HSC Maths Extension 1 past papers: downloads and worked examples

You can find official papers in the NESA Mathematics Extension 1 past-paper archive. Keep each paper with its marking guidelines. If you are sitting the HSC in 2027, start with the sample materials for the new syllabus before choosing questions from older papers.

Download official papers and marking guidelines

We checked these links on 23 September 2026. The sample paper shows how NESA may test the new syllabus. It does not predict the questions in the 2027 exam.

Choose papers for your HSC year

If you are sitting the HSC in 2026, you are studying the Mathematics Extension 1 Stage 6 Syllabus (2017). The first HSC exam for the Mathematics Extension 1 11-12 Syllabus (2024) is in 2027. NESA implementation and examination requirements

The 2025 paper matches the syllabus for the 2026 HSC. For the 2027 HSC, choose questions from it only after checking that they test skills in the new course. A score on a whole older paper will not tell you whether you have covered the new syllabus.

The new exam allows ten minutes of reading followed by two hours of writing. It is worth 70 marks: 10 for objective-response questions and 60 for Section II. NESA provides a reference sheet. Check the official requirements above before practising under timed conditions, and ask your school about any approved exam adjustments that apply to you.

Choose a skill to practise

These three questions from the 2025 paper can help you practise particular skills. The page numbers are the numbers printed on the paper. Each question covers only the skill listed, not the whole focus area.

We matched these questions to the NESA Extension 1 outcomes. NESA has not approved these matches. Before using the questions as a practice assessment, ask your teacher to check them against what you have learnt in class.

Write down which questions you leave out of an older paper. A percentage from a shortened paper is not directly comparable with a result from a whole paper. Our 2027 maths syllabus guide explains why you need to check older resources.

Worked example: show each step in an induction proof

We wrote these exercises for this guide. They are not NESA questions or marked student work, and they do not predict future exam questions.

Prove that, for every integer $n\ge1$,

$$1\cdot2+2\cdot3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}.$$

Base case

For $n=1$, the left side is $2$ and the right side is $1\cdot2\cdot3/3=2$. The statement holds.

Induction assumption

Assume that for some integer $k\ge1$,

$$\sum_{r=1}^{k}r(r+1)=\frac{k(k+1)(k+2)}{3}.$$

We have assumed the formula works for the sum up to $k$. We now need to show that it also works for the sum up to $k+1$.

Add the next term

The next term is $(k+1)(k+2)$. Therefore

$$\begin{aligned} \sum_{r=1}^{k+1}r(r+1) &=\sum_{r=1}^{k}r(r+1)+(k+1)(k+2)\\ &=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\\ &=\frac{(k+1)(k+2)(k+3)}{3}. \end{aligned}$$

In the second line, we replaced the sum using our induction assumption. The last line gives the formula we wanted with $n=k+1$. We have checked the base case and shown that whenever the formula works for $k$, it also works for $k+1$. By mathematical induction, it therefore works for every integer $n\ge1$.

For an arithmetic check, try $n=2$. The sum is $2+6=8$, which matches $2\cdot3\cdot4/3$. Checking a few values can help you spot a calculation error. The induction proof shows that the formula works for every integer in the question.

A common mistake is to write the formula for $k+1$ and say it is true without showing why. Start with the sum up to $k+1$, separate the last term, then use the assumption to replace the earlier part of the sum. Show that step in your working.

Worked example: perpendicular vectors in three dimensions

Let $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(1,t,-2)$. Find $t$ so that the vectors are perpendicular.

Two nonzero vectors are perpendicular when their dot product is zero. Multiply each component of the first vector by the matching component of the second, then add the results:

$$\mathbf{a}\cdot\mathbf{b}=2(1)+(-1)t+3(-2)=-t-4.$$

Set the result equal to zero and solve to get $t=-4$. Check by putting this value back into the original vectors:

$$(2,-1,3)\cdot(1,-4,-2)=2+4-6=0.$$

Both vectors are nonzero, so their zero dot product tells us they meet at a right angle. If you leave out the third components, you get $2-t=0$ and the wrong answer, $t=2$. Include all three components and keep the negative signs in $(-1)t$ and $3(-2)$.

This example covers only part of the three-dimensional vector outcome. You can also ask whether these two vectors could be parallel. To turn the first vector into the second, the first components would need a scale factor of $1/2$, but the third components would need $-2/3$. Parallel vectors need the same scale factor for every component, so no value of $t$ makes these vectors parallel.

Find the first step you could not explain

After trying a question, compare your working with the marking guidelines before reading the full solution. Find the first step you could not explain and write down what you need to practise.

Correct your first attempt, then try a new question on the same skill. Keep both attempts so your teacher can see whether you need help choosing a method, carrying out the steps or explaining why they work.

If you are considering Maths Extension 1 tutoring, bring a question and your working so we can see where you need help. It is fine if your working is unfinished. To plan your study, use the HSC study timetable and resource hub.