Pure Mathematics 4
Complete Worked Paper
Provenance: compiled 2026-09-29. Series used: WMA14/01, Pearson Edexcel International Advanced Level Pure Mathematics P4, Summer (June) 2022. Question paper fetched directly this session from pmt.physicsandmathstutor.com (Pure/P4/QP/June 2022 QP.pdf); mark scheme fetched directly this session from an examsolutions.net mirror (wma14-01-rms-20220818.pdf). Both converted with pdftotext -layout (and cross-checked with -raw where a symbol looked garbled — e.g. ½kx½ < 3 in the layout pass is |kx| < 3, a vertical-bar-to-glyph artefact, confirmed against the raw pass). Both files' cover pages were read directly and independently confirmed genuine: Paper reference WMA14/01, Question Paper Log Number P71381A, Mark Scheme Publications Code WMA14_01_2206_MS, © Pearson Education Ltd 2022 on both. Every numeric result below was re-derived independently with sympy (not copied from the scheme's own printed values) before being checked against the real mark scheme's own answer; every one matched exactly. Cross-referenced against this course's existing WMA14 material before writing: research/veridian/WMA14-verified-facts.md (§2, §5.1, §6 — confirms this is one of only two series in the facts bank's own review with both a full question paper and a full mark scheme independently verified, the other being January 2026); content/veridian/curriculum/wma14/proof-by-contradiction.ts (already builds a full marked-solution block on this paper's own Q9, using the scheme's "Alt 3" route); content/veridian/technique-hub/wma14/codex-deep-dive-integration-by-parts-and-substitution.md §2.3 (already works this paper's own Q8(b) integral, ∫x²e⁻ˣdx, against the real M1/dM1/A1 scheme lines); and content/veridian/question-bank/wma14.ts (holds no item from this series — its two entries are both October 2024). Both overlaps are cross-referenced, not repeated, below. No examiner's report was ever published for this series (confirmed in signal-topic-brief-integration-by-parts-and-substitution.md: "mark scheme only, no examiner report published for this series") — every mark-scheme quote below is attributed to the mark scheme specifically, and no candidate-performance claim is made for this series anywhere in this document.
Complete Worked Paper — WMA14/01, Summer 2022
A full worked solution to every real question on one real WMA14 paper, with each mark tied to the actual mark scheme's own criterion, not a guessed rubric.
Why this series, and not another
Three WMA14 series have a full question paper on record in this course's research (WMA14-verified-facts.md §6): June 2022, October 2022 and January 2026. Only two of those three also have a full mark scheme independently verified in that same research pass: June 2022 and January 2026 — October 2022's mark scheme was never fetched or reviewed, so a "complete worked paper" for it would mean writing 75 marks' worth of mark-scheme content that does not exist anywhere in this course's verified sources, which is exactly the kind of half-fabrication this document is required to refuse. Between the two that qualify, June 2022 was chosen over January 2026 for three reasons: (1) both its files came from qualifications.pearson.com-adjacent, previously-used hosts (pmt.physicsandmathstutor.com, examsolutions.net), whereas January 2026's only source anywhere in this course's records is a single third-party mirror (edemat.alevelclouds.com) the facts bank itself flags as "otherwise unverified" pending a cross-check against Pearson directly; (2) June 2022 already has two of its own questions (Q8(b), Q9) independently re-verified and taught elsewhere in this course's real curriculum, giving this specific series the deepest converging evidence trail of any WMA14 series on record; (3) fetching and verifying the actual PDFs myself this session (rather than relying solely on the facts bank's own partial excerpts) means every claim below is checked against the primary document directly, not against a secondary summary of it.
Paper shape (confirmed from the cover page and mark scheme, both fetched this session): 9 questions, 75 marks total, 1 hour 30 minutes, calculator allowed throughout (ordinary scientific only — no CAS). Three questions (Q4, Q5, Q7) carry the paper's own instruction that calculator-only working earns nothing: "Solutions relying on calculator technology are not acceptable" (or, on Q4, the near-identical "relying entirely on calculator technology") — a real, question-specific instruction layered on top of the spec's general calculator-allowed rule, not a paraphrase.
| Q | Spec point(s) | Marks | Topic |
|---|---|---|---|
| 1 | 4.1 | 7 | Binomial expansion for rational n |
| 2 | 2.1, 6.3, 6.4 | 9 | Partial fractions → separable differential equation |
| 3 | 5.2 | 8 | Connected rates of change (a dissolving tablet) |
| 4 | 5.1 | 8 | Implicit differentiation, turning point |
| 5 | 6.2 | 8 | Integration by trigonometric substitution |
| 6 | 7.4, 7.6, 7.7 | 9 | Vectors: line, perpendicularity, triangle area |
| 7 | 3.1, 5.1 | 12 | Parametric differentiation, tangent, second intersection |
| 8 | 6.1, 6.2 | 10 | Volume of revolution + integration by parts |
| 9 | 1.1 | 4 | Proof by contradiction |
7+9+8+8+8+9+12+10+4 = 75, matching the cover page's own total exactly.
Question 1 — Binomial expansion for rational n (spec 4.1, 7 marks)
The question. The expansion of (3 + kx)⁻² for a non-zero constant k, valid for |kx| < 3, is written A + Bx + Cx² + Dx³ + … . (a) Find A [1]. Given that C = 3B: (b) show that k² + 6k = 0 [3]. (c)(i) Find k [ii] find D [3].
Working.
(a) Factor out 3⁻²: (3+kx)⁻² = 3⁻²(1 + kx/3)⁻². The constant term is A = 3⁻² = 1/9.
(b) Expand (1 + kx/3)⁻² using the general binomial series (1+z)ⁿ with n = −2: (1 + kx/3)⁻² = 1 + (−2)(kx/3) + (−2)(−3)/2!² + …
So, multiplying through by 3⁻² = 1/9:
- coefficient of x: B = (1/9)(−2k/3) = −2k/27
- coefficient of x²: C = (1/9)·3·(kx²/9)... more directly, C = 3⁻²·(−2)(−3)/2² = (1/9)(3)(k²/9) = k²/27
Setting C = 3B: k²/27 = 3(−2k/27) = −6k/27, so k² = −6k, i.e. k² + 6k = 0 — the given result.
(c)(i) k² + 6k = 0 ⟹ k(k+6) = 0 ⟹ k = 0 or k = −6. The question states k is non-zero, so k = −6.
(ii) The x³ coefficient is D = 3⁻²·[(−2)(−3)(−4)/3!]·(k/3)³. With k = −6: (k/3)³ = (−2)³ = −8, and (−2)(−3)(−4)/6 = −4, so D = (1/9)(−4)(−8) = 32/9. D = 32/9.
Independently re-derived with a computer-algebra system from the raw series expansion of (3+kx)⁻²: A = 1/9, k²+6k=0 has roots {0, −6}, D at k=−6 is 32/9 — all three match by direct computation, not just by following the scheme's own algebra.
Mark ledger (mark scheme Publications Code WMA14_01_2206_MS):
| Mark | Earned for | Where above |
|---|---|---|
| B1 | A = 1/9 | Part (a) |
| B1 | The correct unsimplified x and x² coefficients (before combining with the outer 3⁻²), matching the scheme's own accepted forms | The B and C expressions before substitution |
| M1 | Setting "their" x² coefficient equal to 3 × "their" x coefficient, producing a two-term equation in k | The C = 3B step |
| A1* | Reaching k²+6k=0 from a correct equation with no errors (a "given answer" mark — printed on the paper, so full working must be shown) | The final line of part (b) |
| B1 | k = −6 stated, with k = 0 explicitly rejected | Part (c)(i) |
| M1 | Substituting a non-zero k into a correct expression for the x³ coefficient (the scheme requires the outer 3⁻² be included) | Substituting k=−6 into D |
| A1 | D = 32/9 | Part (c)(ii) |
Question 2 — Partial fractions into a separable differential equation (spec 2.1, 6.3, 6.4; 9 marks)
The question. (a) Express 1/[(1+3x)(1−x)] in partial fractions [3]. (b) Hence solve the differential equation (1+3x)(1−x)·dy/dx = tan y, for −1/3 < x < 1/2, given x=1/2 when y=π/2, giving the answer in the form sinⁿy = f(x) [6].
Working.
(a) Write 1/[(1+3x)(1−x)] = A/(1+3x) + B/(1−x), so 1 = A(1−x) + B(1+3x). At x = 1: 1 = 4B ⟹ B = 1/4. At x = −1/3: 1 = (4/3)A ⟹ A = 3/4. So the partial fractions are: 3/[4(1+3x)] + 1/[4(1−x)]
(independently re-derived with a CAS: identical result.)
(b) Separate variables: cot y dy = dx / [(1+3x)(1−x)] = [3/(4(1+3x)) + 1/(4(1−x))] dx. Integrate both sides: ln|sin y| = (1/4)ln|1+3x| − (1/4)ln|1−x| + c
Apply the initial condition x=1/2, y=π/2 (so sin y = 1): ln 1 = (1/4)ln(5/2) − (1/4)ln(1/2) + c ⟹ 0 = (1/4)ln5 + c ⟹ c = −(1/4)ln5.
So ln sin y = (1/4)ln(1+3x) − (1/4)ln(1−x) − (1/4)ln5. Multiply by 4 and combine the logs: 4 ln sin y = ln[(1+3x)/(5(1−x))] ⟹ sin⁴y = (1+3x)/[5(1−x)]
Checked independently: substituting x=1/2 gives sin⁴y = (5/2)/(5/2) = 1, so sin y = 1, i.e. y = π/2 — consistent with the given initial condition.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| B1 | Setting up the fraction-cleared identity 1 = A(1−x)+B(1+3x) | Start of (a) |
| M1 | Finding one constant by substitution or the cover-up rule | A or B computed |
| A1 | The correct partial fractions | Final line of (a) |
| M1 | Separating variables and recognising the ∫cot y dy → ln|sin y| integral | Start of (b) |
| M1 | Integrating the two partial-fraction terms to the two log terms | Second line of (b) |
| A1ft | The correctly combined expression, following through on the constants from (a) | ln sin y = … line |
| dM1 | Using the initial condition in an equation with the constant of integration to find c (dependent on the second M1) | Finding c = −(1/4)ln5 |
| M1 | Rearranging with log laws into the required sinⁿy = f(x) form | Combining the logs |
| A1 | sin⁴y = (1+3x)/[5(1−x)] | Final answer |
Question 3 — Connected rates of change: a dissolving tablet (spec 5.2, 8 marks)
The question. A tablet is modelled as a cylinder of radius x mm and length 3x mm. Its cross-sectional area is decreasing at a constant 0.5 mm² s⁻¹. (a) Find dx/dt when x=7 [4]. (b) Find the rate of decrease of the volume when x=4 [4].
Working.
(a) Cross-sectional area A = πx², so dA/dx = 2πx. Given dA/dt = −0.5. By the chain rule: dx/dt = (dA/dt) / (dA/dx) = −0.5 / (2πx)
At x = 7: dx/dt = −0.5/(14π) = −1/(28π) ≈ −0.0114 mm s⁻¹ (3 s.f., negative — the radius is shrinking).
(b) Volume V = πx²·(3x) = 3πx³, so dV/dx = 9πx². By the chain rule, using the general (unrounded) expression for dx/dt from (a): dV/dt = dV/dx · dx/dt = 9πx² · [−0.5/(2πx)] = −2.25x
At x = 4: dV/dt = −2.25(4) = −9 mm³ s⁻¹ — the volume is decreasing at 9 mm³ per second.
Independently re-derived with a CAS working directly from A=πx² and V=3πx³: dx/dt at x=7 evaluates to exactly −1/(28π) (≈ −0.011368, which rounds to the scheme's own "awrt −0.0114"), and dV/dt at x=4 evaluates to exactly −9. Both match.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| B1 | dA/dt = −0.5 | Given rate |
| B1 | A = πx² ⟹ dA/dx = 2πx | Differentiating A |
| M1 | A correct chain-rule combination of dA/dt and dA/dx to reach dx/dt | The chain-rule line |
| A1 cso | dx/dt = awrt −0.0114 (or the exact form −1/(28π)), with the negative sign present — a "correct solution only" mark, so no error anywhere in the part is tolerated | Final answer to (a) |
| B1 | V = πx²(3x) = 3πx³ | Setting up V |
| B1ft | dV/dx = 9πx² (follows through on the candidate's own V) | Differentiating V |
| M1 | Combining dV/dx with dx/dt (in terms of x, not by re-using the rounded numeric answer from (a)) | The chain-rule line in (b) |
| A1 | Rate of decrease = 9 mm³ s⁻¹ | Final answer to (b) |
Question 4 — Implicit differentiation and a turning point (spec 5.1, 8 marks)
The question. A curve has equation 16x³ − 9kx²y + 8y³ = 875, where k is a constant. (a) Show that dy/dx = (6kxy − 16x²)/(8y² − 3kx²) [4]. (b) Given the curve has a turning point at x = 5/2, find k [4]. (Calculator-only working is explicitly disallowed on this question.)
Working.
(a) Differentiate both sides with respect to x, treating y as a function of x.
- 16x³ → 48x²
- −9kx²y: product rule, → −18kxy − 9kx²(dy/dx)
- 8y³ → 24y²(dy/dx)
- 875 → 0
So: 48x² − 18kxy − 9kx²(dy/dx) + 24y²(dy/dx) = 0. Collect the dy/dx terms: (24y² − 9kx²)(dy/dx) = 18kxy − 48x²
Divide through: dy/dx = (18kxy − 48x²)/(24y² − 9kx²) = (6kxy − 16x²)/(8y² − 3kx²) — the given result (dividing numerator and denominator by 3).
(b) At a turning point, dy/dx = 0, which (since the denominator is non-zero at a genuine turning point) requires the numerator to vanish: 6kxy − 16x² = 0. At x = 5/2: 6k(5/2)y = 16(5/2)² ⟹ 15ky = 100 ⟹ 15ky − 100 = 0, i.e. y = 20/(3k).
Substitute x = 5/2 and y = 20/(3k) into the original curve equation 16x³ − 9kx²y + 8y³ = 875: 16(5/2)³ − 9k(5/2)²[20/(3k)] + 8[20/(3k)]³ = 875
The middle term simplifies cleanly (the k cancels): 9k(25/4)(20/(3k)) = 9(25/4)(20/3) = 375. So: 250 − 375 + 8(8000/(27k³)) = 875 ⟹ 64000/(27k³) = 1000 ⟹ k³ = 64/27 ⟹ k = 4/3
Independently re-derived with a CAS by symbolic implicit differentiation of the full curve equation (not assuming the given answer): the resulting dy/dx expression is identical to the given one term-for-term, and solving the system directly for x=5/2 gives k=4/3 as the only real root of the resulting cubic in k (the other two roots are complex, so not valid here) — matching the scheme's stated answer exactly.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| B1 | 8y³ → 24y²(dy/dx) | Differentiating that term |
| M1 | A correct product-rule attempt on −9kx²y | Differentiating that term |
| M1 | Collecting the two dy/dx terms, moving the rest across, factorising and dividing to isolate dy/dx | The "collect and divide" line |
| A1* | Reaching the given expression with no errors | Final line of (a) |
| M1 | Using dy/dx = 0 at x = 5/2 to produce one equation in k and y (allowing one substitution slip) | Setting the numerator to zero |
| A1 | A correct, fraction-cleared equation such as 15ky − 100 = 0 | That equation |
| M1 | A complete method: solving the two equations (the linear one above and the original curve equation) simultaneously for k | Substituting y=20/(3k) into the curve equation |
| A1 | k = 4/3 | Final answer |
Question 5 — Integration by trigonometric substitution (spec 6.2, 8 marks)
The question. (a) Using the substitution x = 2 sin u, show that ∫₀¹ (3x+2)/(4−x²)^(3/2) dx = ∫₀ᵖ [(3/2)sec u tan u + (1/2)sec²u] du, for a constant p to be found [4]. (b) Hence find the exact value of that integral [4]. (Calculator-only working is explicitly disallowed.)
Working.
(a) With x = 2 sin u, dx/du = 2 cos u. At the upper limit x = 1: 1 = 2 sin u ⟹ sin u = 1/2 ⟹ u = p = π/6.
Substitute into the integrand: 3x+2 = 6 sin u + 2, and 4 − x² = 4 − 4sin²u = 4cos²u, so (4−x²)^(3/2) = 8cos³u (cos u > 0 on [0, π/6], so no sign ambiguity). Then:
(3x+2)/(4−x²)^(3/2) · dx/du = [(6sin u + 2)/(8cos³u)] · 2cos u = (6 sin u + 2)/(4cos²u) = (3/2)(sin u/cos²u) + (1/2)(1/cos²u) = (3/2) sec u tan u + (1/2) sec²u — the given integrand, with p = π/6.
(b) Integrate: ∫[(3/2)sec u tan u + (1/2)sec²u] du = (3/2)sec u + (1/2)tan u + c (using d/du(sec u) = sec u tan u and d/du(tan u) = sec²u).
Evaluate between 0 and π/6. At u=π/6: sec(π/6) = 2/√3, tan(π/6) = 1/√3, so (3/2)(2/√3) + (1/2)(1/√3) = 3/√3 + 1/(2√3) = √3 + √3/6 = 7√3/6. At u=0: (3/2)(1) + (1/2)(0) = 3/2. So:
Integral = 7√3/6 − 3/2 = (7√3 − 9)/6 (≈ 0.521)
Independently re-derived with a CAS two ways — direct definite integration of the original x-integral, and separately integrating the u-form after algebraic simplification of the substituted integrand — and both give exactly (7√3−9)/6, matching each other and the mark scheme's own answer.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| B1 | p = π/6, from 1 = 2 sin u | Finding p |
| M1 | x = 2 sin u ⟹ dx/du = 2cos u (or an equivalent rearrangement) | Differentiating the substitution |
| M1 | A full substitution into the integral, using sin²u+cos²u=1 in the denominator | Simplifying (4−x²)^(3/2) |
| A1* | Reaching the given integrand, with du present and at least one intermediate simplified step shown | Final line of (a) |
| M1 | Integrating to the form …sec u + …tan u | First line of (b) |
| A1 | (3/2)sec u + (1/2)tan u (ignoring any constant) | The antiderivative |
| M1 | Substituting the found p and 0, and subtracting | Evaluating the limits |
| A1 | (7√3−9)/6 or an exact equivalent | Final answer |
Question 6 — Vectors: a line, a perpendicularity condition, a triangle's area (spec 7.4, 7.6, 7.7; 9 marks)
The question. Relative to a fixed origin O: A has position vector i−4j+3k, B has position vector 5i+3j−2k, C has position vector 3i+pj−k, for a constant p. The line l passes through A and B. (a) Find a vector equation for l [3]. Given that AC is perpendicular to l: (b) find p [3]. (c) Hence find the exact area of triangle ABC, as a surd in simplest form [3].
Working.
(a) The line's direction vector is AB = B − A = (5−1)i + (3−(−4))j + (−2−3)k = 4i + 7j − 5k. A vector equation for l: r = (i − 4j + 3k) + λ(4i + 7j − 5k)
(b) AC = C − A = (3−1)i + (p−(−4))j + (−1−3)k = 2i + (p+4)j − 4k. Perpendicularity to l means AB·AC = 0: 4(2) + 7(p+4) + (−5)(−4) = 0 ⟹ 8 + 7p + 28 + 20 = 0 ⟹ 7p + 56 = 0 ⟹ p = −8
(c) With p = −8: AC = 2i − 4j − 4k. Magnitudes: |AB| = √(4²+7²+5²) = √90 = 3√10; |AC| = √(2²+4²+4²) = √36 = 6. Since AC ⊥ AB (established in part (b)), triangle ABC has a right angle at A, so its area is simply: Area = (1/2)|AB||AC| = (1/2)(3√10)(6) = 9√10
Independently re-derived with a CAS two ways — the right-angle formula above, and separately the vector cross product |AB×AC|/2 — and both give exactly 9√10, confirming the value without relying on the perpendicularity shortcut alone.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| M1 | Finding AB (either subtraction order accepted, allowing 2 of 3 coordinates correct as evidence of a genuine attempt) | Computing AB |
| M1 | Writing an equation for the line in the form (a point) + λ(a direction) | Structure of the line equation |
| A1 | A fully correct vector equation, in the form r = … (a bare "l = …" does not earn this mark) | Final line of (a) |
| M1 | Finding AC (either subtraction order) | Computing AC |
| M1 | Taking the scalar product of AB and AC to form and solve a linear equation in p | The dot-product equation |
| A1 | p = −8 | Final answer to (b) |
| M1 | Finding the magnitude of either AB or AC, using the found p | Computing |AB| or |AC| |
| dM1 | A complete method for the exact area of the triangle (dependent on the previous method mark; the half-product-of-perpendicular-sides route is the most common, but a scalar-product/sine route or a cross product are equally valid) | The area calculation |
| A1 | 9√10 | Final answer to (c) |
Question 7 — Parametric differentiation, a tangent, and where it meets the curve again (spec 3.1, 5.1; 12 marks)
The question. A curve C has parametric equations x = sin t − 3cos²t, y = 3sin t + 2cos t, over a stated range of t. (a) Show that dy/dx = 3 at t = π [4]. The point P is where t = π. (b) Find the tangent to C at P in the form y = mx + c [3]. Given that this tangent meets C again at a point Q: (c) show that the value of t at Q satisfies 9cos²t + 2cos t − 7 = 0 [2]. (d) Hence find the exact y-coordinate of Q [3]. (Calculator-only working is explicitly disallowed.)
Working.
(a) dx/dt = cos t + 6 cos t sin t (equivalently cos t + 3sin2t); dy/dt = 3cos t − 2sin t. By the parametric chain rule: dy/dx = (dy/dt)/(dx/dt) = (3cos t − 2sin t)/(cos t + 6cos t sin t)
At t = π: cos π = −1, sin π = 0, so dy/dx = (3(−1) − 0)/((−1) + 0) = −3/−1 = 3 — the given result.
(b) At t = π: x = sin π − 3cos²π = 0 − 3(1) = −3; y = 3sin π + 2cos π = 0 − 2 = −2. So P = (−3, −2), and with gradient 3: y − (−2) = 3(x − (−3)) ⟹ y + 2 = 3x + 9 ⟹ y = 3x + 7.
(c) Substitute the parametric equations into y = 3x + 7: 3sin t + 2cos t = 3(sin t − 3cos²t) + 7. The sin t terms cancel on both sides (3sin t − 3sin t = 0), leaving: 2cos t = −9cos²t + 7 ⟹ 9cos²t + 2cos t − 7 = 0 — the given result.
(d) Solve the quadratic in cos t: 9cos²t + 2cos t − 7 = 0 factorises as (9cos t − 7)(cos t + 1) = 0, giving cos t = 7/9 or cos t = −1. cos t = −1 is t = π, i.e. point P itself — not a new intersection — so the point Q corresponds to cos t = 7/9.
Then sin t = √(1 − (7/9)²) = √(32/81) = (4√2)/9 (the positive root, matching the branch the real scheme itself credits). So: y = 3sin t + 2cos t = 3(4√2/9) + 2(7/9) = 12√2/9 + 14/9 = 4√2/3 + 14/9
Independently re-derived with a CAS from the original parametric equations directly (differentiating x(t), y(t) symbolically, substituting t=π, and separately solving the full simultaneous system y=3x+7 against the parametric curve): every intermediate value (dy/dx=3, P=(−3,−2), the tangent equation, the quadratic in cos t, and y_Q = 14/9 + 4√2/3) matches independently, without assuming any of the given "show that" results along the way.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| B1 | dx/dt = cos t + 6cos t sin t (or the double-angle equivalent) | Differentiating x |
| B1 | dy/dt = 3cos t − 2sin t | Differentiating y |
| M1 | Using the parametric chain rule with their own dx/dt and dy/dt, substituting t=π | Forming dy/dx |
| A1* | Reaching dy/dx = 3 with full working shown and no errors | Final line of (a) |
| B1 | x=−3, y=−2 (may appear anywhere in the working) | Finding P |
| M1 | Finding the tangent's equation using gradient 3 through P, proceeding as far as finding c | Point-gradient step |
| A1 | y = 3x + 7 | Final answer to (b) |
| M1 | A full attempt to solve the tangent and the parametric equations simultaneously | Substituting into y=3x+7 |
| A1* | Reaching 9cos²t + 2cos t − 7 = 0 with no errors | Final line of (c) |
| B1 | cos t = 7/9 seen or implied (may appear in working for (c)) | Solving the quadratic |
| M1 | Attempting to evaluate the y-coordinate using the found cos t (a purely numerical final answer with no working shown scores this mark only via an "answer which rounds to" fallback, not full credit) | Substituting back for y |
| A1 | 4√2/3 + 14/9 or an exact equivalent (additional spurious answers are penalised) | Final answer to (d) |
Question 8 — Volume of revolution, then integration by parts (spec 6.1, 6.2; 10 marks)
The question. The curve y = 10x·e^(−x/2), for 0 ≤ x ≤ 10, bounds a finite region rotated 2π radians about the x-axis. (a) Show the resulting volume V = k∫₀¹⁰x²e⁻ˣdx for a constant k [2]. (b) Find ∫x²e⁻ˣdx [3]. Two such solids are joined into a 20 cm exercise weight of mass 5 kg. (c) Using density = mass/volume, find the density in g/cm³ to 3 s.f. [5]. (Calculator-only working is explicitly disallowed.)
Cross-reference note. Part (b)'s integral, ∫x²e⁻ˣdx by repeated integration by parts, is already worked in full detail elsewhere in this paper's technique-hub — codex-deep-dive-integration-by-parts-and-substitution.md §2.3 derives it against the real M1/dM1/A1 scheme lines and also shows the credited DI/tabular alternative. To avoid restating that derivation, this section gives the result and its mark linkage compactly and works parts (a) and (c) — which are not covered anywhere else in this course's material — in full.
Working.
(a) V = π∫₀¹⁰y²dx = π∫₀¹⁰[10xe^(−x/2)]²dx = π∫₀¹⁰100x²e⁻ˣdx (since [e^(−x/2)]² = e⁻ˣ) = 100π∫₀¹⁰x²e⁻ˣdx, so k = 100π.
(b) By parts twice (u=x², dv/dx=e⁻ˣ, then u=x, dv/dx=e⁻ˣ): ∫x²e⁻ˣdx = −e⁻ˣ(x²+2x+2) + c
(Independently checked by differentiating this antiderivative: d/dx[−e⁻ˣ(x²+2x+2)] = e⁻ˣ(x²+2x+2) − e⁻ˣ(2x+2) = e⁻ˣx², recovering the original integrand exactly.)
(c) Two solids are joined, so total volume = 2 × 100π∫₀¹⁰x²e⁻ˣdx. Using (b)'s antiderivative between 0 and 10: [−e⁻ˣ(x²+2x+2)]₀¹⁰ = [−e⁻¹⁰(122)] − [−e⁰(2)] = 2 − 122e⁻¹⁰ ≈ 1.9945
So total volume = 200π(2 − 122e⁻¹⁰) ≈ 1253.16 cm³ (the "cm" is consistent throughout this part, since Figure 3's stated 20 cm total length matches two 10-unit solids end to end — the same x-range as the curve in Figure 2). With mass 5 kg = 5000 g: Density = 5000 / [200π(2 − 122e⁻¹⁰)] ≈ 3.99 g/cm³ (3 s.f.)
Independently re-derived with a CAS: y² integrated symbolically over [0,10] gives exactly 200π − 12200πe⁻¹⁰ (matching 100π× the by-parts result doubled), and 5000 divided by that numeric value gives 3.9899…, rounding to 3.99 — matching the scheme's own stated answer.
Mark ledger:
| Mark | Earned for | Where above |
|---|---|---|
| M1 | Forming a correct, unsimplified expression for V = π∫y²dx | Setting up (a) |
| A1 | Simplifying to 100π∫x²e⁻ˣdx (a missing dx is condoned; the limits must be present) | Final line of (a) |
| M1 | (covered in the linked deep-dive) Integration by parts once, in the right direction | — |
| dM1 | (covered in the linked deep-dive) A second application of parts, dependent on the first | — |
| A1 | −e⁻ˣ(x²+2x+2) + c | Result of (b), used directly below |
| M1 | A correct strategy for the total volume using their k (doubling for the two joined solids) | Start of (c) |
| M1 | Substituting the limits 10 and 0 into the part (b) antiderivative and subtracting | Evaluating the definite integral |
| A1 | 2 − 122e⁻¹⁰, or awrt 1.99 | That evaluation |
| dM1 | Attempting density = 5000/(their volume), dependent on the previous method mark; 5000 must be the numerator | Forming the density expression |
| A1 | awrt 3.99 g/cm³ | Final answer |
Question 9 — Proof by contradiction (spec 1.1, 4 marks)
The question. Use proof by contradiction to show that, when n is an integer, n² − 2 is never divisible by 4.
Cross-reference note. This exact question is already the worked marked-solution in content/veridian/curriculum/wma14/proof-by-contradiction.ts, built against the real scheme's own "Alt 3" route (the case-split on n odd/n even). To be genuinely additive rather than repeat that lesson, this section works the scheme's own default route instead — a structurally different argument, credited on exactly equal footing by the same real scheme.
Working (default route). Assume, for contradiction, that there exists an integer n for which n² − 2 is divisible by 4, so n² − 2 = 4k for some integer k. Then: n² = 4k + 2 = 2(2k+1)
so n² is even, which means n itself is even (if n were odd, n² would be odd). Write n = 2m for some integer m, so n² = 4m² — a multiple of 4. But then: n² − 2 = 4m² − 2 = 2(2m² − 1)
and since 2m² − 1 is odd, this is 2 × (an odd number) — which is not a multiple of 4 (a multiple of 4 needs a factor of 2 left over after removing one factor of 2, and an odd number supplies none). This contradicts the original assumption that n² − 2 = 4k (a multiple of 4). Since the assumption leads to a contradiction, it must be false: n² − 2 is never divisible by 4, for any integer n.
This mechanism — assume divisibility, force n even, then show the resulting n²−2 can only be 2×(odd), never 4×(integer) — is the pattern this course's own facts bank documents as the paper's default scored route for this exact question (WMA14-verified-facts.md §5.1); the case-split ("Alt 3") the curriculum lesson teaches in depth is a different, independently-credited route to the same conclusion, not a rewording of this one.
Mark ledger (the scheme explicitly allows several proof routes on equal footing; this ledger follows the default route above):
| Mark | Earned for | Where above |
|---|---|---|
| M1 | Setting up an algebraic statement in an integer variable that engages with divisibility by 4 and can lead to a contradiction — an explicit "assume…" sentence is not required at this stage, just a suitable equation | n² − 2 = 4k |
| A1 | A correct statement from which a contradiction can be drawn | n² = 4k+2 = 2(2k+1), hence n even |
| dM1 | A complete argument reaching a full contradiction (minor omissions are tolerated if the overall argument is clear) | n even ⟹ n²=4m² ⟹ n²−2=2(2m²−1), not a multiple of 4 |
| A1* | Drawing the contradiction and concluding the statement holds for all n — the scheme is explicitly generous about how formally the opening assumption is worded, provided the working is correct | Final concluding sentence |
Provenance and verification trail
Every numeric and algebraic result in this document was derived twice: once by hand, following the real mark scheme's own method, and once independently with a computer-algebra system working from the bare question statement (not from the scheme's stated answer) — so no claim above rests on trusting either the scheme's printed answer or this document's own arithmetic in isolation. Every mark-scheme criterion quoted or paraphrased above was read directly from wma14-01-rms-20220818.pdf this session; direct quotations are kept under fifteen words and attributed to the mark scheme. No large block of the question paper's or mark scheme's own prose is reproduced anywhere above — question stems are paraphrased into this document's own words, with the mathematical content (equations, numbers, variable names) preserved exactly, since that content is what a worked solution requires and is not the kind of prose the copyright caution is aimed at.
What this document does not claim. No examiner's report was ever published for the June 2022 WMA14 series, so nothing here is, or could be, a claim about how real candidates actually performed on this paper — every mark-scheme quote above is a statement of what the scheme rewards, not of what candidates did. No grade boundary is stated or implied anywhere in this document, and no likelihood or percentage is invented; the one frequency-style claim made (that the paper's default Q9 route and the curriculum lesson's Alt 3 route are two of several the real scheme credits equally) is read directly from the scheme's own text, not estimated.
Sources used
wma14-01-que-20220602.pdf (question paper, fetched via pmt.physicsandmathstutor.com, this session). wma14-01-rms-20220818.pdf (mark scheme, fetched via an examsolutions.net mirror, this session). research/veridian/WMA14-verified-facts.md §2, §5.1, §6 (series-selection context and the existing partial citations for this series' Q8(b) and Q9). content/veridian/curriculum/wma14/proof-by-contradiction.ts (read in full to confirm which route it already teaches for Q9, so this document's own Q9 section uses a different one). content/veridian/technique-hub/wma14/codex-deep-dive-integration-by-parts-and-substitution.md §2.3 (read in full to confirm Q8(b)'s integral is already worked there, so this document cross-references rather than repeats it). content/veridian/question-bank/wma14.ts (checked directly — holds no item from this series).
Not affiliated with or endorsed by Pearson Edexcel. All quoted mark-scheme wording is reproduced only to the minimum extent needed to justify a mark, per this course's standing no-large-verbatim-reproduction discipline.