New separator ideas

Descriptions of questions that would plausibly separate a cohort, aimed at the weaknesses the examiner reports document year after year.

Compiled 15 September 2026. This document contains no questions — no rules to substitute into, no numbers, no worked answers. Each entry describes a design: what would be given, what would be asked, in what order, which step the middle of the cohort would fail, and what archive evidence supports the prediction.

Companion to 01-study-design.md, 02-functions-graphs.md, 03-algebra-number.md, 04-calculus.md, 05-probability-statistics.md and 07-exam-craft.md. Every percentage is the published pct from corpus/mm/questions.json — the percentage of the state awarded full marks on that part. None is estimated.

Every design is in scope for the 2023–2027 study design. The two retired constructions — matrix representation of transformations (last live 2021, deleted from [SD] 2023) and functional relations such as f(x + y) = … (last live 2017, deleted 2023) — are excluded, and where an archive comparison sits inside a retired construction it is flagged.

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Contents

1. What the evidence says a separator needs
2. The designs, by area of study - 2.1 Functions, relations and graphs — twelve designs - 2.2 Algebra, number and structure — twelve designs - 2.3 Calculus — twelve designs - 2.4 Data analysis, probability and statistics — twelve designs
3. Designs that exploit the 2023–2027 study design specifically
4. Designs that cross areas of study
5. How to use these

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1. What the evidence says a separator needs

1.1 The base rates a design has to beat

A separator in this archive is a part with pct ≤ 50. That threshold is not demanding: 779 of the 1 504 November parts with published statistics are separators. Area by area:

So "under 50%" is the median question, not a hard one. A design intended to separate a strong cohort has to aim lower, and the archive says exactly where the floor is.

Difficulty scales with mark value, almost perfectly. From 04-calculus.md §1.4, across the 441 Calculus parts with statistics:

A three-mark Calculus part is a separator 86% of the time. If you want a separator and you have three marks to spend, you barely have to try. The interesting question is not how to make a part hard but how to make it hard and informative.

Position does most of the work. Across the 146 multi-part questions in the corpus with published statistics (07-exam-craft.md §4.4):

Median first part 67%, median last part 16%. A Section B question loses roughly two-thirds of the cohort between its opening and its close. On Exam 1 the same gradient runs across the paper: Q1 averages 63.5%, Q8–Q9 average 32%.

1.2 The constructions with the lowest full-mark rates

Reading the four area documents together, seven constructions account for nearly every part below 10%.

One more construction deserves separate mention because it is perfect, not merely low: every archive instance of "domain, range or existence of a composite function" is a separator — 2016 Exam 1 Q5aii 15%, 2020 Exam 2 Section A Q20 18%, 2017 Exam 1 Q7bii 20%, 2017 Exam 1 Q7bi 29%, 2017 Exam 1 Q7c 30%, 2023 Exam 2 Section A Q20 30%. Six for six, across nine years, in both papers and both sections.

1.3 The failure mechanisms the reports name most often

From 07-exam-craft.md §6, counted across every report in the corpus:

Two structural confusions sit alongside those and are worth naming separately because they are content failures, not presentation failures:

• Average value versus average rate of change. Named in the reports for 31 Calculus parts. [RPT] 2024 Exam 2: "Some students found the average value when the average rate of change was required." [RPT] 2015 Exam 2 says the same thing nine years earlier. VCAA has deliberately paired them — 2016 Exam 1 Q6a (32%) and Q6b (16%) are the same function over the same interval — and the state still does not separate them.
• The function versus its gradient function. 2019 Exam 2 Section B Q2b asked for the set where the gradient of the hill is strictly decreasing and scored 3%, with a mark distribution of {0: 97%, 1: 3%}. [RPT]: "Most students interpreted the question as asking where the function modelling the hill was strictly decreasing."

1.4 When a question discriminates rather than merely defeats

This is the distinction a question writer actually has to manage, and the mark distributions in questions.json settle it. Compare two groups of low-scoring parts.

Parts that discriminate — the distribution is spread, and partial credit is really being earned:

Parts that defeat — almost everyone scores zero, and the part tells you nothing about the difference between a 35 and a 45 student:

The difference is not the topic. 2025 Exam 1 Q8a and 2016 Exam 2 Section B Q4fii are both hard; one grades the cohort into four bands and the other does not. Reading across the two tables, four conditions separate the discriminating design from the defeating one.

1. Marks above one. Every defeating part above is worth one or two marks; every discriminating part is worth three or four. A 1-mark part cannot express a partial insight, so a 1-mark part with a single non-obvious idea produces a binary outcome. 03-algebra-number.md §4.15 makes the same point from the other direction: in the sub-25% algebra group the dominant failure is non-attempt, and the zero columns confirm it — 94%, 90%, 89%, 87%, 83%, 81%, 80%.

2. A first step every competent student can take. 2025 Exam 1 Q8a gives an explicit density and an explicit probability target, so a student who can antidifferentiate earns a mark before the quadratic arrives. 2016 Exam 2 Section B Q4fii — "Show that 0 < A(k) < 2 for all k > 1" — offers no first step at all, and 94% wrote nothing worth a mark.

3. The difficulty is a decision, not a discovery. The discriminating parts ask students to choose: which tail, which bracket, which of the stationary point and the endpoint, exact or approximate, upper minus lower or the reverse. The defeating parts ask them to invent a bounding argument or spot a symmetry with no prompt.

4. The staircase resets. 07-exam-craft.md §4.4 records the resets inside real questions: 2025 Exam 2 Section B Q4 runs 93 → 78 → 15 → 70 → 65 → 28 → 49 → 23 → 10 → 68 → 29 → 14, and 2024 Exam 2 Section B Q4 runs 79 → 85 → 71 → 52 → 36 → 11 → 33 → 18 → 56 → 8. A "show that" or a fresh sub-scenario re-supplies the input so that a collapse in one part does not condemn the next. A question with no reset converts one hard idea into six dead parts.

Every design below is built to the discriminating pattern: two to four marks where the mark scheme can express a partial route, a first step that is genuinely available, and a difficulty that is a decision rather than an invention. Where a design is inherently a one-mark decision, that is said, and the predicted band reflects it.

1.5 A note on what "predicted full-mark band" means below

Each design carries a predicted percentage band for the proportion of a full state cohort that would earn full marks, resting on a named archive comparison. These are predictions by analogy, not measurements. They assume a November-standard cohort and the placement stated. A school cohort sitting a trial exam will typically score below the state figure on a novel item and above it on a rehearsed one.

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2. The designs, by area of study

2.1 Functions, relations and graphs — twelve designs

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F1. Both orders, one map

The design. Print two graphs, one obtained from the other by a horizontal dilation, a horizontal translation and a vertical translation. Part (a) asks for a sequence of three transformations mapping the first to the second in which the dilation is applied before the horizontal translation; part (b) asks for a sequence mapping the same graph to the same image in which the horizontal translation is applied first. Both parts are two marks, and both stems name the required order explicitly so that the student cannot dodge the issue by choosing a convenient order.

The discriminator. The magnitude of the horizontal translation is different in the two orders, and the middle of the cohort will write the same number twice. Students who have internalised the A f(n(x + b)) + c decomposition as a fixed recipe rather than as a composition will not notice that factoring n out of the bracket is what makes the translation b rather than nb. The vertical translation, which genuinely can go anywhere in the sequence, is the decoy: a student who correctly observes that it is order-free will often generalise the observation to the horizontal pair.

The evidence. 2024 Exam 2 Section B Q1dii asked for exactly this pair of valid orders and scored 5% (2 marks); the report says "The vertical translation could be completed at any stage in the sequence. The other transformations had to be in the correct order." 2024 Exam 1 Q5b, describing the dilation and translation reconciling two population models, scored 2%; "Many students were able to list one transformation, usually the dilation; however, frequently the incorrect axis or direction was specified." 2007 Exam 2 Section B Q3di scored 13%. The 2026 NHT Examination 2 already asks the two-order version as consecutive parts — "a dilation followed by a translation", then "a translation followed by a dilation" — which is VCAA signalling that it knows where the difficulty sits.

Placement and marks. Exam 1 as a standalone two-part question, or Exam 2 Section B as parts (d) and (e) of a graph question. 2 + 2 marks.

Predicted full-mark band (both parts). 8–18%, against 2024 Exam 2 Section B Q1dii at 5% for the harder half alone and 2024 Exam 2 Section B Q1di at 37% for the translation-only half. Naming the order in the stem removes the guessing and should lift the floor above 5%.

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F2. The inverse whose domain is used downstream

The design. Define a many-to-one function and restrict it to one side of its stationary point so that an inverse exists. Part (a) asks for the rule and domain of the inverse. Part (b) asks for something that is only answerable if the domain of part (a) is right — the range of a composite built from the inverse, or the coordinates where the inverse meets a named line, or the interval on which the inverse is defined after a further translation. Part (c) asks for the largest restriction of the original domain for which an inverse exists, phrased as "the largest value of a".

The discriminator. Part (a) fails at the branch choice and at the domain: the algebra produces a ±, and the sign is fixed by the range of the inverse, which is the domain of the original. Part (b) then punishes the omission a second time, which is the design's point — a domain left out in part (a) is usually invisible to the marker in a way that costs one mark, but here it propagates. Part (c) fails on inclusion: the endpoint is the stationary point and it belongs in the interval.

The evidence. 2023 Exam 1 Q7c (equation and domain of an inverse) — 21%; "The most common error was writing the function as the positive arm of the inverse." 2016 Exam 1 Q5bi — 16%; "few students took care to determine the range of the inverse function and select for the negative root." 2009 Exam 1 Q3 — 38%; "Few students realised that the inverse function required the rule and the domain to be specified." The 2007 Exam 2 report states the policy: "When the inverse function is asked for, the domain must be given. Students will be penalised in the future if the domain is left out." The largest-restriction part: 2025 Exam 1 Q5b — 43%; 2013 Exam 2 Section A Q7 — 37%.

Placement and marks. Exam 1, three parts, 2 + 2 + 1 = 5 marks.

Predicted full-mark band. Part (a) 20–30% against 2023 Exam 1 Q7c at 21%; part (b) 12–22% against 2016 Exam 1 Q5aii at 15%; part (c) 35–48% against 2025 Exam 1 Q5b at 43%. All five marks: 10–18%.

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F3. Composite existence, run backwards and forwards

The design. Give an outer function with a genuinely restricted domain — a square root or a logarithm — and an inner function with a parameter in its domain endpoint. Part (a) asks for the largest value of the parameter such that the range of the inner is a subset of the domain of the outer. Part (b) asks for the range of the composite at that parameter value. Part (c) reverses the composition and asks for the largest interval on which both composites exist.

The discriminator. Part (a) requires the existence condition read backwards: compute the range of the inner as a function of the parameter, set its boundary against the boundary of the outer's domain, solve, and then choose the root lying in the stated region. Part (b) requires ran(f ∘ g) = f(ran g), not ran f — a distinction the cohort does not hold. The middle of the cohort will write down the range of the outer function over its whole domain. Part (c) then requires the intersection of two existence conditions running in opposite directions.

The evidence. This is the only construction in the archive where every instance is a separator: 2016 Exam 1 Q5aii 15%, 2020 Exam 2 Section A Q20 18%, 2017 Exam 1 Q7bii 20%, 2017 Exam 1 Q7bi 29%, 2017 Exam 1 Q7c 30%, 2023 Exam 2 Section A Q20 30%. The 2017 report spells out the whole idea in the process of explaining why the state missed it: "Since [that interval] is the domain of g, the range of g is the same as the domain of f. Hence, in this case, the range of f(g(x)) is the same as the range of f." 2016 Exam 1 Q5aii's report: "students appeared to experience difficulty in determining the range of the composite function. A quick sketch over the given domain would have been helpful." The 2023 sample Exam 1 asks "Determine the maximal domain, D, such that g ∘ h exists" — the form VCAA has signposted for the current design.

Placement and marks. Exam 1, three parts, 2 + 1 + 1 = 4 marks. Section A can carry part (c) alone as a single item.

Predicted full-mark band. 15–25% for the four marks together, against the mean of the six archive instances (23.7%) and specifically against 2017 Exam 1 Q7bi at 29% with distribution {0: 53%, 1: 18%, 2: 29%} — a distribution that already shows the partial-credit spread this design wants.

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F4. Maximal domain with two constraints and a hole

The design. Define a function as a difference of two logarithms whose arguments are polynomial, chosen so that each logarithm supplies its own constraint, the constraints overlap only partially, and a point inside the overlap must be excluded because one argument vanishes there. Ask for the maximal domain. Follow with a one-mark part asking for the range over that domain.

The discriminator. Three failures compound. First, students apply the constraint to the simplified quotient rather than to each original logarithm, which silently re-admits an excluded interval. Second, the answer is a union and the archive says the state writes an intersection. Third, the excluded point turns the answer into a set difference, which requires either ∪ around a removed point or the \{ } notation — and 07-exam-craft.md ranks bracket and notation errors fourth and eighth among all named report faults.

The evidence. 2019 Exam 1 Q8b — maximal domain of a difference of logarithms, requiring the intersection of two constraints and an excluded zero — 9% for one mark. The follow-on range part, 2019 Exam 1 Q8c, is the lowest-scoring written part in the archive at 1%, and its report comment is one sentence: "Some students sketched various graphs with limited success." 2022 Exam 1 Q5b — 27%; "A common error was writing the interval as an intersection not a union." 2021 Exam 2 Section B Q3a — 36%.

Placement and marks. Exam 1, 2 + 1 = 3 marks. This must not be placed at Q8 or Q9 with no scaffold; the 1% precedent is what happens when it is.

Predicted full-mark band. 10–20% for the domain part against 2019 Exam 1 Q8b at 9% and 2022 Exam 1 Q5b at 27% — the difference between those two being whether the constraint is single or double. The range part should be written as "state the range" only if the preceding part has forced a sketch; otherwise expect the 1% outcome, which defeats rather than discriminates.

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F5. The family slider, with the bracket decided at the critical value

The design. Print a fixed graph with at least one turning point and one endpoint inside the visible window. Introduce a second graph depending on a single parameter that translates it horizontally. Ask, in consecutive one-mark parts, for the values of the parameter giving exactly one intersection, then exactly two, then none. The parameter values must be chosen so that at least one boundary is attained (closed bracket) and at least one is not (open bracket).

The discriminator. The cohort finds the critical values and then gets the inequality direction or the bracket wrong. That is not a guess — it is what the reports say happens, question after question. The design's second discriminator is the requirement to test the critical case itself rather than to test either side, because the answer at tangency differs between "exactly one" and "exactly two".

The evidence. 2014 Exam 2 Section B Q5ci — 7%; "(1, 3) and 1 < d < 3 were common incorrect answers" where the answer was the half-open interval. 2014 Exam 2 Section B Q5cii — 19%. 2018 Exam 2 Section B Q1hi — 18%; "Common incorrect answers were a ≤ 1, a < 0 or a > 0." 2019 Exam 2 Section B Q1biii — 35%; "Common incorrect answers were d = −1/e, d ≤ −1/e, d > 1/e" where the answer was the strict inequality in the other direction. 2012 Exam 2 Section A Q16 — 34%. This type produces 19 of the 233 separators in the area, more than any other except tangent geometry.

Placement and marks. Exam 2 Section B, three consecutive 1-mark parts late in a graph question. CAS sliders are explicitly endorsed for this work — the 2024 NHT report says "This value can also be found using the slider functionality on CAS" — so the design is honest on a technology-active paper.

Predicted full-mark band. 12–25% per part, against 2014 Exam 2 Section B Q5cii at 19% and 2019 Exam 2 Section B Q1biii at 35%. All three parts: 5–12%, against 2014 Exam 2 Section B Q5ci at 7%.

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F6. Self-symmetry of a circular function

The design. Give a transformed circular function on a restricted domain. Part (a), one mark, asks for the period. Part (b), two marks, asks for the smallest positive shift h such that shifting the graph by h maps it onto its own reflection in the horizontal axis — that is, such that the shifted function equals the negative of the original — and for the largest domain on which that statement holds.

The discriminator. Part (a) is free and is the reset. Part (b) is not a formula. It requires recognising that the shift mapping a sinusoid to its own negative is a half-period, then checking that the restricted domain survives the shift, which it generally does not over the whole of the printed interval. The second half of part (b) — the largest domain — is where the cohort collapses, because it requires reasoning about the intersection of the original domain with its own translate.

The evidence. 2025 Exam 2 Section B Q4c — find k and the largest a such that the shift condition holds on an interval — 15% (2 marks). 2021 Exam 2 Section B Q5c — smallest positive h with a reflection-symmetry condition — 21% (1 mark). 2023 Exam 2 Section A Q18 — the number of local minima of a circular function whose parameter appears in both the rule and the domain — 29%. The contrast that shows the design works: 2021 Exam 1 Q3b, "state the period of g", scored 89%.

Placement and marks. Exam 2 Section B, parts (a) and (b) of a circular-function modelling question, 1 + 2 = 3 marks.

Predicted full-mark band. Part (a) 80–90% against 2021 Exam 1 Q3b at 89%; part (b) 15–28% against 2025 Exam 2 Section B Q4c at 15% and 2021 Exam 2 Section B Q5c at 21%.

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F7. Continuous, smooth, then sketched

The design. Model a physical quantity with a three-branch piecewise function in which two parameters are unknown. Part (a), one mark, imposes continuity at the first join and asks for one parameter. Part (b), two marks, imposes continuity and smoothness at the second join and asks for the other. Part (c), two marks, asks for a sketch of the completed model over the full domain, showing the coordinates of every endpoint of every branch.

The discriminator. Part (b) is the first real separator: continuity gives one equation, smoothness gives a second, and the cohort supplies only the first. The sketch in part (c) is the second, and its failure mode is documented and specific — students draw cusps at joins that the smoothness condition has just made smooth, and they omit endpoint coordinates or mark closed endpoints with square brackets.

The evidence. 2023 Exam 2 Section A Q9 — a hybrid function continuous and smooth at the join, solve for the parameter — 42%. 2023 Exam 2 Section B Q2di — find two parameters making a piecewise wheel model continuous — 40%; 2023 Exam 2 Section B Q2dii, the general-solution follow-on, 12%; 2023 Exam 2 Section B Q2diii, the sketch showing endpoint coordinates, 24%. 2007 Exam 2 Section B Q2a — 44%; "Many students did not know to substitute t = 8 or t = 16 into the equation." 2007 Exam 2 Section B Q2b, the matching sketch — 37%; "The curve had to be continuous and should have been smooth at t = 8 and t = 16… Others drew cusps instead of turning points." 2022 Exam 1 Q7c — determine all four endpoints of two branch functions and confirm they match — 16%.

Placement and marks. Exam 2 Section B, 1 + 2 + 2 = 5 marks. The technology-free variant belongs on Exam 1 without the sketch.

Predicted full-mark band. Part (a) 40–55% against 2023 Exam 2 Section B Q2di at 40%; part (b) 25–40% against 2023 Exam 2 Section A Q9 at 42%, discounted because Section B demands working; part (c) 20–32% against 2023 Exam 2 Section B Q2diii at 24%.

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F8. The inequality you can read off your own sketch

The design. A single one-mark part placed immediately after a by-hand sketch, asking for the values of x satisfying an inequality involving the function just drawn. The function must have a vertical asymptote or a domain break inside the region, so that the answer is a union of two intervals rather than one.

The discriminator. Nothing mathematical. The design works because the cohort restarts algebraically instead of looking at the graph on the facing page, and having restarted, finds one boundary and reports a single interval. The asymptote is what converts a careless answer into a wrong one.

The evidence. Three archive instances, three separators, all worth one mark: 2023 Exam 1 Q3b 38% ("many students did not use their graph from part 3a. to assist them to correctly identify the interval required"); 2021 Exam 1 Q4b 32% ("Most students attempted to solve algebraically instead of using the graph, and only obtained the lower bound of inequality"); 2025 Exam 1 Q7dii 27%. It is the cheapest mark in the area and the state loses it every time it is offered.

Placement and marks. Exam 1, one mark, immediately following the sketch part.

Predicted full-mark band. 25–38%, against 2021 Exam 1 Q4b at 32%. This is a one-mark binary part by construction, so it discriminates only weakly — include it for the diagnostic value, not for the grading.

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F9. Recover the rule, dilation factor included

The design. Print the graph of a quartic with one repeated root (a touch), one simple root (a crossing) and one turning point labelled with its coordinates. Ask for the rule of the function. Follow with a one-mark part asking for the rule of a named transform of it — the reflection in the horizontal axis, or the graph shifted so that the touch point moves to the origin.

The discriminator. Reading multiplicity from shape is the easy half and most of the cohort manages it. The mark is lost on the dilation factor, which requires substituting the labelled turning point into the factored form. The second part then tests whether the recovered rule is being treated as an object or as a memorised shape.

The evidence. 2019 Exam 1 Q8a — 14% for one mark; "This question was well attempted but not done well, with many students overlooking the dilation factor." 2015 Exam 2 Section A Q3 — the multiple-choice version, 20%, and the report records that 61% of the state chose a single distractor differing only in the sign of one linear factor: "Most students chose option A… but the factor (x + b) is incorrect." That is the most lopsided distractor in the corpus.

Placement and marks. Exam 1, 1 + 1 = 2 marks, or Section A as a single item with the sign distractor built in.

Predicted full-mark band. 15–28%, against 2019 Exam 1 Q8a at 14% and 2015 Exam 2 Section A Q3 at 20%. Labelling the turning point rather than a general point lifts the floor; omitting the label drops it below 14%.

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F10. The range of a restricted transform, as a set

The design. Give a function whose graph has an interior stationary point, restrict it to a closed interval whose endpoints are not symmetric about that stationary point, apply a single transformation that changes the orientation — a reflection, or a negative dilation factor — and ask for the range. One mark. Follow with a one-mark part asking for the range of the absolute value of the same function on the same domain.

The discriminator. Three decisions, each of which the reports record the cohort failing: whether the endpoint values or the stationary value bound the range; which of the two is the maximum after the reflection; and whether each bound is attained, which fixes the bracket. The absolute-value follow-on adds a fourth — the new lower bound is zero if and only if the function crosses the axis inside the domain.

The evidence. 2024 Exam 1 Q7bii — state an interval from substituted endpoint values in correct bracket notation — 20%; the report records that "Some students, incorrectly, reversed the order of the interval." 2006 Exam 1 Q7b — range of an absolute value on a restricted domain — 35%; "Common incorrect responses were R, (0, 20), [20, 0], R⁺, and [0, 320]" — note [20, 0], reversed. 2007 Exam 2 Section A Q6 — 39%. 2025 Exam 1 Q3a, the easy comparator with no transformation, scored 81%, and its report gives the notation rule: "incorrectly stating the range of values in terms of y (as in −1 ≤ y ≤ 3) is not acceptable notation."

Placement and marks. Exam 1, 1 + 1 = 2 marks.

Predicted full-mark band. 20–32% per part, against 2024 Exam 1 Q7bii at 20% and 2006 Exam 1 Q7b at 35%.

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F11. The parametric tangent and the family it generates

The design. A six-part Section B question on a single curve. Part (a): "show that" the tangent at a general point x = a has a stated horizontal-axis intercept expressed in terms of a — the answer is printed, so the marks are for the derivation. Part (b): the values of a for which that intercept does not exist. Part (c): what the tangent line is at those values. Part (d): the values of a for which the tangent at a is parallel to the tangent at the intercept. Part (e): the range of intercept values as a varies over a stated interval.

The discriminator. The parameter itself. The mathematics is a tangent equation, which the state handles at 70–87% when the point is a number. Replacing the number by a letter halves the success rate immediately and then halves it again at each subsequent part, because every later part requires the symbolic expression from part (a) to be manipulated rather than evaluated. The design's virtue is that part (a) is a "show that" and therefore a reset — a student who cannot derive the intercept can still quote it and attempt parts (b) onward.

The evidence. 2020 Exam 2 Section B Q5 is this question, and its published staircase is the prediction: Q5a (show the parametric intercept) 49%, Q5b (values where it does not exist) 46%, Q5c (nature of the tangent there) 23%, Q5e (parallel tangents) 7%, Q5g 3%. Supporting: 2023 Exam 2 Section B Q3ci, the tangent at a general point, 52%, against 2025 Exam 2 Section B Q4d, the tangent at a named point, 70%. 2012 Exam 2 Section B Q2c, the "show that" form of a parametric tangent, 10%, with the report naming bracket collapse as the cause. 2014 Exam 2 Section B Q5fi — 22% — where the report prints the wrong version verbatim as a missing pair of brackets around the gradient.

Placement and marks. Exam 2 Section B, parts (c) through (h) of an 11–15 mark question. 2 + 1 + 1 + 1 + 1 = 6 marks.

Predicted full-mark band. A staircase from 45–55% on the "show that" to 8–15% on the last part, against 2020 Exam 2 Section B Q5a at 49% and Q5e at 7%.

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F12. The image point, without the rule

The design. Two Section A items. The first names a point on the graph of an unspecified function and asks for the coordinates of its image on a graph written in A f(n(x + b)) + c form with n ≠ 1 and A < 0. The second inverts it: names a point on the image and asks which point must lie on the original. Distractors are built from the three classic errors — applying the horizontal operations in the wrong order, applying them in the wrong direction, and applying the vertical dilation to the x-coordinate.

The discriminator. No rule is available, so the student must operate on the argument: solve n(x + b) = the old x-value rather than substituting. The second item then requires the inverse map, and 07-exam-craft.md §5.4 documents that "applied forwards instead of inverted" is the single most effective distractor family in Section A — 2013 Exam 2 Section A Q20 had 53% choose the forward-transformation distractor against a 25% key.

The evidence. 2024 Exam 2 Section A Q12 — 47%; the report's working is the method in one line: "The graph of h has been dilated by a factor of ½ from the y-axis and translated a unit left." 2025 Exam 2 Section A Q15 — 44%. 2018 Exam 2 Section A Q4 — 48%. 2024 Exam 2 Section B Q3aii, the written version — 26%; "Many students did not realise they only needed to translate the point… Others translated the local minimum."

Placement and marks. Exam 2 Section A, two items, 1 + 1 = 2 marks. Place the inverted one at Q16 or later; 07-exam-craft.md §5.2 records that the genuine Section A cliff is Q17–Q20.

Predicted full-mark band. Forward item 40–50% against 2018 Exam 2 Section A Q4 at 48%; inverted item 25–35% against 2024 Exam 2 Section B Q3aii at 26% and 2013 Exam 2 Section A Q20 at 25%.

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2.2 Algebra, number and structure — twelve designs

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A1. Two determinant roots, one justification

The design. A system of two linear equations in two unknowns in which a parameter appears in three of the four coefficients and in at least one constant term, arranged so that the determinant condition is a quadratic with two distinct roots, one giving no solution and the other giving infinitely many. Part (a), three marks, asks for the value giving no solution. Part (b), one mark, asks what happens at the other root.

The discriminator. The determinant condition produces candidates; it does not answer the question. The mark that the cohort loses is the justification — substituting each root back and reporting which case it produces. [AG] allocates a separate mark for exactly this in both 2022 and 2024. Part (b) exists to make the omission explicit rather than invisible: a student who never tested the second root cannot answer it.

The evidence. 2024 Exam 1 Q2 — 37% (3 marks); the report is the fullest statement VCAA has made on the type: "Students using the determinant method often arrived at [two values] and then did not justify which answer was the valid solution… Some students incorrectly put [the other value] as the final solution, rejecting [the correct one], indicating confusion about the definition between 'infinite solutions' and 'no solution'." 2022 Exam 1 Q3 — 36% (3 marks); "Those who knew that the two lines needed to be identical were generally successful. Students using the determinant method often arrived at [two values], and then did not justify which value was valid." 2025 Exam 2 Section A Q4 — 55%; 2007 Exam 2 Section A Q5 — 36%, whose distribution shows 24% of the state giving the determinant roots where the complement was wanted.

Placement and marks. Exam 1, 3 + 1 = 4 marks. This type has appeared on the technology-free paper in 2022, 2024 and the 2024 NHT paper, and as Section A multiple choice in 2006, 2007, 2008, 2009, 2010, 2012, 2014 and 2025.

Predicted full-mark band. 22–35% for the four marks, against 2024 Exam 1 Q2 at 37% and 2022 Exam 1 Q3 at 36% for the three-mark half alone. Adding part (b) should cost 5–10 points.

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A2. Logarithm laws to a cubic, with two roots rejected

The design. A single technology-free equation combining a squared logarithmic argument with a second logarithm, in a base other than e, arranged so that the collapse produces a cubic. The cubic must have one obvious integer root reachable by the factor theorem and a quadratic factor whose roots are irrational. The implied domain must exclude the integer root and one of the two irrational roots. Four marks, one part.

The discriminator. The design has four separate failure points and a documented distribution showing where the cohort stops. First, choosing the right combination of laws — students who move the coefficient inside as a power before combining succeed; students who combine first do not. Second, producing the cubic. Third, factorising it rather than reaching for a formula. Fourth, and decisively, applying the implied domain to reject two of the three roots.

The evidence. 2024 Exam 1 Q6 — 9% (4 marks), with distribution {0: 18%, 1: 52%, 2: 8%, 3: 13%, 4: 9%}. Fifty-two per cent of the state scored exactly one mark: they applied a law and stopped. The report: "many students overlooked the fact that the domain of this log function must be [positive] and, as a result, did not reject the two invalid solutions." Supporting: 2009 Exam 1 Q9 — 22% (4 marks); "many did not eliminate x = −1 as a solution." 2020 Exam 1 Q4 — 26% (3 marks); "Those who did end up with the appropriate quadratic equation and solved it correctly did not always check the validity of their answers." 2013 Exam 1 Q5a, the two-mark version with no rejection, scored 54% — the 45-point gap is the whole content of the domain step.

Placement and marks. Exam 1, 4 marks, mid-paper. The distribution above shows this design discriminating across five bands, which is as much as a single question can do.

Predicted full-mark band. 8–18%, against 2024 Exam 1 Q6 at 9% and 2009 Exam 1 Q9 at 22%. Placing the cubic's integer root inside the domain rather than outside would lift the band to roughly 20–30% and lose the second rejection.

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A3. Exactly three solutions, scaffolded by the turning point

The design. Two curves meeting in a number of points that depends on a positive parameter. Part (a), two marks, asks for the coordinates of the turning point of the difference function in terms of the parameter. Part (b), two marks, asks — using the word "hence" — for the values of the parameter giving exactly three solutions.

The discriminator. Part (b) unaided is the hardest construction in the subject. Part (a) is the scaffold that converts it from a defeating part into a discriminating one: it hands the student the object whose vertical position controls the solution count, so that part (b) becomes a decision (set the turning-point ordinate against the other curve, decide whether tangency counts) rather than a discovery. Students who ignore the "hence" and expand to a quartic will run out of time, which is itself the documented failure.

The evidence. 2025 Exam 1 Q9bii — 4% (2 marks); the report is the best single paragraph VCAA has written on the type: "very few students made significant progress… Few students used a 'hence' approach and thus did not identify a connection between the turning point and the number of solutions. Those students who were able to demonstrate clear mathematical communication skills… were the most successful." 2018 Exam 1 Q8b — 3%, distribution {0: 89%, 1: 8%, 2: 3%}; "Most students found the correct quadratic equation to solve but solved for k, rather than the x value." 2016 Exam 2 Section B Q4eiii — 3%. 2025 Exam 1 Q9a, the same configuration at a fixed parameter value, scored 19% — that is the reset this design needs.

Placement and marks. Exam 1, final question, 2 + 2 = 4 marks. Part (a) must come first; without it the design defeats.

Predicted full-mark band. Part (a) 20–32% against 2025 Exam 1 Q9bi at 22%; part (b) 12–22% against 2025 Exam 1 Q9bii at 4% unscaffolded — the scaffold is worth roughly ten to fifteen points on this evidence.

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A4. Discriminant plus a sign condition

The design. A Section A item giving a quadratic equation in which the parameter appears in the coefficient of x², in the coefficient of x, and in the constant. The requirement is not "two real solutions" but "two real solutions of opposite sign", or "two positive real solutions". Four options: the discriminant condition alone, the sign condition alone, their intersection, and their union.

The discriminator. Two conditions must be imposed, and the options are constructed so that each single condition is available as a distractor. There is a third trap: because the parameter sits in the leading coefficient, the degenerate case where that coefficient vanishes must be checked separately — the equation is then linear and has exactly one solution.

The evidence. 2023 Exam 2 Section A Q19 — "has two real solutions for x, one positive and one negative" — 32%; VCAA's published solution names the two-condition structure explicitly. 2017 Exam 2 Section A Q7 — 32%, with distribution A 19%, B 32%, C 12%, D 29%, E 7%: 29% of the state chose the same quadratic with the inequality reversed. 2019 Exam 2 Section A Q2, the single-condition version, scored 59% — the 27-point gap is what the second condition costs.

Placement and marks. Exam 2 Section A, one mark, at Q17 or later.

Predicted full-mark band. 25–35%, against 2023 Exam 2 Section A Q19 and 2017 Exam 2 Section A Q7, both 32%. The four-option format in force since November 2024 raises the guessing floor to 25%, so aim the distractors at the two single-condition answers rather than at arithmetic slips.

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A5. The literal constraint before the calculus

The design. A solid of revolution or an inscribed figure with two dimensions linked by a similarity or a fixed-volume constraint. Part (a), two marks, asks for one dimension in terms of the other. Part (b), one mark, asks for the quantity to be optimised in terms of that single variable and for its domain. Part (c), two marks, asks for the optimising value.

The discriminator. Part (a) is the separator and it is not calculus at all — it is recognising that similar figures are involved and writing the correct ratio. The archive shows the cohort writing the ratio upside down. Part (b) adds the domain, which is the mark students omit and which then decides part (c) if the design places the optimum near an endpoint. Part (c) is routine for anyone who survived (a) and (b), which is the point: the question grades on modelling, not on differentiation.

The evidence. 2010 Exam 1 Q11a — h in terms of r by similar triangles — 11% (2 marks); "Few students realised that similar figures were needed. A common incorrect formulation was h/5 = r/2." 2010 Exam 1 Q11b, the substitution step, 47%; 2010 Exam 1 Q11c, the optimisation, 10%. 2008 Exam 1 Q9a — 35%; "Quite a few students had difficulty finding the volume of the prism… Students are expected to have adequate algebraic facility with these sorts of expressions." 2014 Exam 1 Q10bi — 32%. 2021 Exam 2 Section B Q1b, "state the domain" for a box-volume model, 42%; 2021 Exam 2 Section B Q1fi, the same for the general-width version, 33%.

Placement and marks. Exam 1, 2 + 1 + 2 = 5 marks, as the final question. Exam 2 Section B carries the same skeleton with a numerical answer at the end.

Predicted full-mark band. Part (a) 15–30% against 2010 Exam 1 Q11a at 11% and 2014 Exam 1 Q10bi at 32%; part (b) 30–45% against 2021 Exam 2 Section B Q1b at 42%; part (c) 12–25% against 2010 Exam 1 Q11c at 10%.

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A6. "For all values of", reduced to the endpoints

The design. A quantity defined on a closed interval and depending on a parameter. Part (a), one mark, asks for the values of the quantity at the two interval endpoints, in terms of the parameter. Part (b), two marks, asks for the values of the parameter for which the quantity stays below a stated bound for every point of the interval.

The discriminator. The technique is that a "for all" condition over an interval reduces to a condition at the extreme values of the quantity on that interval. The cohort's documented responses are to substitute particular values, to solve the general equation, or to write nothing. Part (a) is the scaffold: it names the endpoints so that the student is at least holding the right two numbers. Part (b) then requires the leap from "true at the endpoints" to "true everywhere", which is legitimate only when the extremes are at the endpoints — so the design must choose a monotone quantity and should say so in the stem if the cohort is a school one.

The evidence. 2020 Exam 2 Section B Q2f — "for all parts of the river" — 7% (2 marks); "Some students were able to set up an appropriate inequality… Some students did not substitute either x = 0 or x = 200. A common incorrect approach was solving [the general equation]." 2016 Exam 2 Section B Q4d — show a monotonicity statement — 6%, distribution {0: 90%, 1: 4%, 2: 6%}; "Some just substituted in specific values, which was not acceptable." 2016 Exam 2 Section B Q4fii — show a two-sided bound holds for all parameter values — 2%, distribution {0: 94%, 1: 4%, 2: 2%}. Those three form the hardest group in the algebra area, median 6.5%.

Placement and marks. Exam 2 Section B, 1 + 2 = 3 marks. Without part (a) this is a defeating design, and the 90%-and-94%-zero distributions above are what that looks like.

Predicted full-mark band. Part (a) 40–55%, no direct comparison but the substitution is routine; part (b) 12–25% against 2020 Exam 2 Section B Q2f at 7%, with the scaffold worth roughly ten points.

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A7. Transform the domain, then solve

The design. A technology-free trigonometric equation of the guaranteed form — a circular function of a linear argument equal to an exact value — over an interval chosen so that the transformed interval for the argument spans more than two full periods and therefore contains six or more solutions. The interval endpoints must be chosen so that one solution sits exactly at an endpoint.

The discriminator. The step the reports name every single year is the domain transformation, and this design makes skipping it fatal rather than merely careless: a student who solves on the original interval will produce two solutions where six are required. The endpoint solution then tests whether the transformed interval was written as closed.

The evidence. Eight archive instances, eight separators: 2010 Exam 1 Q4b 39%, 2008 Exam 1 Q3 41%, 2009 Exam 1 Q4 41%, 2007 Exam 1 Q8a 45%, 2013 Exam 1 Q4 47%, 2019 Exam 1 Q4a 48%, 2015 Exam 1 Q5b 53%, 2014 Exam 1 Q3 55%. The reports are unanimous: "giving answers outside the given domain… or only finding [one solution]" (2007); "when the given domain was ignored. It resulted in either only one solution or too many solutions" (2008); "Most students correctly chose the initial angle but went on to have problems dividing by 2 and selecting the appropriate angles for the set domain" (2009); "Many students did not account for the restricted domain" (2019).

Placement and marks. Exam 1, 3 marks. This is the single most reliable annual appearance in the subject — present in every November Exam 1 from 2007 to 2025 — so it belongs in any trial paper.

Predicted full-mark band. 33–48%, against 2009 Exam 1 Q4 at 41% (3 marks) and 2007 Exam 1 Q8a at 45%. Widening the interval to three periods rather than two should push it to the lower end.

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A8. The same equation, general then restricted

The design. Part (a), two marks: solve a trigonometric equation and give the general solution, with the quantifier stated. Part (b), two marks: state which of those solutions lie in a given interval — an interval whose endpoints are not multiples of the period, so that the count is not obvious.

The discriminator. The two halves fail in opposite directions, and the archive shows both. A general solution given where an interval was stated is penalised; a list of particular solutions given where a general solution was wanted is penalised. Part (a) additionally fails on the quantifier: the reports record n ∈ R and n ∈ J as submitted answers, and record calculator syntax (@n1) being copied across unedited. Part (b) fails on the arithmetic of selecting integer values of the index.

The evidence. 2021 Exam 1 Q3c — 17% (3 marks); "The construction of a general solution, while attempted, was not done well… or lacked a correct number categorisation of n." 2006 Exam 2 Section B Q1d — 15%; "some used R… or J instead of Z. Some students interpreted the period as 2π rather than π." 2023 Exam 2 Section B Q2dii — 12%; "Some students were able to set up a correct equation. A general solution was required." 2008 Exam 2 Section B Q4dii — 40%; "Some students gave the general solution using calculator syntax." The mirror error: 2025 Exam 2 Section B Q4b scored 78% but its report still says "Others incorrectly gave extra solutions or a general solution, not considering the restricted domain."

Placement and marks. Exam 1, 2 + 2 = 4 marks. On Exam 2 the same pair is fair but the calculator-syntax trap becomes the dominant failure.

Predicted full-mark band. 18–32% for both parts, against 2021 Exam 1 Q3c at 17% and 2008 Exam 2 Section B Q4dii at 40%.

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A9. The solution set as a union, with an empty factor

The design. A Section A item presenting a trigonometric or exponential equation that factorises into two factors, one of which is unsatisfiable over the reals. The options are written as sets: each single factor's solution set, their union, their intersection, and the union with the empty factor retained.

The discriminator. Two ideas at once. The null factor law expressed as a union of sets rather than a list of numbers, and the recognition that one factor contributes nothing because it demands a value outside the function's range. The option retaining the empty factor is the one the cohort chooses, because dropping it feels like losing information.

The evidence. 2007 Exam 2 Section A Q21 — 27%, with a five-way split in which the correct option was tied for the lead: A 27%, B 11%, C 20%, D 27%, E 15%. 2012 Exam 2 Section A Q20 — a probability item whose difficulty is the same partial-complement structure — 19%, with 27% choosing a single distractor. Contrast 2007 Exam 2 Section A Q11, the same null-factor idea with both factors satisfiable, at 82%, and 2016 Exam 2 Section A Q3 at 77%. The empty factor is worth roughly fifty points.

Placement and marks. Exam 2 Section A, one mark, Q17 or later. Under the four-option format, drop the intersection distractor and keep the two single-factor sets plus the retained-empty-factor union.

Predicted full-mark band. 25–40% under four options, against 2007 Exam 2 Section A Q21 at 27% under five.

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A10. Two exponential equations divided, using algebra

The design. A pair of exponential models agreeing with a printed graph at two labelled points. Part (a), three marks, is a "show that" for the two model parameters, with the stem containing the words "using algebra". Part (b), one mark, asks for a value of the model at a third point, exact.

The discriminator. The instruction "using algebra" forbids the CAS on a technology-active paper, and the reports say this is where marks go. The method is to divide the two equations to eliminate the multiplicative constant, not to substitute. A student who solves the system on CAS and writes the answer has the right numbers and no marks, because the answer was printed.

The evidence. 2025 Exam 2 Section B Q2a — 31% (3 marks); "This was a 'show that' question and students were required to show the algebraic steps. Many students were able to set up the two simultaneous equations; but some unnecessarily solved when the values were on the diagram. Some used a combination of their CAS and algebraic steps and were unable to gain full marks." 2008 Exam 2 Section B Q2d — 9% (2 marks); "Many students tried unsuccessfully to solve the simultaneous equations by hand. Other students gave only one solution." 2024 Exam 2 Section B Q3ai — four equations recovered from four stated features — 42%, with [AG] awarding "1M for any 2 equations, 1M for other 2 equations", which is the partial-credit structure this design should copy.

Placement and marks. Exam 2 Section B, 3 + 1 = 4 marks, opening a modelling question.

Predicted full-mark band. 25–35% for part (a), against 2025 Exam 2 Section B Q2a at 31%. Removing "using algebra" from the stem would push it above 60% and destroy the design.

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A11. Two intervals, four bracket decisions

The design. A technology-active inequality whose solution set is the union of two intervals, one closed at its left end and open at its right, the other the reverse, with a stated decimal accuracy. The inequality must come from a real quantity — a distance, a width, a concentration — so that the inequality signs are dictated by the context rather than printed.

The discriminator. Four independent bracket decisions and an ordering decision, none of which carries partial credit. The reports name every failure: values in the wrong order inside an interval; round brackets where square were needed; a single interval where two were required; exact values where a decimal accuracy was stated; solutions to the equation reported instead of the inequality.

The evidence. 2020 Exam 2 Section B Q5dii — 13% (1 mark); "Some students had the values within the interval in the wrong order. Others had incorrect brackets." 2020 Exam 2 Section B Q1f — 23% (2 marks); "Some students gave exact values for their answers… Others had incorrect inequality signs. Some had extra solutions or only gave the values of x for when D = 2 units." 2012 Exam 2 Section B Q1b — 38%; "Others had incorrect notation such as [0, 36] or {0, 36}." 2023 Exam 2 Section B Q3e — 35%; "Round brackets were often seen; these were incorrect as the largest interval of x values was required, which included the interval endpoints." [AG] 2024 Exam 2 Q5d records the marking rule directly: "Two intervals only full marks."

Placement and marks. Exam 2 Section B, 2 marks.

Predicted full-mark band. 15–28%, against 2020 Exam 2 Section B Q1f at 23% and 2012 Exam 2 Section B Q1b at 38%.

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A12. Recover the rule from a stated composition

The design. Give an outer function explicitly and give the composite explicitly; ask for the rule of the inner function. Part (a), one mark, is the rule. Part (b), two marks, asks for the maximal domain on which the composition in the stated order is defined, and whether the reverse composition is defined on the same set.

The discriminator. Part (a) is index-law or logarithm-law manipulation run backwards, which the cohort handles at around 70%. Part (b) is the existence condition, which no cohort handles. The second half — whether the reverse composite exists on the same set — forces the student to compute two ranges and compare them with two domains, which is four objects where the archive shows students holding one.

The evidence. 2023 Exam 2 Section A Q16 — given an outer exponential and a product, find the other factor — 69%: that is the achievable half. 2023 Exam 2 Section A Q20 — "The largest interval of x values for which (f ∘ g)(x) and (g ∘ f)(x) both exist" — 30%. 2016 Exam 1 Q5aiii — the "show that" version of a composite identity — 31%; "Many students were unsure of how to present their working… Poor notation was again evident." 2019 Exam 1 Q9a and Q9c, bare composite rules, scored 94% and 86% — the contrast that locates the difficulty precisely in the domain, not in the substitution.

Placement and marks. Exam 1, 1 + 2 = 3 marks.

Predicted full-mark band. Part (a) 60–75% against 2023 Exam 2 Section A Q16 at 69%; part (b) 20–32% against 2023 Exam 2 Section A Q20 at 30%. Combined: 20–32%.

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2.3 Calculus — twelve designs

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C1. The maximum that is not at the stationary point

The design. An optimisation in a physical context whose natural domain is a closed interval. The constraint must place the single stationary point inside the interval but make it a minimum, so that the maximum is at one of the two endpoints — and the two endpoint values must differ, so that "check both" is a real instruction rather than a formality. Part (a), two marks, asks for the quantity as a function of one variable with its domain. Part (b), two marks, asks for the maximum value and where it occurs.

The discriminator. Part (b) requires evaluating at the stationary point and at both endpoints and comparing three numbers. The cohort differentiates, solves, substitutes, and reports. The study design names this explicitly — "identification of interval endpoint maximum and minimum values" — and it is the single most reliably fatal construction in the archive.

The evidence. 2021 Exam 1 Q9cii — 4% (2 marks), distribution {0: 81%, 1: 15%, 2: 4%}; "Because many students overlooked the domain in their definition of the function, they did not realise that the maximum occurs at an endpoint. Instead, many attempted this question by differentiation and then solved to get the incorrect maximum." 2014 Exam 1 Q10biii — 8% (1 mark), distribution {0: 92%, 1: 8%}; "Many students who attempted this question incorrectly assumed that the value of the maximum area occurred at a local stationary point." 2015 Exam 2 Section B Q5ci and Q5cii — 4% each. 2007 Exam 2 Section B Q1d — 3%; "Many students found the minimum surface area instead of the maximum. Some students chose the wrong endpoint." 2010 Exam 2 Section B Q3e — 17%.

Placement and marks. Exam 1, 2 + 2 = 4 marks. Part (a) must demand the domain; that is the reset, and it is the part that makes the design discriminate rather than defeat. The 81% and 92% zero columns above are what happens when the domain is never asked for.

Predicted full-mark band. Part (a) 30–45% against 2021 Exam 2 Section B Q1fi at 33%; part (b) 15–28% against 2021 Exam 1 Q9cii at 4% unscaffolded and 2010 Exam 2 Section B Q3e at 17%.

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C2. The function, then its gradient function

The design. A modelling context with a named physical object — a hill, a ramp, a temperature curve. Part (a), one mark, asks for the interval on which the model is strictly decreasing. Part (b), one mark, asks for the set of values on which the gradient of the model is strictly decreasing. Both answers must be intervals with finite endpoints so that the bracket type is also in play.

The discriminator. Part (a) is a 50–70% question. Part (b) is the same sentence with one noun changed, and the archive says the state does not read the noun. The second discriminator is the bracket: VCAA accepts and expects the closed interval, and the reports record round brackets as the dominant error on the ones the cohort does read correctly.

The evidence. 2019 Exam 2 Section B Q2b — "State the set of values for which the gradient of the hill is strictly decreasing" — 3%, distribution {0: 97%, 1: 3%}; "Most students interpreted the question as asking where the function modelling the hill was strictly decreasing, rather than the gradient of the hill, and so the most common incorrect response was [the function's interval]." 2023 Exam 2 Section B Q3e — the function version — 35%; "Round brackets were often seen; these were incorrect as the largest interval of x values was required, which included the interval endpoints." 2009 Exam 2 Section B Q1a, the plain function version, scored 72%. [SAMP] Exam 2 asks "Find the interval of x for which the gradient function of the ramp is strictly increasing", so the construction is officially signposted.

Placement and marks. Exam 2 Section B, 1 + 1 = 2 marks, consecutive.

Predicted full-mark band. Part (a) 35–50% against 2023 Exam 2 Section B Q3e at 35%; part (b) 10–25% against 2019 Exam 2 Section B Q2b at 3% — the lift coming entirely from placing part (a) immediately before it, which flags that the two nouns are different.

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C3. Average value and average rate of change, same interval

The design. One function, one interval, three consecutive parts. Part (a), two marks: the average rate of change of the function over the interval. Part (b), three marks: the average value of the function over the same interval. Part (c), one mark: the value of the variable at which the function actually attains that average value. The function should carry exact circular or logarithmic values at both endpoints so that the arithmetic is the second difficulty.

The discriminator. VCAA has already built this pairing twice and the cohort still conflates the two formulas — neither of which is on the formula sheet. Part (a) requires no calculus at all and the state still loses two-thirds of it, because students average the derivatives at the endpoints. Part (b) fails on the placement of the reciprocal interval width, which the reports name as the main error. Part (c) is the reset, and it tests whether the student understands the average value as a height rather than as a computed number.

The evidence. 2016 Exam 1 Q6a (average rate) — 32%, and Q6b (average value, same function, same interval) — 16%, distribution {0: 31%, 1: 23%, 2: 30%, 3: 16%}. [SAMP] Exam 1 Q5 pairs them identically. 2013 Exam 1 Q6 — average value run backwards for a parameter — 16%, distribution {0: 40%, 1: 19%, 2: 25%, 3: 16%}; its report states the whole issue: "The formula for the 'average value' was not on the formula sheet." 2015 Exam 1 Q4c — 41%; "The main error in student responses was the misplacement of ½ in the integrand." 2022 Exam 1 Q8c — average value optimised over an interval endpoint — 18%. The reports name the confusion for 31 Calculus parts.

Placement and marks. Exam 1, 2 + 3 + 1 = 6 marks, as a mid-to-late question. Both distributions quoted above are four-band spreads, which is what this design is for.

Predicted full-mark band. Part (a) 25–40% against 2016 Exam 1 Q6a at 32%; part (b) 12–25% against 2016 Exam 1 Q6b and 2013 Exam 1 Q6, both 16%; part (c) 25–40%, no direct comparison.

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C4. The greatest positive rate of change

The design. A periodic or damped-periodic model in context. Part (a), one mark, asks for the time at which the modelled quantity is greatest. Part (b), two marks, asks for the time at which its rate of change takes its greatest positive value, over a stated later interval containing more than one candidate.

The discriminator. The two-level structure. The rate of change is the derivative, so its maximum requires maximising the derivative — a second-order question that the cohort answers by maximising the function. The word "positive" is doing work in the stem: on a periodic model the greatest negative rate is an equally prominent feature and is the documented wrong answer. Part (a) is the trap and the reset simultaneously, because a student who answers (a) correctly and reuses the answer in (b) is exactly the student this design is designed to catch.

The evidence. 2022 Exam 2 Section B Q2g — "the time… when the rate of change of the rabbit population is at its greatest positive value" — 19%; "Another common incorrect answer was 76 weeks. This is when the rate of change of the rabbit population is at its greatest negative value." 2017 Exam 2 Section B Q2c — 29%; "Many could not find the maximum rate of change… Many found the value of t for the maximum value of h." 2024 Exam 2 Section B Q3biv — 21%; "Many students gave extra values or only one value. Others did not give the maximum instantaneous rate of change or found the minimum."

Placement and marks. Exam 2 Section B, 1 + 2 = 3 marks.

Predicted full-mark band. Part (a) 60–75%; part (b) 15–28% against 2022 Exam 2 Section B Q2g at 19% and 2024 Exam 2 Section B Q3biv at 21%.

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C5. Total area, sign change and symmetry

The design. A region bounded by a curve and the horizontal axis that crosses the axis at least twice inside the stated terminals, and which possesses a symmetry — odd symmetry about a crossing, or even symmetry about a vertical line — that halves the work. The word "total" must appear. Technology-free, two or three marks.

The discriminator. Three decisions. Whether to split at every crossing; whether to negate the piece below the axis or to negate the integral; and whether to exploit the symmetry. The reports record all three failures and also record the opposite error — splitting into three or four integrals where two suffice, which multiplies the arithmetic risk on a paper where arithmetic is the leading cause of lost marks.

The evidence. 2018 Exam 1 Q9d — total area of six shaded regions with a symmetry route available — 4%; "Students who recognised that the graph for this question was simply a combination of translations and reflections of an earlier simpler graph were able to use symmetry to determine the areas… Students who determined the equation of the tangents… were rarely successful." 2023 Exam 1 Q7d — total area of two identical regions — 19%; "Using symmetry eliminated the need to evaluate an additional integration calculation, however, many students did not utilise this property." 2020 Exam 1 Q8c — area below the axis — 11%; "A common oversight was the fact that the required area was below the x-axis." 2024 Exam 1 Q3b — 27%; "Many students arrived at a negative answer and knew that the area needed to be positive, but did not provide correct reasoning steps." 2014 Exam 1 Q5c — 21%; "Some students unnecessarily 'overworked' the problem by creating three or four integrations."

Placement and marks. Exam 1, 3 marks.

Predicted full-mark band. 15–28%, against 2023 Exam 1 Q7d at 19% and 2024 Exam 1 Q3b at 27%.

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C6. The parameter in the terminal, with a parity split

The design. A definite integral whose terminals are consecutive multiples of a period, indexed by a positive integer parameter. The antiderivative is supplied in the stem — there is no other legitimate route in Methods. Part (a), two marks, asks for the value when the parameter is even; part (b), two marks, for the value when it is odd. Both answers must be requested "in simplest form".

The discriminator. Carrying the letter through the substitution instead of replacing it with a number. The reports say the state substitutes a value, obtains a specific answer, and loses the general one. The second discriminator is the parity simplification — recognising that the circular function of an integer multiple of the period takes one of two values according to parity — which is the step that makes the two parts different rather than identical.

The evidence. 2018 Exam 1 Q9ai — 17% (2 marks); "Students in general seemed to find dealing with the parameter n difficult. Many tried substituting a value of n, rather than using n, thus resulting in a specific solution rather than the general [one] required." The same report adds: "Quite a few students included +c in the definite integral." 2015 Exam 2 Section B Q4di — the even case — 12%; Q4dii — the odd case — 9%; "Many students did not realise that when n is odd, cos(nπ) = −1." 2019 Exam 2 Section B Q5d — a total area in terms of a parameter — 22%. This family has a median around 13% and a floor of 2%, and 04-calculus.md names it the hardest construction in the area of study.

Placement and marks. Exam 1, 2 + 2 = 4 marks, final question. The supplied antiderivative is compulsory: without it the design is out of scope, because there is no integration by parts in Methods.

Predicted full-mark band. 10–20% per part, against 2018 Exam 1 Q9ai at 17% and 2015 Exam 2 Section B Q4di at 12%.

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C7. The area equation solved for the parameter

The design. A region bounded by two curves, one of which carries a parameter in a way that changes both the intersection points and the height of the region. Part (a), one mark, asks for the intersections in terms of the parameter. Part (b), two marks, asks for the parameter value making the bounded area equal a stated number.

The discriminator. The integral becomes an equation rather than a computation, and the terminals themselves contain the unknown. There is no numerical answer to check against, so a single sign or bracket error propagates untrapped to the end. Part (a) is the reset and carries most of the available partial credit.

The evidence. 2021 Exam 2 Section B Q2f — "Find the values of a such that the area… is equal to [a number]" — 2% (4 marks). 2018 Exam 1 Q8d — find the parameter from a stated area expression — 13% (3 marks); "While students could equate their answer to part c. to [the target], many students did not use their result from part a." 2007 Exam 1 Q10 — 33% (3 marks); "Some students who had little idea simply set the given expression equal to [the number]." 2013 Exam 1 Q6 — the average-value analogue — 16%, with a four-band distribution.

Placement and marks. Exam 1, 1 + 2 = 3 marks.

Predicted full-mark band. Part (a) 35–50%; part (b) 12–25% against 2018 Exam 1 Q8d at 13% and 2007 Exam 1 Q10 at 33%. The 2% precedent is what this design becomes without part (a).

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C8. Anti-differentiation by recognition, then applied

The design. Part (a), one mark: differentiate a printed product and state the result — a routine product-rule exercise. Part (b), two marks: hence find an antiderivative of a function that cannot otherwise be antidifferentiated in this course. Part (c), one mark: use it to evaluate a definite integral with exact terminals.

The discriminator. The word "hence". The archive records students differentiating in part (b), integrating the given expression, or starting again — anything except integrating both sides of the identity they were just handed. The second discriminator is the rearrangement: the right side of the identity is a sum, and the student must antidifferentiate every term they can and isolate the one they cannot.

The evidence. 2007 Exam 1 Q7 — 28% (3 marks); "Some students differentiated, while others ignored x and proceeded to anti-differentiate regardless. The instruction 'hence' was often ignored. Setting out and notation were very poor in this question and 'dx' often did not appear." 2022 Exam 1 Q8b — 22%. 2020 Exam 1 Q8b — the "show that" form — 36%. 2017 Exam 1 Q2b — 45%; "Students generally were not able to form an integral from their previous answer, ignoring the 'hence' instruction." 2018 Exam 1 Q8c, the "write down a definite integral" reset in the same family, scored 63%.

Placement and marks. Exam 1, 1 + 2 + 1 = 4 marks.

Predicted full-mark band. Part (a) 65–80%; part (b) 20–32% against 2007 Exam 1 Q7 at 28% and 2022 Exam 1 Q8b at 22%; all four marks 18–32%.

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C9. The tangent through a point that is not on the curve

The design. A curve and a fixed external point, with the point chosen so that two distinct tangents pass through it. Part (a), one mark, asks for the gradient of the curve at a general point. Part (b), two marks, asks for the equations of both tangents from the external point, to a stated accuracy.

The discriminator. The point of tangency is unknown, so two conditions are needed: the line passes through the external point and its gradient equals the derivative there. The efficient route equates the chord gradient from the external point to the derivative, giving one equation in one unknown. The archive records students building the full tangent equation first — a valid but much longer route that produces the wrong quadratic under time pressure — and records students using the external point's x-value to compute the gradient. The word "both" is the second discriminator.

The evidence. 2023 Exam 2 Section B Q3cii — 15%; 2020 Exam 1 Q7biii — 31%; "Students who equated gradients tended to score more highly. Many of those who used the 'equation of the tangent' method could not form the correct quadratic equation." 2012 Exam 1 Q10b — 22%. 2010 Exam 2 Section B Q1bv — 30%; "Many students worked out the equation of the tangent, which was unnecessary and very time-consuming, instead of equating the gradient of the segment with the derivative." 2009 Exam 1 Q8 — 37%; the report names "using x = 0 instead of x = a to find the gradient of the tangent." 2016 Exam 2 Section B Q1d — both tangents with a stated gradient — 47%.

Placement and marks. Exam 2 Section B, 1 + 2 = 3 marks.

Predicted full-mark band. 15–30%, against 2023 Exam 2 Section B Q3cii at 15% and 2020 Exam 1 Q7biii at 31%.

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C10. The constant of integration that must reach the rule

The design. Give a derivative that requires the linear-inner-function rule — so that the reciprocal of the inner coefficient is in play — together with a boundary condition stated as a stationary point rather than as a plain point, so that the student must extract the coordinate pair from a description. Ask for the rule of the function. Three marks.

The discriminator. Four documented failure points in three marks: forgetting the reciprocal coefficient; producing a logarithm where a power was needed or the reverse; omitting the constant of integration, which makes the boundary condition unusable; and finding the constant but reporting only the constant rather than the complete rule. The stationary-point phrasing adds a fifth, because the student must realise that a stationary point supplies a point, not a derivative condition, at this stage.

The evidence. 2021 Exam 1 Q8a — find the rule from a derivative and a stated stationary point — 21% (3 marks); the report's general comments: "it was important to write the rule and not just the constant of integration as the final answer." 2018 Exam 1 Q2 — 30%; "Some students found a value of c but did not substitute it back into the final answer to state f(x)." 2014 Exam 1 Q7 — 38%; "The students who omitted a constant of integration (+c)… were then unable to find the specific equation required." 2025 Exam 1 Q2 — 40%. 2013 Exam 1 Q3 — 42%; "Most students could anti-differentiate sin(2x) but many neglected '+ c', which was essential in order to move to the next step."

Placement and marks. Exam 1, 3 marks, at Q2 or Q3 — the archive puts an antidifferentiation question there in ten of twenty years.

Predicted full-mark band. 20–35%, against 2021 Exam 1 Q8a at 21% and 2018 Exam 1 Q2 at 30%.

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C11. Integral properties under a transformation of the integrand

The design. A Section A item giving the value of a definite integral of an unspecified function over one interval, then asking for the value of a definite integral of a transformed version of that function over a different interval. The transformation must combine a horizontal dilation with a horizontal translation, so that the terminals move and the area scales.

The discriminator. No rule is ever supplied, so nothing can be computed; the student must reason about how the region changes. The dilation scales the area by the reciprocal of the dilation factor and the translation does not change it at all — a fact the cohort applies to one of the two operations and not both. The distractors write themselves: the unscaled value, the value scaled by the factor rather than its reciprocal, and the value with the translation incorrectly counted.

The evidence. 2020 Exam 2 Section A Q9 — 35%; the report's published working is one line: "Dilate by a factor of ½ from the y-axis. Translating 2 units to the left does not change the area." 2010 Exam 2 Section A Q20 — 25%. 2015 Exam 2 Section A Q16 — 22%. 2019 Exam 2 Section A Q12 — 38%. 2022 Exam 1 Q2b, the technology-free version — 40%; "Some students incorrectly tried to expand the expression as a product of two integrals." 2008 Exam 2 Section A Q4, the linearity-only version, scored 49% — the transformation is worth 10 to 25 points.

Placement and marks. Exam 2 Section A, one mark, Q17 or later.

Predicted full-mark band. 25–38%, against 2020 Exam 2 Section A Q9 at 35% and 2010 Exam 2 Section A Q20 at 25%.

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C12. The gradient table as the answer

The design. A function with a repeated factor, so that one of its stationary points is a stationary point of inflection and the other is a genuine turning point. Part (a), one mark, asks for the coordinates of both stationary points. Part (b), two marks, prints an empty gradient table with headings and asks the student to complete it with appropriate values to show which of the two is the stationary point of inflection.

The discriminator. Part (b) marks the justification, not the conclusion. A student who knows the answer and writes "stationary point of inflection" scores nothing; the marks are for the chosen test values, the evaluated signs, and the arrow row. The reports record that most students know they must consider the slope on either side and that few show the substitutions.

The evidence. 2025 Exam 2 Section B Q1c — "Complete the following gradient table with appropriate values of x and g′(x) to show that g has a stationary point of inflection" — 55%: the printed table lifts the part above the separator line, which is what makes this design a grading tool rather than a wall. 2021 Exam 1 Q8b — the same task without a printed table — 23%; "Most students had a valid approach, but not all provided convincing arguments that showed the working out of substituting suitable values. Those who tried a second derivative approach met with mixed success." 2019 Exam 2 Section B Q1bi — "State the nature of the stationary point" — 72%; "Common incorrect answers were point of inflection, stationary points and turning points." 2023 Exam 2 Section B Q3d — coordinates of a non-stationary inflection — 58%.

Placement and marks. Exam 2 Section B, 1 + 2 = 3 marks, opening a graph question. Note the study-design gap: [SD] names points of inflection but never mentions the second derivative, and the formula sheet carries no second-derivative notation. The gradient table is therefore the only route guaranteed to be available, and a design that requires f″ is asking for something the course does not promise.

Predicted full-mark band. Part (a) 60–80% against 2025 Exam 2 Section B Q1a at 91% and 2019 Exam 1 Q9e at 30%; part (b) 30–45% against 2025 Exam 2 Section B Q1c at 55% with the table printed and 2021 Exam 1 Q8b at 23% without it.

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2.4 Data analysis, probability and statistics — twelve designs

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S1. The transformed density, with the area invariant named

The design. A probability density function on a bounded support, with its mean or median established in an earlier part. Part (a), one mark: explain why, for a transformed density written as a scaled horizontal dilation of the original, the product of the two parameters must equal one. Part (b), two marks: find both parameters, given the value the mean or median is required to take.

The discriminator. Part (b) alone is the hardest single idea in the area of study and it defeats: all five archive instances are separators and three sit below 7%. Part (a) converts it, because the area condition is the entire mathematical content and naming it as a one-mark part is what lets the cohort start. Part (b) then fails on the second invariant — that a horizontal dilation scales the mean, the median and every quantile by the same factor — and on the integral route's specific trap, which is failing to scale the terminals.

The evidence. 2021 Exam 2 Section B Q4h — 2%, distribution {0: 86%, 1: 12%, 2: 2%}; "Many students were unable to set up the correct equations. The terminals were often incorrect." 2023 Exam 2 Section B Q4j — 6%; "Others were able to recognise that ab = 1 but were unable to find their values. A common incorrect answer was a = 1 and b = 1. Many of those who attempted the second method did not multiply the terminals by b." 2022 Exam 2 Section B Q3biii — 6%; "Many students did not attempt it." 2018 Exam 2 Section A Q20 — 20%. 2025 Exam 1 Q8b — the technology-free linear-transform version — 27%; "many students incorrectly proceeded to factor out m from the entire integral… The students who recognised that the total probability is equal to 1 and applied it, were generally successful."

Placement and marks. Exam 2 Section B, 1 + 2 = 3 marks, as the final two parts of the statistics question.

Predicted full-mark band. Part (a) 30–45%; part (b) 10–25% against 2023 Exam 2 Section B Q4j at 6% unscaffolded. The 86%-zero distribution is what this design looks like without part (a).

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S2. The condition that contains the event

The design. A continuous random variable on a bounded support, with its density given. Part (a), one mark: a plain probability over a sub-interval. Part (b), one mark: write, in set notation, the two events described by a sentence such as "has already waited more than one unit" and "waits more than three units in total". Part (c), two marks: the conditional probability, exact.

The discriminator. Part (b) is the design. The archive's evidence is that the cohort translates "has already waited one unit" as an equality rather than an inequality, and that the whole failure is in the English, not in the integration. Part (c) then requires recognising that one event is a subset of the other, so the intersection collapses to the smaller set — after which the question is one division. Making the set-notation step its own mark converts a binary part into a graded one.

The evidence. 2023 Exam 1 Q8c — 11% (3 marks), distribution {0: 63%, 1: 13%, 2: 13%, 3: 11%}; "Common errors included writing the conditional probability as [an equality condition], where students had incorrectly interpreted the mathematical meaning of 'already queued for one minute'… There were also errors where students incorrectly identified the terminals of integration." 2008 Exam 1 Q4b — 27%; "Many recognised the need for conditional probability but thought the required intersection was between [the wrong pair of bounds]." 2014 Exam 1 Q8b — 24%; "The most common error was assuming that the value of m… was equivalent to Pr(X ≤ m)." 2007 Exam 2 Section B Q5c — 43%. 2019 Exam 2 Section B Q4c, the same idea on a technology-active paper, scored 57%.

Placement and marks. Exam 1, 1 + 1 + 2 = 4 marks.

Predicted full-mark band. Part (c) 12–25% against 2023 Exam 1 Q8c at 11%, with the named set-notation part worth a few points. All four marks: 10–20%.

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S3. A binomial probability in a demanded algebraic form

The design. A binomial with a small number of trials and a probability that is a unit fraction with a power-of-two or power-of-three denominator. Ask for a tail probability requiring two or three terms to be summed, with the answer demanded in a stated algebraic form whose letters are constrained to the integers. Technology-free.

The discriminator. Not the probability — the arithmetic. The student must compute two or three binomial terms by hand, add them over a common denominator, and then recognise the denominator as a power of a prime in order to express the answer in the demanded form. The reports record students computing one term and stopping, and record students giving a decimal because they could not factorise the denominator.

The evidence. 2025 Exam 1 Q6b — answer demanded in the form of an integer over a power of two — 34% (2 marks); "some students were not able to reduce 4096 down to the prime factorisation of 2¹². Instead, many students gave the answer as [a decimal]. Many students calculated only one term." 2019 Exam 1 Q6b — 12% (2 marks), distribution {0: 59%, 1: 29%, 2: 12%}; "few managed to correctly find the two component expressions. Even fewer successfully managed to manipulate these expressions to the format specified by the question." 2020 Exam 1 Q5b — 10%. 2021 Exam 1 Q6c — 29% (3 marks). 2016 Exam 1 Q4c — 61% — the one-term version, which is the reset this design should include as a preceding part.

Placement and marks. Exam 1, 1 + 2 = 3 marks, the first part being a single-term probability in the same form.

Predicted full-mark band. Part (a) 55–70% against 2016 Exam 1 Q4c at 61%; part (b) 15–32% against 2025 Exam 1 Q6b at 34% and 2019 Exam 1 Q6b at 12%.

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S4. The sample-proportion interval, converted to counts

The design. A binomial context with a small sample size. Part (a), two marks: the expected value and standard deviation of the sample proportion. Part (b), two marks: the probability that the sample proportion lies within one standard deviation of its expected value, with the stem carrying the instruction "Do not use a normal approximation."

The discriminator. Part (a) fails on the distinction between the expected value of the count and the expected value of the proportion — the report records students computing the former. Part (b) requires converting a continuous-looking interval on the proportion into an interval on the integer count, rounding the bounds inwards, and then using the binomial. The instruction blocking the normal approximation is what makes the conversion compulsory, and the reports record that the cohort uses the normal anyway.

The evidence. 2019 Exam 2 Section B Q4fiii — mean and standard deviation of the sample proportion — 42%; "Some students found E(X) = [n × p]" — the count, not the proportion — and the general comments add "E(P̂) cannot be greater than one." 2019 Exam 2 Section B Q4fiv — the within-one-standard-deviation part — 19%; "Many students were able to find the first interval. Some students used the normal distribution." 2018 Exam 2 Section B Q4ciii — 8%; "Many students appeared to be confused by the terminology [Pr(P̂ₙ > 1/n)]." 2021 Exam 2 Section B Q4c — 33%; 2021 Exam 2 Section B Q4d — 40%. 2025 Exam 2 Section B Q3biii, the simplest version with a five-day sample, 49%.

Placement and marks. Exam 2 Section B, 2 + 2 = 4 marks.

Predicted full-mark band. Part (a) 35–50% against 2019 Exam 2 Section B Q4fiii at 42%; part (b) 18–32% against 2019 Exam 2 Section B Q4fiv at 19% and 2021 Exam 2 Section B Q4d at 40%.

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S5. The confidence interval inverted, then reasoned about

The design. A 95% confidence interval printed as two endpoints, with the instruction to use the round approximation to the quantile so that the arithmetic is technology-free. Part (a), one mark: the sample proportion used. Part (b), two marks: the sample size. Part (c), one mark: the factor by which the width of the interval changes when the sample size is multiplied by a stated square number and the proportion is unchanged.

The discriminator. Part (a) is the midpoint and most of the cohort gets it. Part (b) requires rearranging the half-width formula for the sample size, which the reports say the state cannot transpose, with a specific documented failure — not extending the square-root sign over the sample size. Part (c) is a pure reasoning mark with no computation at all, and it has the worst return of any one-mark part in the area.

The evidence. 2023 Exam 1 Q6a — recover the proportion — 52%. 2023 Exam 1 Q6b — recover the sample size using the round quantile — 26% (2 marks). 2023 Exam 1 Q6c — the width factor — 23% (1 mark), distribution {0: 77%, 1: 23%}; "This question was not responded to well… A factor of ¼ was a common incorrect answer." 2019 Exam 2 Section B Q4g — 25%; "Many students had the proportion as [wrong values]. Others did not include the 1.96." 2017 Exam 1 Q4 — smallest sample size for a stated standard deviation — 31%; "many were unable to correctly transpose the inequality to solve for n… they did not extend the square root sign to include n." 2022 Exam 2 Section B Q3ciii — the halving version — 28%; "Common incorrect answers were 0, 10, 11, 50 and 101."

Placement and marks. Exam 1, 1 + 2 + 1 = 4 marks.

Predicted full-mark band. Part (a) 45–60% against 2023 Exam 1 Q6a at 52%; part (b) 20–32% against 2023 Exam 1 Q6b at 26%; part (c) 20–32% against 2023 Exam 1 Q6c at 23%. All four marks: 12–22%.

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S6. Interpret the interval against a named proportion

The design. A confidence interval computed in an earlier part, plus a population proportion for a different population established still earlier. Part (a), one mark: state whether the second population's proportion lies inside the interval. Part (b), one mark: explain what that implies about the two populations.

The discriminator. The archive's version of this asks for the explanation alone and scores 11%, with an 89% zero column — it defeats. Splitting it converts the discovery into a decision: part (a) forces the student to make the comparison that is the whole content, and part (b) then asks only for the sentence. The failure this design still catches is the one the report names — an answer that does not quote the interval.

The evidence. 2018 Exam 2 Section B Q4dii — "Explain why this confidence interval suggests that the proportion of adults with a slow heart rate in Statsville could be different from the proportion in Mathsland" — 11% (1 mark), distribution {0: 89%, 1: 11%}; "The confidence interval needed to be referred to in the answer." The model answer VCAA published names both endpoints and the comparison value explicitly. Supporting: 2018 Exam 2 Section B Q4di, recovering the proportion from the same interval, 45%; 2022 Exam 2 Section B Q3ci, "Is the random variable P̂ discrete or continuous? Justify your answer" — 68%, which shows that a justification part can score when the required form of the answer is obvious.

Placement and marks. Exam 2 Section B, 1 + 1 = 2 marks.

Predicted full-mark band. Part (a) 50–65%; part (b) 20–35% against 2018 Exam 2 Section B Q4dii at 11% unsplit.

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S7. A binomial probability as a polynomial, then optimised

The design. A fixed small number of trials with the success probability left as a free parameter. Part (a), two marks: express the probability of a stated compound event as a polynomial in that parameter. Part (b), two marks: find the maximum value of that probability over the unit interval and the parameter value at which it occurs, to a stated accuracy.

The discriminator. Part (a) fails on assembling two or three binomial terms with the parameter symbolic. Part (b) is calculus applied to a statistics object, which the archive shows the cohort not recognising as available; and when they do differentiate, they report only the optimising parameter and not the maximum probability, or they report a root outside the unit interval. The reports name both failures.

The evidence. 2017 Exam 2 Section B Q3f — express the probability as a polynomial — 32%; 2017 Exam 2 Section B Q3gi — the maximum and where it occurs — 23%; "Some students knew to solve q′(p) = 0 if they had an equation in Question 3f. Others found only p. Some gave exact values for their answers" where four decimal places were demanded. 2007 Exam 2 Section B Q5e — 25%; 2007 Exam 2 Section B Q5fi — 15%; "Some students gave p = [a value] which is greater than one." 2012 Exam 2 Section B Q3c — the "show that" version — 13%. 2025 Exam 2 Section B Q1f — a binomial tail probability shown to equal a cubic-quartic defined at the top of a calculus question — 45%.

Placement and marks. Exam 2 Section B, 2 + 2 = 4 marks.

Predicted full-mark band. Part (a) 28–40% against 2017 Exam 2 Section B Q3f at 32%; part (b) 18–32% against 2017 Exam 2 Section B Q3gi at 23%.

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S8. The parameterised table with a feasibility constraint

The design. A Section A item presenting a discrete probability distribution whose entries are linear or quadratic expressions in a single parameter, arranged so that the sum-to-one condition does not determine the parameter uniquely. The question asks for the largest (or the smallest and largest) possible value of the mean.

The discriminator. A probability table carries two constraints, not one: the entries sum to one, and every entry lies between zero and one. The cohort imposes the first, writes the mean as a function of the surviving parameter, optimises it over the reals, and reports a value corresponding to a negative probability. The feasible interval is the whole question, and its endpoints are where the optimum sits.

The evidence. 2016 Exam 2 Section A Q19 — "The smallest and largest possible values of E(X) are respectively" — 15%. 2020 Exam 2 Section A Q19 — the mass function of the complementary count — 15%. 2023 Exam 2 Section A Q12 — "The maximum possible value for the mean of X is" — 29%. 2012 Exam 2 Section A Q20 — 19%. These are the four hardest multiple-choice items in the area and 01-study-design.md §10.6 lists three of them among the ten hardest multiple-choice questions in the whole archive.

Placement and marks. Exam 2 Section A, one mark, Q18 or later. Under four options, build the distractors as: the unconstrained optimum, the value at the wrong endpoint of the feasible interval, and the mean at the parameter value that makes one entry zero.

Predicted full-mark band. 15–30% under four options, against 2016 Exam 2 Section A Q19 and 2020 Exam 2 Section A Q19, both 15%, and 2023 Exam 2 Section A Q12 at 29%.

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S9. The quantile with a root to reject

The design. A linear or simple polynomial density on a bounded support. Part (a), one mark: verify the normalising constant. Part (b), three marks: find the value of the variable with a stated upper-tail probability, exactly and by hand, where the resulting quadratic has one root inside the support and one outside.

The discriminator. Four steps, each documented as a failure point: antidifferentiating a bracketed linear term and dividing by the correct constant; forming the equation with the unknown in a terminal; setting the quadratic to zero before factorising rather than guessing at a non-zero form; and rejecting the root outside the support with a stated reason. The tail direction is a fifth: the archive records the mirror-image answer as the common wrong response.

The evidence. 2025 Exam 1 Q8a — 24% (3 marks), distribution {0: 30%, 1: 23%, 2: 23%, 3: 24%} — a genuine four-band spread; "Some students who chose to integrate the term as a bracketed term raised the power to 2 but then divided by 2 instead of 6. Some students were unable to form the correct quadratic equation or solve it correctly… The quadratic expression was readily factorised by inspection, but a large proportion of students used the quadratic formula." 2012 Exam 1 Q8b — 45%; "The quadratic formula and null factor law or factorisation can be used only if the quadratic is equated to zero. A number of students established a quadratic that was not equal to zero and used guesswork." 2006 Exam 1 Q6b — 39%; "Including [a value outside the domain] was a common error." 2017 Exam 2 Section B Q3d — the technology-active version — 30%; the report names the mirror-tail answer as the common error.

Placement and marks. Exam 1, 1 + 3 = 4 marks.

Predicted full-mark band. Part (a) 40–55% against 2023 Exam 1 Q8a at 43% and 2021 Exam 1 Q7a at 48%; part (b) 20–32% against 2025 Exam 1 Q8a at 24%.

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S10. The smallest integer, with two adjacent values shown

The design. A binomial context. Part (a), one mark: express the probability of at least one success in a sample of unspecified size, in terms of that size. Part (b), two marks: the smallest sample size for which that probability exceeds a stated threshold, with the stem noting that trial and error is acceptable provided the working is shown.

The discriminator. Inverting the complement. Forwards, this is a 60–75% question; backwards it is a separator every time it is asked. The failure modes are an inequality rounded in the wrong direction, an answer reported as a non-integer inequality, and — the one this design tests directly — an answer with no working, because trial and error leaves no trace unless the student writes down both the value that fails and the value that works.

The evidence. 2020 Exam 2 Section B Q3ei — express the complement in terms of the sample size — 24% (1 mark). 2020 Exam 2 Section B Q3eii — the smallest size — 23%; "Some students left their answer as 18.43 or rounded down to 18." 2014 Exam 2 Section B Q4e — 23%. 2019 Exam 2 Section B Q4fii — 23%; "Trial and error is an acceptable method." 2015 Exam 2 Section B Q3dii — 35%; "Others did not state the minimum value, leaving their answer as n ≥ 22.7566." 2025 Exam 2 Section B Q3biv — the double inversion, where the parameter sits inside an integral terminal that produces the probability — 17%; "An integer value was required… Others just gave the answer, without showing appropriate working as required."

Placement and marks. Exam 2 Section B, 1 + 2 = 3 marks.

Predicted full-mark band. Part (a) 22–35% against 2020 Exam 2 Section B Q3ei at 24%; part (b) 20–35% against 2019 Exam 2 Section B Q4fii and 2014 Exam 2 Section B Q4e, both 23%.

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S11. Sketch the density, including where it is zero

The design. A two-piece density on a bounded support, one piece linear and one piece curved, printed axes supplied with a scale on both. Part (a), three marks: sketch the density over a domain that is wider than the support, so that the zero pieces must be drawn. Part (b), one mark: state the value of the density at the join.

The discriminator. Everything the reports name, and none of it is probability. Drawing the density as zero outside the support; using a ruler on the linear piece; the endpoint conventions at the join; and the maximum value, which students report as the variable value rather than as the density value. Part (b) exists to catch the last of these separately.

The evidence. 2017 Exam 2 Section B Q3a — 29% (3 marks); "Many students did not draw their graphs along the t-axis, ignoring f(t) = 0. Some had an open circle at [the join]. Others had an open circle over a closed circle. Many students did not use rulers to draw the line segments. Some graphs looked like parabolas." 2007 Exam 2 Section B Q5a — 22% (2 marks); "A common error was that [the variable value and the density value] were given as the coordinates of the maximum." Three archive instances, three separators. 05-probability-statistics.md calls this "the worst full-mark rate of any 'easy' statistics task in the archive."

Placement and marks. Exam 2 Section B, 3 + 1 = 4 marks, opening the statistics question. Printing axes wider than the support is the design decision that makes it work.

Predicted full-mark band. Part (a) 25–40% against 2017 Exam 2 Section B Q3a at 29%; part (b) 50–70%.

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S12. Find the deviation, then both translated values

The design. A normal model in context. Part (a), two marks: find the standard deviation required so that a stated proportion falls beyond a threshold, the mean being fixed. Part (b), two marks: with the standard deviation fixed instead, find both values of a translation parameter for which a stated proportion falls within a symmetric interval.

The discriminator. Part (a) fails on the tail: the reports record students using the complementary probability, and record students using the confidence level where its half-tail was needed. Part (b) fails on the word "both": the symmetric configuration has two solutions and the cohort reports one. It also rewards a diagram, which the reports repeatedly name as the difference between the students who score and those who do not.

The evidence. 2023 Exam 2 Section B Q4f — find the standard deviation so that more than a stated percentage meet a criterion — 26%; "Trial and error could be used but students must make sure they show some appropriate working. Drawing a diagram and showing the probabilities was acceptable." 2016 Exam 2 Section B Q3e — 32%; the same report names the units disaster in the neighbouring part: "Some students thought 3 hours and 10 minutes was 3.1 hours." 2009 Exam 2 Section B Q3d — 22%; "Many students used 0.99 as the required probability instead of 0.995… Other students used 0.05 instead of 0.005." 2020 Exam 2 Section B Q3c — the both-values part — 6%; "Many students did not find [both values]." 2024 Exam 2 Section B Q4eii — minimum and maximum of an expression — 8%; "Many students were able find the minimum value… but not the maximum value."

Placement and marks. Exam 2 Section B, 2 + 2 = 4 marks.

Predicted full-mark band. Part (a) 22–35% against 2023 Exam 2 Section B Q4f at 26%; part (b) 12–25% against 2020 Exam 2 Section B Q3c at 6%, lifted by the explicit word "both" in the stem — the archive instance did not carry it.

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3. Designs that exploit the 2023–2027 study design specifically

The 2023 design moved three constructions in — Newton's method, the trapezium rule, pseudocode — and left one dot point, simulation of random sampling, that has never been examined in twenty years of written papers. It also restored points of inflection after a seven-year absence, and it retains a named function, log₁₀, that the November papers have never once used. Every design in this section sits in content that is unambiguously examinable and almost entirely unrehearsed, which is the cleanest separator material available to a question writer.

3.0 What the documents permit, and what they constrain

Newton's method. [SD] places it inside an Outcome 1 key skill — "apply a range of analytical, graphical and numerical processes (including the algorithm for Newton's method)… to obtain general and specific solutions (exact or approximate) to equations (including literal equations) over a given domain" — not in a content dot point. The formula sheet carries the iteration. Because it is an Outcome 1 key skill and Examination 1 assesses Outcome 1, it is examinable technology-free, and the 2023 sample Examination 1 Question 6 is a worked technology-free example. In four years of live November papers it has never appeared on Examination 1. Its live appearances are 2023 Exam 2 Section A Q13 (52%), 2023 Exam 2 Section B Q3f (54%) and Q3g (21%), plus 2025 Exam 2 Section B Q4e, the 2024 NHT Exam 2 and the 2026 NHT Exam 2. Constraint: Examination 1 never asks for a decimal accuracy, so a technology-free Newton design must terminate after one iteration in an exact fraction.

The trapezium rule. [SD] names it twice — in the Algebra key knowledge ("the concept of approximation to the area under a curve using the trapezium rule") and in the Calculus content ("approximation of definite integrals using the trapezium rule") — and the formula sheet carries it. It replaced the pre-2023 rectangle approximation. Live November appearances: 2024 Exam 1 Q7a (30%, distribution {0: 28%, 1: 31%, 2: 11%, 3: 30%}), 2023 Exam 1 Q4 (45%) and 2025 Exam 2 Section A Q6 (50%). It has never appeared in Section B of a November paper. Constraint: the 2024 report states that "any attempt to calculate this area using integral calculus was not acceptable" — the phrase "using the trapezium rule" forbids integration, which is what makes the design safe from a CAS.

Pseudocode. [SD] places it in Outcome 2 key knowledge: "key elements of algorithm design, including sequencing, decision-making and repetition, and representations of the ordered steps for an algorithm including through the use of pseudocode." Examination 1 assesses Outcome 1 only, so pseudocode is very likely excluded from it — a reasoned inference from the outcome mapping in [SPEC], backed by every paper since 2023, not a stated VCAA exclusion. Live appearances are three multiple-choice items — 2025 Exam 2 Section A Q7 (72%), the 2024 NHT Exam 2 Section A Q14 and the 2025 NHT Exam 2 Section A Q18 — plus three in the sample Examination 2. It has never appeared in Section B of any paper.

Systems with no solution, one solution, or infinitely many. Retained and sharpened. The dot point's parenthesis is the constraint: "geometric interpretation only required for two equations in two variables". A three-variable system may be asked algebraically; it may not be asked to be interpreted as intersecting planes. The 2023 sample Examination 2 Question 1 asks for the general solution of a three-variable, two-equation system in parameter form and no November paper has ever asked it.

Simulation of random sampling. [SD] lists it twice — as a content dot point and as a key skill — yet grepping every paper and every report in the corpus for simulat returns zero hits, 2006 to 2026 NHT. It is assessed in School-assessed Coursework, not in the written examinations. A trial-exam writer may use it; a prediction that VCAA will is unsupported.

log₁₀. [SD] names it explicitly among the functions whose graphs and key features are examinable: "logarithmic functions, y = logₑ(x) and y = log₁₀(x)". It occurs in exactly one paper in the entire corpus — the 2017 NHT Examination 2 — and in no November paper in twenty years. (Extraction of subscripts is imperfect, so treat that count as a firm upper bound on rarity rather than a proof of zero.)

Points of inflection. Removed by the 2016–2022 design's exclusion clause ("consideration of the second derivative is not required") and restored in 2023. The phrase does not appear in any paper between 2016 and 2022 and appears in 2023, 2024 NHT, 2025 Exam 1, 2025 Exam 2 and the sample papers. Constraint: [SD] names inflection points but never mentions the second derivative, and the formula sheet carries no second-derivative notation — so a design may allow f″ but may not require it.

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N1. Newton's method, technology-free, one exact iteration

The design. A cubic or a cubic-plus-linear function with an easily computed derivative. Part (a), one mark: verify that the function takes opposite signs at two stated integers, and state what that guarantees. Part (b), two marks: apply one iteration of Newton's method from a stated integer starting value, giving the result as an exact fraction. Part (c), one mark: state, with a reason, whether the new estimate lies in the interval identified in part (a).

The discriminator. Nothing about the formula, which is printed on the sheet. The separator is the exact arithmetic — evaluating the function and its derivative at the starting value and then subtracting a fraction from an integer, all by hand, on a paper whose instruction page demands an exact value. Part (c) is the real discriminator: it asks the student to check the output rather than to produce it, which is a habit no cohort has, because the construction has never been examined technology-free.

The evidence. No November Examination 1 instance exists — that is the design's point. The nearest comparisons are [SAMP] Examination 1 Question 6, which pairs a sign check with a single iteration in exactly this shape, and 2023 Exam 2 Section B Q3f at 54% with technology available and three decimal places requested. The exact-arithmetic analogue is 2024 Exam 1 Q7a at 30%, where the formula was known and the surd and fraction arithmetic was not.

Placement and marks. Examination 1, 1 + 2 + 1 = 4 marks.

Predicted full-mark band. 20–35%, against 2023 Exam 2 Section B Q3f at 54% with technology, discounted for exact by-hand arithmetic by roughly the margin visible between 2023 Exam 1 Q4 (45%, two trapeziums) and 2024 Exam 1 Q7a (30%, three trapeziums with exact circular values).

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N2. Why this starting value fails

The design. A function with two stationary points inside the interval of interest and a single root outside them. Part (a), one mark: state the values at which the iteration is undefined. Part (b), two marks: explain why an initial estimate at, or very close to, one of those values will not produce a usable next estimate — the answer must refer to the tangent, not merely to the stationary points.

The discriminator. Part (a) is a derivative equated to zero and is worth banking. Part (b) is an explanation, and the archive shows explanations are where the cohort loses most. The published failure is precise: students identify the stationary points and stop, without saying that the tangent there is horizontal and therefore never meets the horizontal axis.

The evidence. 2023 Exam 2 Section B Q3g — the only archive instance — 21% (1 mark), distribution {0: 79%, 1: 21%}; "The solutions to [f′(x) = 0] will give the x values of the turning points of the graph. The tangents to the graph will be horizontal lines… Hence, [x₁] will be undefined. There were some good explanations. Some students only mentioned the two solutions." The general comparison for explanation parts is 2018 Exam 2 Section B Q4dii at 11% against 2021 Exam 2 Section B Q3c — "Explain why p is not a one-to-one function" — at 66%; the difference is whether the required form of the answer is obvious.

Placement and marks. Examination 2 Section B, 1 + 2 = 3 marks. Splitting part (a) off is what lifts this above the 79%-zero outcome.

Predicted full-mark band. Part (a) 50–65%; part (b) 25–40% against 2023 Exam 2 Section B Q3g at 21% unsplit.

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N3. The iteration drawn, not computed

The design. A printed graph of a function with one visible root. Part (a), one mark: compute the first iterate from a stated starting value, to a stated accuracy. Part (b), one mark: draw, on the printed axes, the tangent to the graph at the first iterate. Part (c), one mark: mark and label the second iterate on the horizontal axis, without further computation.

The discriminator. Part (c). The geometric meaning of the method — each iterate is where the tangent at the previous one crosses the axis — is the only content, and the archive shows that students who can execute the formula cannot draw it. Part (b) is marked on the tangent's position and gradient, not on its equation, so a student who computes the equation and plots two points is doing more work than the mark requires.

The evidence. 2025 Exam 2 Section B Q4eii — "On the axes in part d, draw the tangent to the graph of y = f(x) at the point where x = x₁" — 28% (1 mark). Compare 2025 Exam 2 Section B Q4d, the equation of a tangent at a named point inside the same question, at 70%. That 42-point gap between computing a tangent and drawing one is the whole design. Supporting evidence on drawn-answer parts: 2024 Exam 1 Q7c, sketching a derivative graph with endpoints positioned off the axis, 15%; the 2024 report's general comment — "The graph of a derivative function needs to correspond to key points such as when the gradient is at a maximum, minimum or has a zero gradient value."

Placement and marks. Examination 2 Section B, 1 + 1 + 1 = 3 marks.

Predicted full-mark band. Part (a) 50–65% against 2023 Exam 2 Section B Q3f at 54%; parts (b) and (c) 20–35% each against 2025 Exam 2 Section B Q4eii at 28%.

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N4. The iteration that converges to the wrong root

The design. A function with three distinct roots, two of them close together. Part (a), two marks: three iterations from a starting value chosen so that the sequence converges to the far root rather than the nearest one, reported to a stated accuracy in a printed table. Part (b), one mark: state which root the sequence is approaching, and identify one root that no starting value in a stated interval will reach.

The discriminator. Students are taught that Newton's method finds "the" root and will report the nearest one regardless of what their own table says. Part (b) requires reading the table against the graph — a comparison, not a computation. This has never been examined in any form.

The evidence. No archive instance. The closest comparisons are the two ends of the known Newton range: 2023 Exam 2 Section B Q3f at 54% for a straight three-iteration table, and 2023 Exam 2 Section B Q3g at 21% for a one-mark reasoning part about the method's behaviour. The general pattern for "read your own output and draw a conclusion" is severe: the report for 2016 Exam 2 Section B Q3hii records a student solving the wrong equation and reporting a value "very different from" the correct one without noticing.

Placement and marks. Examination 2 Section B, 2 + 1 = 3 marks.

Predicted full-mark band. Part (a) 45–60% against 2023 Exam 2 Section B Q3f at 54%; part (b) 18–30% against 2023 Exam 2 Section B Q3g at 21%.

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N5. Three trapeziums with exact values, then the direction of the error

The design. A curve whose values at four equally spaced nodes are exact circular or surd quantities. Part (a), three marks: the trapezium-rule approximation over the interval, exact, combined into a single expression. Part (b), one mark: state whether the approximation exceeds or falls short of the exact area, with a reason referring to the gradient or the shape of the curve.

The discriminator. Part (a) is arithmetic and the archive says that is enough on its own — the doubling pattern for interior nodes, the exact node values, and the requirement to combine into one term rather than leaving a sum of separate trapezium areas. Part (b) is concavity, examined once in Section A and never as a written part, and it is the harder half.

The evidence. 2024 Exam 1 Q7a — 30% (3 marks), distribution {0: 28%, 1: 31%, 2: 11%, 3: 30%}; "some students did not apply it correctly, often writing [the sum] with the coefficient '2' missing from the middle two terms… It is expected that students will have a way of remembering the exact values… The arithmetic manipulation of fractions and surds presented a challenge for some students, with some leaving their answer as the sum of two or three separate area parts, instead of combining them into a single term." 2023 Exam 1 Q4 — the two-trapezium version — 45%; "Common errors involved incorrect values of h… Arithmetic manipulation errors (frequently) arose from dealing with the different denominators of the fractions." The concavity half: 2025 Exam 2 Section A Q6 — "For which function will the trapezium rule estimate be larger than the exact area?" — 50%; [SAMP] Examination 2 asks it in words — "Referring to the gradient of the curve, explain why a trapezium rule approximation would be greater than the actual cross-sectional area."

Placement and marks. Examination 1, 3 + 1 = 4 marks.

Predicted full-mark band. Part (a) 25–38% against 2024 Exam 1 Q7a at 30%; part (b) 25–40% against 2025 Exam 2 Section A Q6 at 50%, discounted because a written reason is harder than a four-option choice.

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N6. The trapezium rule with no rule at all

The design. A table of measured values at equally spaced points — a rainfall record, a flow rate, a heart-rate trace — and no function rule anywhere in the question. Part (a), two marks: the trapezium-rule estimate of the area under the implied curve. Part (b), one mark: state, in the units of the context, what that area represents. Part (c), one mark: state one reason the estimate could differ from the true value.

The discriminator. No rule means no CAS, on a technology-active paper. The design is legal precisely because the trapezium rule needs only node values, which is what the study design's phrase "approximation to the area under a curve" permits. Part (b) is the interpretation mark — the reports name units and meaning as a fault for 24 questions — and part (c) requires the concavity idea expressed without a formula.

The evidence. No archive instance: the trapezium rule has never been asked from a table and has never appeared in Section B of a November paper. The interpretation half has clear precedent: 2013 Exam 2 Section B Q1d — "For how many hours during the 24-hour time interval is T ≥ 26?" — 45%; "Some students just gave the values of t and did not find the difference between them." The report for 2009 Exam 2 Section B Q4cii names "Some students had the units as m/s or m." The 2024 Exam 2 general comments: "Some students misinterpreted Questions 2f.ii and 2f.iii and did not consider the area under the curve."

Placement and marks. Examination 2 Section B, 2 + 1 + 1 = 4 marks.

Predicted full-mark band. Part (a) 45–60% — the arithmetic is technology-assisted; part (b) 35–50%; part (c) 20–35% against 2025 Exam 2 Section A Q6 at 50% for the recognition-only version.

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N7. The loop that stops one pass early

The design. A Section A item printing a short algorithm containing a while loop whose body both updates a variable and prints it, with the update and the print in an order that determines whether the final value is printed. The options list the printed sequence with and without the final value, and with one value too many at the start.

The discriminator. When the loop stops. Every distractor is an off-by-one, and the correct answer requires evaluating the loop condition before each pass and noticing whether the print precedes or follows the update. The construction has exactly one November data point and it is comfortable; a design that moves the print to the other side of the update should not be assumed to stay comfortable.

The evidence. 2025 Exam 2 Section A Q7 — 72%, the only November pseudocode item in existence. Its trap is an option containing a value the loop condition prevents from ever being printed. The 2024 NHT Exam 2 Section A Q14 uses a tolerance test inside the loop; the 2025 NHT Exam 2 Section A Q18 uses a printed value. [SAMP] Examination 2 Question 7 is the harder completion form — an incomplete Newton implementation with two missing lines, the candidate lines differing only in which variable is returned and which difference is tested.

Placement and marks. Examination 2 Section A, one mark. Under the four-option format, keep the two off-by-one variants and the one-too-many variant and drop the fourth.

Predicted full-mark band. 50–70% for a trace item against 2025 Exam 2 Section A Q7 at 72%; 30–45% for the completion form, which has no state data and is structurally closer to 2023 Exam 2 Section A Q13, a Newton-in-a-loop item, at 52%.

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N8. The algorithm inside Section B

The design. A Section B question whose opening parts establish a function and a bounded region in the ordinary way. A later part prints an algorithm implementing a numerical approximation to that region's area. Part (i), one mark: state what the algorithm returns for a stated input. Part (ii), two marks: state what single line would have to change for the algorithm to double the number of strips, and what effect that would have on the accuracy.

The discriminator. Pseudocode has never appeared in Section B of any paper, so no cohort has rehearsed reading an algorithm as mathematics rather than as a trace exercise. Part (ii) is the separator: it requires connecting the loop counter to the strip width and the strip width to the error, which is the trapezium-rule concavity idea arriving through a different door.

The evidence. No Section B instance exists. 2025 Exam 2 Section A Q7 (72%) and 2023 Exam 2 Section A Q13 (52%) bound the trace half. The modification half has no precedent at all; the nearest comparisons are reasoning parts that follow computational parts, and those run low — 2023 Exam 2 Section B Q3g 21%, 2018 Exam 2 Section B Q4dii 11%, 2022 Exam 2 Section B Q3ciii 28%.

Placement and marks. Examination 2 Section B, 1 + 2 = 3 marks.

Predicted full-mark band. Part (i) 45–60% against 2023 Exam 2 Section A Q13 at 52%; part (ii) 20–35% against 2022 Exam 2 Section B Q3ciii at 28%.

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N9. Two equations, three unknowns

The design. A system of two linear equations in three variables, arising from a described context so that the variables mean something. Part (a), one mark: explain why the system cannot have a unique solution. Part (b), two marks: give the general solution in parameter form, with the parameter's set stated. Part (c), one mark: state one additional equation that would make the solution unique.

The discriminator. No November paper has ever asked this and no textbook routine covers it. Part (b) requires setting one variable equal to a parameter and back-substituting — mechanically easy, conceptually unfamiliar, and the archive says the state will not leave a letter in an answer. Part (a) is the reset; part (c) checks that the student understands why the family exists rather than how to write it. Note the constraint: the answer to (a) must be algebraic, because [SD] requires geometric interpretation only for two equations in two variables.

The evidence. [SAMP] Examination 2 Question 1 is the only official instance and it is multiple choice, with options differing by a single constant. No November instance exists. The comparisons for parameter-form answers are severe: 2016 Exam 2 Section B Q4ei — coordinates in terms of a parameter — 24%; 2015 Exam 1 Q10a — a point in terms of an angle — 20%; 2022 Exam 2 Section A Q19 — a fully literal optimisation — 34%. 03-algebra-number.md §5.1 names the mechanism: "the difficulty is not algebraic; it is that students are unwilling to leave a letter in an answer."

Placement and marks. Examination 1, 1 + 2 + 1 = 4 marks. Technology-free is the right home; on Examination 2 a CAS returns the parameter form directly.

Predicted full-mark band. Part (a) 45–60%; part (b) 20–35% against 2016 Exam 2 Section B Q4ei at 24%; part (c) 30–45%.

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N10. What happens to the graph when the parameter moves

The design. A probability mass function or density whose defining parameter appears in the stem. Part (a), one mark: state, with a reason, what happens to the mean as the parameter increases over a stated range. Part (b), one mark: state what happens to the variance over the same range. Part (c), one mark: sketch, on one set of axes, the shape of the distribution at the two ends of the range.

The discriminator. The dot point — "effect of variation in the value(s) of defining parameters on the graph of a given probability mass function", and its identical twin for densities — appears in the study design twice and has essentially never been examined. Part (b) is the separator, because the direction in which the variance moves is not the same across the whole parameter range, so a student reasoning from a formula must also reason about which part of the range they are in.

The evidence. The only clean archive instance is 2026 Exam 2 NHT Section A Q10 — "Let X ~ Bi(n, p), where 0.5 < p < 1. For a particular value of n, as p increases…" — which carries no state data. The nearest November comparisons are parameter-reasoning items: 2017 Exam 2 Section A Q18, where the mean and standard deviation are set equal and the sample size must be recovered, at 38%; 2016 Exam 2 Section A Q19 at 15%; 2020 Exam 2 Section A Q19 at 15%. Sketching a distribution: 2017 Exam 2 Section B Q3a at 29%.

Placement and marks. Examination 2 Section B, 1 + 1 + 1 = 3 marks, or Section A as a single "which of the following must be true" item.

Predicted full-mark band. Part (a) 45–60%; part (b) 20–35% against 2017 Exam 2 Section A Q18 at 38%; part (c) 25–40% against 2017 Exam 2 Section B Q3a at 29%.

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N11. The logarithm the papers have never used

The design. A modelling context whose natural scale is base ten — sound intensity, acidity, magnitude, a decade-scaled population. Part (a), one mark: evaluate the model at a value chosen so the base-ten logarithm is an integer. Part (b), two marks: solve an equation in the model, exactly, by hand. Part (c), one mark: state the transformation mapping the base-ten graph onto the natural-logarithm graph of the same argument.

The discriminator. log₁₀ is named in the study design's function list and has appeared in exactly one paper in the corpus — the 2017 NHT Examination 2 — and in no November paper in twenty years. Part (b) fails because students reach for the natural logarithm out of habit. Part (c) is the interesting one: the two logarithm graphs differ by a vertical dilation whose factor is the change-of-base constant, which is a transformation question wearing a logarithm's clothes.

The evidence. No November instance exists. The change-of-base half has precedent: 2019 Exam 2 Section A Q20 — an expression combining two logarithms in different bases — 47%; 2013 Exam 2 Section A Q18 — which logarithm identity holds for all positive values — 35%, with a near-uniform five-way distribution (A 12%, B 35%, C 22%, D 17%, E 13%) that 03-algebra-number.md describes as "a type the state guesses". The by-hand equation half compares to 2013 Exam 1 Q5a, a base-three logarithm equation, at 54%.

Placement and marks. Examination 1, 1 + 2 + 1 = 4 marks.

Predicted full-mark band. Part (a) 55–70%; part (b) 35–50% against 2013 Exam 1 Q5a at 54%, discounted for the unfamiliar base; part (c) 20–35% against 2013 Exam 2 Section A Q18 at 35%.

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4. Designs that cross areas of study

The archive's clearest structural finding is that the cohort's loss is concentrated where a result must be carried from one area into another. 03-algebra-number.md opens with the study design's own sentence about this — "This content is to be incorporated as applicable to the other areas of study" — and records that 89 of the 195 algebra-tagged parts sit inside Section B questions whose headline subject is something else. The reports' sixth standing complaint is that students restart instead of using the previous part.

Ten designs. Each names the two areas and the object that has to cross between them.

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X1. Calculus applied to a probability density — the standard deviation of a hybrid

The areas. Statistics provides the object; calculus does the work.

The design. A two-piece density on a bounded support, with the pieces of different functional type. Part (a), one mark: verify the normalising constant, using two integrals. Part (b), two marks: the mean. Part (c), three marks: the standard deviation, correct to a stated accuracy.

The discriminator. Three carries. The support must be split at the join for every integral, not just the first. The mean requires the variable inside the integrand — the single most-named omission in the area. The standard deviation requires the variance formula and then the square root, and the archive says the cohort stops at the variance.

The evidence. 2021 Exam 2 Section B Q4g — the standard deviation of a hybrid density — 32% (3 marks); "Some students worked out the variance instead of the standard deviation." 2021 Exam 2 Section B Q4f — the median of the same hybrid — 37%. 2013 Exam 2 Section B Q2ci — the mean of a hybrid — 46%. 2025 Exam 2 Section B Q3aii — a standard deviation on a single-piece density — 65%; "while they worked out the variance, they did not proceed to compute the standard deviation." The gap between 65% and 32% is what the join costs.

Placement and marks. Examination 2 Section B, 1 + 2 + 3 = 6 marks.

Predicted full-mark band. Part (c) 25–38% against 2021 Exam 2 Section B Q4g at 32%.

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X2. Calculus applied to a binomial probability

The areas. Statistics provides the polynomial; calculus optimises it.

The design. Open in a calculus register: define a polynomial, find its stationary points, sketch it. Then reveal, in a later part, that the same polynomial is the probability of a stated event for a binomial random variable with a small fixed number of trials and a free success probability, and ask the student to show that the two agree. Close by asking for the success probability maximising that event's probability, using the stationary points already found.

The discriminator. Recognising that a probability can be a polynomial, and that the stationary point found three parts earlier is the answer. The design is a deliberate reset in the sense of §1.4: the calculus is banked before the statistics arrives, so a student who cannot make the crossing still holds the opening marks.

The evidence. VCAA has already built it. 2025 Exam 2 Section B Q1 defines a quartic, works it through four calculus parts, and at part (f) asks students to show that a binomial tail probability equals that quartic — 45%; "This was a 'show that' question and appropriate working needed to be shown. Some students worked out Pr(X = 3) instead of Pr(X ≥ 3)." Its calculus opener Q1a scored 91%. The pure-statistics versions of the same idea are much harder: 2017 Exam 2 Section B Q3f 32%, Q3gi 23%, 2007 Exam 2 Section B Q5e 25%, 2012 Exam 2 Section B Q3c 13%.

Placement and marks. Examination 2 Section B, the whole question, with the crossing part worth 2 marks and the optimisation worth 2.

Predicted full-mark band. Crossing part 35–50% against 2025 Exam 2 Section B Q1f at 45%; optimisation 20–32% against 2017 Exam 2 Section B Q3gi at 23%.

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X3. A transformation applied to a definite integral

The areas. Functions provides the transformation; calculus provides the integral.

The design. Part (a), one mark: the value of a definite integral of a named function over a stated interval, computed or given. Part (b), two marks: the value of the integral of a transformed version of the same function over the correspondingly transformed interval, with the transformation described in words rather than in symbols. Part (c), one mark: the single transformation that would leave the integral's value unchanged.

The discriminator. The area scales by the reciprocal of the horizontal dilation factor and is unaffected by either translation; a vertical dilation scales it directly. The cohort applies one of the three rules and not the others. Part (c) inverts the question and tests whether the rule is understood or memorised.

The evidence. 2016 Exam 2 Section A Q20 — a transformed definite integral combining a reflection, a vertical dilation and a vertical translation — 17%, with 30% of the state choosing a single distractor obtained by adding the vertical translation once instead of across the interval width. 2020 Exam 2 Section A Q9 — 35%; the report's entire working is "Dilate by a factor of ½ from the y-axis. Translating 2 units to the left does not change the area." 2010 Exam 2 Section A Q20 — 25%. 2019 Exam 2 Section B Q3e — find four transformation parameters so that transformed integrals reproduce a given area — 10%.

Placement and marks. Examination 2 Section B, 1 + 2 + 1 = 4 marks, or Section A as a single item.

Predicted full-mark band. Part (b) 20–35% against 2020 Exam 2 Section A Q9 at 35% and 2016 Exam 2 Section A Q20 at 17%; part (c) 30–45%.

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X4. Fit the transformation parameters to a named target

The areas. Functions provides the family; algebra provides the equating.

The design. A cubic or quartic in factored form, and a second function asserted to be the first shifted vertically by an unknown amount, whose factored form has a repeated root. Part (a), two marks: find the vertical shift. Part (b), two marks: find all possible pairs of roots consistent with that shift.

The discriminator. The repeated root is a condition, not a fact to be read off: it says the shifted curve is tangent to the horizontal axis, which pins the shift to a turning-point ordinate. The cohort tries to expand and match, generating a quartic identity it cannot manage. The word "all" in part (b) is the second discriminator — there are two answers, one at each turning point, and the archive says students give one.

The evidence. 2023 Exam 2 Section B Q1d — "Let h(x) = (x − a)(x − b)², where h(x) = f(x) + k. Find the possible values of a and b" — 13% (4 marks), the highest-value separator in the modern functions archive; "Those who used method 1 [equating coefficients] were generally successful. Those who used method 2 [using transformations] often had sign errors… Some students only gave one set of values for a and b." 2019 Exam 1 Q2c — find the translation making a function equal its own inverse — 24%. 2013 Exam 2 Section B Q1fi — 10%; Q1fii — 12%.

Placement and marks. Examination 2 Section B, 2 + 2 = 4 marks.

Predicted full-mark band. 12–25% against 2023 Exam 2 Section B Q1d at 13%, with the split into two named parts worth a few points.

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X5. The solution count decided by a turning point

The areas. Calculus locates the turning point; algebra counts the solutions.

The design. Two curves whose intersections depend on a parameter, with the difference function's turning point computable by differentiation. Part (a), two marks: the coordinates of that turning point in terms of the parameter. Part (b), one mark: the parameter value at which the two curves are tangent. Part (c), two marks: the values of the parameter giving exactly two, and exactly four, intersections.

The discriminator. The crossing itself. A student who treats part (a) as a calculus exercise and part (c) as an algebra exercise will not connect them, and the report for the archive's version says exactly that. Part (b) is the bridge, deliberately isolated as a one-mark part so that the connection is stated rather than left to be discovered.

The evidence. 2025 Exam 1 Q9bii — 4% (2 marks); "Few students used a 'hence' approach and thus did not identify a connection between the turning point and the number of solutions." 2025 Exam 1 Q9bi, the turning-point-in-terms-of-the-parameter part, 22%. 2018 Exam 1 Q8b — 3%. 2025 Exam 2 Section B Q4fiii — 10%. 2023 Exam 2 Section A Q14 — the number of tangents through a named point — 29%.

Placement and marks. Examination 1, final question, 2 + 1 + 2 = 5 marks.

Predicted full-mark band. Part (a) 20–32% against 2025 Exam 1 Q9bi at 22%; part (b) 15–28%; part (c) 10–22% against 2025 Exam 1 Q9bii at 4% unbridged.

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X6. The density that is also a hybrid function

The areas. Functions provides continuity at a join; statistics provides the normalisation.

The design. A density defined in two pieces with two unknown constants — one fixed by continuity at the join, the other by the requirement that the total area is one. Part (a), two marks: both constants. Part (b), two marks: a probability over an interval straddling the join. Part (c), one mark: state whether the density is smooth at the join, with a reason.

The discriminator. Two conditions producing two constants, which the cohort treats as one condition producing one. Part (b) then fails on the split: the archive records students integrating a single piece across the whole interval. Part (c) is a free extra discriminator — a density must be continuous to be sensible but need not be smooth, and that is exactly the distinction 2023 Exam 2 Section A Q9 tests.

The evidence. 2022 Exam 2 Section B Q3bii — find three coefficients of a density from three stated conditions — 34%. 2023 Exam 2 Section A Q9 — continuity and smoothness giving two equations — 42%. 2021 Exam 2 Section B Q4f, the median of a hybrid density, 37%, and Q4g, its standard deviation, 32%. 2015 Exam 2 Section B Q3ai, a probability on a single-piece density, 80% — the join is worth roughly forty points.

Placement and marks. Examination 2 Section B, 2 + 2 + 1 = 5 marks.

Predicted full-mark band. Part (a) 25–40% against 2022 Exam 2 Section B Q3bii at 34%; part (b) 35–50%; part (c) 35–50% against 2023 Exam 2 Section A Q9 at 42%.

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X7. The area between a function and its own inverse

The areas. Functions provides the reflection; calculus provides the integral.

The design. An increasing function on a restricted domain, together with its inverse. Part (a), two marks: the coordinates where the two graphs meet. Part (b), one mark: write down a definite integral giving the area enclosed between them — the expression only. Part (c), two marks: evaluate it exactly, using the symmetry about the line y = x.

The discriminator. Part (a) rewards the shortcut — for an increasing function the intersections lie on y = x — and punishes the student who sets the two rules equal and produces an intractable equation. Part (c) rewards symmetry: the region is symmetric about y = x, so the area is twice the area between one curve and that line. The archive records students integrating both curves separately and failing on the arithmetic.

The evidence. 2017 Exam 2 Section B Q4c — area between a function and its inverse, exact — 49%. 2010 Exam 2 Section B Q1aiv — 41%. 2020 Exam 1 Q6c — the technology-free version in a demanded surd form — 10% (4 marks), distribution {0: 28%, 1: 22%, 2: 23%, 3: 18%, 4: 10%}, a genuine five-band spread. 2019 Exam 2 Section B Q5f, the parameter version — 4%. The intersection shortcut: 2010 Exam 2 Section B Q1aiii — 66%; 2020 Exam 1 Q8dii, the non-intersection version — 3%; "Some students tried to algebraically find the point of intersection of the graphs of function and its inverse function, with limited progress." The "write down the integral" reset: 2020 Exam 2 Section B Q1eii — 76%.

Placement and marks. Examination 1, 2 + 1 + 2 = 5 marks.

Predicted full-mark band. Part (a) 45–60% against 2010 Exam 2 Section B Q1aiii at 66%; part (b) 55–70% against 2020 Exam 2 Section B Q1eii at 76%; part (c) 15–28% against 2020 Exam 1 Q6c at 10% with no scaffold.

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X8. Newton's method on a function the earlier parts built

The areas. Calculus builds the function; algebra runs the algorithm; functions supplies the graph to read it off.

The design. A Section B question whose parts (a)–(c) establish a function through differentiation and a sketch. Part (d), one mark: state why the function has a root in a named interval. Part (e), two marks: two iterations from a stated starting value, to a stated accuracy. Part (f), one mark: mark the second iterate on the sketch from part (c), and state whether the sequence is approaching the root from above or below.

The discriminator. The carry. The derivative needed for the iteration is the one computed in part (b) — a student who differentiates again will usually differentiate correctly and lose time, and one who used a CAS for part (b) without storing the result will lose more. Part (f) requires the graph from part (c) to be used, which the archive's sixth standing complaint says the cohort will not do.

The evidence. 2025 Exam 2 Section B Q4 is the template: part (d) prints axes and asks for the equation of the tangent at a named point (70%), part (e.i) asks for the second Newton iterate (65%), and part (e.ii) asks for that tangent to be drawn on the part (d) axes (28%). 2023 Exam 2 Section B Q3 runs the same shape across parts (c) to (g): 52%, 15%, 58%, 35%, 54%, 21% — note the resets at 58% and 54%. The failure-to-carry evidence: 2018 Exam 1 Q8d — "While students could equate their answer to part c. to [the target], many students did not use their result from part a." — 13%.

Placement and marks. Examination 2 Section B, 1 + 2 + 1 = 4 marks for parts (d)–(f).

Predicted full-mark band. Part (d) 45–60%; part (e) 45–60% against 2023 Exam 2 Section B Q3f at 54%; part (f) 20–35% against 2025 Exam 2 Section B Q4eii at 28%.

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X9. The approximation, the exact value, and the reason for the gap

The areas. Algebra supplies the trapezium rule; calculus supplies the exact integral; functions supplies the concavity.

The design. Part (a), two marks: the trapezium-rule estimate of an area with a stated number of strips. Part (b), two marks: the exact area, by integration. Part (c), one mark: state which is larger and why, referring to the shape of the curve. Part (d), one mark: state what would happen to the gap if the number of strips were doubled.

The discriminator. Part (c) is the crossing and is a reasoning mark, which the archive says is the worst-returning kind. Part (d) requires knowing that the error shrinks but does not vanish, and that it shrinks faster than the strip width does — a statement about the rule rather than about this curve. Parts (a) and (b) are independent resets, so a student who cannot reason can still bank four marks.

The evidence. No archive part combines them, which is the opportunity. The components: 2024 Exam 1 Q7a 30%; 2023 Exam 1 Q4 45%; 2025 Exam 2 Section A Q6 50%; [SAMP] Examination 2 asks the concavity reason in words as its own mark. The definite-integral half: 2016 Exam 1 Q3b 46%, 2009 Exam 1 Q2b 48%. The report constraint that makes part (a) safe from a CAS: "any attempt to calculate this area using integral calculus was not acceptable" — so parts (a) and (b) must be explicitly separated by method.

Placement and marks. Examination 1, 2 + 2 + 1 + 1 = 6 marks.

Predicted full-mark band. Part (a) 25–40% against 2024 Exam 1 Q7a at 30%; part (b) 40–55% against 2016 Exam 1 Q3b at 46%; part (c) 25–40% against 2025 Exam 2 Section A Q6 at 50%; part (d) 20–35%.

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X10. The integer parameter buried inside a probability

The areas. Calculus produces the probability; algebra inverts it; statistics uses it.

The design. A density with a parameter in one of its terminals, modelling a threshold — a delivery time, a tolerance, a cut-off. Part (a), two marks: the probability of exceeding the threshold, in terms of the parameter. Part (b), two marks: the probability of exceeding it at least once in a fixed small number of independent trials, in terms of the same parameter. Part (c), two marks: the integer value of the parameter making that compound probability equal a stated value to a stated accuracy.

The discriminator. A double inversion. The parameter sits inside an integral terminal, which produces a probability, which feeds a binomial complement, which must be solved back for an integer. Nothing can be checked numerically until the very end. The archive's version of part (c) scores 17%, and the report's complaint is not about the mathematics but about the absence of working.

The evidence. 2025 Exam 2 Section B Q3biv — 17% (2 marks); "An integer value was required. Many students tried to solve [the equation directly]. Some students correctly used trial and error. Others just gave the answer, without showing appropriate working as required." 2025 Exam 2 Section B Q3bii, the forward direction with the parameter fixed, 62% — the reset. 2020 Exam 2 Section B Q3ei 24% and Q3eii 23% are the same inversion without the integral. 2017 Exam 2 Section B Q3gii — invert a density quantile using a probability optimised two parts earlier — 7%.

Placement and marks. Examination 2 Section B, 2 + 2 + 2 = 6 marks.

Predicted full-mark band. Part (a) 45–60%; part (b) 50–65% against 2025 Exam 2 Section B Q3bii at 62%; part (c) 15–28% against 2025 Exam 2 Section B Q3biv at 17%.

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5. How to use these

5.1 For a teacher writing a SAC or a trial examination

Build the paper to the archive's difficulty profile, not to instinct. 01-study-design.md §10.4 gives the target: across the current design, 30–40% of every paper consists of marks the state earns less than 40% of, and 8–33% consists of marks the state earns at least 70% of. A trial paper with no gettable marks is not a harder paper; it is an uninformative one. On a 40-mark technology-free trial, budget roughly ten marks the whole class should get, eighteen in the middle, and twelve that separate.

Use the staircase, and build the resets. The measured gradient is 63.5% at a Section B opener and 20.0% at its close, with median 67% and 16%. The archive's real questions are not monotone — 2024 Exam 2 Section B Q4 runs 79 → 85 → 71 → 52 → 36 → 11 → 33 → 18 → 56 → 8 — and the jumps back up are where a "show that" or a fresh sub-scenario re-supplies the input. Every design above with more than two parts has at least one reset marked. Put them in. A question whose parts are strictly dependent converts one hard idea into six blank pages and grades nothing.

Choose the mark value deliberately. The table in §1.1 is the cleanest guide in the corpus: one mark → 36% separator rate, two marks → 64%, three marks → 86%, four marks → 90%. If you want a part that grades, give it three marks and write a mark scheme with three identifiable steps. If you give a single non-obvious idea one mark, you will get a binary outcome and learn nothing — which is what the 97%-zero and 94%-zero distributions in §1.4 are.

Write the discriminating version, not the defeating one. For each design above, the entry names the scaffold that makes it discriminate: part (a) of A3 (the turning point), part (a) of A6 (the endpoint values), part (a) of S1 (the area invariant), part (a) of C1 (the domain), the split in S6, the printed table in C12. Remove those and you reproduce the 2%–8% archive originals, which grade nobody.

Pick from the list by what you want to diagnose.

Mark the working, not the answer, on anything above one mark. Every report in the corpus invokes the instruction "In questions where more than one mark is available, appropriate working must be shown", and 07-exam-craft.md ranks insufficient working ninth among all named faults, with 26 questions plus every "show that". Designs A10, N5 and X9 are worthless if a correct answer with no route earns full marks.

Do not import retired content. Matrix transformations and functional relations are out of scope, and between them they account for eighteen of the 233 separators in the functions area. If you are lifting a question from a pre-2023 paper, check it against 02-functions-graphs.md §4 and 03-algebra-number.md §1.3 first.

On accuracy, be as ruthless as VCAA. Rounding is the most-named fault in the entire corpus — 88 questions — and approximate-where-exact is second with 78. Several designs above (A11, S10, X10, N5) are built specifically around it. Say what you want, once, in the stem, and then mark it.

5.2 For a student who wants to know what they are not prepared for

The arithmetic you are avoiding is the content. Look again at C3, A2, S3 and N5. In each, every report attributes the failure not to the concept but to the execution — exact circular values, surds, prime factorisation, fractions with different denominators, and the cubic that factorises by inspection while half the state reaches for a formula. The 2017 Examination 1 report lists five questions in one paper whose common failure was fractions. If you cannot do the arithmetic without a calculator, you do not know the topic, whatever your SAC score says.

Six sentences you must be able to write on demand. None is on the formula sheet. The average value of a function over an interval. The average rate of change over an interval. The condition for a composite to exist. The condition for an inverse to exist. Why a trapezium-rule estimate exceeds or falls short of the true area. Why a stationary point breaks Newton's method. Designs C3, F3, F2, N5 and N2 exist because the state cannot write these.

The last part is not optional. Section B closing parts average 20.0% full marks and are the most-skipped parts on the paper. 03-algebra-number.md §4.15 shows that on the hardest algebra items 80% to 96% of the state scores zero, and the dominant failure is non-attempt, not error. Consequential-error marking means a partially wrong attempt at a final part frequently scores something. A rough sketch and one sentence of reasoning, on a one-mark part at the end of a fifteen-mark question, is worth ninety seconds.

Three habits that convert more marks than any topic revision.

1. Write the domain before you differentiate, solve or define anything, and check every answer against it. Domain is the third-most-named fault in the corpus and it is the entire content of designs F2, F3, F4, A2, C1 and S9.
2. Read the noun in the question. "The gradient of the hill", not "the hill". "The standard deviation", not "the variance". "The maximum value", not "the coordinates of the turning point". "Both", "all", "total", "exactly", "positive". Design C2 scored 3% in the archive on one noun.
3. Use the previous part. "Hence" is compulsory; a correct independent method can score zero. When the previous part is a graph, look at it. When it is a derivative, it is the derivative you need next. Designs C8, X5, X7 and X8 are built from the archive's records of the state restarting instead.

Know which constructions you have probably never met. If you have done every past paper from 2016 onwards, you have seen Newton's method three times, the trapezium rule three times, pseudocode once, an under-determined system never, a base-ten logarithm never, and a question about the effect of a parameter on the shape of a distribution never. All six are examinable. Section 3 describes eleven designs in that space; if none of them looks familiar, that is the finding.

Use the mark allocation as the instruction it is. [SPEC] says so directly: "Students should use command/task words, other instructional information within questions and corresponding mark allocations to guide their responses." A one-mark part wants an answer. A three-mark part wants three identifiable steps, and if you have written one line you have not finished. A "show that" prints the answer, so every mark is in the route — and the three documented ways to lose all of them are assuming the result, starting from "LHS = RHS", and verifying with one particular value.

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Appendix — every design, with its evidence anchor

Sixty-nine designs. Forty-eight by area of study — twelve each — eleven exploiting the 2023–2027 content, ten crossing areas of study. Every anchor percentage above is the published state full-mark rate from corpus/mm/questions.json.