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HMS, BM EQ-Bank 153 MC

Which energy system produces ATP at the fastest rate?

  1. Glycolytic
  2. Aerobic
  3. ATP-PCr
  4. All systems produce ATP at the same rate
Show Answers Only

\(C\)

Show Worked Solution

Consider Option C: ATP-PCr

  • ATP-PCr has fastest production rate but limited stores

Other Options:

  • A is incorrect: Slower than ATP-PCr but faster than aerobic
  • B is incorrect: Slowest but most efficient system
  • D is incorrect: Systems have different production rates

\(\Rightarrow C\)

HMS, BM EQ-Bank 138

Outline how inefficient jumping technique can affect the skeletal system and require first aid intervention.   (3 marks)

Show Answers Only

Sample Answer

  • Incorrect landing can cause immediate joint compression
  • Impact forces travel incorrectly through bones causing stress
  • First aid required includes RICER and joint stabilisation
Show Worked Solution

Sample Answer

  • Incorrect landing can cause immediate joint compression
  • Impact forces travel incorrectly through bones causing stress
  • First aid required includes RICER and joint stabilisation

HMS, BM EQ-Bank 128 MC

During a cross-country run, an athlete experiences severe abdominal cramping. Which first aid response would be most appropriate?

  1. Continue running at race pace
  2. Increase fluid intake rapidly
  3. Start walking immediately
  4. Stop activity and lie in a comfortable position
Show Answers Only

\(D\)

Show Worked Solution

Consider Option D: Stop activity and lie in a comfortable position

  • Stopping activity and finding a comfortable position allows assessment of digestive system stress and prevents further complications.

Other Options:

  • A is incorrect: Would increase digestive system stress
  • B is incorrect: Rapid fluid intake may worsen symptoms
  • C is incorrect: Immediate walking may exacerbate cramping

\(\Rightarrow D\)

HMS, BM EQ-Bank 122

Outline how the muscular and skeletal systems work together during movement and identify when first aid intervention is required.   (3 marks)

Show Answers Only

Sample Answer

  • Muscles and bones work as a lever system through insertion points and origins to create movement
  • When movement occurs outside normal ranges, strain on attachment points can require first aid intervention
  • Signs requiring intervention include: sudden pain during movement, inability to contract muscle, or loss of joint stability
Show Worked Solution

Sample Answer

  • Muscles and bones work as a lever system through insertion points and origins to create movement
  • When movement occurs outside normal ranges, strain on attachment points can require first aid intervention
  • Signs requiring intervention include: sudden pain during movement, inability to contract muscle, or loss of joint stability

HMS, BM EQ-Bank 109

Outline how the skeletal and muscular systems work together during a squat movement.   (3 marks)

Show Answers Only

Sample Answer

Descent:

  • Hip and knee joints bend while thigh muscles lengthen to control movement
  • Skeleton provides stable framework as muscles control lowering

Bottom Position:

  • Skeleton maintains alignment while muscles hold stable position

Rising:

  • Leg muscles contract to push while skeleton provides leverage for upward movement
Show Worked Solution

Sample Answer

Descent:

  • Hip and knee joints bend while thigh muscles lengthen to control movement
  • Skeleton provides stable framework as muscles control lowering

Bottom Position:

  • Skeleton maintains alignment while muscles hold stable position

Rising:

  • Leg muscles contract to push while skeleton provides leverage for upward movement

HMS, BM EQ-Bank 103 MC

During a netball game, a player performs a layup shot. Which body systems are working together to execute this movement?

  1. Skeletal and respiratory only
  2. Muscular and circulatory only
  3. Skeletal, muscular and nervous
  4. Respiratory and circulatory only
Show Answers Only

\(C\)

Show Worked Solution

Consider Option C:  Skeletal, muscular and nervous

  • skeletal system provides the framework for movement
  • muscles contract to produce the force
  • nervous system coordinates the timing and precision of the layup shot

Other Options:

  • A is incorrect: Misses muscular control needed
  • B is incorrect: Omits skeletal framework required
  • D is incorrect: Misses skeletal and muscular components

\(\Rightarrow C\)

HMS, BM EQ-Bank 26 MC

The muscle group indicated in the image below is primarily responsible for:

  1. Hip flexion
  2. Knee extension
  3. Knee flexion
  4. Hip extension
Show Answers Only

\(B\)

Show Worked Solution

Consider Option B:

  • The quadriceps (specifically rectus femoris, vastus lateralis, vastus medialis, and vastus intermedius) are the primary knee extensors

Other Options:

  • A:  Incorrect because hip flexion is primarily performed by the iliopsoas
  • C:  Incorrect because knee flexion is performed by the hamstring muscle group
  • D:  Incorrect because hip extension is primarily performed by the gluteus maximus and hamstrings

\(\Rightarrow B\)

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.

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

  2. 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.
  3. State a sequence of two transformations that maps \(h\) to this new model \(h_1\).  (2 marks)

  4. In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
  5.  i. 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)

  6. ii. A new sample of \(n\) people results in the same sample proportion.
  7.     Find the smallest value of \(n\) to achieve a standard deviation of  \(\dfrac{\sqrt{2}}{50}\)  for the sample proportion.  (1 mark)

Show Answers Only

a.    \(600\)

b.   \(\text{Transformations:}\)

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

\(\text{2- Translation of 1500 units upwards}\)

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

♦♦♦ Mean mark (a) 17%.

\(\text{Given}\ \ h(0)=h_1(0):\)

\(\dfrac{1500}{0+1} +C= 3000\ \ \Rightarrow\ \ C=1500\)

\(h_1(t)=\dfrac{1500}{t+1}+1500\)
 

\(\text{Transformations:}\)

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

\(\text{2- Translation of 1500 units upwards}\)
 

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.   \(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\dfrac{6}{25n}\) \(=\dfrac{2}{2500}\)
  \(\dfrac{25n}{6}\) \(=1250\)
  \(n\) \(=\dfrac{6}{25}\times 1250\)
    \(=300\)
♦♦ Mean mark (c.ii.) 33%.

BIOLOGY, M8 2024 HSC 31

A study monitored the changes in the body temperature of a kookaburra (an Australian bird) and a human over a 24-hour period. The results of the study are shown in the graph.
 

  1. At what time was the kookaburra's body temperature the lowest?   (1 mark)

  1. Some endothermic organisms can display torpor (a significant decrease in physiological activity).

    With reference to the graph, explain whether the human or the kookaburra was displaying torpor and if so, state the time this occurred.   (3 marks)

  1. Outline an adaptation that may lead to an increase in the kookaburra's body temperature during the inactive period.   (2 marks)

Show Answers Only

a.   4 am

b.   Signs of torpor:

→ The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity

→ In contrast, the kookaburra exhibited classic torpor behaviour.

→ Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.
 

c.   Kookaburra adaptation:

→ Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.

→ This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier

→ The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

Show Worked Solution

a.   4 am
 

b.   Signs of torpor:

→ The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity

→ In contrast, the kookaburra exhibited classic torpor behaviour.

→ Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.

Mean mark (b) 56%.

c.   Kookaburra adaptation:

→ Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.

→ This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier

→ The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

♦ Mean mark (c) 47%.

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}\,\text{(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}

  1. For the data in Table 1, determine:
  2.  i. the maximum \(\textit{Mgold}\) in metres   (1 mark)

  3. ii. the percentage of \(\textit{Mgold}\) values greater than 2.25 m.   (1 mark)

  4. The mean of these \(\textit{Mgold}\) values is 2.23 m, and the standard deviation is 0.15 m.
  5. Calculate the standardised \(z\)-score for the 2000 \(\textit{Mgold}\) of 2.35 m.   (1 mark)

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

     

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

     

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

Show Answers Only

a.i.   \(2.39\)

a.ii.  \(50\%\)

b.    \(0.8\)

c.     

d.   
    

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

\(\text{the}\ Mgold\ \text{that is explained by the variation in the }year.\)

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.   \(\text{Using CAS:}\)


  
 

Mean mark (d) 52%.
Mean mark (e) 52%.

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

\(\text{the}\ Mgold\ \text{that is explained by the variation in the }year.\)

Vectors, EXT2 V1 2024 HSC 12a

The vector \(\underset{\sim}{a}\) is \(\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) and the vector \(\underset{\sim}{b}\) is \(\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\).

  1. Find \(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\).   (1 mark)

  2. Show that  \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\)  is perpendicular to \(\underset{\sim}{b}\).  (2 marks)

Show Answers Only

1.     \(\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

\(\therefore\ \text {Vectors are perpendicular.}\)

Show Worked Solution

i.    \(\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\)
 

\(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

 
ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)
 

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

 
\(\therefore\ \text{Vectors are perpendicular.}\)

Complex Numbers, EXT2 N1 2024 HSC 11e

  1. Write the number  \(\sqrt{3}+i\)  in modulus-argument form.   (2 marks)

  2. Hence, or otherwise, write  \((\sqrt{3}+i)^7\)  in exact Cartesian form.   (2 marks)

Show Answers Only

i.     \(2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)

ii.    \(-64 \sqrt{3}-64 i\)

Show Worked Solution

i.     \(z=\sqrt{3}+i\)

\(|z|=\sqrt{3+1}=2\)

\(\arg (z)=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}\)

\(z=2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)
 

ii.     \((\sqrt{3}+i)^7\) \(=2^7\left(\cos \left(\dfrac{7 \pi}{6}\right)+i \sin \left(\dfrac{7 \pi}{6}\right)\right)\)
    \(=128\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} i\right)\)
    \(=-64 \sqrt{3}-64 i\)

Complex Numbers, EXT2 N1 2024 HSC 11b

Let  \(z=2+3 i\)  and  \(w=1-5 i\).

  1. Find  \(z+\bar{w}\).   (1 mark)

  2. Find  \(z^2\).   (1 mark)

Show Answers Only

i.    \(z+\bar{w}=3+8 i\)

ii.   \(z^2=-5+12 i\)

Show Worked Solution

i.     \(z=2+3 i\)

\(w=1-5 i \ \Rightarrow \ \bar{w}=1+5 i\)

\(z+\bar{w}=2+3 i+1+5 i=3+8 i\)
 

ii.    \(z^2\) \(=(2+3 i)^{2}\)
    \(=4+12 i+9 i^2\)
    \(=-5+12 i\)

Calculus, EXT2 C1 2024 HSC 11a

Find \(\displaystyle \int x e^x\, d x\)   (2 marks)

Show Answers Only

\(x e^x-e^x+c\)

Show Worked Solution

\(u=x \quad \ \ u^{\prime}=1\)

\(v^{\prime}=e^x \quad v=e^x\)

  \(\displaystyle\int x e^x \,d x\) \(=u v^{\prime}-\displaystyle \int v u^{\prime}\, d x\)
    \(=x e^x- \displaystyle \int e^x \cdot 1\, d x\)
    \(=x e^x-e^x+c\)

Vectors, EXT1 V1 2024 HSC 11a

Consider the vectors  \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}\)  and  \(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\).

  1. Find  \(2 \underset{\sim}{a}-\underset{\sim}{b}\).   (1 mark)

  2. Find  \(\underset{\sim}{a} \cdot \underset{\sim}{b}\).   (1 mark)

Show Answers Only

i.    \(\displaystyle \binom{7}{0}\)

ii.   \(5\)

Show Worked Solution

i.     \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}, \ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\)

\(2 \underset{\sim}{a}-\underset{\sim}{b}=2 \displaystyle \binom{3}{2}-\binom{-1}{4}=\binom{6}{4}-\binom{-1}{4}=\binom{7}{0}\)
 

ii.    \(\underset{\sim}{a} \cdot \underset{\sim}{b}=\displaystyle\binom{3}{2}\binom{-1}{4}=3 \times(-1)+2 \times 4=5\).

Measurement, STD1 M4 2024 HSC 2 MC

Four people completed the same fitness activity.

The graph shows the heart rate for each person before and after completing the activity.
 

Which person had the LEAST difference in heart rate?

  1. Jo
  2. Kim
  3. Lee
  4. Mal
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Option A: }\)  \(\text{Jo}\)  \(=120-80=40\) 
\(\text{Option B: }\) \(\text{Kim}\) \(=120-100=20\ \checkmark\) 
\(\text{Option C: }\) \(\text{Lee}\) \(=120-90=30\) 
\(\text{Option D: }\) \(\text{Mal}\) \(=150-100=50\) 

  
\(\Rightarrow B\)

Proof, EXT2 P1 2024 HSC 3 MC

Consider the statement:

'If a polygon is a square, then it is a rectangle.'

Which of the following is the converse of the statement above?

  1. If a polygon is a rectangle, then it is a square.
  2. If a polygon is a rectangle, then it is not a square.
  3. If a polygon is not a rectangle, then it is not a square.
  4. If a polygon is not a square, then it is not a rectangle.
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Statement:}\  P \Rightarrow \ Q\)

\(\text{Converse of statement:}\  Q \Rightarrow \ P\)

\(\Rightarrow A\)

Proof, EXT2 P1 2024 HSC 2 MC

Consider the following statement written in the formal language of proof

\(\forall \theta \in\biggl(\dfrac{\pi}{2}, \pi\biggr) \exists\ \phi \in\biggl(\pi, \dfrac{3 \pi}{2}\biggr) ; \ \sin \theta=-\cos \phi\).

Which of the following best represents this statement?

  1. There exists a \(\theta\) in the second quadrant such that for all \(\phi\) in the third quadrant  \(\sin \theta=-\cos \phi\).
  2. There exists a \(\phi\) in the third quadrant such that for all \(\theta\) in the second quadrant  \(\sin \theta=-\cos \phi\).
  3. For all \(\theta\) in the second quadrant there exists a \(\phi\) in the third quadrant such that  \(\sin \theta=-\cos \phi\).
  4. For all \(\phi\) in the third quadrant there exists a \(\theta\) in the second quadrant such that  \(\sin \theta=-\cos \phi\).
Show Answers Only

\(C\)

Show Worked Solution

\(\Rightarrow C\)

Trigonometry, 2ADV T1 2024 HSC 20

A vertical tower \(T C\) is 40 metres high. The point \(A\) is due east of the base of the tower \(C\). The angle of elevation to the top \(T\) of the tower from \(A\) is 35°. A second point \(B\) is on a different bearing from the tower as shown. The angle of elevation to the top of the tower from \(B\) is 30°. The points \(A\) and \(B\) are 100 metres apart.
 

  1. Show that distance \(A C\) is 57.13 metres, correct to 2 decimal places.  (1 mark)
  1. Find the bearing of \(B\) from \(C\) to the nearest degree.  (3 marks)
Show Answers Only

\(194^{\circ}\)

Show Worked Solution

a.   \(\text{In}\ \Delta TCA:\)

\(\tan 35°\) \( =\dfrac{40}{AC}\)  
\(AC\) \( =\dfrac{40}{\tan 35°}\)  
  \(=57.125…\)  
  \(=57.13\ \text{m (2 d.p.)}\)  

 
b.   \(\text{In}\ \Delta TCB:\)

\(\tan 30°\) \( =\dfrac{40}{BC}\)  
\(BC\) \( =\dfrac{40}{\tan 30°}\)  
  \(=69.28\ \text{m}\)  

  
\( \text{Find} \ \angle BCA \ \text{using cosine rule:}\)

\(\cos \angle B CA\) \( = \dfrac{57.13^2+69.28^2-100^2}{2 \times 57.13 \times 69.28}\)  
  \(= -0.2446 \)…  
\(\angle BCA\) \( = 104.2° \)  

\(\therefore\ \text{Bearing of}\ B\ \text{from}\ C= 90+104=194^{\circ} \text{(nearest degree)} \)

Networks, STD2 N2 2024 HSC 16

A network of towns and the distances between them in kilometres is shown.
 

  1. What is the shortest path from \(T\) to \(H\) ?   (2 marks)

  2. A truck driver needs to travel from \(Y\) to \(G\) but knows that the road from \(C\) to \(G\) is closed.
  3. What is the length of the shortest path the truck driver can take from \(Y\) to \(G\) after the road closure?  (2 marks)

Show Answers Only

a.   \(TYWH\)

b.  \(\text{Length of shortest path}\ (YWHMG) = 89 \text{km}\)

Show Worked Solution

a.    \(TYH=30+38=68, \quad TYWH=30+15+20=65\)

\(\therefore \text{ Shortest Path is}\ TYWH.\)
 

b.    \(Y W C M G=15+25+25+25=90\)

\(YWHMG=15+20+29+25=89\)

\(\Rightarrow \ \text{All other paths are longer.}\)

\(\therefore\text{ Length of shortest path = 89 km}\)

BIOLOGY, M1 EQ-Bank 4 MC

Which of the following structures is present in both prokaryotic and eukaryotic cells?

  1. Golgi apparatus
  2. Mitochondria
  3. Ribosomes
  4. Endoplasmic reticulum
Show Answers Only

\(C\)

Show Worked Solution

→ Both prokaryotic and eukaryotic cells contain ribosomes, which are responsible for protein synthesis.

→ Prokaryotes lack membrane-bound organelles like mitochondria or the Golgi apparatus.

\(\Rightarrow C\)

BIOLOGY, M6 2019 VCE 27 MC

Farmers and supermarkets agree that green beans are bought more frequently than yellow beans. A supermarket has asked a farmer to produce only green beans.

One way this could be achieved is by

  1. condensation polymerisation.
  2. DNA hybridisation.
  3. selective breeding.
  4. adaptive radiation.
Show Answers Only

\(C\)

Show Worked Solution

→ Selective breeding involves choosing parents with desirable traits and breeding them to produce offspring with those traits.

→ In this case, the farmer would selectively breed green bean plants to produce more green beans, as that is the preferred variety by consumers and supermarkets.

→ Selective breeding does not involve genetic engineering techniques like DNA hybridization (option B) and is a common agricultural practice used to enhance desirable traits in crops, unlike options A and D which are not relevant to this scenario.

\(\Rightarrow C\)

BIOLOGY, M5 2019 VCE 5 MC

Which one of the following statements about proteins is correct?

  1. The activity of a protein may be affected by the temperature and pH of its environment.
  2. The primary structure of a protein refers to its three-dimensional protein shape.
  3. Proteins are not involved in the human immune response.
  4. A protein with a quaternary structure will be an enzyme.
Show Answers Only

\(A\)

Show Worked Solution

Consider option A:

→ Changes in temperature or pH can disrupt the interactions that stabilise a protein’s higher-order structure.

→ This causes it to lose its three-dimensional shape and become denatured which usually results in a non-functional protein (Option A is Correct)

\(\Rightarrow A\)

BIOLOGY, M7 2021 VCE 36 MC

A study assessed the effectiveness and safety of a drug called doxycycline. One hundred and fifty adults hospitalised with malaria were involved. These adults were randomly placed into two groups of equal size. One group received doxycycline in addition to standard care. The other group received standard care only.

The group receiving standard care only was the

  1. control group.
  2. variable group.
  3. unsupported group.
  4. experimental group.
Show Answers Only

\(A\)

Show Worked Solution

→ A control group does not receive the experimental treatment or any intervention being tested.

\(\Rightarrow A\)

v1 Algebra, STD2 A2 2012 HSC 13 MC

Conversion graphs can be used to convert from one currency to another.  
  


  

Abbie converted 70 New Zealand dollars into Euros. She then converted all of these Euros into Australian dollars.

How much money, in Australian dollars, should Abbie have? 

  1. $30  
  2. $45
  3. $55
  4. $95
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Using the graphs:}\)

\($70\ \text{New Zealand}\) \(=40\ \text{Euro}\)
\(40\ \text{Euro}\) \(=$55\  \text{Australian}\)

 
\(\Rightarrow C\)

v1 Algebra, STD2 A2 2022 HSC 16

Nicole is 38 years old, and likes to keep fit by doing cross-fit classes.

  1. Use this formula to find her maximum heart rate (bpm).
      
  2.      Maximum heart rate = 220 – age in years
      
  3. Nicole's maximum heart rate is ........................... bpm.   (1 mark)
  4. Nicole will get the most benefit from this exercise if her heart rate is between 50% and 85% of her maximum heart rate.
  5. Between what two heart rates should Nicole be aiming for to get the most benefit from her exercise?  (2 marks)

Show Answers Only

a.   \(182\ \text{bpm}\)

b.   \(91-155\ \text{bpm}\)

Show Worked Solution
a.     \(\text{Max heart rate}\) \(=220-38\)
    \(=182\ \text{bpm}\)

 

b.    \(\text{50% max heart rate}\ = 0.5\times 182 = 91\ \text{bpm}\)

\(\text{85% max heart rate}\ = 0.85\times 182 = 154.7\ \text{bpm}\)

\(\therefore\ \text{Nicole should aim for between 91 and 155 bpm during exercise.}\)

v1 Algebra, STD2 A4 2022 HSC 22

The formula  \(C=80n+b\)  is used to calculate the cost of producing desktop computers, where \(C\) is the cost in dollars, \(n\) is the number of desktop computers produced and \(b\) is the fixed cost in dollars.

  1. Find the cost \(C\) when 2458 desktop computers are produced and the fixed cost is \($18\ 230\).  (1 mark)

  2. Some desktop computers have extra features added. The formula to calculate the production cost for these desktop computers is
  3.     \(C=80n+an+18\ 230\)
  4. where \(a\) is the additional cost in dollars per desktop computer produced.
  5. Find the number of desktop computers produced if the additional cost is $35 per desktop computer and the total production cost is \($103\ 330\).  (2 marks)

Show Answers Only
  1. \($214\ 870\)
  2. \(740\ \text{desktop computers}\)
Show Worked Solution

a.   \(\text{Find}\ C,\ \text{given}\ n=2458\ \text{and}\ b=18\ 230\)

\(C\) \(=80\times 2458+18\ 230\)  
  \(=$214\ 870\)  

 

b.   \(\text{Find}\ n,\ \text{given}\ C=103\ 330\ \text{and}\ a=35\)

\(C\) \(=80n+an+18\ 230\)
\(103\ 330\) \(=80n+35n+18\ 230\)
\(115n\) \(=85\ 100\)
\(n\) \(=\dfrac{85\ 100}{115}\)
  \(=740\ \text{desktop computers}\)

PHYSICS, M6 2019 VCE 1 MC

Magnetic and gravitational forces have a variety of properties.

Which of the following best describes the attraction/repulsion properties of magnetic and gravitational forces?

\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}{|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \textbf{Magnetic forces}\rule[-1ex]{0pt}{0pt}& \ \textbf{Gravitational forces} \\
\hline
\rule{0pt}{2.5ex}\text{either attract or repel}\rule[-1ex]{0pt}{0pt}&\text{only attract}\\
\hline
\rule{0pt}{2.5ex}\text{only repel}\rule[-1ex]{0pt}{0pt}& \text{neither attract nor repel}\\
\hline
\rule{0pt}{2.5ex}\text{only attract}\rule[-1ex]{0pt}{0pt}& \text{only attract} \\
\hline
\rule{0pt}{2.5ex}\text{either attract or repel}\rule[-1ex]{0pt}{0pt}& \text{either attract or repel} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ In magnets, like poles repel and opposite poles attract.

→ Gravitational forces are only attractive.

\(\Rightarrow A\)

Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

  1.   i. Find \(\text{Pr}(X=5)\).   (1 mark)

     ii. Find \(\text{Pr}(X \geq 2).\) (1 mark)

  2. iii. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)

  3.  iv. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

  1.   i. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)

  2.  ii. Given that \(\text{Pr}(H \leq 2)=0.35\) and \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

  3. iii. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
  4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
  5. Find the values of \(r\) and \(s\).   (1 mark)

  6. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  7. Bella flips her coin 25 times in order to estimate \(p\).
  8. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)

    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)

    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`    ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`

`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.

  1. Show that `a=900` and `b=\frac{\pi}{80}`.   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of `y=\sin(t)` under the transformation `Q`. Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where `t>40`, in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)


Show Answers Only

ai.   `r(0)=2500`

aii.  Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

aiii. `160`  weeks

b.    See worked solution.

c.    `~~ 5339` (nearest whole number)

d.    Weeks between the periods is 160

e.    `~~ 4142` (nearest whole number)

f.    Average rate of change `= – 3.6` rabbits/week (1 d.p.)

g.    `t = 156` weeks (nearest whole number)  

h.    ` s → 2500`

Show Worked Solution

ai.  Initial population of rabbits

From graph when `t=0, \ r(0) = 2500`

Using formula when `t=0`

`r(t)` `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500`  
`r(0)` `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500`  rabbits  


aii. From graph,

Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

OR

Using formula

Minimum is when `t = 120`

`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`

Maximum is when `t = 40`

`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`

 
aiii. Number of weeks between maximum populations of rabbits `= 200  –  40 = 160`  weeks

 
b.  Period of foxes = period of rabbits = 160:

`frac{\2pi}{b} = 160`

`:.\  b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.

Using the point `(100 , 2500)`

Amplitude when `b = frac{\pi}{80}`: 

`f(t)` `=a \ sin (pi/80(t-60))+1600`  
`f(100)` `= 2500`  
`2500` `= a \ sin (pi/80(100-60))+1600`  
`2500` `= a \ sin (pi/2)+1600`  
`a` `= 2500  –  1600 = 900`  

 
`:.\  f(t)= 900 \ sin (pi/80)(t  –  60) + 1600`

  

c.    Using CAS find `h(t) = f(t) + r(t)`:

`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`

  
`text{fMax}(h(t),t)|0 <= t <= 160`      `t = 53.7306….`
  

`h(53.7306…)=5339.46`

  
Maximum combined population `~~ 5339` (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

`text{fMax}(h(t),t)|0 <= t <= 320`     `t = 213.7305…`
 
`h(213.7305…)=5339.4568….`
 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

`t^{\prime} = frac{90}{pi}t + 60`   →   `t = frac{pi}{90}(t^{\prime} – 60)`

`y^{\prime} = 900y+1600`   →   `y = frac{1}{900}(y^{\prime} – 1600)`

`frac{y^{\prime} – 1600}{900} = sin(frac{pi(t^{\prime} – 60)}{90})`

`:.\  f(t) = 900\ sin\frac{pi}{90}(t  –  60) + 1600`

  
Average combined population  [Using CAS]   
  
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
  

`= 4142.2646…..  ~~ 4142` (nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted `r(t)`. Others used average rate of change instead of average value.

f.   Using CAS

`s(t):= 1700e^(- 0.003t) dot\sin\frac{pit}{80} + 2500`
 

`text{fMax}(s(t),t)|0<=t<=320`        `x = 38.0584….`
 

`s(38.0584….)=4012.1666….`
 

`text{fMax}(s(t),t)|160<=t<=320`     `x = 198.0584….`
 

`s(198.0584….)=3435.7035….`
 

Av rate of change between the points

`(38.058 , 4012.167)`  and  `(198.058 , 3435.704)`

`= frac{4012.1666…. – 3435.7035….}{38.0584…. – 198.0584….} = – 3.60289….`
 

`:.` Average rate of change `= – 3.6` rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
`frac{s(200) -s(40)}{200 – 40}` was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

`s^(”)(t) = 0` , `t = 80(n – 0.049) \ \forall n \in Z`

After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2` 

`t` `= 80(n  –  0.049)`  
  `= 80(2  –  0.049)`  
  `= 156.08`  

 
`:. \ t = 156` weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved `frac{ds}{dt}=0`.
A common answer was 41.8, as `s(156.11…)=41.79`.
Another common incorrect answer was 76 weeks.

h.   As `t → ∞`, `e^(- 0.003t) → 0`

`:.\ s → 2500`

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
 

  1. State the equation of the axis of symmetry of the graph of `f`.   (1 mark)

  2. State the derivative of `f` with respect to `x`.   (1 mark)

The tangent to `f` at point `M` has gradient `-2` .

  1. Find the equation of the tangent to `f` at point `M`.   (2 marks)

The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
 

 

  1.  i. Find the equation of the line perpendicular to the tangent passing through point `M`.   (1 mark)

  2. ii. The line perpendicular to the tangent at point `M` also cuts `f` at point `N`, as shown in the diagram above.
  3.     Find the area enclosed by this line and the curve `y=f(x)`.   (2 marks)

  4. Another parabola is defined by the rule `g(x)=\frac{x^2}{4 a^2}`, where `a>0`.
  5. A tangent to `g` and the line perpendicular to the tangent at `x=-b`, where `b>0`, are shown below.

  1. Find the value of `b`, in terms of `a`, such that the shaded area is a minimum.   (4 marks)

Show Answers Only

a.    `x=0`

b.    ` f^{\prime}(x)=1/6x`

c.    `x=-12`

di.    `y=1/2x + 18`

dii.   Area`= 375` units²

e.    `b = 2a^2`

Show Worked Solution

a.   Axis of symmetry:  `x=0`
  

b.    `f(x)`  `=\frac{x^2}{12}`  
  ` f^{\prime}(x)` `= 1/6x`  

  
c.   At `M` gradient `= -2`

`1/6x` `= -2`  
`x` `= -12`  

 
When  `x = -12`
 

`f(x) = (-12)^2/12 = 12`
 
Equation of tangent at `(-12 , 12)`:

`y  –  y_1` `=m(x  –  x_1)`  
`y  –  12` `= -2(x + 12)`  
`y` `= -2x -12`  


d.i   Gradient of tangent `= -2`

`:.`  gradient of normal `= 1/2` 

Equation at `M(- 12 , 12)`

`y  –  y_1` `=m(x  –  x_1)`  
`y  –  12` `= 1/2(x + 12)`  
`y` `=1/2x + 18`  


d.ii  Points of intersection of `f(x)` and normal are at  `M` and `N`.

So equate ` y = x^2/12` and  `y = 1/2x + 18` to find `N`

`x^2/12` `=1/2x + 18`  
`x^2  –  6x  –  216` `=0`  
`(x + 12)(x  –  18)` `=0`  

   
`:.\  x = -12` or `x = 18`

Area `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x`  
  `= [x^2/4 + 18x  –  x^3/36]_(-12)^18`  
  `= [18^2/4 +18^2  –  18^3/36] – [12^2/4 + 18 xx (-12)  –  (-12)^3/36]`  
  `= 375` units²  

 

e.   `g(x) = x^2/(4a^2)`   `a > 0`

At `x = – b`   `y = (-b)^2/(4a^2) = b^2/4a^2`

`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`

Gradient of tangent `= (-b)/(2a^2)`

Gradient of normal `= (2a^2)/b`

Equation of normal at `(- b , b^2/(4a^2))`

`y  –  y_1` `= m(x  –  x_1)`  
`y  –  b^2/(4a^2)` `= (2a^2)/b(x – (-b))`  
`y` `= (2a^2x)/b + 2a^2 + b^2/(4a^2)`  
`y` `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)`  

 
Points of intersection of normal and parabola (Using CAS)

solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`

`x = – b`  or  `x = (8a^4+b^2)/b`

 
Calculate area using CAS

`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2} – frac{x^2}{4a^2} \right) dx`

`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
 
Using CAS Solve derivative of `A = 0`  with respect to `b` to find `b`

solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`

 
`b = – 2a^2`   and  `b = 2a^2`

Given `b > 0`

`b = 2a^2`


♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found `\int_{-b}^{\frac{8 a^4+b^2}{b}}\left(y_n\right) d x` and failed to subtract `g(x)` or had incorrect terminals.

Calculus, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g'(x)\), can be expressed in the form \(g'(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
    1. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)
    1. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
Show Answers Only

a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g'(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h'(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h”(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h”(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h'(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f'(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

\(\text{Substitute (1) into (2)}\)

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Calculus, MET2 2023 VCAA 1

Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.

  1. State the coordinates of all axial intercepts of \(f\).   (1 mark)
  2. Find the coordinates of the stationary points of \(f\).   (2 marks)
    1. Let \(g:R\rightarrow R, g(x)=x-2\).
    2. Find the values of \(x\) for which \(f(x)=g(x)\).   (1 mark)
    1. Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\).  (2 marks)
    2. Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places.   (1 mark)
  1. Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
  2. Find the possible values of \(a\) and \(b\).   (4 marks)
Show Answers Only

a.    \((-1, 0), (0, 0), (2, 0)\)

b.    \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)

c.i.  \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

 \(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)

c.iii. \(5.95\)

d.   \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

b.    \(\text{Using CAS solve for}\ x:\)

\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)

\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)

\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)

\(\text{The stationary points of }f\ \text{are}:\)

\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
  

ci    \(\text{Given }f(x)=g(x)\)

\(x(x-2)(x+1)\) \(=x-2\)
\(x(x-2)(x+1)(x-2)\) \(=0\)
\((x-2)(x(x+1)-1)\) \(=0\)
\((x-2)(x^2+x-1)\) \(=0\)

  
\(\therefore\ \text{Using CAS: } \)

\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

cii  \(\text{Area of bounded region:}\)

\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
  

ciii  \(\text{Solve the integral in c.ii above using CAS:}\)
  \(\text{Total area}=5.946045..\approx 5.95\)

  

d.   \(\text{Method 1 – Equating coefficients}\)

\((x-a)(x-b)^2=x(x-2)(x+1)+k\)

\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)

\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)

\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)

\(2ab+b^2=-2\ \ …(2)\)

\(\text{Substitute (1) into (2) and solve for }b.\)

\(2b(1-2b)+b^2\) \(=-2\)
\(3b^2-2b-2\) \(=0\)
\(b\) \(=\dfrac{1\pm \sqrt{7}}{3}\)
\(\text{When }b\) \(=\dfrac{1+\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\)
\(\text{When }b\) \(=\dfrac{1-\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\)

  

\(\text{Method 2 – Using transformations}\)

\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)

\(\text{shows that the turning point is on the }x\ \text{axis}.\)

\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)

\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

 

Data Analysis, GEN2 2023 VCAA 2a

The following data shows the sizes of a sample of 20 oysters rated as small, medium or large.

\begin{array} {ccccc}
\text{small} & \text{small} & \text{large} & \text{medium} & \text{medium} \\
\text{medium} & \text{large} & \text{small} & \text{medium} & \text{medium}\\
\text{small} & \text{medium} & \text{small} & \text{small} & \text{medium}\\
\text{medium} & \text{medium} & \text{medium} & \text{small} & \text{large}
\end{array}

  1. Use the data above to complete the following frequency table.   (1 mark)

  1. Use the percentages in the table to construct a percentage segmented bar chart below. A key has been provided.   (1 mark)

     

Show Answers Only

i.    

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Size} \rule[-1ex]{0pt}{0pt} & \textbf{Number} & \textbf{Percentage (%) } \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 7 & 35 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 10 & 50 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 3 & 15 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 20 & 100 \\
\hline
\end{array}

ii.    
     

Show Worked Solution

i.    

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Size} \rule[-1ex]{0pt}{0pt} & \textbf{Number} & \textbf{Percentage (%) } \\
\hline
\rule{0pt}{2.5ex} \text{small} \rule[-1ex]{0pt}{0pt} & 7 & 35 \\
\hline
\rule{0pt}{2.5ex} \text{medium} \rule[-1ex]{0pt}{0pt} & 10 & 50 \\
\hline
\rule{0pt}{2.5ex} \text{large} \rule[-1ex]{0pt}{0pt} & 3 & 15 \\
\hline
\rule{0pt}{2.5ex} \textbf{Total} \rule[-1ex]{0pt}{0pt} & 20 & 100 \\
\hline
\end{array}

 
ii.    
         

Data Analysis, GEN1 2023 VCAA 1-2 MC

The dot plot below shows the times, in seconds, of 40 runners in the qualifying heats of their 800 m club championship.
 

Question 1

The median time, in seconds, of these runners is

  1. 135.5
  2. 136
  3. 136.5
  4. 137
  5. 137

 
Question 2

The shape of this distribution is best described as

  1. positively skewed with one or more possible outliers.
  2. positively skewed with no outliers.
  3. approximately symmetric with one or more possible outliers.
  4. approximately symmetric with no outliers.
  5. negatively skewed with one or more possible outliers.
Show Answers Only

\(\text{Question 1:}\ B\)

\(\text{Question 2:}\ A\)

Show Worked Solution

\(\text{Question 1}\)

\(\text{40 data points}\ \Rightarrow \ \text{Median = average of 20th and 21st data points}\)

\(\text{Median}\ = \dfrac{136 + 136}{2} = 136\)

\(\Rightarrow B\)
 

\(\text{Question 2}\)

\(\text{Distribution is positive skewed (tail stretches to the right)} \)

\(\text{Q}_1 = \dfrac{135+135}{2} = 135\)

\(\text{Q}_3 = \dfrac{138+138}{2} = 138\)

\(\text{IQR} = 138-135=3 \)

\(\text{Outlier (upper fence)}\ = 138+ 1.5 \times 3 = 142.5\)

\(\Rightarrow A\)

ENGINEERING, TE 2023 HSC 3 MC

Why is pure copper preferred over a copper alloy in telecommunications applications?

  1. It has higher stiffness.
  2. It has better conductivity.
  3. It can be precipitation hardened.
  4. It has a better strength to weight ratio.
Show Answers Only

\( B \)

Show Worked Solution

→ Pure copper offers lower stiffness than copper alloys (eliminate A), is work hardened (eliminate C), and generally has lower strength to weight ratios (eliminate D).

→ In telecommunications, where high conductivity is crucial for transmitting electrical signals, pure copper is the preferred choice to ensure efficient signal transmission.

\(\Rightarrow B \)

ENGINEERING, PPT 2023 HSC 24a

Roller coaster support structures can be made from either timber or steel.

Compare the properties of the two materials in roller coaster support structures.  (2 marks)

Show Answers Only

→ Unlike steel, timber has the ability to flex and bend, which can absorb some of the forces exerted by the roller coaster.

→ However, timber has less mechanical strength than steel and is more susceptible to rot and insect damage over time.
 

Answers could include:

→ Steel frames are much more easily fabricated and assembled than timber frames, which can save time and costs in the construction process.

→ Steel is much more resistant to fire than timber, which makes it a safer material to use in roller coasters.

→ Steel has greater mechanical strength and is a more durable material than timber. It is able to withstand the higher stresses and forces exerted on a roller coaster.

Show Worked Solution

→ Unlike steel, timber has the ability to flex and bend, which can absorb some of the forces exerted by the roller coaster.

→ However, timber has less mechanical strength than steel and is more susceptible to rot and insect damage over time.
 

Answers could include:

→ Steel frames are much more easily fabricated and assembled than timber frames, which can save time and costs in the construction process.

→ Steel is much more resistant to fire than timber, which makes it a safer material to use in roller coasters.

→ Steel has greater mechanical strength and is a more durable material than timber. It is able to withstand the higher stresses and forces exerted on a roller coaster.

ENGINEERING, AE 2023 HSC 21b

You are part of a team of engineers working collaboratively on the design of a new aircraft.

Explain the benefits of collaboration when completing the engineering report.   (3 marks)

Show Answers Only

→ A collaborative approach allows for the pooling of expert knowledge. This will produce a more comprehensive report with a lower risk of errors within the report.

→ Time efficiency – a division of tasks among team members that utilises their specific skill sets can expedite the writing of the report.

→ Validation of decisions across multiple team members who can independently verify and validate design calculations and recommendations is a more comprehensive approach.

→ Collaboration between engineers with differing years of experience is crucial for professional development and training the next generation of engineers.

Show Worked Solution

→ A collaborative approach allows for the pooling of expert knowledge. This will produce a more comprehensive report with a lower risk of errors within the report.

→ Time efficiency – a division of tasks among team members that utilises their specific skill sets can expedite the writing of the report.

→ Validation of decisions across multiple team members who can independently verify and validate design calculations and recommendations is a more comprehensive approach.

→ Collaboration between engineers with differing years of experience is crucial for professional development and training the next generation of engineers.

ENGINEERING, AE 2023 HSC 21a

How can computer graphics be utilised as a tool in aeronautical engineering?   (2 marks)

Show Answers Only

→ Computer graphics can create a detailed and accurate visual models in three dimensions.

→ This technology can be extremely efficient in the design of aircraft components through simulation, allowing ideas to be tested and adjusted in short time frames.

Answers could also include:

  • The creation of interactive user interfaces
  • The provision of large amounts of data in the analysis of flight dynamics and servicing of aircraft
  • Improving manufacturing and prototyping.

Show Worked Solution

→ Computer graphics can create a detailed and accurate visual models in three dimensions.

→ This technology can be extremely efficient in the design of aircraft components through simulation, allowing ideas to be tested and adjusted in short time frames.

Answers could also include:

  • The creation of interactive user interfaces
  • The provision of large amounts of data in the analysis of flight dynamics and servicing of aircraft
  • Improving manufacturing and prototyping.

PHYSICS, M2 2017 VCE 7 MC

A model car of mass 2.0 kg is propelled from rest by a rocket motor that applies a constant horizontal force of 4.0 N, as shown below. Assume that friction is negligible.
 

Which one of the following best gives the magnitude of the acceleration of the model car?

  1. \(0.50 \text{ m s} ^{-2}\)
  2. \(1.0 \text{ m s}^{-2}\)
  3. \(2.0 \text{ m s} ^{-2}\)
  4. \( 4.0\text{ m s} ^{-2}\)
Show Answers Only

\(C\)

Show Worked Solution
\(a\) \(=\dfrac{F}{m}\)  
  \(=\dfrac{4.0}{2.0}\)  
  \(=2\ \text{ms}^{-2}\)  

 

\(\Rightarrow C\)

PHYSICS, M4 2021 VCE 2 MC

The diagram below shows the electric field lines between four charged spheres: \(\text{P, Q, R}\) and \(\text{S}\). The magnitude of the charge on each sphere is the same.
  
 


 

Which of the following correctly identifies the type of charge (+ positive or – negative) that resides on each of the spheres \(\text{P, Q, R}\) and \(\text{S}\)?

  \(\textbf{P}\) \(\textbf{Q}\) \(\textbf{R}\) \(\textbf{S}\)
A.   \(\quad - \quad\) \(\quad + \quad\) \(\quad - \quad\) \(\quad + \quad\)
B. \(\quad + \quad\) \(\quad - \quad\) \(\quad + \quad\) \(\quad - \quad\)
C. \(\quad - \quad\) \(\quad - \quad\) \(\quad + \quad\) \(\quad + \quad\)
D. \(\quad + \quad\) \(\quad + \quad\) \(\quad - \quad\) \(\quad - \quad\)
Show Answers Only

\(B\)

Show Worked Solution

→ Electric field lines travel away from positive charges and towards negative charges.

→ Therefore, \(\text{P}\) and \(\text{R}\) must be positive charges.

\( \Rightarrow B\)

CHEMISTRY, M3 2012 HSC 3 MC

What effect does a catalyst have on a reaction?

  1. It increases the rate.
  2. It increases the yield.
  3. It increases the heat of reaction.
  4. It increases the activation energy.
Show Answers Only

\(A\)

Show Worked Solution

→ A catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.

\(\Rightarrow A\)

CHEMISTRY, M3 2017 VCE 1 MC

A catalyst

  1. slows the rate of reaction.
  2. ensures that a reaction is exothermic.
  3. moves the chemical equilibrium of a reaction in the forward direction.
  4. provides an alternative pathway for the reaction with a lower activation energy.
Show Answers Only

\(D\)

Show Worked Solution

→ A catalyst increases the rate of a chemical reaction by reducing the activation energy of the reaction. Thus, by collision theory, particles require less energy than normal to react to form products.

\(\Rightarrow D\)

CHEMISTRY, M2 2016 VCE 9a

Standard solutions of sodium hydroxide, \(\ce{NaOH}\), must be kept in airtight containers. This is because \(\ce{NaOH}\) is a strong base and absorbs acidic oxides, such as carbon dioxide, \(\ce{CO2}\), from the air and reacts with them. As a result, the concentration of \(\ce{NaOH}\) is changed to an unknown extent.

\(\ce{CO2}\) in the air reacts with water to form carbonic acid, \(\ce{H2CO3}\). This can react with \(\ce{NaOH}\) to form sodium carbonate, \(\ce{Na2CO3}\).

  1. Write a balanced overall equation for the reaction between \(\ce{CO2}\) gas and water to form \(\ce{H2CO3}\).   (1 mark)

  2. Write a balanced equation for the complete reaction between \(\ce{H2CO3}\) and \(\ce{NaOH}\) to form \(\ce{Na2CO3}\).   (1 mark)

Show Answers Only

i.    \(\ce{CO2(g) + H2O(l) \rightarrow H2CO3(aq)} \)

ii.   \(\ce{2NaOH(aq) + H2CO3(aq) \rightarrow Na2CO3(aq) + 2H2O(l)}\)

Show Worked Solution

i.    \(\ce{CO2(g) + H2O(l) \rightarrow H2CO3(aq)} \)

ii.   \(\ce{2NaOH(aq) + H2CO3(aq) \rightarrow Na2CO3(aq) + 2H2O(l)}\)

PHYSICS, M2 2014 HSC 3 MC

A pendulum is used to determine the value of acceleration due to gravity. The length of the pendulum is varied, and the time taken for the same number of oscillations is recorded.

Which of the following could increase the reliability of the results?

  1. Changing the mass of the pendulum
  2. Identifying the independent and dependent variables
  3. Recording all measurements to at least four significant figures
  4. Repeating each measurement several times and recording the average
Show Only

\(D\)

Show Worked Solution

→ Reliability of data can be increased by repeating the same experiment numerous times and recording an average from the data as it eliminates random error.

\(\Rightarrow D\)

CHEMISTRY, M7 2023 HSC 1 MC

What is the safest method for disposing of a liquid hydrocarbon after an experiment?

  1. Pour it down the sink
  2. Place it in a garbage bin
  3. Burn it by igniting with a match
  4. Place it in a separate waste container
Show Answers Only

\(D\)

Show Worked Solution

→ Organic substances should be kept separate for safety (avoiding any possible other reactions).

\(\Rightarrow D\)

PHYSICS, M5 2023 HSC 1 MC

The gravitational field strength acting on a spacecraft decreases as its altitude increases.

This is due to a change in the

  1. mass of Earth.
  2. mass of the spacecraft.
  3. density of the atmosphere.
  4. distance of the spacecraft from Earth's centre.
Show Answers Only

\(D\)  

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\(g=\dfrac{GM}{r^2}\ \ \Rightarrow\ \ g \propto \dfrac{1}{r^2}\)

\(\therefore\) As r increases, the gravitational field strength decreases

\( \Rightarrow D\)

Vectors, EXT2 V1 2023 HSC 11b

Find the angle between the vectors

\(\underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k}\)

\(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\),

giving your answer to the nearest degree.  (3 marks)

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\(87^{\circ} \)

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\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} -1 \\ 4 \\ 2 \end{array}\right) \]

\(\Big{|} \underset{\sim}{a} \Big{|} = \sqrt{1+4+9} = \sqrt{14} \)

\(\Big{|} \underset{\sim}{b} \Big{|} = \sqrt{1+16+4} = \sqrt{21} \)

\( \underset{\sim}{a} \cdot \underset{\sim}{b} = -1 + 8-6=1 \)

\(\cos\ \theta \) \(=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\Big{|}\underset{\sim}{a}\Big{|} \cdot \Big{|}\underset{\sim}{b}\Big{|}} \)  
  \(=\dfrac{1}{\sqrt{294}} \)  
\( \theta\) \(=\cos ^{-1} \Big{(}\dfrac{1}{\sqrt{294}}\Big{)} \)  
  \(=86.65…\)  
  \(=87^{\circ} \)  

Calculus, EXT1 C1 2023 HSC 1 MC

The temperature \(T(t)^{\circ} \text{C}\) of an object at time \(t\) seconds is modelled using Newton's Law of Cooling,

\(T(t)=15+4 e^{-3 t}\)

What is the initial temperature of the object?

  1. \(-3\)
  2. \(4\)
  3. \(15\)
  4. \(19\)
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\(D\)

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\(\text{Initial temperature when}\ \ t=0:\)

\(T=15+4e^0=19\)

\(\Rightarrow D\)

Complex Numbers, EXT2 N1 2023 HSC 1 MC

Which of the following is equal to \((a+i b)^3\)?

  1. \( (a^3-3 a b^2)+i (3 a^2 b+b^3) \)
  2. \( (a^3+3 a b^2)+i (3 a^2 b+b^3) \)
  3. \( (a^3-3 a b^2)+i (3 a^2 b-b^3) \)
  4. \( (a^3+3 a b^2)+i(3 a^2 b-b^3)\)
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\(C\)

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\((a+i b)^3\) \(=a^3+3a^2ib+3a(ib)^2+(ib)^3\)  
  \(=a^3+3a^2ib-3ab^2-ib^3\)  
  \(=(a^3-3ab^2)+i(3a^2b-b^3) \)  

 
\(\Rightarrow C\)