Complete Worked Paper

Provenance (2026-09-29): written against research/veridian/WMA12-verified-facts.md (read in full — §3 confirms this series' real question count and the confirmed absence of a published examiner report for it; §6 supplied no per-question detail for June 2021 specifically, so every mathematical claim below traces to the primary PDFs, not to the facts bank), content/veridian/question-bank/wma12.ts (read in full — its two entries are October 2019 Q10 and October 2021 Q10(i); neither is June 2021, so no cross-reference is needed to avoid duplicating a specific question), and all eleven real lesson files plus paper-anatomy.ts in content/veridian/curriculum/wma12/ (grepped for every "June 2021" mention — two found, both cited by name where relevant below, neither a full worked exemplar of any of this series' nine questions). Primary sources for the mathematics itself: wma12-01-que-20210430.pdf (question paper), WMA12_01_rms_20210604.pdf (the final, results-day Mark Scheme, Publications Code WMA12_01_2106_MS, dated Summer 2021), and WMA12_01_msc_20210517.pdf (the provisional mark scheme it superseded), all three held locally at ~/Documents/SOL/Papers/IAL_Paper2_Complete/IAL_Materials/Mathematics/June_2021/. Extracted with pdftotext -layout; two extractions were then independently re-checked by rendering the source PDF page as an image and reading it directly, because pdftotext silently drops square-root glyphs on this file (confirmed on Q6(b)'s 3√7 and Q9(a)/(b)'s 2000√h / h^(3/2), both of which extracted as bare, radical-free numbers — corrected below against the rendered page, not against the flawed text extraction). No examiner report exists for this series — checked directly: the June_2021 folder in the archive above contains only the question paper and the two mark schemes, no pef/report file, consistent with WMA12-verified-facts.md's own independent finding (§3) that June 2021 is one of only two series in its 20-series archive with no published Principal Examiner Feedback, alongside a documented note that this is a genuine, confirmed absence rather than an unchecked gap.

#WMA12 — Complete Worked Paper: June 2021 (WMA12/01)

#0. Why this series, and what this document is

This document works every real question on one real, complete WMA12 sitting from start to finish, with every mark on the real mark scheme accounted for. June 2021 was chosen over other well-evidenced series (January 2020, October 2023, June 2019 among them) for a specific, checkable reason: it is the series with the least existing full-worked-solution coverage anywhere in this course's WMA12 material. A repo-wide search of content/veridian/curriculum/wma12/*.ts and content/veridian/question-bank/wma12.ts for "June 2021" turns up mentions in exactly two files, both structural rather than a worked solution: paper-anatomy.ts cites it five times (verified by grep -o "June 2021" | wc -l, not estimated), always as the same data point for the paper's real question-count range — "June 2021 and October 2020 both had 9" — and the-trapezium-rule.ts (line 632) cites it once, in passing, as a structural comparison for how a real "state the direction and give a reason" mark is built as a single combined B1 rather than two separate marks, illustrated there with an October 2023 trapezium-rule question, not a June 2021 one. Neither the specific arithmetic-series word problem in June 2021 Q1, nor the circle in Q6, nor the cubic-via-logarithms in Q7, nor any other question below, has a full worked solution anywhere else in this course. This document does not duplicate anything; it fills a real, verifiable gap.

The trade-off, disclosed rather than hidden: June 2021 has no published examiner report, so the "why candidates lost marks" narrative available for series like October 2019 or October 2021 (used in question-bank/wma12.ts) cannot be reproduced here from a genuine primary source. What this document offers instead, for every mark, is the real mark scheme's own stated criterion — including every alternative method it explicitly credits, and the specific wrong-working cases its own notes flag (e.g. "withhold if −4 is listed as a solution" in Q7(c)) — which is real, sourced, mark-earning information even without a report's prose commentary. Where the mark scheme itself names a common error or a specific case to watch for, this document says so and cites the line it comes from; it does not invent examiner commentary that was never published for this series.

Nine questions, 75 marks total (7+8+5+8+10+7+10+10+10=75, matching the paper's own closing line). Question text below is paraphrased throughout, per this course's copyright discipline — short phrases under fifteen words are quoted with attribution where the exact wording matters (e.g. a rubric instruction); the surrounding prose and all mathematics are original working, checked independently against the real mark scheme rather than copied from it.

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#Question 1 — Arithmetic series word problem (7 marks)

Paraphrase. Adina saves money weekly toward a computer: £5 in week 1, then 25p more each following week (£5.25 in week 2, £5.50 in week 3, and so on). (a)(i) State, with a reason, whether this is best modelled as an arithmetic or a geometric series, and (ii) give an expression for the amount saved in week $n$. (b) Given the computer costs £350, find how many weeks of saving are needed to have saved enough.

Cross-reference. The general technique — nth-term and sum formulae for an arithmetic series, and how to distinguish arithmetic from geometric growth from a word problem — is taught in full in content/veridian/curriculum/wma12/sequences-and-arithmetic-series.ts; this is a new real anchor for that lesson, not a restatement of it.

Worked solution.

(a)(i) Consecutive weekly amounts increase by a constant £0.25 each week (a common difference, not a common ratio — £5.25/£5 = 1.05, but £5.50/£5.25 ≈ 1.048, not equal), so an arithmetic series should be used.

(ii) With $a=5$, $d=0.25$: $u_n = 5 + 0.25(n-1)$.

(b) Need $S_n \geqslant 350$, where $S_n = \tfrac{n}{2}\bigl(2(5)+(n-1)(0.25)\bigr)$. Multiplying out and clearing the $\tfrac12$:

$$n(9.75+0.25n) \geqslant 700 \;\Rightarrow\; 0.25n^2+9.75n-700\geqslant0 \;\Rightarrow\; n^2+39n-2800\geqslant0$$

By the quadratic formula, $n \geqslant \dfrac{-39+\sqrt{39^2+4(2800)}}{2} = \dfrac{-39+\sqrt{12721}}{2} \approx 36.89$, so the smallest whole number of weeks is $n=37$.

Mark-by-mark justification (real mark scheme, WMA12_01_rms_20210604.pdf).

A real alternative route the mark scheme explicitly credits. The scheme anticipates a candidate who — despite Q1(a)(i) correctly stating "arithmetic" — nonetheless proceeds with geometric-series formulae in part (b) by mistake, and lays out exactly how far that answer can be credited: if a stated common ratio is used consistently to solve $\dfrac{5(r^n-1)}{r-1}\geqslant350$, the working scores M0A0M1A0 (the middle "solves the equation" M1 only) rather than zero throughout — a genuinely generous piece of real mark-scheme design worth knowing, since it rewards internally consistent method even when the underlying series choice in (a)(i) was already marked wrong.

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#Question 2 — Exponential graphs and their intersection (8 marks)

Paraphrase. Given a sketch of $y=4^x$, (a) sketch, on a copy of the same axes, (i) $y=2^x$ and (ii) $y=4^x-6$, labelling any axis-intercepts. (b) The curves $y=2^x$ and $y=4^x-6$ meet at point $P$; find its exact coordinates algebraically.

Cross-reference. Exponential graph shape and the laws used to solve $a^x=b$-type equations are taught in content/veridian/curriculum/wma12/logarithms-and-exponentials.ts; this pairs an exponential-graph sketch with an algebraic intersection in one question, which is new to this course's material.

Worked solution.

(a)(i) $y=2^x$ has the same shape and the same $y$-intercept $(0,1)$ as the given $y=4^x$, but rises more gently (base $2<4$): it sits above $y=4^x$ for $x<0$ and below it for $x>0$, crossing at $x=0$.

(a)(ii) $y=4^x-6$ is $y=4^x$ translated down 6 units: $y$-intercept at $1-6=-5$; $x$-intercept where $4^x=6$, i.e. $x=\log_4 6$ (exact) $\approx 1.29$.

(b) Let $y=2^x$. Since $4^x=(2^x)^2=y^2$, the second curve is $y=y^2-6$ in terms of the same substitution — so setting the two expressions for the shared quantity equal:

$$2^x = 4^x - 6 \;\Rightarrow\; y = y^2-6 \;\Rightarrow\; y^2-y-6=0 \;\Rightarrow\; (y-3)(y+2)=0 \;\Rightarrow\; y=3 \text{ or } y=-2$$

$y=2^x>0$ always, so $y=-2$ is rejected. $y=3 \Rightarrow 2^x=3 \Rightarrow x=\log_2 3$. Point $P = \left(\log_2 3,\ 3\right)$.

Mark-by-mark justification.

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#Question 3 — Proof: exhaustion, then general algebraic proof (5 marks)

Paraphrase. (i) Prove that for every single-digit prime $p$, $p^3+p$ is a multiple of 10. (ii) Show algebraically that, for any positive integer $n$, $(n+1)^3-n^3$ is never a multiple of 3.

Cross-reference. Both proof methods — exhaustion for a genuinely finite case set, and general algebraic proof for an infinite domain — are taught in full in content/veridian/curriculum/wma12/proof.ts; research/veridian/WMA12-verified-facts.md §6.1 independently anchors exhaustion questions to other real series (October 2022, October 2023, June 2019, June 2022) but not to June 2021, so this is a new real anchor for that topic, not a restatement.

Worked solution.

(i) The single-digit primes are $2,3,5,7$ (four values — genuinely exhaustible, which is exactly what makes exhaustion the valid method here rather than a general algebraic proof). Check each:

$$2^3+2=10,\quad 3^3+3=30,\quad 5^3+5=130,\quad 7^3+7=350$$

All four are multiples of 10, and since 2, 3, 5, 7 are the only single-digit primes, the result holds for all of them.

(ii) Expand directly: $(n+1)^3-n^3 = (n^3+3n^2+3n+1)-n^3 = 3n^2+3n+1 = 3(n^2+n)+1$. This is exactly one more than a multiple of 3 for every integer $n$, so it can never itself be a multiple of 3.

Mark-by-mark justification.

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#Question 4 — Binomial expansion, numerical approximation, and error direction (8 marks)

Paraphrase. (a) Expand $\left(2+\dfrac{x}{8}\right)^{13}$ in ascending powers of $x$, up to and including the $x^3$ term, fully simplified. (b) Use that expansion to approximate $2.0125^{13}$ to 3 decimal places. (c) Without calculating the exact value, state and justify whether that approximation is an overestimate or an underestimate.

Cross-reference. The full binomial-expansion method and the "compare to $(1+x)^n$"/convergence conditions are taught in content/veridian/curriculum/wma12/binomial-expansion.ts; that lesson's own real anchors are drawn from other series (per its header). This question's specific combination — expand, then use the expansion numerically, then justify the direction of the resulting error — is a new real anchor.

Worked solution.

(a) $\left(2+\dfrac{x}{8}\right)^{13} = 2^{13} + \binom{13}{1}2^{12}\left(\dfrac x8\right) + \binom{13}{2}2^{11}\left(\dfrac x8\right)^2 + \binom{13}{3}2^{10}\left(\dfrac x8\right)^3+\dots$

$2^{13}=8192$. Term 1: $13\times4096\times\tfrac18 x = 6656x$. Term 2: $78\times2048\times\tfrac1{64}x^2=2496x^2$. Term 3: $286\times1024\times\tfrac1{512}x^3=572x^3$.

$$\left(2+\frac x8\right)^{13} = 8192+6656x+2496x^2+572x^3+\dots$$

(b) $2.0125 = 2+0.0125$, and $0.0125=\dfrac{x}{8}\Rightarrow x=0.1$. Substituting:

$$2.0125^{13}\approx 8192+6656(0.1)+2496(0.1)^2+572(0.1)^3 = 8192+665.6+24.96+0.572=8883.132$$

(c) Every term kept in the expansion is positive (since $x=0.1>0$ and every coefficient above is positive), and the true value includes infinitely many further positive terms not yet added — so the truncated sum in (b) must be smaller than the true value: an underestimate.

Mark-by-mark justification.

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#Question 5 — Trapezium rule, definite integration, and area between two curves (10 marks)

Paraphrase. Two curves $C_1$ and $C_2$ meet at $x=2.5$ and $x=4$, with $C_1$ given only as a table of seven $y$-values at $x=2.5,2.75,\dots,4$ and $C_2$ given by $y=x^{3/2}-3x+9$. (a) Using the trapezium rule with all seven table values, estimate the area under $C_1$ between $x=2.5$ and $x=4$, to 2 d.p. (b) Find $\displaystyle\int\left(x^{3/2}-3x+9\right)dx$. (c) Combine (a) and (b) to estimate, to 1 d.p., the area of the region between the two curves.

Cross-reference. The trapezium-rule method itself, and definite integration for area, are taught in content/veridian/curriculum/wma12/the-trapezium-rule.ts and definite-integrals-and-areas.ts respectively; this question's combination of a numerical estimate for one curve and an exact integral for the other, subtracted to find the area between them, is a genuinely different question shape from either lesson's own existing anchors and is new to this course's material for this pairing.

Worked solution.

(a) Strip width $h=\dfrac{4-2.5}{6}=0.25$. Trapezium rule: $A\approx\dfrac h2\bigl[(y_0+y_6)+2(y_1+y_2+y_3+y_4+y_5)\bigr]$.

$$A \approx \frac{0.25}{2}\Bigl[(5.453+5)+2(7.764+9.375+9.964+9.367+7.626)\Bigr] = 0.125\times[10.453+88.192]=0.125\times98.645\approx\mathbf{12.33}$$

(b) $\displaystyle\int\left(x^{3/2}-3x+9\right)dx = \frac25x^{5/2}-\frac32x^2+9x\ (+c)$.

(c) Evaluate (b) from $2.5$ to $4$: at $x=4$, $4^{5/2}=32$, so $\tfrac25(32)-\tfrac32(16)+9(4)=12.8-24+36=24.8$. At $x=2.5$, $2.5^{5/2}\approx9.8821$, so $\tfrac25(9.8821)-\tfrac32(6.25)+9(2.5)\approx3.9528-9.375+22.5=17.0778$. Area under $C_2$ $\approx24.8-17.078=7.722$.

$$\text{Area }R \approx 12.33-7.722 = 4.608 \approx \mathbf{4.6}$$

Mark-by-mark justification.

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#Question 6 — Circle: tangency condition and a triangle's exact area (7 marks)

Paraphrase. A circle has equation $x^2-6x+y^2+8y+k=0$ for a positive constant $k$, and the $x$-axis is a tangent to it. (a) Find $k$. The circle meets the coordinate axes at points $R$, $S$, $T$. (b) Find the exact area of triangle $RST$.

Cross-reference. Completing the square to find a circle's centre and radius, and the radius-perpendicular-to-tangent property, are both taught in full in content/veridian/curriculum/wma12/the-circle.ts, whose own header (codex-deep-dive-the-circle.md) catalogues real anchors across fourteen other sittings but not June 2021 — this is a new anchor for that catalogue, and specifically for the "tangent to a coordinate axis" construction route named there.

Worked solution.

(a) Complete the square: $(x-3)^2-9+(y+4)^2-16+k=0 \Rightarrow (x-3)^2+(y+4)^2=25-k$. Centre $(3,-4)$, radius$^2=25-k$. Since the $x$-axis is tangent, the perpendicular distance from the centre to the $x$-axis — which is simply $|-4|=4$ — must equal the radius: $4^2=25-k \Rightarrow k=9$.

(b) With $k=9$: $(x-3)^2+(y+4)^2=16$, radius 4. Tangent to the $x$-axis means the circle touches it at exactly one point, directly below the centre: $R=(3,0)$. For the $y$-axis, set $x=0$: $9+(y+4)^2=16 \Rightarrow (y+4)^2=7 \Rightarrow y=-4\pm\sqrt7$, giving $S=(0,-4+\sqrt7)$ and $T=(0,-4-\sqrt7)$.

Triangle $RST$ has its base $ST$ lying on the $y$-axis, length $(-4+\sqrt7)-(-4-\sqrt7)=2\sqrt7$, and height equal to $R$'s horizontal distance from the $y$-axis, which is 3:

$$\text{Area} = \frac12\times2\sqrt7\times3 = \mathbf{3\sqrt7}$$

Mark-by-mark justification. (Two independently credited routes exist for part (a) in the real mark scheme; both are given, since either is a full, correct method a student might reasonably choose.)

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#Question 7 — Logarithmic equation reducing to a cubic, solved by the factor theorem (10 marks)

Paraphrase. (a) Given $3\log_3(2x-1)=2+\log_3(14x-25)$, show this reduces to $2x^3-3x^2-30x+56=0$. (b) Show that $x=-4$ is a root of that cubic. (c) Hence solve the original logarithmic equation fully, using algebra and showing every step.

Cross-reference. Log laws are taught in logarithms-and-exponentials.ts; the factor theorem and cubic factorisation are taught in factor-and-remainder-theorems.ts. This question chains both topics end-to-end inside a single real question — a genuinely different shape from either lesson's own standalone anchors, and new to this course's material as a combined worked example.

Worked solution.

(a) Write $2=\log_3 9$ (since $3^2=9$). Then $2+\log_3(14x-25)=\log_39+\log_3(14x-25)=\log_3\bigl(9(14x-25)\bigr)$, and $3\log_3(2x-1)=\log_3(2x-1)^3$. Equating the arguments (both sides are $\log_3$ of something, and the equation asserts equality):

$$(2x-1)^3 = 9(14x-25)$$

Expanding the left side: $8x^3-12x^2+6x-1$. So $8x^3-12x^2+6x-1=126x-225 \Rightarrow 8x^3-12x^2-120x+224=0$, and dividing through by 4:

$$2x^3-3x^2-30x+56=0 \quad\checkmark$$

(b) Substitute $x=-4$: $2(-4)^3-3(-4)^2-30(-4)+56 = -128-48+120+56=0$. Since the cubic evaluates to zero, $x=-4$ is confirmed as a root.

(c) Factor out $(x+4)$: $2x^3-3x^2-30x+56=(x+4)(2x^2-11x+14)$ (checked by expanding back: $(x+4)(2x^2+bx+c)=2x^3+(b+8)x^2+(4b+c)x+4c$, matching coefficients gives $b=-11$, $c=14$ — consistent both ways). Factor the quadratic: $2x^2-11x+14=(2x-7)(x-2)$ (discriminant $121-112=9$; roots $\tfrac{11\pm3}{4}=3.5$ or $2$). So:

$$(x+4)(2x-7)(x-2)=0 \Rightarrow x=-4,\ \frac72,\ \text{or } 2$$

But $x=-4$ makes both $2x-1$ and $14x-25$ negative in the original logarithmic equation (undefined domain), so it must be rejected. The two genuine solutions are $x=2$ and $x=\dfrac72$ (both checked: at each, $2x-1>0$ and $14x-25>0$).

Mark-by-mark justification.

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#Question 8 — Trigonometric equations: a linear compound-angle equation, then a cubic-in-sine equation (10 marks)

Paraphrase. (This question carries the paper's "you must show all stages of your working; a calculator-only answer is not acceptable" instruction — confirmed as the exact rubric wording used on several other real series too, per research/veridian/WMA12-verified-facts.md §3.) (i) Solve, for $0<\theta<360°$, $3\sin(\theta+30°)=7\cos(\theta+30°)$, to 1 d.p. (ii)(a) Show that $3\sin^3x=5\sin x-7\sin x\cos x$ can be written as $\sin x(a\cos^2x+b\cos x+c)=0$ for constants $a,b,c$. (b) Hence solve, for $-\tfrac\pi2\leqslant x\leqslant\tfrac\pi2$, the same equation.

Cross-reference. Solving equations of the linear compound-angle shape (i), and of the cubic-in-one-trig-ratio shape (ii), are both taught in content/veridian/curriculum/wma12/trigonometric-equations.ts, whose own real anchors span other series (October 2021, October 2022, June 2025, per that file's header). Both question shapes here are new real anchors for that lesson.

Worked solution.

(i) Divide both sides by $\cos(\theta+30°)$: $3\tan(\theta+30°)=7 \Rightarrow \tan(\theta+30°)=\tfrac73$.

$$\theta+30° = \arctan\left(\frac73\right) \approx 66.8°$$

Since $\tan$ has period $180°$, and $0<\theta<360°\Rightarrow30°<\theta+30°<390°$, the two values of $\theta+30°$ in that extended range are $66.8°$ and $66.8°+180°=246.8°$, giving:

$$\theta \approx 36.8°\ \text{or}\ 216.8°$$

(ii)(a) Substitute $\sin^2x=1-\cos^2x$ into $\sin^3x$: $3\sin x(1-\cos^2x)=5\sin x-7\sin x\cos x$. Expand and gather everything on one side:

$$3\sin x-3\sin x\cos^2x-5\sin x+7\sin x\cos x=0 \;\Rightarrow\; \sin x\bigl(-3\cos^2x+7\cos x-2\bigr)=0$$

Multiplying the bracket by $-1$ (and the whole equation is still $=0$) gives the requested form with $a=3,\ b=-7,\ c=2$:

$$\sin x\bigl(3\cos^2x-7\cos x+2\bigr)=0$$

(b) Either factor $\sin x=0$, or factor the quadratic in $\cos x$: $3\cos^2x-7\cos x+2=(3\cos x-1)(\cos x-2)$ (discriminant $49-24=25$; roots $\tfrac{7\pm5}{6}=2$ or $\tfrac13$).

• $\sin x=0 \Rightarrow x=0$ (the only solution in range; $x=\pi$ lies outside $-\tfrac\pi2\leqslant x\leqslant\tfrac\pi2$).
• $\cos x=2$: impossible, since $|\cos x|\leqslant1$ always — rejected.
• $\cos x=\tfrac13 \Rightarrow x=\pm\arccos\left(\tfrac13\right)\approx\pm1.23$ (both lie inside $\pm\tfrac\pi2\approx\pm1.571$).

$$x = 0,\ \pm1.23\ \text{(awrt)}$$

Mark-by-mark justification.

A genuine real alternative the scheme credits for part (i). Rather than dividing by cosine directly, the scheme also credits squaring both sides of $3\sin(\theta+30°)=7\cos(\theta+30°)$ and applying $\sin^2+\cos^2=1$ to reach an equation purely in $\sin(\theta+30°)$ or $\cos(\theta+30°)$ before inverting — provided any extraneous solution introduced by squaring is explicitly rejected. A second alternative expands $\sin(\theta+30°)$ and $\cos(\theta+30°)$ via the compound-angle formulae first and rearranges directly to $\tan\theta=\dots$ — genuinely different algebra reaching the identical two answers.

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#Question 9 — Optimisation: an open-top box, differentiation, and a second-derivative justification (10 marks)

Paraphrase. A square-based, open-top box has base edge $l$ cm and height $h$ cm, and volume $250\,000\text{ cm}^3$. (a) Show the external surface area is $S=\dfrac{250000}{h}+2000\sqrt h$. (b) Using differentiation, show $S$ has a stationary point when $h=250^{k}$ for a rational constant $k$ to be found. (c) Justify, by further differentiation, that this $h$ gives the minimum surface area.

(Provenance note specific to this question: pdftotext -layout silently drops the square-root glyph in both the question paper and the mark scheme here, rendering the printed result as the radical-free "$S=250000/h+2000h$" and "$h=250^{k}$" with no visible exponent structure. Both were re-checked by rendering the source PDF pages as images and reading them directly — the real printed result is $S=\dfrac{250000}{h}+2000\sqrt h$, and the real final answer is $h=250^{2/3}$, i.e. $k=\tfrac23$. This is exactly the kind of silent extraction error WMA12-verified-facts.md itself already warns about for this archive, and the same discipline — visually confirming anything a text extraction renders ambiguously — is applied here.)

Cross-reference. Differentiation to find and classify a stationary point (including the second-derivative test) is taught in full in content/veridian/curriculum/wma12/stationary-points-and-curve-sketching.ts. This specific "derive the formula, then optimise it" combination — building $S(h)$ from a volume constraint before differentiating — is a new real anchor for that lesson.

Worked solution.

(a) Volume: $V=hl^2=250000 \Rightarrow l^2=\dfrac{250000}h \Rightarrow l=\dfrac{500}{\sqrt h}$. Surface area (open top: one base, four sides) $=l^2+4hl$:

$$S = \frac{250000}{h} + 4h\times\frac{500}{\sqrt h} = \frac{250000}{h}+\frac{2000h}{\sqrt h} = \frac{250000}{h}+2000\sqrt h \quad\checkmark$$

(b) Write $\sqrt h=h^{1/2}$: $S=250000h^{-1}+2000h^{1/2}$.

$$\frac{dS}{dh} = -250000h^{-2}+1000h^{-1/2}$$

Set to zero: $1000h^{-1/2}=250000h^{-2}$. Divide both sides by 1000: $h^{-1/2}=250h^{-2}$. Multiply both sides by $h^2$: $h^{2-1/2}=250 \Rightarrow h^{3/2}=250 \Rightarrow h=250^{2/3}$ ($k=\tfrac23$, a rational constant, as asked).

(c) $\dfrac{d^2S}{dh^2} = 500000h^{-3}-500h^{-3/2}$. At $h=250^{2/3}\approx39.7$, this evaluates to $6>0$, confirming (by the second-derivative test) that this stationary point is a minimum.

Mark-by-mark justification.

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#Summary: marks, topics, and where each question sits in this course's material

Every mathematical claim above traces to one of three sources, named at the point of use: wma12-01-que-20210430.pdf (the real question paper), WMA12_01_rms_20210604.pdf (the real, final mark scheme), or independent arithmetic worked out in this document and checked by hand against the mark scheme's own stated final answers. No likelihood or grade-boundary figure appears anywhere in this document — none was sourced, and none is invented. Where the provisional mark scheme (WMA12_01_msc_20210517.pdf) differed from the final version used throughout, the final version's own "Summary of changes" note records the difference: clarifications were added for how to mark odd/even case-splitting in Q3, for what counts as "at least twice" in Q4's coefficient mark, for where the Q6(b) zero-ordinate may appear, and for whether Q8(ii) may be marked as one combined block rather than strictly by sub-part — all reflected in the mark-by-mark tables above, which follow the final version throughout.