Functions, relations and graphs

The definitive reference, built from the 2006–2026 examination archive. Compiled 15 September 2026. Companion to research/sm/01-study-design.md (read that first — nothing here contradicts it).

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0. Sources and evidence tags

Corpus defects that bite in this area specifically. The 2011 papers extract with a CID-shifted font encoding — every code point is offset and all spaces are lost, so 2011 Exam 1 Q3b–3b, 2011 Exam 1 Q8 and 2011 Exam 2 Section A Q1 cannot be quoted from the text files and are described from the report text only. The 2024 November paper text files are empty (24–36 bytes); 2024 questions below are quoted from the page images in corpus/sm/pages/. The 2025 NHT text files are also empty. Where a rule could not be read reliably, that is stated rather than reconstructed.

One more gap that shapes every table below. Of the 264 tagged parts, 27 are NHT sittings (2017–2025) and none of them carries a pct — VCAA does not publish mark distributions for the Northern Hemisphere Timetable. Every percentage, separator count and median in this document is therefore computed on the 237 November parts only. NHT papers are used here as evidence of wording and scope, never of difficulty. The 2026 NHT papers are in corpus/sm/text/ but not yet in [QJSON].

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1. What the study design puts in this area — and what the exams actually test

1.1 The study design text, verbatim

[SD] overview:

"In this area of study students cover rational functions and other simple quotient functions, curve sketching of these functions and relations, and the analysis of key features of their graphs including intercepts, asymptotic behaviour and the nature and location of stationary points and points of inflection and symmetry."

[SD] content — the entire area of study is three dot points:

This area of study includes:
- rational functions and the expression of rational functions of low degree as sums of partial fractions
- graphs of rational functions of low degree, their asymptotic behaviour, and the nature and location of stationary points and points of inflection
- graphs of simple quotient functions, their asymptotic behaviour, and the nature and location of stationary points and points of inflection.

1.2 The three dot points are not the scope

Read literally, the content list contains no inverse circular functions, no ellipses, no hyperbolas, no absolute value, no trigonometric identities and no loci. All of those are examined every year. They are in scope because they arrive through four other doors, and a 45+ candidate needs to know which door each one comes through:

Two further Outcome 1 phrases are load-bearing for this area and are easy to miss:

• "the form of their sketch graphs and their key features, including linear asymptotes." The word linear is why oblique asymptotes are squarely examinable and why VCAA is content to set y = (x² + a)/(bx + c). It is also why a "curved asymptote" (y = −x², y = x² + x) is never asked for by name even when the function has one — see 2016 Exam 2 Section B Q1c and 2024 Exam 2 Section B Q1a, where only the vertical asymptotes had to be labelled.
• "apply implicit differentiation, by hand in simple cases." "By hand" is an Exam 1 flag. It licenses 2024 Exam 1 Q8 and 2025 Exam 1 Q1.

And one Outcome 3 key skill is the technology-paper counterpart: "use appropriate domain and range specifications to illustrate key features of graphs of functions and relations." [RPT25 E2] turns this into explicit advice: "To improve accuracy, students can sketch the function on their CAS calculator and set the domain, range and scale to match those provided in the question."

1.3 What the archive says about weight

All figures from [QJSON], November sittings 2006–2025 unless stated.

Marks by area of study, November 2006–2025:

And under the current design (November 2023–2025), the mix moves but this area holds:

1.4 Where in the paper it lives — and where it hurts

This is the single most important table in the document. Two-thirds of every Functions-and-graphs part ever set on Examination 1 has been a separator, and the median Exam 1 part in this area is answered completely by 39% of the state. No other combination of paper and area of study in the archive is that hostile. The reason is structural: Exam 1 asks for a sketch with every feature labelled, and a sketch is marked against a checklist. Miss one asymptote equation out of four required labels and you lose a mark; miss two and you lose the question.

Comparison of separator rates across all six areas (November 2006–2025, graded parts only):

Across the whole archive this area sits mid-table, because its 75 Section A items are easy (median 68%). Restrict to the current design and to written work only and it is bottom of the table: median 47% on 27 written parts 2023–2025, below Space and measurement (55%), Calculus (56%), Algebra (59%) and Statistics (65.5%). (01-study-design.md §1.5 quotes 44% on a narrower cut of 22 parts; either cut puts this area last.)

1.5 Where in each paper the questions sit

[QJSON], November sittings, by year:

Three structural facts fall out of that table.

1. Examination 2 Section B Question 1 is the graph question. In 13 of the 20 years (2008, 2013, 2014, 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2023, 2024, 2025) the first extended-response question of Exam 2 is a functions-and-graphs question, usually a rational or quotient function with a sketch worth 3 marks and a volume of revolution attached. If you open Exam 2 Section B and the first question is not about a graph, that is the unusual year.
2. The Section A items cluster at Q1–Q5. In 15 of 20 years, every one of the area's Section A items is numbered 1 to 5. VCAA front-loads Section A with functions and graphs. The exceptions are the paths-of-particles items that drift into the vector block (2019 Q7, 2019 Q11, 2020 Q9, 2021 Q7, 2023 Q9, 2024 Q15, 2025 Q9).
3. On Exam 1 it moves. It has been Q1 (2008, 2023), Q2 (2012, 2019, 2025), mid-paper (Q4–Q7) and last (Q8–Q10, in 2009, 2012, 2017, 2022). There is no positional tell. What is constant is that every November Exam 1 in the archive contains at least one part from this area, and in 17 of 20 years at least one of those parts is a sketch or a graph-derived exact-value question.

A tagging caveat, stated honestly. [QJSON]'s topic tags are assigned per question part, and a handful of parts tagged to this area are really vector or complex-number questions that happen to be about a graph or a path: 2008 Exam 1 Q8b (fourth vertex of a parallelogram), 2009 Exam 2 Section B Q2d (line meets |z| = 2), 2021 Exam 2 Section A Q5 (maximum |z| on a circle), 2021 Exam 2 Section B Q3bii/Q3biii/Q3c (a leaking-vessel differential equation), 2023 Exam 2 Section B Q5b/Q5f (shortest distance to a line segment; a line meeting a plane). They are listed in §4 because the brief asks for every part with pct ≤ 50, but they are flagged there so nobody drills the wrong thing.

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2. The complete catalogue of question types

Thirty types. For each: a name, the literal VCAA wording quoted from a real paper, what it is really testing, the standard method, at least three archive instances with ref and pct, typical marks, and the traps the reports name.

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Type 1 — Sketch a rational function with an oblique asymptote

VCAA wording, 2023 Exam 1 Q1:

"Consider the function f with rule f(x) = (x² + x − 6)/(x − 1).
a. Show that the rule for the function f can be written as f(x) = x + 2 − 4/(x − 1). (1 mark)
b. Sketch the graph of f on the axes below, labelling any asymptotes with their equations. (3 marks)"

Really testing: that you convert (quadratic)/(linear) into quotient-plus-remainder form before you try to sketch, and that you know the quotient is the oblique asymptote.

Standard method. (i) Divide, by long division or by forcing the numerator: x² + x − 6 = (x − 1)(x + 2) − 4. (ii) Vertical asymptote where the denominator vanishes. (iii) Oblique asymptote y = the polynomial part. (iv) Axis intercepts from the original rule. (v) Sign of the remainder term on each side tells you which side of the oblique asymptote each branch sits on. (vi) Draw both asymptotes with a ruler first, then the branches.

Traps the reports name. 2008 Exam 1 Q1 is the canonical failure. [RPT08 E1]: "Many students were able to find the equations of the asymptotes, many were able to find the x-intercept and many were able to find the turning point, but few were able to complete all of these successfully… Those who were able to fulfil all of these requirements often did not show asymptotic behaviour, with some graphs colliding with the asymptotes and others swerving away from them. A large number of students did not give coordinates for the x-intercept or the turning point, and just gave the x-coordinate… The vertical asymptote was often labelled as y = 0 and sometimes as y = x; often there was no equation given. Some students correctly found the oblique asymptote but then drew its graph as if it was that of y = −x or y = −2x." [RPT22 E2] on Section B Q1b: "The oblique asymptote was occasionally sketched hastily without due regard to accurate position." Typical marks: 3–5, and the marks are a checklist.

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Type 2 — "Show that f(x) can be written in the form …" (partial fractions / division)

VCAA wording, 2024 Exam 1 Q3:

"Let f : R \ {−1} → R, f(x) = (x − 1)²/(x + 1)². The rule f(x) can be written in the form f(x) = A + B/(x + 1) + C/(x + 1)², where A, B, C ∈ Z.
a. Show that A = 1, B = −4 and C = 4. (1 mark)"

and 2021 Exam 2 Section B Q1a:

"Let f(x) = (2x − 3)(x − 5) / ((x − 1)(x + 2)). Express f(x) in the form A + (Bx + C)/((x − 1)(x + 2)), where A, B and C are real constants." (1 mark)

Really testing: that partial fractions and polynomial division are setup for the sketch, not an end in themselves — and that a "show that" demands visible, complete algebra.

Standard method. Force the numerator into multiples of the denominator, or equate coefficients after multiplying out. For a repeated linear factor, substitute u = x + 1 and expand. Never quote the answer back.

Traps. [RPT25 E1] on Q9a: "Students needed to be very careful to ensure that their working actually produced the required result. Some poor algebraic working was observed." [RPT23 E1] on Q1a: "Various approaches were seen, including long and synthetic division. Some students made algebraic or arithmetic errors." Typical marks: 1–2. This part is usually the easiest in the question and the one a 45+ candidate must never drop.

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Type 3 — Sketch a (linear)/(quadratic) rational function: two vertical asymptotes and y = 0

VCAA wording, 2018 Exam 1 Q5 (4 marks):

"Sketch the graph of f(x) = (x + 1)/(x² − 4) on the axes provided below, labelling any asymptotes with their equations and any intercepts with their coordinates."

Really testing: that a numerator of lower degree forces y = 0, and that a three-branch graph must actually be drawn with three branches.

Standard method. Factorise the denominator → two vertical asymptotes. Degree of numerator < degree of denominator → horizontal asymptote y = 0. Find both intercepts. Test the sign of f in each of the three intervals to decide whether each branch goes up or down at each asymptote. Check whether the curve crosses y = 0 (it does, at the x-intercept) — a horizontal asymptote may be crossed.

Traps. [RPT18 E1]: "Most students realised that x = −2 and x = 2 were vertical asymptotes, although the horizontal asymptote y = 0 was often not stated. Students who found the axis intercepts were not always able to position them correctly on the axes. Some students showed a stationary point of inflection on their graph or were missing the outer branches." [RPT14 E2] on Q1b: "The most frequent error was the omission of the asymptote y = 0." [RPT21 E2] on Q1c: "A significant number of responses did not include the middle branch." [RPT24 E1] on Q3c: "Some students only drew the right-hand branch of the graph. Students were much more successful in showing the correct behaviour of the graph on the left-hand side if they evaluated the function at several points." Typical marks: 3–4. This is the most reliably brutal single item type in the area.

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Type 4 — Rational function with a removable discontinuity

VCAA wording, 2025 Exam 1 Q9 (6 marks):

"Let f : R \ {1, −1} → R, f(x) = (x³ + x² − 2x)/(1 − x²).
a. Show that f(x) can be written in the form f(x) = −x − 1 + 1/(x + 1), for x ∈ R \ {1, −1}. (2 marks)
b. … Find the value of k such that the graph of g is continuous at x = 1. (1 mark)
c. Sketch the graph of y = f(x) on the axes below. Label the asymptotes with their equations. (3 marks)"

and the multiple-choice version, 2024 Exam 2 Section A Q2:

"Consider the function f with rule f(x) = { (x² + 3x − 10)/(x − 2), x ∈ R \ {2}; 7, x = 2 }. Which of the following statements is correct? … A. The function f is continuous. B. The graph of y = f(x) has a vertical asymptote. C. … horizontal asymptote. D. … a point of discontinuity."

Really testing: that a common factor cancels to a hole, not to an asymptote, and that the hole must be drawn as an open circle with its coordinates.

Standard method. Factorise fully, identify the cancelling factor, note the excluded x-value, evaluate the simplified rule there to get the y-coordinate of the hole. For the continuity part, k is exactly that y-value.

Traps. [RPT25 E1] on Q9c: "Students were required to label the asymptotes with their equations. An open circle to indicate the point of discontinuity at x = 1 needed to be shown. A number of students included an incorrect vertical asymptote or had curves that did not pass through the axis intercepts at x = 0 and x = −2. The point of discontinuity was often missing or was placed incorrectly. Students who were most successful used a ruler to draw the asymptotes and had graphs that did not curve away from the asymptotes." Typical marks: 1–3.

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Type 5 — Simple quotient functions: f(x)/g(x) where neither is a polynomial

VCAA wording, 2017 Exam 2 Section B Q1b (3 marks):

"Sketch the graph of f(x) = x/(1 + x³) from x = −3 to x = 3 on the axes provided below, marking all stationary points, points of inflection and intercepts with axes, labelling them with their coordinates. Show any asymptotes and label them with their equations."

and 2020 Exam 2 Section B Q3c (3 marks):

"Sketch the graph of y = f(x) on the axes provided below, labelling the local maximum stationary point and all points of inflection with their coordinates, correct to two decimal places." (for f(x) = x²e^(−x))

Really testing: the study design's phrase "simple quotient functions" — f/g where f and g need not be polynomials. In practice: x/(1 + x³), (4 + x² + x³)/x, 6√x/(3x² + 1), x²e^(−x), sec(4x), 1/(sec(3x) − 3/2), (x⁴ − x² + 1)/(1 − x²).

Standard method (Exam 2). Use CAS, but set the window to the printed axes. Get stationary points and inflection points to the demanded accuracy, get every asymptote by inspection of the denominator, then transfer the curve point by point onto the printed grid.

Traps. [RPT16 E2]: "Students missed out on marks for ignoring the domain of the function or a lack of accuracy in the placement of the endpoints. Students are advised to use their technology as a tool to support the sketching of an accurate graph rather than simply copying a roughly correct shape from a screen." [RPT24 E2] on Q1a: "Many students did not draw this graph accurately. To improve accuracy, students can sketch the function on their CAS calculator and set the domain, range and scale to match those provided in the question. The graph must be flatter near the intercept, with turning points and end points precisely positioned." [RPT25 E2] on Q1a: "Some responses did not include the coordinates of the maximum in exact form, but rounded to 1 decimal place. Care must be taken with the shape of the graph, with asymptotic behaviour and smooth curves. Several responses, incorrectly, sketched the point of inflection as a stationary one. Many responses did not label the horizontal asymptote y = 0." Typical marks: 2–3.

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Type 6 — Find all asymptotes (no sketch)

VCAA wording, 2019 Exam 2 Section A Q2:

"The asymptote(s) of the graph of f(x) = (x² + 1)/(2x − 8) has equation(s) …"

and 2021 Exam 2 Section B Q1b: "State the equations of the asymptotes of the graph of f." (2 marks); and 2025 Exam 2 Section B Q1c: "Find the equations of the vertical asymptotes of the curve given by y = …" (1 mark).

Really testing: whether you check all three possible kinds — vertical, horizontal, oblique — every time.

Standard method. Denominator = 0 → vertical (after cancelling common factors). Compare degrees: numerator degree < denominator → y = 0; equal → y = ratio of leading coefficients; numerator exactly one higher → divide and read the linear quotient. State each as a separate equation.

Traps. [RPT17 E2] on Q1ai — the lowest of these at 36% — "The majority of students stated the vertical asymptote but significantly fewer stated the horizontal asymptote." [RPT21 E2] on Q1b: "The most common error was to give only the vertical asymptotes, leaving out the horizontal asymptote." [RPT20 E2] on Section B Q3b: "Some students incorrectly gave an additional vertical asymptote." [RPT25 E2] on Q1c: "This question specified that separate equations should be written for each asymptote, but many responses did not include this." Typical marks: 1–2.

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Type 7 — Parameter families: how many asymptotes / stationary points

VCAA wording, 2021 Exam 2 Section B Q1d:

"Let g_k(x) = (2x − 3)(x − 5)/((x − k)(x + 2)), where k is a real constant.
i. For what values of k will the graph of g_k have two asymptotes? (2 marks)
ii. Given that the graph of g_k has more than two asymptotes, for what values of k will the graph of g_k have no stationary points? (2 marks)"

and 2022 Exam 2 Section A Q3:

"The graph of y = (x² + 2x + c)/(x² − 4), where c ∈ R, will always have
A. two vertical asymptotes and one horizontal asymptote. … E. a horizontal asymptote with equation y = 1 and at least one vertical asymptote."

and 2026 NHT Exam 2 Section B Q1f/g: "For what value(s) of k will the graph of f_k have only one straight-line asymptote?" / "For what value(s) of b will the graph of f_b have two straight-line asymptotes?"

Really testing: the cancellation cases. A vertical asymptote disappears when the parameter makes the denominator share a factor with the numerator. This is the single most reliable way VCAA generates a sub-25% item.

Standard method. Write down the values of the parameter that make a denominator root also a numerator root — there are usually two or three of them, and each must be listed separately. Then handle the generic case. For "no stationary points", differentiate symbolically and require the numerator quadratic to have non-positive discriminant, and separately handle the degenerate case where cancellation leaves a linear function (which also has no stationary points).

Traps. [RPT21 E2] on Q1di: "Very few students gave all three values. Many responses included only one value." On Q1dii: "A common error was to include other incorrect values of k. Many students left this question blank." [RPT22 E2] on Section A Q3 explains the cancellation directly: "so only one vertical asymptote in this instance." [RPT24 E2] on Section A Q3: "Solving assuming no cancellation yields … For no turning points we require … If … then … cancels to a linear function which also has no turning points." Typical marks: 1–2 each, and they are the hardest marks in the area.

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Type 8 — Stationary points and points of inflection of a quotient function

VCAA wording, 2017 Exam 2 Section B Q1a:

"ii. Find f′(x) and state the coordinates of any stationary points of the graph of f, correct to two decimal places. (2 marks)
iii. Find the coordinates of any points of inflection of the graph of f, correct to two decimal places. (2 marks)"

Really testing: that f″(x) = 0 is necessary but not sufficient for a point of inflection — concavity must change.

Standard method. Solve f′ = 0 for stationary points, f″ = 0 for candidate inflections, then test the sign of f″ either side of each candidate. Reject candidates where f″ does not change sign. Reject candidates outside the stated domain. Always give coordinates, never just the x-value.

Traps. [RPT17 E2] on Q1aiii is the definitive statement, and it explains a 13%: "The majority of students provided the correct inflection point. A common error was to erroneously include the point (0, 0), which is another point where f″(x) = 0, but it is not a point of inflection as there is no change of concavity; f″(x) does not change sign." [RPT08 E2] on Q1bii: "in some instances extra points were given which were outside the domain of f. A frequent error was to give only the x coordinates of the two points of inflection." [RPT16 E2] on Q1b: "The most common error resulted from substituting a rounded x value, yielding an incorrect y value of −1.58." Typical marks: 1–3.

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Type 9 — Justify a point of inflection by hand

VCAA wording, 2020 Exam 1 Q6 (5 marks):

"Let f(x) = arctan(3x − 6) + π.
a. Show that f′(x) = 3/(9x² − 36x + 37). (1 mark)
b. Hence, show that the graph of f has a point of inflection at x = 2. (2 marks)
c. Sketch the graph of y = f(x) on the axes provided below. Label any asymptotes with their equations and the point of inflection with its coordinates. (2 marks)"

Really testing: that "show that there is a point of inflection" is a two-part argument — f″(2) = 0 and a sign change — and that by hand you must exhibit both.

Standard method. Differentiate again (chain or quotient rule), show f″(2) = 0, then either evaluate f″ at a point either side, or argue from the sign of the factor (x − 2) in f″. State the conclusion in words.

Traps. [RPT20 E1] on Q6b: "Most students showed, by using the chain or quotient rules, that f″ = 0 when x = 2. Few students attempted to justify that a point of inflection occurred at this point." That is a 9% item created entirely by a missing justification sentence. [RPT08 E2] on Q1a: "the failure to verify the maximum turning point using the second derivative or other suitable test." Typical marks: 2.

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Type 10 — Sketch y = 1/f(x) from a given graph of y = f(x)

VCAA wording, 2019 Exam 1 Q5 (6 marks):

"The graph of f(x) = cos²(x) + cos(x) + 1 over the domain 0 ≤ x ≤ 2π is shown below.
a. i. Find f′(x). (1 mark) ii. Hence, find the coordinates of the turning points of the graph in the interval (0, 2π). (2 marks)
b. Sketch the graph of y = 1/f(x) on the set of axes above. Clearly label the turning points and endpoints of this graph with their coordinates. (3 marks)"

Really testing: the reciprocal transformation rules — maxima become minima, zeros become vertical asymptotes, values of 1 are fixed, and the endpoints of the domain map to the reciprocals of the endpoint values.

Standard method. Mark on the given graph: every zero of f (→ vertical asymptote of 1/f), every turning point (→ turning point at the reciprocal height, nature inverted), every point where f = ±1 (→ fixed), and both endpoints. Sketch through those.

Traps. [RPT19 E1] on Q5b: "Common errors included neglecting to label the turning point at (π, 1), their graph not passing through the intersection points, and poor estimation of the location of the heights with respect to the given scale. Some students drew their graphs with an open circle at the endpoints." [RPT06 E1] on Q3a: "Some strange graphs were seen for this routine reciprocal function question; however, most students made a good attempt at the shape of the graph. The main error was a failure to mention and/or label [the asymptotes]… Students should be encouraged to use a pencil to draw graphs." [RPT11 E1] on Q3b: "A large number of students seemed to interpret the denominator as x + 1 or as (x + 1)(x − 1) and consequently had one or two vertical asymptotes… typical errors included omitting the label of the intercept or the asymptote, showing the maximum point as a cusp rather than as a turning point and not showing asymptotic behaviour." Typical marks: 3.

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Type 11 — Absolute-value equations

VCAA wording, 2019 Exam 1 Q2 (3 marks):

"Find all values of x for which |x − 4| = x/2 + 7."

Really testing: case-splitting on the sign of the expression inside the modulus, and — critically — rejecting solutions that fail the case condition.

Standard method. Either split into x − 4 = x/2 + 7 (valid when x ≥ 4) and −(x − 4) = x/2 + 7 (valid when x < 4), solving each and checking the condition; or square both sides and discard extraneous roots; or sketch both graphs and read off the intersections.

Traps. [RPT19 E1]: "Some students drew a graph to support their reasoning. Students who solved linear equations were generally more successful than those who solved a quadratic equation. In the latter case, some students … had difficulty solving the quadratic equation." Typical marks: 3. The 2022 Section A item — "For the interval x ≥ 3, the graph of y = |2x − 1| + |x − 3| is the same as the graph of …" — is the easy multiple-choice version at 85%: on that interval both moduli open positively and the answer is a single straight line.

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Type 12 — Rational inequalities, answer in interval notation

VCAA wording, 2020 Exam 1 Q4 (4 marks):

"Solve the inequality (3 − x)/(x − 4) > 1 for x, expressing your answer in interval notation."

Really testing: that you may not multiply an inequality by x − 4 without knowing its sign; and that the final answer must be an interval, not a set of endpoints.

Standard method. Move everything to one side, combine over a common denominator, factorise, and use a sign table on the critical values (roots of numerator and denominator). Exclude the denominator's zeros. Alternatively sketch y = LHS and y = 1 and read off. Finish by writing the answer as a union of intervals with correct open/closed brackets.

Traps. [RPT20 E1]: "A quick sketch was helpful… Students who approached this problem algebraically were often unsure how to deal with the inequality signs. A number of students who found [the correct set] did not receive full marks as they did not write the final answer in interval notation." [RPT15 E1] on Q7b: "High-scoring students used a graphical argument for this question. Typical errors included incorrect simplification with inequalities (multiplying by a term that could be negative), including endpoint(s) and giving single value answers rather than intervals." Typical marks: 1–4. The lesson is uniform across three different papers: draw the graph, then read the intervals.

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Type 13 — Implied (maximal) domain of a composed inverse circular function

VCAA wording — the exact phrase recurs. 2009 Exam 1 Q10a: "State the implied domain and the range of f." 2016 Exam 2 Section A Q2: "The implied domain of y = arccos((x − a)/b), where b > 0, is …" 2019 Exam 2 Section A Q3: "The implied domain of the function with rule f(x) = √(1 − sec(x + π/4)) is …" 2021 Exam 2 Section A Q2: "The implied domain of the function with rule f(x) = cos⁻¹(log_e(bx)), b > 0, is …" 2023 NHT Exam 2 Section A Q1: "The implied domain and range of f(x) = sin(cos⁻¹(1 − 2x)) are respectively …" Exam 1 uses "maximal domain" interchangeably: 2013 Exam 1 Q4a, 2012 Exam 1 Q10a, 2017 Exam 1 Q10b, 2018 Exam 2 Section B Q1a.

Really testing: solving a compound inequality −1 ≤ (inner) ≤ 1 (for arcsin/arccos) or intersecting several constraints, without sign errors when dividing by a negative.

Standard method. Write the constraint from every component — the arcsin/arccos argument in [−1, 1], any square root ≥ 0, any denominator ≠ 0, any log argument > 0 — solve each, then intersect. When dividing by a negative, reverse both inequality signs. [RPT13 E1] recommends a shortcut: "The most successful technique appeared to be to solve the equation 1 − 2x = ±1 to find the endpoints of the domain."

Traps. [RPT13 E1] on Q4a: "There were many incorrect answers given for the domain (for example, [−1, 1], [−1, 3], [−0.5, 1.5]), with poor attempts to solve the inequality −1 ≤ 1 − 2x ≤ 1. Some students reached −2 ≤ −2x ≤ 0 and then divided by −2 to write 1 ≤ x ≤ 0." [RPT12 E1] on Q10aii: "Many made unfortunate slips with inclusion/exclusion of values at the boundaries." On Q10aiii: "This question involved finding the intersection of the domains found in the previous two parts. Many students seemed not to realise this." [RPT09 E1] on Q10a lists the wrong answers explicitly, including "giving the domain as [0, −4] or [0, 4] and the range as [−2, −4]" — i.e. endpoints written in the wrong order. Typical marks: 1–2.

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Type 14 — Domain and range of a transformed inverse circular function

VCAA wording, 2017 Exam 1 Q10b (2 marks): "State the maximal domain and the range of f(x) = arccos(x/2)." VCAA wording, 2018 Exam 2 Section B Q1a (2 marks): "Consider the function f : D → R, where f(x) = 2arcsin(x² − 1). Determine the maximal domain D and the range of f." VCAA wording, 2020 Exam 2 Section A Q2: "A function f has the rule f(x) = |b cos⁻¹(x) − a| … The range of f is …" (the printed rule is not fully legible in the text extraction; the answer key gives B and pct = 42)

Really testing: the base ranges, which you must know cold — arcsin : [−1, 1] → [−π/2, π/2]; arccos : [−1, 1] → [0, π]; arctan : R → (−π/2, π/2) (open, with horizontal asymptotes y = ±π/2) — and then how a dilation, reflection and translation move them.

Standard method. Domain first, from the inner expression. Range: apply the transformations to the base range endpoints in order, and if the transformation includes a reflection (negative dilation factor), swap the endpoints so the interval reads low-to-high. Use exact values (π, not 3.14) and square brackets unless the function is arctan or the domain endpoint is excluded.

Traps. [RPT17 E1] on Q10b: "The most common errors were to state the domain as (−2, 2) or [0, 2] or other variations; the range was frequently given as [0, π]" — i.e. the base range, unchanged, with the dilation ignored. [RPT18 E2] on Q1a: "Common errors included: giving open endpoints with round brackets on the intervals, decimal approximations rather than exact values and failing to state the range. Students should read questions carefully and ensure that all required information is supplied." [RPT13 E1] on Q4a: "The range was generally given correctly, but many students did not state it."

2014 Exam 1 Q7a deserves its own line. The question is one mark: "Consider f(x) = 3x·arctan(2x). Write down the range of f." Four per cent of the state got it. [RPT14 E1]: "Few realised that x and the arctan function are both positive for the same values, negative for the same values and zero for the same values" — so the product is ≥ 0, giving [0, ∞). "Many students seemed to use the product of the range of each of the 'parts', some ignored one part and others found the product of the range of one part and the variable x." This is the lowest-scoring one-mark question in the entire area.

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Type 15 — Sketch an inverse circular function, labelling endpoints

VCAA wording, 2013 Exam 1 Q4b (2 marks):

"Sketch the graph of y = arccos(1 − 2x) over its maximal domain. Label the endpoints with their coordinates."

and 2018 Exam 2 Section B Q1b (3 marks):

"Sketch the graph of y = f(x) on the axes below, labelling any endpoints and the y-intercept with their coordinates."

and 2010 Exam 2 Section B Q4a (3 marks):

"Sketch the graph of the relation y = f(x) on the axes below. Label the endpoints with their exact coordinates, and label the x and y intercepts with their exact values." (for f(x) = sin⁻¹(2x² − 1))

Really testing: endpoint coordinates, orientation (increasing or decreasing), and the shape at the endpoints — the tangent to arcsin/arccos is vertical at x = ±1, not horizontal.

Standard method. Domain and range first (Type 14). Compute both endpoint coordinates exactly. Decide orientation from the sign of the coefficient of x inside. Mark the point of inflection (the image of the origin of the base curve) and note it is non-stationary. Draw the curve steepening toward each endpoint.

Traps. [RPT13 E1] on Q4b is the complete list of ways to lose the marks: "Many students showed the incorrect orientation of the graph (i.e. pairing the endpoints of the domain and range the wrong way), perhaps using the shape of y = arccos(x) and ignoring the reflection. Several incorrectly had a stationary rather than non-stationary point of inflection in the middle of the graph (where the gradient is 2), while others gave the correct endpoint but incorrect shape (with the gradient near zero at the endpoint). A few students sketched the correct graph but did not label both endpoints with their coordinates as required — usually (0, 0) was missing." [RPT18 E2] on Q1b: "many graphs had an obvious turning point at the y-intercept rather than the required shape." [RPT10 E2] on Q4a: "Common errors included incorrect shape, omission of intercept and end point labels, not using the scale provided, and using decimal approximations for the exact values requested. The behaviour of the curve at x = 0 was the most elusive feature." Typical marks: 2–3.

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Type 16 — Derivative of an inverse circular function and the set where it is defined

VCAA wording, 2017 Exam 1 Q6 (3 marks):

"Let f(x) = 1/arcsin(x). Find f′(x) and state the largest set of values of x for which f′(x) is defined."

and 2010 Exam 1 Q5 (3 marks): "Given that f(x) = arctan(2x), find f″(π/2)."

Really testing: the chain rule on the formula-sheet derivatives, and the fact that the derivative's domain is strictly smaller than the function's — the endpoints x = ±1 go, and so does anything that makes a denominator vanish.

Standard method. Quote the formula-sheet derivative, apply the chain rule (the inner derivative is the most-forgotten factor), then build the domain by excluding: endpoints of the original domain, zeros of any new denominator.

Traps. [RPT17 E1] on Q6: "Some students confused the inverse function with the reciprocal function… Common errors for the domain included R, R \ {−1, 0, 1}, [−1, 1], (−1, 1) and [−1, 1] \ {0}. Many students did not exclude zero." [RPT18 E2] on Q1ei: "The most common error was to include x = 0 in the domain. Another common error was to include the endpoints x = ±√2." [RPT10 E1] on Q5: "A large proportion of students were unable to find the first derivative correctly, despite the assistance of the formula sheet. The chain rule was often not used, with the derivative of arctan(2x) commonly being given as 1/(1 + 4x²). A number of students did not recognise arctan(2x) as being an inverse circular function but interpreted it as being either a reciprocal tan function, a tan function, or tried to use an invented trigonometric identity." [RPT13 E1] on Q4c: "many students not recognising the need to use the chain rule… The 2 was commonly missing." Typical marks: 2–3.

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Type 17 — Hybrid (piecewise) specification forced by a modulus in a derivative

VCAA wording, 2010 Exam 2 Section B Q4d (2 marks):

"Complete the following to specify f′(x) as a hybrid function over the maximal domain of f′."

and 2018 Exam 2 Section B Q1e:

"The derivative f′(x) can be expressed in the form f′(x) = g(x)/√(2 − x²) over its maximal domain. i. Find the maximal domain of f′. (1 mark) ii. Find g(x), expressing your answer as a piecewise (hybrid) function. (1 mark) iii. Sketch the graph of g on the axes below. (2 marks)"

and 2017 Exam 2 Section B Q3b (1 mark): "Specify the piecewise function that describes the edges in the third quadrant."

Really testing: that differentiating arcsin(u) where u involves x² produces √(x²) = |x|, and that |x| must then be split by sign — with the correct strict/non-strict endpoints.

Standard method. Differentiate, notice the √(x²) or |x|, split at x = 0 (which is itself excluded), and write two branches with their exact domains.

Traps. [RPT10 E2] on Q4d: "Common errors included the use of √(1 − x²) for both parts of f′(x) and errors involving domain endpoints such as x ∈ (−1, 0]. Many students could complete only the first part of the hybrid function specification." On Q4e: "The most popular response was a 'U shape' curve with a vertex at (0, 2), sometimes with (0, 2) removed, and sometimes with the correct asymptotes. These students ignored the graph in Question 4a., which clearly had negative gradients to the left of the y-axis. Some students had the correct curves and asymptotes, but did not exclude the y-intercept points." [RPT17 E2] on Q3b: "Many students did not use a hybrid function. Of those who did, domains were frequently incorrect." Typical marks: 1–2.

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Type 18 — Sketch an inverse function by reflection in y = x

VCAA wording, 2015 Exam 1 Q8b :

"The graph of f(x) = ½ arctan(x) is shown below. i. Write down the equations of the asymptotes. (1 mark) ii. On the axes above, sketch the graph of f⁻¹, labelling any asymptotes with their equations. (1 mark)"

and 2015 Exam 2 Section B Q1c/d:

"Find the rule for the inverse function f⁻¹, and state the domain and range of f⁻¹. (3 marks) … Sketch and label the graphs of f and f⁻¹ on the axes below. (2 marks)"

Really testing: that reflecting in y = x swaps domain and range, swaps horizontal and vertical asymptotes, and that the inverse of arctan is a restricted tan.

Standard method. Swap x and y in the asymptote equations: y = ±π/4 becomes x = ±π/4. Swap the endpoint coordinates. If asked for the rule, interchange x and y and solve, then state the domain of f⁻¹ = range of f.

Traps. [RPT15 E1] on Q8bii: "Typical errors included poor attempts at the shape of the inverse (sometimes graphed as y = −tan(2x)), poor positioning of the vertical asymptotes, and either not labelling or incorrect labelling of the vertical asymptotes — for example, y = ±π/4 [instead of x = ±π/4] — and drawing the graph beyond its domain. Asymptotic behaviour was lacking in some of the attempts." [RPT15 E2] on Q1c: "The main errors were the incorrect domain and/or range, or the omission of one or both. A small number of students gave the inverse relation by including ± in front of the square root." On Q1d: "Many students did not accurately transfer the graphs from a CAS screen… Of those who managed to draw the graphs correctly, a significant number did not label them." Typical marks: 1–3.

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Type 19 — Piecewise inverse-circular designs (the "brooch" family)

VCAA wording, 2017 Exam 2 Section B Q3 (10 marks):

"A brooch is designed using inverse circular functions to make the shape shown in the diagram below. … The edges of the brooch in the first quadrant are described by the piecewise function
f(x) = { 3·arcsin(x/2), 0 ≤ x ≤ √2 ; arccos(x/2), √2 < x ≤ 2 }
a. Write down the coordinates of the corner point of the brooch in the first quadrant. (1 mark)
b. Specify the piecewise function that describes the edges in the third quadrant. (1 mark)
c. … find the area of the brooch. (3 marks)
d. Find the acute angle between the edges of the brooch at the origin. (3 marks)"

Really testing: symmetry arguments plus exact handling of the join point, and the fact that "the angle between two curves" means the angle between their tangents (arctan of gradients, then subtract).

Standard method. Corner point: solve the two branches equal. Third-quadrant branch: apply the point reflection (x, y) → (−x, −y), i.e. g(x) = −f(−x) with the domain negated and the inequalities reversed. Angle: differentiate each branch at the origin, take arctan of each gradient, subtract.

Traps. [RPT17 E2] on Q3a: "Some students who knew the correct x-coordinate value did not correctly apply the dilation factor of 3 to obtain the correct y-coordinate value." On Q3b, quoted above: hybrid form omitted, domains wrong. Typical marks: 1–3.

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Type 20 — Exact values from double- and compound-angle formulas

VCAA wording, 2008 Exam 1 Q4 (3 marks): "Given that sin(θ) = (√5 − 1)/4, find sec(2θ) in the form a√5 + b, where a, b ∈ R." VCAA wording, 2007 Exam 1 Q10 (3 marks): "Given that tan(2x) = 4√2/7 where x ∈ (0, π/4), find the exact value of sin(x)." VCAA wording, 2016 Exam 1 Q9 (3 marks): "Given that cos(x − y) = 3/5 and tan(x)·tan(y) = 2, find cos(x + y)." VCAA wording, 2006 Exam 1 Q5a (4 marks): "Show that tan(π/8) = √2 − 1."

Really testing: picking the right identity out of the six on the formula sheet, and — in every one of these — justifying a sign choice from the quadrant.

Standard method. Draw a right triangle (or use cos²+sin²=1) to get the second ratio from the first, fixing its sign from the given interval. Then apply the double- or compound-angle formula. For tan(π/8): use tan(2θ) with θ = π/8, solve the resulting quadratic in t = tan(π/8), and explain in words why the negative root is rejected (π/8 is in the first quadrant).

Traps. [RPT06 E1] on Q5a: "Quite a few students whose working was correct failed to complete their solution, giving no proper explanation as to why the negative answer should be rejected, or not mentioning the negative answer at all. Substituting tan(π/8) = √2 − 1 into either the double angle formula or the compound angle formula was not sufficient to achieve full marks." [RPT07 E1] on Q10: "The most successful method involved first finding cos(2x)… then using the appropriate double angle formula. Another common approach, though not as successful, was to use the double angle formula for tan(2x) to try to find tan(x) first." [RPT16 E1] on Q9: "Of great concern was the number of students who gave answers for sine or cosine that were either less than −1 or greater than 1." [RPT12 E1] on Q10b: "Most students were unable to make any meaningful attempt at this question as most failed to recognise that they were dealing with the sine of a compound angle." [RPT08 E1] on Q4: "Some students made mistakes in stating a double angle formula despite them being on the formula sheet." Typical marks: 3–4. This family is the second-hardest cluster in the area after parameter counting.

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Type 21 — Use a trigonometric identity to convert a parametric path to Cartesian form

VCAA wording, 2024 Exam 2 Section B Q4a (1 mark):

"A model yacht is sailing on a lake between two buoys. Its path from one buoy to the other, relative to an origin O, is given by r_Y(t) = 3sec(t)i + 2tan(t)j, where 2π/3 ≤ t ≤ 4π/3. … Use a trigonometric identity to show that the Cartesian equation of the path is given by x²/9 − y²/4 = 1."

and 2013 Exam 1 Q7a (2 marks): "Show that the cartesian equation of the path of the particle is x²/16 − y²/4 = 1." and 2012 Exam 2 Section B Q1a (2 marks): "Show that the curve can be expressed in the cartesian form (x − 1)² − y²/4 = 1."

Really testing: the two Pythagorean identities sec²θ − tan²θ = 1 and cosec²θ − cot²θ = 1 (both derivable from the formula sheet's cot²(x) + 1 = cosec²(x) and 1 + tan²(x) = sec²(x)), applied in reverse.

Standard method. Make the trig function the subject of each parametric equation, then substitute into the identity. Show the substitution step explicitly — these are "show that" marks.

Traps. [RPT24 E2] on Q4a states the standard for every "show that": "A 'show that' question requires logical steps set out to indicate how the solution could be found. In this case an appropriate trigonometric identity needed to be used." [RPT13 E2] on Q1a: "nearly all students realising that t had to be eliminated. Of the few students who gave a result with y as the subject, most omitted the ±." [RPT16 E2] Section A Q1 is the multiple-choice version — "The cartesian equation of the relation given by x = 3cosec²(t) and y = 4cot(t) − 1" — answered by 61%.

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Type 22 — Ellipses: parametric ↔ Cartesian, and sketching

VCAA wording, 2007 Exam 2 Section A Q2:

"An ellipse has a horizontal semi-axis length of 3 and a vertical semi-axis length of 2. Given that the centre of the ellipse has coordinates (1, 3), a possible parametric form for the ellipse is … E. x = 1 + 3cos(t) and y = 3 + 2sin(t)."

and 2013 Exam 2 Section B Q1a/c:

"A curve is defined by the parametric equations x = 1 + 3cos(t), y = −2 + 2sin(t) for t ∈ [0, 2π]. a. Find the cartesian equation of the curve. (2 marks) … c. Sketch the graph of (x − 1)²/4 + y²/9 = 1 and find the values of the x-axis and y-axis intercepts. (3 marks)"

and 2023 Exam 2 Section B Q1d/e (the elliptical walking track):

"The return track from point D to point O follows an elliptical path given by x = 2cos(t) + 2, y = (e − 2)sin(t), where t ∈ [−π/2, π/2]. d. Find the Cartesian equation of the elliptical path. (2 marks) e. Sketch the elliptical path from D to O on the diagram on page 10. (1 mark)"

Really testing: the mapping between x = h + a·cos(t), y = k + b·sin(t) and (x − h)²/a² + (y − k)²/b² = 1 — note a and b are semi-axis lengths, not their squares — and the ability to restrict the sketch to the arc the parameter interval actually traces.

Standard method. cos(t) = (x − h)/a, sin(t) = (y − k)/b, then cos² + sin² = 1. To sketch: plot the centre, step a horizontally and b vertically to get the four extreme points, then restrict to the parameter range — compute the endpoint coordinates by substituting the parameter endpoints.

Traps. [RPT07 E1] on Q6c: "Most students were unable to apply the domain restrictions correctly, which was also a problem with a similar question on last year's paper. Usually, a full ellipse was drawn, occasionally a quarter-ellipse. Many of the attempts at drawing ellipses were not very elliptical." [RPT08 E2] on Q4b: "the most common error being to show only the top half of the ellipse." [RPT23 E2] on Q1e: "The quarter ellipse was often sketched without sufficient accuracy. While the curves drawn mostly connected point D to the origin, the quarter ellipse curves were often not vertical at the origin." [RPT13 E2] on Q1c: "Many ellipses were drawn roughly, with 'points' at the x-intercepts and imprecise semi-axis lengths. Some students gave decimal approximations for the y-intercepts instead of exact values." Typical marks: 1–3.

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Type 23 — Hyperbolas: Cartesian form, asymptote equations, intercepts

VCAA wording, 2010 Exam 1 Q9a (3 marks):

"On the axes below sketch the graph with equation x²/4 − (y + 2)² = 1. State all intercepts with the coordinate axes and give the equations of any asymptotes."

and 2014 Exam 2 Section A Q1: "The asymptotes of the hyperbola given by (x − 3)²/9 − y²/4 = 1 intersect the coordinate axes at …" and 2015 Exam 2 Section A Q4: "The two asymptotes of a particular hyperbola have gradients 2/3 and −2/3 respectively and intersect at the point (2, 1). One branch of the hyperbola passes through the point (5, 5). The equation of the hyperbola is …"

Really testing: that the asymptotes of (x − h)²/a² − (y − k)²/b² = 1 are y − k = ±(b/a)(x − h) — lines through the centre, not through the origin — and that a hyperbola has no horizontal or vertical asymptotes.

Standard method. Read h, k, a, b off the equation. Centre (h, k). Vertices at (h ± a, k). Asymptotes y = k ± (b/a)(x − h). Intercepts by setting x = 0 and y = 0 — and if that has no real solution, say so. Draw the asymptotes first with a ruler.

Traps. [RPT10 E1] on Q9a: "Most were able to correctly identify the x-intercepts, but quite a few seemed to have little idea of the approximate size of √2, placing their intercepts a long way away from the true value… Many graphs did not show asymptotic behaviour, with some graphs looking like horizontal parabolas or semi-circles. Common errors included stating equations of the asymptotes as y = 2x or y = 4x + 2. Some students seemed to believe that all hyperbolas have vertical and horizontal asymptotes. Also seen were circles, ellipses and parabolas." 2008 Exam 2 Section A Q4 at 26% is the deepest of these: "P is any point on the hyperbola … If m is the gradient of the hyperbola at P, then m could be … E. any real number in the interval R \ [−2, 2]" — the gradient approaches the asymptote gradients but never equals them, which is why the excluded interval is closed. Typical marks: 1–3.

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Type 24 — Hyperbolas in parametric form (sec/tan, cosec/cot), sketched on one branch

VCAA wording, 2013 Exam 1 Q7b (2 marks):

"Sketch the path of the particle on the axes below, labelling any asymptotes with their equations." (for r(t) = 4sec(t)i + 2tan(t)j, t ∈ [0, π/2))

and 2012 Exam 2 Section B Q1b (4 marks): "Sketch the curve defined by the parametric equations x = cosec(θ) + 1, y = 2cot(θ), labelling any asymptotes with their equations."

Really testing: that the parameter interval selects a branch or part of a branch, not the whole hyperbola — and that you must find the starting point by substituting the parameter endpoint.

Standard method. Convert to Cartesian (Type 21). Then determine the range of x and y over the given parameter interval: for t ∈ [0, π/2), sec(t) ≥ 1 so x ≥ 4 — the right branch only, from the vertex upward. Mark the starting point with its coordinates, add an arrow for the direction of motion if asked, and draw the asymptote.

Traps. [RPT13 E1] on Q7b: "A large number of students drew a complete hyperbola or the complete right-hand branch. A few students had the branch in the first quadrant but inexplicably indicated that it stopped at about x = 8. The equation of the asymptote was often correctly found, although a few students did not label it as required. There were, however, many incorrect equations, including y = ¼x, y = ½ and y = 2x. There were also several graphs that did not exhibit asymptotic behaviour. A variety of graphs other than hyperbolas were often seen, including ellipses." [RPT12 E2] on Q1b: "Most students knew what to sketch for this question, but many did not show the correct asymptotic behaviour of each branch of the hyperbola. Other problems were asymptotes that were not ruled straight or not labelled." [RPT24 E2] on Q4b: "Students often did not draw this graph well, and it was often not symmetrical over the x-axis. Negative signs were often left off the coordinates of the end points. The direction of the path of the yacht was often left out or in the wrong direction." Typical marks: 2–4.

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Type 25 — Domain and range of the Cartesian relation implied by a parameter restriction

VCAA wording, 2019 Exam 2 Section B Q1 (11 marks):

"A curve is defined parametrically by x = sec(t) + 1, y = tan(t), where t ∈ [0, π/2).
a. Show that the curve can be represented in cartesian form by the rule y = √(x² − 2x). (2 marks)
b. State the domain and range of the relation given by y = √(x² − 2x). (2 marks)"

Really testing: that the parametric restriction and the Cartesian implied domain are different things, and the question asks for the one it says.

Standard method. Find the image of the parameter interval under x(t) — that is the domain of the path. Find the image under y(t) — that is the range. Do not just state the maximal domain of the Cartesian rule.

Traps. [RPT19 E2] on Q1b: "While most students stated the correct range, a significant number gave a domain which did not account for the restriction on t." [RPT15 E2] on Q3cii: "Common errors included (−√3/2, √3/2), (−1, 1) and (−π/3, π/3)" — three different ways of confusing the parameter range with the x-range. Typical marks: 1–2.

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Type 26 — Implicitly defined curves: gradient and tangent

VCAA wording, 2024 Exam 1 Q8 (4 marks):

"Consider the relation x²y² + xy = 2, where x, y ∈ R.
a. Using implicit differentiation, show that dy/dx = −y/x given that 2xy ≠ −1. (2 marks)
b. Find all points on the graph of x²y² + xy = 2 where the slope of the tangent is equal to −1. (2 marks)"

and 2010 Exam 1 Q9b (3 marks): "Find the gradient of the curve with equation x²/4 − (y + 2)² = 1 at the point where x = 2 and y < 0." and 2025 Exam 1 Q1 (4 marks): "Consider the curve with equation xe^(2y) + y²e^x = 8e⁴. Find the equation of the tangent to the curve at the point (4, −2)."

Really testing: product rule plus chain rule with dy/dx carried through, and then — the part that separates — using the constraint that the point lies on the curve.

Standard method. Differentiate both sides with respect to x, treating y as a function of x. Collect dy/dx terms, factor, divide. For "find all points where the slope is m": set the gradient expression equal to m, get a relation between x and y, then substitute back into the original equation and solve the resulting single-variable equation.

Traps. [RPT24 E1] on Q8b: "Some students, while recognising the relationship y = x, neglected to consider that the points lay on the graph. Some students who did find an equation such as … had difficulty solving it or found incorrect coordinates in addition to the correct ones." [RPT09 E1] on Q5a: "some students simply verified (0, 4) and then did not proceed any further. Others solved the relevant quadratic and found that y = 1 or y = 4 but did not give the coordinates of the second point or gave the coordinates as (1, 0) instead of (0, 1)." Typical marks: 2–4.

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Type 27 — Loci in the plane: perpendicular bisectors, circles and rays

VCAA wording, 2020 Exam 2 Section B Q2 (11 marks):

"Two complex numbers, u and v, are defined as u = −2 − i and v = −4 − 3i.
a. Express the relation |z − u| = |z − v| in the cartesian form y = mx + c, where m, c ∈ R. (3 marks)
b. Plot the points that represent u and v and the relation |z − u| = |z − v| on the Argand diagram below. (2 marks)
c. State a geometrical interpretation of the graph of |z − u| = |z − v| in relation to the points that represent u and v. (1 mark)"

and 2024 Exam 2 Section B Q2a (2 marks), the same construction under the current design; and 2019 Exam 2 Section B Q2bii (1 mark): "Find the cartesian equation of the circle |z + m| = n."

Really testing: two interchangeable routes — algebraic (substitute z = x + iy, square both moduli, expand, cancel x² and y²) and geometric (perpendicular bisector: midpoint plus negative-reciprocal gradient).

Standard method. Algebraic: (x − a₁)² + (y − b₁)² = (x − a₂)² + (y − b₂)², expand, the quadratic terms cancel, rearrange to y = mx + c. Geometric: midpoint of uv, gradient of uv, take the negative reciprocal, use point–gradient form. Name the locus in words when asked.

Traps. [RPT20 E2] on Q2c: "The line is the perpendicular bisector of the line segment joining the points represented by u and v. A variety of reasonable responses were accepted. Insufficiently precise responses such as 'a linear …' ** [were not]." [RPT24 E2] on Q2a: "Some students did not provide sufficient working to gain both marks. Several students made sign errors when substituting into the distance formula. Students who used the perpendicular bisector method mostly found the gradient correctly. However, some of those students did not correctly use the coordinates of the midpoint." [RPT19 E2] on Q2bii: "Incorrect responses included sign errors and algebraic errors resulting from unnecessary attempts to isolate y**." [RPT09 E2] on Q2d: "The most common error was the cartesian equation of the circle being written as x² + y² = 2" — squaring omitted. Typical marks: 1–3.

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Type 28 — Paths of particles: sketch the path, label the start, show the direction

VCAA wording, 2020 Exam 2 Section B Q4bii (3 marks):

"Sketch the path of the aeroplane on the axes provided below. Label the position of the aeroplane when t = 0, using coordinates, and use an arrow to show the direction of motion of the aeroplane."

and 2024 Exam 2 Section A Q15:

"The position of a moving body is given by r(t) = sin(t)i + cos(2t)j, where t is measured in seconds, for t ≥ 0. The motion of the body can be described as moving along a parabolic path given by … A. y = 1 − 2x², starting at (0, 1), reversing direction at (1, −1) and then again at (−1, −1), then returning to (0, 1) after 2π seconds."

Really testing: that a vector function traces a portion of a curve, back and forth, with a period — and that the sketch must record where it starts, where it turns, and which way it goes.

Standard method. Eliminate t for the Cartesian rule (use cos(2t) = 1 − 2sin²t here). Compute the range of each component over the given interval → the arc actually traced. Substitute t = 0 for the start. Differentiate to find where the motion reverses (dx/dt = 0). Determine the period.

Traps. [RPT24 E2] on Section A Q15 sets out the full reasoning: "The motion has period 2π. Starting at (0, 1) at t = 0, after visiting (1, −1) at t = π/2 and (−1, −1) at t = 3π/2, the body returns to (0, 1) at t = 2π." [RPT24 E2] on Section B Q4b, quoted above: endpoints unsigned, direction missing or reversed, graph not symmetric. Typical marks: 1–3.

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Type 29 — Transformations: express a new result in terms of an old one

VCAA wording, 2021 Exam 1 Q4 (4 marks):

"a. The shaded region … is bounded by the graph of y = sin(x) and the x-axis between the first two non-negative x-intercepts … rotated about the x-axis … Find the volume, V_s, of the solid formed. (3 marks)
b. Now consider the function y = sin(kx), where k is a positive real constant. … Find the volume of this solid in terms of V_s. (1 mark)"

and 2022 Exam 2 Section B Q1cii (2 marks): "Find the distance between the two turning points of the graph of f(x) = x²/(x − k) in terms of k." and 2016 Exam 2 Section B Q1e (1 mark): "The volume of the solid formed is given by V = π∫_c^a b·x² dy. Find the values of a, b and c. Do not attempt to evaluate this integral."

Really testing: whether you can see a dilation rather than redo the integral, and whether you answer in the form requested.

Standard method. Identify the transformation (x → kx is a horizontal dilation by factor 1/k). A volume of revolution about the x-axis scales by the horizontal factor. State the answer in the demanded symbols.

Traps. [RPT21 E1] on Q4b: "Very few students recognised that dilating the graph (and hence the solid) from part a. by a factor 1/k yields the graph and solid for part b. Of those who were successful, many did not write their answer in terms of V_s, as instructed." [RPT22 E2] on Q1cii: "Students generally applied a distance formula successfully, but many did not restrict their final answer to positive values" — i.e. 2√5|k|, not 2√5k. [RPT16 E2] on Q1e: "A number of students incorrectly gave decimal approximations for the value of b. Students must note and follow the general instructions given at the start of Section B." Typical marks: 1–2.

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Type 30 — Attached integral work: volumes, areas and arc lengths hung off the sketched graph

VCAA wording, 2014 Exam 2 Section B Q1d:

"The region bounded by the coordinate axes, the graph of f and the line x = 3, is rotated about the x-axis to form a solid of revolution. i. Write down a definite integral in terms of x that gives the volume of this solid of revolution. (2 marks) ii. Find the volume of this solid, correct to two decimal places. (1 mark)"

and 2010 Exam 2 Section B Q4bi: "Write down a definite integral in terms of y, which when evaluated will give the volume of the solid of revolution formed by rotating the graph drawn above about the y-axis." and 2023 Exam 2 Section B Q1fi (1 mark): "Write down a definite integral in terms of t that gives the length of the elliptical path from D to O."

Really testing: setup, not evaluation. Almost every Exam 2 Section B graph question ends this way, and the "write down the integral" mark is the most reliably available mark in the whole area.

Standard method. About the x-axis: V = π∫ y² dx with x-terminals. About the y-axis: V = π∫ x² dy with y-terminals — rearrange the rule to give x in terms of y first. Parametric arc length: ∫√((dx/dt)² + (dy/dt)²) dt with t-terminals (this is the only arc-length formula on the current sheet — the Cartesian form was removed in 2023).

Traps. [RPT10 E2] on Q4bi: "Common errors involved omission of brackets, incorrect terminals, and integrands involving x [when y was required]. A number of students left out π." [RPT13 E2]: "The main errors were the omission of π, incorrect rearrangement of the equation of the ellipse for the integrand and some integrals missing the dx." [RPT15 E1] on Q5: "Typical errors included finding an area rather than a volume and rotating about the incorrect axis. Several students who rotated about the correct axis integrated from 0 to 5 rather than −3 to 5." [RPT06 E1] on Q3b: "A large proportion of students did not account for the fact that the area specified was below the x-axis… Another common error was to neglect the modulus (absolute value) sign in ∫1/x dx = log|x| + c. This led to numbers such as log_e(−5)… Too many students gave an answer involving arctan or even the logarithm of a quadratic expression." Typical marks: 1–4.

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3. The standard wordings

These are VCAA's recurring sentences. Learn them as instructions with a mark attached, because that is what they are.

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4. The separators

Every question part in this area with pct ≤ 50. 107 items, grouped by type. Format: ref — pct% — description.

4.1 Rational-function sketching and asymptotes (18)

• 2006 Exam 1 Q3a — 39% — sketch y = 36/(2x² − 18), indicating asymptotes and intercepts
• 2008 Exam 1 Q1 — 19% — sketch y = (2 − x²)/(2x): exact turning point, x-intercept, vertical and oblique asymptotes
• 2011 Exam 1 Q3b–3b — 49% — sketch a reciprocal-quadratic graph; asymptote and turning point (0, 3) required (2011 paper text unreadable; described from [RPT11 E1]. The duplicated part label is VCAA's own pagination artefact, preserved in [QJSON])
• 2018 Exam 1 Q5 — 15% — sketch f(x) = (x + 1)/(x² − 4), asymptotes with equations and intercepts with coordinates
• 2023 Exam 1 Q1b — 39% — sketch f(x) = (x² + x − 6)/(x − 1), labelling asymptotes (oblique y = x + 2)
• 2024 Exam 1 Q3c — 10% — sketch f(x) = (x − 1)²/(x + 1)², asymptotes and axial intercepts
• 2025 Exam 1 Q9a — 45% — show f(x) = −x − 1 + 1/(x + 1) for (x³ + x² − 2x)/(1 − x²)
• 2025 Exam 1 Q9b — 46% — value of k making the hybrid continuous at the removable discontinuity
• 2025 Exam 1 Q9c — 16% — sketch it: oblique and vertical asymptotes plus an open circle at the hole
• 2012 Exam 2 Section B Q1b — 42% — sketch the cosec/cot hyperbola labelling asymptotes
• 2017 Exam 2 Section B Q1ai — 36% — equations of the asymptotes of f(x) = x/(1 + x³) (horizontal one omitted)
• 2021 Exam 2 Section B Q1c — 22% — sketch (2x − 3)(x − 5)/((x − 1)(x + 2)): asymptotes, max TP, inflection to 2 dp, intercepts
• 2022 Exam 2 Section B Q1b — 43% — sketch y = x²/(x − 1) with turning points and oblique asymptote
• 2024 Exam 2 Section B Q1a — 17% — sketch (x⁴ − x² + 1)/(1 − x²), vertical asymptotes and stationary points
• 2025 Exam 2 Section B Q1a — 40.9% — sketch a quotient function: asymptotes, turning point, inflection to 1 dp
• 2016 Exam 2 Section B Q1c — 46% — sketch f(x) = (4 + x² + x³)/x on [−3, 3], TP and inflection to 2 dp
• 2008 Exam 2 Section B Q1c — 43% — sketch 6√x/(3x² + 1) showing intercepts, maximum TP and both inflection points
• 2022 Exam 1 Q10a — 13% — sketch f(x) = sec(4x) on [−π/4, π/4]: asymptotes with equations, turning points and endpoints with coordinates

4.2 Stationary points, inflections and continuity (5)

• 2008 Exam 2 Section B Q1bii — 46% — coordinates of two points of inflection (extra roots outside the domain rejected)
• 2017 Exam 2 Section B Q1aiii — 13% — inflection coordinates, rejecting x = 0 where f″ = 0 but no sign change
• 2020 Exam 1 Q6b — 9% — "hence show that the graph has a point of inflection at x = 2" (sign change not argued)
• 2020 Exam 2 Section B Q3c — 45% — sketch x²e^(−x) labelling the local maximum and all inflections to 2 dp
• 2024 Exam 2 Section A Q2 — 48% — hybrid rational function: continuous, or a point of discontinuity?

4.3 Parameter families — counting asymptotes and stationary points (5)

• 2020 Exam 2 Section B Q3eii — 2% — table of values of n for which xⁿe^(−x) has 0, 1, 2, 3 inflections
• 2021 Exam 2 Section B Q1di — 6% — values of k for which g_k has exactly two asymptotes (three values needed)
• 2021 Exam 2 Section B Q1dii — 13% — values of k for which g_k has no stationary points
• 2022 Exam 2 Section A Q3 — 38% — what (x² + 2x + c)/(x² − 4) "will always have" (cancellation case)
• 2022 Exam 2 Section B Q1cii — 45% — distance between the two turning points of x²/(x − k) in terms of k

4.4 Inverse circular functions — domain, range, derivative (11)

• 2009 Exam 1 Q10a — 34% — implied domain and range of a transformed arcsin
• 2012 Exam 1 Q10aii — 33% — maximal domain of 3/√(25x² − 1)
• 2012 Exam 1 Q10aiii — 26% — intersection of the two maximal domains
• 2013 Exam 1 Q4c — 47% — gradient of the tangent to y = arccos(1 − 2x) at x = ¼ (chain rule)
• 2014 Exam 1 Q7a — 4% — "write down the range of f(x) = 3x·arctan(2x)"
• 2017 Exam 1 Q6 — 18% — derivative of 1/arcsin(x) and the largest set where it is defined
• 2017 Exam 1 Q10b — 40% — maximal domain and range of arccos(x/2)
• 2010 Exam 1 Q5 — 35% — second derivative of arctan(2x) evaluated at π/2
• 2018 Exam 2 Section B Q1ei — 21% — maximal domain of f′ (must exclude 0 and both endpoints)
• 2020 Exam 2 Section A Q2 — 42% — range of a dilated, translated, modulused arccos
• 2020 Exam 2 Section A Q4 — 28% — rule and range of f(g(x)) with f(x) = √(x − 1)/x, g(x) = cosec²x

4.5 Inverse circular functions — sketching and hybrid derivatives (9)

• 2010 Exam 2 Section B Q4a — 44% — sketch arcsin(2x² − 1) with exact endpoints and intercepts
• 2010 Exam 2 Section B Q4c — 37% — use calculus to show the derivative form and find a
• 2010 Exam 2 Section B Q4d — 30% — specify f′ as a hybrid function over its maximal domain
• 2010 Exam 2 Section B Q4e — 19% — sketch that hybrid derivative showing asymptotes and excluded points
• 2013 Exam 1 Q4b — 46% — sketch y = arccos(1 − 2x) labelling endpoints with coordinates
• 2018 Exam 2 Section B Q1b — 50% — sketch 2arcsin(x² − 1) labelling endpoints and y-intercept
• 2018 Exam 2 Section B Q1eiii — 39% — sketch the hybrid g
• 2015 Exam 2 Section B Q1d — 36% — sketch and label f and f⁻¹ on the same axes
• 2017 Exam 2 Section B Q3b — 49% — piecewise function for the third-quadrant edges of the brooch

4.6 Reciprocal graphs (2)

• 2019 Exam 1 Q5aii — 42% — turning points of cos²x + cos x + 1 in the open interval (0, 2π)
• 2019 Exam 1 Q5b — 27% — sketch y = 1/f(x) labelling turning points and endpoints

4.7 Trigonometric identities, exact values and equations (11)

• 2006 Exam 1 Q5a — 22% — show tan(π/8) = √2 − 1, with the negative root explicitly rejected
• 2007 Exam 1 Q10 — 20% — exact sin(x) given tan(2x) = 4√2/7, x ∈ (0, π/4)
• 2008 Exam 1 Q4 — 23% — sec(2θ) in the form a√5 + b given sin(θ) = (√5 − 1)/4
• 2011 Exam 1 Q8 (3 marks) — 27% — coordinates satisfying a cosec/sin relation, exact values required
• 2011 Exam 1 Q8 (2 marks) — 7% — the m² = 7 part: both ± roots required (2011 text unreadable)
• 2012 Exam 1 Q2 — 11% — all real solutions of 2cos(x) = 3cot(x) (general solution)
• 2012 Exam 1 Q10b — 13% — sin(arcsin(1/8) + arcsin(3/4)) via the compound-angle formula
• 2015 Exam 1 Q7a — 34% — solve sin(2x) = sin(x) on [0, 2π]
• 2016 Exam 1 Q9 — 44% — cos(x + y) given cos(x − y) = 3/5 and tan(x)tan(y) = 2
• 2023 Exam 2 Section B Q2fi — 47% — express cis(2π/7) + cis(12π/7) as A·cos(B)
• 2024 Exam 2 Section A Q4 — 27% — cos(x/2) in terms of a where sin(x) = a, x ∈ (3π/2, 2π)

4.8 Inequalities (3)

• 2015 Exam 1 Q7b — 11% — {x : cosec(2x) < cosec(x)} on (0, π/2) ∪ (π/2, π)
• 2017 Exam 2 Section A Q2 — 37% — solutions of cos(x) > ¼·cosec(x) on (0, 2π) \ {π}
• 2020 Exam 1 Q4 — 12% — (3 − x)/(x − 4) > 1, answer in interval notation

4.9 Graph-identification and classification multiple choice (11)

• 2007 Exam 2 Section A Q1 — 49% — product of the gradients of a hyperbola's asymptotes
• 2007 Exam 2 Section A Q5 — 48% — identify the cosec rule from a printed graph
• 2008 Exam 2 Section A Q2 — 42% — x² + ax + y² + 1 = 0 represents a circle when a < −2 or a > 2
• 2008 Exam 2 Section A Q4 — 26% — possible gradients on a hyperbola: R \ [−2, 2] (closed interval)
• 2009 Exam 2 Section A Q2 — 43% — number of points common to an ellipse and a hyperbola
• 2011 Exam 2 Section A Q1 — 34% — 45% of the state chose the same wrong option (2011 text unreadable; stem not recoverable)
• 2013 Exam 2 Section A Q3 — 47% — find a, b, c for y = 1/(ax² + bx + c) from two vertical asymptotes and a stationary point
• 2013 Exam 2 Section A Q4 — 43% — y = ax and y = arctan(bx) meet exactly three times when b < a < 0
• 2015 Exam 2 Section A Q3 — 50% — a²x² + (1 − a²)y² = c² cannot represent a single straight line
• 2015 Exam 2 Section A Q4 — 43% — equation of a hyperbola from asymptote gradients, centre and one point
• 2018 Exam 2 Section A Q4 — 49% — cosec(−x) in terms of a, b given cos(x) = −a, cot(x) = b

4.10 Conics, parametric paths and particle motion (10)

• 2010 Exam 1 Q9a — 44% — sketch x²/4 − (y + 2)² = 1, stating all axis intercepts and the equations of both asymptotes
• 2007 Exam 1 Q6c — 22% — sketch the elliptical path of v = −4sin(2t)i + 6cos(2t)j with the domain restriction applied
• 2013 Exam 1 Q7b — 22% — sketch one branch of 4sec(t)i + 2tan(t)j, asymptote labelled
• 2019 Exam 2 Section B Q1b — 38% — domain and range of y = √(x² − 2x) given t ∈ [0, π/2)
• 2020 Exam 2 Section A Q9 — 35% — slope-field matching for dy/dx = y/(x − y)
• 2021 Exam 2 Section A Q7 — 39% — shortest arc length along a parametric circle from A to B
• 2023 Exam 2 Section B Q1e — 22% — sketch the quarter-elliptical return path from D to O
• 2024 Exam 2 Section A Q15 — 36% — describe r(t) = sin(t)i + cos(2t)j as a parabolic path with reversals and period
• 2024 Exam 2 Section B Q4b — 41% — sketch the yacht's hyperbolic arc with endpoints and direction of motion
• 2013 Exam 2 Section B Q3e — 39% — sketch N(t) showing endpoints, inflection and change of concavity

4.11 Implicit differentiation and loci (3)

• 2009 Exam 2 Section B Q2d — 44% — Cartesian intersection points of a line and |z| = 2
• 2024 Exam 1 Q8b — 32% — all points on x²y² + xy = 2 where the tangent slope is −1
• 2023 Exam 2 Section B Q4f — 43% — sketch the logistic curve, labelling the asymptote and the Q-intercept

4.12 Transformations and parameter-form answers (3)

• 2016 Exam 2 Section B Q1e — 34% — find a, b, c in V = π∫_c^a b·x² dy (exact b required)
• 2021 Exam 1 Q4b — 30% — volume for y = sin(kx) expressed in terms of V_s
• 2022 Exam 2 Section B Q3bii — 31% — sketch x = log_e(tan⁻¹(2t) + 1) with the horizontal asymptote and a labelled point

4.13 Attached integral work (7)

• 2006 Exam 1 Q3b — 22% — exact area below the axis, by partial fractions, with modulus signs kept
• 2010 Exam 2 Section B Q4bi — 47% — volume integral in terms of y for rotation about the y-axis
• 2010 Exam 2 Section B Q4bii — 43% — evaluate that volume exactly
• 2015 Exam 1 Q5 — 37% — volume about the y-axis for the region bounded by y = 2x² − 3, y = 5 and the y-axis
• 2015 Exam 2 Section B Q3e — 20% — "hence" antiderivative plus the area of the parametric logo
• 2020 Exam 1 Q9a — 49% — show a = 1, b = −2, c = 16 in the decomposition of (dy/dt)² for an arc-length integrand
• 2025 Exam 2 Section A Q9 — 49% — surface-area integral for a parametric curve rotated about the x-axis

4.14 Items tagged here that are really vectors, complex numbers or differential equations (9)

Listed for completeness; do not drill these as functions-and-graphs.

• 2008 Exam 1 Q8b — 27% — fourth vertex of a parallelogram, coordinates required
• 2008 Exam 2 Section B Q4cii — 39% — read the fox population at the rabbits' minimum (unit confusion)
• 2008 Exam 2 Section B Q4e — 9% — minimum and maximum rabbit numbers from the implicit solution curve
• 2021 Exam 2 Section A Q5 — 32% — maximum |z| on the circle |z − 2 − √3 i| = 1
• 2021 Exam 2 Section B Q3bii — 22% — maximum rate of decrease of depth, and the corresponding depth
• 2021 Exam 2 Section B Q3biii — 23% — maximum inflow rate without overflow
• 2021 Exam 2 Section B Q3c — 12% — refill time from depth 25
• 2023 Exam 2 Section B Q5b — 29% — shortest distance from B to the line segment AC
• 2023 Exam 2 Section B Q5f — 34% — coordinates of D where the normal line meets the plane

Count check: 107 separators — §4.1 (18) + §4.2 (5) + §4.3 (5) + §4.4 (11) + §4.5 (9) + §4.6 (2) + §4.7 (11) + §4.8 (3) + §4.9 (11) + §4.10 (10) + §4.11 (3) + §4.12 (3) + §4.13 (7) + §4.14 (9) = 107. 2011 Exam 1 Q8 appears twice because [QJSON] carries two distinct parts under that single ref (a 3-mark part at 27% and a 2-mark part at 7%) — this is a report-pagination artefact in the source, not a duplicate.

4.15 Which types separate most

Ranked by the median pct of the separators in each group, and by how deep the worst items go:

What the reports say went wrong, in VCAA's own words, condensed to the five recurring causes:

1. Labels omitted. "The most frequent error was the omission of the asymptote y = 0" ([RPT14 E2]); "Many responses did not label the horizontal asymptote y = 0" ([RPT25 E2]); "Many responses lacked at least one of the required details such as coordinates of the point of inflection or coordinates of one of the axial intercepts" ([RPT21 E2]); "A few students sketched the correct graph but did not label both endpoints with their coordinates as required" ([RPT13 E1]).
2. Asymptotic behaviour not shown. "some graphs colliding with the asymptotes and others swerving away from them" ([RPT08 E1]); "some gained only partial marks where … the curve swung away from an asymptote" ([RPT14 E2]); "Some responses did not show appropriate asymptotic behaviour" ([RPT22 E2]).
3. Domain restrictions ignored. "Most students were unable to apply the domain restrictions correctly… Usually, a full ellipse was drawn" ([RPT07 E1]); "A large number of students drew a complete hyperbola or the complete right-hand branch" ([RPT13 E1]); "Students missed out on marks for ignoring the domain of the function" ([RPT16 E2]); "A significant number of responses did not include the middle branch" ([RPT21 E2]).
4. CAS transcription. "Setting the calculator screen to match the grid provided will help students sketch graphs correctly" ([RPT22 E2]); "Students are advised to use their technology as a tool to support the sketching of an accurate graph rather than simply copying a roughly correct shape from a screen" ([RPT16 E2]); "Students are reminded to take care when transferring graphical information from technology to a graph with a given scale" ([RPT10 E2]).
5. Presentation. "Students should be encouraged to use a pencil to draw graphs; some graphs were very messy and hard to read, with multiple lines drawn in some cases" ([RPT06 E1]); "A number of graphs were not smooth and were roughly sketched with multiple lines. The worst of these were drawn in ink" ([RPT10 E1]); "A number of students drew their graphs in pen rather than pencil, which made it very messy when corrections had to be made" ([RPT12 E2]); "Students who were most successful used a ruler to draw the asymptotes" ([RPT25 E1]).

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5. What makes a hard one hard

5.1 Domain and range of inverse circular functions

Three base facts, and every transformed question is these three plus arithmetic:

The hardness comes from three places. First, dividing a compound inequality by a negative number — [RPT13 E1] records students reaching −2 ≤ −2x ≤ 0 and writing 1 ≤ x ≤ 0. VCAA's own recommended fix is to solve the equation at each endpoint instead. Second, the range endpoints must be re-ordered after a reflection; [RPT09 E1] lists "giving the domain as [0, −4]" as an actual submitted answer. Third, intersecting constraints: 2012 Exam 1 Q10aiii is one mark for recognising that the domain of a sum is the intersection of the domains of its parts, and 74% of the state missed it.

The derivative's domain is a fourth trap and a separate one. f(x) = arcsin(x) is defined on [−1, 1]; f′(x) = 1/√(1 − x²) is defined on (−1, 1). If the function is 1/arcsin(x), you additionally lose x = 0. [RPT17 E1]: "Many students did not exclude zero."

5.2 Branch choices

Four distinct branch decisions recur, and each one is a mark:

• Quadrant sign. Given sin(θ), the sign of cos(θ) is fixed by the stated interval. 2024 Exam 2 Section A Q4 turns entirely on x ∈ (3π/2, 2π) forcing cos(x) > 0 while x/2 ∈ (3π/4, π) forces cos(x/2) < 0. Twenty-seven per cent.
• Rejecting the negative root. 2006 Exam 1 Q5a requires you to say that π/8 is in the first quadrant so tan(π/8) > 0. [RPT06 E1]: students "gave no proper explanation as to why the negative answer should be rejected."
• Which branch of a conic. x = 4sec(t), t ∈ [0, π/2) gives x ≥ 4: the right branch only, and only the upper half of it.
• The ± when inverting. [RPT15 E2]: "A small number of students gave the inverse relation by including ± in front of the square root" — an inverse function takes one sign, determined by the range of the original.

5.3 Endpoint conventions

5.4 Labelling requirements on sketches

The single highest-leverage thing to internalise: a sketch is marked against the instruction sentence, one mark per feature-set. On a 3-mark sketch the typical allocation is (1) correct shape and asymptotic behaviour, (2) asymptotes drawn and labelled with equations, (3) named points plotted accurately and labelled with coordinates. You can draw a perfect curve and score 1 of 3.

What "labelled" means, precisely, from the reports:

• An asymptote needs a dashed ruled straight line and an equation ([RPT12 E2]: "asymptotes that were not ruled straight or not labelled").
• A point needs an ordered pair, not an x-value ([RPT08 E1]: "just gave the x-coordinate"; [RPT08 E2]: "A frequent error was to give only the x coordinates").
• Accuracy is part of labelling: the point must be plotted where the label says it is ([RPT16 E2]: "these points were not always plotted with appropriate accuracy. Careful attention to the axes scale is required"; [RPT20 E2]: "having the x-value of the left-most point of inflection rounded to 0.58 instead of 0.59").
• Exactness: "Give the exact coordinates" means surds and π, not decimals ([RPT10 E2], [RPT13 E2], [RPT25 E2]).
• If a scale is printed, use the whole of it ([RPT20 E1]: "when a grid is provided … sufficient area should be utilised so that all features of the graph can be shown. Some students drew their graphs on such a limited domain that the asymptotic behaviour was not shown").

5.5 Parameter ranges in parametric form

The parameter interval is not decoration. It determines:

• Which portion of the curve exists. t ∈ [−π/2, π/2] on x = 2cos t + 2, y = (e − 2) sin t gives the right half of the ellipse; 2023 Exam 2 Section B Q1e asks for a quarter of it and 22% got it.
• The domain and range of the Cartesian relation (Type 25). 2019 Exam 2 Section B Q1b at 38%: "a significant number gave a domain which did not account for the restriction on t."
• The direction of travel, and hence the arrow. 2024 Exam 2 Section B Q4b at 41%: the direction "was often left out or in the wrong direction."
• Whether the path repeats. 2024 Exam 2 Section A Q15 at 36%: r(t) = sin(t)i + cos(2t)j traverses the parabolic arc back and forth with period 2π, not π.
• The terminals of any attached integral. An arc-length or surface-area integral in t takes the parameter endpoints; a volume integral in x takes the image of those endpoints under x(t).

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6. A worked method sheet — the ten highest-yield types

Ranked by (marks available) × (frequency) × (how recoverable the marks are).

M1. Rational function sketch, Exam 1, 3–5 marks

1. Divide or decompose first. Write f(x) = quotient + remainder/(denominator).
2. Vertical asymptotes: denominator = 0 after cancelling any common factor. If a factor cancelled, that x-value is a hole, not an asymptote.
3. Non-vertical asymptote: the quotient. Degree difference 0 → horizontal; 1 → oblique; ≥ 2 → a curve, which VCAA will not ask you to label ("straight-line asymptotes").
4. Intercepts: y-intercept f(0); x-intercepts from the numerator.
5. Turning points: differentiate the decomposed form (easier), set to zero, get coordinates.
6. Side test: the sign of the remainder term tells you which side of the non-vertical asymptote each branch lies on, and whether the curve crosses it.
7. Draw: asymptotes with a ruler, dashed, each carrying its equation. Then one smooth pencil curve per branch — all branches.
8. Annotate every intercept and turning point with an ordered pair.

Must be on the page to earn every mark: every asymptote as a ruled dashed line with its equation; every axis intercept as (a, 0) / (0, b); every turning point as (p, q); correct number of branches; curves approaching, not touching or fleeing, the asymptotes.

M2. Quotient function sketch, Exam 2, 3 marks

1. Set the CAS window to exactly the printed axes — domain, range and scale.
2. Get stationary points and inflection points at the stated accuracy, and check the instruction for exact vs decimal separately for each feature.
3. Test each f″ = 0 candidate for a sign change; discard the ones that fail.
4. Identify every asymptote by hand from the denominator; CAS will not label them.
5. Transfer point by point. Flat where the graph is flat; steep where it is steep.
6. Restrict to the printed domain and put closed dots with coordinates at both endpoints.

Must be on the page: the named features only — but all of them, plotted accurately, with coordinates at the stated precision; asymptotes labelled; endpoints as closed dots.

M3. "Show that f(x) can be written as …", 1–2 marks

Force the numerator: write numerator = (denominator)(quotient) + remainder and verify by expansion, or equate coefficients after multiplying the target form out. Show at least two intermediate lines. Never start from the answer.

M4. State the implied/maximal domain and range, 1–2 marks

1. List every constraint: arcsin/arccos argument in [−1, 1]; square-root radicand ≥ 0; denominator ≠ 0; log argument > 0.
2. Solve each. For −1 ≤ g(x) ≤ 1, solve g(x) = −1 and g(x) = 1 to find the endpoints, then decide which way round.
3. Intersect them.
4. Range: transform the base range endpoints; reverse the order if there is a reflection; use exact values.
5. Write both, with the right brackets. arctan ranges are open.

M5. Sketch an inverse circular function, 2–3 marks

Domain → range → the two endpoint coordinates (exact) → orientation (sign of the inner coefficient) → the non-stationary point of inflection at the image of the base origin → vertical tangents at the endpoints for arcsin/arccos. Closed dots, both labelled.

M6. Parametric → Cartesian → sketch the arc, 2–4 marks

1. Make the trig function the subject of each equation.
2. Apply cos² + sin² = 1, sec² − tan² = 1, or cosec² − cot² = 1. Show the substitution.
3. Compute x(t) and y(t) at the parameter endpoints — those are your endpoint coordinates.
4. Determine the range of x and of y over the parameter interval; that fixes which arc.
5. Draw only that arc, with closed dots at the endpoints (coordinates labelled), the asymptote(s) if it is a hyperbola, and an arrow for direction if the question mentions motion.

M7. Exact trigonometric values, 3–4 marks

1. From the given ratio, build a right triangle or use cos² + sin² = 1 to get the companion ratio.
2. Fix its sign from the stated interval, and write the reason down.
3. Choose the identity that gets you from what you have to what you want in one step (the formula sheet carries all of sin(x ± y), cos(x ± y), tan(x ± y), and the double-angle forms).
4. Simplify surds fully; rationalise.
5. Check plausibility: any sine or cosine outside [−1, 1] is wrong ([RPT16 E1]).

M8. Rational or trigonometric inequality, 2–4 marks

1. Move everything to one side; single fraction; factorise.
2. Sign table on the critical values (numerator zeros and denominator zeros).
3. Sketch if the algebra is getting messy — VCAA's reports say repeatedly that the graphical route wins.
4. Exclude denominator zeros.
5. Write the answer as a union of intervals with the correct brackets.

M9. Parameter family: count the asymptotes / stationary points, 1–2 marks

1. Write the function with the parameter in place.
2. Ask: for which parameter values does a denominator root coincide with a numerator root? List every such value separately. These are the cases with fewer asymptotes.
3. Handle the generic case.
4. For "no stationary points": differentiate symbolically, require the numerator (usually a quadratic) to have discriminant ≤ 0; then add the degenerate cases where cancellation leaves a linear or constant function, which also has none.
5. Answer as a set or union, covering every case.

M10. Attached volume / area / arc length, 1–3 marks

• About the x-axis: V = π∫_a^b [f(x)]² dx.
• About the y-axis: V = π∫_c^d [x(y)]² dy — rearrange for x first, and use y-terminals.
• Parametric arc length: ∫_{t₁}^{t₂} √((dx/dt)² + (dy/dt)²) dt. This is the only arc-length form on the current formula sheet.
• Surface of revolution about the x-axis for a parametric curve: S = 2π∫ y √((dx/dt)² + (dy/dt)²) dt, and a substitution may convert it to a u-integral (2025 Exam 2 Section A Q9).
• Area below the x-axis needs a sign or a modulus ([RPT06 E1]).
• Always write π, the squared integrand, the correct differential and the correct terminals. That is four separate ways to drop the mark.

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7. One-page summary for a 45+ candidate

• This area is 16.8% of the marks and has been on every Examination 1 and every Examination 2 since 2006.
• On Examination 1, two-thirds of its parts are separators and the median part is answered completely by 39% of the state. This is where a 45 is won or lost.
• On Examination 2 Section A it is easy (median 68%) and clustered at Questions 1–5. Bank those.
• On Examination 2 Section B it is usually Question 1, worth 9–12 marks, with a 3-mark sketch and a volume of revolution.
• The sketch is a checklist. Read the instruction sentence twice, write out the required labels as a list before you draw, and tick them off.
• The three deepest wells are: counting asymptotes in a parameter family (as low as 2%), exact trigonometric values with a sign justification (as low as 7%), and domain/range of inverse circular compositions (as low as 4%).
• Use a pencil and a ruler. VCAA has said so in the reports of 2006, 2010, 2012 and 2025.