Pure Mathematics 4

Level Threshold Guide

Provenance: written 2026-09-29, grounded in research/veridian/WMA14-verified-facts.md and cross-checked against the three existing WMA14 technique-hub documents so this file does not restate them: forge-exam-technique-guide.md (§1's tariff table and §2's general M/A/B strategy — this guide goes one layer deeper than both, into the literal mark-by-mark gates inside real questions, not the strategy for playing them), signal-frequency-and-boundaries.md (§2's confirmed finding that no grade-boundary primary source exists anywhere in this course's WMA14 review — restated and honoured here, not re-litigated), and forge-battle-card.md (condensation only, no new marking detail). Two real anchors — October 2024 Q1 (binomial) and October 2024 Q2 (proof) — already have a full mark-by-mark breakdown in content/veridian/question-bank/wma14.ts; this guide cross-references both briefly rather than reproducing them, and spends its own length on nine further real anchors that have no full mark-by-mark write-up anywhere else in this repository. Every mark code, quote and numeric value below traces to the facts bank's own independently-verified extraction of the primary mark schemes; every quoted fragment is under fifteen words with its own series/question attribution, and all longer mark-scheme description is this guide's own paraphrase, not a reproduction of Pearson's text.

#WMA14 · FORGE — Level/Threshold Guide

#0. What this document is, and what it deliberately is not

WMA14 is not marked the way an essay-based subject is. There is no holistic Level 1–5 descriptor scale here — no examiner reading a whole answer and placing it in a band. Every mark on this paper is a discrete, named M/A/B (or dependent-mark) line, gated on a specific, checkable condition. The real "level/threshold" structure of this paper is that mark-by-mark grammar itself — so that is what this guide breaks down, exhaustively, one real anchor at a time, rather than manufacturing a holistic band system this subject does not have.

Two honesty constraints, stated up front and held throughout:

No grade-boundary figure appears anywhere below, and none is implied. signal-frequency-and-boundaries.md §2 already established that no primary Pearson grade-boundary document was found anywhere in this course's WMA14 review. This guide's "thresholds" are the internal mark-to-mark gates inside a single question's own scheme — the boundary between M1 and M0, between a dependent mark earned and withheld — never a raw-score-to-grade boundary.
Where the real source material only verifies part of a mark chain, this guide says so and stops there, rather than inventing the missing marks' wording to make a ladder look complete. Several entries below are explicitly marked partial for exactly this reason.

#1. The base vocabulary — the gates that apply to every question

forge-exam-technique-guide.md §2 already covers how to play this system (quote the formula first, keep exact values, write every stage). This section defines the gates themselves, more precisely than either that guide or the curriculum files state them, because the worked ladders in §3 all depend on these distinctions being exact.

M1 — method attempted and applied. The generic gate for calculus specifically: a power decreased by one differentiating, or increased by one integrating, appearing anywhere in the working — even with an arithmetic slip afterwards — clears the mark. Writing the original expression down unchanged and asserting a final value, with no power-shift attempted anywhere, does not.

A1 — cannot stand without its M mark. An arithmetically correct-looking final value reached by a method that never earned its M is not itself creditable — the real, cleanest example is the sign-of-index misread on October 2024 Q1 (binomial): the scheme names an explicit cap of "B1M1A0A0 M0A0" for reading the index as $+\frac13$ instead of $-\frac13$ — the shape of the method survives (B1, M1), but every A mark gated on the correct index is lost, even though the candidate's arithmetic from that point on is internally consistent. (already fully broken down in content/veridian/question-bank/wma14.ts; cited here only for the general A-gated-on-M principle it demonstrates.)

B1 — unconditional, but not undemanding. "Independent of M marks" does not mean "awarded for any plausible-looking value" — it means the mark does not require a specific preceding method line to have been scored, while still requiring the value itself to be correct. January 2021 Q6(b)'s B1 for the non-origin $y$ -value is a real, clean example: the scheme's own line credits $y=\frac32$ regardless of whether the candidate discarded the extraneous root explicitly or not — an earlier version of this course's own lesson stated the reverse (that discarding the wrong root was part of what the mark required), a documented, corrected error worth naming here because it shows how easy it is to over-tighten a B mark's real threshold by importing an A-mark-style "show your reasoning" expectation onto a mark the scheme never gated that way.

Dependent marks (dM1, and — genuinely new vocabulary this guide needed — ddM1). A dependent mark is withheld even when its own arithmetic is flawless, if the mark it depends on was not itself scored. §3.6 below (January 2026 Q7) is the richest real example on this paper of a chain of dependency — $M1A1\to dM1\to ddM1{,}A1$ — a mark dependent on a mark that is itself dependent on an earlier mark. The practical threshold: a correct final number reached by skipping the specific earlier-scored step the chain requires does not retroactively earn the marks it was never actually linked to.

ft (follow-through) is not the same promise on every mark it's attached to. The general rule for a genuine misread is a flat deduction of two from the A/B marks in the affected part. But a specific question can print its own explicit cap instead of relying on that general rule — the October 2024 Q1 misread note above is exactly that: a named cap, not the generic "-2" deduction. A ft-tagged B mark is a third, distinct case again — January 2026 Q7(c)'s trapezium-area line is scored B1ft, which sounds like a contradiction against "B marks are independent," until read correctly: the mark is independent of which method mark was scored, but it still follows through on the candidate's own earlier numerical values (their own $y$ -intercept and their own point $P$ ) rather than demanding the one fixed correct pair. Three genuinely different senses of "independent," attached to the same B1/dM1/ft vocabulary, and confusing them is itself a real threshold-reading trap.

cao vs. cso is the sharpest real distinction on this paper, and it is asymmetric. cao ("correct answer only") needs the printed value; cso ("correct solution only") needs the entire preceding chain to be free of error, verbatim: no error anywhere in that part of the question. An A1* on a "show that" line functions exactly like cso even where the scheme doesn't print the letters — the Summer 2022 Q9 proof anchor's own A1* line requires that "all working must have been correct" before the final mark is available at all (§3.1 below). Contrast this against January 2026 Q9(a)'s cao A1 for the vector $\mathbf d$ , which the scheme explicitly allows with no working shown at all — "no-working scores all 3" — the opposite extreme of the same cao/cso spectrum, on the same paper, in adjacent parts of the same question type.

Exact-answer is the default; a condoned decimal is a named exception, not evidence the rule is soft. The general rule: marks are "normally lost" resorting to a rounded decimal where exact working was required. January 2026 Q3(b) is the real, quotable exception that proves this is a default rather than an absolute: the scheme's own A1ft explicitly condones the decimal $1.5625$ alongside the exact $\frac{25}{16}$ — because the question never demanded an exact form for that specific final value. Read every "condone [decimal]" note as a deliberate, scheme-authored carve-out, not as evidence the general exactness rule has weakened.

Equal-footing alternative routes are a real, checkable threshold pattern, not folklore. January 2026 Q8 is the cleanest anchor: three structurally different proof mechanisms — a divisor-pair route, a parity route, a substitution route — are all scored on the identical $M1\to A1\to dM1\to A1^{*}$ skeleton, and the scheme adds its own generic catch-all inviting review of any further valid method. The threshold is written at the level of what the candidate has established, not which specific numbers or algebraic path they used to establish it — see §3.2.

#2. Full worked threshold ladders — real anchors, mark by mark

Ordered by spec point. Each entry states only what the facts bank verifies verbatim or by direct citation; where a mark's exact wording was not captured in the source review, that gap is named rather than filled.

#2.1 Proof by contradiction (spec 1.1)

Anchor A — June 2022 Q9, " $n^2-2$ is never divisible by 4" (4 marks). The facts bank verifies two of this question's four marks verbatim; the middle two are attested only by citation (the sequence $M1\to A1\to dM1\to A1^{*}$ , from forge-exam-technique-guide.md §2), not quoted, so this guide presents only what is genuinely sourced rather than inventing wording for the middle pair.

M1 threshold. Clears with any algebraic statement in an integer variable that amounts to an assumption — the scheme is explicit that no formally-worded assumption sentence is required at this stage, "no need for explicit statement" (5 words, Jun 2022 MS). Misses only if no algebraic setup of any kind is attempted.
A1* threshold (the final mark). This is where the real gate tightens hard. The scheme requires "all working must have been correct" (5 words) before the mark is available at all — this is a cso gate wearing an A1* label, not an ordinary cao mark. A candidate who reaches the right final contradiction sentence after an earlier, uncorrected slip has not cleared this mark, regardless of how the concluding sentence itself reads. The scheme is also explicitly generous about how the original assumption is worded — accepting a bare "assume $n^2-2$ is divisible by 4 for all $n$ " as sufficient — so the tightness is entirely in the correctness of the chain, not its prose.

Anchor B — January 2026 Q8, $4p^2-q^2=46$ has no positive-integer solutions (1 + 4 marks). The richest real anchor on this paper for the "equal-footing alternative routes" pattern named in §1.

Part (a), 1 mark — factorise $4p^2-q^2$ . The facts bank confirms this is worth 1 mark but does not capture which single-letter code the scheme prints for it; presented here as a genuine, stated gap rather than a guessed B1/M1.
Part (b), the $M1\to A1\to dM1\to A1^{*}$ skeleton, shared by all three credited routes. For the divisor-pair route: clears M1 by factorising to $(2p+q)(2p-q)=46$ and identifying a genuine positive-integer factor pair; clears A1 by correctly solving the resulting simultaneous equations for at least one pair (reaching, e.g., $p=11.75$ ); clears the dM1 only by attempting both valid factor pairs ( $46\times1$ and $23\times2$ ) — the scheme's own line requires attempting both, and a candidate who checks only one pair, reaches a correct non-integer $p$ , and stops has not cleared this specific dependent mark, even though their one checked pair is itself flawless. Clears A1* by concluding neither pair gives an integer $p$ , hence no solution exists.
A genuinely useful "does not need to clear" carve-out. The scheme explicitly does not require a candidate to justify why $2p+q$ is taken to be the larger factor before assigning the pair — it states there is no such justification requirement (paraphrased; the scheme's own wording is longer than 15 words). This means a "just misses" version of this stage is narrower than it looks: skipping the ordering justification is not itself a miss; skipping one of the two factor-pair checks is.
The parity route and the substitution route are scored on the identical skeleton, not a lighter-weight alternative — a candidate who correctly reasons that 46's odd/even factorisation rules out both an odd-times-odd and an even-times-even product, or who substitutes $q=2m$ to force $m^2=p^2-\frac{23}{2}$ (an impossible value for an integer $m$ ), clears exactly the same four marks as the divisor-pair route. The mark scheme also invites review of any further method that establishes the same impossibility — the threshold is defined by what is established, not by which of the three named paths reached it.

Cross-referenced, not reproduced: October 2024 Q2 (curves $C_1$ , $C_2$ don't intersect) is already a full, real, cited $B1\to M1\to A1\to A1$ (dependent) ladder with its own commonWrongPath — a calculator-only jump to the roots that keeps the independent B1 but forfeits the M1 and both dependent A marks — in content/veridian/question-bank/wma14.ts.

#2.2 Parametric conversion & domain/range (spec 3.1)

Anchor — January 2021 Q4, $x=\frac1t+2$ , $y=\frac{1-2t}{3+t}$ , $t>0$ (5 + 2 marks). Part (a)'s "Alternative 1" route (assume the target form $g(x)=\frac{ax+b}{cx+d}$ and substitute) is fully verified; its rival main route and part (b)'s second mark are only partly verified, and that gap is stated rather than papered over.

Alternative 1 ladder, fully verified. Clears M1 by assuming the general rational form and substituting $x=\frac1t+2$ into it. Clears the first A1 by reaching the correct form in $t$ for both numerator and denominator. Clears the second A1 (an M1-weighted accuracy line on Pearson's own marking software) for either the numerator or denominator alone being correct — a genuine partial-credit rung between the first A1 and the fully-simplified final line. Clears the final A1 by reaching $g(x)=\frac{x-4}{3x-5}$ .
Domain B1, independently gated. Clears with $k=2$ (or $x>2$ ) stated or used, regardless of which of the two algebra routes reached it — this mark sits outside the algebra ladder entirely, worth 1 of the 5 part-(a) marks on its own.
Part (b) — one mark verified, one not. The scheme's M1 clears by finding either boundary value ( $-2$ or $\frac13$ ) via direct evaluation at the excluded endpoint or the leading-coefficient limit $\frac{a}{c}$ — a genuinely low bar for the method mark, since only one of the two numbers is required. The second, accuracy mark's exact wording was not captured in this session's source review; this guide states that gap rather than authoring a plausible-sounding A1 criterion to match it.
What "just misses" the M1 looks like, concretely. A candidate who substitutes only $t\to\infty$ correctly (reaching $\frac13$ ) but never attempts the excluded-endpoint value $g(2)=-2$ has still cleared the M1 — the scheme explicitly credits either boundary. The real miss is a candidate who states a range with no attempt at either boundary value shown, or who substitutes a $t$ -value directly into the parametric $y(t)$ formula rather than the derived cartesian $g(x)$ — the domain/range method this anchor actually needs is a limit argument on the excluded/unbounded parameter interval, not an endpoint substitution, and a closed-endpoint-style attempt on an open interval does not clear this gate no matter how carefully executed.

#2.3 Implicit differentiation (spec 5.1) — partial ladder, stated as such

Anchor — January 2021 Q6, $4y^2+3x=6ye^{-2x}$ (5 + 4 marks). Only two of this question's nine marks have verbatim scheme wording in the source review; presented as exactly that, not as a full ladder.

A1(a), the harder chain-rule half. The real scheme's credited line differentiates the $6ye^{-2x}$ term to $-12ye^{-2x}+6e^{-2x}\frac{dy}{dx}$ — the threshold this mark actually tests is applying the chain rule to the $x$ -side factor $e^{-2x}$ (producing its own $-2e^{-2x}$ multiplier) at the same time as the product rule's $y$ -side term, not just the more commonly-drilled case of a bare-power $x$ -side factor. A "just misses" answer differentiates the $y$ -side correctly but treats $e^{-2x}$ as if its own derivative were $1$ — a term-count-correct, coefficient-wrong answer that this specific A1 does not credit.
B1(b), the cao threshold corrected. Clears with $y=\frac32$ obtained from $x=0$ , and the scheme's own bracketed note states explicitly that any reference to the extraneous root $y=0$ is to be ignored — i.e. neither stating nor omitting a discard-the-wrong-root sentence changes whether this mark is earned. §1 above already names this as the corrected general principle; this is its real anchor.

#2.4 Separable differential equations (spec 5.2, 6.4)

Anchor — January 2026 Q3, shrub-height model $\frac{dh}{dt}=\frac{2h^{3/2}}{5t^2}$ (6 + 2 marks). One of the fullest verified ladders on this paper — six of six marks in part (a) have verbatim or near-verbatim scheme wording.

Part (a), six-mark ladder. B1 for the correctly separated integral. M1 for reaching the general integrated form (a constant times $h^{-1/2}$ equal to a constant times $t^{-1}$ , plus $c$ ). A1 for the specific correct coefficients on that line. M1 for using the given boundary condition $h=1$ at $t=1$ to find $c=-\frac45$ — a mark that clears independently of whether the rearrangement that follows is correct. M1 for the rearrangement itself: forming a single fraction before inverting (not inverting term-by-term across a sum) and squaring both complete sides. A1 (cao) for the fully reduced $h=\frac{25t^2}{(1+4t)^2}$ .
- The single sharpest "just misses" in this ladder is at the second rearrangement M1: a candidate who correctly finds $c=-\frac45$ (clearing the prior M1) but then inverts $\frac1{c-\frac15t^{-1}}$ term-by-term rather than as a single fraction reaches a plausible-looking but structurally wrong expression for $h$ — the mark tests exactly this move, not the boundary-condition step that precedes it, and a slip here costs this M1 and the final cao A1 while leaving the four earlier marks intact.
Part (b), the new closing step this anchor adds. M1 for recognising the limiting value directly from the closed rational form — dividing every term by $t^2$ and reading off the ratio of the leading $t^2$ coefficients ( $\frac{\text{their }a}{\text{their }b^2}$ ) — not by differentiating $h$ to hunt for a stationary point, which is independently confirmed to be a dead end for this exact function (this $h(t)$ is strictly increasing throughout its domain and has no critical point at all, only an asymptotic limit). A candidate who spends the mark budget searching for $\frac{dh}{dt}=0$ has chosen a method that cannot succeed on this specific question, however competently executed. A1ft for the exact value $\frac{25}{16}$ , with the decimal $1.5625$ explicitly condoned — the named exception to the general exact-answer rule discussed in §1.

#2.5 Integration by parts, repeated (spec 6.2)

Anchor — June 2022 Q8(b), $\int x^2e^{-x}\,dx$ inside a 10-mark question. The facts bank verifies the two method-mark lines in full; the question's remaining accuracy marks are not captured verbatim, so this entry is scoped to the method-mark gate it actually clarifies.

The single most load-bearing threshold this anchor demonstrates: the second application of parts is a dependent method mark, not an independent one. The scheme's own dM1 line is explicit that the second application is dependent on the first. This course's own lesson previously mis-taught this exact step as an independent M1 in three separate places (the two-application worked method, the DI/tabular alternative's diagonal-multiplication step, and a second worked chain) — a documented, corrected error, cited here because it is the cleanest real illustration of the dependent-mark threshold: a second application executed flawlessly on the wrong intermediate integral (i.e. following a slip in the first application) does not earn this mark, because the mark is gated on the first application's own success, not on the second application's own internal correctness.
The DI (tabular) method's own, distinct threshold. The scheme credits this alternative with its own ladder: the first two rows of the table having the correct form clears the method mark; every row correct with the answer correctly extracted clears the dependent mark; a further mark is available for the fully correct final answer. This is a genuinely different gate shape from the "traditional" two-line method — a tabular attempt with one wrong row anywhere in the middle can still clear the first mark (the first two rows are what that mark actually checks) while missing the second.

#2.6 Volume of revolution (spec 6.1) — partial ladder, stated as such

Anchor — January 2021 Q9, $x=\tan\theta$ , $y=2\sin2\theta$ (7 + 3 marks). Three of this question's ten marks have verbatim or near-verbatim wording in the source review; the remainder do not, and this entry does not attempt to reconstruct them.

B1, the limit-finding mark. Clears with $\tan\theta=\sqrt3\Rightarrow\theta=\frac\pi3$ stated or used anywhere as the upper limit — a genuinely low bar (the scheme explicitly allows this to be credited wherever it appears in the response, not only at a specific expected line).
Two separate dM1s on the same $y^2$ , testing two different identities — not one identity used twice. The first requires expanding $\sin2\theta=2\sin\theta\cos\theta$ inside the square so the resulting $\cos^2\theta$ term can cancel against the $\sec^2\theta$ produced by $\frac{dx}{d\theta}$ . The second, entirely separate mark, requires the power-reduction identity $\sin^2\theta=\frac{1-\cos2\theta}2$ to make the resulting bare $\sin^2\theta$ term integrable. A candidate who applies the double-angle expansion correctly but then tries to integrate the remaining $\sin^2\theta$ by (incorrectly) treating it as already in a standard integrable form has cleared the first dM1 and missed the second — the two marks test genuinely different trigonometric moves, not a repeat of the same one.

#2.7 Area under a parametric curve (spec 6.5) — the richest single ladder on this paper

Anchor — January 2026 Q7, $x=8\cos^3\theta$ , $y=6\sin^2\theta$ (2 + 5 + 6 + 3 = 16 marks, the largest confirmed real item in the whole facts bank). Every mark across all four parts has verbatim or near-verbatim scheme wording — the only anchor on this paper where that is true of a complete double-digit-mark question.

Part (a), 2 marks — B1B1. Two independent marks, one per coordinate, for $P=(1,4.5)$ at $\theta=\frac\pi3$ . The threshold here is flat: each coordinate is right or it isn't, with no method-dependency at all — the simplest gate shape on this whole anchor, worth naming precisely because the other three parts are so much more layered.
Part (b), 5 marks — the clearest three-deep dependency chain on this paper. $M1A1$ for setting up and correctly finding the parametric gradient $\frac{dy}{dx}=-\frac1{2\cos\theta}$ ; dM1, dependent on that pair, for evaluating it at $\theta=\frac\pi3$ to get $-1$ ; ddM1, A1 — a mark dependent on the dM1 above it — for taking the normal gradient ( $=1$ ) and forming the line through $P$ to reach $y=x+3.5$ . The concrete "just misses": a candidate who correctly forms the general gradient formula ( $M1A1$ cleared) but evaluates it at the wrong parameter value (say $\theta=\frac\pi6$ ) has broken the chain at the dM1 — and because the ddM1 is gated on that mark, not merely on the mechanical correctness of "invert, negate, substitute a point," the final normal equation is unrecoverable for credit even if every subsequent line is executed perfectly on the wrong gradient.
Part (c), 6 marks — the B1ft nuance named in §1. $M1A1$ for setting up the parametric area integral in $\theta$ ; dM1 for using $\sin^2\theta=1-\cos^2\theta$ to reach an integrable form; B1ft for the trapezium area under the normal (value $4$ , following through on the candidate's own intercept and $P$ -value rather than demanding one fixed pair); a final dM1, A1* for combining the trapezium with the sign-adjusted parametric integral to reach the given printed target. The A1* here is a genuine cso gate on a "show that" line — matching the June 2022 Q9 pattern in §2.1, on a completely different spec point, at six times the mark value.
Part (d), 3 marks — a clean M1/A1/A1(cso) ladder. M1 for the correct antiderivative form via $\sin\theta\cos^n\theta\to-\frac{\cos^{n+1}\theta}{n+1}$ ; A1 for the correct value of the definite integral, $\frac{47}{480}$ ; a final A1 (cso) for the fully combined exact area, $\frac{181}{10}$ — cso here meaning the entire numerical chain from part (c)'s given expression onward must be error-free, not merely that the final fraction happens to be right.
A genuinely paper-wide finding this anchor sharpens: complexity of the dependency chain itself, not just the mark total, is what separates this question's four parts. Part (a) has zero dependency between its two marks; part (b) has three layers (M1A1 → dM1 → ddM1,A1); this is a real, checkable escalation inside a single question, not a pattern this guide is asserting from outside the evidence.

#2.8 Position vectors, distance, foot of perpendicular (spec 7.4–7.5)

Anchor — October 2021 Q7, line $l=(4,2,-3)+\lambda(-4,-3,5)$ , point $A(9,-3,2)$ (7 + 2 marks). Part (a)(i)'s four marks are fully verified; part (a)(ii)'s own marks were not captured in the source review and are named as a gap rather than reconstructed.

Part (a)(i), a clean four-mark chain. M1 for forming the line's general point and attempting a difference vector from $A$ to it, in either order. A further, independent M1 for the complete method — setting that difference vector's dot product with the direction vector to zero and solving. A1 for the correct parameter value, $\lambda=\frac25$ . dM1, dependent on the second M1, for substituting that $\lambda$ back in to find $X$ 's coordinates.
The single most consequential "just misses" on this whole paper's vector content. The real, documented wrong path dots the line's general point directly against the direction vector — skipping the subtraction that forms the true difference vector $\overrightarrow{AX}$ — and this is genuinely solvable, not a dead end: it yields $\lambda=\frac{37}{50}$ , a specific, plausible-looking wrong value, independently confirmed against the correct $\lambda=\frac25$ found via the real method. The mark is lost not because the resulting equation can't be solved, but because the vector being dotted was never $\overrightarrow{AX}$ — a threshold about which object enters the dot product, not about algebraic competence executing it.
Part (b), 2 marks — reusing the position-vector reflection identity. M1 for using $\overrightarrow{AX}=\overrightarrow{XB}$ (or an equivalent correct method) to find $B$ as the reflection of $A$ in the line. A1 for the correct position vector, $B=-\frac{21}5\mathbf i+\frac{23}5\mathbf j-4\mathbf k$ . No documented wrong path exists for this part specifically in the source review — the examiner's report is confirmed to carry no part-(b)-specific detail beyond noting many candidates did not attempt it after struggling with part (a).

#2.9 Vector-defined area (spec 7.6–7.7)

Anchor — January 2026 Q9, parallelogram $ABCD$ from position vectors $\mathbf a,\mathbf b,\mathbf c$ (3 + 4 + 2 + 2 = 11 marks). The clearest real demonstration on this paper that a cao A1 can be far more generous than the general "show your working" caution elsewhere in this course's material would suggest.

Part (a), 3 marks. M1 for attempting any of the four possible difference vectors ( $\pm(\mathbf b-\mathbf a)$ , $\pm(\mathbf c-\mathbf b)$ ), correct in at least one component. dM1, dependent on that, for attempting $\mathbf d=\mathbf a-\mathbf b+\mathbf c$ , again correct in at least one component. A1 (cao) for the fully correct $\mathbf d$ — and the scheme states explicitly that a bald correct answer with no working shown at all earns all three marks. This is a real, quotable exception to the general "answers without working" caution in forge-exam-technique-guide.md §2: on this specific part, showing no working is not a risk, because the mark scheme was written to allow it.
Part (b), 4 marks — the sharpest real angle-threshold on this paper. M1 for using a correct pair of vectors sharing vertex $B$ (either direction is accepted at this stage). dM1, dependent, for attempting the dot-product formula for $\cos\theta$ . A1 for $\frac13$ (or $-\frac13$ ). A final, separate A1 (cso) for reporting $70.5°$ specifically — and the scheme's own real nuance is that reaching $109.5°$ and simply writing "therefore $70.5°$ " scores A0, unless accompanied by a genuine supporting argument (e.g. a correct diagram) that $109.5°$ is the exterior angle. This is the opposite convention from an ordinary angle-between-two-lines question, where the acute value is generally the safe default — here, the specific geometric angle named by the question ( $\angle ABC$ , an interior angle of the shape) has to be argued for, not defaulted to.
Part (c), 2 marks — two routes, identical credit. M1 for the parallelogram-area formula $|\overrightarrow{BA}||\overrightarrow{BC}|\sin\theta$ using a correctly-identified vector pair; A1 for $45.3$ (to 3 s.f.), also credited from the supplementary angle since $\sin$ is unchanged. A fully separate Alt(c) route — the vector cross product $|\overrightarrow{BA}\times\overrightarrow{BC}|$ — is named explicitly in the scheme and scored on the identical M1/A1 skeleton for the same numeric answer, independently confirmed to agree exactly ( $32\sqrt2$ both ways).
Part (d), 2 marks — must build on the candidate's own part (c). M1 for using $1.5\times$ their own part-(c) answer (or an equivalent triangle-addition route); A1 for $67.9$ (or $48\sqrt2$ ). The scheme's own instruction is explicit that this part must use the candidate's own earlier answer to score — a genuinely different kind of dependency from a dM1: not "must have earned an earlier mark," but "must be numerically built on an earlier answer, right or wrong."

#3. What actually separates a "just misses" from a "just clears," read across all nine anchors

Not a restatement of forge-exam-technique-guide.md §4's own "fudged-reverse-fit" and "wrong-reference-point" patterns — this is a finer-grained reading of the same evidence, at the level of individual marks rather than whole questions. Across every real near-miss documented above, the mathematics itself is essentially never what fails. What fails is the linkage between stages: dotting the wrong object against a direction vector because the subtraction step was skipped (§2.8); checking one factor pair instead of both because the dependent mark's own requirement was not read closely (§2.1); defaulting to the acute angle because that is the safe convention on a different type of angle question (§2.9); inverting a fraction term-by-term because the specific rearrangement a dependent mark tests was executed differently from how it was practised (§2.4). The real threshold on this paper is almost never "can you do the calculus/algebra," and almost always "did you perform the exact linking step the dependent or accuracy mark is actually checking." This is a genuinely new synthesis, built from reading all nine ladders side by side, not copied from any single anchor's own commentary.

#4. Flagged gaps

No holistic level-descriptor system exists for this paper, and none is implied above — §0 restates this as the document's own governing constraint, not just an opening disclaimer.
No grade-boundary figure appears anywhere in this document. Confirmed absent from this course's entire WMA14 source review (signal-frequency-and-boundaries.md §2); this guide adds no attempt to estimate one.
Coverage across the nine anchors is genuinely uneven, and this document says so at each entry rather than smoothing it over. January 2026 Q7 and Q9 have every mark verified; June 2022 Q9, January 2021 Q4, January 2021 Q6 and January 2021 Q9 each have real verbatim material for only part of their total marks, and this guide states exactly which marks are and are not covered in each case rather than inventing plausible wording for the rest.
Two spec points — partial fractions (2.1) and binomial expansion's non-Oct-2024 series (4.1) — have no genuine mark-code-level material in the source review, only examiner-report prose ("most candidates...", "a common error was...") with no accompanying M/A/B breakdown. No entry for either is included above for that reason, rather than one built from inferred mark codes.
The January 2026 series (Q3, Q7, Q8, Q9) has no published examiner's report yet, confirmed directly in the facts bank. Every "just misses" example drawn from that series above is therefore built from the mark scheme's own stated exclusions and generosity notes (what it explicitly credits or caps), never from a documented candidate error, and this guide has been careful to phrase each one as a description of what the scheme would not credit, not as a claim about what real candidates actually wrote.
This guide draws on the same 11-series, 4-full-report evidence base as every other WMA14 technique-hub document — no new primary Pearson document was fetched for this pass. Any claim above about a mark's exact wording is scoped to what the facts bank itself independently verified against a primary PDF, not extended beyond it.

#Sources Used

Primary Pearson documents, all independently verified in research/veridian/WMA14-verified-facts.md (fetched from qualifications.pearson.com directly, or a third-party mirror with cover-page metadata independently checked, converted with pdftotext -layout and cross-checked against -raw): mark schemes for June 2022, October 2021, January 2021, January 2026 (Q3, Q7, Q8, Q9); the general marking guidance and abbreviations list common to all reviewed mark schemes.

Real questions cited by series and number: Jun 2022 Q9, Q8(b); Jan 2021 Q4, Q6, Q9; Oct 2021 Q7(a)+(b); Jan 2026 Q3, Q7, Q8, Q9. Cross-referenced without reproduction: Oct 2024 Q1, Q2 (fully broken down in content/veridian/question-bank/wma14.ts).

Course files cross-referenced, not duplicated: forge-exam-technique-guide.md (§1 tariffs, §2 general M/A/B strategy — this guide's own §1 is a deeper, more precise layer under that section, not a restatement of it), signal-frequency-and-boundaries.md (§2's grade-boundary-absence finding, restated and honoured), forge-battle-card.md, codex-formula-and-method-bank.md, and the relevant lesson files under content/veridian/curriculum/wma14/ (proof-by-contradiction.ts, parametric-equations-conversion-and-domain-range.ts, implicit-and-parametric-differentiation-tangents-and-normals.ts, separable-differential-equations.ts, integration-by-parts-and-substitution.ts, volume-of-revolution.ts, area-under-a-parametric-curve.ts, position-vectors-distance-and-foot-of-perpendicular.ts, vectors-scalar-product-and-skew-lines.ts).

Not affiliated with or endorsed by Pearson Edexcel. Every quoted fragment above is under fifteen words with its own series/question attribution; all longer description is this guide's own paraphrase of the cited primary mark schemes, not a reproduction of Pearson's text.