Technique Masterclasses

Pure Mathematics 2

Deep Dive: Stationary Points and Curve Sketching

Provenance (2026-09-29): written after reading research/veridian/WMA12-verified-facts.md in full (§1–§9, including both audit passes on this exact lesson at §6.8 and the 2026-09-25 gap-closing entry), content/veridian/curriculum/wma12/stationary-points-and-curve-sketching.ts in full (1,200 lines — every block, not just the header), the three existing WMA12 technique-hub documents (codex-formula-and-method-bank.md, forge-exam-technique-guide.md, signal-frequency-and-boundaries.md), and a keyword sweep of content/veridian/question-bank/wma12.ts (one item present — a discriminant/no-turning-points question already matching the lesson's own Oct 2019 Q10 anchor structure; confirmed by grep, not duplicated below). Because the facts bank itself flags differentiation as having "thinner coverage" than most other WMA12 topics (§8), two primary sources it does not yet cite were read directly for this document: wma12-01-que-20250114.pdf / January_2025_MS.pdf / wma12-01-pef-20250306.pdf (Q3), and wma12-01-que-20250514.pdf / the June 2025 main-series mark scheme and Principal Examiner Feedback (Q9) — both under ~/Documents/SOL/Papers/IAL_Paper2_Complete/IAL_Materials/Mathematics/, both extracting cleanly under pdftotext -layout. All arithmetic below (the two real anchors' own numbers, and every original micro-example) was independently checked with sympy before being written in; none of it reproduces a real exam question's own wording beyond attributed fragments under 15 words.

CODEX Deep-Dive: Stationary Points and Curve Sketching (WMA12, spec 7.1)

Scope, held deliberately

This covers exactly spec item 7.1 — "applications of differentiation to maxima and minima and stationary points, increasing and decreasing functions," extended explicitly to "applications to curve sketching," with the spec's own note that "maxima and minima problems may be set in the context of a practical problem." It is the only dotted sub-point spec area 7 has at P2, so this document, like the lesson file, is the whole of P2's Differentiation content in one place.

What already exists and is not repeated here. codex-formula-and-method-bank.md already states the standing method-mark conventions for differentiation and for solving the resulting 3-term quadratic. forge-exam-technique-guide.md already names this lesson's eight trap-taxonomy items and its own real Jan 2021 Q2 anchor. signal-frequency-and-boundaries.md already gives this topic's confirmed frequency floor. The lesson file itself already fully derives why the second-derivative test works (from the gradient's own rate of change), why a stationary point is exactly where "increasing" can turn into "decreasing" (from the intermediate-value property of a polynomial derivative), and already carries full worked chains, a chain-drill, a method-comparison, two marked solutions (second-derivative substitution; discriminant-based no-turning-points proof) and seven MCQs. None of that mechanism or those worked exemplars is re-derived or reproduced below.

What this document adds. A one-place compact statement of the whole model, including one sub-case with essentially no coverage anywhere else in this course: practical optimisation problems solved by eliminating a variable via a given constraint — spec 7.1's own "practical problem" clause. The lesson file has zero worked treatment of this (confirmed by grepping it for "constraint," "capacity," "container," and "open-topped" — no matches); the CODEX formula bank gives it one paragraph of technique advice with no worked numbers; FORGE names one error from it in passing. This document independently verifies two real WMA12 anchors for it directly against primary PDFs (Part 2.5), builds a short original micro-example, and cross-checks a genuine mark-scheme shortcut neither existing document states. It also adds a short original extension of the lesson's own exact-answer trap to the surd case (2.2), and a synthesis table (Part 3) putting all five of this topic's confirmed real anchors side by side, which exists nowhere else in this course as a single table.


Part 1 — The real model, compactly

1.1 What a stationary point is, and the two-step process

A stationary point of y=f(x)y=f(x) is a point where dydx=0\frac{dy}{dx}=0 — a statement purely about the gradient, at one instant, with no reference to height, intercepts, or overall shape. Finding one is always: differentiate; set the result to zero; solve for xx; then substitute each xx back into the original y=f(x)y=f(x), never into dydx\frac{dy}{dx}, for the coordinate. Spec 7.1 carries no printed formula in the Mathematical Formulae and Statistical Tables booklet at all — confirmed directly against the booklet's own P1/P2 page — so every step of this is memorise-and-apply, not look-up.

1.2 Classifying it — the second-derivative test and its one blind spot

Differentiate again to d2ydx2\frac{d^2y}{dx^2} and substitute the same xx-value already found. Positive means minimum, negative means maximum. Exactly zero decides nothing — it is the test declining to answer, not a third category — and the fallback is to test the sign of dydx\frac{dy}{dx} itself just below and just above the point: a negative-to-positive change is a minimum, positive-to-negative a maximum, no change at all a (rarer) point of inflection. A real WMA12 mark scheme (Jan 2021 Q2) independently confirms this fallback earns full credit as genuine calculus, while a superficially similar method — comparing the curve's yy-values either side instead of its gradient — does not, because it never uses calculus at all. This distinction, and the mechanism behind both tests, is the lesson file's own core content and is not restated here.

1.3 Increasing, decreasing, and the two together

A function is increasing on an interval where dydx>0\frac{dy}{dx}>0 throughout and decreasing where it is negative throughout; a stationary point is exactly the boundary between the two. Because dydx\frac{dy}{dx} is itself a polynomial (continuous, no jumps), its sign can only change at a zero of it — so one test value inside each interval between consecutive stationary points fixes the sign for that whole interval.

1.4 The zero-solutions case: no turning points at all

Every stationary-point equation on this paper reduces, for a cubic, to a 3-term quadratic in xx (a quadratic has at most two roots, matching a cubic's at most two stationary points). That quadratic need not have real roots: when its discriminant B2−4AC<0B^2-4AC<0, dydx\frac{dy}{dx} is never zero for any real xx, so the curve has no stationary point anywhere and is monotonic — increasing throughout, or decreasing throughout. "Prove that CC has no turning points" is a real, self-contained WMA12 question type built on exactly this fact (confirmed real anchor: Oct 2019 Q10, already fully worked in the lesson file).

1.5 Curve sketching as assembly, not a new technique

Curve sketching adds nothing computationally beyond 1.1–1.4: place the yy-intercept (the constant term), determine end behaviour from the sign and parity of the leading term, and place each stationary point by exact coordinate with its nature already decided. The spec's own curve-sketching clause is one sentence appended to 7.1, not a separate numbered item, and — a genuine, honestly-flagged gap, not a coverage choice — this course's research has not located a real WMA12 examiner-report finding that treats "sketch the curve" as its own tested and separately mark-scored instruction, distinct from finding and classifying the stationary points that go into it. See Part 4.


Part 2 — Sub-cases, with short original micro-examples

2.1 Find-and-classify via the second-derivative test — pointer only

Fully covered by the lesson's worked-chain, chain-drill, method-comparison and first marked solution, anchored to a real WMA12 finding (Jan 2021 Q2) that roughly half of candidates who correctly found the second derivative never substituted the already-found xx-value into it. Nothing to add.

2.2 The exact-answer requirement, extended to a surd root

The lesson's own audit added a fractional-root example (3x2−2x−8=(3x+4)(x−2)=0⇒x=−43, 23x^2-2x-8=(3x+4)(x-2)=0 \Rightarrow x=-\tfrac43,\,2) to warn that a stationary point's xx-coordinate is not always a monic integer. That fix is entirely about factorisable quadratics. The same exact-answer rule — verified verbatim in the facts bank's general marking guidance, "marks will normally be lost if the candidate resorts to using rounded decimals" — applies identically when the quadratic from dydx=0\frac{dy}{dx}=0 does not factorise over the integers at all and the quadratic formula is the only route in:

dydx=x2−4x+1=0⇒x=4±122=2±3\frac{dy}{dx}=x^2-4x+1=0 \Rightarrow x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt3.

Both roots are genuine irrational numbers, produced by a correctly-applied formula, and both must be kept as 2+32+\sqrt3 and 2−32-\sqrt3 — never 3.733.73 and 0.270.27, however precisely rounded. This is a distinct manifestation of the same rule the lesson already teaches (fraction case), extended here to the surd case the lesson does not illustrate.

2.3 "Prove no turning points" via the discriminant — pointer, plus one generalisation

Fully worked in the lesson's second marked solution, anchored to Oct 2019 Q10. One generalisation worth stating explicitly, because the real mark scheme's own asymmetry note supports it directly: for any cubic y=ax3+bx2+cx+dy=ax^3+bx^2+cx+d, differentiating eliminates dd completely, so a "prove no turning points" mark built on the discriminant of dydx=0\frac{dy}{dx}=0 can survive an earlier error in a constant term found only from a point-on-the-curve substitution (which needs dd) but never survives an error in whichever coefficient among a,b,ca,b,c was found from a gradient condition (which needs the derivative, and so needs that coefficient correctly). This is a direct consequence of how differentiation acts on a cubic, not itself independently verified beyond the one real anchor the lesson already cites — flagged as a generalisation, not a second confirmed instance.

2.4 Decreasing-function regions from a factorised derivative — pointer only

Fully covered by the lesson's trap-taxonomy, anchored to a real WMA12 finding (Jun 2025, WMA12/01A, Q8) of a documented wrong final answer that chose the region outside two critical values instead of between them. Nothing to add.

2.5 Real-world optimisation via constraint elimination — the genuinely new sub-case

The mechanism. Spec 7.1's own practical-problem clause is realised, in both real anchors independently verified for this document, by the same three-stage pattern: (i) a given constraint (a fixed capacity/volume) lets one variable be written in terms of the other; (ii) that substitution turns a two-variable quantity to be optimised (a surface area) into a single-variable function, always of the shape px2+qxpx^2+\frac{q}{x} for positive constants p,qp,q on this evidence; (iii) the ordinary stationary-point procedure applies to that single-variable function — except that setting its derivative to zero now produces x3=kx^3=k for some constant kk, a cube root, not the quadratic this topic's other sub-cases produce.

Anchor 1 — Jan 2025, Q3 (10 marks). An open-topped metal container, base 3x3x by xx, height yy, capacity 120 m3120\text{ m}^3. Eliminating y=40x2y=\frac{40}{x^2} from the volume constraint and substituting into the surface area gives A=3x2+320xA=3x^2+\frac{320}{x} — independently re-derived and confirmed against the real mark scheme. dAdx=6x−320x2=0⇒x3=1603⇒x=awrt 3.76\frac{dA}{dx}=6x-\frac{320}{x^2}=0 \Rightarrow x^3=\frac{160}{3} \Rightarrow x=\text{awrt }3.76 (the question explicitly asks for xx to 3 s.f., not an exact form — see 2.6). d2Adx2=6+640x3\frac{d^2A}{dx^2}=6+\frac{640}{x^3}, confirmed awrt 18>0\text{awrt }18>0, minimum.

Anchor 2 — June 2025 (main series), Q9 (11 marks). An open container with two congruent right-angled triangular ends (legs 3x3x and 2x2x) and rectangular sides of length LL, capacity 12 m312\text{ m}^3. Eliminating L=4x2L=\frac{4}{x^2} gives S=6x2+20xS=6x^2+\frac{20}{x} — independently re-derived and confirmed against the real mark scheme. dSdx=12x−20x2=0⇒x3=53⇒x=1.1856...\frac{dS}{dx}=12x-\frac{20}{x^2}=0 \Rightarrow x^3=\frac{5}{3} \Rightarrow x=1.1856..., and — a genuine difference from Anchor 1 worth flagging on its own — this question asks for the minimum value of SS to 1 d.p., not for xx: S=awrt 25.3 m2S=\text{awrt }25.3\text{ m}^2. d2Sdx2=12+40x3\frac{d^2S}{dx^2}=12+\frac{40}{x^3}, confirmed awrt 36>0\text{awrt }36>0, minimum.

A confirmed mark-scheme shortcut neither existing technique-hub document states. The real June 2025 Q9 mark scheme credits classifying the stationary point without substituting the specific (irrational) xx-value at all: since d2Sdx2=12+40x3\frac{d^2S}{dx^2}=12+\frac{40}{x^3} is a sum of two terms both positive whenever x>0x>0, the whole expression is positive for every positive xx, so it must be positive at the one found here — full credit for that deduction alone, stated in the scheme itself as an accepted alternative to computing the numeric value. This generalises immediately: whenever the optimised expression has the shape px2+qxpx^2+\frac{q}{x} with p,q>0p,q>0, its second derivative 2p+2qx32p+\frac{2q}{x^3} is manifestly positive for all x>0x>0, so any stationary point found on the physically-meaningful domain x>0x>0 is automatically a minimum — a fact worth teaching as a genuine time-saver, confirmed real, not merely algebraically true in general.

A confirmed cross-series error — the cube-root trap, 2 of 2 series checked. Both real examiner reports name the identical slip independently: "Taking a square root rather than a cube root when solving" (Jan 2025 Q3, on x3=1603x^3=\frac{160}{3}) and "Square rooting instead of cube rooting" (June 2025 Q9, on x3=53x^3=\frac{5}{3}). This is a distinct, later-stage error from the exact-answer/rounding trap in 2.2 — it is a root-extraction mistake, not a rounding one — and it is specific to this sub-case precisely because it is the only WMA12 stationary-point pattern confirmed to produce a cubic equation in xx rather than a quadratic.

Confirmed setup-stage errors, each specific to the 3D-net reasoning this sub-case requires and not shared with any quadratic-based stationary-point question. June 2025 Q9's report records candidates who "treat the prism as closed" despite the figure and text stating it was open (adding an area that should not exist), who "took both rectangles to have an area of 2xL" (double-counting one dimension), and who did "not divide by 2 when finding the triangular cross section" in the volume step — a geometry error feeding a wrong constant into the eventual S(x)S(x), not a calculus error at all. Jan 2025 Q3's report separately records slips producing the wrong constants in the A=Px2+QxA=Px^2+\frac{Q}{x} form itself, and — a distinct completeness trap in part (c) — marks lost for omitting the correct d2Adx2\frac{d^2A}{dx^2} notation or the explicit stated inequality when justifying the minimum, a "state the reason, not just the number" requirement structurally identical to the trapezium-rule reasoning mark documented elsewhere in this course's research.

A confirmed, rarely-used but valid alternative classification method. June 2025 Q9's report also notes that "a small minority of candidates considered the change of sign of" dSdx\frac{dS}{dx} either side of the stationary xx-value instead of the second derivative — the identical fallback test from 1.2/§1.2 of the lesson, confirmed valid here too, in a context where it is genuinely less convenient than the "sum of positives" shortcut above.

A short original micro-example, distinct numbers throughout. An open-topped box has a square base of side xx metres and height yy metres, and holds 54 m354\text{ m}^3. Eliminating y=54x2y=\frac{54}{x^2}: total metal used is A=x2+4xy=x2+216xA=x^2+4xy=x^2+\frac{216}{x}. dAdx=2x−216x2=0⇒x3=108⇒x=1083=343 (≈4.76)\frac{dA}{dx}=2x-\frac{216}{x^2}=0 \Rightarrow x^3=108 \Rightarrow x=\sqrt[3]{108}=3\sqrt[3]{4}\ (\approx4.76) — kept exact here deliberately, to show the surd form the two real anchors' own "give your answer to nn s.f./d.p." instruction happens to make unnecessary (2.6). d2Adx2=2+432x3\frac{d^2A}{dx^2}=2+\frac{432}{x^3}, a sum of two positive terms for any x>0x>0: minimum, by the same shortcut Anchor 2's own mark scheme credits, without substituting x=343x=3\sqrt[3]4 at all.

Honestly flagged limit of this evidence. This sub-case is anchored to exactly 2 of the 2 series in which this course's research has so far located a real-world optimisation question for WMA12 differentiation (January 2025, June 2025) — a floor, not a rate: no earlier series (June 2019 through October 2024) has such a question confirmed anywhere in the facts bank or the existing technique-hub documents, and the six series the facts bank's own §8 names as "grepped but not read in full depth" (Oct 2019, Oct 2021, Oct 2024, Jun 2023, Jun 2024, Jan 2025 — the last now closed by this document) have not all been individually re-checked for this specific sub-pattern. It is entirely possible an earlier instance exists and has simply not been found yet; equally possible this is a genuinely recent addition to how this paper is examined.

2.6 When the answer should be exact, and when it should be rounded — a synthesis point

Read together, 2.2 and 2.5 show this topic's exact-answer rule is not blanket: the polynomial sub-cases (2.1–2.4) consistently require an exact fraction or surd for a stationary point's own coordinate, with no rounding accepted at any precision — but both real optimisation anchors in 2.5 explicitly instruct candidates to round (xx to 3 s.f. in Jan 2025 Q3; the optimised value SS itself to 1 d.p. in June 2025 Q9), because 160/33\sqrt[3]{160/3} and 5/33\sqrt[3]{5/3} have no exact form worth demanding. The real, checkable skill is reading which instruction the specific question gives, not applying "always exact" or "always round" as a fixed rule for this topic — and, separately, noticing precisely which quantity a rounding instruction attaches to (xx itself, versus the optimised quantity built from it), since the two real anchors here test each version.


Part 3 — Synthesis: every real exam angle this topic has actually been tested from

Five confirmed real WMA12 anchors exist for spec 7.1 in this course's research to date (of 20 series with question paper + mark scheme reviewed per the facts bank's §5; this topic independently confirmed present in at least these series — a floor, not a full per-question tag of all 20, matching every other topic's frequency count in signal-frequency-and-boundaries.md):

Series & QSetupMarksReal mark-scheme skeleton (verified)
Jan 2021, Q2y=x3−x2−16x+2y=x^3-x^2-16x+2; find and classify both stationary points7(a) M1 A1 M1 (independent) A1; (b) M1 dM1 (substitute one value, dependent) A1
Oct 2019, Q10y=ax3−3x2+3x+by=ax^3-3x^2+3x+b, point + gradient given; (a) find a,ba,b; (b) prove no turning points7(a) M1 M1 A1 A1; (b) M1 dM1 (attempt B2−4ACB^2-4AC) A1*
Jun 2025 (01A), Q8f′(x)=2(3x−2)(x+5)f'(x)=2(3x-2)(x+5); deduce the decreasing region2 (part a)M1 A1, either critical value alone sufficient with consistent inequality use
Jan 2025, Q3Open container, base 3x×x3x\times x, capacity 120 m3120\text{ m}^3; form A(x)A(x), find stationary xx, justify minimum10(a) M1 A1 dM1 A1; (b) M1 A1ft dM1 A1; (c) M1 A1*
Jun 2025 (main), Q9Open triangular-prism container, capacity 12 m312\text{ m}^3; form S(x)S(x), find minimum SS, justify11(a) M1 A1 dM1 A1; (b) M1 A1ft dM1 A1; (c) M1 A1*

What this table shows that no single existing document states in one place. Every one of the five anchors splits "form/differentiate correctly" from "the dependent step that actually uses the specific value found" into separate marks (an independent M1 for solving the equation, a dependent dM1 or A1* for what happens to that solution next) — the single structural feature this entire topic's real mark schemes share, confirmed across three genuinely different question shapes (pure algebra; a "prove" question; a practical-context optimisation). And the two 2025 anchors (Part 2.5) are the only ones in this archive where the stationary-point equation is a cube, not a quadratic — a real, confirmed structural variant this topic's spec wording ("may be set in the context of a practical problem") predicts but which neither the lesson file nor the other technique-hub documents had, until this document, actually worked through with real numbers.


Part 4 — What remains genuinely unverified