§-Past paper

VICMath Methods2025

VCE Math Methods 2025

Walkthrough of the 2025 VCE math methods exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks: 80
Time: 120 min
Authority: VCAA
Updated: 2026-05-29

Open the official PDF ->Official marking notes

What this paper assessed

This guide covers the VCE Mathematical Methods Units 3 and 4 technology-active examination (Exam 2, 80 marks, 120 minutes), where an approved CAS calculator is permitted. It combines a multiple-choice section (Section A) with extended-response questions (Section B), and assesses the VCAA study design areas of study:

Functions, relations and graphs: polynomial, power, exponential, logarithmic, circular and inverse functions, and transformations.
Algebra: solving equations (including with CAS), composite and inverse functions.
Calculus: differentiation and applications (rates, optimisation, tangents), antidifferentiation, definite integrals and areas, and the relationship between a function and its derivative graph.
Probability and statistics: discrete and continuous random variables (including the normal distribution), the binomial distribution, and sample proportions and confidence intervals.

The technology-active paper rewards correct setup (defining the function, stating the rule or integral) before using the CAS, and answers that address the original context.

Structure and timing

The paper is 80 marks in 120 minutes (plus 15 minutes reading time) - about 1.5 minutes per mark.

Section A - Multiple choice (~20 marks, 20 questions): budget about 30 minutes (1.5 min each).
Section B - Extended response (~60 marks): budget about 90 minutes at ~1.5 min/mark, with multi-part context questions getting a moment to read the whole structure first.

Use the 15 minutes reading time to set up functions on the CAS and to identify the long modelling question. The biggest time loss is over-investing in one hard part; if a part stalls past ~4 minutes, move on and return. Reserve ~8 minutes to check that final answers address the context (and units).

Worked practice questions (exam-style)

Worked example

Question 1 (4 marks): For $f(x) = 2x^3 - 9x^2 + 12x$ , find the coordinates of the stationary points and classify them.

Stationary points are located where the derivative is zero. We differentiate and factor to solve, substitute to find coordinates, then use the second derivative to confirm the nature of each point.

Step 1: Differentiate and factor

Differentiating term by term and factoring out the common factor:

f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2).

Setting $f'(x) = 0$ gives stationary $x$ -values: $x = 1$ and $x = 2$ .

Step 2: Find the $y$ -coordinates

Substitute back into $f(x)$ :

f(1) = 2 - 9 + 12 = 5, \qquad f(2) = 16 - 36 + 24 = 4.

The stationary points are $(1, 5)$ and $(2, 4)$ .

Step 3: Classify using the second derivative

Compute $f''(x) = 12x - 18$ and evaluate at each stationary point. The sign of $f''$ reveals the concavity and hence the nature of each point:

At $x=1$ : $f''(1) = 12 - 18 = -6 < 0$ , so $(1, 5)$ is a local maximum.
At $x=2$ : $f''(2) = 24 - 18 = 6 > 0$ , so $(2, 4)$ is a local minimum.

Final answer: local maximum at $(1, 5)$ and local minimum at $(2, 4)$ .

Worked example

Question 2 (5 marks): The continuous random variable $X$ has probability density function $f(x) = kx$ for $0 \le x \le 4$ , and $0$ otherwise. (a) Find $k$ . (b) Find $P(X \le 2)$ .

(a) A valid pdf integrates to 1 over its support:

\int_0^4 kx\,dx = k\left[\frac{x^2}{2}\right]_0^4 = k\cdot\frac{16}{2} = 8k = 1 \Rightarrow k = \frac{1}{8}.

(b) $P(X \le 2) = \displaystyle\int_0^2 \frac{1}{8}x\,dx = \frac{1}{8}\left[\frac{x^2}{2}\right]_0^2 = \frac{1}{8}\cdot\frac{4}{2} = \frac{1}{8}\cdot 2 = \frac{1}{4}.$ $

So $P(X \le 2) = 0.25$ . The setup mark depends on recognising that the total probability (the integral of the pdf over its domain) must equal 1; the CAS then handles the integration.

Worked example

Question 3 (6 marks): A factory's machine produces components whose lengths are normally distributed with mean $\mu = 50.0\ \text{mm}$ and standard deviation $\sigma = 0.4\ \text{mm}$ . (a) Find the proportion of components shorter than $49.5\ \text{mm}$ . (b) Components are rejected if outside $50.0 \pm 0.8\ \text{mm}$ . Find the proportion rejected.

(a) Standardise: $z = \dfrac{49.5 - 50.0}{0.4} = -1.25$ . Then $P(X < 49.5) = P(Z < -1.25)$ . Using the CAS (or tables), this is about 0.1056, so roughly 10.6% are shorter than 49.5 mm.

(b) The acceptance interval is $49.2 \le X \le 50.8$ , i.e. $z$ from $\dfrac{49.2-50}{0.4} = -2.0$ to $\dfrac{50.8-50}{0.4} = +2.0$ .

P(49.2 \le X \le 50.8) = P(-2 \le Z \le 2) \approx 0.9545.

So the proportion rejected is

1 - 0.9545 = 0.0455

, about 4.6%. The recurring error is finding the accepted proportion and forgetting to subtract from 1 for the rejected proportion.

Common errors students made

The VCAA examiners' report patterns flagged chain-rule slips on composite functions, sign errors when integrating, skipped domain or restriction statements, and probability questions answered with the wrong conditional setup. Further recurring traps (technology-active):

Trusting a CAS answer without checking setup - a mis-entered function or integral terminal produces a confident wrong answer.
Forgetting to subtract from 1 when a question asks for the complementary probability (e.g. "rejected", "at least one").
Omitting domain or restriction statements for inverse functions and transformations.
Final answers that ignore context or units, leaving a probability as a $z$ -score or a length without millimetres.
Over-rounding intermediate values before a final calculation.

How to use this paper

Sit Section A in 30 minutes and Section B in 90 minutes with your approved CAS, then mark against the official VCAA examiners' report and answers at the links in the frontmatter, noting where method (setup) marks are awarded separately from the final answer. Rework every item that lost more than half its marks, writing the rule, function or integral on its own line before using the CAS. Keep an error log sorted by question class (stationary points, pdf normalisation, normal-distribution tails) and re-test that class technology-active next session.

Use this paper well

Sit the paper under exam conditions (120 minutes, 80 marks).
Mark yourself against the official VCAA marking notes.
Compare against the Math Methods hub to find the syllabus dot points this paper tested.

What this paper assessed

Structure and timing

Worked practice questions (exam-style)

Common errors students made

How to use this paper

Use this paper well

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