Complete Worked Paper

Provenance (2026-09-29). Series used: October 2020, Pearson Edexcel IAL Mathematics (WMA13), Pure Mathematics P3 — Publications Code WMA13_01_2010_MS (question-paper serial P65758RA), the single most heavily provenance-checked series in research/veridian/WMA13-verified-facts.md (cited there against primary sources at least 30 times across almost every spec topic, more than any other series in the bank). The question paper and mark scheme were fetched fresh for this pass from the PhysicsAndMathsTutor mirror of the Pearson originals and converted to text locally with pdftotext -layout, rather than trusted from the research file's own excerpts — every one of this document's 75 marks was checked against that freshly-extracted primary text, not copied from WMA13-verified-facts.md. One discrepancy is flagged rather than quietly resolved: the mirrored question-paper PDF's own cover page prints "Wednesday 3 June 2020", while the mark scheme's cover page prints "October 2020" and the shared Publications Code on both files reads WMA13_01_2010_MS (the "10" denoting October, matching Pearson's own code convention). The paper's content matches, word for word, the Oct 2020 Q1/Q5/Q7/Q9 anchors already independently verified in the research file (the "sin 3x ≡ 3 sin x − 4 sin³x" proof, the 2cos2x=7cosx equation, the seabird harmonic-form model, the algebraic-division "show Q=5"), so the mismatch is treated here as a mislabelled cover page on the mirror site, not evidence of a different series — but it is recorded here rather than silently corrected, per this course's standing rule against quietly overwriting a provenance question. Every numeric answer below was independently recomputed by hand and cross-checked with Python (math/sympy) before being written in; none is copied from the mark scheme without an independent check. Also read for this pass: content/veridian/curriculum/wma13/ (all eleven lesson files, to find exactly which parts of this series are already taught elsewhere) and content/veridian/question-bank/wma13.ts (grepped — contains no October 2020 material). Exam text is paraphrased throughout; where the mark scheme's own wording is quoted, it is kept under fifteen words and attributed. No Pearson textbook was consulted.

#CODEX Complete Worked Paper — WMA13 October 2020

#Why this series, and what "complete" means here

A "Complete Worked Paper" needs every question in one series solved and mark-justified end to end — not the best-evidenced fragments of several series stitched together. October 2020 is the only WMA13 series in this course's research where that is honestly possible: it is the series WMA13-verified-facts.md returns to most often, and re-verifying it against a freshly downloaded primary source (rather than trusting that file's own transcriptions) confirmed all nine questions, not just the ones already quoted there. No other series in the archive has that property — most are represented by one or two anchor questions, not a full nine-question set independently re-checked from a primary PDF.

The trade-off is that five of this paper's marks-worth of questions already have a fuller pedagogical treatment somewhere in content/veridian/curriculum/wma13/ — this document does not re-teach those from first principles. What it adds that nothing else in this course does: all 75 marks of one real paper, in paper order, each one individually justified against the real mark scheme, plus four questions (Q2, most of Q4, Q6, and most of Q8 and Q9(b)/(c)) that have never been fully worked anywhere in this course before.

Totals: 5+6+7+8+8+9+9+9+14 = 75, matching the paper's own printed total and the spec's fixed mark allocation.

---

#Question 1 (5 marks) — spec 2.3

Paraphrased. Solve, for 0° ⩽ x < 360°, the equation 2cos2x = 7cosx, to one decimal place, without relying on a graphical or numerical calculator method.

Worked solution. Write cos2x as 2cos²x − 1 (the form that turns the equation into a polynomial purely in cosx):

2(2cos²x − 1) = 7cosx ⟺ 4cos²x − 7cosx − 2 = 0

This factorises as (4cosx + 1)(cosx − 2) = 0, giving cosx = −¼ or cosx = 2. The second root is impossible (cosine is bounded by ±1), so cosx = −¼. Taking arccos(−¼) = 104.4775…° gives one solution in range; cosine's symmetry about 180° gives the second as 360° − 104.4775° = 255.5225°. To one decimal place: x = 104.5°, 255.5°.

Mark scheme in full.

• M1 — for using the 2cos²x−1 form of the double-angle identity to turn the equation into a quadratic in cosx at all. This mark is guarded closely: writing 2×2cos²x−1 = 7cosx with no bracket around (2cos²x−1) scores nothing unless the identity was stated correctly first, because the un-bracketed line is algebraically a different (wrong) equation.
• M1 — for attempting to solve the resulting three-term quadratic by any legitimate method, whether or not that method is executed correctly.
• A1 — for reaching cosx = −¼ specifically. A candidate who substitutes y = cosx and stops at y = −¼ without converting back to cosx does not earn this mark — the answer must be back in terms of the original variable.
• dM1 (dependent on the second M1) — for taking arccos of at least one valid root to reach at least one value of x in range.
• A1 — for both final values, 104.5° and 255.5°, with nothing extra given outside the stated range.

Cross-reference. content/veridian/curriculum/wma13/compound-and-double-angle-formulae.ts teaches this exact question as its own real-evidence worked chain (see its lines ~717 onward) and drills the specific bracket-loss error this mark scheme guards against; integration-by-recognition.ts uses the same error as a trap for a different skill. Neither file states the mark-by-mark map exactly as above; that is this document's addition.

---

#Question 2 (6 marks) — spec 3.3

Paraphrased. A bacteria colony's coverage area N mm² on a dish, t days after monitoring began (0 ⩽ t ⩽ 30), is modelled by log₁₀N = 0.0646t + 1.478. Part (a): show the model can be rewritten N = ab^t, giving a to the nearest integer and b to 3 s.f. (4 marks). Part (b): use the model to find the coverage after 30 days, to 2 s.f. (2 marks).

Worked solution.

(a) Undo the log by raising 10 to both sides of log₁₀N = 0.0646t + 1.478:

N = 10^(0.0646t + 1.478) = 10^1.478 × 10^(0.0646t) = 10^1.478 × (10^0.0646)^t

So a = 10^1.478 and b = 10^0.0646. Numerically, 10^1.478 = 30.06… → a = 30 (nearest integer), and 10^0.0646 = 1.1601… → b = 1.16 (3 s.f.). The model is N = 30 × 1.16^t.

(b) At t = 30: the cleanest route avoids rounding a and b early — substitute t = 30 directly into the original log form: log₁₀N = 0.0646(30) + 1.478 = 1.938 + 1.478 = 3.416, so N = 10^3.416 = 2606.8… → N ≈ 2600 mm² (2 s.f.).

Mark scheme in full. (a) M1 — for correctly splitting the exponent, i.e. writing 10^(0.0646t+1.478) as a product of a constant power of 10 and a variable power of 10 (seeing 10^1.478 or 10^0.0646 anywhere in the working is enough to imply this step). A1 — for either awrt 30 or awrt 1.16 appearing on its own. dM1 (dependent) — for applying the index laws correctly and reaching numerical values for both constants, with no incorrect index work anywhere in that line. A1 — for both a = awrt 30 and b = awrt 1.16 stated together (or the model written out with both rounded values in place). (b) M1 — for substituting t = 30 into a version of the model that uses the candidate's own values of a and b (or the equivalent direct log route). A1 — for the final answer, awrt 2600, ignoring any subsequent working after a correct answer is reached.

Why this question is worth including in full. It is the paper's only pure "read a log-linear model, split it, then use it" question with no differentiation, transformation, or geometric context attached — the cleanest available worked example in this entire series for spec 3.3's own core skill (log-linear parameter extraction), distinct from the log-log-graph-reading skill already taught via Jun 2023 Q2 in exponentials-and-logarithmic-graphs.ts. This question gives a and b directly from an already-linear log equation rather than from reading a graph's gradient and intercept — a genuinely different entry point into the same spec point, and one this course has not used before.

---

#Question 3 (7 marks) — spec 4.2

Paraphrased. A curve y = f(x) has f(x) = (2x+3)/√(4x−1), for x > ¼. Part (a): find f′(x) in simplest form (4 marks). Part (b): hence find the range of f (3 marks).

Worked solution.

(a) Apply the quotient rule with u = 2x+3, v = (4x−1)^½. Then u′ = 2 and, by the chain rule, v′ = 2(4x−1)^(−½):

f′(x) = [u′v − uv′] / v² = [2(4x−1)^½ − (2x+3)·2(4x−1)^(−½)] / (4x−1)

Multiplying every term by (4x−1)^½ to clear the negative power from the numerator:

f′(x) = [2(4x−1) − 2(2x+3)] / (4x−1)^(3/2) = (8x − 2 − 4x − 6) / (4x−1)^(3/2) = (4x − 8) / (4x−1)^(3/2)

(b) The only stationary point is where the numerator vanishes: 4x − 8 = 0 ⟹ x = 2 (inside the domain x > ¼). At x = 2: f(2) = (2·2+3)/√(4·2−1) = 7/√7 = √7. Since f(x) → +∞ as x → ¼⁺ (denominator → 0⁺ with a positive numerator) and f(x) → +∞ as x → ∞ too, x = 2 is a minimum, so the range of f is f(x) ⩾ √7.

Mark scheme in full. (a) M1 — for attempting the quotient rule (or an equivalent product-rule route on the rewritten form) and reaching the correct general shape, coefficient slips condoned. A1 — for the correct, unsimplified derivative in that form. dM1 (dependent) — for multiplying through by the matching power of (4x−1) to clear the negative index and reach a linear numerator. A1 — for the final single-fraction answer, (4x−8)/(4x−1)^(3/2) or an exact equivalent, as one combined term. (b) M1 — for setting the derivative to zero and solving for x. dM1 (dependent) — for substituting that x-value back into f itself to find the corresponding y-value. A1 — for the range stated correctly in function notation (f(x) ⩾ √7, or equivalent); a range written as "x ⩾ √7" is not accepted, because the inequality describes the output of f, not the input.

Cross-reference. content/veridian/curriculum/wma13/differentiation-rules.ts teaches part (a) of this exact question in full (its "a technique this lesson used to leave out" block), including the real examiner-report finding that only around a third of candidates scored the simplification mark and only a quarter scored all four — this document does not repeat that derivation. Part (b) — the range — sits outside that lesson's declared scope (it only covers finding dy/dx) and has not been worked anywhere else in this course; it is new here.

---

#Question 4 (8 marks) — spec 1.3 / 1.2 / 1.4

Paraphrased. A sketched graph y = f(x), x ⩾ 0, has f(x) = 21 − 2|2−x|. Part (a): find ff(6) (2 marks). Part (b): solve f(x) = 5x (2 marks). Part (c): given that f(x) = k has exactly two roots, state the set of values k can take (2 marks). Part (d): the graph y = f(x) is transformed to y = af(x−b), with vertex (6,3); find a and b (2 marks).

Worked solution. First unpack the modulus into its two linear pieces: for 0 ⩽ x ⩽ 2, |2−x| = 2−x, so f(x) = 21 − 2(2−x) = 17 + 2x; for x ⩾ 2, |2−x| = x−2, so f(x) = 21 − 2(x−2) = 25 − 2x. This is a "tent" shape: it rises from f(0)=17 to a peak f(2)=21, then falls without bound.

(a) f(6): since 6 ⩾ 2, f(6) = 25 − 2(6) = 13. Then f(13): since 13 ⩾ 2, f(13) = 25 − 2(13) = −1. So ff(6) = f(13) = −1.

(b) Solve f(x) = 5x on each piece. Rising piece (0⩽x⩽2): 17+2x = 5x ⟹ x = 17/3 ≈ 5.67, which falls outside 0⩽x⩽2, so this root is rejected. Falling piece (x⩾2): 25−2x = 5x ⟹ x = 25/7 ≈ 3.57, which does satisfy x⩾2, so it is kept. x = 25/7 only.

(c) A horizontal line y=k meets this tent shape: for k > 21 (above the peak) there are zero intersections; at k = 21 (exactly the peak) there is exactly one; for 17 ⩽ k < 21 the line crosses both the rising piece (once) and the falling piece (once), giving exactly two; for k < 17 only the falling piece reaches that low, giving exactly one. So exactly two roots requires 17 ⩽ k < 21.

(d) The vertex of y=f(x) is its peak, (2, 21). Under y = af(x−b), a horizontal shift of b and a vertical scaling by a, the peak maps to (2+b, 21a). Matching this to the given vertex (6,3): 2+b = 6 ⟹ b = 4, and 21a = 3 ⟹ a = 1/7, b = 4.

Mark scheme in full. (a) M1 — for attempting the composition twice (finding f(6) then feeding that value back into f). A1 — for −1. (b) M1 — for setting up the correct equation on either the correct or an incorrect piece and reaching x=. A1 — for x = 25/7, and only that value; if extra (rejected) solutions are also shown, the mark still stands only if they are explicitly identified and discarded rather than left ambiguous. (c) M1 — for identifying either boundary condition (k < 21 or k ⩾ 17) on its own. A1 — for the full combined range 17 ⩽ k < 21, which may be written as two separate inequalities or in interval notation. (d) B1 — for either a = 1/7 or b = 4 found correctly on its own. B1 — for both found correctly together.

Why parts (a), (b), (d), and the mechanics of (c) are new to this course. modulus-and-graph-transformations.ts cites this question only once, for part (c)'s general "solve one branch, then pick the wrong region" failure mode — it does not work through the tent-shape mechanics, the two-branch composition in part (a), the branch-rejection logic in part (b), or the vertex-mapping arithmetic in part (d). All four are genuinely new worked material for this course, and part (c) in particular is a clean, real example of a "how many roots" question built entirely from modulus branch logic rather than a quadratic discriminant — a question shape this course has not otherwise modelled.

---

#Question 5 (8 marks) — spec 2.3 / 5.2

Paraphrased. Part (a): show that sin3x ≡ 3sinx − 4sin³x (4 marks). Part (b): hence, using algebraic integration, find ∫₀^(π/3) sin³x dx (4 marks).

Worked solution.

(a) Write 3x as 2x+x and expand with the compound-angle formula for sine:

sin3x = sin(2x+x) = sin2x·cosx + cos2x·sinx

Substitute sin2x ≡ 2sinx·cosx, and — choosing this specific form because the target expression is entirely in sinx — cos2x ≡ 1−2sin²x:

= (2sinx·cosx)·cosx + (1−2sin²x)·sinx = 2sinx·cos²x + sinx − 2sin³x

One cos²x remains; eliminate it with cos²x ≡ 1−sin²x:

= 2sinx(1−sin²x) + sinx − 2sin³x = 2sinx − 2sin³x + sinx − 2sin³x = 3sinx − 4sin³x ∎

(b) Rearranging part (a)'s identity for sin³x: 4sin³x = 3sinx − sin3x, so sin³x = ¾sinx − ¼sin3x. Integrate term by term:

∫sin³x dx = ∫(¾sinx − ¼sin3x) dx = −¾cosx + (1/12)cos3x + c

Evaluate from 0 to π/3. At x=π/3: cos(π/3)=½, cos(3·π/3)=cosπ=−1, giving −¾(½) + (1/12)(−1) = −3/8 − 1/12 = −11/24. At x=0: cos0=1 (twice), giving −¾ + 1/12 = −2/3. Subtracting: −11/24 − (−2/3) = −11/24 + 16/24 = 5/24.

Mark scheme in full. (a) M1 — for expanding sin3x as sin(2x+x) using the compound-angle formula. M1 — for substituting correct double-angle identities for both sin2x and cos2x. ddM1 (dependent on both preceding method marks) — for the final elimination step, using cos²x ≡ 1−sin²x to collapse everything to a single-variable expression in sinx. A1* — for reaching the given, printed answer with every line correctly notated; the scheme specifically penalises sloppy notation here even when the mathematics is right (for instance, writing a bare "sin" with no argument in place of "sinx"). (b) M1 — for using part (a)'s identity, rearranged, to rewrite sin³x as a combination of sinx and sin3x. dM1 (dependent) — for integrating that rewritten form term by term. A1 — for the correct antiderivative, −¾cosx + (1/12)cos3x, unsimplified forms allowed. A1 — for the final evaluated answer, 5/24, correct and simplified, with the working fully shown (the scheme is explicit that a bare "5/24" with no supporting algebra scores nothing here, since the question specifically asks for algebraic integration).

Cross-reference. Part (a) is taught in full, with the identical four-step derivation and full mark map, in content/veridian/curriculum/wma13/compound-and-double-angle-formulae.ts (its "beyond spec" closing section) — not repeated in depth here. Part (b) — the hence-integration — is new to this course: no existing file uses this question's second half, which is a genuinely useful worked example of "hence integrate an odd power of sine using a triple-angle identity," a technique distinct from the even-power (cos2x-substitution) route integration-by-recognition.ts already teaches for sin²x/cos²x.

---

#Question 6 (9 marks) — spec 3.1 / 6.1 / 6.2

Paraphrased. Figure 3 shows curve C₁: y = 5e^(x−1) + 3, and curve C₂: y = 10 − x². Point P lies on C₁ with y-coordinate 18. Part (a): find P's x-coordinate in the form ln k (3 marks). C₁ meets C₂ at x=α and x=β. Part (b): using a stated interval and function, show α = 1.134 to 3 d.p. (3 marks). Part (c): using the iteration xₙ₊₁ = −√(7 − 5e^(xₙ−1)) with x₁ = −3, find x₂ and the value of β, each to 6 d.p. (3 marks).

Worked solution.

(a) Set y=18 in C₁'s equation: 5e^(x−1) + 3 = 18 ⟹ e^(x−1) = 3 ⟹ x−1 = ln3 ⟹ x = ln3 + 1. Since 1 = ln e, this is x = ln(3e).

(b) Where C₁ meets C₂: 5e^(x−1) + 3 = 10 − x², i.e. define h(x) = x² + 5e^(x−1) − 7 (zero exactly where the curves meet). Evaluate at the two ends of a tight interval bracketing the claimed root: h(1.1335) = 1.1335² + 5e^(0.1335) − 7 = 1.28482 + 5(1.14283) − 7 = 1.28482 + 5.71415 − 7 = −0.00103 h(1.1345) = 1.1345² + 5e^(0.1345) − 7 = 1.28709 + 5(1.14397) − 7 = 1.28709 + 5.71985 − 7 = +0.00694 h(1.1335) < 0 and h(1.1345) > 0 — a sign change. Since h is a sum of a polynomial and an exponential, both continuous everywhere, h is continuous on [1.1335, 1.1345]. By the sign-change/continuity test, a root lies in this interval, so, to 3 d.p., α = 1.134.

(c) Substitute x₁ = −3 into the iteration: x₂ = −√(7 − 5e^(−3−1)) = −√(7 − 5e^(−4)) = −√(7 − 5(0.0183156)) = −√(7 − 0.091578) = −√6.908422 = −2.628388 (6 d.p.). Iterating further to find where the sequence settles (x₃ = −√(7−5e^(x₂−1)) = −2.620535, x₄ = −2.620336, x₅ = −2.620330, x₆ = −2.620330, …) shows the sequence converging to six decimal places by the fifth iterate: β = −2.620330.

Mark scheme in full. (a) M1 — for progressing from the given equation to e^(x−1)=3 or an equivalent isolated-exponential form. A1 — for x = ln3+1 (or the equivalent e^x=3e). A1 — for the final answer converted into the required single-logarithm form, x = ln(3e). (b) B1 — for correctly setting the two curve equations equal and rearranging to a single function suitable for a sign-change test. M1 — for substituting a suitable pair of x-values either side of the claimed root and reaching at least one function value correct to 1 s.f. A1 — for both values correct (to at least 1 s.f.), together with a reason that explicitly names both the sign change and the continuity of the function, and a minimal concluding statement. (c) M1 — for substituting x₁ into the given iterative formula (this may be implied by the correct numerical answer appearing with no algebra shown). A1 — for x₂ = awrt −2.628388. A1 — for β = awrt −2.620330, contingent on genuine evidence of continued iteration rather than an unexplained final value.

Why this question is worth including in full. numerical-methods.ts cites only part (c)'s premature-rounding trap in passing; the full three-part arc — an exact log-form solve, a genuine sign-change proof on a mixed polynomial-exponential function, and an iteration converging over five steps to a six-decimal-place answer — has not been worked anywhere else in this course. It is also the paper's clearest illustration of why the sign-change interval in a real question is often far narrower than the requested decimal precision would suggest (a 0.001-wide interval to justify a 3-d.p. answer, not the 0.05-wide interval a first attempt might reach for) — a point about exam technique this document's worked answer demonstrates directly rather than merely asserting.

---

#Question 7 (9 marks) — spec 2.3

Paraphrased. Part (a): express cosx + 4sinx in the form Rcos(x−α), R>0, 0<α<π/2, giving R exactly and α in radians to 3 d.p. (3 marks). A seabird's height H metres above sea level, t seconds after leaving the nest (0⩽t⩽6.5), is modelled by H = 24/(3 + cos(½t) + 4sin(½t)). Part (b): find the minimum height, to the nearest cm (2 marks). Part (c): find t, to 2 d.p., when H=10 (4 marks).

Worked solution.

(a) Match coefficients: Rcos(x−α) = Rcosα·cosx + Rsinα·sinx, so Rcosα = 1 and Rsinα = 4. Then R² = 1²+4² = 17, so R = √17; and tanα = 4/1 = 4, so α = arctan4 = 1.326 (3 d.p., radians).

(b) The denominator 3 + cos(½t) + 4sin(½t) has the wave part cos(½t)+4sin(½t) = √17·cos(½t−1.326) by part (a) (with x replaced by ½t). H is minimised when the denominator is maximised — a fraction is smallest when its denominator is largest — which happens when the cosine term reaches its greatest value, +√17. So minimum H = 24/(3+√17) = 24/7.1231… = 3.369…m ≈ 337 cm (3.37m to the nearest cm).

(c) Set H=10: 24/(3 + √17cos(½t−α)) = 10 ⟹ 3 + √17cos(½t−α) = 2.4 ⟹ cos(½t−α) = −0.6/√17 = −0.14550. Carrying α to full precision here (1.325818 rad) rather than the 3-d.p. answer already rounded for part (a) — rounding this intermediate value early is exactly the kind of premature-rounding slip this course's numerical-methods research flags as a real, recurring mark-loser elsewhere in this paper's own family of questions, and it is avoided here on principle. Taking arccos: ½t − α = ±1.716836 (radians). Since t∈[0,6.5], ½t∈[0,3.25], so ½t−α ranges over about [−1.326, 1.924] — only the positive branch, +1.716836, fits inside that range. So ½t = 1.325818+1.716836 = 3.042654 ⟹ t = 6.085308, which rounds to t = 6.09 (2 d.p.).

Mark scheme in full. (a) B1 — for R=√17 exactly; a correct decimal such as 4.12 earns nothing here, because the question specifically asks for the exact value. M1 — for a correct method reaching tanα (or an equivalent sinα/cosα route) from the matched coefficients. A1 — for α = awrt 1.326 in radians specifically; the mark scheme explicitly rejects the correct angle given in degrees (75.96°) even though it is numerically the same angle. (b) M1 — for identifying that the minimum height comes from 24/(3+R). A1 — for 3.37 (metres), with the unit required; 3.37cm is not credited, though 337cm is, provided the unit is stated. (c) M1 — for using the candidate's own R and α from part (a) to reach an equation of the form cos(βt±α) = k inside the valid range for k. A1 — for the correct value of k (−0.6/√17, awrt −0.146). M1 — for a complete method converting that equation back to t (doubling the half-angle argument correctly, since the wave's argument here is ½t, not a bare t). A1 — for t = awrt 6.09, provided any extra solutions inside the domain are not also given (an extra solution outside the domain may be ignored, but one inside it, left un-rejected, costs the mark).

Cross-reference. This exact question, including this same mark map, is already taught in full — with a worked derivation, a marked solution, and two dedicated traps built directly from its own mark scheme (the exact-value-of-R trap and the maximise-not-minimise trap) — in content/veridian/curriculum/wma13/harmonic-form.ts. It is included here in condensed form only for the paper's completeness; readers wanting the full teaching treatment should go to that file, not this one.

---

#Question 8 (9 marks) — spec 4.2 / 4.3

Paraphrased. Part (i): a curve has g(x) = e^(3x)sec2x, −π/4 < x < π/4. (a) Find g′(x) (2 marks). (b) Hence find the x-coordinate of the stationary point (3 marks). Part (ii): a different curve has x = ln(siny), 0<y<π/2. Show that dy/dx = eˣ/f(x) for some function f(x) of eˣ (4 marks).

Worked solution.

(i)(a) Apply the product rule to g(x) = e^(3x)·sec2x. The derivative of e^(3x) is 3e^(3x); the derivative of sec2x is 2sec2x·tan2x (a standard result printed in the formula booklet). So:

g′(x) = 3e^(3x)·sec2x + e^(3x)·2sec2x·tan2x = e^(3x)sec2x(3 + 2tan2x)

(i)(b) Since e^(3x) is never zero and sec2x is never zero, g′(x)=0 exactly when 3+2tan2x=0, i.e. tan2x = −3/2. Taking arctan: 2x = arctan(−1.5) = −0.98279 (this lies inside the valid range for 2x, since −π/4<x<π/4 means −π/2<2x<π/2). So x = −0.491 (3 d.p.).

(ii) Differentiate x = ln(siny) with respect to y: dx/dy = (1/siny)·cosy = cosy/siny. Take the reciprocal: dy/dx = siny/cosy. Now convert back into x: from x=ln(siny), siny = eˣ; and since 0<y<π/2, cosy is positive, so cosy = √(1−sin²y) = √(1−e^(2x)). Substituting:

dy/dx = eˣ / √(1 − e^(2x)), so f(x) = √(1−e^(2x)).

Mark scheme in full. (i)(a) M1 — for a correct attempt at the product rule (or an equivalent quotient-rule route on 1/cos2x), correct general shape. A1 — for the correct derivative, e^(3x)sec2x(3+2tan2x) or an equivalent unfactorised form. (i)(b) M1 — for setting g′(x)=0 and factoring out the common e^(3x)sec2x term to isolate a linear equation in tan2x. dM1 (dependent) — for the correct order of operations converting that equation into a value of x. A1 — for x = awrt −0.491, and only that value; a second, spurious solution arising from incorrectly also setting sec2x=0 is not credited. (ii) B1 — for correctly finding dx/dy = cosy/siny (or the equivalent eˣ=siny route restated as a derivative relationship). M1 — for one of the two operations needed to complete the proof: either expressing cosy in terms of eˣ via sin²y+cos²y≡1, or taking the reciprocal to form dy/dx. dM1 (dependent on both) — for applying both operations together to reach an expression purely in eˣ. A1 — for the final answer, dy/dx = eˣ/√(1−e^(2x)), in terms of x throughout (not left partly in y).

Why this question is worth including in full. differentiation-rules.ts quotes this question's part (ii) only for its examiner-report wording about candidates stopping one line early (reaching eˣ/cosy and never converting cosy back into x) — it does not work through either part's full mechanics. Part (i) in particular is a clean worked example of "set a product-rule derivative to zero and solve," a mark-scheme shape (isolate a linear equation in tanθ inside a larger factored expression) this course has not modelled in full before; both parts are new here.

---

#Question 9 (14 marks) — spec 1.1 / 4.2 / 5.2

Paraphrased. Part (a): given (x⁴−x³−10x²+3x−9)/(x²−x−12) ≡ x²+P+Q/(x−4), for x>−3, find P and show that Q=5 (4 marks). A curve y=g(x) uses the same expression, for −3<x<3.5. Part (b): find the tangent to the curve at x=2, in the form y=mx+c (5 marks). Part (c): the region R is bounded by the curve, the y-axis, the x-axis, and the line x=2; find the exact area of R in the form a+b ln2 (5 marks).

Worked solution.

(a) Divide x⁴−x³−10x²+3x−9 by x²−x−12. The first step of long division: x⁴÷x² = x², and x²(x²−x−12) = x⁴−x³−12x²; subtracting from the original leaves 2x²+3x−9. The second step: 2x²÷x² = 2, and 2(x²−x−12) = 2x²−2x−24; subtracting leaves 5x+15. So the quotient is x²+2 with remainder 5x+15, giving:

(x⁴−x³−10x²+3x−9)/(x²−x−12) = x²+2 + (5x+15)/(x²−x−12)

Factorise the remainder's denominator: x²−x−12 = (x−4)(x+3). The remainder's numerator also factorises: 5x+15 = 5(x+3). The (x+3) factor cancels top and bottom (legitimate since x≠−3 is already excluded):

= x² + 2 + 5(x+3)/[(x−4)(x+3)] = x² + 2 + 5/(x−4)

So P = 2, and the remainder collapses to exactly 5/(x−4), confirming Q = 5.

(b) g(x) = x²+2+5/(x−4). Differentiate term by term: g′(x) = 2x − 5/(x−4)² (using the chain rule on the second term, treating it as 5(x−4)^(−1)). At x=2: g′(2) = 2(2) − 5/(2−4)² = 4 − 5/4 = 11/4. The point of tangency: g(2) = 2²+2+5/(2−4) = 4+2−5/2 = 7/2. The tangent line through (2, 7/2) with gradient 11/4:

y − 7/2 = (11/4)(x−2) ⟹ y = (11/4)x − 11/2 + 7/2 = (11/4)x − 2 ⟹ y = 2.75x − 2

(c) The area of R is ∫₀² g(x) dx (checking first that g(x) stays positive on [0,2]: g(0) = 0+2+5/(0−4) = 2−1.25 = 0.75 > 0, and g(2) = 3.5 > 0, so the curve does not cross the x-axis on this interval and a plain integral gives the area directly). Integrating term by term:

∫(x²+2+5/(x−4)) dx = x³/3 + 2x + 5ln|x−4| + c

Evaluating from 0 to 2: at x=2, this is 8/3 + 4 + 5ln|2−4| = 8/3 + 4 + 5ln2 (since |2−4|=2). At x=0, this is 0+0+5ln|0−4| = 5ln4. Subtracting:

Area = (8/3 + 4 + 5ln2) − 5ln4

Since ln4 = 2ln2, 5ln4 = 10ln2, so:

Area = 8/3 + 4 + 5ln2 − 10ln2 = 20/3 − 5ln2

matching the required a+b ln2 form with a = 20/3, b = −5.

Mark scheme in full. (a) M1 — for dividing to a quadratic quotient and a linear remainder by any full, legitimate process. A1 — for the correct quotient x²+2 and remainder 5x+15. M1 — for writing the remainder's denominator in factorised form (x−4)(x+3), or equivalently dividing the remainder by (x+3) directly. A1 — for the fully justified final answer; because Q=5 is a given value the question asks the candidate to show, this mark is not awarded for simply asserting Q=5 without the cancellation step shown. (b) M1 — for differentiating the 5/(x−4) term correctly (recognising the −1 power and applying the chain rule). A1 — for g′(x) = 2x − 5/(x−4)². M1 — for substituting x=2 to find the numerical gradient at that point. dM1 (dependent) — for a complete method finding the equation of the tangent through the correct point with the correct gradient, proceeding as far as finding the constant term if using y=mx+c. A1 — for y = 2.75x−2 (or an exact equivalent), with subsequent working after a correct answer ignored. (c) M1 — for integrating the 5/(x−4) term correctly to a log form (condoning a missing modulus sign at this stage). A1ft (follow-through on the candidate's own value of P) — for the correct antiderivative, x³/3+2x+5ln|x−4|. dM1 (dependent on the first M) — for substituting the limits 0 and 2 into that antiderivative and subtracting in the correct order. ddM1 (dependent on both preceding method marks) — for correctly combining the two resulting log terms using legitimate log laws. A1 — for the final answer, 20/3 − 5ln2, exact and fully simplified. A candidate who drops the modulus sign and tries to evaluate ln(−2) and ln(−4) directly (both undefined) cannot reach this mark by that route — this is precisely the real error this series' examiner report records happening to a substantial number of candidates on this exact part.

Cross-reference. Part (a) is taught in full — including this same mark map and the real examiner-report finding about candidates stopping before the final cancellation — in content/veridian/curriculum/wma13/rational-expressions-and-algebraic-division.ts; not repeated here beyond the summary above. Part (c)'s modulus-dropping trap is cited (with the same undefined-ln(−2)/ln(−4) detail) in integration-by-recognition.ts, but that file does not work through the full area calculation. Part (b) — the tangent-line mechanics — and the full worked mechanics of part (c) are new to this course.

---

#Reconciliation and limitations

Mark total: 5 + 6 + 7 + 8 + 8 + 9 + 9 + 9 + 14 = 75, matching both the question paper's own printed total and the spec's fixed 75-mark allocation for this unit.

What this document does not claim. It does not claim October 2020 is a "typical" or "representative" WMA13 paper, or that any of its questions recur with any stated frequency across other series — those are claims for the cross-series research file (WMA13-verified-facts.md), not for a single-series worked-paper document, and none is made here. It reports no grade boundary and no likelihood percentage for any question resurfacing on a future paper, because neither is knowable from a single series and neither is stated anywhere in the source material used. Where this document's account of why a mark is earned goes beyond a bare paraphrase of the mark scheme's own wording (for instance, the reasoning about which branch of an arccos/arctan solution survives a stated domain), that reasoning was independently derived and checked against the mark scheme's stated final answer, not read out of the scheme's own notes, which are typically terser than the explanation given here.

One item flagged, not resolved. The mirrored question-paper PDF's cover date ("Wednesday 3 June 2020") does not match its mark scheme's own cover date ("October 2020") or the shared Publications Code (WMA13_01_2010_MS). This document treats the mark scheme's cover and the Publications Code as authoritative, on the strength of the paper's content matching the Oct 2020 anchors already independently verified elsewhere in this course's research — but a future pass with access to a second, independent mirror of the same paper should re-check this rather than treat it as settled.