§-Past paper

NSWMaths Extension 12021

HSC Maths Extension 1 2021

Walkthrough of the 2021 HSC Mathematics Extension 1 exam: what the paper assessed, how to split your 120 minutes, six original exam-style worked questions on induction, projectiles, inverse trig integration, differential equations, sample proportions and combinatorics, and common errors from NESA's 2021 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 2021 HSC Mathematics Extension 1 paper sampled the full Year 11 and Year 12 Extension 1 course. Reading the paper alongside NESA's mapping grid, the main areas were:

Vectors (ME-V1): adding and subtracting vectors in component form, interpreting the sign of a dot product, a projectile modelled with a displacement vector, relative velocity with a wind (a bearing problem), and a vector proof about a trapezium with equal diagonals using $\underline{v} \cdot \underline{v} = |\underline{v}|^2$ .
Calculus (ME-C1, C2, C3): related rates for a growing sphere, integration by substitution, integrals that give an inverse sine or inverse tangent, volumes of solids of revolution about the $y$ -axis, an area involving an absolute value function, and the derivative of a product that contains an inverse function.
Differential equations: a direction field sketch, a Newton's law of cooling (warming) model solved by separating variables, and a logistic population model using partial fractions.
Proof by mathematical induction (ME-P1): a summation identity with a three-factor product in each denominator.
Functions and polynomials (ME-F1, F2): the remainder theorem, sums and products of roots of a quartic, parametric equations of a semicircle, and a restricted-domain parabola with its inverse function.
Trigonometry (ME-T1, T2, T3): $\sin^2$ as a double-angle integral, the graph of $\sin^{-1}(\sin x)$ , auxiliary-angle style graphs, a cubic in $\sin x$ , and sum-to-product identities used to solve an equation.
Combinatorics and statistics (ME-A1, S1): the binomial expansion, selecting a committee, the pigeonhole principle, a binomial probability and a normal approximation to a sample proportion.

Structure and timing

The cover gives 10 minutes reading time and 2 hours working time, for 70 marks in total. NESA-approved calculators are allowed and a reference sheet is supplied.

Section I (10 marks): Questions 1 to 10, multiple choice. Allow about 15 minutes.
Section II (60 marks): Questions 11 to 14, each in its own writing booklet. Allow about 1 hour and 45 minutes. Question 11 is worth 16 marks, Question 12 is 14, Question 13 is 14 and Question 14 is 16.

The arithmetic: $120 \div 70 \approx 1.7$ minutes per mark overall. Section I gets $15 \div 10 = 1.5$ minutes per mark, and Section II gets $105 \div 60 = 1.75$ minutes per mark.

A suggested split for Section II is about 28 minutes each for Questions 11 and 14 ( $16 \times 1.75$ ) and about 24 to 25 minutes each for Questions 12 and 13 ( $14 \times 1.75 = 24.5$ ). Question 11 is mostly routine, so try to bank a few minutes there for the harder vector and statistics parts at the end of Question 14.

Worked practice questions (exam-style)

These are original questions written in the same syllabus areas as the 2021 paper. They are not NESA questions.

Worked example

Question 1 (3 marks): Use mathematical induction to prove that $\displaystyle\sum_{r=1}^{n} r \cdot 2^r = (n-1)\,2^{n+1} + 2$ for all integers $n \ge 1$ .

Step 1: Base case, $n = 1$

\text{LHS} = 1 \times 2^1 = 2, \qquad \text{RHS} = (1-1)\,2^{2} + 2 = 0 + 2 = 2

So the statement is true for $n = 1$ .

Step 2: Inductive hypothesis

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

\sum_{r=1}^{k} r \cdot 2^r = (k-1)\,2^{k+1} + 2

Step 3: Prove it for $n = k + 1$

We need to show $\displaystyle\sum_{r=1}^{k+1} r \cdot 2^r = k \cdot 2^{k+2} + 2$ . Start from the left-hand side and use the assumption:

\sum_{r=1}^{k+1} r \cdot 2^r = \underbrace{(k-1)\,2^{k+1} + 2}_{\text{by the assumption}} + (k+1)\,2^{k+1}

= \big[(k-1) + (k+1)\big]\,2^{k+1} + 2 = 2k \cdot 2^{k+1} + 2 = k \cdot 2^{k+2} + 2

This is the right-hand side for $n = k + 1$ , so if the statement is true for $n = k$ it is true for $n = k + 1$ .

Step 4: Conclusion

Since it is true for $n = 1$ , and truth for $n = k$ implies truth for $n = k + 1$ , the statement is true for all integers $n \ge 1$ by mathematical induction.

Final answer: Proven: $\displaystyle\sum_{r=1}^{n} r \cdot 2^r = (n-1)\,2^{n+1} + 2$ for all integers $n \ge 1$ .

Worked example

Question 2 (4 marks): A golf ball is struck from ground level with speed $25$ m/s at $40^\circ$ to the horizontal. Its displacement vector after $t$ seconds is $\underline{r}(t) = (25t\cos 40^\circ)\,\underline{i} + (-5t^2 + 25t\sin 40^\circ)\,\underline{j}$ . A tree $11$ m tall stands $40$ m away in the direction of the shot. Find the maximum height of the ball and show that it passes over the tree.

Step 1: Components

25\cos 40^\circ \approx 19.151, \qquad 25\sin 40^\circ \approx 16.070

Step 2: Maximum height

The vertical velocity is $\dfrac{d}{dt}\left(-5t^2 + 25t\sin 40^\circ\right) = -10t + 25\sin 40^\circ$ . This is zero when

t = \frac{25\sin 40^\circ}{10} \approx 1.607 \text{ s}

y_{\max} = -5(1.607)^2 + 16.070(1.607) \approx 12.91 \text{ m}

Step 3: Time to reach the tree

Set the horizontal displacement to $40$ :

25t\cos 40^\circ = 40 \quad\Rightarrow\quad t = \frac{40}{19.151} \approx 2.089 \text{ s}

Step 4: Height at the tree

y = -5(2.089)^2 + 16.070(2.089) \approx -21.81 + 33.56 \approx 11.75 \text{ m}

The ball is still in the air (the height is positive) and $11.75 > 11$ , so it clears the tree by about $0.75$ m. It is already descending here, which is consistent: $2.089 > 1.607$ and $11.75 < 12.91$ .

Final answer: Maximum height $\approx 12.9$ m; at the tree ( $t \approx 2.09$ s) the ball is about $11.75$ m high, so it passes over the $11$ m tree.

Worked example

Question 3 (3 marks): Find the exact area of the region bounded by $y = \dfrac{12}{9 + x^2}$ , the $x$ -axis and the lines $x = -\sqrt{3}$ and $x = 3$ .

The curve is always above the $x$ -axis, so the area is the definite integral. Use the reference sheet result $\displaystyle\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c$ with $a = 3$ .

Step 1: Integrate

A = \int_{-\sqrt{3}}^{3} \frac{12}{9 + x^2}\,dx = 12 \cdot \frac{1}{3}\Big[\tan^{-1}\frac{x}{3}\Big]_{-\sqrt{3}}^{3} = 4\Big[\tan^{-1}\frac{x}{3}\Big]_{-\sqrt{3}}^{3}

Step 2: Evaluate in radians

\tan^{-1}(1) = \frac{\pi}{4}, \qquad \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}

A = 4\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = 4 \cdot \frac{5\pi}{12} = \frac{5\pi}{3}

Final answer: $A = \dfrac{5\pi}{3}$ square units.

Worked example

Question 4 (4 marks): A bowl of soup at $90^\circ$ C is left in a kitchen kept at $20^\circ$ C. Its temperature $T$ after $t$ minutes satisfies $\dfrac{dT}{dt} = k(T - 20)$ . After $5$ minutes the soup is at $70^\circ$ C. By solving the differential equation, find when the soup reaches $40^\circ$ C, to the nearest minute.

Step 1: Separate variables and integrate

\int \frac{dT}{T - 20} = \int k\,dt \quad\Rightarrow\quad \ln|T - 20| = kt + c

Since $T > 20$ throughout, $T - 20 = Ae^{kt}$ , so $T = 20 + Ae^{kt}$ .

Step 2: Use the initial condition

At $t = 0$ , $T = 90$ : $\;90 = 20 + A$ , so $A = 70$ and $T = 20 + 70e^{kt}$ .

Step 3: Find $k$

At $t = 5$ , $T = 70$ :

50 = 70e^{5k} \quad\Rightarrow\quad e^{5k} = \frac{5}{7} \quad\Rightarrow\quad k = \frac{1}{5}\ln\frac{5}{7} \approx -0.0673

Step 4: Solve for $T = 40$

20 = 70e^{kt} \quad\Rightarrow\quad e^{kt} = \frac{2}{7} \quad\Rightarrow\quad t = \frac{\ln\frac{2}{7}}{k} = \frac{5\ln\frac{2}{7}}{\ln\frac{5}{7}} \approx \frac{5(-1.2528)}{-0.3365} \approx 18.6

The answer is a decimal number of minutes (not degrees and minutes), so round directly.

Final answer: $t \approx 18.6$ , so the soup reaches $40^\circ$ C after about $19$ minutes.

Worked example

Question 5 (4 marks): A seed supplier states that $80\%$ of its seeds germinate. (a) Five seeds are planted. Find the probability that exactly four germinate. (b) A sample of $400$ seeds is tested and the sample proportion $\hat{p}$ that germinate is approximately normal with mean $p$ and variance $\dfrac{p(1-p)}{n}$ . Use the empirical rule to estimate $P(\hat{p} \le 0.76)$ . (c) What sample size would make the standard deviation of $\hat{p}$ equal to $0.01$ ?

Step 1: Binomial probability for (a)

With $X \sim \text{Bin}(5, 0.8)$ :

P(X = 4) = \binom{5}{4}(0.8)^4(0.2)^1 = 5 \times 0.4096 \times 0.2 = 0.4096

Step 2: Standard deviation for (b)

The variance is $\dfrac{0.8 \times 0.2}{400} = 0.0004$ , so the standard deviation (not the variance) is

\sigma = \sqrt{0.0004} = 0.02

Step 3: $z$ -score and empirical rule

z = \frac{0.76 - 0.8}{0.02} = -2

About $95\%$ of a normal distribution lies within $2$ standard deviations of the mean, leaving $5\%$ split equally between the two tails:

P(\hat{p} \le 0.76) \approx \frac{5\%}{2} = 2.5\%

Step 4: Sample size for (c)

\sqrt{\frac{0.16}{n}} = 0.01 \quad\Rightarrow\quad \frac{0.16}{n} = 0.0001 \quad\Rightarrow\quad n = 1600

Final answer: (a) $0.4096$ ; (b) approximately $0.025$ (that is, $2.5\%$ ); (c) $n = 1600$ seeds.

Worked example

Question 6 (3 marks): (a) A debating squad of $4$ is chosen from $7$ boys and $6$ girls. How many squads contain at least $2$ girls? (b) Show that in any group of $100$ people, at least $9$ were born in the same month.

Step 1: Split (a) into cases

Choose the girls and the boys separately and multiply within each case, then add the cases:

\binom{6}{2}\binom{7}{2} + \binom{6}{3}\binom{7}{1} + \binom{6}{4}\binom{7}{0} = 15 \times 21 + 20 \times 7 + 15 \times 1 = 315 + 140 + 15 = 470

Step 2: Check (a) with the complement

\binom{13}{4} - \binom{7}{4} - \binom{6}{1}\binom{7}{3} = 715 - 35 - 210 = 470 \;\checkmark

Step 3: Pigeonhole principle for (b)

There are $12$ months (pigeonholes) and $100$ people (pigeons). If every month had at most $8$ people, there would be at most $12 \times 8 = 96 < 100$ people, a contradiction. So some month has at least

\left\lceil \frac{100}{12} \right\rceil = 9 \text{ people}

Final answer: (a) $470$ squads; (b) since $12 \times 8 = 96 < 100$ , at least $9$ people share a birth month.

Common errors students made

NESA's 2021 marking feedback for Mathematics Extension 1 pointed to these recurring problems (paraphrased):

Binomial expansion signs. Students needed to handle the negative sign correctly when expanding a binomial with a subtraction inside, whether using Pascal's triangle or squaring twice.
Substitution integrals left unfinished. Areas to improve were integrating fractional powers correctly, converting back from $u$ to $x$ , and including the constant for an indefinite integral.
Adding instead of multiplying in counting. When forming a group from two separate pools, students needed to multiply the two selection counts rather than add them.
Related rates. Students needed to recall the volume of a sphere, apply the chain rule and round correctly to one decimal place.
Inverse trig integrals. Students were encouraged to use the reference sheet for primitives that give inverse trig functions and to evaluate them in radians.
Trig equations. Areas to improve were spotting the common factor, using both signs of the square root, not dividing through by a factor (which loses solutions), and giving answers in radians.
Roots and coefficients. Students needed to know the quartic relationships and combine four reciprocals into one fraction.
Differential equations. Areas to improve included solving the separable equation, finding the constant from initial conditions, converting logs to exponentials, applying log laws, and not using the degrees-minutes key on a decimal time answer. Graphs needed to start at $t = 0$ (time cannot be negative) and show the intercept and asymptote.
Induction set-out. Better responses showed the $n = 1$ substitution, stated the assumption and the target clearly, and showed exactly where the assumption was used. The feedback also warned against manipulating both sides at once and against algebra errors with fractions.
Inverse functions on a restricted domain. The feedback stressed that finding the inverse takes more than swapping $x$ and $y$ : students must choose the correct sign of the square root, and link the range of $f$ to the domain of $f^{-1}$ .
Volumes and areas. Areas to improve were identifying the inner and outer volumes, choosing correct limits, using symmetry, and handling regions below the $x$ -axis and absolute values.
Projectiles. Students needed to account for the launch height (so doubling the time taken to reach the top does not give the time of flight), pair the right time with the right component, and act on contradictory results.
Vectors and statistics. Students were advised to distinguish vectors from their magnitudes in proofs, use the link between parts of a question, and separate the variance from the standard deviation when finding a $z$ -score.
Derivative of an inverse. Students were reminded not to confuse the gradient of an inverse with the gradient of a normal, and to use the fact that if $g(a) = b$ then $g^{-1}(b) = a$ .

How to use this paper

Download the official 2021 paper from the link above and sit it under exam conditions: 10 minutes reading, then 2 hours working, with about 15 minutes on Section I.
Mark your work against NESA's 2021 marking guidelines (https://www.nsw.gov.au/sites/default/files/noindex/2025-05/2021-hsc-maths-ext-1-mg.pdf), noting which criterion each lost mark came from.
Read NESA's 2021 marking feedback (linked above) for every question you dropped marks on and compare your set-out with what better responses did.
Redo the practice questions on this page without looking, especially the induction proof and the differential equation, writing every step as a marker would expect.
Keep an error log of sign slips, missing constants, degree/radian mix-ups and variance versus standard deviation, and check it before your next timed 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)

Question 1 (3 marks): Use mathematical induction to prove that ∑r=1nr⋅2r=(n−1) 2n+1+2\displaystyle\sum_{r=1}^{n} r \cdot 2^r = (n-1)\,2^{n+1} + 2r=1∑n​r⋅2r=(n−1)2n+1+2 for all integers n≥1n \ge 1n≥1.

Step 1: Base case, n=1n = 1n=1

Step 2: Inductive hypothesis

Step 3: Prove it for n=k+1n = k + 1n=k+1

Step 4: Conclusion

Step 1: Components

Step 2: Maximum height

Step 3: Time to reach the tree

Step 4: Height at the tree

Question 3 (3 marks): Find the exact area of the region bounded by y=129+x2y = \dfrac{12}{9 + x^2}y=9+x212​, the xxx-axis and the lines x=−3x = -\sqrt{3}x=−3​ and x=3x = 3x=3.

Step 1: Integrate

Step 2: Evaluate in radians

Step 1: Separate variables and integrate

Step 2: Use the initial condition

Step 3: Find kkk

Step 4: Solve for T=40T = 40T=40

Step 1: Binomial probability for (a)

Step 2: Standard deviation for (b)

Step 3: zzz-score and empirical rule

Step 4: Sample size for (c)

Question 6 (3 marks): (a) A debating squad of 444 is chosen from 777 boys and 666 girls. How many squads contain at least 222 girls? (b) Show that in any group of 100100100 people, at least 999 were born in the same month.

Step 1: Split (a) into cases

Step 2: Check (a) with the complement

Step 3: Pigeonhole principle for (b)

Common errors students made

How to use this paper

Use this paper well

Keep going

Question 1 (3 marks): Use mathematical induction to prove that $\displaystyle\sum_{r=1}^{n} r \cdot 2^r = (n-1)\,2^{n+1} + 2$ for all integers $n \ge 1$ .

Step 1: Base case, $n = 1$

Step 3: Prove it for $n = k + 1$

Question 3 (3 marks): Find the exact area of the region bounded by $y = \dfrac{12}{9 + x^2}$ , the $x$ -axis and the lines $x = -\sqrt{3}$ and $x = 3$ .

Step 3: Find $k$

Step 4: Solve for $T = 40$

Step 3: $z$ -score and empirical rule

Question 6 (3 marks): (a) A debating squad of $4$ is chosen from $7$ boys and $6$ girls. How many squads contain at least $2$ girls? (b) Show that in any group of $100$ people, at least $9$ were born in the same month.