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NSWMaths Extension 12022

HSC Maths Extension 1 2022

Walkthrough of the 2022 HSC Mathematics Extension 1 exam: what it assessed, timing, original exam-style worked questions on auxiliary angles, volumes, differential equations, induction, vectors and the normal approximation, and common errors from NESA's 2022 marking feedback.

Marks: 70
Time: 120 min
Authority: NESA
Updated: 2026-09-28

Open the official PDF Official marking notes

What this paper assessed

The 2022 HSC Mathematics Extension 1 paper sampled every strand of the Year 11 and Year 12 Extension 1 course, with a strong lean towards the Year 12 topics of vectors, calculus and statistics. In broad terms it covered:

Trigonometric and inverse functions: exact values from compound angles, writing a sum of sine and cosine terms as a single wave (the auxiliary angle method), and deciding when an inverse trigonometric function really is the inverse of a trigonometric function on a given domain.
Functions and graphs: transformations involving absolute value and reflections, combining graphs by addition, parametric curves, inequalities with the variable in a denominator, and properties of a function and its inverse (where the graphs meet and how their tangents relate).
Polynomials: the degree of a divisor given a remainder, and relationships between roots and coefficients of a monic cubic, combined with its derivative.
Combinatorics and proof: counting triangles from points on the sides of a figure, the pigeonhole principle, the binomial theorem (specific coefficients of an expansion with a fractional negative term), and proof by mathematical induction of a divisibility result.
Calculus: integration by substitution, the derivative of a product involving inverse tangent and the tangent line it gives, volumes of solids of revolution requiring a double angle identity, and separable differential equations, including a Newton's law of cooling model, a direction field and a differential equation whose particular solution needed careful handling of an absolute value.
Vectors: component arithmetic, dot products and perpendicularity, projections, a geometric vector proof on a circle, a proof that the projection gives the shortest distance to a line, and projectile motion against a moving target.
Statistics: expected value of a game built on a binomial distribution, and the normal approximation to the binomial (a hypothesis-style check of a manufacturer's claim, a question on whether the method was valid, and an overbooking problem that reduced to a quadratic in the number of tickets).

Structure and timing

The paper cover gives 10 minutes reading time and 2 hours working time for 70 marks. A NESA-approved calculator and the reference sheet are allowed, and a standard normal table was printed at the back of the paper for the statistics questions.

Section	Questions	Marks	Suggested time
Section I: multiple choice	1 to 10	10	about 15 minutes
Section II: extended response	11 (15), 12 (16), 13 (14), 14 (15)	60	about 1 hour 45 minutes

The working-time arithmetic is $\dfrac{120}{70} \approx 1.7$ minutes per mark. A practical split:

Section I: 15 minutes, which is $1.5$ minutes per question. Flag any graph-reading item that stalls you and return later.
Questions 11 and 12 (31 marks): about 50 minutes. These are mostly routine skills, so bank the marks with complete working.
Questions 13 and 14 (29 marks): about 50 minutes. The last parts of Question 14 (the projectile and the overbooking problem) are the hardest items on the paper.
Final check: 5 minutes for re-reading "show that" lines and rounding instructions.

Worked practice questions (exam-style)

Worked example

Question 1 (3 marks): Express $2\cos x - 2\sqrt{3}\sin x$ in the form $R\cos(x + \alpha)$ , where $R > 0$ and $0 < \alpha < \dfrac{\pi}{2}$ . Hence solve $2\cos x - 2\sqrt{3}\sin x = 2$ for $0 \le x \le 2\pi$ .

Step 1: Expand and equate coefficients

R\cos(x + \alpha) = R\cos x\cos\alpha - R\sin x\sin\alpha

Matching with $2\cos x - 2\sqrt{3}\sin x$ gives

R\cos\alpha = 2, \qquad R\sin\alpha = 2\sqrt{3}.

Step 2: Find $R$ and $\alpha$

R^2 = 2^2 + (2\sqrt{3})^2 = 4 + 12 = 16 \implies R = 4

\tan\alpha = \frac{2\sqrt{3}}{2} = \sqrt{3}

Both $\cos\alpha$ and $\sin\alpha$ are positive, so $\alpha$ is in the first quadrant and $\alpha = \dfrac{\pi}{3}$ .

Step 3: Solve the equation

4\cos\left(x + \frac{\pi}{3}\right) = 2 \implies \cos\left(x + \frac{\pi}{3}\right) = \frac{1}{2}

Since $0 \le x \le 2\pi$ , the argument satisfies $\dfrac{\pi}{3} \le x + \dfrac{\pi}{3} \le \dfrac{7\pi}{3}$ . In that interval cosine equals $\dfrac{1}{2}$ at $\dfrac{\pi}{3}$ , $\dfrac{5\pi}{3}$ and $\dfrac{7\pi}{3}$ , so

x = 0, \quad \frac{4\pi}{3}, \quad 2\pi.

Final answer: $2\cos x - 2\sqrt{3}\sin x = 4\cos\left(x + \dfrac{\pi}{3}\right)$ , and the solutions are $x = 0, \dfrac{4\pi}{3}, 2\pi$ .

Worked example

Question 2 (3 marks): The region under the curve $y = \sin x + \cos x$ between $x = 0$ and $x = \dfrac{\pi}{2}$ is rotated about the $x$ -axis. Find the exact volume of the solid formed.

Step 1: Set up the volume integral

V = \pi\int_0^{\pi/2} (\sin x + \cos x)^2\,dx

Step 2: Simplify the integrand

(\sin x + \cos x)^2 = \sin^2 x + 2\sin x\cos x + \cos^2 x = 1 + \sin 2x

using $\sin^2 x + \cos^2 x = 1$ and $2\sin x\cos x = \sin 2x$ .

Step 3: Integrate and substitute the limits

V = \pi\left[x - \frac{1}{2}\cos 2x\right]_0^{\pi/2} = \pi\left[\left(\frac{\pi}{2} - \frac{1}{2}\cos\pi\right) - \left(0 - \frac{1}{2}\cos 0\right)\right]

= \pi\left(\frac{\pi}{2} + \frac{1}{2} + \frac{1}{2}\right) = \pi\left(\frac{\pi}{2} + 1\right)

Final answer: $V = \pi\left(\dfrac{\pi}{2} + 1\right) = \dfrac{\pi^2 + 2\pi}{2}$ cubic units (about $8.08$ ).

Worked example

Question 3 (4 marks): A tray of food is taken from a fridge at $5^\circ$ C and left in a kitchen kept at $25^\circ$ C. Its temperature $T$ after $t$ minutes satisfies $\dfrac{dT}{dt} = k(T - 25)$ . After 10 minutes the food is at $13^\circ$ C. (a) Solve the differential equation by separating variables to find $T$ as a function of $t$ . (b) Find, to the nearest minute, when the food reaches $20^\circ$ C.

Step 1: Separate the variables

The food is always colder than the room, so $T - 25 < 0$ and $|T - 25| = 25 - T$ .

\int \frac{dT}{T - 25} = \int k\,dt \implies \ln(25 - T) = kt + c

25 - T = Ae^{kt}, \quad A = e^{c} \implies T = 25 - Ae^{kt}

Step 2: Use the initial condition

At $t = 0$ , $T = 5$ : $5 = 25 - A$ , so $A = 20$ and $T = 25 - 20e^{kt}$ .

Step 3: Use the second condition to find $k$

At $t = 10$ , $T = 13$ :

13 = 25 - 20e^{10k} \implies e^{10k} = \frac{12}{20} = 0.6 \implies k = \frac{1}{10}\ln 0.6 \approx -0.0511

Step 4: Find when $T = 20$

20 = 25 - 20e^{kt} \implies e^{kt} = \frac{1}{4} \implies t = \frac{\ln 0.25}{k} = \frac{10\ln 0.25}{\ln 0.6} \approx 27.14

Final answer: (a) $T = 25 - 20e^{\frac{t}{10}\ln 0.6}$ ; (b) the food reaches $20^\circ$ C after about $27$ minutes.

Worked example

Question 4 (3 marks): Use mathematical induction to prove that $4^n + 6n - 1$ is divisible by $9$ for all integers $n \ge 1$ .

Step 1: Base case

For $n = 1$ : $4^1 + 6(1) - 1 = 9 = 9 \times 1$ , which is divisible by $9$ . So the statement is true for $n = 1$ .

Step 2: Inductive hypothesis

Assume the statement is true for some integer $n = k \ge 1$ , that is

4^k + 6k - 1 = 9M \quad \text{for some integer } M, \quad\text{so}\quad 4^k = 9M - 6k + 1.

Step 3: Inductive step

Prove it for $n = k + 1$ :

4^{k+1} + 6(k+1) - 1 = 4 \cdot 4^k + 6k + 5

= 4(9M - 6k + 1) + 6k + 5 = 36M - 24k + 4 + 6k + 5

= 36M - 18k + 9 = 9(4M - 2k + 1)

Since $4M - 2k + 1$ is an integer, the expression is divisible by $9$ .

Step 4: Conclusion

The statement is true for $n = 1$ , and if it is true for $n = k$ then it is true for $n = k + 1$ . Hence, by mathematical induction, it is true for all integers $n \ge 1$ .

Final answer: $4^{k+1} + 6(k+1) - 1 = 9(4M - 2k + 1)$ , so by induction $4^n + 6n - 1$ is divisible by $9$ for all integers $n \ge 1$ .

Worked example

Question 5 (3 marks): Let $\underline{a} = 2\underline{i} - \underline{j}$ and $\underline{b} = \underline{i} + 3\underline{j}$ . (a) Find the projection of $\underline{a}$ onto $\underline{b}$ . (b) Find the real number $t$ for which $\underline{a} - t\underline{b}$ is perpendicular to $\underline{b}$ , and explain the connection with part (a).

Step 1: Dot products

\underline{a} \cdot \underline{b} = 2(1) + (-1)(3) = -1, \qquad \underline{b} \cdot \underline{b} = 1^2 + 3^2 = 10

Step 2: Projection

\text{proj}_{\underline{b}}\,\underline{a} = \frac{\underline{a} \cdot \underline{b}}{\underline{b} \cdot \underline{b}}\,\underline{b} = -\frac{1}{10}\left(\underline{i} + 3\underline{j}\right) = -0.1\underline{i} - 0.3\underline{j}

Step 3: Perpendicular condition

Perpendicular vectors have a zero dot product:

\left(\underline{a} - t\underline{b}\right) \cdot \underline{b} = \underline{a} \cdot \underline{b} - t\,\underline{b} \cdot \underline{b} = -1 - 10t = 0 \implies t = -\frac{1}{10}

This is the same scalar as in the projection: subtracting the projection of $\underline{a}$ onto $\underline{b}$ leaves the component of $\underline{a}$ that is perpendicular to $\underline{b}$ (and that component is also the shortest vector of the form $\underline{a} - t\underline{b}$ ).

Final answer: (a) $-\dfrac{1}{10}\underline{i} - \dfrac{3}{10}\underline{j}$ ; (b) $t = -\dfrac{1}{10}$ , the scalar multiple of $\underline{b}$ in the projection.

Worked example

Question 6 (3 marks): A seed supplier claims that 90% of its seeds germinate. A gardener plants 400 randomly chosen seeds and only 345 germinate. (a) Assuming the claim is true, use a normal approximation to the binomial distribution (without a continuity correction) to estimate the probability that fewer than 345 of the 400 seeds germinate. (b) Comment on whether the approximation is reasonable and what the result suggests about the claim.

Step 1: Binomial parameters

Let $X$ be the number of seeds that germinate, so $X \sim \text{Bin}(400, 0.9)$ .

E(X) = np = 400 \times 0.9 = 360, \qquad \text{Var}(X) = np(1-p) = 400 \times 0.9 \times 0.1 = 36, \qquad \sigma = 6

Step 2: Standardise

z = \frac{345 - 360}{6} = -2.5

P(X < 345) \approx P(Z < -2.5) = 1 - P(Z < 2.5) = 1 - 0.9938 = 0.0062

Step 3: Interpret

The approximation is reasonable because $n$ is large and both $np = 360$ and $n(1-p) = 40$ are well above the usual threshold, so the binomial distribution is close to symmetric. A result this low would happen only about $0.6\%$ of the time if the claim were true, which is strong evidence that the true germination rate is below 90%.

Final answer: $P(X < 345) \approx 0.0062$ ; the approximation is valid since $np$ and $n(1-p)$ are both large, and the result casts serious doubt on the 90% claim.

Common errors students made

NESA's 2022 marking feedback for Mathematics Extension 1 noted the following, paraphrased here:

Vectors: some students did not treat the dot product as a scalar, used vector notation carelessly, or made arithmetic slips when combining components. In the perpendicular-vectors part, errors came from expanding brackets and factorising the resulting quadratic. The feedback recommended stating up front that perpendicular vectors have a zero dot product.
Integration by substitution: students lost marks by not changing the limits to match the new variable, by not using $du$ properly to rebuild the integrand, and by mishandling negative fractional powers when finding the primitive.
Binomial expansion: negative terms and fractions raised to odd and even powers caused sign errors. NESA encouraged using the general term to find the required coefficients directly instead of expanding everything.
Auxiliary angle: students were advised to expand the compound angle form from the reference sheet, equate coefficients, and pay attention to the quadrant of the angle.
Inequalities with a variable in the denominator: multiplying both sides by the denominator itself (whose sign is unknown) was a common flaw; NESA pointed to multiplying by its square, or finding critical values and testing regions, and reminded students that the value making the denominator zero must be excluded.
Pigeonhole principle and direction fields: weaker answers did not show the division and remainder that justifies the conclusion, and did not draw the relative steepness of slopes correctly or show the substitution that produced each slope.
Differential equations: separating variables, integrating $\dfrac{1}{y}$ to $\ln|y|$ , and deciding and justifying which sign of the absolute value applies were identified as weaknesses. For the cooling model, better responses identified the room temperature constant early and used the two given conditions to find both constants.
Volumes of revolution: students needed to recall the volume formula, use the double angle result to replace a squared sine, take constants outside the integral and integrate trigonometric functions correctly.
Inverse functions: weaker explanations did not address the domain and range conditions, or did not use a specific value where composing the functions fails to return the input.
Induction: some students did not verify the base case properly or left out steps of the inductive argument and the conclusion.
Normal approximation: students needed to know what the reference sheet provides, identify the correct mean and standard deviation, read the $z$ -table correctly (including all its columns), and give specific reasons about sample size when judging validity rather than general comments. In the overbooking question, NESA flagged the algebra of rearranging the $z$ -score condition into a quadratic in the number of tickets as an area to improve.
Projectile and vector proofs: NESA pointed to planning the approach, telling apart the horizontal and vertical components and simplifying the trigonometric algebra, and in the projection proof, the value of a large labelled diagram showing where the projection lies.

How to use this paper

Sit the official paper (linked above) under exam conditions: 10 minutes reading time, then 2 hours, spending about 15 minutes on Section I.
Mark it against the official marking guidelines at https://www.nsw.gov.au/sites/default/files/noindex/2025-05/2022-hsc-maths-ext-1-mg.pdf, paying attention to how the marks are split between setting up, the key middle step and the final answer.
Read the marking feedback (linked above) for each question you dropped marks on, and list the specific habit you need to change, for example changing limits in a substitution or justifying the sign of an absolute value.
Redo the worked questions on this page without looking, then write your own variant of each by changing the numbers or context.
Give the statistics and vector proof questions (Question 13 (e) and Question 14) a second timed attempt a week later, since these were the most demanding parts of the paper.

Use this paper well

Sit the paper under exam conditions (120 minutes, 70 marks).
Mark yourself against the official NESA marking notes.
Compare against the Maths Extension 1 hub to find the syllabus dot points this paper tested.

What this paper assessed

Structure and timing

Worked practice questions (exam-style)

Step 1: Expand and equate coefficients

Step 2: Find RRR and α\alphaα

Step 3: Solve the equation

Question 2 (3 marks): The region under the curve y=sin⁡x+cos⁡xy = \sin x + \cos xy=sinx+cosx between x=0x = 0x=0 and x=π2x = \dfrac{\pi}{2}x=2π​ is rotated about the xxx-axis. Find the exact volume of the solid formed.

Step 1: Set up the volume integral

Step 2: Simplify the integrand

Step 3: Integrate and substitute the limits

Step 1: Separate the variables

Step 2: Use the initial condition

Step 3: Use the second condition to find kkk

Step 4: Find when T=20T = 20T=20

Question 4 (3 marks): Use mathematical induction to prove that 4n+6n−14^n + 6n - 14n+6n−1 is divisible by 999 for all integers n≥1n \ge 1n≥1.

Step 1: Base case

Step 2: Inductive hypothesis

Step 3: Inductive step

Step 4: Conclusion

Step 1: Dot products

Step 2: Projection

Step 3: Perpendicular condition

Step 1: Binomial parameters

Step 2: Standardise

Step 3: Interpret

Common errors students made

How to use this paper

Use this paper well

Keep going

Step 2: Find $R$ and $\alpha$

Question 2 (3 marks): The region under the curve $y = \sin x + \cos x$ between $x = 0$ and $x = \dfrac{\pi}{2}$ is rotated about the $x$ -axis. Find the exact volume of the solid formed.

Step 3: Use the second condition to find $k$

Step 4: Find when $T = 20$

Question 4 (3 marks): Use mathematical induction to prove that $4^n + 6n - 1$ is divisible by $9$ for all integers $n \ge 1$ .