2026 VCE Mathematical Methods Exam: What to Learn From Last Year’s Mistakes

2025 median exam grade: B in both exams. VCAA does not publish an exact median mark.

Exam 1 average: 22.35/40, or 55.9%. Exam 2 average: 44.15/80, or 55.2%. VCAA reports these as 44.7/80 and 88.3/160; the figures here are converted to each paper's mark scale.

16,040 students were assessed in Exam 1 and 16,044 in Exam 2. A+ went to 9.0% and 9.2% respectively. The 2025 A+ cutoffs were equivalent to 36/40 and 70/80. Those are last year's cutoffs, not predictions for 2026.

Figures: VCAA's 2025 grade distribution, Exam 1 on page 2 and Exam 2 on page 3.

Question averages come from the Exam 1 report and Exam 2 report. Percentages are rounded. The working below illustrates the fixes.

Exam 1

Exam 1, Q2: Integrating 1/(2x + 3)

Average: 1.2/2. 22% received zero. 40% received full marks.

Where students went wrong. They missed the factor of 1/2 when integrating, or did not recognise the logarithm.

What they should have done. Account for the derivative of the expression inside the logarithm, then use the initial condition.

g(x) = (1/2)ln(2x + 3) + C.

Since g(1) = 0, C = −(1/2)ln(5).

So g(x) = (1/2)ln((2x + 3)/5), on the given domain x > −3/2. Differentiate your answer: the factor of 2 from the chain rule cancels the 1/2.

Exam 1, Q4a: Showing the possible values of k

Average: 1.3/2. 26% received zero. 56% received full marks.

Where students went wrong. They substituted the supplied answers to verify them. That did not show how the values followed from the probability distribution.

What they should have done. Set the sum of the probabilities equal to 1 and solve.

4/k + 2k/75 + k/75 + 2/k = 1.

This becomes 6/k + k/25 = 1. Multiplying by 25k gives k² − 25k + 150 = 0, so (k − 10)(k − 15) = 0. Therefore k = 10 or 15. The working produces the result instead of starting with it.

Exam 1, Q5b: Restricting a function so it has an inverse

Average: 1.0/2. 42% received zero. 43% received full marks.

Where students went wrong. Some substituted a new variable for eˣ, found its value and forgot to convert back to x.

What they should have done. Find the turning point in the original variable and use it as the endpoint of the largest allowed interval.

For g(x) = e²ˣ − 8eˣ + 7, g′(x) = 2eˣ(eˣ − 4). Since eˣ is positive, the stationary point satisfies eˣ = 4, giving x = ln(4). The largest upper bound is therefore a = ln(4), not 4.

Exam 1, Q9a: Getting all four solutions

Average: 1.0/3. 47% received zero. 19% received full marks.

Where students went wrong. They expanded into a quartic and stalled, or lost one sign when taking square roots.

What they should have done. Keep both cases before solving the quadratics.

With w = −3, the equation becomes 9 = [(x − 1)(x + 3)]², with x ≠ 1.

So (x − 1)(x + 3) = 3 or −3. The first gives x = −1 ± √7. The second gives x = 0 or −2. All four satisfy the domain restriction. Using only the positive square root loses half the solutions.

Exam 2

Exam 2, B1d: Average value is not average rate of change

Average: 1.5/2. 25% received zero. 72% received full marks.

Where students went wrong. They calculated the average rate of change. Some entered the correct integral into CAS incorrectly.

What they should have done. Integrate the function over the interval and divide by the interval's width.

Here g(x) = 4x³ − 3x⁴ and the interval is [0, 2]. Its definite integral is −16/5. Divide by 2 to get the average value, −8/5. Calculating [g(2) − g(0)]/2 answers a different question.

Exam 2, B1e: Transformation order

Average: 1.2/3. 30% received zero. 20% received full marks.

Where students went wrong. They gave transformations in the wrong order or used the reciprocal of the required dilation factor.

What they should have done. State the axis for each transformation and check the key points after applying them in order.

One valid sequence is reflection in the y-axis, dilation by factor 2 from the y-axis, then translation 1 unit right. The original stationary points (0, 0) and (1, 1) become (1, 0) and (−1, 1), as required. Check both points; one correct image does not establish the whole sequence.

Exam 2, B3a(ii): Standard deviation in exact form

Average: 1.5/2. 17% received zero. 65% received full marks.

Where students went wrong. They stopped at the variance, supplied no working, or rounded to 5.34… when an exact value was required.

What they should have done. Calculate E(T²) − [E(T)]², then take its square root and retain the exact answer.

For the supplied density, E(T) = 39 and E(T²) = 10847/7. The variance is 200/7, so the standard deviation is √(200/7) minutes. The square root and the unit finish the answer.

Use the paper alongside this

I run ATARMAXXING. Our VCE Mathematical Methods hub has notes and extra practice.

Both 2025 official exams and the assessment guides and Exam 1 report are in the 2025 row of our past-paper index. These links open the index, then the official VCAA documents. The separate Exam 2 report is linked above. Official past-paper links are free.

Try Exam 1 Q2, Q4a and Q9a without technology before reading the solutions. Together they are worth 7 marks. Check the working, not just whether the final answer matches.

If you want to see which of these areas keep appearing, the 2026 exam predictors page lists the topics to prioritise. You can also submit a practice response for free AI marking, open to every account until Sunday 1 November 2026 to check whether your working, not just your answer, would earn the marks.

This post first appeared on r/ATARMAXXING.