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2026 VCE Mathematical Methods exam: What to learn from last year's mistakes
Hi all, in preparation for the upcoming exam, here's 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: source.
Question averages come from the Exam 1 report: source and Exam 2 report: source. Percentages are rounded. The working below illustrates the fixes.
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, 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.
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.
Use the VCE Mathematical Methods hub and past papers and examiner reports alongside this.
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