P a polynomial of degree n), and why does the "DI" table always terminate for a polynomial factor? Back: ∫P(x)e^(kx)dx = e^(kx)·Σᵢ₌₀ⁿ (−1)ⁱ P⁽ⁱ⁾(x)/kⁱ⁺¹ + c — it terminates after exactly n+1 terms because P⁽ⁿ⁺¹⁾(x)≡0, which is why the D-column always reaches 0 for a polynomial factor. Source: PARTS §2.3.

Card 91 Front: WMA14 June 2022 Q8(b): find ∫x²e^(−x)dx. State the two applications and the result. Back: First (u=x², dv/dx=e^(−x)): ∫x²e^(−x)dx = −x²e^(−x) + 2∫xe^(−x)dx. Second (u=x, dv/dx=e^(−x)): ∫xe^(−x)dx = −xe^(−x) − e^(−x) + c. Combined: ∫x²e^(−x)dx = −e^(−x)(x²+2x+2) + c. Source: PARTS §2.3; PAPER, Question 8.

Card 92 Front: State the general cyclic-case formulas for ∫e^(ax)sin(bx)dx and ∫e^(ax)cos(bx)dx. Back: ∫e^(ax)sin(bx)dx = [e^(ax)(a sin bx − b cos bx)]/(a²+b²) + c. ∫e^(ax)cos(bx)dx = [e^(ax)(a cos bx + b sin bx)]/(a²+b²) + c. Source: PARTS §2.4.

Card 93 Front: Evaluate ∫e^(2x)cos x dx using the cyclic method. Back: I = (e^(2x)/5)(2cos x + sin x) + c — matches the general cosine formula at a=2, b=1. Source: PARTS §2.4.

Card 94 Front: Evaluate ∫₀¹ x(1−x)³dx by substitution u=1−x, tracking the limit reversal. Back: u=1−x ⟹ x=1−u, du/dx=−1. When x=0, u=1; when x=1, u=0 — the limits swap order because u decreases as x increases. ∫₀¹(u³−u⁴)du = 1/4 − 1/5 = 1/20. Source: PARTS §2.5.

Card 95 Front: What does spec 6.2's guidance state about more complicated substitution integrands? Back: "The substitution will be given in more complicated integrals." Source: PARTS §2.6.

Card 96 Front: What real, documented error occurs when converting dx during substitution, rather than converting properly? Back: Candidates "ignored the dx and simply wrote it as du" instead of converting via du=g'(x)dx (Oct 2021, Q6). Source: PARTS Part 3.

Card 97 Front: What real, documented error occurs with definite-integral limits under substitution? Back: "Use of the x limits...instead of the u limits" — reusing the original x-limits with a u-integrand instead of converting them (Oct 2022, Q7i). Source: PARTS Part 3.

Card 98 Front: What real, documented coefficient-loss error recurs across both parts and substitution questions? Back: On parts: "some candidates failed to realise they required a factor of 3" (Oct 2022, Q7ii). On substitution: "losing the factor 4 in the expression" (Oct 2021, Q6) — the same failure mode confirmed on both techniques independently. Source: PARTS Part 3.

Card 99 Front: What real, documented unjustified shortcut broke a marked derivation chain? Back: "Many candidates assumed that ln(x²) was identical to 2 ln x" mid-derivation (Oct 2021, Q8b) — the identity is true, but using it as an invisible jump breaks a line-by-line mark scheme. Source: PARTS Part 3.

Card 100 Front: What real, credited alternative method for repeated integration by parts is confirmed on two independent series? Back: The tabular "DI" approach — "a few used the tabular 'DI' approach" (Jan 2024, Q5a), against the majority who used the traditional method, both scored on equal footing; independently reconfirmed by the June 2022 mark scheme's own explicit DI-method note for Q8(b). Source: PARTS Part 3.

Card 101 Front: What is a genuinely repeated application of parts credited as in the mark scheme, and how is this confirmed directly? Back: A dependent method mark (dM1), never an independent one — confirmed directly from a real mark scheme, June 2022, WMA14/01, Q8(b): "Dependent on the previous method mark." Source: PARTS Part 1.

9. Separable Differential Equations & Connected Rates (spec 6.4, 5.2)

Card 102 Front: State the general method for solving a separable first-order differential equation. Back: dy/dx = f(x)g(y) ⟹ ∫1/g(y) dy = ∫f(x) dx (+c). Both general and particular solutions are spec-required (a boundary condition solves for c). Source: FORMULA §1; SEPDE §1.

Card 103 Front: Why does a first-order separable equation need exactly one constant of integration, not two? Back: The derivation (via G(y), an antiderivative of 1/g(y), and the chain rule) shows the arbitrary constant enters on only one side of the argument (the x-side) — a direct consequence of where integration actually happens in the proof, not a convention. Source: SEPDE §1.

Card 104 Front: What is the real anchor question for the "pure reverse-power-rule, no logarithms" family, and what mechanical risk does it carry? Back: January 2026, Q3: dh/dt = 2h^(3/2)/(5t²), separating to ∫(1/2)h^(−3/2)dh = ∫(1/5)t^(−2)dt — every exponent is a plain rational power. The risk is handling a negative, fractional exponent correctly under the power rule (h^(−3/2) → −2h^(−1/2)). Source: SEPDE §2.1.

Card 105 Front: What real, documented finding confirms the "logarithmic integration" family genuinely involves logs, and what is the exact difficulty? Back: January 2021, Q10: candidates who found the constant "had difficulty dealing with the logarithms and rearranging." Source: SEPDE §2.2.

Card 106 Front: What is the mirror-image trap on the same logarithmic family — manufacturing a log where none belongs? Back: October 2021, Q9: candidates "integrating 1/320 to ln(320t)" — inventing a logarithm on a side of the equation where the correct antiderivative of a plain constant with respect to time is simply a linear term, (constant)×t. Source: SEPDE §2.2.

Card 107 Front: What real, documented separation-step algebra error occurs before any integration starts? Back: October 2021, Q2: "moved the 4" — an error in the rearrangement that produces the two integrals in the first place, before either is even written down. Source: SEPDE §2.3.

Card 108 Front: What real, documented final-rearrangement error distributes a reciprocal incorrectly? Back: October 2021, Q2: converting a/y = b√(4x+5)+c to y = 1/(ab√(4x+5)) + 1/c distributes a reciprocal across a sum — since 1/(p+q) ≠ 1/p + 1/q. Source: SEPDE §2.4.

Card 109 Front: On the January 2026 Q3(a) closed form, what two-step final-rearrangement discipline does the real mark scheme require (and explicitly condone an arithmetic slip within)? Back: Combine the right-hand side into a single fraction before inverting (not inverting term-by-term), then square both full sides of the resulting equation, not each term separately. The scheme condones an arithmetic slip in the squaring step (landing on 5t² instead of 25t²) provided the method — squaring a whole side — is the one used. Source: SEPDE §2.4.

Card 110 Front: What is the "fudged-reverse-fit" trap on a "show that" separable-DE target, quoted directly? Back: A "show that" target "persuaded many candidates to 'adjust' their working following obvious mistakes" (Jan 2021, Q10) — an "ag" line is scored on the genuineness of the derivation, and quietly changing an intermediate number to land on the printed target is not credited even when the final line matches. Source: SEPDE §2.5; FORMULA §2.0.

Card 111 Front: On January 2026 Q3(b), how is the limit of h as t→∞ found from the closed form h=25t²/(1+4t)², and why does "differentiate and set to zero" fail here? Back: Divide every term by t² (equivalently, read off the ratio of the two leading t² coefficients) — every lower-order term vanishes as t→∞, leaving 25/16. "Differentiate and set to zero" fails because dh/dt is strictly positive for every finite t>0, so h(t) never has a stationary point at all, only an asymptote it approaches but never reaches. Source: SEPDE §4.

Card 112 Front: State spec 5.2's own guidance clause, verbatim. Back: "Formation of simple differential equations. Guidance: Questions involving connected rates of change may be set." Source: SEPDE scope section.

Card 113 Front: What is confirmed as this bundled topic's (5.2+6.4) real frequency, per this course's own SIGNAL cross-check? Back: Seen in 4 of 7 series that received real question-level mining — stated as a count against a small, named denominator, never as a percentage or predictive likelihood. Source: SEPDE §5.

10. Area Under a Parametric Curve (spec 6.5)

Card 114 Front: State the formula for the area under a parametric curve, and what must happen to the limits. Back: A = ∫y dx = ∫y (dx/dt) dt — limits converted from x-values into t-values before integrating. Source: FORMULA §1; AREA §2.1.

Card 115 Front: What is the "one hidden requirement" for A=∫y(dx/dt)dt to compute a genuine geometric area? Back: The map from t to x has to be one-to-one across the interval of t-values used — in practice, dx/dt keeps one sign (or is zero only at isolated points) across the whole interval, so x(t) is monotonic there. Source: AREA §2.2.

Card 116 Front: Find the area under x=t+1, y=t², t≥0, between x=1 and x=3 (x increasing with t). Back: dx/dt=1; x=1⟹t=0, x=3⟹t=2. A=∫₀² t²(1)dt = [t³/3]₀² = 8/3. Source: AREA §3, Case A.

Card 117 Front: Find the area under x=9−t², y=t, 0≤t≤3, between x=0 and x=9 (x decreasing with t — limits reverse). Back: dx/dt=−2t; x=9⟹t=0, x=0⟹t=3. A=∫_{t=3}^{t=0} t(−2t)dt = [−2t³/3] from 3 to 0 = 0−(−18) = 18 — confirmed by eliminating the parameter, y=√(9−x), ∫₀⁹√(9−x)dx=18. Source: AREA §3, Case B.

Card 118 Front: When a bounded region includes a straight edge as well as the curve (e.g. a normal, a chord), how should the area be found? Back: Split the region into a straight-edge component (computed directly via elementary coordinate geometry — a trapezium or triangle area) plus the curve's own parametric-integral component over its portion of the boundary, then add them. Do not force the whole region into a single parametric integral. Source: FORMULA §2.9; AREA §3, Case C.

Card 119 Front: Find the area under x=t²+2, y=3t, t≥0, between x=2 and x=6, by eliminating the parameter first. Back: Eliminating, t=√(x−2), y=3√(x−2); ∫₂⁶ 3√(x−2)dx = 16 — matching the parametric route ∫₀² 3t(2t)dt = ∫₀² 6t²dt = 16 exactly. Source: AREA §3, Case D.

Card 120 Front: What is the real anchor question for spec 6.5, and how are its 16 marks split across spec points? Back: WMA14/01A, January 2026, Q7 (16 marks): 2 (parametric coordinates, spec 3.1) + 5 (tangent/normal via parametric differentiation, spec 5.1) + 6+3 (the area itself, spec 6.5, split across a "show that" part and a "hence find" part). Source: AREA §4.

Card 121 Front: What does the general "ag" (answer given) marking convention require, applied to spec 6.5's two-component (trapezium + parametric integral) structure? Back: cso — "there must be no errors in this part of the question." Both the trapezium component and the parametric-integral component have to be individually correct, added in the right order, and left in the exact printed form for the final mark. Source: AREA §4.

Card 122 Front: What are the two genuinely new failure modes area-under-a-parametric-curve could introduce, reasoned as inferences from the adjacent substitution topic? Back: Dropping the dx/dt factor entirely (reasoned from "ignored the dx and simply wrote it as du," Oct 2021 Q6), and using the x-limits directly as t-limits without converting them (reasoned from "use of the x limits... instead of the u limits," Oct 2022 Q7i). Both are disclosed as reasoned inferences from an adjacent spec point, not direct evidence for 6.5 itself. Source: AREA §4.

Card 123 Front: (Illustrative, not a past-paper question) Curve C has x=t², y=t for 0≤t≤2, running from the origin to P=(4,2); a straight chord runs from P to (6,0). Find the total bounded area. Back: Area under C: ∫₀² t(2t)dt = [2t³/3]₀² = 16/3. Area under the chord (a right triangle, base 2, height 2): (1/2)(2)(2)=2. Total = 16/3 + 2 = 22/3. Source: AREA §3, Case C.

11. Vectors (spec 7.1–7.7)

Card 124 Front: State the magnitude formula for a vector a=(a₁,a₂,a₃), and the unit vector formula. Back: |a| = √(a₁²+a₂²+a₃²); unit vector â = a/|a|. Source: FORMULA §1 (spec 7.2 guidance).

Card 125 Front: State the position-vector identity for AB in terms of position vectors a and b. Back: OB − OA = AB = b − a. Source: FORMULA §1; POSVEC §1.

Card 126 Front: State the distance formula between two points (x₁,y₁,z₁) and (x₂,y₂,z₂). Back: d² = (x₁−x₂)² + (y₁−y₂)² + (z₁−z₂)². Source: FORMULA §1; POSVEC §1.

Card 127 Front: Why is the distance formula not a second, independent fact alongside position vectors — what is it really? Back: It's the position-vector identity AB=b−a composed with the ordinary 3D-Pythagoras magnitude formula: distance = |AB| = |b−a| = √((x₂−x₁)²+(y₂−y₁)²+(z₂−z₁)²); squaring recovers the distance formula exactly. Source: POSVEC §2.2.

Card 128 Front: State the two forms for a vector equation of a line. Back: r = a + tb (point + direction), or r = c + t(d−c) (two points). Source: FORMULA §1.

Card 129 Front: What is the ONE item on spec P4.2.3's "must memorise, will not appear in the booklet" list for the whole of P4? Back: The component-form scalar (dot) product: for (x,y,z) and (a,b,c), (x,y,z)·(a,b,c) = xa+yb+zc. Source: FORMULA §1; VECSKEW §1.

Card 130 Front: State the angle-between-two-vectors formula, and why does arccos never fail to produce a value from it? Back: cos∠AOB = (a·b)/(|a||b|). It never fails because for any two nonzero vectors, |(a·b)/(|a||b|)| ≤ 1 always (Cauchy–Schwarz specialised to 3 real dimensions) — so cosθ never lands outside [−1,1]. Source: FORMULA §1; VECSKEW §2.

Card 131 Front: State the perpendicularity and parallel conditions for two nonzero vectors a, b. Back: Perpendicular: a·b=0. Parallel: a=kb for some real scalar k — not merely a≠b (a confirmed real trap shows two direction vectors aren't identical and wrongly treats that as proof they aren't parallel, Oct 2022 Q9). Source: FORMULA §1.

Card 132 Front: State the skew-lines condition, verbatim from spec guidance. Back: Spec 7.6 guidance: "Conditions for two lines to be parallel, intersecting or skew" — two lines are skew iff they are not parallel AND do not intersect; both conditions required. "Not perpendicular" is never a valid substitute for either condition (Jan 2021, Q8). Source: FORMULA §1.

Card 133 Front: State the two real-mark-scheme-credited routes for the area of a vector-defined parallelogram. Back: Area = |a||b|sinθ, or equivalently Area = |a×b| — both credited on equal footing (Jan 2026, Q9, "Alt(c)"). Source: FORMULA §1.

Card 134 Front: What are the four (not three) possible classifications of two lines in 3-space once their direction vectors are checked? Back: Coincident (same line, described twice), distinct parallel (never meet, share a direction), intersecting (meet at exactly one point), or skew (not parallel and don't intersect) — "parallel" in the spec's three-way guidance actually splits into two sub-cases. Source: VECSKEW §4.

Card 135 Front: Once two lines are confirmed parallel, what single test distinguishes coincident from distinct-parallel? Back: Take any one known point from the first line's equation and substitute it into the second line's equation. If a consistent parameter value exists across all components, the lines are coincident; if the system is inconsistent, they are distinct parallel lines. Source: VECSKEW §4.

Card 136 Front: Are l₁: r=(1,2,3)+t(2,−1,4) and l₂: r=(5,0,11)+s(2,−1,4) coincident or distinct parallel? Back: Same direction vector, so parallel. Does (5,0,11) lie on l₁? x: 1+2t=5⟹t=2; y: 2−t=0⟹t=2; z: 3+4t=11⟹t=2 — all consistent at t=2, so the lines are coincident (the same line, written two ways). Source: VECSKEW §4.

Card 137 Front: State the closed-form solution for the parameter t* at the foot of the perpendicular from external point a to line l: r=c+td. Back: t* = [(a−c)·d] / (d·d) = [(a−c)·d] / |d|². Source: POSVEC §2.3.

Card 138 Front: State the closed-form solution for the shortest distance squared from external point a to line l: r=c+td, at the foot of the perpendicular. Back: |AX|²_min = |a−c|² − [(a−c)·d]² / |d|² — a Pythagorean decomposition: |a−c|² is the squared distance from A to known point C on the line, minus the squared length of C's own projection onto d. Source: POSVEC §2.3.

Card 139 Front: What is the single most robustly attested error pattern in the whole WMA14 vectors facts bank, confirmed on how many independent series? Back: Using the wrong pair of vectors for a named angle (or substituting a position vector for the needed direction vector) — confirmed three times independently (Jan 2021 Q2a; Oct 2022 Q3b; Jan 2024 Q6c). Source: FORMULA §2.10; VECSKEW §6.

Card 140 Front: What is the documented foot-of-perpendicular trap, and what does it actually compute (structural observation)? Back: Dotting the line's raw general point directly against d — (c+td)·d=0 — instead of first subtracting a to form AX. This wrong equation is not gibberish: it is exactly the correct condition for the foot of the perpendicular from the origin, not from the actual external point A. Real quote: "most incorrect responses attempted to set (c+λd)·d=0 rather than AX·d=0" (Oct 2021, Q7). Source: POSVEC §2.4.

Card 141 Front: Find the foot of the perpendicular X from A=(5,2,0) to line l: r=(1,−2,3)+t(2,1,−2), and the shortest distance. Back: General point X=(1+2t,−2+t,3−2t); AX=(2t−4, t−4, 3−2t). AX·d=0: 2(2t−4)+1(t−4)−2(3−2t)=9t−18=0⟹t=2. X=(5,0,−1); shortest distance = |AX| = |(0,−2,−1)| = √5. Source: POSVEC §3, Case B.

Card 142 Front: State the reflection formula for the reflection B of point A in line l, given the foot of the perpendicular X. Back: B = 2X − a — the same position-vector midpoint identity from spec 7.4 (AX=XB), not a new formula. Source: FORMULA §2.10; POSVEC §3, Case D.

Card 143 Front: WMA14 October 2021 Q7 real anchor: state the mark-scheme's dependency chain for part (a) (finding the foot of perpendicular). Back: M1 (form AX from the general point) → M1 (the full method: dot with d, solve for λ) → A1 (λ=2/5) → dM1 (dependent on the preceding M1, for substituting λ back in to get X's coordinates). Source: POSVEC §4.

Card 144 Front: On Oct 2021 Q7, what wrong value of λ does the documented trap equation produce, versus the correct value? Back: The wrong equation ((c+λd)·d=0) gives λ=37/50 — a specific, plausible-looking wrong value — against the correct λ=2/5. Source: POSVEC §4.

Card 145 Front: For what value of k are (k,3,−2) and (4,k,1) perpendicular? Back: Set the dot product to zero: 4k+3k+(−2)(1)=0 ⟹ 7k−2=0 ⟹ k=2/7. Source: VECSKEW §5.

Card 146 Front: What real, documented finding concerns which angle (acute vs obtuse) to report between two named sides of a shape? Back: "Many candidates gave the acute angle... as their final answer" when the obtuse angle was required (Jan 2021, Q2a). Reaching the supplementary angle and simply taking its acute complement, without a convincing geometric justification, does not earn the final accuracy mark on a between-two-sides-of-a-shape question (Jan 2026, Q9b). Source: FORMULA §2.10; VECSKEW §6.

Card 147 Front: WMA14 June 2022 Q6: A=(1,−4,3), B=(5,3,−2). Find the vector equation of line l through A and B. Back: Direction AB = B−A = (4,7,−5). Vector equation: r = (i−4j+3k) + λ(4i+7j−5k). Source: PAPER, Question 6.

Card 148 Front: In the same Q6, with C=(3,p,−1) and AC perpendicular to l, find p, then the exact area of triangle ABC. Back: AC=(2,p+4,−4); AB·AC=0 ⟹ 4(2)+7(p+4)+(−5)(−4)=0 ⟹ 7p+56=0 ⟹ p=−8. With p=−8, AC=(2,−4,−4), |AB|=√90=3√10, |AC|=√36=6; since AC⊥AB, area = (1/2)|AB||AC| = (1/2)(3√10)(6) = 9√10. Source: PAPER, Question 6.

What this document does not add

Consistent with this course's standing no-fabrication discipline: no card above states a likelihood, a grade boundary, or a frequency claim beyond what its source document already states as a plain "N of M series checked" count. Where a source flags a mechanism, formula, or micro-example as course-original reasoning rather than an exam-board-sourced fact (the foot-of-perpendicular closed form and distance-squared identity, the Maclaurin derivation of the binomial coefficient, the four-way line-classification proof, the general integration-by-parts reduction formulas, the disk-method cross-sectional-area theorem), this deck's corresponding card is drawn from that same flagged content and does not present it as a Pearson-sourced fact. No card introduces a worked example, a numeric answer, or a mark-scheme quotation that is not already present in one of the thirteen source files listed at the top of this document. Every direct quotation above is reproduced under fifteen words with its own series/question attribution, exactly as its source document already quotes it.

Sources used

codex-formula-and-method-bank.md (Parts 1 and 2 in full); codex-deep-dive-proof-by-contradiction.md; codex-deep-dive-partial-fractions-decomposition-and-integration.md; codex-deep-dive-parametric-equations-conversion-and-domain-range.md; codex-deep-dive-binomial-expansion-for-rational-n.md; codex-deep-dive-implicit-and-parametric-differentiation-tangents-and-normals.md; codex-deep-dive-separable-differential-equations.md; codex-deep-dive-volume-of-revolution.md; codex-deep-dive-integration-by-parts-and-substitution.md; codex-deep-dive-area-under-a-parametric-curve.md; codex-deep-dive-position-vectors-distance-and-foot-of-perpendicular.md; codex-deep-dive-vectors-scalar-product-and-skew-lines.md; codex-complete-worked-paper.md — all thirteen files read in full before this document was written, all in content/veridian/technique-hub/wma14/. No primary Pearson PDF, examiner's report, or mark scheme was fetched or re-read for this document; every claim traces to the reformatting sources above, which had already independently verified it.

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