Band 6

Mechanics, EXT2 EQ-Bank 36

An experimental rocket is at a height of 5000 m, ascending at a speed of $50\sqrt{2}$ m s$^{-1}$ at an angle of 45° to the horizontal, when its engine stops. The rocket is then subject to gravity and to air resistance proportional to its velocity. Take $g$ = 10 m s$^{-2}$.

The velocity vector of the rocket, $t$ seconds after the engine stops, is

$\mathbf{v}(t) = 50e^{-0.2t}\,\mathbf{i} + (100e^{-0.2t}-50)\mathbf{j}.$ (Do NOT prove this.)

Show that the rocket reaches its greatest height when $t =5\ln 2$ seconds, and calculate its greatest height. (3 marks)

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The pilot can only operate the ejection seat while the rocket is descending at an angle between 45° and 60° to the horizontal. Find the earliest and latest times at which the pilot can eject. (3 marks)

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As the rocket continues to fall, its speed approaches a limiting value. Find this terminal speed, justifying your answer. (1 mark)

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a. $\text{See Worked Solutions}$

b. $\text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{Terminal speed}=50 \ \text{ms}^{-1}$

Show Worked Solution

a. $\mathbf{v}(t)=50 e^{-0.2 t}\,\mathbf{i}+\left(100 e^{-0.2 t}-50\right)\mathbf{j}$

$\text{Max height occurs when} \ \ \dot{y}=0:$

$100 e^{-0.2 t}-50$	$=0$
$e^{-0.2 t}$	$=\dfrac{1}{2}$
$-0.2 t$	$=-\ln 2$
$t$	$=5 \ln 2$

$\text{Find} \ y \ \text{when}\ \ t=5 \ln 2:$

$y(t)=\displaystyle \int 100 e^{-0.2 t}-50\, d t=-500 e^{-0.2 t}-50 t+c$

$\text{When} \ \ t=0, y=5000:$

$5000=-500 e^{\circ}+c \ \Rightarrow \ c=5500$

$y=5500-500 e^{-0.2 t}-50 t$

$\text{At} \ \ t=5\ln 2:$

$y=5500-500 e^{-\ln 2}-50 \times 5 \ln 2=5076.71 \ldots=5077 \ \text{m}$.

b. $\text {On descent,} \ \ \dot{y}<0.$

$\text{Let} \ \ \theta=\text{angle below the horizontal}$

$\tan \theta=\dfrac{\abs{\dot{y}}}{\dot{x}}$

$\text{Let}\ \ a=e^{-0.2 t}$

$\tan \theta=\dfrac{50-100 a}{50 a}=\dfrac{1}{a}-2 \ \Rightarrow \ \theta=\tan ^{-1}\left(\dfrac{1}{a}-2\right)$

$\text{When} \ 45^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=1 \ \Rightarrow \ a=3$

$e^{-0.2 t}=\dfrac{1}{3} \ \Rightarrow \ t=\dfrac{\ln 3}{0.2} \approx 5.5 \ \text{s (1 d.p.)}$

$\text{When} \ 60^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=\sqrt{3} \ \Rightarrow \ a=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$e^{-0.2 t}$	$=2-\sqrt{3}$
$-0.2 t$	$=\ln (2-\sqrt{3})$
$t$	$=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s (1 d.p.)}$

$\therefore \ \text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{As} \ \ t \rightarrow \infty:$

$e^{-0.2 t} \rightarrow 0\ \ \Rightarrow\ \ \dot{x} \rightarrow 0, \ \ \dot{y} \rightarrow -50$

$\therefore \ \text{Terminal speed}=\sqrt{0^2+50^2}=50 \ \text{ms}^{-1}$

Mechanics, EXT2 M1 2022 HSC 8 MC*

As a projectile of mass $m$ kilograms travels through air, it experiences a frictional force. The magnitude of this force is proportional to the speed $v$ of the projectile. The constant of proportionality is the positive number $k$. The position of the particle at time $t$ is denoted by $\displaystyle \binom{x}{y}$. The acceleration due to gravity is $g \ \text{m s}^{-2}$.

Based on Newton's laws of motion, which equation models the motion of this projectile?

$\displaystyle\binom{0}{-m g}+k\binom{\dot{x}}{\dot{y}}=m\binom{\ddot{x}}{\ddot{y}}$
$\displaystyle\binom{0}{-m g}-k\binom{\dot{x}}{\dot{y}}=m\binom{\ddot{x}}{\ddot{y}}$
$\displaystyle\binom{0}{-m g}+k v\binom{\dot{x}}{\dot{y}}=m\binom{\ddot{x}}{\ddot{y}}$
$\displaystyle\binom{0}{-m g}-k v\binom{\dot{x}}{\dot{y}}=m\binom{\ddot{x}}{\ddot{y}}$

Show Answers Only

$B$

Show Worked Solution

$\text{Friction} \ (F) \ \text{works against velocity}$

$\Rightarrow \ \text{the resistance term is subtracted (eliminate A and C)}$

$\text{Force vector has magnitude and direction.}$

$\text{Magnitude:} \ \ \abs{F} \propto v \ \Rightarrow \ \abs{F}=kv\ \ \ (k>0)$

$\text{Direction is in line with unit vector:} \ \ \dfrac{(\dot{x}, \dot{y})}{\abs{(\dot{x}, \dot{y})}}=\dfrac{(\dot{x}, \dot{y})}{v}$

$\text{Velocity vector} \ (\dot{x}, \dot{y}) \ \text{has magnitude equal to the speed:}$

$\abs{(\dot{x}, \dot{y})}=\sqrt{(\dot{x})^2+(\dot{y})^2}=v$

$\abs{-k(\dot{x}, \dot{y})}=k\abs{\dot{x}, \dot{y}}=k v$

$\text{Magnitude of resistance} \propto \text{speed}\ \ \Rightarrow\ \ -k \displaystyle\binom{\dot{x}}{\dot{y}}\ \text{(no extra}\ v \ \text{needed)}$

$\Rightarrow B$

Proof, EXT2 EQ-Bank 34

Consider a sequence of rectangles with side lengths $a_{ n }$ and $b_{ n }$.

The first rectangle has $a_1=2$ and $b_1=1$.

For integers $n \geq 1,\ \ a_{n+1}=\dfrac{a_n+b_n}{2}$ and $b_{n+1}=\dfrac{2}{a_{n+1}}.$

Show that every rectangle in the sequence has an area of 2 square units. (1 mark)

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Use the relationship between the arithmetic mean and the geometric mean to prove that $a_n \geq \sqrt{2}$ for any integer $n \geq 1$. (2 marks)

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Use mathematical induction to prove that $a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2})$ for any integer $n \geq 1$. (4 marks)

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Use the squeeze theorem to show that the rectangles approach a square as $n$ approaches infinity. (2 marks)

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Show Worked Solution

a. $\text{Since}\ \ a_1 b_1=2\ \ \text{and}\ \ a_{n+1} b_{n+1}=a_{n+1} \times \dfrac{2}{a_{n+1}}=2$

$\Rightarrow\ \text{Each rectangle has area 2.}$

b. $a_1=2\ \ \Rightarrow\ \ a_1 \geq \sqrt{2}\ \ \text{(true for 1st rectangle)}$

$\text{AM/GM inequality:}\ \ \dfrac{a_n+b_n}{2} \geq \sqrt{a_nb_n} $

$a_nb_n=2\ \ \text{(from part (a))}$

$\dfrac{a_n+b_n}{2} \geq \sqrt{2}\ …\ (1)$

$\text{Since}\ \ a_{n+1}=\dfrac{a_n+b_n}{2}:$

$\ a_{n+1} \geq \sqrt{2}\ \ \ \text{(using (1) above)}$

$\therefore a_n \geq \sqrt{2}$

c. $\text{Prove}\ \ a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}),\ \ \text{for}\ \ n \geq 1$

$\text{If}\ \ n=1:$

$\ a_1-\sqrt{2}=2-\sqrt{2} \leq \dfrac{1}{2^{0}}(2-\sqrt{2})$.

$\Rightarrow\ \text{True for}\ \ n=1.$

$\text{Assume true for}\ \ n=k:$

$a_k-\sqrt{2} \leq \dfrac{1}{2^{k-1}}(2-\sqrt{2})\ …\ (1) $

$\text{Prove true for}\ \ n=k+1:$

$\text{i.e.}\ \ a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\text{By definition,} \ \ a_{k+1}=\dfrac{1}{2}\left(a_k+b_k\right)$

$a_{k+1}-\sqrt{2}$	$=\dfrac{1}{2}\left(a_k-\sqrt{2}\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) $
	$\leq \dfrac{1}{2}\left(\dfrac{1}{2^{k-1}}(2-\sqrt{2})\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right)\ \ \text{(see (1) above)}\ $

$\text{Since}\ \ a_k \geq \sqrt{2}\ \ \text{and}\ \ a_kb_k=2:$

$\ b_k \leq \sqrt{2}\ \ \text{and}\ \ b_k-\sqrt{2} \leq 0$

$a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\Rightarrow\ \text{True for}\ \ n=k+1.$

$\therefore\ \text{Since true for}\ \ n=1,\ \text{by PMI, true for integers}\ \ n \geq 1.$

d. $\text {Combining parts (b) and (c):}$

$0 \leq a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}).$

$\text{As}\ \ n \rightarrow \infty, \ \dfrac{1}{2^{n-1}} \rightarrow 0$

$\Rightarrow a_n-\sqrt{2} \rightarrow 0\ \ \text{(by squeeze theorem)}$

$\text{Since rectangles have an area = 2:}$

$\text{As}\ \ n \rightarrow \infty,\ b_n \rightarrow \sqrt{2}$

$\text{i.e. rectangles approach a square.}$

Financial Maths, STD2 F1 EQ-Bank 42

Sienna buys a new car and pays a total of $2,200 in stamp duty.

Stamp duty is calculated on the vehicle as follows:

- 3% of market value up to $\$45\,000$
- 5% of market value over $\$45\,000$

Determine the market value of Sienna's car. (2 marks)

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$\$62\,000$

Show Worked Solution

$\text{Stamp duty on first}\ \$45\,000=0.03\times 45\,000=\$1350$

$\text{Remaining stamp duty}=2200-1350=\$850$

$\text{Amount over}\ \$45\,000=\dfrac{850}{0.05}=\$17\,000$

$\therefore\ \text{Market value}=45\,000+17\,000=\$62\,000$

Financial Maths, STD1 F1 EQ-Bank 17 MC

The total cost of a tradesperson's invoice, including 10% GST, is $462.

What was the cost of the job before GST was added?

$415.80
$420.00
$462.00
$508.20

Show Answers Only

$B$

Show Worked Solution

$\text{GST-inclusive price is 110\% of original cost}$

$\text{Original cost}=\dfrac{462}{1.1}=\$420.00$

$\Rightarrow B$

Financial Maths, STD1 F1 EQ-Bank 32

Zoe purchased a new ute for $36 000. The straight-line depreciation model used by her accountant assumes the ute decreases in value by $4800 each year.

Use the straight-line depreciation formula $S = V_0-Dn$ to find the salvage value of the ute after 5 years. (1 mark)

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After how many full years will the model predict the ute is worth less than $10 000? (2 marks)

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Zoe's accountant suggests that this straight-line model may not be appropriate for predicting the ute's value beyond 7 years. Give ONE reason why the model may not be suitable for long-term predictions. (1 mark)

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a. $\$12\,000$

b. $\text{After 6 full years}$

c. $\text{Model limitations}$

$\text{Consider the expected value of the ute at}\ \ t=8:$

$S=36\,000-4800 \times 8=-\$2400$

$\text{The model predicts negative values in the long term which is unrealistic.}$

Show Worked Solution

a. $\text{Find salvage value:}$

$S$	$= V_0-Dn$
	$= 36\,000-4800 \times 5$
	$= \$12\,000$

b. $\text{Find}\ n\ \text{when}\ \ S<10\,000:$

$S$	$< 10\,000$
$36\,000-4800n$	$< 10\,000$
$26\,000$	$< 4800n$
$n$	$> \dfrac{26\,000}{4800}= 5.4166…$

$\therefore\ n=6\ \text{full years}$

c. $\text{Model limitations}$

$\text{Consider the expected value of the ute at}\ \ t=8:$

$S=36\,000-4800 \times 8=-\$2400$

$\text{The model predicts negative values in the long term which is unrealistic.}$

Financial Maths, STD1 F1 2021 HSC 19 v1

Maya purchased a motorbike for $12 000. The value of the motorbike decreases according to a linear model. The graph shows the value of the motorbike, $$V$, against the time, $t$ months, since it was purchased.

By how much does the value of the motorbike decrease every 10 months? (1 mark)

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Find the value of the motorbike after 4 years. (1 mark)

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Identify ONE problem with using this model to determine the value of Maya's motorbike over time. (1 mark)

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a. $\$1000$

b. $\$7200$

c. $\text{Motorbike will have a negative value after 120 months.}$

Show Worked Solution

a. $\text{Total decrease over 100 months}= 12\,000-2000= \$10\,000$

$\text{Decrease per 10 months}= \dfrac{10\,000}{10}= \$1000$

b. $\text{4 years} = 4 \times 12 = 48\ \text{months}$

$\text{Depreciation rate}= \$100\ \text{per month}$

$V = 12\,000-(100 \times 48)=\$7200$

c. $\text{Model limitations:}$

$\text{The linear model will eventually predict a value of \$0 (at 120 months) and negative}$

$\text{values beyond that, which is unrealistic.}$

Algebra, STD1 EQ-Bank 36

Two cyclists, Aiko and Ben, are riding along the same straight track in the same direction.

Aiko starts 2000 m ahead of Ben. Aiko rides at a constant speed of 250 metres/minute and Ben rides at a constant speed of 500 metres/minute.

Let $d$ = distance from the starting point in metres, and

$t$ = time in minutes.

The equation $d=2000+250t$ models Aiko's distance from the starting point.
Write an equation to model Ben's distance. (1 mark)

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Use the equations from (a) to graph Aiko and Ben's journeys on the grid below. (2 marks)

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After how many minutes does Ben catch Aiko? (1 mark)

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How far has each cyclist travelled from their own starting point when they meet? (2 marks)

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a. $d=500t$

c. $8\ \text{minutes}$

d. $\text{Aiko travelled 2000 m, Ben travelled 4000 m.}$

Show Worked Solution

a. $d=500t$

b. $\text{Table of values:}$

$\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 \\ \hline \text{Aiko} & 2000 & 2500 & 3000 & 3500 & 4000 & 4500 \\ \hline \text{Ben} & 0 & 1000 & 2000 & 3000 & 4000 & 5000 \\ \hline \end{array}$

c. $\text{From the graph, the lines intersect at }\ t=8.$

$\therefore\ \text{Ben catches Aiko after 8 minutes}$

d. $\text{When}\ \ t=8, d=4000:$

$\text{Aiko started 2000 m ahead, so the distance travelled}$

$=4000-2000=2000\ \text{m}$

$\text{Ben started at the origin, so the distance travelled =4000 m}$

Algebra, STD1 EQ-Bank 34

Two water tanks sit side by side on a farm.

Tank A has a capacity of 1000 litres and is full. It is being emptied at a constant rate of 60 litres per minute.

At the same time, Tank B is empty and is being filled at a constant rate of 40 litres per minute.

Let $V$ = volume of water in litres, and

$t$ = time in minutes.

The equation $V=1000-60t$ models the volume of water in Tank A.
Write an equation to model the volume of water in Tank B. (1 mark)

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Complete the table below and use the values to graph the volume of water in Tank A and Tank B on the grid below. (3 marks)

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline \text{Tank A} & \ \ \ \ \ \ \ \ & 880 & 760 & 640 & 520 & 400 & 280 \\ \hline \text{Tank B} & 0 & \ \ \ \ \ \ \ \ \ & 160 & 240 & 320 & 400 & \ \ \ \ \ \ \ \ \\ \hline \end{array}$

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After how many minutes do both tanks contain the same volume of water? (1 mark)

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What is the volume of water in each tank at this time? (1 mark)

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a. $V=40t$

b. $\text{Table of values:}$

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline \text{Tank A} & 1000 & 880 & 760 & 640 & 520 & 400 & 280 \\ \hline \text{Tank B} & 0 & 80 & 160 & 240 & 320 & 400 & 480 \\ \hline \end{array}$

c. $10\ \text{minutes}$

d. $400\ \text{litres}$

Show Worked Solution

a. $V=40t$

b. $\text{Table of values:}$

c. $\text{From the graph, the lines intersect at }\ t=10.$

$\therefore\ \text{Both tanks contain the same volume after 10 minutes}$

d. $\text{Method 1: Graphically}$

$\text{From the graph, when }\ t=10,\ \text{the volume in both tanks = 400 litres}$

$\text{Method 2: Algebraically}$

$\text{When }\ t=10:$

$\text{Tank A volume:}\ \ V=1000-60\times10=400\ \text{L}$

$\text{Tank B volume:}\ \ V=40\times10=400\ \text{L}\ \checkmark$

$\therefore\ \text{Each tank contains } 400\ \text{litres}$

ENGINEERING, AE 2025 HSC 27c

The orthographic views of a security camera cover, manufactured from sheet metal, are shown. Complete a half pattern development of the camera cover. (6 marks)

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Show Worked Solution

♦♦♦ Mean mark 34%.

ENGINEERING, AE 2025 HSC 27b

Describe the process of artificial age hardening of aluminium alloys. (3 marks)

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The aluminium alloy is first heated to a high temperature to create a homogeneous solid solution, then rapidly quenched to trap alloying elements in the lattice.
The alloy is then reheated to a moderate temperature for a controlled period.
This allows fine precipitate particles to form within the lattice, significantly increasing the strength and hardness of the alloy.

Show Worked Solution

The aluminium alloy is first heated to a high temperature to create a homogeneous solid solution, then rapidly quenched to trap alloying elements in the lattice.
The alloy is then reheated to a moderate temperature for a controlled period.
This allows fine precipitate particles to form within the lattice, significantly increasing the strength and hardness of the alloy.

♦♦♦ Mean mark 29%.

Calculus, MET2 2025 VCAA 4

Consider the function $f:\left[0, \dfrac{5 \pi}{2}\right] \rightarrow R, f(x)=\sin (x)+1$.

The graph of $y=f(x)$ is shown below.

Evaluate $f\left(\dfrac{2 \pi}{3}\right)$. (1 mark)

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Find the exact values of $x$ for which $f(x)=\dfrac{3}{2}$. (1 mark)

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There exist real numbers $a$ and $k$ in the interval $\left(0, \dfrac{5 \pi}{2}\right)$, such that $f(x+k)=f(x)$ for all $x \in[0, a]$.
Find the value of $k$ and the largest possible value of $a$. (2 marks)

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Consider the tangent to the graph of $y=f(x)$ at the point $A$ where $x=\dfrac{2 \pi}{3}$, as shown on the axes below.

Find the equation of the tangent to the graph of $y=f(x)$ at the point where $x=\dfrac{2 \pi}{3}$. (1 mark)

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Apply two iterations of Newton's method to $f$ with $x_0=\dfrac{2 \pi}{3}$.
1. Write down $x_2$, correct to one decimal place. (1 mark)
  
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2. On the axes in part d, draw the tangent to the graph of $y=f(x)$ at the point where $x=x_1$.
3. Answer on the graph in part d. (1 mark)
  
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Now consider the line $y=t(x)$, which is the tangent to the graph of $y=f(x)$ at the point $(p, f(p))$, where $p \in\left(0, \dfrac{5 \pi}{2}\right)$.

1. 1. Show that $t(x)=\cos (p)(x-p)+\sin (p)+1$. (2 marks)
    
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  2. Determine the minimum and maximum possible values for the $y$-intercept of $y=t(x)$, for $p \in\left(0, \dfrac{5 \pi}{2}\right)$. (2 marks)
    
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  3. Determine the values of $p$ for which $y=t(x)$ has a unique $x$-intercept that is equal to the $x$-intercept of $y=f(x)$.
  4. Give your answers correct to two decimal places. (2 marks)
    
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Let $g:\left[0, \dfrac{5 \pi}{2}\right] \rightarrow R, g(x)=a x^3+b x^2+c x+d$ be a polynomial function, where $a, b, c, d \in R$.
Suppose $g(0)=f(0)$ and $g^{\prime}(0)=f^{\prime}(0)$.
1. Show that $c=1$ and $d=1$. (2 marks)
  
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2. If $g(2 \pi)=f(2 \pi)$ and $g^{\prime}(2 \pi)=f^{\prime}(2 \pi)$, determine the area bounded by the graphs of $y=f(x)$ and $y=g(x)$, for $x \in[0,2 \pi]$.
3. Give your answer correct to two decimal places. (2 marks)
  
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4. Let $a=0, c=1, d=1$.
5. Find $b$ and $r$, such that $g(r)=f(r)$ and $g^{\prime}(r)=f^{\prime}(r)$, where $b \in R$ and $r \in\left(0, \dfrac{5 \pi}{2}\right)$. (2 marks)
  
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a. $f\left(\dfrac{2 \pi}{3}\right)=\dfrac{2+\sqrt{3}}{2}$

b. $x=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}, \dfrac{13 \pi}{6}$

c. $k=2 \pi, \ \ a=\dfrac{\pi}{2}$

d. $y=-\dfrac{x}{2}+\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2}+1$

e.i. $x_2=5.2$

e.ii.

f.i. $f^{\prime}(p)=\cos (p)$

$\text{Equation of tangent at}\ (p, \sin (p)+1):$

$y-(\sin (p)+1)$	$=\cos (p)(x-p)$
$t(x)$	$=\cos (p)(x-p)+\sin (p)+1$

f.ii. $y \text{-int (min)}=1-2 \pi \ \text { (at } p=2 \pi)$

$y \text{-int (max)}=1+\pi \ \text { (at } p= \pi)$

f.iii. $p=2.38 \ \text{or} \ 7.04$

g.i. $g(x)=a x^3+b x^2+c x+d, \ f(x)=\sin (x)+1$

$\text{Given} \ \ f(0)=g(0):$

$d=\sin (0)+1=1$

$g^{\prime}(x)=3 a x^2+2 b x+c, \ f^{\prime}(x)=\cos (x)$

$\text{Given} \ \ f^{\prime}(0)=g^{\prime}(0):$

$c=1$

g.ii. $\text {Area}=1.53$

g.iii. $r=\pi, \ b=-\dfrac{1}{\pi}$

Show Worked Solution

a. $f(x)=\sin (x)+1$

$f\left(\dfrac{2 \pi}{3}\right)=\sin \left(\dfrac{2 \pi}{3}\right)+1=\dfrac{2+\sqrt{3}}{2}$

b. $\text{Solve \(\ f(x)=\dfrac{3}{2} \ $ for $x$ (by CAS):}\)

$\sin (x)+1=\dfrac{3}{2} \ \Rightarrow \ \sin (x)=\dfrac{1}{2}$

$\text{Solve} \ \ \sin (x)=\dfrac{1}{2} \ \text { for } \ x \in\left[0, \dfrac{5 \pi}{2}\right]:$

$x=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}, \dfrac{13 \pi}{6}$

c. $f(x+k)=f(x) \ \ \text{for} \ \ x \in[0, a]$

$\text{By inspection of graph:}$

$k=2 \pi, \ \ a=\dfrac{\pi}{2}$

♦♦ Mean mark (c) 28%.

d. $f(x)=\sin (x)+1 \ \Rightarrow \ f^{\prime}(x)=\cos (x)$

$f^{\prime}\left(\dfrac{2 \pi}{3}\right)=-\dfrac{1}{2}$

$\text{Find equation of line} \ \ m_2=-\dfrac{1}{2} \ \ \text{through}\ \ \left(\dfrac{2 \pi}{3}, \dfrac{2+\sqrt{3}}{2}\right):$

$y=-\dfrac{x}{2}+\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2}+1$

e.i. $x_0=\dfrac{2 \pi}{3}$

$x_1=\dfrac{2 \pi}{3}-\dfrac{f\left(\dfrac{2 \pi}{3}\right)}{f^{\prime}\left(\dfrac{2 \pi}{3}\right)}=5.8264 \ldots$

$x_2=5.8264 \ldots-\dfrac{f(5.8264)}{f^{\prime}(5.8264)}=5.2 \ \text{(1 d.p.)}$

e.ii.

♦♦ Mean mark (e.ii) 28%.

f.i. $f^{\prime}(p)=\cos (p)$

$\text{Equation of tangent at}\ (p, \sin (p)+1):$

$y-(\sin (p)+1)$	$=\cos (p)(x-p)$
$t(x)$	$=\cos (p)(x-p)+\sin (p)+1$

f.ii. $t(x)=\cos (p) x+\sin (p)+1-p \times \cos (p)$

$y\text{-intercept}=\sin (p)+1-p \times \cos (p)$

$\text{Find max/min of} \ y\text{-int for} \ p \in\left[0, \dfrac{5 \pi}{2}\right] \ \ \text{(by CAS):}$

$y \text{-int (min)}=1-2 \pi \ \text { (at } p=2 \pi)$

$y \text{-int (max)}=1+\pi \ \text { (at } p= \pi)$

♦♦ Mean mark (f.ii) 26%.
♦♦♦ Mean mark (f.iii) 23%.

f.iii. $x \text{-intercept of} \ f(x) \ \text{occurs at} \ \ x=\dfrac{3 \pi}{2}$

$\text{Solve} \ \ t\left(\dfrac{3 \pi}{2}\right)=0 \ \ \text {for}\ p:$

$p=2.38 \ \text{or} \ 7.04$

g.i. $g(x)=a x^3+b x^2+c x+d, \ f(x)=\sin (x)+1$

$\text{Given} \ \ f(0)=g(0):$

$d=\sin (0)+1=1$

$g^{\prime}(x)=3 a x^2+2 b x+c, \ f^{\prime}(x)=\cos (x)$

$\text{Given} \ \ f^{\prime}(0)=g^{\prime}(0):$

$c=1$

g.ii. $g(x)=a x^3+b x^2+x+1 \ \Rightarrow \ g^{\prime}(x)=3 a x^2+2 b x+1$

$\text{Given \(\ g(2 \pi)=f(2 \pi)=1\ $ and $\ \ g^{\prime}(2 \pi)=f^{\prime}(2 \pi)=1$}\)

$\text{Solve \(\ g(2 \pi)=1 \ $ and $\ g^{\prime}(2 \pi)=1$ simultaneously for $a, b$:}\)

$a=\dfrac{1}{2 \pi^2}, \ b=-\dfrac{3}{2 \pi}$

♦♦ Mean mark (g.ii) 33%.
♦♦♦ Mean mark (g.iii) 24%.

$\text{Find intersection of}\ f(x)\ \text{and}\ g(x)\ \text{(by CAS)}:$

$f(x)=g(x)\ \ \Rightarrow\ \ x=\pi$

$\text {Area}=\displaystyle \int_0^\pi f(x)-g(x)\, d x+\int_\pi^{2 \pi} g(x)-f(x)\, d x=1.53$

g.iii. $a=0, c=1, d=1$

$g(x)=b x^2+x+1, \ f(x)=\sin (x)+1$

$g^{\prime}(x)=2 b x+1, \ f^{\prime}(x)=\cos (x)$

$\text{Solve simultaneous equations for \(b$ and $r$:}\)

$br^2+r=\sin (r)\ \ldots\ (1)$

$2 b r+1=\cos (r)\ \ldots\ (2)$

$r=\pi, \ b=-\dfrac{1}{\pi}$

ENGINEERING, TE 2025 HSC 15 MC

A data signal and a carrier wave are combined to form the modulated signal shown.

What type of modulation is depicted?

Amplitude modulation
Frequency modulation
Phase shift key modulation
Frequency shift key modulation

Show Answers Only

$D$

Show Worked Solution

The data signal is digital (binary 1s and 0s) — this immediately eliminates analogue AM and FM.
In frequency shift key (FSK) modulation, the carrier frequency switches between two frequencies corresponding to binary 1 and 0.
The modulated signal shows higher frequency during the data signal high (1) and lower frequency during the low (0) — the defining characteristic of FSK.
Phase shift key modulation changes the phase, not the frequency — the waveform shown has no phase shifts, only frequency changes.

$\Rightarrow D$

♦♦♦ Mean mark 28%.

ENGINEERING, PPT 2025 HSC 12 MC

An image of a screw jack is shown. The tommy bar is used to operate the screw jack

The screw jack has a pitch of 8 mm, and the force on the tommy bar is applied 400 mm from the axis of the screw.

What is the velocity ratio (VR) of the screw jack?

Show Answers Only

$C$

Show Worked Solution

$VR$	$= \dfrac{2\pi L}{p}$
	$= \dfrac{2\pi \times 400}{8}$
	$= \dfrac{2513.3}{8}$
	$= 314$

$\Rightarrow C$

♦♦♦ Mean mark 24%.

Calculus, 2ADV C3 2025 MET2 16 MC

Consider the function $h(x)=a\, \log _e(b x)$, where $a, b \in R$.

Given that its derivative $h^{\prime}(x)$ has range $(0, \infty)$, which of the following must be true?

$ a>0$ only
$ a>0$ and $b<0$
$a>0$ and $b>0$
$a b>0$

Show Answers Only

$D$

Show Worked Solution

$h(x)=a\, \log _e(b x)$

$h^{\prime}(x)=a\, \times \dfrac{b}{bx}=\dfrac{a}{x}$

$\text{Consider} \ \ h(x)=a\, \log _e(b x) \ \Rightarrow \ b x>0$

$\text{Case 1:} \ \ b>0 \ \Rightarrow \ x>0 \ (\text{since} \ \ b x>0 )$

$\dfrac{a}{x} \ \ \text{is only positive when}\ \ a>0$

$\text {Case 2:} \ \ b<0 \ \Rightarrow \ x<0 \ \ (\text{since} \ \ b x>0)$

$\dfrac{a}{x} \ \ \text{is only positive when} \ \ a<0$

$\text{In both cases,} \ ab>0$

$\Rightarrow D$

Calculus, MET2 2025 VCAA 19 MC

Let $A$ be a point on the line $y=x+c$ and $B$ be a point on the curve $y=\log _e(x-1)$.

If $A$ and $B$ are placed such that the line segment $A B$ has the minimum possible length, and this length is $\sqrt{2}$, the value of $c$ must be

$\sqrt{2}-2$
$\sqrt{2}$
$1$
$0$

Show Answers Only

$D$

Show Worked Solution

$f(x)=x+c, \ g(x)=\log _e(x-1)$

$\text {Shortest distance between line and curve is perpendicular distance.}$

$\text{Consider the example when}\ \ c=0:$

$\text{At point \(B$, gradient $=1$}\)

$\text{Solve} \ \ g^{\prime}(x)=1:$

$\dfrac{1}{x-1}=1 \ \Rightarrow \ x=2$

♦♦♦ Mean mark 14%.

$\text{Find \(\perp$ distance $(d)$ between $(2,0)$ and $(x, y)$ which lies on $f(x)$:}\)

$d=\sqrt{(x-2)^2+(f(x)-0)^2}=\sqrt{(x-2)^2+(x+c)^2}$

$\text{Minimise} \ d\ \text{(by CAS)}\ \Rightarrow \ x=\dfrac{2-c}{2}$

$d\left(\dfrac{2-c}{2}\right)=\sqrt{2} \ \Rightarrow \ c=0\ \ (c \neq-4)$

$\Rightarrow D$

Calculus, MET2 2025 VCAA 16 MC

Consider the function $h(x)=a\, \log _e(b x)$, where $a, b \in R \backslash\{0\}$.

Given that its derivative $h^{\prime}(x)$ has range $(0, \infty)$, which of the following must be true?

$ a>0$ only
$ a>0$ and $b<0$
$a>0$ and $b>0$
$a b>0$

Show Answers Only

$D$

Show Worked Solution

$h(x)=a\, \log _e(b x)$

$h^{\prime}(x)=a\, \times \dfrac{b}{bx}=\dfrac{a}{x}$

$\text{Consider} \ \ h(x)=a\, \log _e(b x) \ \Rightarrow \ b x>0$

♦♦♦ Mean mark 18%.

$\text{Case 1:} \ \ b>0 \ \Rightarrow \ x>0 \ (\text{since} \ \ b x>0 )$

$\dfrac{a}{x} \ \ \text{is only positive when}\ \ a>0$

$\text {Case 2:} \ \ b<0 \ \Rightarrow \ x<0 \ \ (\text{since} \ \ b x>0)$

$\dfrac{a}{x} \ \ \text{is only positive when} \ \ a<0$

$\text{In both cases,} \ ab>0$

$\Rightarrow D$

ENGINEERING, PPT 2025 HSC 14 MC

An image of a shaft and its end view are shown.

Which drawing correctly represents section $\text{A–A}$?

Show Answers Only

$C$

Show Worked Solution

Per AS 1100, splines and shafts are not hatched in section — only the solid flange hub receives section hatching.
The correct section must show the central bore as an unhatched void and the spline teeth as unhatched solid features.
Option $\text{C}$ correctly applies hatching only to the solid flange material, with splines and bore shown without hatching.
Options $\text{A, B}$ and $\text{D}$ incorrectly apply hatching across spline teeth or omit the correct representation of the bore.

$\Rightarrow C$

♦♦♦ Mean mark 25%.

Calculus, SPEC2 2025 VCAA 3

A tank initially contains 5 kg of salt dissolved in 3000 litres of water. Salty water that contains 0.1 kg of salt per litre of water enters the tank at a rate of 20 litres per minute. The solution is kept thoroughly mixed and drains from the tank via a tap at the same rate of 20 litres per minute.

By considering concentration, explain whether the quantity of salt in the tank increases with time. (1 mark)

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Let $Q$ denote the quantity of salt, in kilograms, in the tank at time $t$ minutes.
Show that $Q$ satisfies the differential equation $\dfrac{d Q}{d t}=\dfrac{300-Q}{150}$. (1 mark)

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Using Euler's method with a step size of 15 minutes, find $Q(30)$, the approximate quantity of salt in the tank after 30 minutes.
Give your answer in kilograms, correct to two decimal places. (2 marks)

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Use calculus to solve the differential equation $\dfrac{d Q}{d t}=\dfrac{300-Q}{150}$, expressing $Q$ in terms of $t$. (3 marks)

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What value does the quantity of salt in the tank approach as time approaches infinity?
Give your answer in kilograms. (1 mark)

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Find the time taken for the quantity of salt in the tank to reach 100 kg. (1 mark)

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When the quantity of salt in the tank reaches 100 kg , the tap draining the tank is turned off. Assume that the tank does not overflow and there is no change to the inflow rate.
After the tap is turned off, how many minutes does it take for the concentration of salt in the tank to reach $\dfrac{1}{20} \ \text{kg L}^{-1}$? (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Initial concentration}=\dfrac{5}{3000}=0.001\dot{6} \ \text{kg/L}$

$\text{Concentration entering tank}=0.1 \ \text{kg/L}$

$\text{Since} \ \ 0.1>0.001\dot{6}, \ \text{salt concentration increases with time.}$

b. $Q=\text{salt in tank at time}\ t$

$\dfrac{d Q}{d t}=\dfrac{d Q}{d t_{\text {in }}}-\dfrac{d Q}{d t_{\text {out }}}$

$\dfrac{d Q}{d t}=0.1 \times 20-\dfrac{Q}{3000} \times 20=2-\dfrac{Q}{150}=\dfrac{300-Q}{150}$

c. $61.05 \ \text{kg}$

d. $Q=300-295 e^{-\tfrac{t}{150}}$

e. $\text{As} \ \ t \rightarrow \infty, Q \rightarrow 300$

f. $t=150\, \log _e\left(\dfrac{59}{40}\right) \ \text {minutes }$

g. $t=50 \ \text{minutes}$

Show Worked Solution

a. $\text{Initial concentration}=\dfrac{5}{3000}=0.001\dot{6} \ \text{kg/L}$

$\text{Concentration entering tank}=0.1 \ \text{kg/L}$

$\text{Since} \ \ 0.1>0.001\dot{6}, \ \text{salt concentration increases with time.}$

♦♦♦ Mean mark (a) 20%.

b. $Q=\text{salt in tank at time}\ t$

$\dfrac{d Q}{d t}=\dfrac{d Q}{d t_{\text {in }}}-\dfrac{d Q}{d t_{\text {out }}}$

$\dfrac{d Q}{d t}=0.1 \times 20-\dfrac{Q}{3000} \times 20=2-\dfrac{Q}{150}=\dfrac{300-Q}{150}$

c. $Q_1=5+15 \times \dfrac{300-5}{150}=34.5 \ \text{kg}$

$Q_2=34.5+15 \times \dfrac{300-34.5}{150}=61.05 \ \text{kg}$

♦ Mean mark (c) 46%.

d.	$\dfrac{d Q}{d t}$	$=\dfrac{300-Q}{150}$
	$\dfrac{d t}{d Q}$	$=\dfrac{150}{300-Q}$
	$\displaystyle \int d t$	$=\displaystyle \int \frac{150}{300-Q} d Q$
	$ t$	$=-150\, \log _e(300-Q)+c$

$\text{When} \ \ t=0, Q=5:$

$0=-150\, \log _e 295+c \ \ \Rightarrow \ \ c=150\, \log _e 295$

$ t$	$=150\, \log _e 295-150\, \log _e(300-Q)$
$ t$	$=150\, \log _e\left(\dfrac{295}{300-Q}\right)$

$\text{Solve for} \ Q \ \text{ by (CAS):}$

$Q=300-295 e^{-\tfrac{t}{150}}$

e. $\text{As} \ \ t \rightarrow \infty, Q \rightarrow 300$

f. $\text{Find \(t$ when $Q=100$ (using part d):}\)

$t=150\, \log _e\left(\dfrac{295}{300-100}\right)=150\, \log _e\left(\dfrac{59}{40}\right) \ \text {minutes }$

g. $\text{After the tap is turned off:}$

$Q=100+0.1 \times 20 t=100+2 t$

$\text{Volume in tank}=3000+20 t$

$\text{Solve for \(t$:}\)

$\dfrac{1}{20}=\dfrac{100+2 t}{3000+20 t}$

$t=50 \ \text{minutes}$

♦♦♦ Mean mark (g) 18%.

Vectors, SPEC1 2025 VCAA 5

The position vectors of particles $P$ and $Q$ at time $t$ seconds are given by

\begin{align*}
{\underset{\sim}{r}}_P(t)=\left(t^3+a t^2\right) \underset{\sim}{i}-\underset{\sim}{j} \ \ \text {and} \ \ {\underset{\sim}{r}}_Q(t)=(b t+2 t) \underset{\sim}{i}+\left(2 t^2+c t+t\right) \underset{\sim}{j},
\end{align*}

where $t \geq 0$ and $a, b, c \in R$.

The particles collide when $t=1$.

Show that for collision to occur when $t=1$, the value of $c$ is $-4$. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---

When the particles collide, their velocities are at right angles to each other.

Find the two possible values of $a$ for collision to occur when $t=1$. (2 marks)

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When the particles collide at $t=1$, the magnitudes of their accelerations are equal.
Find the values of $a$ and $b$ for collision to occur when $t=1$. (1 mark)

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Show Answers Only

a. $\text{\(P$ and $Q$ collide at $\ t=1$.}\)

$\text{Equate co-efficients of} \ \underset{\sim}{j}:$

$2+c+1=-1 \ \ \Rightarrow \ \ c=-4$

b. $a=-\dfrac{3}{2},-1$

c. $a=-1, \ b=-2$

Show Worked Solution

a. $\text{\(P$ and $Q$ collide at $\ t=1$.}\)

$\text{Equate co-efficients of} \ \underset{\sim}{j}:$

$2+c+1=-1 \ \ \Rightarrow \ \ c=-4$

b. $r_P=\left(t^3+a t^2\right)\underset{\sim}{i}-\underset{\sim}{j} \ \ \Rightarrow \ \ \dot{r}_P=\left(3 t^2+2 a t\right) \underset{\sim}{i}$

$r_Q=(b t+2 t) \underset{\sim}{i}+\left(2 t^2-3 t\right)\underset{\sim}{j} \ \ \Rightarrow \ \ \dot{r}_Q=(b+2)\underset{\sim}{i}+(4 t-3)\underset{\sim}{j}$

$\text {Velocities are perpendicular at} \ \ t=1:$

$\displaystyle \dot{r}_p \cdot \dot{r}_Q=\binom{3+2 a}{0}\binom{b+2}{1}=(3+2 a)(b+2)=0\ \ldots\ (1)$

♦ Mean mark (b) 49%.

$\text{Equating co-efficients of \(\underset{\sim}{i}$ at $\ t=1$:}\)

$1+a=b+2$

$\text{Substitute into (1) above:}$

$(3+2 a)(1+a)=0$

$a=-\dfrac{3}{2},-1$

c. $\ddot{r}_P=(6 t+2 a)\underset{\sim}{i}, \ \ddot{r}_Q=4 \underset{\sim}{j}$

$\text{At \(\ t=1$, magnitudes of $\ddot{r}_P$ and $\ddot{r}_Q$ are equal.}\)

$6+2 a=4 \ \ \Rightarrow \ \ a=-1$

$1-1=b+2 \ \ \Rightarrow \ \ b=-2$

♦♦♦ Mean mark (c) 25%.

Algebra, MET1 2025 VCAA 9

Consider the functions

$f: R \backslash\{1\} \rightarrow R, f(x)=\dfrac{w^2}{(x-1)^2}$

and

$g: R \rightarrow R, g(x)=(x-w)^2$

where $w \in R$.

If $w=-3$, find the four solutions to $f(x)=g(x)$. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---
Consider the case where $w>0$.
1. Find, in terms of $w$, the coordinates of the minimum point of the graph of $y=(x-1)(x-w)$. (2 marks)
  
  --- 6 WORK AREA LINES (style=lined) ---
2. Hence, or otherwise, find the positive values of $w$ for which $f(x)=g(x)$ has exactly three solutions. (2 marks)
  
  --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):$

$\dfrac{(-3)^2}{(x-1)^2}=(x+3)^2$

$(x+3)^2(x-1)^2=9$

$(x+3)(x-1)= \pm 3$

$\text{Case 1}$

$x^2+2 x-3=3$

$x^2+2 x-6=0$

$x=-1 \pm \sqrt{7}$

$\text{Case 2}$

$x^2+2 x-3$	$=-3$
$x(x+2)$	$=0$
$x=0 \ \ \text{or} \ -2$

b.i. $\text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)$

b.ii. $\text{See worked solutions.}$

Show Worked Solution

a. $\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):$

$\dfrac{(-3)^2}{(x-1)^2}=(x+3)^2$

$(x+3)^2(x-1)^2=9$

$(x+3)(x-1)= \pm 3$

$\text{Case 1}$

$x^2+2 x-3=3$

$x^2+2 x-6=0$

$x=-1 \pm \sqrt{7}$

♦♦ Mean mark (a) 34%.

$\text{Case 2}$

$x^2+2 x-3$	$=-3$
$x(x+2)$	$=0$
$x=0 \ \ \text{or} \ -2$

b.i. $\text{Find minimum TP of} \ \ y=(x-1)(x-w)$

$\text{Axis of quadratic:}\ \ x=\dfrac{w+1}{2},\ \ (w>0)$

$y$	$=\left(\dfrac{w+1}{2}-1\right)\left(\dfrac{w+1}{2}-w\right)$
	$=\left(\dfrac{w-1}{2}\right)\left(\dfrac{1-w}{2}\right)$
	$=-\dfrac{1}{4}(w-1)^2$

$\therefore \ \text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)$

♦♦ Mean mark (b.i) 39%.
♦♦♦ Mean mark (b.ii) 11%.

b.ii. $\text{Solve for \(w\ (w>0)$ where $f(x)=g(x)$:}\)

$\dfrac{w^2}{(x-1)^2}$	$=(x-w)^2$
$(x-w)^2(x-1)^2$	$=w^2$
$[(x-w)(x-1)]^2-w^2$	$=0$

$\text{Difference between two squares:}$

$[(x-w)(x-1)-w][(x-w)(x-1)+w]=0$

$\text{Case 1}$

$(x-w)(x-1)-w$	$=0$
$x^2-x-w x+w-w$	$=0$
$x^2-(w+1) x$	$=0$
$x[x-(w+1)]$	$=0$

$x=0 \ \ \text{or} \ \ w+1$

$\text{Case 2}$

$(x-w)(x-1)+w$	$=0$
$x^2-(w+1) x+2 w$	$=0$

$\text{1 solution if} \ \ \Delta=0:$

$\Delta$	$=[-(w+1)]^2-4 \times 1 \times 2 w$
	$=w^2+2 w+1-8 w$
	$=w^2-6 x+1$

$\text{If} \ \ \Delta=0:$

$w$	$=\dfrac{6 \pm \sqrt{(-6)^2-4 \times 1 \times 1}}{2}$
	$=\dfrac{6 \pm \sqrt{32}}{2}$
	$=3+2 \sqrt{2}\ \ (\text{reject}\ w=3-2 \sqrt{2}\ \ \text{as}\ \ w>0)$

$\therefore f(x)=g(x) \ \ \text{has exactly 3 solutions}\ (w>0).$

Networks, GEN2 2025 VCAA 18

Frances is constructing a home gym. This project requires 12 activities, $A$ to $L$, to be completed.

The activity network below shows each activity and its completion time in days.

This network contains two critical paths.
State the activities that are common to both critical paths. (1 mark)

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Determine the latest start time, in days, for activity $E$. (1 mark)

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Which activity has the longest float time? (1 mark)

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The table below shows five activities that can have their completion time reduced.
It shows the maximum reduction time (days) and the additional cost per day, for each of the five activities.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \textbf{Activity} \ \ & \textbf{Maximum reduction time} & \textbf{Additional cost per day }\\
\textbf{} & \textbf{(days)} \rule[-1ex]{0pt}{0pt} & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} \textit{A} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{500} \\
\hline
\rule{0pt}{2.5ex} \textit{F} \rule[-1ex]{0pt}{0pt} & \text{4} \rule[-1ex]{0pt}{0pt} & \text{150} \\
\hline
\rule{0pt}{2.5ex} \textit{G} \rule[-1ex]{0pt}{0pt} & \text{4} \rule[-1ex]{0pt}{0pt} & \text{150} \\
\hline
\rule{0pt}{2.5ex} \textit{H} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{300} \\
\hline
\rule{0pt}{2.5ex} \textit{K} \rule[-1ex]{0pt}{0pt} & \text{1} \rule[-1ex]{0pt}{0pt} & \text{100} \\
\hline
\end{array}

Frances would like to construct the home gym in three days less than was previously possible.
What is the minimum additional amount Frances will need to pay? (1 mark)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Critical paths: }ADHJL, ADIKL$

b. $\text{LST for Activity}\ E=9 \ \text{days}$

c. $\text{Activity \(F$.}\)

d. $\$1400$

Show Worked Solution

a. $\text{Scan network forwards/backwards:}$

$\text{Critical paths: }ADHJL, ADIKL$

b. $\text{LST for Activity}\ E$

$=20-4-4-3$

$=9 \ \text{days}$

♦ Mean mark (a) 44%.
♦ Mean mark (b) 46%.

c.	$\begin{array}{\|l\|c\|c\|c\|c\|c\|} \hline \rule{0pt}{2.5ex}\ \ \text{Activity} \ \ \rule[-1ex]{0pt}{0pt}& \ \ B \ \ &\ \ C \ \ & \ \ E \ \ & \ \ F \ \ & \ \ G \ \ \\ \hline \rule{0pt}{2.5ex} \ \ \text{Float} \ \ \rule[-1ex]{0pt}{0pt}& 5 & 1 & 5 & 6 & 1 \\ \hline \end{array}$

$\therefore\ \text{Activity \(F$ has the longest float time.}\)

♦ Mean mark (c) 47%.

d. $\text{Current time = 25 days}$

$\text{New time = 22 days.}$

$\text{Reduce:} \ A(2 \ \text {days}), H(1 \ \text{day}), K(1 \ \text{day})$

$\text{Minimum cost}=2 \times 500+1 \times 300+1 \times 100=\$1400$

♦♦♦ Mean mark (d) 21%.

Matrices, GEN2 2025 VCAA 14

An early learning centre runs seven different activities during its 40-day holiday program.

The activities are cooking $(C)$, drama $(D)$, gardening $(G)$, lunch $(L)$, music $(M)$, reading $(R)$ and sport $(S)$.

The timetabled order of the activities for day one of the holiday program is shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \text{9 am} \ \ \rule[-1ex]{0pt}{0pt} & \text{10 am} \rule[-1ex]{0pt}{0pt} & \text{11 am} \rule[-1ex]{0pt}{0pt} & \text{12 pm} \rule[-1ex]{0pt}{0pt} & \ \ \text{1 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{2 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{3 pm} \ \ \\
\hline
\rule{0pt}{2.5ex} \textit{C} \rule[-1ex]{0pt}{0pt} & \textit{D} \rule[-1ex]{0pt}{0pt} & \textit{G} \rule[-1ex]{0pt}{0pt} & \textit{L} \rule[-1ex]{0pt}{0pt} & \textit{M} \rule[-1ex]{0pt}{0pt} & \textit{R} \rule[-1ex]{0pt}{0pt} & \textit{S}\\
\hline
\end{array}

The timetabled order of the activities for day one is also shown in matrix $X$ below.

\begin{aligned}
X = & \begin{bmatrix}
C \\
D \\
G \\
L \\
M \\
R\\
S \\
\end{bmatrix}
\end{aligned}

Matrix $P$, shown below, is a permutation matrix used to determine the timetabled order of activities from one day to the next.

\begin{aligned}
P = & \begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{aligned}

A column matrix containing the timetabled order of activities on one day is multiplied by matrix $P$ to determine the timetabled order of activities for the next day.

State the activities that are always held at the same time on each day of the program. (1 mark)

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Determine the timetabled order of the seven activities on day three of the program. (1 mark)

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$P^4$ is an identity matrix.
Explain what this means for the timetabled order of the activities over the 40-day holiday program. (1 mark)

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Show Answers Only

a. $\text{Gardening, lunch, music.}$

b. $\text{Drama, cooking, gardening, lunch, music, sport, reading.}$

c. $\text{Order of activities rotate on a 4 day cycle.}$

$\text{Over 40 days, there will be 10 cycles of activities}$

Show Worked Solution

a. $\text{Activities held af the same time:}$

$\Rightarrow \ \text{correspond to 1’s in leading diagonal}$

$\text{Gardening, lunch, music.}$

\begin{aligned}X_2=P \times X=\begin{bmatrix}
R \\
S \\
G \\
L \\
M \\
D \\
C
\end{bmatrix}, \quad X_3=P \times X_2=\begin{bmatrix}
D \\
C \\
G \\
L \\
M \\
S \\
R
\end{bmatrix}
\end{aligned}

$\text{Order on day 3:}$

$\text{Drama, cooking, gardening, lunch, music, sport, reading.}$

♦♦ Mean mark (b) 27%.

c. $P^4 \ \Rightarrow \ \text{identity matrix}$

$\text{Order of activities rotate on a 4 day cycle.}$

$\text{Over 40 days, there will be 10 cycles of activities}$

♦♦♦ Mean mark (c) 11%.

Matrices, GEN2 2025 VCAA 13

An early learning centre offers a 10-week activity program for four-year-old children. There are 27 children enrolled in the program. They participate in three different activities over the 10 weeks. The activities are cooking $(C)$, gardening $(G)$ and music $(M)$.

The transition matrix $K$, shown below, gives the expected proportion of children in the program who will change activities from one week to the next.

\begin{aligned}
& \quad \quad \ \ \ \textit{this week} \\
& \quad \ \ C \quad \quad G \ \quad \ \ M \\
K = & \begin{bmatrix}
0 & 0.76 & 0.36 \\
0.55 & 0 & 0.64 \\
0.45 & 0.24 & 0\\
\end{bmatrix}\begin{array}{l}
C\\
G\\
M
\end{array} \ \textit{next week} \\
\end{aligned}

What do the values on the leading diagonal in matrix $K$ indicate? (1 mark)

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In Week 1 of the program, all 27 children participate in cooking $(C)$.
1. Calculate the expected percentage of children who will participate in cooking in Week 10 of the program. Round your answer to one decimal place. (1 mark)
  
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2. Find the expected number of children who will participate in gardening $(G)$ in Week 3 of the program and then move across to music $(M)$ in Week 4 of the program. Round your answer to the nearest whole number. (2 marks)
  
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Show Answers Only

a. $\text{No child does the same activity in two consecutive weeks.}$

b.i. $37.4\%$

b.ii. $2 \ \text{children}$

Show Worked Solution

a. $\text{Leading diagonal made up of 0’s:}$

$\text{No child does the same activity in two consecutive weeks.}$

b.i. $\text{Find number of children cooking in week 10:}$

\begin{aligned}K^9\begin{bmatrix}
27 \\
0 \\
0
\end{bmatrix}=\begin{bmatrix}
10.08 \ldots\\
9.96 \ldots \\
6.94 \ldots
\end{bmatrix}
\end{aligned}

$\text{% children cooking}=\dfrac{10.08 \ldots}{27} \times 100=37.35 \ldots=37.4\% \ \text{(1 d.p.)}$

♦♦♦ Mean mark (b.i) 9%.

b.ii. $\text{At the start of week 3:}$

\begin{aligned}K^2\begin{bmatrix}
27 \\
0 \\
0
\end{bmatrix}=\begin{bmatrix}
15.66 \\
7.776 \\
3.564
\end{bmatrix}
\end{aligned}

$\text{Children who move from \(G$ (week 3) to $M$ (week 4)}\)

$=7.776 \times 0.24=1.886 \ldots=2 \ \text{children}$

♦♦♦ Mean mark (b.ii) 21%.

Matrices, GEN2 2025 VCAA 12

The early learning centre contains three rooms, Nursery $(N)$, Toddler $(T)$ and Pre-kinder $(P)$ .

From one year to the next, children can move between rooms, stay in the same room, or may leave $(L)$ the centre. The following transition matrix, $M$, shows the expected proportion of children who will move between categories or stay in the same category from one year to the next.

\begin{aligned}
& \quad \quad \quad \quad \quad \textit{this year} \\
& \quad \quad \ \ \ \ N \quad \quad \ \ T \quad \quad P \quad \ L \\
M&=\begin{bmatrix}
0.25 & 0 & 0 & 0 \\
0.625 & 0.25 & 0 & 0 \\
0 & 0.625 & 0.1 & 0 \\
0.125 & 0.125 & 0.9 & 1
\end{bmatrix}\begin{array}{l}
N\\
T \\
P \\
P
\end{array}\quad \textit{next year}
\end{aligned}

The number of children expected to be in each of the four categories, from one year to the next, can be calculated using the matrix recurrence relation
1. $S_{n+1}=M S_n$
where $S_n$ represents the expected number of children in each of the four categories at the start of year $n$.
The state matrix $S_{2024}$, shown below, gives the number of children in each category at the start of 2024.
1. 1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
Find $S_{2023}$. (1 mark)

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From the start of 2025, new children commenced in the early learning centre at the start of each year.
A new matrix recurrence relation for determining the expected number of children in each of the four categories from one year to the next is
1. 1. \begin{align*}
    S_{n+1}=M S_n+B
    \end{align*}
where
1. 1. \begin{align*}B=\left[\begin{array}{c}12 \\5 \\10 \\0\end{array}\right] \begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
gives the number of new children enrolled in each room of the early learning centre at the start of each year.
Given the state matrix
1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
find the expected total number of children to be enrolled in the early learning centre at the start of 2026. Round your answer to the nearest whole number. (1 mark)

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Show Answers Only

$S_{2023}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}$

b. $\text{Children enrolled (2026) =50}$

Show Worked Solution

a. $S_{n+1}=M S_n \ \Rightarrow \ S_n=M^{-1} S_{n+1}$

$S_{2023}=\begin{bmatrix}0.25 & 0 & 0 & 0 \\ 0.625 & 0.25 & 0 & 0 \\ 0 & 0.625 & 0.1 & 0 \\ 0.125 & 0.125 & 0.9 & 1\end{bmatrix}^{-1}\begin{bmatrix}4 \\ 15 \\ 15 \\ 27\end{bmatrix}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}$

♦ Mean mark (a) 53%.

$S_{2025}=M S_{2024} + B=\begin{bmatrix}13 \\ 11.25 \\ 20.87 \\ 42.87\end{bmatrix}$

$S_{2026}=M S_{2025} + B=\begin{bmatrix}15.25 \\ 15.93 \\ 19.11 \\ 64.69\end{bmatrix}$

$\text{Children enrolled (2026)\(=15.25+15.93+19.11=50$ (nearest whole)}\)

♦♦♦ Mean mark (b) 13%.

Financial Maths, GEN2 2025 VCAA 10

Using profits from his recent film, Declan invests $650 000 into a 10-year annuity.

The annuity earns interest at 6.4% per annum, compounding quarterly.

Declan receives a regular quarterly payment from the annuity.

Halfway through the 10-year annuity, Declan writes a recurrence relation to represent the quarter-to-quarter balance for the remainder of the annuity.

Let $D_n$ be the balance of Declan's annuity $n$ quarters after the halfway point of the annuity.

Complete the recurrence relation below in terms of $D_0, D_{n+1}$ and $D_n$ that can model this balance. (2 marks)

$D_0=\begin{array}{|c|}\hline \rule{0pt}{4ex} \quad \quad \quad \quad \quad & \quad \quad \quad \quad \quad \\ \hline \end{array}, D_{n+1}=1.016 \times D_n+\begin{array}{|c|}\hline \rule{0pt}{4ex} \quad \quad \quad \quad \quad & \quad \quad \quad \quad \quad \\ \hline \end{array}$

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$D_0=376\,159.43, D_{n+1}=1.016 \times D_n-22\,126.27$

Show Worked Solution

$\text{Find annuity quarterly payment.}$

$\text{Solve for \(PMT$ (by CAS):}\)

$N$	$=40$
$I(\%)$	$=6.4$
$PV$	$=-650\,000$
$PMT$	$=\boldsymbol{22\,126.27}$
$FV$	$=0$
$PY$	$=CY=4$

♦♦♦ Mean mark 17%.

$\text{Find annuity balance at halfway:}$

$\text{Solve for \(FV$ (by CAS):}\)

$N$	$=20$
$I(\%)$	$=6.4$
$PV$	$=-650\,000$
$PMT$	$=22\,126.27$
$FV$	$=\boldsymbol{376\,159.428}$
$PY$	$=CY=4$

$\text{Recurrence relation:}$

$D_0=376\,159.43, D_{n+1}=1.016 \times D_n-22\,126.27$

Financial Maths, GEN2 2025 VCAA 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)} \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline
\end{array}

What amount, in dollars, did Declan borrow? (1 mark)
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Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
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The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (1 mark)

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The last payment required to fully repay the loan is $15 730.71, correct to the nearest cent.
How many payments of $15 730.88 did Declan make before this final payment? (1 mark)

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a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

d. $\text{59 payments made before the final payment.}$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

d. $\text{Solve for \(N$ (by CAS):}\)

$N$	$=\boldsymbol{59.99 \ldots}$
$I(\%)$	$=4.2$
$PV$	$=850\,000$
$PMT$	$=-15\,730.88$
$FV$	$=0$
$PY$	$=CY=12$

$\therefore \ \text{59 payments made before the final payment.}$

♦♦♦ Mean mark (d) 21%.

Data Analysis, GEN2 2025 VCAA 6

The time series plot below shows the number of homes sold in a town each month over a four‑year period.

Month 1 is January 2016 and month 48 is December 2019.

Excluding any possible outliers, identify two qualitative features of the time series plot. (2 marks)

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The total number of sales in each of the four years is given in the table below.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \textbf{Year} \quad \rule[-1ex]{0pt}{0pt}& \textbf{Total number of sales}\\
\hline \rule{0pt}{2.5ex}2016 \rule[-1ex]{0pt}{0pt}& 361 \\
\hline \rule{0pt}{2.5ex}2017 \rule[-1ex]{0pt}{0pt}& 354 \\
\hline \rule{0pt}{2.5ex}2018 \rule[-1ex]{0pt}{0pt}& 358 \\
\hline \rule{0pt}{2.5ex}2019 \rule[-1ex]{0pt}{0pt}& 357 \\
\hline
\end{array}

A seasonal index can be calculated for each month based on the four-year period.
Calculate this seasonal index for September, the ninth month in the calendar year. Round your answer to three decimal places. (2 marks)

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Show Answers Only

a. $\text{Seasonality, irregular fluctuations.}$

b. $\text{Seasonal index(Sep)} =0.587$

Show Worked Solution

a. $\text{Qualitative features of time series plot:}$

$\text{Seasonality}$

$\text{Irregular fluctuations}$

b. $\text{Sept avg}=\dfrac{15+20+20+15}{4}=17.5$

$\text {Monthly avg}=\dfrac{361+354+358+357}{48}=29.79 \ldots$

$\text{Seasonal index (Sep)}=\dfrac{17.5}{29.79 \ldots}=0.5874 \ldots=0.587\ \text{(3 d.p.)}$

Data Analysis, GEN2 2025 VCAA 4

The scatterplot below shows the sale price of a home, in dollars, against the distance of the home from the city centre of Melbourne, in kilometres, distance from city centre.

The sample consists of three‑bedroom homes sold between 2016 and 2018

The equation of the least squares line for the data in the scatterplot is

sale price$=1\,765\,353-35\,054 \times$distance from city centre

The coefficient of determination is 0.0806

Identify the explanatory variable in the least squares equation. (1 mark)

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Calculate the value of the correlation coefficient $r$. Round your answer to three decimal places. (1 mark)

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Use the equation of the least squares line to predict the sale price for a three-bedroom home, located in the city centre of Melbourne, sold between 2016 and 2018. (1 mark)

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Jocelyn wants to sell her three-bedroom home located two kilometres from the city centre of Melbourne.
Would the predicted sale price be an example of interpolation or extrapolation?
Briefly explain your answer. (1 mark)

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Describe the linear association between sale price and distance from city centre in terms of its strength and direction. Answer in the table below. (2 marks)

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\begin{array}{|l|l|}
\hline \rule{0pt}{2.5ex}\text {strength} \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\
\hline \rule{0pt}{2.5ex}\text {direction} \rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

A residual plot associated with the least squares line is shown below.
It is missing one point.

The residual associated with the home that is furthest from the city centre of Melbourne is missing from the residual plot. The home is 15.5 km from the city centre and sold for $1 250 000.
1. Show that the value of the missing residual is 27 984. (1 mark)
  
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2. Plot the residual from part i by placing an $\text{X}$ on the residual plot above. (1 mark)
  
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a. $\text{Distance from city centre}$

b. $r=-0.284$

c. $\text{sale price}=1\,765\,353$

d. $\text{Extrapolation as 2 km lies outside the explanatory variable data range.}$

e. $\text{Strength: weak. Direction: negative}$

f.i. $\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016$

$\text{Actual sale price}=1\,250\,000$

$\text{Residual}=1\,250\,000-1\,222\,016=27\,984$

f.ii.

Show Worked Solution

a. $\text{Distance from city centre}$

b. $\text{Since slope is negative}$

$r=-\sqrt{0.0806}=-0.284$

♦♦♦ Mean mark (b) 21%.

c. $\text{Find sale price when distance fran city centre}=0:$

$\text{sale price}=1\,765\,353$

d. $\text{Extrapolation as 2 km lies outside the explanatory variable data range.}$

$\text{(Note: interpolation/extrapolation should be referenced to the}$

$\text{explanatory variable range).}$

♦♦ Mean mark (d) 27%.

e. $\text{Strength: weak}$

$\text{Direction: negative}$

f.i. $\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016$

$\text{Actual sale price}=1\,250\,000$

$\text{Residual}=1\,250\,000-1\,222\,016=27\,984$

♦ Mean mark (f.i) 41%.
♦ Mean mark (f.ii) 45%

f.ii.

HMS, TIP 2025 HSC 26

Analyse the relationship between training thresholds and TWO physiological adaptations. In your answer, provide examples of both aerobic and resistance training. (8 marks)

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Overview Statement:

Training thresholds represent critical intensity levels that trigger specific physiological adaptations.
Understanding how aerobic and resistance thresholds connect to metabolic and muscular changes can open up pathways to enhanced athletic performance.

Aerobic Threshold and Cardiovascular Adaptations:

The aerobic training threshold occurs at approximately 70% of maximum heart rate. Training at this intensity influences fuel utilisation and cardiovascular function.
This causes the body to shift from primarily using fat to using carbohydrates for energy. For example, a marathon runner training at 70% max heart rate stimulates this metabolic adaptation.
The threshold also triggers increased stroke volume through enhanced left ventricle filling capacity. This relationship results in improved cardiac output and oxygen delivery to working muscles.
Consequently, athletes sustain effort over extended periods with greater efficiency. This shows that aerobic threshold training enables both metabolic and cardiovascular improvements for endurance performance.

Resistance Threshold and Muscular Adaptations:

Resistance training thresholds involve working at 70-85% of one-rep maximum with 6-12 repetitions. This intensity creates sufficient mechanical stress to stimulate muscle hypertrophy.
For instance, a weightlifter performing squats at 80% of 1RM bmicroscopic muscle fibre damage. This initiates increased protein synthesis and muscle repair processes.
The threshold works through progressive overload that leads to enlarged muscle fibres with increased actin and myosin filaments. As a result, greater force production capacity develops.
The significance is that athletes gain strength and power output essential for explosive movements.

Implications and Synthesis:

These thresholds work together as intensity markers that determine adaptation type. Aerobic thresholds influence metabolic and cardiovascular systems whilst resistance thresholds affect muscular structure.
Therefore, coaches must apply appropriate threshold intensities to achieve specific performance goals. This reveals that training success depends on understanding the precise relationship between intensity levels and resulting physiological changes.

Show Worked Solution

Overview Statement:

Training thresholds represent critical intensity levels that trigger specific physiological adaptations.
Understanding how aerobic and resistance thresholds connect to metabolic and muscular changes can open up pathways to enhanced athletic performance.

Aerobic Threshold and Cardiovascular Adaptations:

The aerobic training threshold occurs at approximately 70% of maximum heart rate. Training at this intensity influences fuel utilisation and cardiovascular function.
This causes the body to shift from primarily using fat to using carbohydrates for energy. For example, a marathon runner training at 70% max heart rate stimulates this metabolic adaptation.
The threshold also triggers increased stroke volume through enhanced left ventricle filling capacity. This relationship results in improved cardiac output and oxygen delivery to working muscles.
Consequently, athletes sustain effort over extended periods with greater efficiency. This shows that aerobic threshold training enables both metabolic and cardiovascular improvements for endurance performance.

Resistance Threshold and Muscular Adaptations:

Resistance training thresholds involve working at 70-85% of one-rep maximum with 6-12 repetitions. This intensity creates sufficient mechanical stress to stimulate muscle hypertrophy.
For instance, a weightlifter performing squats at 80% of 1RM bmicroscopic muscle fibre damage. This initiates increased protein synthesis and muscle repair processes.
The threshold works through progressive overload that leads to enlarged muscle fibres with increased actin and myosin filaments. As a result, greater force production capacity develops.
The significance is that athletes gain strength and power output essential for explosive movements.

Implications and Synthesis:

These thresholds work together as intensity markers that determine adaptation type. Aerobic thresholds influence metabolic and cardiovascular systems whilst resistance thresholds affect muscular structure.
Therefore, coaches must apply appropriate threshold intensities to achieve specific performance goals. This reveals that training success depends on understanding the precise relationship between intensity levels and resulting physiological changes.

♦♦ Mean mark 36%.

CHEMISTRY, M6 2025 HSC 34

A 0.010 L aliquot of an acid was titrated with 0.10 mol L$^{-1} \ \ce{NaOH}$, resulting in the following titration curve.

Calculate the $K_a$ for the acid used. (3 marks)
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The concentration of the $\ce{NaOH}$ was 0.10 mol L$^{-1}$.
Explain why the pH of the final solution never reached 13. (2 marks)
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a. $\text{Strategy 1:}$

$\text{From the shape of the titration curve, the acid was weak.}$

$\text{Equivalence point}\ \ \Rightarrow \ \ \ce{NaOH}\ \text{added}\ = 0.24\ \text{L} $

$\text{Halfway to the equivalent point = 0.012 L}, \ce{[ HA ]=\left[ A^{-}\right]}$

$\text{Here the pH} \approx 4.4 , \text{or}\ \ce{\left[H^{+}\right] is 4.0 \times 10^{-5}}$.

$K_a=10^{-\text{pH}}=\ce{\left[H+\right]} \times \dfrac{\ce{\left[A^{-}\right]}}{\ce{[HA]}}$

$\text{At this pH,} \ \ce{\left[A^{-}\right]=[HA] so} \ K_a=4.0 \times 10^{-5}$.

$\text{Strategy 2:}$

$\text{Equivalence point is at} \ \ce{0.024 L NaOH added}$.

$\text{Shape of curve shows acid is monoprotic.}$

$\ce{[HA] \times 0.010=0.1 \times 0.024}$

$\ce{[HA]=0.24 mol L^{-1}}$

$\text{pH at start is approximately 2.5}$

$\text{So,} \ \ce{\left[H+\right]=3.16 \times 10^{-3}}$

$K_a=\dfrac{\ce{\left[H^{+}\right]\left[A^{-}\right]}}{\ce{[HA]}} \quad \ce{\left[A^{-}\right]=\left[H^{+}\right]}$

$\ce{[HA]=0.24-3.16 \times 10^{-3}=0.237}$

$K_a=\dfrac{\left(3.16 \times 10^{-3}\right)^2}{0.237}=4.2 \times 10^{-5}$

b. pH of the final solution < 13:

Some of the hydroxide was neutralised by the acid.
The 10 mL of acid also diluted the NaOH.
So the $\ce{NaOH}$ concentration of the mixture will be less than 0.1 mol L$^{-1}$ and the pH will be less than 13.

Show Worked Solution

a. $\text{Strategy 1:}$

♦♦ Mean mark 40%.

$\text{From the shape of the titration curve, the acid was weak.}$

$\text{Equivalence point}\ \ \Rightarrow \ \ \ce{NaOH}\ \text{added}\ = 0.24\ \text{L} $

$\text{Halfway to the equivalent point = 0.012 L}, \ce{[ HA ]=\left[ A^{-}\right]}$

$\text{Here the pH} \approx 4.4 , \text{or}\ \ce{\left[H^{+}\right] is 4.0 \times 10^{-5}}$.

$K_a=10^{-\text{pH}}=\ce{\left[H+\right]} \times \dfrac{\ce{\left[A^{-}\right]}}{\ce{[HA]}}$

$\text{At this pH,} \ \ce{\left[A^{-}\right]=[HA] so} \ K_a=4.0 \times 10^{-5}$.

$\text{Strategy 2:}$

$\text{Equivalence point is at} \ \ce{0.024 L NaOH added}$.

$\text{Shape of curve shows acid is monoprotic.}$

$\ce{[HA] \times 0.010=0.1 \times 0.024}$

$\ce{[HA]=0.24 mol L^{-1}}$

$\text{pH at start is approximately 2.5}$

$\text{So,} \ \ce{\left[H+\right]=3.16 \times 10^{-3}}$

$K_a=\dfrac{\ce{\left[H^{+}\right]\left[A^{-}\right]}}{\ce{[HA]}} \quad \ce{\left[A^{-}\right]=\left[H^{+}\right]}$

$\ce{[HA]=0.24-3.16 \times 10^{-3}=0.237}$

$K_a=\dfrac{\left(3.16 \times 10^{-3}\right)^2}{0.237}=4.2 \times 10^{-5}$

b. pH of the final solution < 13:

Some of the hydroxide was neutralised by the acid.
The 10 mL of acid also diluted the NaOH.
So the $\ce{NaOH}$ concentration of the mixture will be less than 0.1 mol L$^{-1}$ and the pH will be less than 13.

♦♦♦ Mean mark 19%.

CHEMISTRY, M5 2025 HSC 32

The following three solids were added together to 1 litre of water:

$\ce{0.006\ \text{mol}\ Mg(NO3)2}$
$\ce{0.010\ \text{mol}\ NaOH}$
$\ce{0.002\ \text{mol}\ Na2CO3}$.

Which precipitate(s), if any, will form? Justify your answer with appropriate calculations. (5 marks)

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Show Answers Only

All sodium and nitrate salts are soluble $\Rightarrow$ possible precipitates are $\ce{Mg(OH)2}$ and $\ce{MgCO3}$.

$\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}$

$\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}$

$\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad K_{\textit{sp}}=6.82 \times 10^{-6}}$

$\ce{Mg(OH)2}$ is a lot less soluble than $\ce{MgCO3}$ and will precipitate preferentially.

$\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}$

$\Rightarrow \ce{Mg(OH)2}$ will precipitate.

$\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}$

Since $K_{\textit{sp}}$ is small, assume reaction goes to completion.

$\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}$

Using stoichiometric ratio $(1:2)$

$0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}$ is in excess.

Concentration drops to: $6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}$.

Check if $\ce{MgCO3}$ will precipitate:

$\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}$ becomes $2 \times 10^{-6} <K_{\textit{sp}}$.

$\ce{\Rightarrow\ MgCO3}$ won’t precipitate.

Show Worked Solution

All sodium and nitrate salts are soluble $\Rightarrow$ possible precipitates are $\ce{Mg(OH)2}$ and $\ce{MgCO3}$.

$\ce{\left[Mg^{2+}\right]=6 \times 10^{-3} \quad\left[ OH^{-}\right]=1 \times 10^{-2} \quad\left[ CO3^{2-}\right]=2 \times 10^{-3}}$

$\ce{\left[Mg^{2+}\right]\left[ OH^{-}\right]^2=6 \times 10^{-7} \quad \quad \quad \ \ \ K_{\textit{sp}}=5.61 \times 10^{-12}}$

$\ce{\left[ Mg^{2+}\right]\left[ CO3^{2-}\right]=1.2 \times 10^{-5} \quad \quad K_{\textit{sp}}=6.82 \times 10^{-6}}$

♦♦ Mean mark 40%.

$\ce{Mg(OH)2}$ is a lot less soluble than $\ce{MgCO3}$ and will precipitate preferentially.

$\ce{\left[Mg^{2+}\right]\left[OH^{-}\right]^2 > K_{\textit{sp}}}$

$\Rightarrow \ce{Mg(OH)2}$ will precipitate.

$\ce{Mg^{2+}(aq) + 2OH-(aq) \rightarrow Mg(OH)2(s)}$

Since $K_{\textit{sp}}$ is small, assume reaction goes to completion.

$\ce{n(Mg^{2+})=0.006\ mol, n(OH^{-})=0.010\ mol}$

Using stoichiometric ratio $(1:2)$

$0.006 > \dfrac{0.010}{2}=0.005\ \ \Rightarrow \ce{Mg^{2+}}$ is in excess.

Concentration drops to: $6 \times 10^{-3}-5 \times 10^{-3}=1 \times 10^{-3}$.

Check if $\ce{MgCO3}$ will precipitate:

$\ce{\left[Mg^{2+}\right]\left[CO3^{2-}\right]}$ becomes $2 \times 10^{-6} <K_{\textit{sp}}$.

$\ce{\Rightarrow\ MgCO3}$ won’t precipitate.

PHYSICS, M8 2025 HSC 20 MC

Consider the possibility of an electron and a positron colliding in a particle accelerator to produce a proton and an antiproton, as shown in the equation below.

$\text{electron}+ \text{positron}\ \rightarrow \ \text{proton}+ \text {antiproton}$

Which statement makes the correct conclusion about the possibility of such a reaction, and provides a plausible reason for this conclusion?

The reaction is impossible because electrons and positrons will combine to produce a single neutral particle.
The reaction is possible because the masses of the proton and antiproton are the result of their relativistic velocities.
The reaction is possible because the masses of the proton and antiproton come mainly from energy supplied by the accelerator.
The reaction is impossible because protons are much more massive than electrons and hence the reaction violates the law of conservation of mass.

Show Answers Only

$C$

Show Worked Solution

Option $C$ is correct:

The electron and positron have very small rest masses compared to protons and antiprotons.
The particle accelerator supplies enormous kinetic energy to the electron and positron by accelerating them to speeds close to the speed of light.
This kinetic energy can be converted into mass to create the heavier proton-antiproton pair according to Einstein’s $E = mc^2$ equation.
This means conservation of energy is satisfied, even though the products have much greater rest mass than the reactants.

♦ Mean mark 47%.

Other options:

The reaction is possible – see above (eliminate $A$ and $D$).
Option $B$ is incorrect. The proton and antiproton’s mass must come from the reactants’ energy (electron + positron + accelerator energy), not from their own relativistic velocities.

PHYSICS, M5 2025 HSC 18 MC

The escape velocity from the surface of a planet, which has no atmosphere, is $v$. A mass is launched at 45° to the planet's surface at $v$.

What will be the subsequent motion of the mass?

A circular orbit around the planet
An elliptical orbit around the planet
A parabolic trajectory, returning to land with velocity $v$
A trajectory reaching zero velocity at an infinite distance

Show Answers Only

$D$

Show Worked Solution

The mass is launched at escape velocity which means the object has just enough energy to escape the planet’s gravity.
As the mass travels away from the planet, kinetic energy converts into gravitational potential energy.
Since its initial velocity was exactly escape velocity, at infinite distance, all kinetic energy has been converted to potential energy. This means the velocity reaches zero exactly when the distance becomes infinite.
The 45° launch angle does not matter – escape velocity depends only on speed, not direction.

$\Rightarrow D$

♦♦♦ Mean mark 29%.

PHYSICS, M5 2025 HSC 36

A satellite with velocity $v$, is in a geostationary orbit as shown in Figure 1.

At point $Y$, the satellite explodes and splits into two pieces $m_{ a }$ and $m_{ b }$, of identical mass. As a result of the explosion, the velocity of one piece, $m_{ a }$, changes from $v$ to $2 v$ as shown in Figure 2.

Analyse the subsequent motion of BOTH $m_{ a }$ and $m_{ b }$ after the explosion. Include reference to relevant conservation laws and formulae in your answer. (8 marks)

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At the point in time of the explosion:

By the law of conservation of momentum, the momentum changes of $m_a$ and $m_b$ caused by the explosion must be equal in magnitude but opposite in direction.
Since their masses are equal, both pieces experience the same change in velocity, but in opposite directions.
Because the velocity of $m_a$ increases by $v$ in the direction it was already travelling, the velocity of $m_b$ must also change by $v$ but in the opposite direction to its original motion.
Hence, relative to Earth, the velocity of $m_b$ becomes zero at that instant.

Motion after the explosion:

The satellite’s orbital velocity before the explosion is $v_{\text{orb}}=\sqrt{\dfrac{GM}{r}}$ and the instantaneous velocity of $m_a$ becomes twice this value.
Since $2v_{\text{orb}} = 2 \times \sqrt{\dfrac{GM}{r}} >\sqrt{\dfrac{2GM}{r}} \left(v_{\text{esc}}\right)$, the speed of $m_a$ after the explosion exceeds escape velocity.
After the explosion, $m_a$ continues moving away from Earth with a decreasing speed, but because its initial speed is greater than escape velocity from that point, it will never return. Its total mechanical energy (kinetic + potential) remains constant and positive.
Meanwhile, $m_b$ begins accelerating from an initial velocity of zero toward Earth’s centre. Its initial acceleration is less than $\text{ 9.8 m s}^{-2}$ because it is not at Earth’s surface. As gravitational force increases, its acceleration also increases, and it continues gaining speed at an increasing rate until it reaches Earth’s atmosphere.
In accordance with the conservation of energy, until $m_b$ reaches the atmosphere, the sum of its kinetic and potential energy remains constant and equal to its initial potential energy immediately after the explosion.

Show Worked Solution

At the point in time of the explosion:

By the law of conservation of momentum, the momentum changes of $m_a$ and $m_b$ caused by the explosion must be equal in magnitude but opposite in direction.
Since their masses are equal, both pieces experience the same change in velocity, but in opposite directions.
Because the velocity of $m_a$ increases by $v$ in the direction it was already travelling, the velocity of $m_b$ must also change by $v$ but in the opposite direction to its original motion.
Hence, relative to Earth, the velocity of $m_b$ becomes zero at that instant.

♦ Mean mark 39%.

Motion after the explosion:

The satellite’s orbital velocity before the explosion is $v_{\text{orb}}=\sqrt{\dfrac{GM}{r}}$ and the instantaneous velocity of $m_a$ becomes twice this value.
Since $2v_{\text{orb}} = 2 \times \sqrt{\dfrac{GM}{r}} >\sqrt{\dfrac{2GM}{r}} \left(v_{\text{esc}}\right)$, the speed of $m_a$ after the explosion exceeds escape velocity.
After the explosion, $m_a$ continues moving away from Earth with a decreasing speed, but because its initial speed is greater than escape velocity from that point, it will never return. Its total mechanical energy (kinetic + potential) remains constant and positive.
Meanwhile, $m_b$ begins accelerating from an initial velocity of zero toward Earth’s centre. Its initial acceleration is less than $\text{ 9.8 m s}^{-2}$ because it is not at Earth’s surface. As gravitational force increases, its acceleration also increases, and it continues gaining speed at an increasing rate until it reaches Earth’s atmosphere.
In accordance with the conservation of energy, until $m_b$ reaches the atmosphere, the sum of its kinetic and potential energy remains constant and equal to its initial potential energy immediately after the explosion.

PHYSICS, M8 2025 HSC 33

Analyse the role of experimental evidence and theoretical ideas in developing the Standard Model of matter. (6 marks)

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Overview Statement

The Standard Model’s development depends on the cyclical relationship between experimental evidence and theoretical predictions.
These components interact with each other, where theory guides experiments and results validate or refine theory.

Particle Discovery

Theoretical predictions lead to targeted experimental searches for specific particles.
The Higgs Boson was theoretically proposed decades before its experimental discovery to explain particle mass.
Cloud chambers discovered antimatter after theory predicted its existence.
Particle accelerators verified quarks existed by revealing the internal structure of protons and neutrons.
This pattern shows theory provides the framework while experiments confirm reality.
Consequently, successful verification enables confidence in theoretical models and guides further predictions.

Experimental Tools Driving Theoretical Refinement

High-energy particle accelerators create small wavelength ‘matter probes’ allowing high-resolution investigation of matter’s structure.
These experiments verified electroweak theory by demonstrating electromagnetic and weak nuclear forces result from the same underlying interaction.
Unexpected experimental results sometimes cause theoretical modifications or new predictions.
The significance is that increasingly powerful experimental tools reveal deeper layers of matter structure.

Implications and Synthesis

This reveals the Standard Model emerged from iterative cycles where theory and experiment continuously influence each other.
Neither component alone could have produced the model.
Together, they form a self-correcting system advancing our understanding of fundamental matter.

Show Worked Solution

Overview Statement

The Standard Model’s development depends on the cyclical relationship between experimental evidence and theoretical predictions.
These components interact with each other, where theory guides experiments and results validate or refine theory.

♦♦♦ Mean mark 37%.

Particle Discovery

Theoretical predictions lead to targeted experimental searches for specific particles.
The Higgs Boson was theoretically proposed decades before its experimental discovery to explain particle mass.
Cloud chambers discovered antimatter after theory predicted its existence.
Particle accelerators verified quarks existed by revealing the internal structure of protons and neutrons.
This pattern shows theory provides the framework while experiments confirm reality.
Consequently, successful verification enables confidence in theoretical models and guides further predictions.

Experimental Tools Driving Theoretical Refinement

High-energy particle accelerators create small wavelength ‘matter probes’ allowing high-resolution investigation of matter’s structure.
These experiments verified electroweak theory by demonstrating electromagnetic and weak nuclear forces result from the same underlying interaction.
Unexpected experimental results sometimes cause theoretical modifications or new predictions.
The significance is that increasingly powerful experimental tools reveal deeper layers of matter structure.

Implications and Synthesis

This reveals the Standard Model emerged from iterative cycles where theory and experiment continuously influence each other.
Neither component alone could have produced the model.
Together, they form a self-correcting system advancing our understanding of fundamental matter.

PHYSICS, M7 2025 HSC 32

Analyse the consequences of the theory of special relativity in relation to length, time and motion. Support your answer with reference to experimental evidence. (8 marks)

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Overview Statement

Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

Time–Length Relationship in Muon Observations

Muon-decay experiments show how time dilation and length contraction depend on relative motion.
Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
In the muon’s frame of reference, the atmosphere is length-contracted according to the equation $l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}$, so the distance they travel is much shorter.
Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

Relativistic momentum explains how objects behave as they approach light speed.
Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
At high velocities, motion fundamentally alters measurements of time, distance and momentum.
Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

Show Worked Solution

Overview Statement

Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

♦ Mean mark 53%.

Time–Length Relationship in Muon Observations

Muon-decay experiments show how time dilation and length contraction depend on relative motion.
Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
In the muon’s frame of reference, the atmosphere is length-contracted according to the equation $l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}$, so the distance they travel is much shorter.
Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

Relativistic momentum explains how objects behave as they approach light speed.
Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
At high velocities, motion fundamentally alters measurements of time, distance and momentum.
Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

CHEMISTRY, M6 2025 HSC 19 MC

0.1 mol of solid sodium acetate is dissolved in 500 mL of 0.1 mol L$^{-1}$ $\ce{HCl}$ in a beaker. This solution has a pH of 4.8 .

500 mL of distilled water is then added to the beaker.

What is the pH of the final solution?

Show Answers Only

$C$

Show Worked Solution

Strategy 1

Sodium acetate + $\ce{HCl}$ produces a buffer solution (equal moles of weak acid $\ce{CH3COOH}$ and its conjugate base $\ce{CH3COO-}$).
A buffer’s pH is independent of dilution (the ratio of [acid]:[conjugate base] stays constant).
Therefore pH remains 4.8

Strategy 2 (Quantitative Analysis “trap”)

The initial reaction between sodium acetate and hydrochloric acid runs to completion as $\ce{HCl}$ is a strong acid:
$\ce{CH3COO-(aq) + HCl(aq) -> CH3COOH(aq) + Cl-(aq)}$
$n(\ce{CH3COO-}) = 0.1\ \text{mol}$
$n(\ce{HCl}) = 0.5\ \text{L} \times 0.1\ \text{mol L}^{-1} = 0.05\ \text{mol}$
As they react in a $1:1$ ratio, $\ce{HCl}$ is the limiting reagent.
$n(\ce{CH3COO-_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{initial}}})-n(\ce{CH3COO_{\text{reacted}}}) = 0.1-0.05 = 0.05\ \text{mol}$
$n(\ce{CH3COOH_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{reacted}}}) = 0.05\ \text{mol}$
The following equilibrium reaction is then established below dilution
$\ce{CH3COOH(aq) + H2O(l) \leftrightharpoons CH3COO-(aq) + H3O+(aq)}$
Therefore the following ice table can be constructed:

♦♦♦ Mean mark 19%.
COMMENT: The key insight here is conceptual not computational.

\begin{array} {|c|c|c|c|}
\hline & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 & 0.1 & 0 \\
\hline \text{Change} & -10^{-4.8} & +10^{-4.8} & +10^{-4.8} \\
\hline \text{Equilibrium} & 0.1-10^{-4.8} & 0.1+10^{-4.8} & +10^{-4.8} \\
\hline \end{array}

$\therefore K_a = \dfrac{(0.1+10^{-4.8})(10^{-4.8})}{0.1-10^{-4.8}} = 1.585395 \times 10^{-5}$

Then considering the dilution which would shift the equilibrium position to the right.

\begin{array} {|c|c|c|c|}
\hline & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.05-x & 0.05+x & +x \\
\hline \end{array}

As $K_{eq}$ is small, $0.05 -x \approx 0.05$ and $0.05 + x \approx 0.05$.
$\therefore K_{eq} = \dfrac{0.05x}{0.05} = x = 1.585395 \times 10^{-5}$
$\text{pH} = -\log_{10}\ce{[H3O+]} = -\log_{10}(1.585395 \times 10^{-5}) = 4.79962 = 4.8\ \text{(1 d.p.)}$

$\Rightarrow C$

CHEMISTRY, M8 2025 HSC 18 MC

The concentration of silver ions in a solution is determined by titrating it with aqueous sodium chloride, using yellow potassium chromate as the indicator.

Which row of the table correctly identifies the colour change at the endpoint and the more soluble salt?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \text{Colour change} \ \ & \text{More soluble salt} \\
\text{at endpoint}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}\text{Red to yellow}\rule[-1ex]{0pt}{0pt}& \ce{Ag2CrO4}\\
\hline
\rule{0pt}{2.5ex}\text{Yellow to red}\rule[-1ex]{0pt}{0pt}& \ce{Ag2CrO4}\\
\hline
\rule{0pt}{2.5ex}\text{Red to yellow}\rule[-1ex]{0pt}{0pt}& \ce{AgCl} \\
\hline
\rule{0pt}{2.5ex}\text{Yellow to red}\rule[-1ex]{0pt}{0pt}& \ce{AgCl} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

The two key reactions that take place in this titration are
$\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$
$\ce{2Ag+(aq) + CrO4^{2-}(aq) -> Ag2CrO4(s)}$
For the titration to occur effectively all of the $\ce{Cl-}$ ions must react with $\ce{Ag+}$ ion and be precipitated out of the solution as the sodium chloride is Titrant (known volume and known concentration).
Only once all of the $\ce{Cl-}$ ions are used up will silver ions begin to react the chromate ions signally the endpoint of the titration.
Hence, $\ce{Ag2CrO4}$ is the more soluble salt and as the potassium chromate is initially yellow, the colour change must be from yellow to red.

$\Rightarrow B$

♦♦♦ Mean mark 14%.

Vectors, EXT2 V1 2025 HSC 16c

Consider the point $B$ with three-dimensional position vector $\underset{\sim}{b}$ and the line $\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}$, where $\underset{\sim}{a}$ and $\underset{\sim}{d}$ are three-dimensional vectors, $\abs{\underset{\sim}{d}}=1$ and $\lambda$ is a parameter.

Let $f(\lambda)$ be the distance between a point on the line $\ell$ and the point $B$.

Find $\lambda_0$, the value of $\lambda$ that minimises $f$, in terms of $\underset{\sim}{a}, \underset{\sim}{b}$ and $\underset{\sim}{d}$. (2 marks)

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Let $P$ be the point with position vector $\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}$.
Show that $PB$ is perpendicular to the direction of the line $\ell$. (1 mark)

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Hence, or otherwise, find the shortest distance between the line $\ell$ and the sphere of radius 1 unit, centred at the origin $O$, in terms of $\underset{\sim}{d}$ and $\underset{\sim}{a}$.
You may assume that if $B$ is the point on the sphere closest to $\ell$, then $O B P$ is a straight line. (3 marks)

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i. $\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})$

ii. $\text{See Worked Solutions.}$

iii. $d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}$

Show Worked Solution

i. $\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1$

$\text{Vector from point \(B$ to a point on $\ell$}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

$f(\lambda)=\text{distance between \(\ell$ and $B$}\)

$f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}$

$\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}$

♦♦ Mean mark (i) 33%.

$f(\lambda)^2$	$=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2$
	$=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})$
	$=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot \underset{\sim}{d}$
	$=\lambda^2\|\underset{\sim}{d}\|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2$
	$=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2$

$f(\lambda)^2 \ \ \text{is a concave up quadratic.}$

$f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}$

$\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})$

ii. $P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}$

$\text{Show} \ \ \overrightarrow{PB} \perp \ell:$

♦♦♦ Mean mark (ii) 22%.

$\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}$

$\overrightarrow{P B} \cdot \underset{\sim}{d}$	$=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}$
	$=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}$
	$=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2$
	$=0$

$\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. $

iii. $\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1$)}\)

$=\ \text{(shortest distance \(\ell$ to $O$)}-1\)

♦♦♦ Mean mark (iii) 4%.

$f\left(\lambda_0\right)=\text{shortest distance \(\ell$ to point $B$}\)

$\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell$ to $0$}\)

$\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}$

$f\left(\lambda_0\right)$	$=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}$
$f\left(\lambda_0\right)^2$	$=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2$
	$=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2$
$f\left(\lambda_0\right)$	$=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}$

$\text {Shortest distance of \(\ell$ to sphere $\left(d_{\min }\right)$:}\)

$d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}$

Mechanics, EXT2 M1 2025 HSC 16b

A particle of mass 1 kg is projected from the origin with a speed of 50 ms$^{-1}$, at an angle of $\theta$ below the horizontal into a resistive medium.

The position of the particle $t$ seconds after projection is $(x, y)$, and the velocity of the particle at that time is $\underset{\sim}{v}=\displaystyle \binom{\dot{x}}{\dot{y}}$.

The resistive force, $\underset{\sim}{R}$, is proportional to the velocity of the particle, so that $\underset{\sim}{R}=-k \underset{\sim}{v}$, where $k$ is a positive constant.

Taking the acceleration due to gravity to be 10 ms$^{-2}$, and the upwards vertical direction to be positive, the acceleration of the particle at time $t$ is given by:

$\underset{\sim}{a}=\displaystyle \binom{-k \dot{x}}{-k \dot{y}-10}$. (Do NOT prove this.)

Derive the Cartesian equation of the motion of the particle, given $\sin \theta=\dfrac{3}{5}$. (5 marks)

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Show Answers Only

$y=\left(\dfrac{1-3 k}{4 k}\right) x+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}$

Show Worked Solution

$\sin \theta=\dfrac{3}{5} \ \Rightarrow \ \cos \theta=\dfrac{4}{5}$

$\text{Components of initial velocity:}$

$\dot{x}(0)=50\, \cos \theta=50 \times \dfrac{4}{5}=40 \ \text{ms}^{-1}$

$\dot{y}(0)=50\, \sin \theta=50 \times \dfrac{3}{5}=-30\ \text{ms}^{-1}$

♦♦ Mean mark 35%.

$\text{Horizontal motion:}$

$\dfrac{d \dot{x}}{dt}$	$=-k \dot{x} \ \ \text{(given)}$
$\dfrac{dt}{d \dot{x}}$	$=-\dfrac{1}{k \dot{x}}$
$\displaystyle \int dt$	$=-\dfrac{1}{k} \int \dfrac{1}{\dot{x}}\, d x$
$t$	$=-\dfrac{1}{k} \ln \dot{x}+c$

$\text{When} \ \ t=0, \ \dot{x}=40 \ \Rightarrow \ c=\dfrac{1}{k} \ln 40$

$t$	$=\dfrac{1}{k} \ln 40-\dfrac{1}{k} \ln \abs{\dot{x}}=\dfrac{1}{k} \ln \abs{\dfrac{40}{\dot{x}}}$
$k t$	$=\ln \abs{\dfrac{40}{\dot{x}}}$
$e^{k t}$	$=\dfrac{40}{\dot{x}}$
$\dot{x}$	$=40 e^{-k t}$
$x$	$\displaystyle=\int 40 e^{-k t}\, d t=-\dfrac{40}{k} \times e^{-k t}+c$

$\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{40}{k}$

$x=\dfrac{40}{k}-\dfrac{40}{k} e^{-k t}=\dfrac{40}{k}\left(1-e^{-k t}\right)\ \ldots\ (1)$

$\text{Vertical Motion }$

$\dfrac{d \dot{y}}{dt}$	$=-k \dot{y}-10 \quad \text{(given)}$
$\dfrac{d t}{d \dot{y}}$	$=-\dfrac{1}{k} \times \dfrac{1}{\dot{y}+\frac{10}{k}}$
$t$	$=-\dfrac{1}{k} \displaystyle \int \dfrac{1}{\dot{y}+\frac{10}{k}} \, d \dot{y}=-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}+c$

$\text{When} \ \ t=0, \, \dot{y}=-30 \ \ \Rightarrow\ \ c=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}$

$t$	$=\dfrac{1}{k} \ln \abs{-30+\frac{10}{k}}-\dfrac{1}{k} \ln \abs{\dot{y}+\frac{10}{k}}$
	$=\dfrac{1}{k} \ln \abs{\frac{-30+\frac{10}{k}}{\dot{y}+\frac{10}{k}}}$
$e^{k t}$	$=\abs{\dfrac{-30+\frac{10}{k}}{y+\frac{10}{k}}}$
$\dot{y}$	$=\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10}{k}$
$y$	$=\displaystyle \left(-30+\dfrac{10}{k}\right) \int e^{-kt}\, d t-\int \dfrac{10}{k}\, dt$
	$=-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10 t}{k}+c$

$\text{When} \ \ t=0, y=0 \ \Rightarrow \ c=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)$

$y$	$=\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right)-\dfrac{1}{k}\left(-30+\dfrac{10}{k}\right) e^{-k t}-\dfrac{10t}{k}$
	$=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right)\left(1-e^{-k t}\right)-\dfrac{10 t}{k}\ \ldots\ (2)$

$\text {Cartesian equation (using (1) above):}$

$x$	$=\dfrac{40}{k}\left(1-e^{-k t}\right)$
$\dfrac{k x}{40}$	$=1-e^{-k t}$
$e^{-k t}$	$=1-\dfrac{k x}{40}$
$-k t$	$=\ln \abs{1-\dfrac{k x}{40}}$
$t$	$=-\dfrac{1}{k} \ln \abs{1-\dfrac{kx}{40}}$

$y$	$=\left(\dfrac{10}{k^2}-\dfrac{30}{k}\right) \times \dfrac{k x}{40}+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}$
	$=\left(\dfrac{1-3 k}{4 k}\right) x+\dfrac{10}{k^2} \times \ln \abs{1-\dfrac{k x}{40}}$

Proof, EXT2 P1 2025 HSC 16a

Consider the equation

$z^n \cos\left[n \theta\right]+z^{n-1} \cos \left[(n-1) \theta\right]+z^{n-2} \cos \left[(n-2) \theta\right]+\cdots+z\, \cos\left[\theta\right]=1$

where $z \in \mathbb{C} , \theta \in \mathbb{R} $, and $n$ is a positive integer.

Using a proof by contradiction and the triangle inequality, or otherwise, prove that all the solutions to the equation lie outside the circle $\abs{z}=\dfrac{1}{2}$ on the complex plane. (4 marks)

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$\text{Proof by contradiction}$

$\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}$

$z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1$

$\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):$

$\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|$

$\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|$

$1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|$

$1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)$

$1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) $

$1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\ a=2^{-n}, r=2}$

$1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}$

$1 \leqslant 1-2^{-n}$

$2^{-n} \leqslant 0 \ \ \text {(which is not true)}$

$\therefore \ \text{By contradiction, the original statement is correct}$

Show Worked Solution

$\text{Proof by contradiction}$

$\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}$

$z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1$

♦♦♦ Mean mark 10%.

$\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):$

$\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|$

$\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|$

$1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|$

$1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)$

$1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) $

$1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\ a=2^{-n}, r=2}$

$1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}$

$1 \leqslant 1-2^{-n}$

$2^{-n} \leqslant 0 \ \ \text {(which is not true)}$

$\therefore \ \text{By contradiction, the original statement is correct}$

Complex Numbers, EXT2 N2 2025 HSC 15c

Show that
$\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.$ (3 marks)

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Use De Moivre's theorem to show that the sixth roots of $-1$ are given by
$\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)$ for $k=0,1,2,3,4,5$. (2 marks)

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Hence, or otherwise, show the solutions to $\left(\dfrac{z-1}{z+1}\right)^6=-1$ are
$z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)$, and $i \cot \left(\dfrac{11 \pi}{12}\right)$. (2 marks)

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i. $\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\frac{\theta}{2}\right)$

$\text{Method 1:}$

$\text{LHS}$	$=\dfrac{2\cos^{2}\frac{\theta}{2}+2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}-2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}} $
	$=\dfrac{2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\,\sin\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right) } $
	$=\dfrac{i\,\cos\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}\left( \sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)} $
	$=i\, \cot \left(\frac{\theta}{2}\right)$

$\text{Method 2 (exponential form – ex-syllabus in 2027):}$

$\text{LHS}$	$=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}$
	$=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}$
	$=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}$
	$=i \cot \left(\frac{\theta}{2}\right)$

ii. $z=\cos \theta+i \sin \theta$

$\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}$

$z^6=\cos (6 \theta)+i \sin (6 \theta)=-1$

$\cos (6 \theta)$	$=-1 \ \text{and} \ \ \sin (6 \theta)=0$
$6 \theta$	$=\pi, 3 \pi, 5 \pi, \ldots$
$\theta$	$=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5$

$\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for} \ \ k=0,1, \ldots, 5$

iii. $\left(\dfrac{z-1}{z+1}\right)^6=-1$

$\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1$

$\text {Using part (ii):}$

$\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)$

$\alpha=\dfrac{z-1}{z+1}\ \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}$

$\text{Consider} \ \ \alpha=\operatorname{cis}\left(\frac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}$

$z=\dfrac{1+\operatorname{cis}\left(\frac{\pi}{6}\right)}{1-\operatorname{cis}\left(\frac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{\pi}{12}\right) \quad \text{(using part (i))}$

$\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\frac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\frac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{3 \pi}{12}\right)$

$\therefore z=i \cot \left(\frac{\pi}{12}\right), \, i \cot \left(\frac{3 \pi}{12}\right), \, i \cot \left(\frac{5 \pi}{12}\right), \ldots, i \cot \left(\frac{11 \pi}{12}\right)$

Show Worked Solution

i. $\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\frac{\theta}{2}\right)$

$\text{Method 1:}$

$\text{LHS}$	$=\dfrac{2\cos^{2}\frac{\theta}{2}+2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}-2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}} $
	$=\dfrac{2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\,\sin\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right) } $
	$=\dfrac{i\,\cos\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}\left( \sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)} $
	$=i\, \cot \left(\frac{\theta}{2}\right)$

$\text{Method 2 (exponential form – ex-syllabus in 2027):}$

$\text{LHS}$	$=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}$
	$=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}$
	$=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}$
	$=i \cot \left(\frac{\theta}{2}\right)$

ii. $z=\cos \theta+i \sin \theta$

$\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}$

$z^6=\cos (6 \theta)+i \sin (6 \theta)=-1$

$\cos (6 \theta)$	$=-1 \ \text{and} \ \ \sin (6 \theta)=0$
$6 \theta$	$=\pi, 3 \pi, 5 \pi, \ldots$
$\theta$	$=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5$

$\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for} \ \ k=0,1, \ldots, 5$

iii. $\left(\dfrac{z-1}{z+1}\right)^6=-1$

$\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1$

♦♦♦ Mean mark (iii) 23%.

$\text {Using part (ii):}$

$\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)$

$\alpha=\dfrac{z-1}{z+1}\ \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}$

$\text{Consider} \ \ \alpha=\operatorname{cis}\left(\frac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}$

$z=\dfrac{1+\operatorname{cis}\left(\frac{\pi}{6}\right)}{1-\operatorname{cis}\left(\frac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{\pi}{12}\right) \quad \text{(using part (i))}$

$\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\frac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\frac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{3 \pi}{12}\right)$

$\therefore z=i \cot \left(\frac{\pi}{12}\right), \, i \cot \left(\frac{3 \pi}{12}\right), \, i \cot \left(\frac{5 \pi}{12}\right), \ldots, i \cot \left(\frac{11 \pi}{12}\right)$

BIOLOGY, M8 2025 HSC 32

A population lives across three regions, $A, B$ and $C$.

People in community $B$ developed an environmental disease. An epidemiological study was carried out to determine the risk of developing the disease due to age at exposure. The results of this study are shown in the graph.

Design an epidemiological study that could be used to produce the results shown in the graph. Justify the features of your design. (7 marks)

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Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Participants: Recruit individuals from community B across three age groups: 10-year-olds, 20-year-olds and 30-year-olds at the time of exposure to the environmental factor. This is justified because the graph displays separate curves for exposure at these three ages.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

Longitudinal Follow-up: Monitor all participants annually for disease development over 60 years. This is justified because the graph tracks disease risk across this timeframe and shows when risk peaks and declines.

Data Collection: Document whether each participant develops the disease and calculate the percentage of each age cohort affected at yearly intervals. This is justified because the y-axis shows risk as a percentage.

Control Variables: Ensure all participants experience similar levels of environmental exposure in community B. This is justified because the study isolates age at exposure as the only variable affecting disease risk.

Statistical Analysis: Calculate risk percentages for each time point after exposure for each age group. This is justified because it produces the three distinct curves showing risk declining differently based on initial exposure age.

Show Worked Solution

Study Type: A prospective cohort study would be used. This is justified because it follows participants over extended time periods (up to 60 years) to observe disease development naturally.

Baseline Data: Record each participant’s exact age at first exposure. This is justified because age at exposure is the independent variable being tested.

♦♦♦ Mean mark 46%.

Measurement, STD2 M6 2025 HSC 35

The triangle $PTA$ is shown. The length of $PA$ is 75 m and the length of $PT$ is 51 m.

The angle of depression from $T$ to $A$ is 36°, and the angle $PTA$ is obtuse.

Find the length of $TA$. Give your answer correct to 2 decimal places. (3 marks)

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$TA=35.03 \ \text{m}$

Show Worked Solution

♦♦♦ Mean mark 29%.

$\angle TAP=36^{\circ} \ \text {(alternate)}$

$\text{Using sine rule in} \ \triangle TAP:$

$\dfrac{\sin \angle PTA}{75}$	$=\dfrac{\sin 36^{\circ}}{51}$
$\sin \angle PTA$	$=75 \times \dfrac{\sin 36^{\circ}}{54}=0.864 \ldots$
$\angle PTA$	$=\sin ^{-1}(0.864 \ldots)=180-59.81=120.19^{\circ}\ \ \text{(obtuse)}$

$\angle PTX=120.19-54=66.19^{\circ}$

$\angle TPA=90-66.19=23.81^{\circ}\ \left(180^{\circ}\ \text{in}\ \triangle \right)$

$\text{Using sine rule in} \ \triangle TAP:$

$\dfrac{TA}{\sin 23.81^{\circ}}$	$=\dfrac{51}{\sin 36^{\circ}}$
$TA$	$=\dfrac{51 \times \sin 23.81^{\circ}}{\sin 36^{\circ}}$
	$=35.03 \ \text{m (2 d.p.)}$

Measurement, STD1 M4 2025 HSC 28

The table provides information about a $2 coin and a $5 note.

\begin{array} {|c|c|c|}
\hline \text{Coin/note} & \text{Quantity needed to} & \text{Mass of this number} \\ & \text{make \$1000} & \text{of notes/coins} \\& & \text{(kg)} \\
\hline \$2 & 500 & 3.3 \\
\hline \$5 & 200 & 0.157 \\
\hline \end{array}

Calculate the mass of a $2 coin in grams, correct to 1 decimal point? (2 marks)
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Suppose the $2 coin is to be replaced with a note that has the same mass as a $5 note.
What is the mass of $1000 in $2 notes in grams? Give your answer correct to the nearest gram. (2 marks)
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a. $ 6.6\ \text{g}$

b. $ 393\ \text{g (nearest gram)}$

Show Worked Solution

a. $\text{Using 1 kg = 1000 grams:}$

$\text{Mass of \$2 coin}=\dfrac{3.3\times 1000}{500}=6.6\ \text{g}$

♦♦ Mean mark (a) 34%.

b. $ \text{Number of \$2 notes}=\dfrac{1000}{2}=500$

$\text{Mass of one \$5 note} = \dfrac{0.157}{200}=0.000785\ \text{kg}$

$\text{Since \$2 note weighs the same as \$5 note:}$

$\text{Mass of 500 \$2 notes}$	$=500\times 0.000785$
	$=0.3925\ \text{kg}$
	$=393\ \text{g (nearest gram)}$

♦♦♦ Mean mark (b) 28%.

Financial Maths, STD1 F3 2025 HSC 27

The graph shows the salvage value of a car over 5 years.

The salvage values are based on the declining-balance method.

By what amount will the car’s value depreciate during the 10th year? (4 marks)

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$$1476.40$

Show Worked Solution

$\text{Find}\ r:$

$\text{When}\ \ n=1, \ S=$44\ 000\ \ \text{(see graph)}$

$S$	$=V_0(1-r)^n$
$44\ 000$	$=55\ 000(1-r)^1$
$\dfrac{44\ 000}{55\ 000}$	$=1-r$
$1-r$	$=0.8$
$r$	$=1-0.8=0.20$

$\text{Find \(S$ when}\ \ n=9\ \ \text{and}\ \ n=10:\)

$S_9=55\ 000(1-0.20)^{9}=$7381.97504$

$S_{10}=55\ 000(1-0.20)^{10}=$5905.5800$

$S_9-S_{10}=$7381.9750-$5905.580=$1476.40\ \text{(nearest cent)}$

$\therefore\ \text{The car’s value will depreciate by \$1476.40 in the 10th year.}$

♦♦♦ Mean mark 23%.

Algebra, STD1 A3 2025 HSC 23

It costs $465 to register a passenger car and $350 to register a motorcycle.

Let	$P$	$\ =\ \text{the number of passenger cars, and}$
	$B$	$\ =\ \text{the number of motorcycles}$

Write TWO linear equations that represent the relationship below.

There are 11 times as many passenger cars as motorcycles.
The total registration fees for passenger cars and motorcycles is $494 million. (2 marks)

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$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Show Worked Solution

$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

♦♦♦ Mean mark 11%.

Measurement, STD1 M3 2025 HSC 22

An isosceles triangle is drawn inside a circle as shown. The base of the triangle is 4.8 cm long, the length of other sides is 4 cm and the height is $h$ cm.

Calculate the height, $h$, of triangle $ABC$. (2 marks)
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The area of triangle $ABC$ is 7.68 cm².
The radius of the circle is 2.5 cm.
Express the area of triangle $ABC$ as a percentage of the area of the circle, correct to 1 decimal place. (2 marks)
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a. $h=3.2\ \text{cm}$

b. $39.1\%$

Show Worked Solution

a. $\text{Since}\ \Delta ABC\ \text{is isosceles:}$

$BM\ \text{bisects}\ AC\ \ \Rightarrow\ \ AM=MC=2.4$

$\text{Using Pythagoras:}$

$h^2=4^2-2.4^2=16-5.76=10.24$

$h = \sqrt{10.24} = 3.2\ \text{cm}$

♦♦ Mean mark (a) 31%.

b. $\text{Area of circle}=\pi\times 2.5^2$

$\text{Percentage}$	$=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%$
	$=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%$
	$=39.11\ …\%$
	$\approx 39.1\%\ \text{(1 decimal place)}$

♦♦♦ Mean mark (b) 22%.

Measurement, STD1 M3 2025 HSC 20

A map of a park containing a duck pond is shown.

A fence is built passing through the points $A$, $B$ and $C$ around the duck pond.

Using the scale provided on the map, calculate the length of the fence $AB$. (2 marks)
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The length of $AB$ is equal to the length of $BC$.
Use Pythagoras’ theorem to calculate the length of $AC$ in metres. Give your answer correct to 3 significant figures. (3 marks)
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What is the true bearing of point $A$ from point $C$ ? (2 marks)
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a. $75\ \text{metres}$

b. $106\ \text{metres}$

c. $225^\circ$

Show Worked Solution

a. $\text{Scale: 1 grid width = 5 metres}$

$AB = 15 \times 5 = 75\ \text{metres}$

Mean mark (a) 54%.

b. $\text{Using Pythagoras:}$

$AC^2=AB^2+BC^2$

$AC^2=75^2+75^2=11250$

	$\therefore\ AC$	$=\sqrt{11250}$
		$=106.066\ …$
		$\approx 106\ \text{m (3 sig fig)}$

♦♦ Mean mark (b) 38%.

c. $\text{Since }AB=BC:$

$\angle BAC=\angle BCA=45^\circ$

$\text{Bearing of \(A$ from $C$}\ =180+45=225^\circ\)

♦♦♦ Mean mark (c) 24%.

Algebra, STD1 A2 2025 HSC 16

The mass $(M )$ of a box with a square base, in grams, is directly proportional to the area of its base, in cm².

A box with a square base of side length 5 cm has a mass of 500 g.

What is the mass of a similar box with a square base of side length 3 cm? (3 marks)

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$M=180\ \text{grams}$

Show Worked Solution

$\text{Area of square base }= s^2$

$M \propto s^2\ \ \Rightarrow \ \ M=k\times s^2$

$\text{Find \(k$ given $\ s=5\ $ when $\ M=500$:}\)

$500$	$=k\times 5^2$
$25k$	$=500$
$k$	$=\dfrac{500}{25}=20$

$\text{Find \(M$ when $s=3$:}\)

$M=20\times 3^2=180\ \text{grams}$

♦♦♦ Mean mark 21%.

Statistics, STD1 S3 2025 HSC 15

A researcher is using the statistical investigation process to investigate a possible relationship between average number of minutes per day a person spends watching television, and the average number of minutes per day the person spends exercising.

State the statistical question being posed. (1 mark)
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Participants were asked to record the number of minutes they spent watching television each day and the number of minutes they spent exercising each day. The averages for each participant were recorded and graphed, and a line of best fit was included.

From the graph, identify the dependent variable. (1 mark)

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Describe the bivariate dataset in terms of its form and direction. (2 marks)

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The points $(0, 70)$ and $(60, 10)$ lie on the line of best fit. By first plotting these points on the graph, find the gradient and the $y$-intercept of the line of best fit. (3 marks)
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Explain why it is NOT appropriate to extrapolate the line of best fit to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day. (1 mark)
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a. $\text{How does the average daily time spent watching television}$

$\text{relate to the average daily time spent exercising?}$

b. $\text{Dependent variable: Average minutes per day exercising, or }y.$

c. $\text{Form: Linear}$

$\text{Direction: Negative}$

d. $\text{Gradient}=-1$

e. $\text{The extrapolation of the graph past 70 minutes produces}$

$\text{negative average minutes per day exercising (impossible).}$

Show Worked Solution

a. $\text{How does the average daily time spent watching television}$

$\text{relate to the average daily time spent exercising?}$

b. $\text{Dependent variable:}$

$\text{Average minutes per day exercising, or }y.$

♦♦♦ Mean mark (b) 21%.

c. $\text{Form: Linear}$

$\text{Direction: Negative}$

♦♦ Mean mark (c) 45%.

$y-\text{intercept = 70}$

$\text{Gradient}=\dfrac{\text{rise}}{\text{run}}=\dfrac{-60}{60}=-1$

♦♦♦ Mean mark (d) 30%.

e. $\text{The extrapolation of the graph past 70 minutes produces}$

$\text{negative average minutes per day exercising (impossible).}$

♦♦♦♦ Mean mark (e) 14%.

Measurement, STD1 M1 2025 HSC 13

A trapezium is shown.

Write an expression for the area of this trapezium. (1 mark)

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Find the value of $a$, given that the area of the trapezium is 330 cm². (2 marks)
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a. $\text{A}=6(a+30)\ \text{or}\ A=6a+180$

b. $a=25$

Show Worked Solution

a. $A=\dfrac{h}{2}\Big(a+b\Big)=\dfrac{12}{2}\Big(a+30\Big)=6a+180$

b. $\text{Solve for}\ a\ \text{when} \ A=330:$

$6a+180$	$=330$
$6a$	$=330-180$
$6a$	$=150$
$a$	$=25$

♦♦♦ Mean mark (a) 24%.
♦♦♦ Mean mark (b) 25%.

Financial Maths, STD1 F2 2025 HSC 9 MC

An amount of $90 000 is invested at 4% per annum, compounded quarterly.

Which expression gives the value of this investment, in dollars, after 6 years?

$90\ 000(1+0.04)^6$
$90\ 000(1+0.04)^{24}$
$90\ 000(1+0.01)^6$
$90\ 000(1+0.01)^{24}$

Show Answers Only

$D$

Show Worked Solution

$PV=90\ 000,\ r=\dfrac{4\%}{4}=1\%=0.01,\ n=4\times 6=24$

$FV$	$=PV(1+r)^n$
	$=90\ 000(1+0.01)^{24}$

$\Rightarrow D$

♦♦♦ Mean mark 22%.

Measurement, STD1 M1 2025 HSC 4 MC

What is $6\ 280\ 000$ in standard form?

$628\times 10^4$
$62.8\times 10^5$
$6.28\times 10^6$
$0.628\times 10^7$

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$C$

Show Worked Solution

$6\ 280\ 000=6.28\times 10^6$

$\Rightarrow C$

♦♦♦ Mean mark 29%.

Statistics, STD1 S1 2025 HSC 1 MC

Which of the following could be classified as discrete data?

Colour of a car
Time taken to swim 200 m
Temperature of an ice block
Number of children in a class

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$D$

Show Worked Solution

$\text{Discrete data refers to individual and countable items that can be listed.}$

$\text{Consider Options:}$

$\text{A. Colour → categorical}$

$\text{B. Time → measurement → continuous}$

$\text{C. Temperature → measurement → continuous}$

$\text{D. Number of children → countable → discrete}$

$\Rightarrow D$

♦♦♦ Mean mark 29%.

Calculus, EXT1 C2 2025 HSC 10 MC

For the function $f(x)$, it is known that $f(3)=1, f^{\prime}(3)=2$ and $f^{\prime \prime}(3)=4$.

Let $g(x)=f^{-1}(x)$.

What is the value of $g^{\prime \prime}(1)$ ?

$\dfrac{1}{4}$
$-\dfrac{1}{4}$
$-\dfrac{1}{2}$
$-1$

Show Answers Only

$C$

Show Worked Solution

$f(3)=1, f^{\prime}(3)=2, f^{\prime \prime}(3)=4$

$\text{Given} \ \ g(x)=f^{-1}(x):$

$f(g(x))=x \ \ \text{(Definition of an inverse fn)}$

♦♦♦ Mean mark 22%.

$\text{Differentiate both sides:}$

$g^{\prime}(x) \cdot f^{\prime}(g(x))=1 \ \ \Rightarrow \ \ g^{\prime}(x)=\dfrac{1}{f^{\prime}(g(x))}$

$g^{\prime \prime}(x)=\dfrac{d}{d x}\left(\dfrac{1}{f^{\prime}(g(x))}\right)=-\dfrac{f^{\prime \prime}(g(x)) \cdot g^{\prime}(x)}{\left[f^{\prime}(g(x))\right]^2}$

$\text{When}\ \ x=1:$

$g^{\prime}(1)$	$=\dfrac{1}{f^{\prime}(g(1))}=\dfrac{1}{f^{\prime}(3)}=\dfrac{1}{2}$
$g^{\prime \prime}(1)$	$=-\dfrac{f^{\prime \prime}(g(1)) \cdot g^{\prime}(1)}{\left[f^{\prime}(g(1))\right]^2}=-\dfrac{f^{\prime \prime}(3) \cdot \dfrac{1}{2}}{\left[f^{\prime}(3)\right]^2}=-\dfrac{4 \times \dfrac{1}{2}}{2^2}=-\dfrac{1}{2}$

$\Rightarrow C$

Functions, EXT1 F2 2025 HSC 14e

It is given that $\tan \alpha, \tan \beta$ and $\tan \gamma$ are the three real roots of the polynomial equation $x^3+b x^2+c x-1+b+c=0$, where $b$ and $c$ are real numbers and $c \neq 1$.

Find the smallest positive value of $\alpha+\beta+\gamma$. (3 marks)

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$\alpha+\beta+\gamma=\dfrac{3 \pi}{4}$

Show Worked Solution

$x^3+b x^2+c x-1+b+c=0$

$\text{Roots:} \ \tan \alpha, \tan \beta, \tan \gamma$

$\tan \alpha+\tan \beta+\tan \gamma=-\dfrac{b}{a}=-b$

$\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma=c$

$\tan \alpha \cdot \tan \beta \cdot \tan \gamma=-\dfrac{d}{a}=1-b-c$

♦♦♦ Mean mark 26%.

$\text { Find smallest +ve value of} \ \ \alpha+\beta+\gamma:$

$\tan (\alpha+\beta+\gamma)$	$=\dfrac{\tan (\alpha+\beta)+\tan \gamma}{1-\tan (\alpha+\beta) \tan \gamma}$
	$=\dfrac{\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}+\tan \gamma}{1-\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} \times \tan \gamma}$
	$=\dfrac{\tan \alpha+\tan \beta+\tan \gamma(1-\tan \alpha \cdot \tan \beta)}{1-\tan \alpha \cdot \tan \beta-(\tan \alpha+\tan \beta) \tan \gamma}$
	$=\dfrac{\tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \cdot \tan \beta \cdot \tan \gamma}{1-(\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma)}$
	$=\dfrac{-b-(1-b-c)}{1-c}$
	$=\dfrac{-1+c}{1-c}$
	$=-1$

$\therefore \ \text{Smallest +ve value of} \ \ \alpha+\beta+\gamma=\dfrac{3 \pi}{4}$

Calculus, 2ADV C3 2025 HSC 10 MC

The graph of $y=f(x)$, with all its stationary points, is shown.

How many stationary points does the graph of $y=f\left(e^x\right)$ have?

Show Answers Only

$C$

Show Worked Solution

$y=f(e^{x})\ \ \Rightarrow\ \ y^{\prime}=e^{x} \times f(e^{x}) $

$\text{Find number of \(x$ values where}\ \ y^{\prime}=0.\)

$\text{Since}\ e^{x} \in (0, \infty)\ \text{for all}\ x: $

$\text{Stationary points of \(f(e^x)$ = 2 (SP’s of $f(x)$ for}\ x \in (0, \infty)).\)

$\Rightarrow C$

♦♦♦ Mean mark 29%.

Calculus, 2ADV C3 2025 HSC 9 MC

The diagram shows the graph of $y=f^{\prime}(x)$.

Given $f(1)=6$, which interval includes the best estimate for $f(1.1)$ ?

$[6.2,6.4)$
$[6.0,6.2)$
$[5.8,6.0)$
$[5.6,5.8)$

Show Answers Only

$B$

Show Worked Solution

$\text{Gradient of \(f(x)$ at $x=1$ is 2 (see graph).}\)

$\text{Gradient of \(f(x)$ at $x=1.1$ is slightly below 2 (see graph).}\)

$\text{As \(x$ increases 0.1 (from 1.0 to 1.1), $y$ will increase less than 0.2 units.}\)

$\therefore f(1.1) \in [6.0,6.2)$

$\Rightarrow B$

♦♦♦ Mean mark 27%.

\(100 e^{-0.2 t}-50\)	\(=0\)
\(e^{-0.2 t}\)	\(=\dfrac{1}{2}\)
\(-0.2 t\)	\(=-\ln 2\)
\(t\)	\(=5 \ln 2\)

\(e^{-0.2 t}\)	\(=2-\sqrt{3}\)
\(-0.2 t\)	\(=\ln (2-\sqrt{3})\)
\(t\)	\(=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s (1 d.p.)}\)

\(S\)	\(= V_0-Dn\)
	\(= 36\,000-4800 \times 5\)
	\(= \$12\,000\)

\(S\)	\(< 10\,000\)
\(36\,000-4800n\)	\(< 10\,000\)
\(26\,000\)	\(< 4800n\)
\(n\)	\(> \dfrac{26\,000}{4800}= 5.4166…\)

\(y-(\sin (p)+1)\)	\(=\cos (p)(x-p)\)
\(t(x)\)	\(=\cos (p)(x-p)+\sin (p)+1\)

d.	\(\dfrac{d Q}{d t}\)	\(=\dfrac{300-Q}{150}\)
	\(\dfrac{d t}{d Q}\)	\(=\dfrac{150}{300-Q}\)
	\(\displaystyle \int d t\)	\(=\displaystyle \int \frac{150}{300-Q} d Q\)
	\( t\)	\(=-150\, \log _e(300-Q)+c\)

\(x^2+2 x-3\)	\(=-3\)
\(x(x+2)\)	\(=0\)
\(x=0 \ \ \text{or} \ -2\)

\(x^2+2 x-3\)	\(=-3\)
\(x(x+2)\)	\(=0\)
\(x=0 \ \ \text{or} \ -2\)

\(y\)	\(=\left(\dfrac{w+1}{2}-1\right)\left(\dfrac{w+1}{2}-w\right)\)
	\(=\left(\dfrac{w-1}{2}\right)\left(\dfrac{1-w}{2}\right)\)
	\(=-\dfrac{1}{4}(w-1)^2\)