Algebra, number and structure

The definitive reference. Built from corpus/mm/questions.json (every graded question part 2006–2025 with its state mark distribution), the paper texts in corpus/mm/text/, the VCAA assessment reports and assessment guides, and research/mm/01-study-design.md. Nothing here is inferred from general mathematical knowledge; every claim is tied to a ref, a quoted report, or a quoted study-design clause.

Written for students targeting 45+, and for the people building their material.

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0. Provenance and the numbers you can trust

Three data-quality rules carried over from 01-study-design.md §0.1 and applied throughout:

1. 2011 is unusable. Both 2011 reports were OCR'd from degraded scans; question labels are duplicated (2011 Exam 1 Q3b–3b appears twice with different percentages) and half the mark distributions are missing. The seven 2011 algebra separators are listed in §4 for completeness and excluded from every statistic.
2. NHT papers carry no state statistics. pct is null for 2024 NHT and 2025 NHT. They are quoted as content evidence only.
3. The 2024 November paper texts are empty in the corpus (the source PDFs are image-only). 2024 November wordings below are reconstructed from [RPT] and [AG], and are flagged as such.

pct throughout means the percentage of the state that earned full marks on that part. A "separator" is a part with pct ≤ 50.

0.1 How hard this area is, relative to the rest of the course

Algebra is the hardest of the four areas on every measure in the archive. It is also the smallest: about 21% of the marks, but it supplies a disproportionate share of the parts where the state collapses.

By era (November papers only, 2011 excluded):

The 2016–2022 design was the low point. The 2023 design has not made this area easier — it has changed what is hard (see §1.3).

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1. What the study design puts in this area

1.1 The overview and the content dot points, verbatim

[SD], the area-of-study overview in full:

"In this area of study students cover the algebra of functions, including composition of functions, inverse functions and the solution of equations. They also study the identification of appropriate solution processes for solving equations, and systems of simultaneous equations, presented in various forms. Students also cover recognition of equations and systems of equations that are solvable using inverse operations or factorisation, and the use of graphical and numerical approaches for problems involving equations where exact value solutions are not required, or which are not solvable by other methods. This content is to be incorporated as applicable to the other areas of study."

The final sentence — "This content is to be incorporated as applicable to the other areas of study" — is the most consequential sentence in the area. It is VCAA saying that algebra is not a topic with its own questions; it is the thing every other question is made of. The [QJSON] tagging bears this out: of the 195 algebra-tagged parts, 89 sit inside Section B extended-response questions whose headline subject is a calculus or functions context, 48 are Section A multiple choice, and 58 are Exam 1 short-answer. There is no such thing as an "algebra question" in the way there is a "statistics question".

The content dot points, verbatim:

This area of study includes:
- solution of polynomial equations with real coefficients of degree n having up to n real solutions, including numerical solutions
- functions and their inverses, including conditions for the existence of an inverse function, and use of inverse functions to solve equations involving exponential, logarithmic, circular and power functions
- composition of functions, where f composite g, f ∘ g, is defined by (f ∘ g)(x) = f(g(x)) given r_g ⊆ d_f
- solution of equations of the form f(x) = g(x) over a specified interval, where f and g are functions of the type specified in the 'Functions, relations and graphs' area of study, by graphical, numerical and algebraic methods, as applicable
- solution of literal equations and general solution of equations involving a single parameter
- solution of simple systems of simultaneous linear equations, including consideration of cases where no solution or an infinite number of possible solutions exist (geometric interpretation only required for two equations in two variables).

1.2 The key knowledge and key skills that live here

These sit in Outcome 1. [SD], key knowledge:

• "exponent laws and logarithm laws"
• "analytical, graphical and numerical approaches to solving equations and the nature of corresponding solutions (real, exact or approximate) and the effect of domain restrictions"
• "the concept of an inverse function, connection between domain and range of the original function and its inverse relation and the conditions for existence of an inverse function, including the form of the graph of the inverse function for specified functions"
• "the concept of combined functions, and the connection between domain and range of the functions involved and the domain and range of the combined functions"
• "the concept of approximation to the area under a curve using the trapezium rule, the ideas underlying the fundamental theorem of calculus and the relationship between the definite integral and area"
• "representations of points and transformations of the plane"

[SD], key skills:

• "apply a range of analytical, graphical and numerical processes (including the algorithm for Newton's method), as appropriate, to obtain general and specific solutions (exact or approximate) to equations (including literal equations) over a given domain and be able to verify solutions to a particular equation or equations over a given domain"
• "solve by hand equations of the form sin(ax + b) = c, cos(ax + b) = c and tan(ax + b) = c with exact value solutions over a given interval"
• "apply algebraic, logarithmic and circular function properties to the simplification of expressions and the solution of equations"
• "find the rule of an inverse function and give its domain and range"
• "find the rule of a composite function and give its domain and range"
• "evaluate approximations to the area under a curve using the trapezium rule, find and verify anti-derivatives of specified functions and evaluate definite integrals"

Outcome 2 contributes one key-knowledge point, examinable only in Examination 2:

• "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"

Four phrases in that list do nearly all the work in the examinations:

• "general and specific solutions (exact or approximate)" — the exact/approximate distinction, which the reports name more often than any other single fault in this area (22 of the 195 parts have a report comment containing the word "exact").
• "(including literal equations)" — the licence for every "find h in terms of r" question.
• "including the algorithm for Newton's method" — the entire study-design authority for Newton's method, sitting inside a key skill rather than a content dot point.
• "solve by hand … with exact value solutions over a given interval" — the technology-free trigonometric guarantee. It names three forms and nothing else.

1.3 What changed in 2023 — the two movements that matter

Numerical algorithms moved in. The 2023 design added "including numerical solutions" to the polynomial dot point, added "including the algorithm for Newton's method" to the equation-solving key skill, replaced "approximation to the area under a curve using rectangles" with "using the trapezium rule", and added the pseudocode key-knowledge point to Outcome 2. [FS] gained the two matching formulas at the same time:

Newton's method x_n+1 = x_n − f(x_n) / f′(x_n)

trapezium rule approximation Area ≈ ((x_n − x₀)/2n) [ f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(x_n−2) + 2f(x_n−1) + f(x_n) ]

— [FS], as printed on the 2026 NHT Exam 1 formula sheet

Where these have actually been examined, from a complete scan of [PAPERS]:

Two facts in that table are worth isolating. Newton's method has never yet appeared as a question in a live November Examination 1. In the 2023, 2025 and 2026 NHT Examination 1 papers the word "Newton" occurs only on the formula sheet. (The 2023 hits for "trapezium rule" in the Exam 1 and Exam 2 papers are likewise formula-sheet lines.) It is nonetheless examinable there — it is an Outcome 1 key skill, Examination 1 assesses Outcome 1, and [SAMP] Exam 1 Q6 is a worked technology-free example. Treat it as overdue rather than excluded. Pseudocode has never appeared in any Examination 1, consistent with it being Outcome 2 key knowledge and Examination 1 being Outcome 1 only — a reasoned inference, not a stated VCAA exclusion.

Matrix transformations moved out. [SD] 2016–2022 said "the matrix representation of points and transformations of the plane" and carried the key skill "apply matrices to transformations of functions and their graphs". The 2023 design deletes the word "matrix" from the first and deletes the second outright. The exam record matches:

• Matrix representations of a linear system were a recurring multiple-choice type: 2006 Exam 2 Section A Q4 (pct 59) and 2010 Exam 2 Section A Q5 (pct 80) both read "an equivalent matrix equation is"; 2012 Exam 2 Section A Q17 (pct 43) opens "A system of simultaneous linear equations is represented by the matrix equation".
• Matrix transformations were still live in 2020 and 2021: 2020 Exam 2 Section B Q5h defines T : R² → R², T([x, y]) = [[m, 0], [0, n]][x, y] + [h, k]; 2021 Exam 1 Q9 defines T([x, y]) = [[1, 0], [0, q]][x, y] and then asks, in part b.i (pct 6), for "the values of q for which the graph of h intersects with the unit circle at least once".
• No paper from 2022 onwards contains a matrix in any form. If you are drilling pre-2023 papers, skip every matrix item. The underlying questions survive — "for what k does this system have no solution" is still asked every year — but the determinant-as-matrix presentation does not.

One further deletion, carried over from 01-study-design.md §3.2 and worth repeating because it removes a whole multiple-choice genre: the dot point on functional relations (f(x + k) = f(x), f(xy) = f(x)f(y), and so on) is gone. 2018 Exam 2 Section A Q10 — "The function f has the property f (x + f (x)) = f (2x) for all non-zero real numbers x. Which one of the following is a possible rule for the function?" — is out of scope. So is 2015 Exam 2 Q18.

Retained and examined much harder: systems with no solution or infinitely many. [SAMP] Exam 2 Q1 goes beyond anything the pre-2023 papers asked, requiring the general solution of a three-variable, two-equation system in parameter form:

"x − 2y = 3
2y − z = 4
Which one of the following correctly describes the general solution to the system of linear equations given above?
A. x = k, y = ½(k − 3), z = k − 1, for all k ∈ R …"
— [SAMP] Exam 2, Question 1

Note the parenthesis in the dot point: "geometric interpretation only required for two equations in two variables". You can be asked to read a 2×2 system as two parallel or coincident lines. You cannot be asked to interpret three variables as planes.

1.4 What the exams actually test, with counts

All figures November-only, 2011 excluded, from [QJSON].

Read the volatility, not the average. The area is worth anything from 1 to 19 marks on a single paper. 2018 and 2022 Exam 2 each had a single algebra-tagged mark; 2016 Exam 2 had 19. What is stable is the Exam 1 presence: since 2012 every November Examination 1 has carried between 1 and 7 algebra parts, and in the current design that has settled at 4–5 parts for 10–11 marks — roughly a quarter of the technology-free paper.

The stable annual fixtures across the whole archive:

• A technology-free equation-solving question in Exam 1 — logarithmic, exponential, index-law or trigonometric. Present in every November Exam 1 from 2007 to 2025 without exception.
• A parameter-condition multiple-choice item in Exam 2 Section A — simultaneous equations, or a discriminant, or a solution count. 2006, 2007, 2008, 2009, 2010, 2012, 2014, 2017, 2019, 2022, 2023 and 2025 all carry at least one.
• A literal-equation step inside a Section B modelling question — "find h in terms of r", "show that V = …". Present in essentially every Section B optimisation or geometric context in the archive.

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2. The complete catalogue of question types

Thirty types. For each: the literal VCAA wording template quoted from a real paper, what is being tested underneath, the standard method, archive instances with ref and pct, typical mark value, and the traps the reports name.

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Type 1 — Simultaneous linear equations with a parameter: "no solution"

Wording template (2025 Exam 2 Section A Q4, verbatim):

"Consider the system of equations below containing the parameter k, where k ∈ R.
kx − 3y = k + 2
2x + (2k − 1)y = 6 + 2k
Find the value(s) of k for which this system has no real solutions.
A. k = −2 only B. k = 3/2 only C. k = −2 or 3/2 D. k ∈ R \ {−2, 3/2}"

Older phrasing (2006 Exam 2 Section A Q19): "The simultaneous linear equations (m − 2)x + 3y = 6 and 2x + (m − 3)y = m − 1 have no solution for …". 2014 Exam 2 Section A Q17: "The simultaneous linear equations ax − 3y = 5 and 3x − ay = 8 − a have no solution for …".

What it is really testing. Whether you know that "no solution" means parallel and distinct, and that the determinant condition alone is insufficient — it gives you candidate values, and each must then be tested.

Standard method. Write the system as a₁x + b₁y = c₁, a₂x + b₂y = c₂. Set a₁b₂ − a₂b₁ = 0 and solve for the parameter. For each root, substitute back: if the two equations become inconsistent (same left side, different right side) you have no solution; if they become identical you have infinitely many. Report only the roots giving the case asked for.

Instances. 2025 Exam 2 Section A Q4 — 55%. 2014 Exam 2 Section A Q17 — 50%. 2006 Exam 2 Section A Q19 — 35%. 2024 Exam 1 Q2 — 37% (3 marks). [SAMP] Exam 1 Q4 — "Find the values of a and b for which the simultaneous equations have no solutions", 4 marks.

Typical marks. 1 mark in Section A; 3–4 marks in Exam 1.

Traps. The 2024 Exam 1 report [RPT] is the fullest statement VCAA has made about this type:

"Students using the determinant method often arrived at [two values] and then did not justify which answer was the valid solution. Students who set the two initial equations equal to one another commonly found they had multiple variables to deal with and did not know how to solve for [the parameter]. … 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'."

And on the algebra itself:

"The quadratic was readily factorised by inspection, yet a large proportion of students used the quadratic formula or other techniques such as splitting the middle term and grouping."

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Type 2 — Simultaneous linear equations with a parameter: "infinitely many solutions"

Wording template (2022 Exam 1 Q3, verbatim):

"Consider the system of equations
kx − 5y = 4 + k
3x + (k + 8)y = −1
Determine the value of k for which the system of equations above has an infinite number of solutions."

Multiple-choice form (2008 Exam 2 Section A Q6): "ax + 3y = 0, 2x + (a + 1)y = 0 where a is a real constant, have infinitely many solutions for …". 2010 Exam 2 Section A Q7: "(m − 1)x + 5y = 7 and 3x + (m − 3)y = 0.7m have infinitely many solutions for …".

What it is really testing. The same determinant condition as Type 1, but now the surviving case is the consistent one — the two equations must be scalar multiples of each other, coefficients and constant term.

Standard method. Determinant zero gives candidates; substitute each; keep the value that makes the two equations identical. Equivalently, require all three ratios a₁/a₂ = b₁/b₂ = c₁/c₂ to be equal and solve; the value satisfying all three is the answer.

Instances. 2022 Exam 1 Q3 — 36% (3 marks). 2010 Exam 2 Section A Q7 — 47%. 2008 Exam 2 Section A Q6 — 45%. 2024 NHT Exam 1 Q2a: "Find the value of a for which there are infinitely many solutions" (no state data).

Typical marks. 1 mark MC; 3 marks Exam 1.

Traps. The 2022 Exam 1 report:

"There were multiple ways to approach this question. Students generally approached this by either equating gradients and y-intercepts separately, using a matrix/determinant method, forming ratios or attempting to solve simultaneously. These methods were met with varying degrees of success. 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."

Justification is part of the mark. In both 2022 and 2024 the assessment guide allocates one mark for producing the quadratic and a separate mark for identifying which root is valid.

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Type 3 — Simultaneous linear equations with a parameter: "unique solution"

Wording template (2007 Exam 2 Section A Q5, verbatim):

"The simultaneous linear equations
mx + 12y = 24
3x + my = m
have a unique solution only for
A. m = 6 or m = −6 B. m = 12 or m = 3 C. m ∈ R \ {−6, 6} D. m = 2 or m = 1 E. m ∈ R \ {−12, −3}"

2009 Exam 2 Section A Q1: "kx − 3y = 0, 5x − (k + 2)y = 0 where k is a real constant, have a unique solution provided …".

What it is really testing. That the answer is a set complement, not a value. The distractors are always the determinant roots themselves.

Standard method. Determinant ≠ 0. Solve a₁b₂ − a₂b₁ = 0, then write R \ {roots}.

Instances. 2007 Exam 2 Section A Q5 — 36%. 2009 Exam 2 Section A Q1 — 49%. 2024 NHT Exam 1 Q2b: "Find the values of a for which there is a unique solution."

Typical marks. 1 mark.

Traps. The distribution for 2007 Q5 is the diagnosis: A 18%, B 24%, C 36%, D 10%, E 10%. Nearly a quarter of the state gave the determinant roots (B) rather than the complement. The word "only" in "have a unique solution only for" is doing the set-complement work and is routinely read past.

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Type 4 — The general solution of an under-determined system

Wording template ([SAMP] Exam 2 Q1, verbatim):

"x − 2y = 3
2y − z = 4
Which one of the following correctly describes the general solution to the system of linear equations given above?
A. x = k, y = ½(k − 3), z = k − 1, for all k ∈ R …"

Reasoning form (2025 NHT Exam 2 Section A Q20, from [RPT] commentary; paper text unavailable):

"To have one unique solution, the gradients have to be different … For an infinite number of solutions, the gradients and y-intercepts have to be the same … If these four linear equations are combined into a new system of four linear equations, there will be one solution."

What it is really testing. That a system with more unknowns than equations has a one-parameter family of solutions, and that the family is written by setting one variable to k and expressing the others in terms of k.

Standard method. Set the free variable equal to k. Back-substitute. Check every component of each option against a single substituted value — the distractors differ only in a constant.

Instances. [SAMP] Exam 2 Q1 (the only official worked example). 2025 Exam 2 Section A Q20 (NHT) — no state data. No November paper has yet asked it.

Typical marks. 1 mark.

Traps. This is the clearest example in the subject of content that is in the study design and in the sample questions but has not yet appeared in a November paper. Options differ by a single constant (z = k − 1 versus z = k − 7); substituting one value of k into all three components settles it in ten seconds.

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Type 5 — Simultaneous equations recovered from points on a curve ("show that")

Wording template (2025 Exam 2 Section B Q2a, verbatim):

"Let f : R → R, f (x) = x/2 + 7 and g : R → R, g(x) = Ae^{kx} where A, k ∈ R. The graphs of y = f (x) and y = g(x) intersect at the points (−12, 1) and (2, 8) …
a. Write down two simultaneous equations in terms of A and k. Solve them, using algebra, to show that A = […] and k = (3/14) log_e(2)." (the exact form of A is an index expression lost in the text extraction; the command words and k are legible)

What it is really testing. Eliminating a multiplicative constant by dividing the two equations rather than substituting — and doing it by hand when the question says "using algebra".

Standard method. Ae^{kx₁} = y₁, Ae^{kx₂} = y₂. Divide: e^{k(x₂ − x₁)} = y₂/y₁, so k = log_e(y₂/y₁)/(x₂ − x₁). Substitute back for A. Show every line.

Instances. 2025 Exam 2 Section B Q2a — 31% (3 marks). 2024 Exam 2 Section B Q3ai — 42% (3 marks): [AG] shows four simultaneous equations recovered from four stated features, with "1M for any 2 equations, 1M for other 2 equations". 2008 Exam 2 Section B Q2d — 9% (2 marks): two simultaneous equations in m and n with an exponential/logarithmic structure and two valid solution pairs.

Typical marks. 2–3 marks.

Traps. 2025 Exam 2 report:

"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 report on Q2d:

"Many students tried unsuccessfully to solve the simultaneous equations by hand. Other students gave only one solution, stating that m and n had to be positive but [only] mn had to be positive. Some students gave the second solution only, possibly because this was the first solution given on the calculator and they did not scroll across to get the second solution."

A "show that" in Exam 2 forbids the CAS. This is the single most reliable way to lose marks on a technology-active paper.

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Type 6 — Simultaneous equations from continuity and smoothness of a hybrid function

Wording template (2023 Exam 2 Section A Q9, verbatim):

"The function f is given by
f (x) = −tan(x/2) − 4, x < 2π
= sin(ax), 2π ≤ x ≤ 8
The value of a for which f is continuous and smooth at x = 2π is
A. −2 B. π/2 C. −1/2 …"

What it is really testing. That "continuous" gives one equation (equal values at the join) and "smooth" gives a second (equal derivatives), and the pair is then solved simultaneously for the parameter.

Standard method. Set f₁(c) = f₂(c) and f₁′(c) = f₂′(c); solve the system. [RPT] for 2023: "For f to be continuous at [x = 2π], … For f to be smooth at [x = 2π], … So, solving [both] for a".

Instances. 2023 Exam 2 Section A Q9 — 42%. [SAMP] Exam 2 Section B Q2: "f (x) is both smooth and continuous at x = 5" for a two-piece ramp model. Related: 2023 Exam 2 Section B Q2dii — 12% — recovering n for a three-piece piecewise circular model.

Typical marks. 1 mark MC; 2–3 marks Section B.

Traps. Students supply the continuity equation and stop. A single equation cannot determine the parameter when the join also has to be smooth.

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Type 7 — The discriminant: conditions for real roots

Wording template (2017 Exam 2 Section A Q7, verbatim):

"The equation (p − 1)x² + 4x = 5 − p has no real roots when
A. p² − 6p + 6 < 0 B. p² − 6p + 1 > 0 C. p² − 6p − 6 < 0 D. p² − 6p + 1 < 0 E. p² − 6p + 6 > 0"

2019 Exam 2 Section A Q2: "The set of values of k for which x² + 2x − k = 0 has two real solutions is …". 2023 Exam 2 Section A Q19: "Find all values of k, such that the equation x² + (4k − 3)x + (4k² − 9)/4 = 0 has two real solutions for x, one positive and one negative."

What it is really testing. Rearranging to ax² + bx + c = 0 first — the constant and the coefficient of x² both contain the parameter — and then converting a solution-count requirement into an inequality in the discriminant. The 2023 variant adds a sign requirement, which is a product-of-roots condition, not a discriminant condition.

Standard method. Collect to standard form. Δ = b² − 4ac. Δ < 0 no real roots; Δ = 0 one; Δ > 0 two. For "one positive and one negative", additionally require c/a < 0. Answer as an inequality or interval, not a value.

Instances. 2017 Exam 2 Section A Q7 — 32%. 2023 Exam 2 Section A Q19 — 32%. 2019 Exam 2 Section A Q2 — 59%. Related (tagged Functions): 2014 Exam 2 Section A Q18 — 53% — "The graph of y = kx − 4 intersects the graph of y = x² + 2x at two distinct points for".

Typical marks. 1 mark.

Traps. 2017 Q7's distribution — A 19%, B 32%, C 12%, D 29%, E 7% — shows the failure exactly: 29% chose D, the same quadratic with the inequality reversed. The VCAA solution for 2023 Q19 spells out the two-condition structure:

"Δ > 0 for two unique solutions. One solution has to be positive and the other negative. Solve [the product-of-roots condition] for k."

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Type 8 — "The values of k for which the equation has exactly n solutions"

Wording templates, three real ones:

"Find the value of k for which the graphs of y = f (x) and y = f ′(x) have exactly one point of intersection." — 2018 Exam 1 Q8b

"Hence, or otherwise, find the positive values of w for which f (x) = g(x) has exactly three solutions." — 2025 Exam 1 Q9b.ii

"The number of tangents to this curve that pass through the positive x-intercept is" — 2023 Exam 2 Section A Q14

What it is really testing. That a solution count is a statement about where a family of curves sits relative to a fixed curve, and that the boundary cases are the interesting ones: tangency (repeated root), a turning point landing on a line, or a factor degenerating.

Standard method. Reduce f(x) = g(x) to a single equation in x carrying the parameter. Then one of three routes: (i) if it is quadratic, use the discriminant and count; (ii) if it factorises, count the solutions each factor contributes and force the total; (iii) if neither, find the turning point of one side in terms of the parameter and set its y-value equal to the other side. Route (iii) is what "Hence, or otherwise" points at in 2025.

Instances. 2025 Exam 1 Q9bii — 4% (2 marks). 2018 Exam 1 Q8b — 3% (2 marks). 2021 Exam 1 Q9bi — 6% (1 mark). 2016 Exam 2 Section B Q4eiii — 3% (2 marks). 2023 Exam 2 Section A Q14 — 29%. 2025 Exam 2 Section B Q4fiii — 10% (2 marks).

Typical marks. 1–2 marks. Almost always the last part of a question.

This is the hardest type in the area and one of the hardest in the subject. All six instances above sit at or below 29%; four sit at or below 6%.

Traps. The 2025 Exam 1 report on Q9b.ii is the best single paragraph VCAA has written about how to attack the type:

"There were many ways this question could be approached, however very few students made significant progress. Those students who approached the solution by forming [the equation] and then using a discriminant approach … usually progressed further towards a complete solution. Some students were able to obtain an expanded quartic equation but then did not proceed any further. 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 (listing all the possible solutions, compare them, discuss the nature of turning points or discuss the impact of the discriminant on the number of solutions) were the most successful."

The 2018 report on Q8b:

"Many students found this question challenging. Most students found the correct quadratic equation to solve but solved for k, rather than the x value that satisfied the quadratic equation. Few students realised that x = 0 was the unique solution. Incorrect use of the null factor law and/or the incorrect discriminant of the quadratic were the main sources of error."

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Type 9 — Counting solutions of a combined-function equation

Wording template (2019 Exam 1 Q9f, verbatim):

"State the number of solutions to g ( f (x)) + f ( g(x)) = 0."

What it is really testing. Whether you will sketch. There is no algebra that resolves e^{3+2x−x²} + 3 + 2e^x − e^{2x} = 0; the answer comes from adding ordinates of two graphs built in earlier parts.

Standard method. Sketch both component functions roughly on one set of axes, using the ranges, stationary points and asymptotes established earlier in the question. Add ordinates. Count crossings.

Instances. 2019 Exam 1 Q9f — 19% (1 mark). 2023 Exam 2 Section A Q14 — 29%. 2016 Exam 2 Section B Q4fii — 2% (2 marks), the bounding version: "Show that 0 < A(k) < 2 for all k > 1."

Typical marks. 1 mark, at the very end of a long question.

Traps. 2019 Exam 1 report:

"This question was not well done. Few students attempted to draw a rough sketch of each equation and use addition of ordinates."

A one-mark part at the end of a nine-mark question is worth ninety seconds of sketching, not zero.

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Type 10 — Literal equations: "express x in terms of y" (inverse rearrangement)

Wording template (2017 Exam 2 Section A Q8, verbatim):

"If y = a b^{−4x} + 2, where a > 0, then x is equal to
A. ¼(b − log_a(y − 2)) B. ¼(b − log_a(y + 2)) C. b − log_a(¼(y + 2)) D. (b − log_a(y − 2))/4 E. ¼(b + 2 − log_a(y))"

2007 Exam 2 Section A Q3: "If y = log_a(7x − b) + 3, then x is equal to". 2006 Exam 2 Section A Q9: "If y = 3a^{2x} + b, then x is equal to".

What it is really testing. Inverse operations applied in the right order, and keeping a subtracted constant outside the logarithm.

Standard method. Undo in reverse order of construction: the constant applied last comes off first, then the coefficient, then the log/exponential, then the inner linear expression. Check by substituting a convenient value.

Instances. 2017 Exam 2 Section A Q8 — 64%. 2007 Exam 2 Section A Q3 — 78%. 2006 Exam 2 Section A Q9 — 84%. 2010 Exam 2 Section A Q8 — 79% — "The function f has rule f (x) = 3 log_e(2x). If f (5x) = log_e(y) then y is equal to".

Typical marks. 1 mark.

Traps. The classic distractor is log_a(y) − 2 for log_a(y − 2). These items sit in the 64–84% band — they are not separators, and that is the point: this is the one algebra type the state reliably gets right, so it is free marks you must not drop.

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Type 11 — Literal equations: formulating a constraint from a geometric context

Wording template (2010 Exam 1 Q11, verbatim):

"A cylinder fits exactly in a right circular cone so that the base of the cone and one end of the cylinder are in the same plane … The height of the cone is 5 cm and the radius of the cone is 2 cm. The radius of the cylinder is r cm and the height of the cylinder is h cm. For the cylinder inscribed in the cone as shown above
a. find h in terms of r
b. find S in terms of r
c. find the value of r for which S is a maximum."

2023 Exam 2 Section A Q17: "A cylinder of height h and radius r is formed from a thin rectangular sheet of metal of length x and width y, by cutting along the dashed lines shown below. The volume of the cylinder, in terms of x and y, is given by …"

What it is really testing. Similar triangles or a mensuration formula, then elimination of the unwanted variable. It is pure algebra dressed as geometry, and it is the standard first part of every optimisation question in the archive.

Standard method. Write the constraint (similar triangles, fixed volume, fixed perimeter, Pythagoras). Solve it for the unwanted variable. Substitute into the quantity to be optimised. Simplify before differentiating.

Instances. 2010 Exam 1 Q11a — 11% (2 marks). 2010 Exam 1 Q11b — 47% (1 mark). 2014 Exam 1 Q10bi — 32%. 2008 Exam 1 Q9a — 35% (2 marks). 2008 Exam 1 Q9b — 31%. 2023 Exam 2 Section A Q17 — 28%. 2022 Exam 2 Section A Q19 — 34% — the fully literal box problem, "The maximum volume of the box occurs when x is equal to", with a and b in every option. 2012 Exam 2 Section B Q1a — 62%. 2014 Exam 2 Section B Q2a — 75%.

Typical marks. 1–2 marks, always at the start.

Traps. 2010 Exam 1 report on Q11a:

"Students found this question difficult. Few students realised that similar figures were needed. A common incorrect formulation was h/5 = r/2."

2008 Exam 1 report on Q9a:

"Quite a few students had difficulty finding the volume of the prism; some used the volume of the triangular pyramid given on the formula sheet, others attempted to use surface area or perimeter. … Students are expected to have adequate algebraic facility with these sorts of expressions."

2012 Exam 2 report on Q1a: "In this question, the relevant formulation was required" — that is, the mark is for the derivation, not the answer.

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Type 12 — Index laws: equations with a common base

Wording template (2014 Exam 1 Q4, verbatim):

"Solve the equation 2^{3x − 3} = 8^{2 − x} for x."

2013 Exam 1 Q5b: "Solve the equation 3^{−4x} = 9^{6 − x} for x." 2022 Exam 1 Q5a: "Solve 10^{3x} − 13 = 100 for x."

What it is really testing. Recognising that both sides are powers of the same base, rewriting, equating exponents. Two lines of work.

Standard method. Express both sides with the same base. Equate indices. Solve the linear equation. Simplify the fraction.

Instances. 2014 Exam 1 Q4 — 76% (2 marks). 2013 Exam 1 Q5b — 64% (2 marks). 2022 Exam 1 Q5a — 76% (2 marks).

Typical marks. 2 marks.

Traps. 2013 report: "The majority of incorrect responses involved 9 = 3³." 2014 report: "Some students chose to work with a common base of 8. Students are reminded to simplify their final answer, especially for fraction answers." 2022 report: "a number of students incorrectly wrote 100 as 10^{10}."

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Type 13 — Logarithm laws: collapse to a single logarithm, then solve

Wording template (2012 Exam 1 Q7, verbatim):

"Solve the equation 2 log_e(x + 2) − log_e(x) = log_e(2x + 1), where x > 0, for x."

2013 Exam 1 Q5a: "Solve the equation 2 log₃(5) − log₃(2) + log₃(x) = 2 for x." 2015 Exam 1 Q7a: "Solve log₂(6 − x) − log₂(4 − x) = 2 for x, where x < 4." 2020 Exam 1 Q4: "Solve the equation 2 log₂(x + 5) − log₂(x + 9) = 1." 2007 Exam 1 Q2a: "Solve the equation log_e(3x + 5) + log_e(2) = 2, for x."

What it is really testing. The three laws — n log(a) = log(aⁿ), log(a) + log(b) = log(ab), log(a) − log(b) = log(a/b) — applied in the right order, then removal of the logarithm, then a polynomial equation, then a domain check.

Standard method. Move coefficients inside as powers first. Combine to one log on each side. If both sides are logs with the same base, equate arguments. If one side is a number c, write log_b(A) = c ⟹ A = b^c. Solve. Reject any root that makes any original argument non-positive.

Instances. 2013 Exam 1 Q5a — 54% (2 marks). 2012 Exam 1 Q7 — 52% (3 marks). 2015 Exam 1 Q7a — 65% (2 marks). 2007 Exam 1 Q2a — 58% (2 marks). 2020 Exam 1 Q4 — 26% (3 marks). 2024 Exam 1 Q6 — 9% (4 marks).

Typical marks. 2–4 marks. The mark value is the difficulty signal: 2 marks means one law and a linear equation; 4 marks means a cubic and a domain rejection.

Traps. 2007 report: "Common mistakes included simply cancelling the logarithms, adding to obtain 3x + 7, or multiplying to get 6x + 5." 2013 report: "A disappointing number of students could not combine all parts of the logarithms into a single expression." 2015 report: "Many students correctly applied logarithm laws, but others incorrectly cancelled logarithms." 2012 report: "Most students could apply at least one of the logarithm rules and many of those were able to successfully eliminate the logarithms and produce an easily factorised quadratic expression."

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Type 14 — Logarithmic and exponential equations with domain rejection

Wording template (2009 Exam 1 Q9, verbatim):

"Solve the equation 2 log_e(x) − log_e(x + 3) = log_e(1/2) for x."

2014 Exam 1 Q6: "Solve log_e(x) − 3 = log_e(√x) for x, where x > 0." 2024 Exam 1 Q6 (reconstructed from [AG]): solve log₃((x − 4)²) + log₃(x) = 2, leading to x³ − 8x² + 16x − 9 = 0, (x − 1)(x² − 7x + 9) = 0, with only x = (7 + √13)/2 surviving x > 4.

What it is really testing. The implied domain. The algebra produces two or three roots; the question has one answer.

Standard method. Before solving anything, write down the domain restriction implied by every logarithm in the original equation. Solve. Test each root against that restriction. State the rejection explicitly.

Instances. 2024 Exam 1 Q6 — 9% (4 marks). 2009 Exam 1 Q9 — 22% (4 marks). 2020 Exam 1 Q4 — 26% (3 marks). 2014 Exam 1 Q6 — 44% (2 marks).

Typical marks. 2–4 marks.

Traps. The 2024 report is unusually explicit about where the state stopped:

"Most students demonstrated a knowledge of the logarithm laws needed to simplify this question; however, many did not employ the correct combination of these laws. … Although some students were able to find three possible solutions, many students overlooked the fact that the domain of this log function must be [x > 4] and, as a result, did not reject the two invalid solutions."

The mark distribution tells the same story: 18% scored 0, 52% scored exactly 1, 8% two, 13% three, 9% four. Half the state got the first law right and then stalled.

2009 report:

"Only a few students attained full marks on this question because many did not eliminate x = −1 as a solution; 2 log_e(−1) is not defined as a real value. Some students incorrectly cancelled one x in the denominator with the x in the numerator."

2020 report:

"Students confidently attempted this question; however, many incorrect uses of the logarithmic laws were observed. Those who did end up with the appropriate quadratic equation and solved it correctly did not always check the validity of their answers; these students failed to reject the [invalid] solution."

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Type 15 — Exponential equations that are quadratic in e^x

Wording template (2023 Exam 1 Q2, verbatim):

"Solve e^{2x} − 12 = 4e^x for x ∈ R."

2025 Exam 1 Q5a: "Solve e^{2x} − 8e^x + 7 = 0 for x." 2015 Exam 1 Q7b: "Solve 3e^t = 5 + 8e^{−t} for t." 2019 Exam 1 Q9d: "Solve f ( g(x)) = 0" where f(x) = 3 + 2x − x² and g(x) = e^x, giving 3 + 2e^x − e^{2x} = 0.

What it is really testing. Seeing e^{2x} as (e^x)², substituting a = e^x, factorising (not the quadratic formula), then rejecting any non-positive root because e^x > 0 for all real x.

Standard method. Multiply through by e^t if there is a negative index. Collect to a² + ba + c = 0 with a = e^x. Factorise by the null factor law. Discard a ≤ 0. Take logarithms of the survivors.

Instances. 2023 Exam 1 Q2 — 56% (3 marks). 2025 Exam 1 Q5a — 61% (2 marks). 2019 Exam 1 Q9d — 48% (2 marks). 2015 Exam 1 Q7b — 36% (3 marks). 2007 Exam 2 Section A Q11 — 82% — "The solution set of the equation e^{4x} − 5e^{2x} + 4 = 0 over R is". 2006 Exam 2 Section B Q3b — 66%.

Typical marks. 2–3 marks.

Traps. 2023 report:

"This question required students to solve an exponential equation by treating it as a quadratic equation in terms of [e^x]. The most direct approach was to form a quadratic equation set equal to zero, use the null factor law to factorise and then solve. Some students chose to use the quadratic formula, albeit not always successfully. … Some students arrived at both [roots] and then gave [both logs] as their solutions without discarding [the negative root]. Students need to keep in mind that [log_e(a)] only exists for [a > 0]."

2025 report: "Some students incorrectly discarded the solution [e^x = 1]." Both directions of error appear in the same type: keeping a negative root, and discarding a valid root equal to 1.

2015 report: "Many students were unable to create the quadratic equation evolved from manipulating e^t. Many students solved via the quadratic formula rather than using simpler factorising techniques." 2019 report: "The inclusion of x = log_e(−1) was a common error."

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Type 16 — Logarithm identities and change of base (multiple choice)

Wording template (2013 Exam 2 Section A Q18, verbatim):

"Let g(x) = log₂(x), x > 0. Which one of the following equations is true for all positive real values of x?
A. 2g(8x) = g(x²) + 8 B. 2g(8x) = g(x²) + 6 C. 2g(8x) = (g(x) + 8)² D. 2g(8x) = g(2x) + 6 E. 2g(8x) = g(2x) + 64"

2019 Exam 2 Section A Q20: "The expression log_x(y) + log_y(z), where x, y and z are all real numbers greater than 1, is equal to …". 2020 Exam 2 Section A Q10: "Given that log₂(n + 1) = x, the values of n for which x is a positive integer are …".

What it is really testing. Log laws used as identities rather than as steps in an equation, plus the change-of-base relation log_a(b) = 1/log_b(a).

Standard method. Expand the left side fully using the laws; expand each option; compare. Or substitute a convenient value (x = 1, x = 2) and eliminate — legitimate and fast in multiple choice, but check a second value because more than one option can survive the first.

Instances. 2013 Exam 2 Section A Q18 — 35%. 2019 Exam 2 Section A Q20 — 47%. 2020 Exam 2 Section A Q10 — 62%.

Typical marks. 1 mark.

Traps. 2013 Q18's distribution — A 12%, B 35%, C 22%, D 17%, E 13% — is close to uniform. This is a type the state guesses. Note also 2016 Exam 1 Q5aiii — 31% (2 marks) — the proof version: "Show that h(x) + h(−x) = f ((g(x))²)" where h(x) = log_e(x² + 1). The 2016 report:

"Many students were unsure of how to present their working. In the sample working above, both sides were operated on separately to arrive at the same expression and the conclusion that one side was in fact equivalent to the other. Poor notation was again evident, in particular log_e(−x² + 1) [written for] log_e((−x)² + 1)."

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Type 17 — Trigonometric equations over a restricted domain, by hand

This is the single most reliable annual appearance in Examination 1.

Wording templates, all verbatim:

"Solve the equation sin(2x/3) = −√3/2 for x ∈ [0, 3π]." — 2007 Exam 1 Q8a

"Solve the equation tan(2x) = −√3 for x ∈ [−π/4, 3π/4] […]" — 2009 Exam 1 Q4

"Solve the equation sin(x/2) = −1/2 for x ∈ [2π, 4π]." — 2013 Exam 1 Q4

"Solve 2cos(2x) = −√3 for x, where 0 ≤ x ≤ π." — 2014 Exam 1 Q3

"Solve the equation 2cos(x) + 1 = 0 for 0 ≤ x ≤ 2π." — 2018 Exam 1 Q3a

"Solve 1 − cos(x/2) = cos(x/2) for x ∈ [−2π, π]." — 2019 Exam 1 Q4a

"Solve the equation √3 sin(x) = cos(x) for x ∈ [−π, π]." — 2010 Exam 1 Q4b

The study design guarantees this form: "solve by hand equations of the form sin(ax + b) = c, cos(ax + b) = c and tan(ax + b) = c with exact value solutions over a given interval".

What it is really testing. Four separate skills, each of which the reports single out: the exact value (base angle), the quadrant rule, transforming the domain, and dividing back.

Standard method. 1. Isolate the circular function: sin(θ) = c where θ = ax + b. 2. Transform the domain. If x ∈ [p, q] then θ ∈ [ap + b, aq + b]. Write it down. This is the step that is skipped. 3. Base angle from the exact value; sign of c gives the quadrants. 4. List all θ in the transformed domain — usually more than two. 5. Solve back for x. Count your answers against the width of the domain: a domain of length 2π with sin(2x) has four solutions, not two.

Instances. 2018 Exam 1 Q3a — 72%. 2014 Exam 1 Q3 — 55%. 2015 Exam 1 Q5b — 53%. 2019 Exam 1 Q4a — 48%. 2013 Exam 1 Q4 — 47%. 2007 Exam 1 Q8a — 45%. 2008 Exam 1 Q3 — 41%. 2009 Exam 1 Q4 — 41% (3 marks). 2010 Exam 1 Q4b — 39%.

Typical marks. 2–3 marks.

Traps. Every report names the same three faults. 2007: "The biggest hurdle for many students was not knowing the correct basic angle for the exact value and therefore being unable to obtain the correct result for x. Other common errors were giving answers outside the given domain, including negative values, or only finding [one solution]." 2008: "Solving this expression caused some students problems when the given domain was ignored. It resulted in either only one solution or too many solutions. Occasionally students used incorrect values for cos⁻¹(0.5)." 2009: "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." 2013: "Many students identified a base angle of π/6 but many could not identify the correct quadrants and domain restriction." 2014: "Many students were unsure of exact values and did not ascertain the correct basic angle … or produced solutions beyond the specified domain." 2018: "some students gave solutions beyond the given domain or incorrect values (confusing π/6 with π/3 as the reference angle)." 2010, on the √3 sin(x) = cos(x) variant: "This question was very poorly done and most students who identified the equation using tan(x) were not able to solve it."

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Type 18 — General solution of a trigonometric equation

Wording template (2020 Exam 2 Section A Q4, verbatim):

"The solutions of the equation 2 cos(2(x − π/3)) + 1 = 0 are
A. x = π(6k − 2)/6 or x = π(6k − 3)/6, for k ∈ Z
B. x = π(6k − 2)/6 or x = π(6k + 5)/6, for k ∈ Z …"

Section B form (2023 Exam 2 Section B Q2d.ii): "Find all possible values of n", where a three-piece piecewise circular model must match at a join.

What it is really testing. That when no interval is given, the answer is a family indexed by k ∈ Z, and that both branches of the inverse trigonometric solution must be carried through the ax + b rearrangement.

Standard method. sin(θ) = c ⟹ θ = 2kπ + α or θ = (2k + 1)π − α. cos(θ) = c ⟹ θ = 2kπ ± α. tan(θ) = c ⟹ θ = kπ + α. Substitute θ = ax + b, solve for x, and keep k ∈ Z attached.

Instances. 2020 Exam 2 Section A Q4 — 67%. 2023 Exam 2 Section B Q2dii — 12% (2 marks). 2008 Exam 2 Section B Q4dii — 40% (2 marks).

Typical marks. 1 mark MC; 2 marks Section B.

Traps. 2023 report on Q2d.ii: "This question was not done well. Some students were able to set up a correct equation. A general solution was required." 2008 report on Q4d.ii: "Some students gave the general solution using calculator syntax." Both errors are about presentation: the CAS returns @n1 or constn(1); that is not a mathematical answer.

Conversely, 2025 Exam 2 Section B Q4b (78%) shows the reverse error — "Others incorrectly gave extra solutions or a general solution, not considering the restricted domain."

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Type 19 — Trigonometric solution counting and "the sum of the solutions"

Wording template (2017 Exam 2 Section A Q12, verbatim):

"The sum of the solutions of sin(2x) = −√3/2 over the interval [−π, d] is −π. The value of d could be
A. 0 B. π/6 C. 3π/4 D. 7π/6 E. 3π/2"

2019 Exam 2 Section A Q19: "Given that tan(α) = d, where d > 0 and 0 < α < π/2, the sum of the solutions to tan(2x) = d, where 0 < x < 5π/4, in terms of α, is …"

What it is really testing. Symmetry of the circular functions — solutions come in pairs summing to a constant — and careful inclusion/exclusion at the endpoint of a variable domain.

Standard method. Find the solutions in the largest plausible domain. Note that consecutive tan(2x) solutions differ by π/2, and that sin solutions pair around a line of symmetry. Add progressively and find the d at which the running total is the stated value.

Instances. 2017 Exam 2 Section A Q12 — 45%. 2019 Exam 2 Section A Q19 — 25%. 2006 Exam 2 Section A Q2 — 84% — the easy relative: "The smallest positive value of x for which tan(2x) = 1 is".

Typical marks. 1 mark.

Traps. 2019 Q19's distribution — A 4%, B 12%, C 27%, D 30%, E 25% — has the most popular option wrong. These items reward drawing the solutions on a number line rather than manipulating symbols.

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Type 20 — Solution sets written in set notation

Wording template (2007 Exam 2 Section A Q21, verbatim):

"{x : cos²(x) + 2cos (x) = 0} =
A. {x : cos (x) = 0}
B. {x : cos (x) = −1/2}
C. {x : cos (x) = 1/2}
D. {x : cos (x) = 0} ∪ {x : cos (x) = −1/2}
E. {x : cos (x) = 1/2} ∪ {x : cos (x) = −1/2}"

Related: 2007 Exam 2 Section A Q11 — "The solution set of the equation e^{4x} − 5e^{2x} + 4 = 0 over R is A. {1, 4} … E. {0, log_e(2)}". 2016 Exam 2 Section A Q3 — "f (x) < 0 for A. x ∈ (−2, 0) ∪ (1/3, ∞) …".

What it is really testing. The null factor law expressed as a union of sets, and the difference between the set of factor-roots and the set of x-values.

Standard method. Factorise. Each factor gives one set. The solution set is the union. Then check whether either factor is unsatisfiable over the reals (here cos(x) = −2 is empty) and drop it.

Instances. 2007 Exam 2 Section A Q21 — 27%. 2007 Exam 2 Section A Q11 — 82%. 2016 Exam 2 Section A Q3 — 77%. 2020 Exam 2 Section B Q5dii — 13% (1 mark), the interval-notation version.

Typical marks. 1 mark.

Traps. 2007 Q21's distribution — A 27%, B 11%, C 20%, D 27%, E 15% — is a five-way split with the correct answer tied for the lead. The 2020 report on Q5d.ii: "Some students had the values within the interval in the wrong order. Others had incorrect brackets." Notation is the mark on these.

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Type 21 — The remainder theorem

Wording template (2020 Exam 2 Section A Q2, verbatim):

"Let p(x) = x³ − 2ax² + x − 1, where a ∈ R. When p is divided by x + 2, the remainder is 5. The value of a is
A. 2 B. −7/4 C. 1/2 D. −3/2 E. −2"

2015 Exam 2 Section A Q6: "For the polynomial P(x) = x³ − ax² − 4x + 4, P(3) = 10, the value of a is".

What it is really testing. That the remainder on division by (x − k) is P(k), and that the resulting equation in the parameter is linear.

Standard method. P(−2) = 5. Expand, collect, solve for a.

Instances. 2020 Exam 2 Section A Q2 — 56%. 2015 Exam 2 Section A Q6 — 91%.

Typical marks. 1 mark.

Traps. Sign of k: dividing by x + 2 means evaluating at x = −2. The 2015 version, which hands you P(3) = 10 directly, sits at 91%; the 2020 version, which requires the theorem, drops 35 points. That 35-point gap is the entire content of the remainder theorem.

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Type 22 — The factor theorem and "verify that x = a is a solution"

Wording template (2025 Exam 1 Q7, verbatim):

"Let f : R → R, f (x) = x³ […] 16x […] 20.
a. Verify that x = 5 is a solution of f (x) = 0. (1 mark)
b. Express f (x) in the form (x + d)²(x − 5), where d ∈ R. (2 marks)"

The signs on the cubic's terms are lost in the extraction and are not reconstructed here; the two command words and the target form are legible and are what matters.

2023 Exam 1 Q9a: "Given that f (0) = 12 and g(1) = 9, verify that a = 12 and b = −3."

What it is really testing. The difference between verify (substitute and show it works) and solve (find it). And, in part b, that "express in the form" fixes the factored shape so only the missing constant is needed.

Standard method. For "verify": substitute, evaluate, state the conclusion. For "express in the form": long division, synthetic division, equating coefficients, or grouping — [AG] 2025 lists five acceptable methods (long division, equate coefficients, synthetic division, factorising by grouping, find missing terms).

Instances. 2025 Exam 1 Q7a — 90% (1 mark). 2025 Exam 1 Q7b — 76% (2 marks). 2023 Exam 1 Q9a — 90% (1 mark). 2017 Exam 1 Q3a — 79% (1 mark) — "Show that (x + 2)²(x − 1) = x³ + 3x² − 4." 2014 Exam 2 Section B Q5a — 48% (2 marks) — factorise x⁴ − 8x fully.

Typical marks. 1 mark to verify, 2 to factorise.

Traps. 2023 report: "Most students knew that in order to 'verify' the values they needed to show working to support this." 2025 report on Q7a: "Some students chose to use a factor theorem approach to show that [x = 5] was a solution and, although not necessary, this approach was appropriate." On Q7b:

"This question was well answered, with students using a variety of valid methods such as long division, synthetic division, and expanding and equating coefficients … However, some students gave the correct answer without showing any working to support the answer. Students are reminded that for any question worth more than one mark, working must be shown in order to be awarded full marks."

2014 report on Q5a: "Some students attempted to factorise by hand and obtained two correct linear factors but an incorrect quadratic factor … Others used their technology to factorise appropriately." On a technology-active paper the CAS is the right tool — except where the question says "show that" (Type 5).

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Type 23 — "Express your answer in the form …"

Wording templates:

"Find the distance OP. Express your answer in the form (a√b)/b, where a and b are positive integers." — 2018 Exam 1 Q7b

"Express f (x) = ¼(x + 2)²(x − 2)² in the form f (x) = ¼x⁴ + bx² + c, where b and c are integers." — 2020 Exam 2 Section B Q1b

"Express your answer in the form a/b⁴ − 3/c⁴, where a, b, c ∈ Z⁺." — 2020 Exam 1 Q5b

What it is really testing. That the stated form is a specification, not a suggestion. The marker is checking the shape.

Standard method. Do the mathematics, then force the answer into the named shape — rationalise, expand, or collect as required. Read what the letters are allowed to be ("positive integers", Z⁺, "integers") and check yours comply.

Instances. 2018 Exam 1 Q7b — 27% (2 marks). 2025 Exam 1 Q7b — 76% (2 marks). 2020 Exam 2 Section B Q1b — the paired part ([QJSON] tags it Functions).

Typical marks. 2 marks.

Traps. 2018 report: "This question was attempted well. Some students misquoted the distance formula or made arithmetic errors in their calculations." The type fails on surd arithmetic, not on concept. Rationalising to 4√5/5 is the whole of 2018 Q7b and 73% of the state did not produce it.

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Type 24 — Solving f(x) = g(x) with technology, to a stated accuracy

The workhorse of Exam 2 Section B. Roughly a third of the algebra-tagged Section B parts are this.

Wording templates:

"Find all values of x for which D is at most 2 units. Give your answers correct to two decimal places." — 2020 Exam 2 Section B Q1f

"Find the possible values of x, for Jepzibah's pyramid, correct to two decimal places." — 2010 Exam 2 Section B Q3f

"For how many hours during the 24-hour time interval is T ≥ 26?" — 2013 Exam 2 Section B Q1d

"At what time would the statue first touch the acid?" — 2009 Exam 2 Section B Q4e.ii

"Assuming he takes a capsule at the same time each day, on how many days does he need to take a capsule so that he will no longer be affected by the snake toxin?" — 2008 Exam 2 Section B Q3h

What it is really testing. Setting up the correct equation from a described condition, entering it correctly, and reporting to the stated accuracy in the stated units.

Standard method. Write the equation as LHS = RHS explicitly in your working — the method mark is for the statement, not the number. Solve with technology over the stated domain. Round last, to exactly the accuracy asked. If the question asks for a duration or a difference, subtract.

Instances. 2010 Exam 2 Section B Q3f — 12% (2 marks). 2008 Exam 2 Section B Q3h — 16% (1 mark). 2009 Exam 2 Section B Q4eii — 16% (2 marks). 2015 Exam 2 Section B Q2d — 21% (3 marks). 2015 Exam 2 Section B Q2c — 33% (3 marks). 2013 Exam 2 Section B Q1d — 45% (2 marks). 2016 Exam 2 Section B Q1f — 47% (2 marks). 2015 Exam 2 Section B Q2e — 49% (3 marks). 2013 Exam 2 Section B Q1c — 64%.

Typical marks. 2–3 marks.

Traps. The accuracy instruction is the most-cited fault in the entire area. 2015 report on Q2c: "Some students did not give their answers correct to two decimal places. Some worked to one decimal place and others rounded their answers incorrectly." On Q2e: "Others rounded to 27.00 instead of 28.00." 2010 report on Q3f: "Some students did not give the answers correct to two decimal places or they gave one answer only." 2013 report on Q1d: "Some students just gave the values of t and did not find the difference between them. Others used approximate values in their calculations."

And the recurring syntax warning, 2010 Exam 2 report on Q4b:

"A number of students did not use the appropriate syntax to separate the coefficient a from the variable x when using technology … Others solved the second bracket equal to zero."

2016 report on Q4e.i: "Some did not include a multiplication sign between k and x when using technology."

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Type 25 — Inequalities and interval answers

Wording templates:

"Find all values of x for which D is at most 2 units." — 2020 Exam 2 Section B Q1f

"State the values of a for which 1 ≤ b < 1.1. Give your answers correct to three decimal places." — 2020 Exam 2 Section B Q5d.ii

"Find the values of k such that s(k) ≥ 1." — 2016 Exam 2 Section B Q4e.iii

"Find the largest interval of x values for which h is strictly decreasing. Give your answer correct to two decimal places." — 2023 Exam 2 Section B Q3e

What it is really testing. Converting an equation solution into an interval, with the right bracket types and the right ordering.

Standard method. Solve the corresponding equation to find the boundaries. Test a point inside each candidate interval, or read the graph. Write the answer as an interval or a union of intervals with correct open/closed brackets. Match the inequality: ≤ gives [ ], < gives ( ).

Instances. 2016 Exam 2 Section B Q4eiii — 3% (2 marks). 2020 Exam 2 Section B Q5dii — 13% (1 mark). 2020 Exam 2 Section B Q1f — 23% (2 marks). 2012 Exam 2 Section B Q1b — 38% (2 marks).

Typical marks. 1–2 marks.

Traps. 2012 report on Q1b is the canonical statement:

"Some students correctly solved V(x) = 0 but left their answer as x ≠ 36, x > 0 or x < 36, without considering the inequality. Others had incorrect notation such as [0, 36] or {0, 36}. Some solved the total surface area = 0 instead of V(x) = 0."

2020 report on Q1f: "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." 2016 report on Q4e.iii: "This question was not answered well. Some students had 1 ≤ k ≤ 5/4" — the wrong bracket at one end. [AG] 2024 Exam 2 Q5d: "Two intervals only full marks."

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Type 26 — Newton's method

Wording templates, all verbatim:

"f. Apply Newton's method, with an initial estimate of x₀ = 0, to find an approximate x-intercept of h. Write the estimates x₁, x₂ and x₃ in the table below, correct to three decimal places.
g. For the function h, explain why a solution to the equation log_e(2) · (2^x) − 2x = 0 should not be used as an initial estimate x₀ in Newton's method." — 2023 Exam 2 Section B Q3f, Q3g

"e. Apply two iterations of Newton's method to f with x₀ = 2π/3. i. Write down x₂, correct to one decimal place. ii. On the axes in part d, draw the tangent to the graph of y = f (x) at the point where x = x₁." — 2025 Exam 2 Section B Q4e

(Part d of that question asked for the equation of a tangent on axes already printed — 70% — not for a sketch. The axes are what part e borrows.)

"Newton's method is used to estimate the x-intercept of the function f (x) = ⅓x³ + 2x + 4. a. Verify that f (−1) > 0 and f (−2) < 0. b. Using an initial estimate of x₀ = −1, find the value of x₁." — [SAMP] Exam 1 Q6

"Newton's method can be applied to find an approximate solution to e^{−x} − 2 = 0. If the initial value is x₀ = a, where a is a real number, then the value of x₁ is A. 2e^a − 1 …" — 2026 NHT Exam 2 Section A Q5

"The following algorithm applies Newton's method using a For loop with 3 iterations. … The Return value of the function newton(x³ + 3x − 3, 3x² + 3, 1) is closest to A. 0.83333 B. 0.81785 C. 0.81773 D. 1 E. 3" — 2023 Exam 2 Section A Q13

What it is really testing. The formula x_{n+1} = x_n − f(x_n)/f′(x_n) (given on [FS]), the tangent-line interpretation, and — in the hard parts — the failure condition f′(x_n) = 0.

Standard method. Differentiate. Iterate from x₀, keeping full precision internally and rounding only at the reported step. For a symbolic x₁, substitute a into the formula and simplify. For the "why not" question, argue that at a stationary point the tangent is horizontal, so the next estimate is undefined.

Instances. 2023 Exam 2 Section B Q3f — 54% (2 marks). 2023 Exam 2 Section B Q3g — 21% (1 mark). 2023 Exam 2 Section A Q13 — 52% (tagged Calculus). 2025 Exam 2 Section B Q4e — paired parts. 2024 Exam 2 Section A Q14 (NHT) — no state data.

Typical marks. 1–2 marks per part; typically two consecutive parts.

Traps. 2023 report on Q3f: "Many students were familiar with Newton's method. Answers were required to three decimal places. Some students only had one correct answer. Others had rounding errors." On Q3g:

"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" — that is, they found the stationary points but did not say why that breaks the method.

The 2023 report's general comments confirm the method landed well overall: "most students were able to respond effectively to the questions involving the introduced concepts, such as Newton's method in Questions 3f. and 3g."

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Type 27 — Pseudocode: tracing and completion

Wording template (2025 Exam 2 Section A Q7, verbatim):

"Consider the algorithm below.
n ← 17
k ← 5
while n > k
n ← n − k
print n
end while
In order, the values printed by the algorithm are
A. 12 B. 12, 7 C. 12, 7, 2 D. 12, 7, 2, −3"

Tolerance/termination form (2024 NHT Exam 2 Section A Q14, verbatim):

"The pseudocode below describes a simple algorithm to find the point where the two functions intersect.
define intersect(x0, x1):
x ← x0
while x < x1
if −0.01 < cos(x) − x < 0.01 then
return x
end if
x ← x + 0.01
end while
return "No intersection found""

Completion form ([SAMP] Exam 2 Q7): an incomplete Newton's-method implementation, "with two missing lines indicated by an empty box", and four If … Then … Return candidates differing only in which variable is returned and which difference is tested.

What it is really testing. The three constructs [SD] names — sequencing, decision-making, repetition — and, above all, when the loop stops. Every distractor is an off-by-one.

Standard method. Build a trace table with one column per variable and one row per pass. Evaluate the loop condition before each pass. Note whether print comes before or after the update. Stop when the condition fails, and check whether the final value was printed.

Instances. 2025 Exam 2 Section A Q7 — 72%. 2024 Exam 2 Section A Q14 (NHT) — no state data; the published reasoning is "The algorithm stops when [the tolerance is met]". 2025 Exam 2 Section A Q18 (NHT) — no state data; published answer "Print 4.5". [SAMP] Exam 2 Q7.

Typical marks. 1 mark. Pseudocode has not appeared in Section B of any November paper.

Traps. In 2025 Q7 the trap is option D: n becomes −3, but the loop condition n > k already fails at n = 2, so −3 is never printed. 72% is a comfortable score for a new construct, but it is the only November data point in existence — do not assume the type stays easy.

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Type 28 — The trapezium rule, by hand

Wording template ([SAMP] Exam 1 Q5c, verbatim):

"Four trapeziums of equal width are used to approximate the area between the functions f (x) = 2 − x² and the x-axis from x = −1 to x = 1. The heights of the left and right edges of each trapezium are the values of y = f (x), as shown in the graph below. Find the total area of the four trapeziums." (3 marks)

2024 NHT Exam 1: "Using the trapezium rule approximation method and three trapeziums of equal width, […]". 2024 Exam 1 Q7a (reconstructed from [RPT]/[AG]): three trapeziums approximating the area between f(x) = x sin(x) and the x-axis over [0, π].

Conceptual form (2025 Exam 2 Section A Q6, verbatim):

"The trapezium rule is used, with two trapeziums, to estimate the area bounded by the graph of y = f (x), the x-axis and the lines x = 0 and x = 1. For which function will the trapezium rule estimate be larger than the exact area?"

What it is really testing. The formula on [FS] with the correct coefficient pattern 1, 2, 2, …, 2, 1, exact values of the function at the nodes, and — in the conceptual version — concavity.

Standard method. Compute the strip width (x_n − x₀)/n. Evaluate f at every node exactly. Apply the formula, doubling every interior value. Combine into a single exact expression. For the concavity version: the rule overestimates where the curve is concave up and underestimates where it is concave down.

Instances. 2024 Exam 1 Q7a — 30% (3 marks). 2025 Exam 2 Section A Q6 — tagged Calculus. [SAMP] Exam 1 Q5c and [SAMP] Exam 2 Q3.

Typical marks. 3 marks in Exam 1; 1 mark MC.

Traps. The 2024 report is the most detailed VCAA has published on this construct:

"This question required that students use three trapeziums to approximate the area between the curve and the x-axis over the interval [0, π], as per the trapezium rule. Therefore, any attempt to calculate this area using integral calculus was not acceptable. Although many students knew the trapezium rule, some students did not apply it correctly, often writing [the sum] with the coefficient '2' missing from the middle two terms. Some students gave the formula as stated on the formula sheet with values relevant to the question, however, many did not proceed to calculate [the node values] correctly. It is expected that students will have a way of remembering the exact values of [sin] for values of [x] between 0 and [π] inclusive, as specified in the key knowledge of the study design. … 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."

Distribution: 28% zero, 31% one, 11% two, 30% three. The middle of that distribution is students who wrote the formula and could not finish the arithmetic.

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Type 29 — Inverse functions used to solve equations

Wording templates:

"The function g(x) has exactly one stationary point, a local minimum. Find the largest value of a such that when g is restricted to the domain (−∞, a] it has an inverse function." — 2025 Exam 1 Q5b

"Let k : (−∞, 0] → R, where k(x) = log_e(x² + 1). i. Find the rule for k⁻¹. ii. State the domain and range of k⁻¹." — 2016 Exam 1 Q5b

"Find the value of a for which the graphs of g and g⁻¹ have the same endpoints." — 2018 Exam 2 Section B Q5f

"What is the smallest value of k such that h will intersect with the inverse of h₁? Give your answer correct to two decimal places." — 2023 Exam 2 Section B Q5e

What it is really testing. The condition for existence (one-to-one), the domain/range swap, and the fact that graphs of f and f⁻¹ meet on y = x — which converts an intersection problem into solving f(x) = x.

Standard method. For existence: restrict at a stationary point. For a rule: swap x and y, solve, choose the branch consistent with the original range. For intersections: solve f(x) = x first, then check whether off-line intersections also exist.

Instances. 2018 Exam 2 Section B Q5f — 8% (1 mark). 2023 Exam 2 Section B Q5e — 4% (1 mark). 2025 Exam 1 Q5b — paired with Q5a (61%). 2016 Exam 1 Q5b — paired parts.

Typical marks. 1–2 marks.

Traps. 2018 report on Q5f: "This question was not answered well. Many students did not attempt this question." 2023 report on Q5e: near-identical wording. These are one-mark parts at the end of long questions, and the dominant failure is non-attempt — 92% and 96% respectively scored zero.

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Type 30 — Composite functions: rule, domain, and identities

Wording templates:

"a. State the rule of g ( f (x)). b. Find the values of x for which the derivative of g ( f (x)) is negative. c. State the rule of f ( g(x)). d. Solve f ( g(x)) = 0." — 2019 Exam 1 Q9

"Let f : [0, ∞) → R, f (x) = √(x + 1). b. Let g : (−∞, c] → R, g(x) = x² + 4x + 3, where c < 0. i. Find the largest possible value of c such that the range of g is a subset of the domain of f." — 2017 Exam 1 Q7

"Let f (x) = e^{x−1}. Given that the product function f (x) · g(x) = e^{(x−1)²}, the rule for the function g is" — 2023 Exam 2 Section A Q16

What it is really testing. The condition r_g ⊆ d_f from the study-design dot point, and index-law manipulation of the resulting expression.

Standard method. For the rule: substitute and simplify. For the domain: the domain of f ∘ g is the domain of g, restricted so that g's outputs land inside f's domain. For product identities: use index laws — e^A · e^B = e^{A+B} — and factorise the exponent.

Instances. 2023 Exam 2 Section A Q16 — 69%. 2019 Exam 1 Q9d — 48% (2 marks). 2016 Exam 1 Q5aiii — 31% (2 marks). 2019 Exam 1 Q9f — 19% (1 mark).

Typical marks. 1–2 marks per part, usually three or four parts.

Traps. The study design bans one case explicitly: "not including composite functions that result in reciprocal or quotient functions". This restricts composites only; quotients themselves are fully in scope for differentiation. The other trap is the domain: the composite inherits g's domain, not f's.

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3. The standard wordings

VCAA reuses sentences. These are the recurring ones, with the years they appear and exactly what each demands.

Two more, from the front matter of every paper, that the reports invoke constantly:

"In all questions where a numerical answer is required, an exact value must be given unless otherwise specified." — Exam 1 instruction page

"A correct answer scores 1; an incorrect answer scores 0. Marks will not be deducted for incorrect answers." — Exam 2 Section A instruction page

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4. The separators in this area

Every algebra-tagged part with pct ≤ 50. 116 parts. Grouped by type, each as ref — pct% — description.

4.1 Simultaneous linear equations with a parameter (10)

• 2006 Exam 2 Section A Q19 — 35% — (m−2)x + 3y = 6, 2x + (m−3)y = m−1 have no solution for which m.
• 2007 Exam 2 Section A Q5 — 36% — mx + 12y = 24, 3x + my = m have a unique solution only for which m.
• 2008 Exam 2 Section A Q6 — 45% — homogeneous system ax + 3y = 0, 2x + (a+1)y = 0 with infinitely many solutions.
• 2009 Exam 2 Section A Q1 — 49% — homogeneous system kx − 3y = 0, 5x − (k+2)y = 0, unique solution provided which k.
• 2010 Exam 2 Section A Q7 — 47% — (m−1)x + 5y = 7, 3x + (m−3)y = 0.7m with infinitely many solutions.
• 2012 Exam 2 Section A Q17 — 43% — the same condition presented as a matrix equation; no solution when.
• 2014 Exam 2 Section A Q17 — 50% — ax − 3y = 5, 3x − ay = 8 − a have no solution for which a.
• 2022 Exam 1 Q3 — 36% — 3 marks, technology-free: infinite number of solutions for kx − 5y = 4 + k, 3x + (k+8)y = −1.
• 2024 Exam 1 Q2 — 37% — 3 marks, technology-free: no solution; [AG] lists four acceptable methods.
• 2023 Exam 2 Section A Q9 — 42% — continuity and smoothness at a hybrid join give two simultaneous equations for a.

4.2 Simultaneous equations recovered from stated conditions (5)

• 2008 Exam 2 Section B Q2d — 9% — 2 marks: two equations in m and n; two valid solution pairs, both required.
• 2014 Exam 2 Section B Q5eii — 10% — 1 mark: solve u³ + v³ = 4 simultaneously with the relation from part e.i; only one solution pair valid because m > 0, exact values required.
• 2014 Exam 2 Section B Q5ei — 24% — 2 marks: set up g′(u) = m, g′(v) = m, add, show u³ + v³ = 4.
• 2025 Exam 2 Section B Q2a — 31% — 3 marks: "show that" A and k by dividing the two exponential equations, using algebra.
• 2024 Exam 2 Section B Q3ai — 42% — 3 marks: four equations recovered from four stated features of a model.

4.3 Discriminant and "exactly n solutions" (11)

• 2016 Exam 2 Section B Q4eiii — 3% — 2 marks: values of k with s(k) ≥ 1, where s is the square of a triangle's area.
• 2018 Exam 1 Q8b — 3% — 2 marks: k such that y = f(x) and y = f ′(x) meet exactly once.
• 2023 Exam 2 Section B Q5e — 4% — 1 mark: smallest k for which h meets the inverse of h₁.
• 2025 Exam 1 Q9bii — 4% — 2 marks: positive w for which f(x) = g(x) has exactly three solutions.
• 2021 Exam 1 Q9bi — 6% — 1 mark: values of q for which the image line meets the unit circle at least once (matrix transformation; out of scope from 2023).
• 2025 Exam 2 Section B Q4fiii — 10% — 2 marks: values of p for which the tangent's unique x-intercept equals the function's.
• 2025 Exam 1 Q9a — 19% — 3 marks: the four solutions of f(x) = g(x) when w = 3, by difference of perfect squares.
• 2019 Exam 1 Q9f — 19% — 1 mark: the number of solutions to g(f(x)) + f(g(x)) = 0.
• 2023 Exam 2 Section A Q14 — 29% — number of tangents through the positive x-intercept of y = x(3x−1)(x+3)(x+1).
• 2017 Exam 2 Section A Q7 — 32% — (p−1)x² + 4x = 5 − p has no real roots when.
• 2023 Exam 2 Section A Q19 — 32% — k for two real solutions, one positive and one negative.

4.4 Technology-free trigonometric equations over a restricted domain (8)

• 2017 Exam 1 Q6a — 25% — 1 mark: state all possible values of tan(θ) from a factorised trigonometric product.
• 2017 Exam 1 Q6b — 25% — 2 marks: "Hence" solve the squared version over 0 ≤ θ ≤ π.
• 2010 Exam 1 Q4b — 39% — 2 marks: √3 sin(x) = cos(x) on [−π, π].
• 2008 Exam 1 Q3 — 41% — 2 marks: cos(3x/2) = 1/2 […] over a stated interval.
• 2009 Exam 1 Q4 — 41% — 3 marks: tan(2x) = −√3 over a stated interval.
• 2007 Exam 1 Q8a — 45% — 2 marks: sin(2x/3) = −√3/2 on [0, 3π].
• 2013 Exam 1 Q4 — 47% — 2 marks: sin(x/2) = −1/2 on [2π, 4π].
• 2019 Exam 1 Q4a — 48% — 2 marks: 1 − cos(x/2) = cos(x/2) on [−2π, π].

4.5 Trigonometric general solutions, solution counts and exact values (5)

• 2023 Exam 2 Section B Q2dii — 12% — 2 marks: all possible values of the phase constant n (general solution required).
• 2019 Exam 2 Section A Q19 — 25% — sum of the solutions to tan(2x) = d on 0 < x < 5π/4, in terms of α.
• 2021 Exam 2 Section A Q16 — 31% — exact value of sin(x) + cos(y) given cos(x), sin²(y) and quadrant restrictions.
• 2008 Exam 2 Section B Q4dii — 40% — 2 marks: h′(x) = 0 giving x = 2; general solutions offered in calculator syntax.
• 2017 Exam 2 Section A Q12 — 45% — sum of solutions of sin(2x) = −√3/2 on [−π, d] equals −π; find d.

4.6 Logarithm and exponential equations (9)

• 2024 Exam 1 Q6 — 9% — 4 marks: log laws to a cubic, factor theorem, then reject two of three roots on the domain.
• 2009 Exam 1 Q9 — 22% — 4 marks: 2 log_e(x) − log_e(x + 3) = log_e(1/2); reject the negative root.
• 2020 Exam 1 Q4 — 26% — 3 marks: 2 log₂(x + 5) − log₂(x + 9) = 1; validity check omitted by most.
• 2016 Exam 1 Q5aiii — 31% — 2 marks: "Show that h(x) + h(−x) = f((g(x))²)" for h(x) = log_e(x² + 1).
• 2013 Exam 2 Section A Q18 — 35% — which log identity is true for all positive real x.
• 2015 Exam 1 Q7b — 36% — 3 marks: 3e^t = 5 + 8e^{−t}; quadratic in e^t.
• 2014 Exam 1 Q6 — 44% — 2 marks: log_e(x) − 3 = log_e(√x).
• 2019 Exam 2 Section A Q20 — 47% — log_x(y) + log_y(z) rewritten by change of base.
• 2019 Exam 1 Q9d — 48% — 2 marks: f(g(x)) = 0 reduces to a quadratic in e^x.

4.7 Literal equations and formulation in terms of a parameter (20)

• 2013 Exam 2 Section B Q4c — 6% — 2 marks: find m in terms of k for a tangent from a fixed point.
• 2017 Exam 2 Section B Q2f — 8% — 3 marks: equate two gradient expressions in u, solve to two decimal places.
• 2018 Exam 2 Section B Q5f — 8% — 1 mark: value of a for which g and g⁻¹ share endpoints.
• 2010 Exam 1 Q11a — 11% — 2 marks: h in terms of r by similar triangles.
• 2016 Exam 2 Section B Q4fi — 18% — 2 marks: the rule for A(k), the area between g and y = x.
• 2015 Exam 1 Q10a — 20% — 1 mark: coordinates of a point in terms of θ.
• 2008 Exam 2 Section B Q3g — 22% — 2 marks: the equation of the curve CD, in the variables the question names.
• 2016 Exam 2 Section B Q4ei — 24% — 2 marks: coordinates of X in terms of k.
• 2010 Exam 2 Section B Q3c — 26% — 2 marks: h in surd form from two trigonometric expressions.
• 2018 Exam 1 Q7b — 27% — 2 marks: distance OP expressed in the form a√b/b.
• 2023 Exam 2 Section A Q17 — 28% — volume of a cylinder in terms of the sheet dimensions x and y.
• 2008 Exam 1 Q9b — 31% — 2 marks: total area as a sum of rectangles and triangles in one variable.
• 2014 Exam 1 Q10bi — 32% — 1 mark: v in terms of u by similar triangles.
• 2022 Exam 2 Section A Q19 — 34% — the maximising x for an open box of width a and length b, fully literal.
• 2008 Exam 1 Q9a — 35% — 2 marks: volume of a triangular prism in terms of x and y, equated to 1000.
• 2016 Exam 2 Section B Q4eii — 38% — 2 marks: value of k giving specified coordinates for X.
• 2009 Exam 2 Section B Q4aii — 40% — 1 mark: V in terms of h for an inverted cone.
• 2010 Exam 1 Q11b — 47% — 1 mark: S in terms of r after substituting part a.
• 2015 Exam 1 Q10cii — 47% — 1 mark: d in terms of θ.
• 2024 Exam 1 Q8c — 47% — 1 mark: k = m/3 from a stated feature (reconstructed from [AG]).

4.8 Solving f′(x) = 0 and optimisation where the algebra is the obstacle (14)

• 2014 Exam 1 Q10bii — 9% — 2 marks: minimise an area in one variable, testing endpoints as well as turning points.
• 2010 Exam 1 Q11c — 10% — 2 marks: value of r maximising S.
• 2015 Exam 1 Q10d — 11% — 3 marks: minimum area of a trapezium in terms of θ, exact.
• 2014 Exam 2 Section B Q5d — 17% — 1 mark: solve g′(x) = 0, then report n = g(x) exactly.
• 2008 Exam 1 Q9c — 30% — 3 marks: differentiate the area expression and solve for the stationary value.
• 2015 Exam 2 Section B Q2c — 33% — 3 marks: minimise a distance function, two decimal places.
• 2020 Exam 2 Section B Q2c — 33% — 2 marks: minimum distance from a point to a curve, one decimal place.
• 2021 Exam 2 Section B Q1fii — 33% — 3 marks: maximum volume from a derived formula.
• 2010 Exam 2 Section B Q4b — 38% — 3 marks: solve f ′(x) = 0 where f carries two parameters a and b.
• 2013 Exam 2 Section B Q3f — 44% — 2 marks: solve V′(x) = 0; exact value required.
• 2014 Exam 2 Section A Q15 — 44% — the cut-corners box; value of x maximising the volume, "closest to".
• 2012 Exam 2 Section B Q1d — 45% — 2 marks: dV/dx = 0, then find h; exact values required.
• 2021 Exam 2 Section B Q1d — 49% — 3 marks: solve for the maximising x and then the maximum volume.
• 2014 Exam 2 Section B Q2c — 50% — 2 marks: solve dS/dd = 0 for d, then find S; exact.

4.9 Solving f(x) = g(x) with technology to a stated accuracy (15)

• 2010 Exam 2 Section B Q3f — 12% — 2 marks: values of x satisfying T(x) = 2000√3/27, two decimal places.
• 2008 Exam 2 Section B Q3h — 16% — 1 mark: number of whole days of capsules needed; answer 6.
• 2009 Exam 2 Section B Q4eii — 16% — 2 marks: solve 14 − t = 3t^{1/3}; report as a clock time.
• 2015 Exam 2 Section B Q2d — 21% — 3 marks: three related lengths, all to two decimal places.
• 2014 Exam 2 Section B Q2f — 22% — 3 marks: related rates via dh/dt = (dh/dV)(dV/dt).
• 2009 Exam 2 Section B Q4dii — 25% — 1 mark: integrate, apply the boundary condition, invert t = h³/27 to h = 3∛t.
• 2015 Exam 2 Section B Q2a — 33% — 2 marks: angle of inclination from a gradient, to the nearest degree.
• 2015 Exam 2 Section B Q2b — 44% — 2 marks: gradient of a curve at a point, exact.
• 2013 Exam 2 Section B Q1d — 45% — 2 marks: for how many hours T ≥ 26; the difference, not the endpoints.
• 2021 Exam 2 Section B Q1e — 46% — 2 marks: express a cut-out as a percentage of the sheet area.
• 2016 Exam 2 Section B Q1f — 47% — 2 marks: solve for x; the wrong equation was solved by many.
• 2014 Exam 2 Section B Q5a — 48% — 2 marks: factorise x⁴ − 8x completely, including the irreducible quadratic.
• 2007 Exam 2 Section B Q3a — 49% — 2 marks: solve g(x) = 0 exactly; surd answer.
• 2015 Exam 2 Section B Q2e — 49% — 3 marks: the two x-values where f(x) = g(x), two decimal places.
• 2013 Exam 2 Section B Q3cii — 50% — 2 marks: an exact distance built from a solved x.

4.10 Inequalities and solution sets (4)

• 2020 Exam 2 Section B Q5dii — 13% — 1 mark: values of a with 1 ≤ b < 1.1, three decimal places.
• 2020 Exam 2 Section B Q1f — 23% — 2 marks: all x for which the vertical distance D is at most 2 units.
• 2007 Exam 2 Section A Q21 — 27% — the solution set {x : cos²(x) + 2cos(x) = 0} as a union.
• 2012 Exam 2 Section B Q1b — 38% — 2 marks: solve V(x) > 0 and write it as an interval.

4.11 "For all values of" — proofs and bounding arguments (4)

• 2016 Exam 2 Section B Q4fii — 2% — 2 marks: "Show that 0 < A(k) < 2 for all k > 1."
• 2016 Exam 2 Section B Q4d — 6% — 2 marks: "Show that x₁ < x₂ implies g(x₁) < g(x₂)."
• 2020 Exam 2 Section B Q2f — 7% — 2 marks: values of k for which the width is strictly less than 20 m for all parts of the river.
• 2024 Exam 2 Section B Q5bii — 34% — 2 marks: "show that" there are no solutions, by a range argument on sin(2x).

4.12 Numerical algorithms (2)

• 2023 Exam 2 Section B Q3g — 21% — 1 mark: explain why a stationary point must not be used as x₀ in Newton's method.
• 2024 Exam 1 Q7a — 30% — 3 marks: the trapezium rule with three trapeziums, exact surd arithmetic, by hand.

4.13 2024 November parts whose paper text is unavailable (2)

• 2024 Exam 2 Section B Q1bi — 10% — 2 marks; [RPT] comment in full: "Some of the values were often missing."
• 2024 Exam 2 Section A Q17 — 27% — multiple choice; [RPT] publishes only the worked line of symbols.

4.14 2011 — listed for completeness, excluded from all statistics (7)

Per 01-study-design.md §0.1, the 2011 reports are OCR-corrupt: labels duplicated, distributions incomplete.

• 2011 Exam 1 Q10–10c — 2% — 2 marks: "show that" dL/dθ = 0 when BD = 2CD.
• 2011 Exam 1 Q3b–3b — 3% — 2 marks: sin(2x + π/3) = 1/2 over a restricted domain.
• 2011 Exam 2 Section A Q22 — 4% — multiple choice; wording OCR-destroyed.
• 2011 Exam 1 Q3b–3b — 7% — 2 marks: composite function factorised to find c and d (duplicate label).
• 2011 Exam 2 Section B Q3dii — 9% — 3 marks: k such that a stationary point lies at an intersection of y = g(x) and y = g⁻¹(x).
• 2011 Exam 1 Q5b–5b — 25% — 3 marks: k = −4 for infinitely many solutions, with justification; three methods published.
• 2011 Exam 2 Section B Q4a — 39% — 3 marks: "show that" a pipeline length equals √(m⁴ + 3m² + 4) […].

4.15 Which types separate most

Ranked by the median pct of the separators in each group, hardest first (2011 and the two unreadable 2024 parts excluded):

Two readings of this table matter.

The abstract end of the area is where the state collapses. Every group in the top two is about a condition on a parameter rather than a number. The bottom of the table — solve this trigonometric equation, solve this system, solve f(x) = g(x) — sits at 41–45%, which is bad in absolute terms but is the ordinary difficulty of the subject. The top sits at 6–10%, which is a different phenomenon.

The dominant failure is non-attempt, not error. Reading the report comments for the sub-25% group: "Many students did not attempt this question" (2016 Exam 2 Section B Q4d, 2016 Exam 2 Section B Q4fii, 2018 Exam 2 Section B Q5f, 2008 Exam 2 Section B Q3h, 2011 Exam 2 Section B Q4a), "Many students did not attempt the question" (2023 Exam 2 Section B Q5e), "very few students made significant progress" (2025 Exam 1 Q9bii), "Few students realised" (2010 Exam 1 Q11a, 2018 Exam 1 Q8b), "Many students did not attempt this question or had difficulty in deriving the area function" (2015 Exam 1 Q10d). The zero columns confirm it: 2016 Exam 2 Section B Q4fii — 94% scored zero; 2016 Exam 2 Section B Q4d — 90%; 2013 Exam 2 Section B Q4c — 90%; 2018 Exam 1 Q8b — 89%; 2020 Exam 2 Section B Q5dii — 87%; 2014 Exam 2 Section B Q5d — 83%; 2019 Exam 1 Q9f — 81%; 2015 Exam 1 Q10a — 80%.

4.16 What the reports say went wrong, by theme

Counting report comments across all 195 algebra parts:

The five sentences VCAA writes most often, each quoted from a real report:

1. "Exact values were required." — 2012 Exam 2 Section B Q1d, 2013 Exam 2 Section B Q1c, 2014 Exam 2 Section B Q2c, 2016 Exam 2 Section B Q4bii, 2020 Exam 2 Section B Q1ei, 2020 Exam 2 Section B Q2b, 2021 Exam 2 Section B Q1d, and more. On 2012 Exam 2 Section B Q1d: "Some gave an approximate answer for x and h, such as x = 20.78, h = 29.69. Exact values were required."
2. "Some students did not work to the required number of decimal places or rounded incorrectly." — 2015 Exam 2 Section B Q2d, 2015 Exam 2 Section B Q2e, 2010 Exam 2 Section B Q3f, 2023 Exam 2 Section B Q3f.
3. "Students are reminded that for any question worth more than one mark, working must be shown in order to be awarded full marks." — 2025 Exam 1 report, Q7b. Compare 2013 Exam 2 Section B Q3f: "Some students gave the answer without showing any method."
4. On domain: "many students overlooked the fact that the domain of this log function must be [x > 4] and, as a result, did not reject the two invalid solutions" — 2024 Exam 1, Q6. And 2016 Exam 2 on Q4e.i: "Some students did not consider the domain and chose the incorrect value for x."
5. On technology: "A number of students did not use the appropriate syntax to separate the coefficient a from the variable x when using technology" — 2010 Exam 2, Q4b. And 2016 Exam 2 on Q4e.i: "Some did not include a multiplication sign between k and x when using technology."

One more that is specific to this area and recurs in the 2015, 2022, 2024 and 2025 reports: method-choice waste. 2024 Exam 1 on Q2: "The quadratic was readily factorised by inspection, yet a large proportion of students used the quadratic formula." 2015 Exam 1 on Q7b: "Many students solved via the quadratic formula rather than using simpler factorising techniques." 2025 Exam 1 on Q9a: "Many students expanded the expression and formed a quartic equation but did not proceed further. Students who took the more efficient approach of taking square roots to form two quadratic equations generally reached the correct solutions."

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5. What makes a hard one hard

Six mechanisms. Every separator below 25% uses at least one; the ones below 10% usually use three.

5.1 A parameter instead of a number

The moment the equation contains a letter that is not the variable, the state's success rate roughly halves. Compare two questions of near-identical mathematical content:

• 2015 Exam 2 Section A Q6 — 91% — P(x) = x³ − ax² − 4x + 4, P(3) = 10, find a. One parameter, one equation, one numerical answer.
• 2016 Exam 2 Section B Q4ei — 24% — find the coordinates of X in terms of k. One parameter, but the answer must contain it.

The difficulty is not algebraic; it is that students are unwilling to leave a letter in an answer. 2015 Exam 1 report on Q10a: "The most common error in responses to this question was the oversight of 2 for the x coordinate" — students substituted a number where a symbolic expression was required. 2019 Exam 2 on Q2e.i: "The answer had to be given in terms of a."

2010 Exam 2 Section B Q4b (38%) is the cleanest demonstration: f(x) = (ax − 1)³(b − 3x) + 1 with two parameters, and the report's entire fault list is about parameter handling — "did not use the appropriate syntax to separate the coefficient a from the variable x".

5.2 "For all values of" — quantification

A question that says for all is asking for a proof or a bounding argument, and the state responds with examples. The four instances in §4.11 have a median of 6.5% and their report comments are unanimous.

2016 Exam 2 Section B Q4d — 6% — "Show that x₁ < x₂ implies that g(x₁) < g(x₂)":

"Many students did not attempt this question. Some just substituted in specific values, which was not acceptable."

2020 Exam 2 Section B Q2f — 7% — "Find the values of k for which the distance north across the river, for all parts of the river, is strictly less than 20 m":

"Some students were able to set up an appropriate inequality. There was no need to write out the expression … Some students did not substitute either x = 0 or x = 200 … A common incorrect approach was solving [the general equation]."

The technique both questions want is the same: a "for all" condition over an interval reduces to a condition at the extreme values of that interval. Find where the quantity is largest or smallest, impose the condition there, and it holds everywhere.

5.3 Solution sets, set notation, and intervals

The answer is not a number but a set, and the notation is marked.

• R \ {…} for "unique solution" questions (2007 Exam 2 Section A Q5, 36%).
• Union of sets for null-factor solution sets (2007 Exam 2 Section A Q21, 27%).
• Intervals with correct brackets for inequalities (2012 Exam 2 Section B Q1b, 38%: "incorrect notation such as [0, 36] or {0, 36}").
• Ordered intervals (2020 Exam 2 Section B Q5dii, 13%: "Some students had the values within the interval in the wrong order. Others had incorrect brackets").
• Two intervals where one was given ([AG] 2024 Exam 2 Q5d: "Two intervals only full marks").

There is no partial credit for the right numbers in the wrong container.

5.4 Exactness

The Exam 1 instruction page says it every year: "In all questions where a numerical answer is required, an exact value must be given unless otherwise specified." Twenty-two of the 195 algebra parts have a report comment invoking exactness. The rule in practice:

• Exam 1 default: exact. Surds, fractions, log_e(…), π. Never a decimal.
• Exam 2 default: also exact, unless the question specifies an accuracy. 2012 Exam 2 Section B Q4ci (60%): "Some students gave an approximate answer when an exact value was required." 2006 Exam 2 Section B Q3b (66%): "The main errors were giving an approximate instead of an exact value for a or incorrectly writing down the value from the CAS."
• When an accuracy is given, it is exact in the other direction. 2015 Exam 2 Section B Q2f (54%): "Some gave their answers in exact form" — and lost the mark.

The failure mode nobody expects is premature rounding. 2013 Exam 2 Section B Q1d (45%): "Others used approximate values in their calculations." 2015 Exam 2 Section B Q2c (33%): "Some worked to one decimal place." Carry full precision; round once, at the end.

5.5 Domain restrictions

Two distinct traps, and the reports separate them.

Implied domain in logarithmic and radical equations. The algebra generates roots the original expression cannot accept. 2024 Exam 1 Q6 (9%), 2009 Exam 1 Q9 (22%) and 2020 Exam 1 Q4 (26%) are three versions of the same failure. The 2024 report: "Some students managed to engage with the implied domain of the problem" — phrased as though it were remarkable.

Stated domain in trigonometric equations. The domain is given, and must be transformed before solving. Every technology-free trigonometric report names it:

"giving answers outside the given domain, including negative values, or only finding x = 2π" — 2007, Q8a

"Solving this expression caused some students problems when the given domain was ignored. It resulted in either only one solution or too many solutions." — 2008, Q3

"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, Q4

"Many students did not account for the restricted domain." — 2019, Q4a

And in Exam 2, the mirror error — a general solution given where a domain was stated. 2025 Exam 2 on Q4b (78%): "Others incorrectly gave extra solutions or a general solution, not considering the restricted domain."

A third variant: domain restriction as the subject of the question. 2022 Exam 1 Q5b — "Find the maximal domain of f, where f (x) = log_e(x² […] 2x […] 3)" (signs lost in extraction) — and 2025 Exam 1 Q5b — "Find the largest value of a such that when g is restricted to the domain (−∞, a] it has an inverse function."

5.6 Solution counting as a geometric question

The hardest type in the area (§4.3, median 10%) is hard because it cannot be done by manipulation. 2025 Exam 1 Q9b.ii — 4% — asks for the positive w giving exactly three solutions to w²/(x−1)² = (x − w)². [RPT] publishes five distinct methods; the report says the successful route was "a connection between the turning point and the number of solutions". The mathematics: three solutions occur when a quartic's local maximum touches the horizontal line, i.e. at a repeated root, i.e. at a boundary of the discriminant.

The general principle, and it covers 2018 Exam 1 Q8b, 2016 Exam 2 Section B Q4eiii, 2023 Exam 2 Section A Q14, 2021 Exam 1 Q9bi and 2025 Exam 2 Section B Q4fiii as well:

A question about how many solutions exist is a question about where the boundary case sits. Find the parameter value at which two solutions merge into one (tangency, repeated root, turning point on a line). That value is the answer, or an endpoint of it.

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6. A worked method sheet

The ten highest-yield types, with the technology-free version of each. These are the procedures to be able to execute without thinking.

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M1. Simultaneous linear equations with a parameter

Technology-free, and the only method worth using.

Given a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

1. Compute D = a₁b₂ − a₂b₁. Set D = 0 and solve for the parameter. Factorise by inspection before reaching for the quadratic formula — the 2024 report singles this out.
2. You now have one or two candidate values. Test each by substituting back into both original equations. - Equations become identical → infinitely many solutions. - Same left-hand sides, different right-hand sides → no solution.
3. Answer the question actually asked. "Unique solution" is R \ {all roots of D = 0}.
4. State the justification in one line. In both 2022 and 2024 a mark was allocated to it.

Worked shape (2024 Exam 1 Q2, [AG] Method 3): equal gradients give the quadratic; solving it gives two values; checking the y-intercepts eliminates the one giving infinitely many solutions; the survivor is the answer.

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M2. Trigonometric equations over a restricted domain, by hand

The guaranteed Exam 1 type. Five steps, never four.

1. Isolate the circular function: sin(θ) = c, where θ = ax + b.
2. Transform the domain. x ∈ [p, q] ⟹ θ ∈ [ap + b, aq + b]. Write this line down explicitly; it is where the marks are lost.
3. Base angle from the exact value; quadrants from the sign of c. - sin positive: Q1, Q2. sin negative: Q3, Q4. - cos positive: Q1, Q4. cos negative: Q2, Q3. - tan positive: Q1, Q3. tan negative: Q2, Q4.
4. List every θ in the transformed domain. Add or subtract 2π (or π for tan) until you leave the interval.
5. Solve back for x and sanity-check the count: a domain of length L with sin(kx) contains roughly kL/π solutions.

Memorise the exact values. The 2024 report states the expectation directly: "It is expected that students will have a way of remembering the exact values of [sin and cos] for values of [x] between 0 and [π] inclusive, as specified in the key knowledge of the study design."

For √3 sin(x) = cos(x): divide both sides by cos(x) to get tan(x) = 1/√3. The 2010 report shows most of the state that recognised this could not finish it.

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M3. Logarithmic equations

Technology-free.

1. Write the implied domain first. Every log_e(A) requires A > 0. Record the intersection of all such conditions before touching the algebra.
2. Move coefficients inside as powers: n log(A) = log(Aⁿ).
3. Combine to a single logarithm on each side: log(A) + log(B) = log(AB), log(A) − log(B) = log(A/B).
4. Remove the logarithms. Same base both sides → equate arguments. One side a constant c → A = base^c.
5. Solve the resulting polynomial. Factorise; use the factor theorem if it is a cubic.
6. Reject every root failing step 1, and say so.

Worked shape (2024 Exam 1 Q6, [AG]): log₃((x−4)²) + log₃(x) = 2 ⟹ log₃(x³ − 8x² + 16x) = 2 ⟹ x³ − 8x² + 16x − 9 = 0. P(1) = 1 − 8 + 16 − 9 = 0, so (x − 1) is a factor: (x − 1)(x² − 7x + 9) = 0, giving x = 1 and x = (7 ± √13)/2. The domain is x > 4, so only (7 + √13)/2 survives.

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M4. Exponential equations quadratic in e^x

Technology-free.

1. If there is a negative index, multiply through by e^x to clear it.
2. Substitute a = e^x. The equation becomes a² + ba + c = 0.
3. Factorise — do not use the quadratic formula unless it genuinely will not factor.
4. Discard any root a ≤ 0, because e^x > 0 for all real x. Say why.
5. Take logarithms of the survivors: x = log_e(a).
6. Check you have not discarded a = 1, which gives the perfectly valid x = 0.

Worked shape (2015 Exam 1 Q7b): 3e^t = 5 + 8e^{−t} ⟹ 3e^{2t} − 5e^t − 8 = 0 ⟹ (e^t + 1)(3e^t − 8) = 0. e^t = −1 is impossible; e^t = 8/3 gives t = log_e(8/3).

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M5. Index-law equations

Technology-free, two lines.

1. Rewrite both sides over the same base. 8 = 2³, 9 = 3², 100 = 10², 27 = 3³.
2. Equate the exponents.
3. Solve the linear equation and simplify the fraction.

Worked shape (2014 Exam 1 Q4): 2^{3x−3} = 8^{2−x} = 2^{3(2−x)} ⟹ 3x − 3 = 6 − 3x ⟹ x = 3/2.

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M6. The discriminant

Technology-free.

1. Collect everything to one side in the form ax² + bx + c = 0. The parameter usually appears in more than one of a, b, c — this is the step people skip.
2. Δ = b² − 4ac, expanded and simplified.
3. Translate the requirement: - no real solutions → Δ < 0 - exactly one → Δ = 0 - two → Δ > 0 - two of opposite sign → Δ > 0 and c/a < 0 - two positive → Δ > 0, c/a > 0, −b/a > 0
4. Solve the inequality in the parameter and present it as an inequality or interval.
5. If a itself contains the parameter, check the degenerate case a = 0 separately — the equation is then linear and has exactly one solution.

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M7. Literal equations and formulation

Technology-free, and the standard opening of every modelling question.

1. Name the constraint. Similar triangles, Pythagoras, a fixed volume, a fixed perimeter, a formula from [FS].
2. Solve the constraint for the unwanted variable.
3. Substitute into the quantity of interest.
4. Simplify before differentiating. Combine fractions, cancel, expand where it helps.
5. Leave the parameter in the answer, and use the variables the question named — 2008 Exam 2 Section B Q3g: "Some students used y and x, instead of z and d … Students should use the variables as given in the question."

For similar triangles specifically, the 2010 Q11a error is instructive: the correct relation there was (2 − r)/2 = h/5, not h/5 = r/2. Draw the two triangles separately and label corresponding sides before writing the ratio.

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M8. Newton's method

Technology-free version (as in [SAMP] Exam 1 Q6).

1. Differentiate f.
2. Formula from [FS]: x₁ = x₀ − f(x₀)/f ′(x₀).
3. Evaluate f(x₀) and f ′(x₀) exactly. Keep the fraction.
4. Present x₁ as an exact fraction unless a decimal accuracy is demanded.

Technology-active version.

1. Store f and f ′. Iterate, holding full precision.
2. Round only the reported values, to exactly the stated accuracy.
3. If asked to draw the tangent at x = x₁, draw a straight line through (x₁, f(x₁)) with gradient f ′(x₁); it crosses the x-axis at x₂.

The failure condition. If f ′(x_n) = 0 the tangent is horizontal and x_{n+1} is undefined — the algorithm halts. This is the whole of 2023 Exam 2 Section B Q3g (21%). If the estimate is near a stationary point, the tangent is nearly horizontal and the next estimate is thrown far away. Saying "these are the turning points" is not an answer; saying "the tangent is horizontal, so it never meets the x-axis, and x₁ is undefined" is.

Verification framing. [SAMP] Exam 1 Q6a pairs the method with a sign check: f(−1) > 0 and f(−2) < 0 guarantees a root between them. Expect the pairing.

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M9. The trapezium rule

Technology-free — and integration is explicitly not acceptable.

1. Strip width w = (x_n − x₀)/n.
2. Node values. Evaluate f at x₀, x₁, …, x_n, exactly. For circular functions this means exact values.
3. Apply the formula from [FS], doubling every interior value: Area ≈ (w/2)[ f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(x_{n−1}) + f(x_n) ]
4. Combine into a single exact expression. The 2024 report penalises leaving the answer as separate area pieces.
5. For the conceptual version: the rule overestimates on a concave-up arc and underestimates on a concave-down arc.

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M10. "Exactly n solutions"

The hardest type; here is the decision procedure.

1. Reduce f(x) = g(x) to a single equation in x carrying the parameter. Do not expand if a factored form is available — 2025 Exam 1 Q9a's efficient route was taking square roots of a difference of squares, not expanding to a quartic.
2. Choose a route: - Quadratic → discriminant. Count via Δ. - Factorable into two pieces → count what each contributes and force the total. Three solutions = one factor giving one, the other giving two. - Neither → the "hence" route: find the turning point of one side in terms of the parameter and set its y-value equal to the other side. Tangency is the boundary.
3. Identify the boundary case explicitly. Two solutions merging into one is always where the count changes.
4. Write the reasoning out. The 2025 report: the successful students were those "able to demonstrate clear mathematical communication skills (listing all the possible solutions, compare them, discuss the nature of turning points or discuss the impact of the discriminant on the number of solutions)."
5. If the question offers "Hence, or otherwise", use the hence. Across 2017 Q6b, 2019 Q9f and 2025 Q9b.ii, the reports say the same thing: the students who ignored the previous part did not finish.

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7. The one-page version

• Algebra is 21% of the marks and the hardest of the four areas: mean pct 43.0, separator rate 61.4%.
• It is never a topic on its own. 89 of 195 parts sit inside Section B questions about something else; the study design says so — "This content is to be incorporated as applicable to the other areas of study."
• Exam 1 carries 4–5 algebra parts for 10–11 marks in the current design — a quarter of the technology-free paper. Exam 2 carries 6–11 parts for 9–19 marks.
• The guaranteed annual items: a technology-free equation (log, index, exponential or trigonometric) in Exam 1; a parameter-condition multiple choice in Section A; a literal-equation formulation inside Section B.
• Newton's method, pseudocode and the trapezium rule entered in 2023. Newton's method has never yet appeared in a live November Exam 1 but is examinable there. Pseudocode is Exam 2 only. The trapezium rule has one November data point: 2024 Exam 1 Q7a, 30%.
• Matrix transformations left in 2023. No paper since 2022 contains a matrix. Functional relations (f(x + y) = …) left too.
• The five things that cost marks, in order of report frequency: exactness, domain, technology syntax, working not shown, decimal-place accuracy.
• The hardest question types are the ones where the answer is a condition on a parameter rather than a number: "for all values of" (median 6.5%), "exactly n solutions" (median 10%), interval and set answers (median 25%).
• On those, the dominant failure is non-attempt. Between 80% and 96% of the state scores zero. A rough sketch and one sentence of reasoning is worth more than a blank page, and on a one-mark part at the end of a fifteen-mark question it is worth ninety seconds.