Band 3

HMS, HIC 2022 HSC 2 MC

To reduce the number of young people smoking, the sale of tobacco products to people under 18 years of age was made illegal.

Which action area of the Ottawa Charter is this strategy an example of?

Developing personal skills
Reorienting health services
Building healthy public policy
Strengthening community action

Show Answers Only

$ C$

Show Worked Solution

$\text{Consider option C: Building healthy public policy}$

$\rightarrow\ \text{A legislative measure}$

$\rightarrow\ \text{Implemented through government policy}$

$\rightarrow\ \text{Creates systemic change through regulation}$

$\rightarrow\ \text{Uses laws to promote health }$

$\rightarrow\ \text{Enforced at a governmental level not through individual }$

$\text{skills, healthcare services, or community action}$

$\Rightarrow C$

HMS, BM 2022 HSC 7 MC

Which of the following is an example of an athlete using negative, intrinsic motivation?

Receiving criticism from their coach to help them improve
Ignoring social media comments from club supporters after a loss
Seeking to avoid repeating the feeling of despair following a defeat
Training harder to enhance their chances of selection in a representative team

Show Answers Only

$C$

Show Worked Solution

$\text{Consider option C}$

$\text{Seeking to avoid repeating the feeling of despair following a defeat}$

$\text{This is intrinsic (manifested by an internal feeling of despair)}$

$\text{and negative (avoiding an unpleasant emotional state)}$

$\Rightarrow C$

An Olympic weightlifter continues to train despite a minor injury because they are afraid of losing their sponsorship deals and disappointing their parents who have invested significant money in their career. This athlete's motivation is primarily:

Negative and intrinsic
Negative and extrinsic
Positive and intrinsic
Positive and extrinsic

Show Answers Only

$B$

Show Worked Solution

$\text{The athlete demonstrates:}$

$\text{Extrinsic motivation}$
$\rightarrow\ \text{driven by external factors: sponsorship deals and parental pressure}$

$\text{Negative motivation}$
$\rightarrow\ \text{fear of loss and disappointment, rather than positive goal achievement}$

$\Rightarrow B$

HMS, BM EQ-Bank 1 MC

A swimmer trains consistently at 5 a.m. every morning, setting personal goals to improve their technique and times, with the aim of qualifying for the Olympic team. This athlete's motivation is primarily:

Negative and extrinsic
Positive and extrinsic
Negative and intrinsic
Positive and intrinsic

Show Answers Only

$D$

Show Worked Solution

$\text{The athlete demonstrates:}$

$\text{Intrinsic motivation}$
$\rightarrow\ \text{self-driven, personal goal setting, focus on self-improvement}$

$\text{Positive motivation}$
$\rightarrow\text{consistent commitment to training, constructive goal-setting approach}$

$\Rightarrow D$

PHYSICS, M5 2017 HSC 22a

A torque is applied to a nut, using a wrench, as shown.

Suggest TWO ways that the applied torque could be increased. (2 marks)

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The applied torque can be increased by:

→ Increasing the magnitude of the applied force.

→ Increasing the distance between the nut and the point of applied force.

Show Worked Solution

The applied torque can be increased by:

→ Increasing the magnitude of the applied force.

→ Increasing the distance between the nut and the point of applied force.

ENGINEERING, CS 2024 HSC 27a

A towbar hitch and shear pin is shown.

The hitch is exposed to various environmental conditions.

Outline a method to protect against corrosion. (2 marks)

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→ Applying protective finishes like paint or powder coating provides a barrier that shields the component from corrosive elements.

→ This resistance is a result of preventing the hitch from exposure to oxygen, moisture, and environmental contaminants.

Show Worked Solution

→ Applying protective finishes like paint or powder coating provides a barrier that shields the component from corrosive elements.

→ This resistance is a result of preventing the hitch from exposure to oxygen, moisture, and environmental contaminants.

ENGINEERING, TE 2024 HSC 26a

Outline ways in which satellite services have benefited people living and working in rural areas. (2 marks)

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→ Satellite technology enables communication in remote regions beyond traditional cellular coverage.

→ Because satellites operate independently of ground-based infrastructure, they can deliver services across vast territories without the distance limitations of cellular networks.

ENGINEERING, AE 2024 HSC 25a

Outline how 'Black Box' flight data recorders have led to improved aviation safety. (2 marks)

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→ Flight data analysis allows engineers to investigate the causes of flight incidents.

→ This enables them to identify necessary safety improvements and implement preventive measures based on their findings.

Show Worked Solution

→ Flight data analysis allows engineers to investigate the causes of flight incidents.

→ This enables them to identify necessary safety improvements and implement preventive measures based on their findings.

ENGINEERING, TE 2024 HSC 23c

An orthogonal drawing of a wi-fi router is shown.

Construct a freehand pictorial sketch of the wi-fi router as viewed in the direction of the arrow. (3 marks)

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\begin{array} {|l|}
\hline
\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\ \\
\hline
\end{array}

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Answers could include:

→ Oblique view in the direction of the arrow

→ Perspective view in the direction of the arrow

→ Planometric view in the direction of the arrow.

Show Worked Solution

Answers could include:

→ Oblique view in the direction of the arrow

→ Perspective view in the direction of the arrow

→ Planometric view in the direction of the arrow.

ENGINEERING, TE 2024 HSC 23b

Telecommunications engineers face many work health and safety risks while carrying out their duties.

With reference to ONE issue, describe strategies that can be implemented to reduce the risk for the engineer. (3 marks)

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→ Telecommunications engineers work on satellite dishes and other structures at height.

→ To minimise the risk, they must ensure that they wear safety harnesses that prevent injuries from falling from heights.

→ The engineer should have completed relevant safety training before undertaking the task so that they are aware of the possible risks and how to avoid them.

ENGINEERING, AE 2024 HSC 22c

Discuss the use of computer-aided drawing (CAD) in aeronautical engineering. (3 marks)

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→ CAD enables precise geometrical modelling that can be translated into physical objects through methods like 3D printing and other more traditional manufacturing processes.

→ CAD can be used to analyse aircraft component performance under various conditions, including bending stress of airframes.

→ Simulation results refine CAD designs, ensuring compliance with safety and performance standards.

→ While CAD provides accurate specifications for computer-controlled machining, it has drawbacks – it requires significant investment in both training time to develop expertise and the substantial cost of implementing CAD systems.

Show Worked Solution

→ CAD enables precise geometrical modelling that can be translated into physical objects through methods like 3D printing and other more traditional manufacturing processes.

→ CAD can be used to analyse aircraft component performance under various conditions, including bending stress of airframes.

→ Simulation results refine CAD designs, ensuring compliance with safety and performance standards.

ENGINEERING, AE 2024 HSC 22a

Outline the environmental responsibilities of an aeronautical engineer. (2 marks)

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Answers could include two of the following:

→ Engineers have a professional obligation to operate within environmental laws and regulations.

→ Engineers should design and implement proper waste management protocols to minimise environmental harm.

→ Ensuring aircraft components are designed, tested and certified to meet environmental requirements, particularly regarding noise emission standards.

Show Worked Solution

Answers could include two of the following:

→ Engineers have a professional obligation to operate within environmental laws and regulations.

→ Engineers should design and implement proper waste management protocols to minimise environmental harm.

→ Ensuring aircraft components are designed, tested and certified to meet environmental requirements, particularly regarding noise emission standards.

ENGINEERING, CS 2024 HSC 21a

Two piers from different eras, used to support civil structures, are shown.

Outline how innovations in engineering materials have improved the inservice properties of piers. (2 marks)

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Improvements in pier properties :

→ Concrete better than timber at resisting insect attack

→ Concrete less susceptible to environmental conditions.

→ Concrete properties include increased strength and durability

→ Less maintenance with concrete blocks

Show Worked Solution

Improvements in pier properties :

→ Concrete better than timber at resisting insect attack

→ Concrete less susceptible to environmental conditions.

→ Concrete properties include increased strength and durability

→ Less maintenance with concrete blocks

ENGINEERING, PPT 2024 HSC 19 MC

Which is the correct angle of repose for rock fill, used on a highway construction, that has a coefficient of friction of 0.84?

20°
30°
40°
50°

Show Answers Only

$C$

Show Worked Solution

$\mu = 0.84$

$\tan \theta$	$=0.84$
$\theta$	$= \tan^{-1} 0.84$
	$=40^{\circ}$

$\Rightarrow C$

ENGINEERING, TE 2024 HSC 13 MC

What is the primary function of GPS satellites?

Receiving magnetic field data for accuracy of location
Receiving signals from GPS devices to determine distance
Transmitting signals that provide accurate timing and positioning data
Transmitting weather system data to ground-based stations for tracking

Show Answers Only

$C$

Show Worked Solution

→ GPS satellites continuously broadcast precisely timed radio signals containing their exact orbital position data, which GPS receivers use to calculate their location through trilateration.

$\Rightarrow C$

PHYSICS, M6 2024 HSC 28

An electron gun fires a beam of electrons at 2.0 × 10$^6$ m s$^{-1}$ through a pair of parallel charged plates towards a screen that is 30 mm from the end of the plates as shown.

There is a uniform electric field between the plates of 1.5 × 10$^4$ N C$^{-1}$. The plates are 5.0 mm wide and 20 mm apart. The electron beam enters mid-way between the plates. $X$ marks the spot on the screen where an undeflected beam would strike.

Ignore gravitational effects on the electron beam.

Show that the acceleration of an electron between the parallel plates is 2.6 × 10$^{15}$ m s$^{-2}$. (2 marks)

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Show that the vertical displacement of the electron beam at the end of the parallel plates is approximately 8.1 mm. (2 marks)

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How far from point $X$ will the electron beam strike the screen? (3 marks)

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a. $\text{Using}\ \ F=qE\ \ \text{and}\ \ F=ma:$

$a=\dfrac{qE}{m} = \dfrac{1.5 \times 10^{4} \times 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} = 2.6 \times 10^{15}\ \text{m s}^{-2} $

b. $\text{Time of beam between the plates:}$

$\text{Horizontal velocity}\ (v)\ = 2 \times 10^{6}\ \text{m s}^{-1} $

$\text{Distance to screen}\ (s)\ = 5.0\ \text{mm}\ = \dfrac{5}{1000} = 0.005\ \text{m}$

$t=\dfrac{s}{v} = \dfrac{0.005}{2 \times 10^{6}} = 2.5 \times 10^{-9}\ \text{s} $

$\text{Find vertical displacement at end of plates:}$

$s=\dfrac{1}{2} at^{2} = 0.5 \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^{2} = 0.008125\ \text{m}\ = 8.1\ \text{mm} $

c. $\text{Find vertical velocity of beam when leaving the plates:}$

$v=at=2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^{6}\ \text{m s}^{-1} $

$\text{Time for beam to hit screen:}$

$\text{Distance to screen}\ (s)\ = 30\ \text{mm}\ = \dfrac{30}{1000} = 0.03\ \text{m}$

$t=\dfrac{s}{v} = \dfrac{0.03}{2 \times 10^{6}} = 1.5 \times 10^{-8}\ \text{s} $

$\text{Vertical displacement (from end of plates):}$

$s=vt=6.5 \times 10^{6} \times 1.5 \times 10^{-8} = 0.0975\ \text{m} $

$\text{Distance from}\ X = 0.0081 + 0.0975 = 0.11\ \text{m} $

Show Worked Solution

a. $\text{Using}\ \ F=qE\ \ \text{and}\ \ F=ma:$

$a=\dfrac{qE}{m} = \dfrac{1.5 \times 10^{4} \times 1.602 \times 10^{-19}}{9.109 \times 10^{-31}} = 2.6 \times 10^{15}\ \text{m s}^{-2} $

b. $\text{Time of beam between the plates:}$

$\text{Horizontal velocity}\ (v)\ = 2 \times 10^{6}\ \text{m s}^{-1} $

$\text{Distance to screen}\ (s)\ = 5.0\ \text{mm}\ = \dfrac{5}{1000} = 0.005\ \text{m}$

$t=\dfrac{s}{v} = \dfrac{0.005}{2 \times 10^{6}} = 2.5 \times 10^{-9}\ \text{s} $

$\text{Find vertical displacement at end of plates:}$

$s=\dfrac{1}{2} at^{2} = 0.5 \times 2.6 \times 10^{15} \times (2.5 \times 10^{-9})^{2} = 0.008125\ \text{m}\ = 8.1\ \text{mm} $

c. $\text{Find vertical velocity of beam when leaving the plates:}$

$v=at=2.6 \times 10^{15} \times 2.5 \times 10^{-9} = 6.5 \times 10^{6}\ \text{m s}^{-1} $

$\text{Time for beam to hit screen:}$

$\text{Distance to screen}\ (s)\ = 30\ \text{mm}\ = \dfrac{30}{1000} = 0.03\ \text{m}$

$t=\dfrac{s}{v} = \dfrac{0.03}{2 \times 10^{6}} = 1.5 \times 10^{-8}\ \text{s} $

$\text{Vertical displacement (from end of plates):}$

$s=vt=6.5 \times 10^{6} \times 1.5 \times 10^{-8} = 0.0975\ \text{m} $

$\text{Distance from}\ X = 0.0081 + 0.0975 = 0.11\ \text{m} $

♦ Mean mark (c) 40%.

Functions, MET1 2024 VCAA 5

The function $h:[0, \infty) \rightarrow R, h(t)=\dfrac{3000}{t+1}$ models the population of a town after $t$ years.

Use the model $h(t)$ to predict the population of the town after four years. (1 mark)

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A new function, $h_1$, models a population where $h_1(0)=h(0)$ but $h_1$ decreases at half the rate of $h$ at any point in time.
State a sequence of two transformations that maps $h$ to this new model $h_1$. (2 marks)

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In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
1. Determine an approximate 95% confidence interval for the proportion of people in the town who are unhappy with the roads. Use $z=2$ for this confidence interval. (2 marks)
  
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2. A new sample of $n$ people results in the same sample proportion.
3. Find the smallest value of $n$ to achieve a standard deviation of $\dfrac{\sqrt{2}}{50}$ for the sample proportion. (1 mark)
  
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a. $600$

b. $\text{Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}$

$\text{followed by a translation of 1500 units upwards}$

ci. $\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\ \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)$

cii. $300$

Show Worked Solution

a.	$h(4)$	$=\dfrac{3000}{4+1}$
		$=600$

b.	$h(t)$	$=3000(t+1)^{-1}$
	$h^{\prime}(t)$	$=\dfrac{-3000}{(t+1)^2}$
	$h_1^{\prime}(t)$	$=\dfrac{1}{2}h^{\prime}(t)$
		$=\dfrac{-1500}{(t+1)^2}$
	$\therefore\ h_1(t)$	$=\dfrac{1500}{t+1}$

$\text{Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}$

$\text{followed by a translation of 1500 units upwards}$

ci. $\hat{p}=\dfrac{60}{100}=\dfrac{3}{5},\quad 1-\hat{p}=\dfrac{2}{5},\quad z=2$

$\text{Approx CI}$	$=\left(\dfrac{3}{5}-2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}},\ \dfrac{3}{5}+2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}}\right)$
	$=\left(\dfrac{3}{5}-\dfrac{2\sqrt{6}}{50},\quad \dfrac{3}{5}+\dfrac{2\sqrt{6}}{50}\right)$
	$=\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\quad \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)$

cii.	$\sqrt{\dfrac{\hat{P}(1-\hat{P}}{n}}$	$=\dfrac{\sqrt{2}}{50}$
	$\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}$	$=\dfrac{\sqrt{2}}{50}$
	$\dfrac{\dfrac{6}{25}}{n}$	$=\dfrac{2}{2500}$
	$n$	$=\dfrac{6}{25}\times\dfrac{2500}{2}$
		$=\dfrac{600}{2}=300$

Calculus, MET1 2024 VCAA 1

Let $y=e^x \cos (3 x)$.
Find $\dfrac{d y}{d x}$ (1 mark)

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Let $f(x)=\log _e\left(x^3-3 x+2\right)$.
Find $f^{\prime}(3)$ (2 marks)

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a. $e^x (\cos(3x)-3\sin(3x))$

b. $\dfrac{6}{5}$

Show Worked Solution

a.	$y$	$=e^x \cos(3x)$
	$\dfrac{dy}{dx}$	$=e^x.(-3\sin(3x))+\cos(3x).e^x$
		$=e^x(\cos(3x)-3\sin(3x))$

b.	$f(x)$	$=\log_{e}(x^3-3x+2)$
	$f^{\prime}(x)$	$=\dfrac{3x^2-3}{x^3-3x+2}$
	$f^{\prime}(3)$	$=\dfrac{3(3)^2-3}{(3)^3-3(3)+2}$
		$=\dfrac{24}{20}=\dfrac{6}{5}$

Calculus, MET2 2024 VCAA 3

The points shown on the chart below represent monthly online sales in Australia.

The variable $y$ represents sales in millions of dollars.

The variable $t$ represents the month when the sales were made, where $t=1$ corresponds to January 2021, $t=2$ corresponds to February 2021 and so on.

A cubic polynomial $p ;(0,12] \rightarrow R, p(t)=a t^3+b t^2+c t+d$ can be used to model monthly online sales in 2021.
The graph of $y=p(f)$ is shown as a dashed curve on the set of axes above.

It has a local minimum at (2,2500) and a local maximum at (11,4400).
1. Find, correct to two decimal places, the values of $a, b, c$ and $d$. (3 marks)

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1. Let $q:(12,24] \rightarrow R, q(t)=p(t-h)+k$ be a cubic function obtained by translating $p$, which can be used to model monthly online sales in 2022.
  Find the values of $h$ and $k$ such that the graph of $y=q(t)$ has a local maximum at $(23,4750)$. (2 marks)

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Another function $f$ can be used to model monthly online sales, where
$f:(0,36] \rightarrow R, f(t)=3000+30 t+700 \cos \left(\dfrac{\pi t}{6}\right)+400 \cos \left(\dfrac{\pi t}{3}\right)$

Part of the graph of $f$ is shown on the axes below.

1. Complete the graph of $f$ on the set of axes above until December 2023, that is, for $t \in(24,36]$.Label the endpoint at $t=36$ with its coordinates. (2 marks)
2. The function $f$ predicts that every 12 months, monthly online sales increase by $n$ million dollars.
  Find the value of $n$. (1 mark)

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1. Find the derivative $f^{\prime}(t)$ (1 mark)

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1. Hence, find the maximum instantaneous rate of change for the function $f$, correct to the nearest million dollars per month, and the values of $t$ in the interval $(0,36]$ when this maximum rate occurs, correct to one decimal place. (2 marks)

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ai. $a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18$

aii. $h=12, k=350$

bi.

bii. $ n=360$

biii. $f^{\prime}(t)=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)$

biv. $\text{Maximum rate occurs at }t=10.2, 22.2, 34.2$

$\text{Maximum rate}\ \approx 725\ \text{million/month}$

Show Worked Solution

ai. $\text{Given}\ p(2)=2500, p(11)=4400, p^{\prime}(2)=0\ \text{and}\ p^{\prime}(11)=0$

$p(t)=at^3+bt^2+ct+d\ \Rightarrow\ p^{\prime}(t)=3at^2+2bt+c$

$\text{Using CAS:}$

$\text{Solve}
\begin{cases}
8a+4b+2c+d=2500 \\
1331a+121b+11c+d=4400 \\
12a+4b+c=0 \\
363a+22b+c=0
\end{cases}$

$a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18$

aii. $\text{Local maximim }p(t)\ \text{is}\ (11, 4400)$

$\therefore\ h$	$=23-11=12$
$k$	$=4750-4400=350$

bi. $\text{Plotting points from CAS:}$

$(24,4820), (26, 3930), (28, 3290), (30, 3600), (32, 3410), (34, 4170), (36, 5180)$

bii. $\text{Using CAS: }$

$f(12)-f(0)$	$=4460-4100=360$
$f(24)-f(12)$	$=4820-4460=360$
$f(36)-f(24)$	$=5180-4820=360$

$\therefore\ n=360$

biii.	$f(t)$	$=3000+30t+700\cos\left(\dfrac{\pi t}{6}\right)+400\cos\left(\dfrac{\pi t}{3}\right)$
	$f^{\prime}(t)$	$=30-\dfrac{700\pi}{6}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)$
		$=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)$

biv. $\text{Max instantaneous rate of change occurs when }f^{\prime\prime}(t)=0$

$\text{Maximum rate occurs at }t=10.2, 22.2, 34.2$

$\text{Maximum rate using CAS:}$

$f{\prime}(10.2)=f{\prime}(22.2)=f{\prime}(34.2)=725.396\approx 725\ \text{million/month}$

Calculus, MET2 2024 VCAA 2

A model for the temperature in a room, in degrees Celsius, is given by

$f(t)=\left\{
\begin{array}{cc}12+30 t & \quad \quad 0 \leq t \leq \dfrac{1}{3} \\
22 & t>\dfrac{1}{3}
\end{array}\right.$

where $t$ represents time in hours after a heater is switched on.

Express the derivative $f^{\prime}(t)$ as a hybrid function. (2 marks)

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Find the average rate of change in temperature predicted by the model between $t=0$ and $t=\dfrac{1}{2}$.
Give your answer in degrees Celsius per hour. (1 mark)

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Another model for the temperature in the room is given by $g(t)=22-10 e^{-6 t}, t \geq 0$.
1. Find the derivative $g^{\prime}(t)$. (1 mark)

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1. Find the value of $t$ for which $g^{\prime}(t)=10$.
  Give your answer correct to three decimal places. (1 mark)

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Find the time $t \in(0,1)$ when the temperatures predicted by the models $f$ and $g$ are equal.
Give your answer correct to two decimal places. (1 mark)

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Find the time $t \in(0,1)$ when the difference between the temperatures predicted by the two models is the greatest.
Give your answer correct to two decimal places. (1 mark)

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The amount of power, in kilowatts, used by the heater $t$ hours after it is switched on, can be modelled by the continuous function $p$, whose graph is shown below.

$p(t)=\left\{
\begin{array}{cl}1.5 & 0 \leq t \leq 0.4 \\
0.3+A e^{-10 t} & t>0.4
\end{array}\right.$

The amount of energy used by the heater, in kilowatt hours, can be estimated by evaluating the area between the graph of $y=p(t)$ and the $t$-axis.

1. Given that $p(t)$ is continuous for $t \geq 0$, show that $A=1.2 e^4$. (1 mark)

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1. Find how long it takes, after the heater is switched on, until the heater has used 0.5 kilowatt hours of energy.
  
  Give your answer in hours. (1 mark)

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1. Find how long it takes, after the heater is switched on, until the heater has used 1 kilowatt hour of energy.
  
  Give your answer in hours, correct to two decimal places. (2 marks)

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Show Answers Only

a. $f^{\prime}(t)=\left\{
\begin{array}{cc}30 & \quad \quad 0 \leq t <\dfrac{1}{3} \\
0 & t>\dfrac{1}{3}
\end{array}\right.$

b. $20^{\circ}\text{C/h}$

ci. $g^{\prime}(t)=60e^{-6t}$

cii. $0.299\ \text{(3 d.p.)}$

d. $0.27\ \text{(2 d.p.)}$

e. $0.12\ \text{(2 d.p.)}$

fi. $\text{Because function is continuous}$

$0.3+Ae^{-10t}$	$=1.5$
$Ae^{-10\times 0.4}$	$=1.2$
$A$	$=\dfrac{1.2}{e^{-10\times 0.4}}$
	$=1.2e^4$

fii. $\dfrac{1}{3}\ \text{hours}$

fiii. $1.33\ \text{(2 d.p.)}$

Show Worked Solution

a. $f^{\prime}(t)=\left\{
\begin{array}{cc}30 & \quad \quad 0 \leq t < \dfrac{1}{3} \\
0 & t>\dfrac{1}{3}
\end{array}\right.$

b. $\text{When }\ t=0, f(t)=12\ \ \text{and when }\ t=\dfrac{1}{2}, f(t)=22$

$\therefore\ \text{Average rate of change}$	$=\dfrac{22-12}{\frac{1}{2}}$
	$=20^{\circ}\text{C/h}$

ci.	$g(t)$	$=22-10 e^{-6 t}$
	$g^{\prime}(t)$	$=60e^{-6t},\ \ t\geq 0$

cii.	$60e^{-6t}$	$=10$
	$-6t\,\ln{e}$	$=\ln{\dfrac{1}{6}}$
	$t$	$=\dfrac{\ln{\frac{1}{6}}}{-6}$
		$=0.2986…\approx 0.299\ \text{(3 d.p.)}$

$\left\{
\begin{array}{cc}12+30 t & \ \ 0 \leq t \leq \dfrac{1}{3} \\
22 & t>\dfrac{1}{3}
\end{array}\right.$

$=22-10e^{-6t}$

$\text{Using CAS:}$

$\text{Temps equal when}\ t\approx 0.27\ \text{(2 d.p.)}$

e.	$\text{Difference }(D)$	$=\|g(t)-f(t)\|$
		$=\left(22-10e^{-6t}\right)-(12+30t)$
	$\dfrac{dD}{dt}$	$=60e^{-6t}-30$

$\text{Max time diff when}\ \dfrac{dD}{dt}=0$

$\therefore\ 60e^{-6t}-30$	$=0$
$e^{-6t}$	$=0.5$
$-6t$	$=\ln{0.5}$
$t$	$=0.1155\dots$
	$\approx 0.12\ \text{(2 d.p.)}$

fi. $\text{Because function is continuous}$

$0.3+Ae^{-10t}$	$=1.5$
$Ae^{-10\times 0.4}$	$=1.2$
$A$	$=\dfrac{1.2}{e^{-10\times 0.4}}$
	$=1.2e^4$

fii. $\text{Using CAS:}$

$\text{Or, considering the graph, the area from 0 to 0.4 }=0.6\ \rightarrow t<0.4$

$\therefore\ \text{Solving}\ 1.5t=0.5\ \rightarrow\ t=\dfrac{1}{3}$

$\therefore\ \text{It takes }\dfrac{1}{3}\ \text{hours for heater to use 0.5 kilowatts.}$

fiii. $\text{Using CAS:}$

$1.5\times 4+\displaystyle\int_{0.4}^{t}0.3+1.2e^4.e^{-10t}dt$	$=1$
$\Bigg[0.3t+0.12e^4.e^{-10t}\Bigg]_{0.4}^{t}$	$=0.4$
$t$	$=1.3333\dots$
	$\approx 1.33\ \text{(2 d.p.)}$

Functions, MET2 2024 VCAA 1

Consider the function $ f: R \rightarrow R, f(x)=(x+1)(x+a)(x-2)(x-2 a) \text { where } a \in R \text {. } $

State, in terms of $a$ where required, the values of $x$ for which $f(x)=0$. (1 mark)

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Find the values of $a$ for which the graph of $y=f(x)$ has
1. exactly three $x$-intercepts. (2 marks)

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1. exactly four $x$-intercepts. (1 mark)

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Let $g$ be the function $g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2$, which is the function $f$ where $a=1$.
1. Find $g^{\prime}(x)$ (1 mark)

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1. Find the coordinates of the local maximum of $g$. (1 mark)

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1. Find the values of $x$ for which $g^{\prime}(x)>0$. (1 mark)

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1. Consider the two tangent lines to the graph of $y=g(x)$ at the points where
  
  $x=\dfrac{-\sqrt{3}+1}{2}$ and $x=\dfrac{\sqrt{3}+1}{2}$.
  
  Determine the coordinates of the point of intersection of these two tangent lines. (2 marks)

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Let $g$ remain as the function $g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2$, which is the function $f$ where $a=1$.

Let $h$ be the function $h: R \rightarrow R, h(x)=(x+1)(x-1)(x+2)(x-2)$, which is the function $f$ where $a=-1$.
1. Using translations only, describe a sequence of transformations of $h$, for which its image would have a local maximum at the same coordinates as that of $g$. (1 mark)

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1. Using a dilation and translations, describe a different sequence of transformations of $h$, for which its image would have both local minimums at the same coordinates as that of $g$. (2 marks)

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Show Answers Only

a. $x=-1, x=a, x=2, x=2a$

bi. $a=0, -2, -\dfrac{1}{2}$

bii. $R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}$

ci. $g^{\prime}(x)=2(x-2)(x+1)(2x-1)$

cii. $\left(\dfrac{1}{2} , \dfrac{81}{16}\right)$

ciii. $x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)$

civ. $\left(\dfrac{1}{2}, \dfrac{27}{4}\right)$

di. $\text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}$

dii. $\text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }$

$\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units}$

Show Worked Solution

a. $x=-1, x=a, x=2, x=2a$

bi. $a=0, -2, -\dfrac{1}{2}$

bii. $\text{The solution must be all }R\ \text{except those that give 3 or less solutions.}$

$\therefore\ R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}$

ci.	$g^{\prime}(x)$	$=2(2-x)(x+1)^2+(x-2)^22(x+1)$
		$=2(x-2)(x+1)(x+1+x+2)$
		$=2(x-2)(x+1)(2x-1)$

cii. $\text{When }g^{\prime}(x)=0, x=2, -1, \dfrac{1}{2}$

$\text{From graph local maximum occurs when }x=\dfrac{1}{2}$

$g\left(\dfrac{1}{2}\right)$	$=\left(\dfrac{1}{2}+1\right)^2\left(\dfrac{1}{2}-2\right)^2$
	$=\dfrac{9}{4}\times \dfrac{9}{4}=\dfrac{81}{16}$

$\therefore\ \text{Local maximum at}\ \left(\dfrac{1}{2} , \dfrac{81}{16}\right)$

ciii. $\text{From graph}\ g^{\prime}(x)>0\ \text{when }x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)$

civ. $\text{Use CAS to find tangent lines and solve to find intersection.}$

$\text{Point of intersection of tangent lines}\ \left(\dfrac{1}{2}, \dfrac{27}{4}\right)$

di. $\text{Local maximum of }g(x)\ \rightarrow\left(\dfrac{1}{2}, \dfrac{81}{16}\right)$

$\text{From CAS local maximum of }h(x)\ \rightarrow \left(0, 4\right)$

$\therefore\ \text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}$

dii. $\text{Using CAS to solve }g^{\prime}(x)=0\ \text{and }h^{\prime}(x)=0$

$\text{Local Minimums for }g(x)\ \text{at }(-1, 0)\ \text{and }(2, 0)\ \text{which are 3 apart.}$

$\text{Local minimums for at }h(x)\ \text{at }\left(-\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\ \text{and }\left(\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)$

$\therefore\ \text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }$

$\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units.}$

Functions, MET2 2024 VCAA 8 MC

Some values of the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \ x \quad \rule[-1ex]{0pt}{0pt} & \quad \ 1 \quad \rule[-1ex]{0pt}{0pt} & \quad \ 2 \quad \rule[-1ex]{0pt}{0pt} & \quad \ 3 \quad \\
\hline
\rule{0pt}{2.5ex} \quad f(x) \quad \rule[-1ex]{0pt}{0pt} & \quad \ 0 \quad \rule[-1ex]{0pt}{0pt} & \quad 4 \quad \rule[-1ex]{0pt}{0pt} & \quad 5 \quad \\
\hline
\rule{0pt}{2.5ex} \quad g(x) \quad \rule[-1ex]{0pt}{0pt} & \quad 3 \quad \rule[-1ex]{0pt}{0pt} & \quad 4 \quad \rule[-1ex]{0pt}{0pt} & \quad -5 \quad \\
\hline
\end{array}

The graph of the function $h(x)=f(x)-g(x)$ must have an $x$-intercept at

$(2,0)$
$(3,0)$
$(4,0)$
$(5,0)$

Show Answers Only

$A$

Show Worked Solution

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt} & \quad 1 \quad \rule[-1ex]{0pt}{0pt} & \quad 2 \quad \rule[-1ex]{0pt}{0pt} & \quad 3 \quad \\
\hline
\rule{0pt}{2.5ex} h(x) \rule[-1ex]{0pt}{0pt} & \ \ -3 \ \ \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ \rule[-1ex]{0pt}{0pt} & \ \ 10 \ \ \\
\hline
\end{array}

$\therefore\ x\text{-intercept}\ \rightarrow\ (2, 0)$

$\Rightarrow A$

Calculus, MET2 2024 VCAA 4 MC

If $ { \displaystyle \int_a^b f(x) d x=-5 } $ and $ { \displaystyle \int_a^c f(x) d x=3 } $, where $a<b<c$, then $ { \displaystyle \int_b^c 2 f(x) d x } $ is equal to

$-16$
$16$
$-4$
$4$

Show Answers Only

$B$

Show Worked Solution

${ \displaystyle \int_a^b f(x) d x} +{ \displaystyle \int_b^c f(x) d x}={ \displaystyle \int_a^c f(x) d x} $

$-5+{ \displaystyle \int_b^c f(x) d x}=3$

${ \displaystyle \int_b^c f(x) d x}=3+5=8$

$\therefore{ \displaystyle \int_b^c 2f(x) d x}=16$

$\Rightarrow B$

Calculus, MET2 2024 VCAA 2 MC

A function $g: R \rightarrow R$ has the derivative $ { \displaystyle g^{\prime}(x)=x^3-x } $.

Given that $g(0)=5$, the value of $g(2)$ is

Show Answers Only

$D$

Show Worked Solution

${ \displaystyle g^{\prime}(x)}$	$=x^3-x$
$g(x)$	$=\dfrac{x^4}{4}-\dfrac{x^2}{2}+c$

$\text{Given }g(0)=5,\ c=5$

$\therefore\ g(x)=\dfrac{x^4}{4}-\dfrac{x^2}{2}+5$

$\text{Then }\ g(2)=\dfrac{2^4}{4}-\dfrac{2^2}{2}+5=7$

$\Rightarrow D$

Graphs, MET2 2024 VCAA 1 MC

The asymptote(s) of the graph of $y=\log _e(x+1)-3$ are

$x=-1$ only
$x=1$ only
$y=-3$ only
$x=-1$ and $y=-3$

Show Answers Only

$A$

Show Worked Solution

$\text{Asymptotes occur when }x+1=0$

$\therefore\ \text{Asymptote at }x=-1$

$\Rightarrow A$

PHYSICS, M6 2024 HSC 21

To tighten a nut, a force of 75 N is applied to a spanner at an angle, as shown.

Calculate the magnitude of the torque produced by the applied force. (2 marks)

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Explain TWO ways in which torque can be increased in a simple DC motor. (4 marks)

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Show Answers Only

a. $8.7\ \text{Nm (2 sig.fig)}$

b. Torque $(\tau=nIAB)$ in a simple DC motor can be increased as follows:

→ Increasing the current through the motor by increasing the voltage of the power source. This will increase the force on each arm of the motor due to the motor effect and increase the overall torque of the motor.

→ Increasing the number of coils in the DC motor. This will increase the torque as each individual coil will experience a force due to the motor effect. The forces on each coil will add together to increase the overall force on each arm of the motor increasing the torque of the motor.

→ Other answers could have included: increasing the area of the coil, increasing the strength of the external magnetic field by using electromagnets or adding a radial magnetic field as it increases the average torque applied to the motor.

Show Worked Solution

a.	$\tau$	$=Fr\sin\theta$
		$=75 \times 0.18 \times \sin 40$
		$=8.7\ \text{Nm (2 sig.fig)}$

b. Torque $(\tau=nIAB)$ in a simple DC motor can be increased as follows:

PHYSICS, M7 2024 HSC 6 MC

The photoelectric effect is mathematically modelled by the following relationship:

$K_{\max }=h f-\phi$

In this model, the symbol $\phi$ represents the amount of energy

supplied by a photon to an electron.
retained by an electron after being hit.
required to release an electron from a material.
left over after a collision of a photon with an electron.

Show Answers Only

$C$

Show Worked Solution

→ $\phi$ is the symbol for the work function of the metal.

→ The work function of a metal is defined as the minimum energy require for an electron to break free from the metal.

$\Rightarrow C$

PHYSICS, M8 2024 HSC 5 MC

A star cluster is a group of stars that form at the same time. Hertzsprung-Russell diagrams for three star clusters, $X, Y$ and $Z$ are shown.

Which row of the table correctly shows the three star clusters from youngest to oldest?

Show Answers Only

$A$

Show Worked Solution

→ Most stars begin on the main sequence when they begin to fuse hydrogen to helium in their core. As the stars get older and run out of hydrogen to fuse they will begin to fuse heavier elements and move off the main sequence. When stars stop fusing elements they become white dwarf stars and move to the bottom left side of the Hertzsprung-Russell diagram.

→ Hotter stars with a higher luminosity will move off the main sequence before cooler and less luminous stars.

→ All stars in cluster $Y$ are on the main sequence, some stars in cluster $X$ and $Z$ have moved off the main sequence to fuse heavier elements and some stars in cluster $Z$ have also stopped fusing elements and turned into white dwarf stars.

→ Cluster $Y$ is the youngest, cluster $X$ is in the middle and cluster $Z$ is the oldest.

$\Rightarrow A$

PHYSICS, M7 2024 HSC 2 MC

Which of the following provides evidence for the model of light proposed by Huygens?

Emission spectra
Diffraction of light
Black body radiation
The photoelectric effect

Show Answers Only

$B$

Show Worked Solution

→ Huygens’s proposed that light was a wave and light waves spread out in all directions as spherical wave fronts from a point source.

→ Therefore the diffraction of light provided evidence for Huygens model of light as diffraction is a wave phenomena.

$\Rightarrow B$

PHYSICS, M5 2024 HSC 1 MC

The diagram shows an object, $P$, undergoing uniform circular motion.

Which arrow shows the direction of the net force acting on $P$ ?

$W$
$X$
$Y$
$Z$

Show Answers Only

$C$

Show Worked Solution

→ During uniform circular motion, the direction of the net force acting on an object will always be towards the centre of the circle.

$\Rightarrow C$

CHEMISTRY, M2 EQ-Bank 3

Balance the following chemical equations:

$\ce{HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$ (1 mark)
$\ce{CuSO4(aq) + AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$ (1 mark)

Show Answers Only

a. $\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$

b. $\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$

Show Worked Solution

a. $\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$

b. $\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$

BIOLOGY, M5 2024 HSC 28

Cystic fibrosis is an inherited disorder that causes damage to the lungs, digestive system and other organs in the body. A person with cystic fibrosis will have two faulty recessive alleles for the cystic fibrosis gene (CFTR) on chromosome 7.

Two healthy parents, heterozygous for cystic fibrosis, have a child that does not have cystic fibrosis. They are planning to have a second child.

Using a Punnett square, determine the probability of their second child being born with the condition. Use $R$ for the normal CFTR allele, and $r$ for the faulty CFTR allele. (3 marks)

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The defect that creates the faulty CFTR allele is often caused by the deletion of three nucleotides. The following diagram illustrates a small section of the CFTR gene and the corresponding amino acid sequence of the CFTR protein.

The following codon chart displays all the codons and the corresponding amino acids. The chart translates mRNA sequences into amino acids.

Explain how the deletion of nucleotides in the CFTR gene removes only one amino acid. Include reference to the nucleotides that code for isoleucine and phenylalanine amino acids. (4 marks)

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

Show Answers Only

a. Punnett square:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \boldsymbol{R} & \boldsymbol{r} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{R} \rule[-1ex]{0pt}{0pt} & \textit{RR} & \textit{Rr} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{r}\rule[-1ex]{0pt}{0pt} & \textit{Rr} & \textit{rr} \\
\hline
\end{array}

Probability of 2nd child having cystic fibrosis = 25%.

b. Deletion of nucleotides in CFTR gene:

→ Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.

→ At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.

→ In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.

→ Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

Show Worked Solution

a. Punnett square:

Probability of 2nd child having cystic fibrosis = 25%.

b. Deletion of nucleotides in CFTR gene:

→ Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.

→ At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.

→ In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.

→ Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

♦ Mean mark (b) 56%.

BIOLOGY, M6 2024 HSC 25

One-Eyed Jack was a rescue dog that had been injured and lost an eye before his owner adopted him. One-Eyed Jack was cloned and the clone was born with two eyes.
Explain why the cloned dog was born with two eyes. (2 marks)

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Describe how animals like dogs can be cloned. (4 marks)

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

Show Answers Only

a. Reason(s) the cloned dog had two eyes:

→ The DNA used for cloning came from One-Eyed Jack’s somatic (body) cells which contained the complete genetic code for normal eye development.

→ The physical injury that caused Jack to lose his eye did not alter his genes, so it wasn’t passed on to the clone.

b. Animal cloning process:

→ The cloning process begins by removing the nucleus from a host egg cell.

→ This is replaced with the nucleus from a body cell of the animal to be cloned.

→ This produces a zygote.

→ After electrical stimulation causes the reconstructed cell to begin dividing, it is implanted into a surrogate mother’s uterus.

→ It then develops into a cloned offspring.

Show Worked Solution

a. Reason(s) the cloned dog had two eyes:

→ The DNA used for cloning came from One-Eyed Jack’s somatic (body) cells which contained the complete genetic code for normal eye development.

→ The physical injury that caused Jack to lose his eye did not alter his genes, so it wasn’t passed on to the clone.

b. Animal cloning process:

→ The cloning process begins by removing the nucleus from a host egg cell.

→ This is replaced with the nucleus from a body cell of the animal to be cloned.

→ This produces a zygote.

→ After electrical stimulation causes the reconstructed cell to begin dividing, it is implanted into a surrogate mother’s uterus.

→ It then develops into a cloned offspring.

♦ Mean mark (b) 52%.

BIOLOGY, M8 2024 HSC 24

Outline the cause of a disease due to environmental exposure. (2 marks)

--- 3 WORK AREA LINES (style=lined) ---
Explain how an educational program or campaign can be used to decrease the incidence of a disease caused by environmental exposure. (3 marks)

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

Show Answers Only

a. Answers could include one of the following:

→ Lung cancer can be caused by exposure to inhalation of smoke from cigarettes.

→ Environmental exposure to asbestos can result in some microscopic fibres becoming lodged in the lining of a person’s lungs and cause mesothelioma (cancer).

b. Educational program on the dangers of UV radiation:

→ An educational program in schools could teach students about the link between UV exposure and skin cancer, emphasising protection through hats and sunscreen.

→ This preventative approach helps students understand the risks and increases the chances of them modifying their behaviour to be more sun safe.

→ Overall, the program should contribute to less DNA damage in cells of participants and a reduction in their chances of developing skin cancer.

Show Worked Solution

a. Answers could include one of the following:

→ Lung cancer can be caused by exposure to inhalation of smoke from cigarettes.

→ Environmental exposure to asbestos can result in some microscopic fibres becoming lodged in the lining of a person’s lungs and cause mesothelioma (cancer).

b. Educational program on the dangers of UV radiation:

→ An educational program in schools could teach students about the link between UV exposure and skin cancer, emphasising protection through hats and sunscreen.

→ This preventative approach helps students understand the risks and increases the chances of them modifying their behaviour to be more sun safe.

→ Overall, the program should contribute to less DNA damage in cells of participants and a reduction in their chances of developing skin cancer.

BIOLOGY, M7 2024 HSC 22

A student designed and conducted a practical investigation to test for the presence of microbes in water and food samples.

Justify a safety precaution required to prevent infection when conducting the investigation. (2 marks)

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

Explain how the student could ensure the reliability of the investigation. (2 marks)

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

Show Answers Only

a. Answers could include one of the following:

→ Wearing gloves when handling samples is essential to stop pathogens in the water/food entering through cuts or abrasions on bare hands, potentially causing infection.

→ Disinfecting all equipment after use is necessary because microbes can multiply rapidly and contaminated materials could spread infectious organisms to others in the laboratory.

b. To ensure the reliability of the investigation:

→ Student should prepare multiple food/water samples for testing to ensure consistent results across repeated trials.

→ This process allows them to verify if their findings are reproducible and reliable rather than due to chance.

Show Worked Solution

a. Answers could include one of the following:

→ Wearing gloves when handling samples is essential to stop pathogens in the water/food entering through cuts or abrasions on bare hands, potentially causing infection.

→ Disinfecting all equipment after use is necessary because microbes can multiply rapidly and contaminated materials could spread infectious organisms to others in the laboratory.

b. To ensure the reliability of the investigation:

→ Student should prepare multiple food/water samples for testing to ensure consistent results across repeated trials.

→ This process allows them to verify if their findings are reproducible and reliable rather than due to chance.

BIOLOGY, M5 2024 HSC 21

A diagram of the different parts of a flower is shown.

Identify the structures where pollen and ovules are located. (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Complete the table to compare features of sexual and asexual reproduction. (3 marks)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Feature} \rule[-1ex]{0pt}{0pt} & \quad \quad \text{Sexual}\quad \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \text{Asexual}\quad \quad\\
\hline
\rule{0pt}{2.5ex} \text{Genetic variability} \\
\text{(yes/no)} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \text{Number of parents } \\
\text{required} \rule[-1.5 ex]{0pt}{0pt}\\
\hline
\rule{0pt}{2.5ex} \text{Example of an organism }\\
\text{which uses this type of }& \text{}\\
\text{reproduction} \rule[-1ex]{0pt}{0pt}\\
\hline
\end{array}

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

Show Answers Only

a. Pollen: anther

Ovule: ovary

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Feature} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \ \text{Sexual}\ \ \ \ \ \ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \ \ \text{Asexual}\ \ \ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} \text{Genetic variability} \rule[-1ex]{0pt}{0pt} & \text{Yes} \rule[-1ex]{0pt}{0pt} & \text{No} \\
\rule{0pt}{2.5ex} \text{(yes/no)} \rule[-1ex]{0pt}{0pt} & \text{} \rule[-1ex]{0pt}{0pt} & \text{} \\
\hline
\rule{0pt}{2.5ex} \text{Number of parents } \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{1} \\
\rule{0pt}{2.5ex} \text{required} \rule[-1ex]{0pt}{0pt} & \text{} \rule[-1ex]{0pt}{0pt} & \text{} \\
\hline
\rule{0pt}{2.5ex} \text{Example of an organism } \rule[-1ex]{0pt}{0pt} & \text{} \rule[-1ex]{0pt}{0pt} & \text{} \\
\rule{0pt}{2.5ex} \text{which uses this type of } \rule[-1ex]{0pt}{0pt} & \text{Humans} \rule[-1ex]{0pt}{0pt} & \text{Bacteria} \\
\rule{0pt}{2.5ex} \text{reproduction} \rule[-1ex]{0pt}{0pt} & \text{} \rule[-1ex]{0pt}{0pt} & \text{} \\
\hline
\end{array}

Show Worked Solution

a. Pollen: anther

Ovule: ovary

CHEMISTRY, M7 2024 HSC 24

The boiling points for two series of compounds are listed.

Plot the boiling points for each series of compounds against the number of carbon atoms per molecule. (3 marks)

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With reference to hydrogen bonding and dispersion forces, explain the trends in the boiling point data of these compounds, within each series and between the series. (4 marks)

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

Show Answers Only

b. The alcohols have higher boiling points than amines of the same chain length.

→ Both alcohols and amines have polar hydrogen bonding as a result of the $\ce{OH}$ and $\ce{NH2}$ functional groups respectively.

→ Oxygen molecules have higher electronegativity than nitrogen molecules making the hydrogen bonding in the alcohols significantly stronger than the hydrogen bonding in amines.

→ Since the dispersion forces in amines and alcohols of the same chain length are very similar, the difference in the strength of the intermolecular bonding is dependent on the strength of the hydrogen bonding.

→ Therefore, a larger amount of thermal energy is required to separate the alcohol molecules resulting in higher boiling points than amines.

As the chain length of the alcohols and amines increase, the boiling points also increase.

→ When the chain length increases, the number of electrons in the molecules also increase which corresponds to a larger number of dispersion forces between neighbouring molecules.

→ The stronger dispersion forces between the molecules increase the overall strength of the intermolecular forces in both the alcohols and amines therefore leading to higher boiling points as chain length increases.

Show Worked Solution

b. The alcohols have higher boiling points than amines of the same chain length.

→ Both alcohols and amines have polar hydrogen bonding as a result of the $\ce{OH}$ and $\ce{NH2}$ functional groups respectively.

→ Oxygen molecules have higher electronegativity than nitrogen molecules making the hydrogen bonding in the alcohols significantly stronger than the hydrogen bonding in amines.

→ Therefore, a larger amount of thermal energy is required to separate the alcohol molecules resulting in higher boiling points than amines.

As the chain length of the alcohols and amines increase, the boiling points also increase.

→ When the chain length increases, the number of electrons in the molecules also increase which corresponds to a larger number of dispersion forces between neighbouring molecules.

CHEMISTRY, M6 2024 HSC 21

A solution of acetic acid reacts with magnesium metal.

Write the names of the products of this reaction in the boxes provided. (2 marks)

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Show Answers Only

→ Products: hydrogen gas $\ce{H2(g)}$ and magnesium acetate/ethanoate $\ce{Mg(CH3COO)2(aq)}$

Show Worked Solution

→ Acid + active metal $\ce{->}$ hydrogen + salt

→ Products: hydrogen gas $\ce{H2(g)}$ and magnesium acetate/ethanoate $\ce{Mg(CH3COO)2(aq)}$

CHEMISTRY, M5 2024 HSC 10 MC

The following system is at equilibrium.

$\underset{\text { propan-2-ol }}{\ce{CH_3CHOHCH_3(g)}} \rightleftharpoons \underset{\text {propan-2-one}}{\ce{CH_3COCH_3(g)}}$$\ce{+ H_2(g)}$

A catalyst is added to the system.

Which row of the table correctly identifies the change in the yield of propan-2-one and the reaction rates?

\begin{align*}
\begin{array}{l}
\ & \\
\\
\textbf{A.}\\
\\
\textbf{B.}\\
\\
\textbf{C.}\\
\\
\textbf{D.}\\
\\
\end{array}
\begin{array}{|l|l|}
\hline
\quad\quad \textit{Yield of } & \quad \quad \quad \quad \textit{Reaction Rates} \\
\quad\textit{propan-2-one} & \textit{} \\
\hline
\text{Remains the same} & \text{Both forward and reverse rates} \\
\text{} & \text{are unchanged.} \\
\hline
\text{Remains the same} & \text{Both forward and reverse rates} \\
\text{} & \text{increase equally} \\
\hline
\text{Decreases} & \text{Reverse rate increases more than} \\
\text{} & \text{the forward rate increases.} \\
\hline
\text{Increases} & \text{Forward rate increases more than}\\
\text{} & \text{the reverse rate increases.}\\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

→ Adding a catalyst to an equilibrium system will decrease the activation energy of both the forward and reverse reactions equally.

→ Thus, the reaction rate of the forward and reverse reactions will both increase and cause no change to the yield of the reaction.

$\Rightarrow B$

CHEMISTRY, M8 2024 HSC 9 MC

Which of the following is the mass spectrum of ethanamine?

Show Answers Only

$B$

Show Worked Solution

→ The molar mass of ethanamine, $\ce{CH3CH2NH2}$, is 45.086 g mol$^{-1}$

→ The peak with the largest mass to charge ratio displays the molar mass of the substance.

→ Therefore, the parent ion peak will be at 45.

$\Rightarrow B$

CHEMISTRY, M8 2024 HSC 8 MC

Which pair of ions produce different colours in a flame test?

$\ce{Br^{-} and Cl^{-}}$
$\ce{Ag^{+} and OH^{-}}$
$\ce{Cu^{2+} and Ca^{2+}}$
$\ce{CH_3OOO^{-} and H_2 PO_4^{-}}$

Show Answers Only

$C$

Show Worked Solution

→ Only metal cations produce unique colours during a flame test due to their electron configurations.

→ As the electrons ‘fall back’ down into their shells from an excited state, they emit a specific light wave (colour).

$\Rightarrow C$

CHEMISTRY, M8 2024 HSC 4 MC

An infrared spectrum of an organic compound is shown.

Which of the following compounds would produce the spectrum shown?

Show Answers Only

$A$

Show Worked Solution

→ The broad absorption peak between 3500 to 3250 indicates the presence of an $\ce{O-H}$ alcohol group.

$\Rightarrow A$

Matrices, GEN2 2024 VCAA 9

Vince works on a construction site.

The amount Vince gets paid depends on the type of shift he works, as shown in the table below.

\begin{array}{|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Shift type} \rule[-1ex]{0pt}{0pt}& \textbf{Normal} & \textbf{Overtime} & \textbf {Weekend} \\
\hline
\rule{0pt}{2.5ex} \textbf{Hourly rate of pay} \rule[-1ex]{0pt}{0pt} \ \text{(\$ per hour)} & 36 & 54 & 72 \\
\hline
\end{array}

This information is shown in matrix $R$ below.

\begin{align*}
R=\left[\begin{array}{lll}
36 & 54 & 72
\end{array}\right] \end{align*}

Matrix $R^T$ is the transpose of matrix $R$.
Write down matrix $R^T$ in the space below. (1 mark)

$R^T=$

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During one week, Vince works 28 hours at the normal rate of pay, 6 hours at the overtime rate of pay, and 8 hours at the weekend rate of pay.

Complete the following matrix calculation showing the total amount Vince has been paid for this week. (1 mark)

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Vince will receive $90 per hour if he works a public holiday shift.

Matrix $Q$, as calculated below, can be used to show Vince's hourly rate for each type of shift.

\begin{align*}
\begin{aligned}
Q & =n \times\left[\begin{array}{llll}
1 & 1.5 & 2 & p
\end{array}\right] \\
& =\left[\begin{array}{llll}
36 & 54 & 72 & 90
\end{array}\right] \end{aligned}
\end{align*}

Write the values of $n$ and $p$ in the boxes below. (1 mark)

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

Show Answers Only

a. $R^T=\begin{bmatrix}
36 \\
54 \\
72
\end{bmatrix}$

b. $[28\quad 6\quad 8]\times R^T = [1908]$

c. $n=36\ ,\ p=2.5$

Show Worked Solution

a. $R^T=\begin{bmatrix}
36 \\
54 \\
72
\end{bmatrix}$

b. $\begin{bmatrix}
28 & 6 & 8
\end{bmatrix}\times\ R^T=\begin{bmatrix}
28\times36 + 6\times54+ 8\times 72
\end{bmatrix}=[1908]$

c. $n=\ \text{Normal hourly rate}\ =36$

$p=\ \text{Overtime rate}\ =\dfrac{90}{36}=2.5$

Recursion and Finance, GEN2 2024 VCAA 5

Emi operates a mobile dog-grooming business.

The value of her grooming equipment will depreciate.

Based on average usage, a rule for the value, in dollars, of the equipment, $V_n$, after $n$ weeks is

$V_n=15000-60 n$

Assume that there are exactly 52 weeks in a year.

By what amount, in dollars, does the value of the grooming equipment depreciate each week? (1 mark)

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Emi plans to replace the grooming equipment after four years.
What will be its value, in dollars, at this time? (1 mark)

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$V_n$ is the value of the grooming equipment, in dollars, after $n$ weeks.
Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$ that can model this value from one week to the next. (1 mark)

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The value of the grooming equipment decreases from one year to the next by the same percentage of the original $\$ 15000$ value.
What is this annual flat rate percentage? (1 mark)

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Show Answers Only

a. $$60$

b. $$2520$

c. $V_0=15\,000 , \ \ V_{n+1}=V_n-60$

d. $20.8\%$

Show Worked Solution

a. $$60$

b. $n=4\times 52=208$

$V_{208}$	$=15\,000-60\times208$
	$=$2520$

c. $V_0=15\,000 , \ \ V_{n+1}=V_n-60$

d. $\text{Flat rate}\ =\dfrac{60}{15\,000}\times 52\times 100\%=20.8\%$

Data Analysis, GEN2 2024 VCAA 3

The Olympic gold medal-winning height for the women's high jump, $\textit{Wgold}$, is often lower than the best height achieved in other international women's high jump competitions in that same year.

The table below lists the Olympic year, $\textit{year}$, the gold medal-winning height, $\textit{Wgold}$, in metres, and the best height achieved in all international women's high jump competitions in that same year, $\textit{Wbest}$, in metres, for each Olympic year from 1972 to 2020.

A scatterplot of $\textit{Wbest}$ versus $\textit{Wgold}$ for this data is also provided.

year	1972	1976	1980	1984	1988	1992	1996	2000	2004	2008	2012	2016	2020
Wgold (m)	1.92	1.93	1.97	2.02	2.03	2.02	2.05	2.01	2.06	2.05	2.05	1.97	2.04
Wbest (m)	1.94	1.96	1.98	2.07	2.07	2.05	2.05	2.02	2.06	2.06	2.05	2.01	2.05

When a least squares line is fitted to the scatterplot, the equation is found to be:

$Wbest =0.300+0.860 \times Wgold$

The correlation coefficient is 0.9318

Name the response variable in this equation. (1 mark)

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Draw the least squares line on the scatterplot above. (1 mark)

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Determine the value of the coefficient of determination as a percentage. (1 mark)
Round your answer to one decimal place.
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Describe the association between $\textit{Wbest}$ and $\textit{Wgold}$ in terms of strength and direction. (1 mark)

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

Referring to the equation of the least squares line, interpret the value of the slope in terms of the variables $\textit{Wbest}$ and $\textit{Wgold}$. (1 mark)

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In 1984, the $\textit{Wbest}$ value was 2.07 m for a $\textit{Wgold}$ value of 2.02 m .
Show that when this least squares line is fitted to the scatterplot, the residual value for this point is 0.0328. (2 marks)

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The residual plot obtained when the least squares line was fitted to the data is shown below. The residual value from part f is missing from the residual plot.

1. Complete the residual plot by adding the residual value from part f, drawn as a cross ( X ), to the residual plot above. (1 mark)
  
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2. In part b, a least squares line was fitted to the scatterplot. Does the residual plot from part g justify this? Briefly explain your answer. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---

In 1964, the gold medal-winning height, $\textit{Wgold}$, was 1.90m . When the least squares line is used to predict $\textit{Wbest}$, it is found to be 1.934 m .
Explain why this prediction is not likely to be reliable. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $Wbest$

c. $86.8\%$

d. $\text{Strong, positive}$

e. $Wbest\ \text{will increase, on average by 0.86 metres for every metre of increase in}\ Wgold.$

f.	$Wbest$	$=0.300 +0.86\times 2.02$
		$=2.0372$

$\therefore\ \text{Residual}\ =2.07-2.0372=0.0328$

g.i.

g.ii. $\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}$

h. $\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}$

$\text{and therefore cannot be relied upon.}$

Show Worked Solution

a. $Wbest$

b. $\text{Using points:}\ (1.90, 1.934)\ \text{and}\ (2.00, 2.02)$

c. $r=0.9318$

$r^2=0.9318^2=0.8682\dots$

$\therefore\ \text{Coefficient of determination} \approx 86.8\%$

d. $\text{Strong, positive}$

e. $Wbest\ \text{will increase, on average by 0.86 metres for every metre of increase in}\ Wgold.$

f.	$Wbest$	$=0.300 +0.86\times 2.02$
		$=2.0372$

$\therefore\ \text{Residual}\ =2.07-2.0372=0.0328$

g.i.

g.ii. $\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}$

h. $\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}$

$\text{and therefore cannot be relied upon.}$

Data Analysis, GEN2 2024 VCAA 2

The boxplot below displays the distribution of all gold medal-winning heights for the women's high jump, $\textit{Wgold}$, in metres, for the 19 Olympic Games held from 1948 to 2020.

Describe the shape of this data distribution. (1 mark)

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For this boxplot, what is the smallest possible number of $\textit{Wgold}$ heights lower than 1.85 m (1 mark)

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i. Using the boxplot, show that the lower fence is 1.565 m and the upper fence is 2.325 m. (1 mark)

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ii. Referring to the boxplot, the lower fence and the upper fence, explain why no outliers exist. (1 mark)

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Show Answers Only

a. $\text{Negatively skewed}$

b. $1$

c.i. $Q_1=1.85,\ Q_3=2.04,\ IQR=2.04-1.85=0.19$

$\text{Lower Fence}$	$=Q_1-1.5\times IQR$
	$=1.85-1.5\times 0.19$
	$=1.565$

$\text{Upper Fence}$	$=Q_1+1.5\times IQR$
	$=2.04+1.5\times 0.19$
	$=2.325$

c.ii. $\text{No values exist below the lower fence or above the upper fence.}$

$\therefore\ \text{No outliers exist.}$

Show Worked Solution

a. $\text{Negatively skewed}$

b. $25\%\ \text{of the games had heights}\ <1.85\ \text{m}$

$25\%\times 19= 4.75\rightarrow 4\ \text{games}$

$\therefore\ \text{The smallest number of games is 1}$

c.i. $Q_1=1.85,\ Q_3=2.04,\ IQR=2.04-1.85=0.19$

$\text{Lower Fence}$	$=Q_1-1.5\times IQR$
	$=1.85-1.5\times 0.19$
	$=1.565$

$\text{Upper Fence}$	$=Q_1+1.5\times IQR$
	$=2.04+1.5\times 0.19$
	$=2.325$

c.ii. $\text{No values exist below the lower fence or above the upper fence.}$

$\therefore\ \text{No outliers exist.}$

Data Analysis, GEN2 2024 VCAA 1

Table 1 lists the Olympic year, $\textit{year}$, and the gold medal-winning height for the men's high jump, $\textit{Mgold}$, in metres, for each Olympic Games held from 1928 to 2020. No Olympic Games were held in 1940 or 1944, and the 2020 Olympic Games were held in 2021.

Table 1

\begin{array}{|c|c|}
\hline \quad \textit{year} \quad & \textit{Mgold}(m) \\
\hline 1928 & 1.94 \\
\hline 1932 & 1.97 \\
\hline 1936 & 2.03 \\
\hline 1948 & 1.98 \\
\hline 1952 & 2.04 \\
\hline 1956 & 2.12 \\
\hline 1960 & 2.16 \\
\hline 1964 & 2.18 \\
\hline 1968 & 2.24 \\
\hline 1972 & 2.23 \\
\hline 1976 & 2.25 \\
\hline 1980 & 2.36 \\
\hline 1984 & 2.35 \\
\hline 1988 & 2.38 \\
\hline 1992 & 2.34 \\
\hline 1996 & 2.39 \\
\hline 2000 & 2.35 \\
\hline 2004 & 2.36 \\
\hline 2008 & 2.36 \\
\hline 2012 & 2.33 \\
\hline 2016 & 2.38 \\
\hline 2020 & 2.37 \\
\hline
\end{array}

For the data in Table 1, determine:
1. the maximum $\textit{Mgold}$ in metres (1 mark)
  
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2. the percentage of $\textit{Mgold}$ values greater than 2.25 m. (1 mark)
  
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The mean of these $\textit{Mgold}$ values is 2.23 m , and the standard deviation is 0.15 m .
Calculate the standardised $z$-score for the 2000 $\textit{Mgold}$ of 2.35 m. (1 mark)

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Construct a boxplot for the $\textit{Mgold}$ data in Table 1 on the grid below. (2 marks)

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A least squares line can also be used to model the association between $\textit{Mgold}$ and $\textit{year}$.
Using the data from Table 1, determine the equation of the least squares line for this data set.
Use the template below to write your answer.
Round the values of the intercept and slope to three significant figures. (2 marks)

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The coefficient of determination is 0.857
Interpret the coefficient of determination in terms of $\textit{Mgold}$ and $\textit{year}$. (1 mark)

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

Show Answers Only

a.i. $2.39$

a.ii. $50\%$

b. $0.8$

e. $\text{A coefficient of determination of 85.7% shows the variation in the}\ Mgold$

$\text{that is explained by the variation in the }year.$

Show Worked Solution

a.i. $2.39$

a.ii. $\dfrac{11}{22}=50\%$

b.	$z$	$=\dfrac{x-\overline x}{s_x}$
		$=\dfrac{2.35-2.23}{0.15}$
		$=0.8$

c. $Q_2=\dfrac{2.33+2.25}{2}=2.29$

$Q_1=2.12, \ Q_3=2.36$

$\text{Min}\ =1.94, \ \text{Max}\ =2.39$

d. $\text{Using CAS:}$

e. $\text{A coefficient of determination of 85.7% shows the variation in the}\ Mgold$

$\text{that is explained by the variation in the }year.$

BIOLOGY, M8 2024 HSC 11 MC

The data shows the proportion of adults living in Australia who are obese.

Which of the following can be observed from the data?

The proportion of obese adults always increases with age.
There is a greater percentage of men who are obese than women in all age groups.
The proportion of women who are obese increases from 13% at 18–24 to 38% at 65–74.
The proportion of men who are obese increases from 18% at 18–24 to 35% at 45–54, then decreases to 23% at age 85 and over.

Show Answers Only

$C$

Show Worked Solution

→ Options $A, B$ and $D$ can be shown to be incorrect with reference to the table.

$\Rightarrow C$

BIOLOGY, M8 2024 HSC 7 MC

How do stomata maintain water balance in plants?

They close in hot weather to decrease transpiration.
They open in cold weather to decrease transpiration.
They open in hot weather to decrease evaporative cooling.
They close in cold weather to decrease evaporative cooling.

Show Answers Only

$A$

Show Worked Solution

→ Stomata close in hot weather to decrease transpiration.

→ This adaptive response helps prevent excessive water loss during high temperatures and maintain proper water balance.

$\Rightarrow A$

BIOLOGY, M5 2024 HSC 4 MC

Which row of the table correctly identifies components of DNA?

\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}\textit{Phosphate} \rule[-1ex]{0pt}{0pt}& \quad \textit{Ribose} \quad & \textit{Deoxyribose} & \quad \textit{Uracil} \quad & \ \ \textit{Thymine}\ \ \\
\hline
\rule{0pt}{2.5ex}\checkmark \rule[-1ex]{0pt}{0pt}& \text{} & \checkmark & \checkmark & \text{} \\
\hline
\rule{0pt}{2.5ex}\text{} \rule[-1ex]{0pt}{0pt}& \checkmark & \text{} & \checkmark & \checkmark \\
\hline
\rule{0pt}{2.5ex}\text{} \rule[-1ex]{0pt}{0pt}& \checkmark & \checkmark & \text{} & \checkmark \\
\hline
\rule{0pt}{2.5ex}\checkmark \rule[-1ex]{0pt}{0pt}& \text{} & \checkmark & \text{} & \checkmark \\
\hline
\end{array}
\end{align*}

Show Answers Only

$D$

Show Worked Solution

→ DNA contains phosphate groups, deoxyribose sugar (not ribose), and thymine (not uracil which is found in RNA instead).

→ The last row with checkmarks for phosphate, deoxyribose, and thymine is therefore the only correct combination.

$\Rightarrow D$

BIOLOGY, M6 2024 HSC 3 MC

The image shows a chromosome that has undergone mutation. Each letter represents a gene.

What type of mutation has occurred?

Deletion
Duplication
Inversion
Substitution

Show Answers Only

$B$

Show Worked Solution

→ Looking at the sequence, we can see that genes C and E appear twice in the mutated chromosome (creating a repeat of genetic material).

→ This is characteristic of a duplication mutation rather than deletion (loss), inversion (reversal), or substitution (replacement).

$\Rightarrow B$

BIOLOGY, M6 2024 HSC 2 MC

Resin produced by spinifex grass has long been used by Aboriginal Peoples. Spinifex resin is currently used to produce medicinal creams.

What is this an example of?

Biotechnology
Selective breeding
Artificial insemination
Genetically modified organisms

Show Answers Only

$A$

Show Worked Solution

→ This is an example of biotechnology because it involves using traditional knowledge to harness a biological resource (spinifex resin) for medicinal purposes.

→ This fits the broad definition of biotechnology as using biological systems or processes to create useful products.

$\Rightarrow A$

BIOLOGY, M7 2024 HSC 1 MC

Which of the following are non-cellular pathogens?

Bacteria
Fungi
Prions
Protozoa

Show Answers Only

$C$

Show Worked Solution

→ Bacteria, fungi, and protozoa are all cellular organisms.

→ Prions are misfolded proteins that can cause disease by inducing normal proteins to misfold, making them non-cellular pathogens.

$\Rightarrow C$

CHEMISTRY, M6 2024 HSC 3 MC

Which of the following compounds can be correctly described as an Arrhenius base when dissolved in water?

Sodium nitrate
Sodium sulfate
Sodium chloride
Sodium hydroxide

Show Answers Only

$D$

Show Worked Solution

→ An Arrhenius base is a compound that increases the concentration of $\ce{OH-}$ ions when it is dissolved in water.

→ Sodium hydroxide is the only compound that dissolves in water to produce hydroxide ions.

$\ce{NaOH(s) -> Na+(aq) + OH-(aq)}$

$\Rightarrow D$

CHEMISTRY, M5 2024 HSC 2 MC

Aboriginal and Torres Strait Islander Peoples have used leaching in flowing water over several days to prepare various foods from plants that can be toxic to humans.

What was the reason for this?

To react with toxins
To dissolve low solubility toxins
To prevent the food from decomposing
To break down compounds that are difficult to digest

Show Answers Only

$B$

Show Worked Solution

→ The toxins form a equilibrium system: $\ce{toxins(s)\rightleftharpoons toxins(aq)}$

→ As the water flows away, the concentration of aqueous toxins decreases which shifts the equilibrium system to the right, causing more low solubility toxins to break down.

$\Rightarrow B$

Networks, GEN1 2024 VCAA 34 MC

Consider the following graph.

A Eulerian trail through this graph could be

ABCDEF
ACBDCFDEF
BACBDCFDEF
BDCABCDFCDEF

Show Answers Only

$C$

Show Worked Solution

$\text{An Eulerian trail has exactly 2 vertices of odd degree.}$

$\text{You must start on one and finish on the other ie B to F.}$

$\rightarrow\ \text{Eliminate A and B}$

$\text{A path cannot be repeated}\ \rightarrow\ \text{Eliminate D}$

$\Rightarrow C$

CHEMISTRY, M7 2024 HSC 1 MC

Which two substances are members of the same homologous series?

Show Answers Only

$A$

Show Worked Solution

→ A homologous series refers to compounds with the same or similar functional group and same general formula.

→ Both substances in $A$ contain a hydroxyl group.

$\Rightarrow A$

Matrices, GEN1 2024 VCAA 25 MC

Matrix $J$ is a $2 \times 3$ matrix.

Matrix $K$ is a $3 \times 1$ matrix.

Matrix $L$ is added to the product $J K$.

The order of matrix $L$ is

$1 \times 3$
$2 \times 1$
$2 \times 3$
$3 \times 2$

Show Answers Only

$B$

Show Worked Solution

$J\ \text{is order}\ 2\times 3\ \text{and}\ K\ \text{is order}\ \3\times 1$

$\therefore\ JK\ \text{is order}\ \2\times 1$

$\Rightarrow B$

Recursion and Finance, GEN1 2024 VCAA 19 MC

Liv bought a new car for $35 000. The value of the car will be depreciated by 18% per annum using the reducing balance method.

A recurrence relation that models the year-to-year value of her car is of the form

$L_0=35\,000, \quad L_{n+1}=k \times L_n$

The value of $k$ is

0.0082
0.18
0.82
1.18

Show Answers Only

$C$

Show Worked Solution

$\text{Reducing balance interest rate}\ =1-0.18=0.82$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 15 MC

The table below shows the total number of cans of soft drink sold each month at a suburban cafe in 2023.

\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Month } \rule[-1ex]{0pt}{0pt}& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\rule{0pt}{2.5ex} \textbf{Cans sold } \rule[-1ex]{0pt}{0pt}& 316 & 321 & 365 & 306 & 254 & 308 & 354 & 357 & 381 & 355 & 365 & 324 \\
\hline
\end{array}

The six-mean smoothed value of the number of cans sold, with centring, for month 5 is closest to

Show Answers Only

$C$

Show Worked Solution

$\text{Six-mean smoothed:}$

$\text{Months:}\ 2-7$	$=\dfrac{321+365+306+254+308+354}{6}=318$
$\text{Months:}\ 3-8$	$=\dfrac{365+306+254+308+354+357}{6}=324$

$\therefore\ \text{Centred smoothed mean}$	$=\dfrac{318+324}{2}$
	$=321$

$C$

\(\tan \theta\)	\(=0.84\)
\(\theta\)	\(= \tan^{-1} 0.84\)
	\(=40^{\circ}\)

a.	\(h(4)\)	\(=\dfrac{3000}{4+1}\)
		\(=600\)

b.	\(h(t)\)	\(=3000(t+1)^{-1}\)
	\(h^{\prime}(t)\)	\(=\dfrac{-3000}{(t+1)^2}\)
	\(h_1^{\prime}(t)\)	\(=\dfrac{1}{2}h^{\prime}(t)\)
		\(=\dfrac{-1500}{(t+1)^2}\)
	\(\therefore\ h_1(t)\)	\(=\dfrac{1500}{t+1}\)

\(\text{Approx CI}\)	\(=\left(\dfrac{3}{5}-2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}},\ \dfrac{3}{5}+2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}}\right)\)
	\(=\left(\dfrac{3}{5}-\dfrac{2\sqrt{6}}{50},\quad \dfrac{3}{5}+\dfrac{2\sqrt{6}}{50}\right)\)
	\(=\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\quad \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

cii.	\(\sqrt{\dfrac{\hat{P}(1-\hat{P}}{n}}\)	\(=\dfrac{\sqrt{2}}{50}\)
	\(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\)	\(=\dfrac{\sqrt{2}}{50}\)
	\(\dfrac{\dfrac{6}{25}}{n}\)	\(=\dfrac{2}{2500}\)
	\(n\)	\(=\dfrac{6}{25}\times\dfrac{2500}{2}\)
		\(=\dfrac{600}{2}=300\)

a.	\(y\)	\(=e^x \cos(3x)\)
	\(\dfrac{dy}{dx}\)	\(=e^x.(-3\sin(3x))+\cos(3x).e^x\)
		\(=e^x(\cos(3x)-3\sin(3x))\)

b.	\(f(x)\)	\(=\log_{e}(x^3-3x+2)\)
	\(f^{\prime}(x)\)	\(=\dfrac{3x^2-3}{x^3-3x+2}\)
	\(f^{\prime}(3)\)	\(=\dfrac{3(3)^2-3}{(3)^3-3(3)+2}\)
		\(=\dfrac{24}{20}=\dfrac{6}{5}\)

\(\therefore\ h\)	\(=23-11=12\)
\(k\)	\(=4750-4400=350\)

\(f(12)-f(0)\)	\(=4460-4100=360\)
\(f(24)-f(12)\)	\(=4820-4460=360\)
\(f(36)-f(24)\)	\(=5180-4820=360\)