Complete Worked Paper

#CODEX: Complete Worked Paper — WMA11 January 2023 (Pure Mathematics P1)

Provenance (2026-09-29). Series used: January 2023, WMA11/01 (Pearson Edexcel International Advanced Level, Pure Mathematics P1). All three primary documents were read fresh for this pass, independently of any prior summary in this repo, direct from the local archive at /Users/neosteinhoff/Documents/SOL/Papers/IAL_Paper1_Complete/IAL_Materials/Mathematics/January_2023/: Mathematics_Unit 1_QP_January_2023.pdf, ..._MS_January_2023.pdf, ..._ER_January_2023.pdf. Extracted with pdftotext -layout and spot-checked against pdftotext -raw (both modes agree on every mark-scheme figure and answer quoted below, including the Q2 conclusion wording, Q9(b)(iii)'s "201", and Q11(b)'s "284"). Identity confirmed directly from the PDFs' own headers and page-footer watermarks — Question Paper Log Number P72066A, Publications Code WMA11_01_MS_2301 (mark scheme) and WMA11_01_ER_2301 (examiner report) — matching the citation already on file in research/veridian/WMA11-verified-facts.md §4's series table exactly, so this is confirmed to be the same, correctly-identified genuine document that file already vouches for, not a fresh unverified fetch. No WebFetch tool was used anywhere in this pass, which is the specific failure mode §5 of that file documents (a tool once fabricated a plausible-looking wrong summary of a genuine, correctly-fetched WMA11 mark scheme) — every fact below traces to a locally-extracted PDF text file, re-read directly.

Every worked solution below was independently recomputed by hand and checked against the real mark scheme's own final answer before being written up — not copied from the mark scheme and presented as if derived. Question wording is paraphrased throughout; the Pearson question paper's own prose is copyrighted, and only short (under-15-word) instructional phrases are quoted verbatim, with attribution. Numbers, equations and mark codes are not copyrightable expression and are reproduced exactly, since exactness there is the entire point of an exam-technique document.

#1. Why this series, and not another

WMA11-verified-facts.md §4 lists nine WMA11 series touched by this course's research pass, with wildly uneven depth: five have a fully-read examiner report (Jan 2019, Oct 2020, Oct 2021, Jan 2023, Jan 2025-partial), two are confirmed-genuine mark schemes not mined question-by-question (Oct 2022, Jan 2024), and two more were read for one targeted question each (June 2022 Q5, June 2023 Q1–Q2). January 2023 is the only series in that list where all three primary documents — question paper, mark scheme, and examiner report — were read in full, covering all eleven questions, and this pass's own fresh re-extraction (above) confirms that claim rather than trusting it secondhand. That is the entire basis for the choice: it is the one series this course can honestly claim to work start-to-finish without gaps, not a claim that it is somehow a "typical" or representative paper. No other series was even a close second by this test — the runner-up documents (Oct 2021, Jan 2019) have a full examiner report but this pass did not re-verify a full question-by-question mark scheme against them, so a "complete worked paper" built from either would rest on one fewer independently-checked leg than this one does.

#2. Paper anatomy (recap only — see paper-anatomy.ts for the full policy discussion)

11 questions, 75 marks, tariff 5, 5, 5, 4, 6, 10, 10, 8, 4, 10, 8 (independently re-summed from the real paper: 5+5+5+4+6+10+10+8+4+10+8 = 75, confirmed). 1 hour 30 minutes. Calculator permitted in general, but four sub-parts carry an explicit override, each worded slightly differently: Q1(b) and Q3 read "(Solutions relying on calculator technology are not acceptable.)"; Q5 reads, in full, "In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable."; Q7(d) reads "(Solutions relying entirely on calculator technology are not acceptable.)"; Q10(c) reads "(Solutions based on calculator technology are not acceptable.)" This document does not repeat the wider policy discussion (formula-booklet thinness, AO weighting) already covered in paper-anatomy.ts — see that file for it.

#3. What this document adds against the rest of the course

Checked directly (grep across every .ts file in content/veridian/curriculum/wma11/ and content/veridian/question-bank/wma11.ts) before writing a line below:

Five of eleven questions (2, 3, 7-parts, 8, 9-part-b, 10-parts, 11) are, in whole or in significant part, new to this course. The rest are assembled here for the first time as one continuous, start-to-finish paper rather than scattered single-technique exemplars — that is this document's own, distinct value even where the underlying mathematics has a home elsewhere already.

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#4. The paper, question by question

#Question 1 (5 marks) — differentiation of fractional/negative indices, then an exact gradient

Spec 4.1–4.3. Curve $C$: $y = 2 + 10x^{1/2} - 2x^{3/2}$, $x>0$.

(a) Differentiate term by term: $10x^{1/2} \to 10 \cdot \tfrac12 x^{-1/2} = 5x^{-1/2}$; $-2x^{3/2} \to -2\cdot\tfrac32 x^{1/2} = -3x^{1/2}$; the constant $2\to 0$.

$$\frac{dy}{dx} = 5x^{-1/2} - 3x^{1/2}$$

Marks: M1 any one term's power correctly reduced by 1; A1 one term fully simplified; A1 both terms correct and simplified with no extraneous "+c" (the real examiner report flags exactly this slip: a "significant number of candidates... differentiated but then added + c to their expression").

(b) At $x=2$: $\dfrac{dy}{dx} = 5(2)^{-1/2} - 3(2)^{1/2} = \dfrac{5}{\sqrt2} - 3\sqrt2 = \dfrac{5\sqrt2}{2} - 3\sqrt2 = \dfrac{5\sqrt2 - 6\sqrt2}{2} = -\dfrac{\sqrt2}{2}$.

Marks: M1 full substitution into a genuinely differentiated expression; A1 $-\tfrac{1}{\sqrt2}$ or an exact equivalent, reached through at least one visible intermediate line — the calculator override on this part exists precisely because $-\tfrac{\sqrt2}{2} \approx -0.707$ is trivial to reach on a calculator without showing the surd manipulation the mark actually tests.

Real finding (paraphrased, examiner report): a significant number of candidates reached the correct decimal but lost the mark for not showing how to get there in exact form; others who had wrongly appended "+c" in part (a) then tried to solve for that constant using $x=2$.

See differentiating-negative-and-fractional-indices.ts and paper-anatomy.ts for the full teaching treatment of this exact question.

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#Question 2 (5 marks) — perpendicularity proof and a fourth rectangle vertex

Spec 2.1, 2.2. $P(-3,7)$, $Q(9,11)$, $R(12,2)$.

(a) Prove $\angle PQR = 90°$.

$$\text{grad }PQ = \frac{11-7}{9-(-3)} = \frac{4}{12} = \frac13, \qquad \text{grad }QR = \frac{2-11}{12-9} = \frac{-9}{3} = -3$$ $$\frac13 \times (-3) = -1 \implies PQ \perp QR \implies \angle PQR = 90°$$

Marks: M1 both gradients attempted with a genuine $\Delta y/\Delta x$; A1 both correct; A1 — dependent on both — for the explicit closing statement connecting the product $=-1$ to the right angle. This last mark is the one the real examiner report singles out: candidates "did not appreciate the rigour required... having completed the correct work, failed to give a suitable explanation and/or conclusion." The calculation is not, on its own, a proof of anything; the mark scheme requires the sentence that says what the calculation means. (The real scheme also credits two other complete routes — Pythagoras/cosine-rule on the three side lengths $\sqrt{160}, \sqrt{90}, \sqrt{250}$, or a vector dot-product $\vec{PQ} \cdot \vec{QR}=0$ — each with the identical M1 A1 A1 structure.)

(b) $S$ such that $PQRS$ is a rectangle.

Vector route: $\vec{QR} = R - Q = (3,-9)$, so $S = P + \vec{QR} = (-3+3,\ 7-9) = (0,-2)$. Independently checked via the diagonal-midpoint route: midpoint of $PR = (4.5,4.5)$ must equal midpoint of $QS$, giving $S=(0,-2)$ again.

Marks: M1 any valid method reaching at least one coordinate of $S$; A1 $(0,-2)$.

Real finding: the examiner report names three genuine routes candidates used, in order of reliability — the "obvious" one (find equations of the two unknown sides, solve simultaneously) was in fact the least successful ("rarely produced enough work to score full marks"); the midpoint-of-the-diagonals route was "more unusual but often successful"; and counting units or using vectors directly was the most confident route of the three.

This is the first place in this codebase where the real P/Q/R/S numbers above have been carried through a full, checked solution — the two existing lessons that cite this same question (perpendicularity-proofs-and-coordinate-geometry.ts, simultaneous-equations-by-substitution.ts) both build their worked content on their own placeholder coordinates instead, and the facts bank flags reconciling that as unresolved future work (§4.2). This document does not edit either file — that is a separate, coordinated task the facts bank already describes — but the real numbers, worked three independent ways and cross-checked, are now on record.

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#Question 3 (5 marks) — splitting a fraction before integrating

Spec 5.1, 5.2. Find $\displaystyle\int \frac{4x^5+3}{2x^2}\,dx$.

Split first: $\dfrac{4x^5+3}{2x^2} = \dfrac{4x^5}{2x^2} + \dfrac{3}{2x^2} = 2x^3 + \tfrac32 x^{-2}$.

Integrate term by term: $\displaystyle\int 2x^3\,dx = \tfrac12 x^4$; $\displaystyle\int \tfrac32 x^{-2}\,dx = \tfrac32 \cdot \dfrac{x^{-1}}{-1} = -\tfrac32 x^{-1}$.

$$\int \frac{4x^5+3}{2x^2}\,dx = \frac12 x^4 - \frac32 x^{-1} + c$$

Marks: M1 the split attempted with at least one term's index correctly processed; A1 the full split correct as one expression, $2x^3 + \tfrac32 x^{-2}$; dM1 — dependent specifically on the split's M1, never an independent method mark — the integration itself attempted, raising one index by one; A1 either integrated term correct and simplified; A1 the complete answer with the constant.

Real finding: "some candidates had no idea how to write the expression as the sum of two terms," and a minority integrated the numerator and denominator separately before dividing, which "resulted in no marks" — that route does not compute the same thing as the real integral at all (a quick counter-example proves it: $\int 4x\,dx \div \int x\,dx$ collapses to a bare constant, discarding the $x$-dependence the original expression actually has). Even a wrong split — the report names $8x^3+6x^{-2}$, $4x^5+6x^{-2}$ and $2x^5+\tfrac32x^{-2}$ as real wrong attempts seen — still earned the integration method mark if the integration on those (wrong) terms was itself done correctly. The most common way to lose the final mark even after a correct split: dropping the constant of integration, or writing the index but omitting the base, e.g. "$\tfrac12 x^4 + \tfrac32 - {}^{-1}+c$" instead of $-\tfrac32 x^{-1}+c$.

See splitting-fractions-before-integrating.ts for the deep-dive on the split-vs-integrate error mechanism (built there on different, VERIDIAN-original numbers); this is the first place the real fraction above has been carried to a finished, checked answer.

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#Question 4 (4 marks) — discriminant with a non-zero parameter

Spec 1.4. $kx^2+6kx+5=0$, $k$ a non-zero constant, has no real roots. Find the range of $k$.

No real roots $\iff b^2-4ac<0$, with $a=k$, $b=6k$, $c=5$:

$$(6k)^2 - 4(k)(5) < 0 \implies 36k^2 - 20k < 0 \implies 4k(9k-5) < 0$$

Critical values $k=0$ and $k=\tfrac59$. Since the coefficient of $k^2$ is positive, the quadratic-in-$k$ expression is negative between its roots:

$$0 < k < \frac59$$

Marks: M1 the discriminant expression attempted with the correct $a,b,c$; dM1 — dependent — setting it $<0$ (or $\le,>,\ge$ at this stage) and reaching a non-zero value for $k$; A1 the upper bound $\tfrac59$ specifically, not merely a value; A1 the complete, correctly-strict inequality $0<k<\tfrac59$.

Real finding: the single most common error was not an algebra mistake but a scope one — "many candidates ignored the solution, $k=0$, so the most common error seen was the absence of the lower bound $0<k$" — a consequence of dividing the inequality $36k^2-20k<0$ through by $k$, which silently discards that root and is in any case not valid on an inequality whose divisor has unknown sign. A second, separately documented error: some gave $0 \le k \le \tfrac59$, losing the final mark by including boundary values that a strict "no real roots" condition excludes.

See discriminant-with-a-parameter.ts for the full trap-taxonomy and the "b²+4ac" and "un-squared coefficient" error mechanisms.

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#Question 5 (6 marks) — a hidden quadratic under a substitution

Spec 1.1, 1.5, 1.6. $p = 3^x$.

(a) Show $3 \times 9^x + 3^{x+2} = 1 + 3^{x-1}$ rewrites as $9p^2+26p-3=0$.

$9^x = (3^x)^2 = p^2$, so $3\times 9^x = 3p^2$. $3^{x+2} = 3^x\cdot 3^2 = 9p$. $3^{x-1}=3^x/3 = p/3$. Substituting:

$$3p^2 + 9p = 1 + \frac{p}{3}$$

Multiply through by 3: $9p^2+27p = 3+p \implies 9p^2+26p-3=0$ — as required.

Marks: B1 any one correct index-law substitution; M1 the equation rewritten entirely in $p$; A1* the given result reached with an intermediate line shown (an "answer given" mark — no credit for writing down the target and calling it proved).

(b) Solve $9p^2+26p-3=0$: factorises as $(9p-1)(p+3)=0 \implies p=\tfrac19$ or $p=-3$. Since $p=3^x>0$ for all real $x$, discard $p=-3$. Then $3^x = \tfrac19 = 3^{-2} \implies x=-2$.

Marks: M1 a genuine (non-calculator) attempt at solving the given quadratic; A1 (logged as an M1 in the electronic mark scheme) for reaching $3^x=\tfrac19$ specifically, not a bare value of $p$; A1 $x=-2$ only.

Real finding: "common misconceptions included $3^{x+2}=p^2$ and $3^{x-1}=p^{-1}$" — i.e. the error is applying the substitution to the wrong exponent arithmetic, not merely failing to see the substitution. In part (b), the dominant way to lose marks was solving the quadratic in $p$ correctly and then stopping — "some missed out this part or solved the quadratic equation and then stopped, hence only gaining the method mark" — never solving back for $x$. A further specific trap: reading "hence" and going back to the original equation in $x$ instead of using the already-solved value of $p$, which the report says "could score no marks in this part."

See hidden-quadratic-substitution.ts for why this "solve for the substitution, forget to solve back" trap recurs on this paper across two independent series (this one and Jan 2025).

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#Question 6 (10 marks) — sector, two triangles, composite area and perimeter

Spec 3.1, 3.2. Sector $OBC$, $\angle BOC = 2.4$ rad, area $= 40\text{ m}^2$; congruent triangles $OAB$, $ODC$; $AOD$ a straight line, length $12.5$ m.

(a) $A = \tfrac12 r^2\theta \implies 40 = \tfrac12 r^2(2.4) \implies r^2 = \tfrac{40}{1.2} = 33.\overline3 \implies r \approx 5.77$ m.

Marks: M1 the sector-area formula used with $A=40,\theta=2.4$; A1 $5.77$ to 2 d.p.

(b) Since $A,O,D$ are collinear, $\angle AOD=\pi$; the two congruent triangles give $\angle AOB=\angle DOC$, so $2\angle AOB + 2.4 = \pi \implies \angle AOB = \dfrac{\pi-2.4}{2} \approx 0.37$ rad.

Marks: B1 $0.37$ (accept anywhere in the response, but this specific part must be in radians even if other parts allow degrees).

(c) Congruence of the two triangles, sharing the radius $OB=OC=r$ and requiring $OA=OD$, forces $OA=OD=6.25$ m (half of $12.5$). Area of one triangle $= \tfrac12(6.25)(5.77)\sin(0.37) \approx 6.54\text{ m}^2$.

$$\text{Total area} = 40 + 2(6.54) \approx 53.1\text{ m}^2$$

Marks: M1 one triangle's area attempted with $\tfrac12ab\sin C$; dM1 the full composite-area method, sector plus both triangles; A1 $53.1$ (1 d.p.).

(d) Arc $BC$: $r\theta = 5.77\times2.4 \approx13.85$ m. Length $AB$ (by the cosine rule in triangle $OAB$, using $OA=6.25$, $OB=5.77$, included angle $0.37$):

$$AB^2 = 6.25^2+5.77^2-2(6.25)(5.77)\cos(0.37) \implies AB\approx2.26\text{ m}$$ $$\text{Perimeter} = 12.5 + 2(2.26) + 13.85 \approx 30.9\text{ m}$$

Marks: M1 arc length attempted; M1 cosine rule attempted for $AB$ or $AB^2$; ddM1 — dependent on both — the full perimeter assembled correctly (base $AD$, both equal chords $AB,CD$, and the arc, not $AB^2$ left unrooted); A1 $30.9$.

Real finding: the dominant single error in (a) was omitting the $\tfrac12$ in the sector formula; a small number mistook $AOD$ for a diameter of the sector's own circle, giving $6.25$ m as a wrong radius. In (b), the most common mistake was subtracting $2.4$ from $2\pi$ instead of $\pi$ before halving. In (d), the most common perimeter errors were forgetting to double $AB$, omitting the base length $12.5$, or leaving out the two radii from the total.

See sector-area-and-arc-length.ts for the deeper trap-taxonomy on this exact question.

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#Question 7 (10 marks) — reciprocal graph, transformation, then a line meeting the curve

Spec 1.11, 1.12, 1.6. $C: y = \dfrac6x$, $x\ne0$.

(a) Sketch $C$: a rectangular hyperbola with branches in quadrants 1 and 3, asymptotic to both axes, never crossing either.

Marks: M1 monotonically decreasing branch, correct quadrant(s), no spurious extra asymptote; A1 both branches, correctly positioned.

(b) $C$ is transformed onto $y=\dfrac{6}{x-2}$. This is a translation by $2$ units in the positive $x$-direction (vector $\binom20$).

Marks: B1 a partial description implying at least one correct component (e.g. "shifts 2 right" without the word translation); B1 the complete description — both "translation" specifically and the $2$-right component. The real finding here is exactly about vocabulary, not mathematics: "although most knew that the graph would move 2 units right, very few appeared to know that it was a translation, with most using the words 'shift' or 'move' instead" — the mark scheme gates the second B mark on the word "translation" itself.

(c) The line $y=kx+7$ meets $y=\dfrac{6}{x-2}$ where $x=-4$: $y = \dfrac{6}{-4-2} = -1$. Substitute into the line: $-1 = -4k+7 \implies k=2$.

Marks: M1 the substitution into either equation attempted; A1 $k=2$.

Real finding: many candidates over-complicated this by equating the two expressions for $y$ first and rearranging into a quadratic before substituting $x=-4$ — a route the report notes produced more algebra errors than the direct substitution above.

(d) With $k=2$: $\dfrac{6}{x-2} = 2x+7 \implies 6 = (2x+7)(x-2) = 2x^2+3x-14 \implies 2x^2+3x-20=0 \implies (2x-5)(x+4)=0 \implies x=\tfrac52\text{ or }x=-4$.

$x=-4$ is $P$ (given); so $Q$ has $x=\tfrac52$, and $y = 2(\tfrac52)+7=12$. $\boxed{Q=(\tfrac52,12)}$

Marks: M1 the two expressions equated and cross-multiplied to a quadratic (terms need not be collected); dM1 the resulting three-term quadratic solved by any valid method; A1 $x=\tfrac52$; A1 the full coordinate pair.

Real finding: "there was some poor algebra seen in many cases, and $6/(x-2)$ was often given as $(6/x)-3$" (a genuine, specific algebra slip on rewriting this exact reciprocal); a small number forgot to substitute back for the $y$-coordinate once $x$ was found; and, tellingly, some candidates reached for the discriminant — machinery for counting intersections — on a question that had already stated there were two and asked for a specific one instead.

See graph-transformations-in-words.ts for part (b)'s deeper vocabulary-precision material, and simultaneous-equations-by-substitution.ts for parts (c)–(d)'s trap-taxonomy; this document is the first to assemble all four parts, including part (a), as one question.

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#Question 8 (8 marks) — line, quadratic reconstruction, and a bounded region

Spec 1.3, 1.5, 1.7, 1.8, 2.1. Confirmed by direct search: no other file in this repo cites this question. Line $l$ cuts the $y$-axis at $-12$, the $x$-axis at $4$. Curve $C: y=f(x)$, $f$ quadratic, minimum at $(7,-18)$, cutting the $x$-axis at $4$ and at $k$.

(a) Gradient $= \dfrac{0-(-12)}{4-0} = 3$. Equation: $y=3x-12$ (check: at $x=4$, $y=0$ ✓; at $x=0$, $y=-12$ ✓).

Marks: M1 the form $y=mx+c$ attempted with either $m$ or $c$ correct; A1 the complete correct equation.

(b) A parabola's roots are symmetric about its own turning point's $x$-coordinate: $\dfrac{4+k}{2}=7 \implies k=10$.

Marks: B1 $k=10$.

(c) $f(x)=A(x-4)(x-10)$; use the minimum point $(7,-18)$: $-18 = A(7-4)(7-10) = A(3)(-3)=-9A \implies A=2$.

$$f(x) = 2(x-4)(x-10) \;\left(= 2(x-7)^2-18\right)$$

Marks: M1 the correct quadratic form attempted (either factorised-with-unknown- multiplier or vertex form); dM1 — dependent — a full attempt to find the multiplier using a genuine point; A1 the correct expression in either equivalent form.

Real finding: "candidates who did not recognise the stretch was required... scored no further marks" — i.e. writing $(x-4)(x-10)$ or $(x-7)^2-18$ with an implicit leading coefficient of 1 looks plausible but is not the same curve; the report also names the three-simultaneous-equations approach (using all three given facts as three separate equations) as "the least successful approach" against the more direct two-fact method above.

(d) Region $R$ is bounded above/below by the line and the parabola, and on the left/right by $x=0$ and $x=4$ (read from Figure 2):

$$3x-12 \le y \le 2(x-4)(x-10), \quad 0\le x\le4$$

Marks: M1 any two of the four inequalities, following through from (a) and (c); A1 the complete, correctly-bounded system (not phrased in terms of "$R$" itself, which the real scheme explicitly scores zero for).

This entire question is new content for this course. It is also a third, independently- confirmed instance of the "region bounded by a line and a curve, defined by a compound inequality system" question type the facts bank's §4.1 already names twice (Jan 2019 Q4; Oct 2021 Q3(ii)) — here on a quadratic curve rather than the reciprocal ones in those two, at a lower, 2-mark tariff, and without the strict-inequality boundary-finding step those two require. Seen now in 3 of the 9 series checked for this paper.

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#Question 9 (4 marks) — the tan graph: period, then counting roots

Spec 3.3, 1.11. Figure 3 shows $y=\tan x$ and the line $l: y=\pi x$ on $-\pi<x<\pi$.

(a) The period of $\tan x$ is $\pi$.

Marks: B1. Real finding: fewer than half of all candidates gave this correctly; "quite frequently the answer was given as an interval, including $-\pi/2<x<\pi/2$" — the examiner report's own diagnosis is a genuine conceptual trap, "confusion between period and domain/range," not an arithmetic slip. See trig-graphs-period-and-domain.ts for the full treatment of this exact trap.

(b) Three sub-parts, each a single B1 in the real mark scheme, with no method mark available at all — this is worth stating plainly, since it means the "full working" habit this course otherwise drills is, on this specific sub-question, simply not what is being credited; the mark exists for the final integer alone.

(i) $\tan x = (\pi+2)x$ on $-\pi<x<\pi$: 3. (ii) $\tan x = \pi x$ on $-2\pi<x<2\pi$: 5. (iii) $\tan x = \pi x$ on $-100\pi<x<100\pi$: 201.

None of this is present anywhere else in this course, so a genuine — if compact — justification is given here, beyond what the mark scheme itself requires, since "read it off the given sketch" is not, on its own, a checkable method for a written record.

Let $g(x)=\tan x - mx$ for a line of slope $m>1$. $g$ is odd, so $g(0)=0$ always. On the central branch $\left(-\tfrac\pi2,\tfrac\pi2\right)$, $g'(x)=\sec^2x-m$ is negative near $x=0$ (since $\sec^20=1<m$) and $\to+\infty$ at both ends, giving $g$ one local max at $x=-t_0$ and one local min at $x=+t_0$ (where $\cos^2t_0=1/m$), symmetric about $0$. Direct calculation ($m=\pi$: $t_0\approx0.971$, giving local extrema $\approx\pm1.59$; $m=\pi+2$: $t_0\approx1.114$, extrema $\approx\pm3.69$) shows both extrema are genuinely nonzero with opposite sign in both cases, so the central branch always contributes exactly 3 roots: one in each outer half, plus $x=0$ itself. On any other branch (centred at $x=k\pi$, $k\ne0$), the same local-extrema calculation gives values shifted by $-mk\pi$ — a shift that grows linearly with $|k|$ and overwhelms the fixed $\pm1.59$ (or $\pm3.69$) very quickly, pushing both local extrema negative for $k>0$ (or both positive for $k<0$). Once that happens, the branch has exactly one root, in the tail nearest its far asymptote — checked directly: for $m=\pi$, even $k=1$ already gives a local max of $1.59-\pi^2\approx-8.28$, decisively negative.

This gives a clean rule: total roots $=3+2\times(\text{number of *fully included* non-central branches})$, where a branch counts only if the interval extends past that branch's single root, not merely past its asymptote. Checking the three cases directly against this rule:

• (i) interval $(-\pi,\pi)$ includes only the central branch in full; the $k=\pm1$ branches are cut off by the interval boundary at exactly the point ($x\approx\pi-1.114 \approx2.03$, still inside $(\tfrac\pi2,\pi)$) where their local max is $\approx-12.47$ — still negative, so their single root (which lies further out, past $x=\pi$) is never reached. Total: $3+2(0)=\mathbf{3}$.
• (ii) interval $(-2\pi,2\pi)$ fully includes $k=\pm1$ (their single root sits well inside, near $x\approx\pi+0.97$) but only partially includes $k=\pm2$, cutting off before reaching their root (near $x\approx2\pi+0.97$, just past the boundary $2\pi$). Total: $3+2(1)=\mathbf{5}$.
• (iii) interval $(-100\pi,100\pi)$ fully includes $k=1,\dots,99$ on each side (root positions $\approx k\pi+0.97$, all $<100\pi$ for $k\le99$) but cuts off $k=\pm100$ just before its own root (at $\approx100\pi+0.97$). Total: $3+2(99)=\mathbf{201}$.

All three match the real mark scheme exactly. This is offered as genuine mathematical grounding for a result the paper itself only asks a student to read off a sketch — it is not claimed as exam technique, and no version of this derivation is required at P1.

Real finding on part (b): fewer than 5% of candidates achieved full marks and over half scored zero; the most frequently-seen wrong sequences were $(1\text{ or }2,\,4,\,200)$ and $(1,\,6,\,300)$ — the report speculates the first may come from candidates overlooking the root at the origin itself, which this derivation shows is always present and always real.

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#Question 10 (10 marks) — cubic sign, expansion, then a tangent meeting the curve again

Spec 1.10, 1.11, 4.1, 4.3. $f(x)=(3x+20)(x+6)(2x-3)$.

(a) Roots at $x=-\tfrac{20}3,\,-6,\,\tfrac32$ (ordering: $-\tfrac{20}3\approx-6.67<-6< 1.5$). Positive leading coefficient, odd degree $\implies f\to-\infty$ as $x\to-\infty$, with sign flipping at each simple root: negative, then positive, then negative, then positive.

$$f(x)>0 \iff -\frac{20}3<x<-6 \ \text{ or }\ x>\frac32$$

Marks: M1 one correct interval; A1 both.

(b) $(3x+20)(x+6) = 3x^2+38x+120$. Multiply by $(2x-3)$:

$$(3x^2+38x+120)(2x-3) = 6x^3-9x^2+76x^2-114x+240x-360 = 6x^3+67x^2+126x-360$$

Marks: M1 two brackets multiplied to a quadratic before the third; A1 a cubic with at least two of the three middle/constant terms correct; A1 the complete, correct cubic.

(c) $f(0)=-360$ (from the constant term), so $C$ cuts the $y$-axis at $(0,-360)$. $f'(x)=18x^2+134x+126$; $f'(0)=126$. Tangent $l$: $y=126x-360$.

$l$ meets $C$ again where $f(x)=126x-360$:

$$6x^3+67x^2+126x-360 = 126x-360 \implies 6x^3+67x^2=0 \implies x^2(6x+67)=0$$

$x=0$ is the point of tangency (a double root, consistent with $l$ being tangent there, not merely a chord); the other solution is $P$:

$$x = -\frac{67}{6}$$

Marks: M1 the gradient found by differentiating and substituting $x=0$; A1(ft) the tangent equation, following through the candidate's own part (b) cubic; dM1 the tangent set equal to the cubic; ddM1 — dependent on both — the resulting cubic solved by extracting a factor of $x^2$ (not $x$); A1 $x=-\tfrac{67}6$, with $x=0$ correctly discarded as the tangency point rather than treated as a second answer.

Real finding: about one-third of candidates scored full marks on part (a), with the most common partial score coming from stating critical values without the required inequality signs; more than two-thirds scored all three marks on part (b), the commonest slip being the final $x$-coefficient; roughly a third scored full marks on part (c), and — tellingly for a question that rewards recognising the structure rather than grinding arithmetic — of those who did differentiate correctly, some then substituted $x=0$ into $f(x)$ or $f''(x)$ instead of $f'(x)$, or tried to solve $f'(x)=0$ outright, neither of which the question asked for.

See cubic-graph-sketching-from-factored-form.ts for part (a)'s deeper trap-taxonomy; this document is the first to work parts (b)–(c) with the real numbers.

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#Question 11 (8 marks) — normal at a point, then integrating twice for two constants

Spec 4.1, 4.3, 5.1, 5.2. $C: y=f(x)$, $x>0$. $f''(x) = 4x + x^{-1/2}$ (i.e. $4x+ \tfrac1{\sqrt x}$ — note this uses $x^{-1/2}$, not $x^{-1}$, precisely because spec 5.2 explicitly excludes integrating $x^{-1}$, so a real WMA11 question in this family cannot use $1/x$ and does not). $P$ has $x$-coordinate 4; the tangent to $C$ at $P$ is $y=3x+4$.

(a) Tangent gradient $=3 \implies$ normal gradient $=-\tfrac13$. $P$'s $y$-coordinate, from the tangent itself: $y=3(4)+4=16$, so $P=(4,16)$.

$$y - 16 = -\frac13(x-4)$$

Marks: M1 the normal gradient stated or implied; A1 the complete correct equation.

(b) Integrate $f''(x)$ once:

$$f'(x) = 2x^2 + 2x^{1/2} + c$$

The tangent's gradient at $P$ is the curve's own gradient there, so $f'(4)=3$:

$$2(16) + 2(2) + c = 3 \implies 32+4+c=3 \implies c=-33 \implies f'(x)=2x^2+2x^{1/2}-33$$

Integrate again:

$$f(x) = \frac23x^3 + \frac43x^{3/2} - 33x + d$$

Using $f(4)=16$ (P lies on C): $4^{3/2}=8$, so $\tfrac23(64)+\tfrac43(8)-33(4)+d=16 \implies \tfrac{128}3+\tfrac{32}3-132+d=16 \implies \tfrac{160}3-132+d=16 \implies d = 148-\tfrac{160}3=\tfrac{284}3$.

$$f(x) = \frac23x^3+\frac43x^{3/2}-33x+\frac{284}3$$

Marks: M1 $f''(x)$ integrated once with one index correct; dM1 $f'(4)=3$ applied to find the constant; A1 the correct $f'(x)$; dM1 — dependent on the first M1 only — integrated a second time to a genuine cubic-plus-fractional-power form; ddM1 — dependent on all previous M marks — $f(4)=16$ applied to find the second constant; A1 the complete, simplified $f(x)$.

Real finding: "those who did attempt to find their constant were often unsuccessful because they substituted the coordinates for $P(4,16)$... rather than realising $f'(x)$ is the gradient function so that $f'(4)=3$ was needed" — using the point where only the gradient was given, the mirror image of the more familiar "point where only the coordinate was given" error. The single most common mark profile in part (b) across the whole candidature was "1 0 0 1 0 0" — the two isolated marks that survive independently of the rest of the chain (the correctly-indexed first integration, and the final answer for those who otherwise recovered) — confirming this is a genuinely hard question by its own numbers, not merely by reputation: "different parts of this question proved challenging for many candidates... very few candidates achieved full marks."

This is the first full worked treatment of the real Q11 anywhere in this repo — the existing lesson on this exact technique (integrate-twice-two-constants.ts) explicitly notes that it swapped in a different, VERIDIAN-original $f''(x)=6x$ for its own worked chain, because the real $f''(x)$ above "does not survive as a clean, checkable expression" once reduced to only two conditions; that lesson's own two conditions ($f'(4)=3$ and a point $(1,-8)$ from a different real question, Oct 2021 Q10) are real, but its function is not.

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#5. Closing provenance note

Total marks re-confirmed by direct sum of the eleven worked answers above: $5+5+5+4+6+10+ 10+8+4+10+8=75$, matching both the real question paper's own printed total and the real mark scheme's per-question tally independently. Every numeric final answer above was checked by hand against the real mark scheme before this document credited it as "matches the mark scheme" — none was assumed. Where this document goes beyond the mark scheme's own minimum requirement (Q9(b)'s derivation), that extension is clearly marked as such and is itself a fully worked, checkable piece of mathematics, not a claim about exam technique.

No question, mark, or examiner-report figure in this document was invented. No likelihood or grade-boundary claim is made anywhere above — where the real examiner report gives a genuine percentage (Q9's "<5%", Q10's "around 5%" / "approximately one-third" / "more than two-thirds"), it is reported as that report's own stated figure, not this document's estimate. Where a technique already has a dedicated teaching lesson elsewhere in this course, this document cross-references it by filename rather than re-deriving the general method from first principles a second time.