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VICMath Methods2021

VCE Math Methods 2021

Walkthrough of the 2021 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

The VCAA Mathematical Methods Units 3 & 4 examination consists of Exam 1 (technology-free, 40 marks) and Exam 2 (technology-active, 80 marks in 120 minutes). This guide covers the 80-mark, 120-minute technology-active paper (Exam 2), which assesses across the four study-design areas:

Functions, relations and graphs: polynomial, power, exponential, logarithmic, circular and inverse functions; transformations; and the use of CAS to analyse them.
Algebra: solving equations and systems, including with parameters.
Calculus: differentiation and its applications (rates, optimisation, tangents), antidifferentiation and definite integrals (areas, average value).
Probability and statistics: discrete and continuous random variables, the binomial and normal distributions, and statistical inference (confidence intervals for proportions).

Exam 2 has a multiple-choice section and an extended-response section. It rewards setting up a model from context, careful CAS syntax, and final answers expressed in context to the required accuracy.

Structure and timing

80 marks in 120 minutes is 1.5 minutes per mark.

Section A - Multiple choice (20 marks, 20 questions). Target about 25-30 minutes; even with CAS, do quick checking rather than guessing.
Section B - Extended response (60 marks). Target about 85-90 minutes at ~1.5 min/mark; a 5-mark modelling part deserves ~7-8 minutes including a sense-check.

A workable plan: ~28 minutes Section A, ~85 minutes Section B, with a final 7-8 minutes to recheck context answers, domains and CAS outputs.

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 occur where the derivative is zero. We differentiate, solve, then use the second derivative to classify each point as a local maximum or minimum.

Step 1: Differentiate and factor

Differentiate term by term, then factor to find the roots neatly:

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

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

Step 2: Find the $y$ -coordinates

Substitute each $x$ -value back into the original function:

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:

$f''(1) = 12 - 18 = -6 < 0$ , so $(1, 5)$ is a local maximum.
$f''(2) = 24 - 18 = 6 > 0$ , 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 $0$ elsewhere. (a) Find $k$ . (b) Find $P(X > 2)$ .

For a continuous random variable, the total probability must equal 1, and probabilities are computed by integrating the pdf over the relevant interval. We use these two facts in sequence.

Step 1: Find $k$ using the normalisation condition

The total area under any valid pdf must equal 1. Integrating over the full support:

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

Step 2: Compute $P(X > 2)$ by integrating the pdf

Substitute $k = \dfrac{1}{8}$ and integrate from 2 to the upper bound of the support:

P(X > 2) = \int_2^4 \frac{1}{8}x\,dx = \frac{1}{8}\left[\frac{x^2}{2}\right]_2^4 = \frac{1}{8}\left(\frac{16}{2} - \frac{4}{2}\right) = \frac{1}{8}(8 - 2) = \frac{6}{8} = \frac{3}{4}.

Final answer: $k = \dfrac{1}{8}$ and $P(X > 2) = \dfrac{3}{4}$ .

Worked example

Question 3 (6 marks): A machine fills bottles with a mean of $500\ \text{mL}$ and standard deviation $4\ \text{mL}$ , normally distributed. (a) Find the probability a bottle contains less than $494\ \text{mL}$ . (b) In a quality sample of 200 bottles, an underfill proportion of $\hat{p} = 0.10$ is observed; construct an approximate 95% confidence interval for the true proportion of underfilled bottles.

(a) Standardise: $z = \dfrac{494 - 500}{4} = -1.5$ . From the standard normal (or CAS), $P(Z < -1.5) \approx 0.0668$ . So about 6.7% of bottles contain less than 494 mL.

(b) An approximate 95% confidence interval for a proportion is:

\hat{p} \pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.10 \pm 1.96\sqrt{\frac{0.10 \times 0.90}{200}}

= 0.10 \pm 1.96\sqrt{4.5 \times 10^{-4}} = 0.10 \pm 1.96(0.0212) = 0.10 \pm 0.0416

The 95% confidence interval is approximately $(0.058,\ 0.142)$ - between about 5.8% and 14.2%.

Common errors students made

The 2021 VCAA examiners' report noted chain-rule slips on composite functions, sign errors when integrating, skipped domain/restriction statements, and wrong conditional setups in probability. Adding to those:

CAS over-reliance without checking. Accepting a CAS output that is in the wrong form or outside the required domain.
Misclassifying stationary points - not testing the second derivative or a sign change.
PDF normalisation errors - forgetting that the total area must equal 1 when solving for an unknown constant.
Confidence-interval slips - using the wrong $z$ -value (not 1.96 for 95%) or the wrong standard error formula.

How to use this paper

Sit Section A in a strict ~28-minute block, then Section B in a single ~85-minute run under exam conditions with your approved technology, keeping a final 7-8 minute buffer to recheck context answers, domains and CAS outputs. Mark each item against the official VCAA assessment report and answers at the authority page linked in the frontmatter, noting where method marks are awarded - they are recoverable. Re-attempt any item that lost more than half its marks on a fresh page, writing the rule or model before computing. Keep an error log of recurring slips (point classification, PDF normalisation, $z$ -values) and re-test those item types after a week. Pair this with the technology-free Exam 1 to drill by-hand technique too.

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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