Band 4

Data Analysis-GEN2 2024 VCAA 4

The time series plot below shows the gold medal-winning height for the women's high jump, $\textit{Wgold}$, in metres, for each Olympic year, year, from 1952 to 1988.

A five-median smoothing process will be used to smooth the time series plot above.

The first two points have been placed on the graph with crosses (X) and joined by a dashed line (---).

Complete the five-median smoothing by marking smoothed values with crosses (X) joined by a dashed line (---) on the time series plot above. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Identify two qualitative features that best describe the time series plot above. (1 mark)

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

Show Answers Only

b. $\text{Random fluctuations, increasing trend.}$

Show Worked Solution

a. $\text{Medians are:}$

$\textbf{1960}: 1.67, 1.76, \colorbox{lightblue}{1.82}, 1.85, 1.90$

$\textbf{1964}: 1.76, 1.82, \colorbox{lightblue}{1.85}, 1.90, 1.92$

$\textbf{1968}: 1.82, 1.85, \colorbox{lightblue}{1.90}, 1.92, 1.93$

$\textbf{1972}: 1.82, 1.90, \colorbox{lightblue}{1.92}, 1.93, 1.97$

$\textbf{1976}: 1.82, 1.92, \colorbox{lightblue}{1.93}, 1.97, 2.02$

$\textbf{1980}: 1.92, 1.93, \colorbox{lightblue}{1.97}, 2.01, 2.02$

b. $\text{Time series shows random fluctuations with an increasing trend.}$

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)

--- 1 WORK AREA LINES (style=lined) ---
Draw the least squares line on the scatterplot above. (1 mark)

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

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

\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)

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

--- 4 WORK AREA LINES (style=lined) ---
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)
  
  --- 0 WORK AREA LINES (style=lined) ---
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 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)
  
  --- 1 WORK AREA LINES (style=lined) ---
2. the percentage of $\textit{Mgold}$ values greater than 2.25 m. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
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)

--- 3 WORK AREA LINES (style=lined) ---
Construct a boxplot for the $\textit{Mgold}$ data in Table 1 on the grid below. (2 marks)

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

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)

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

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, M7 2024 HSC 13 MC

Which of the following identifies plant responses to pathogens?

Increased phagocytosis and programmed cell death
Increased number of stomata and programmed cell death
Production of antihistamines and increased thickness of cell walls
Production of antimicrobial substances and increased thickness of cell walls

Show Answers Only

$D$

Show Worked Solution

→ Plants respond to pathogens by producing antimicrobial substances and strengthening their cell walls for enhanced physical defence.

→ The other options either describe animal immune responses (phagocytosis, antihistamines) or incorrect plant responses (increased stomata).

$\Rightarrow D$

BIOLOGY, M7 2024 HSC 12 MC

Robert Koch produced a set of criteria to establish whether a particular organism is the cause of a disease in an animal. The criteria are listed below but not in the correct order.

Which of the following correctly shows the order of steps required to determine the cause of a particular disease in an animal?

2, 3, 1, 4
2, 4, 1, 3
4, 2, 1, 3
4, 3, 2, 1

Show Answers Only

$C$

Show Worked Solution

→ Koch’s postulates follow a logical sequence of first identifying the microorganism in sick but not healthy animals (4).

→ Second step is isolating and culturing it (2), using it to infect a healthy animal to prove it causes disease (1), and finally re-isolating the same organism from the newly infected animal (3) to conclusively prove causation.

$\Rightarrow C$

BIOLOGY, M5 2024 HSC 6 MC

The diagram shows the karyotypes of a body cell for a male and a female fruit fly.

How many chromosomes will the egg of a female fruit fly have?

Show Answers Only

$B$

Show Worked Solution

→ In the diagram, the female fruit fly has 8 chromosomes in body cells (2 sex chromosomes + 6 autosomes across pairs I, II, and III)

→ During gamete formation, the female fruit fly’s cells undergo meiosis which halves the total chromosome number.

→ Therefore her eggs will contain half that number, resulting in 4 chromosomes total.

$\Rightarrow B$

BIOLOGY, M7 2024 HSC 10 MC

Francesco Redi challenged the idea that maggots arose spontaneously from rotting meat. A modern version of his experiment was set up as shown.

Which of the following is correct for this experimental set up?

The sealed jar improves the validity of the experiment.
The independent variable is whether the meat spoils or not.
The use of three jars improves the reliability of the experiment.
The dependent variable is the use of different covers for the jars.

Show Answers Only

$A$

Show Worked Solution

→ The sealed jar improves the validity of Redi’s experiment because it serves as a proper control that completely prevents flies from reaching the meat.

→ This allows Redi to conclusively demonstrate that maggots come from flies laying eggs rather than spontaneous generation.

$\Rightarrow A$

BIOLOGY, M6 2024 HSC 9 MC

The diagram shows a section of a chromosome in an insect. It represents three genes amongst non-coding DNA. The crosses mark locations of four separate mutations.

Which location could produce a new allele for eye colour?

$P$
$Q$
$R$
$S$

Show Answers Only

$C$

Show Worked Solution

→ Since $R$ is located within the gene region for eye colour, a mutation at this location could alter the DNA sequence of this gene and produce a new allele for eye colour,.

→ Mutations at other locations ($P, Q$ and $S$) are either in non-coding regions or different genes.

$\Rightarrow C$

Networks, GEN1 2024 VCAA 40 MC

A project has 15 activities, $A-O$, that need to be completed.

The directed network that represents this project is shown below.

The activities are not labelled.

The activity table that could represent this project is

Show Answers Only

$B$

Show Worked Solution

$\text{Using trial and error:}$

$\Rightarrow B$

BIOLOGY, M5 2024 HSC 5 MC

The diagram shows a cell reproducing.

Which row of the table correctly identifies the method of reproduction and the type of organism shown in the diagram?

\begin{align*}
\begin{array}{l}
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|l|l|}
\hline
\textit{Method of reproduction} & \textit{Type of organism} \\
\hline
\text{Budding} & \text{Fungi} \\
\hline
\text{Binary fission} & \text{Bacteria} \\
\hline
\text{Production of spores} & \text{Plant} \\
\hline
\text{Gamete production} & \text{Protist} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$A$

Show Worked Solution

→ The diagram shows a cell with both mitochondria and a cell wall, and is reproducing via an outgrowth from the parent cell.

→ This represents budding in fungi rather than the other reproductive methods listed for different organisms.

$\Rightarrow A$

BIOLOGY, M5 2024 HSC 4 MC

Which row of the table correctly identifies components of DNA?

\begin{align*}
\begin{array}{l}
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|c|c|}
\hline
\textit{Phosphate} & \textit{Ribose} & \textit{Deoxyribose} & \textit{Uracil} & \textit{Thymine}\\
\hline
\checkmark & \text{} & \checkmark & \checkmark & \text{} \\
\hline
\text{} & \checkmark & \text{} & \checkmark & \checkmark \\
\hline
\text{} & \checkmark & \checkmark & \text{} & \checkmark \\
\hline
\checkmark & \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).

$\Rightarrow B$

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.

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

$\Rightarrow D$

Networks, GEN1 2024 VCAA 36 MC

Eight houses in an estate are to be connected to the internet via underground cables.

The network below shows the possible connections between the houses.

The vertices represent the houses.

The numbers on the edges represent the length of cable connecting pairs of houses, in metres.

The graph that represents the minimum length of cable needed to connect all the houses is

Show Answers Only

$D$

Show Worked Solution

$\text{Consider all options}$

$\text{Option A: contains a circuit}\ \rightarrow\ \text{Eliminate A}$

$\text{Option B:}\ 19+18+16+15+16+14+18=116$

$\text{Option C:}\ 20+19+18+16+15+16+14=118$

$\text{Option D:}\ 19+18+16+15+16+14+17=115$

$\Rightarrow D$

Matrices, GEN1 2024 VCAA 28 MC

A primary school is hosting a sports day.

Students represent one of four teams: blue $(B)$, green $(G)$, red $(R)$ or yellow $(Y)$.

Students compete in one of three sports: football $(F)$, netball $(N)$ or tennis $(T)$.

Matrix $W$ shows the number of students competing in each sport and the team they represent.

\begin{aligned} \\
& \quad B \quad \ G \quad \ R \quad \ Y \\
W = & \begin{bmatrix}
85 & 60 & 64 & 71 \\
62 & 74 & 80 & 64 \\
63 & 76 & 66 & 75
\end{bmatrix}\begin{array}{l}
F\\
N\\
T
\end{array}
\end{aligned}

Matrix $W$ is multiplied by the matrix $\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}$ to produce matrix $X$.

Element $x_{31}$ indicates that

210 students represent the blue team.
210 students compete in netball.
280 students compete in tennis.
280 students compete in football.

Show Answers Only

$C$

Show Worked Solution

\begin{aligned} \\
X = & \begin{bmatrix}
280 \\
280 \\
280
\end{bmatrix}\begin{array}{l}
F\\
N\\
T
\end{array}
\end{aligned}

$\Rightarrow C$

Matrices, GEN1 2024 VCAA 27 MC

Consider the following matrix, where $h \neq 0$.

\begin{bmatrix}
4 & g \\
8 & h
\end{bmatrix}

The inverse of this matrix does not exist when $g$ is equal to

$-2 h$
$\dfrac{h}{2}$
$h$
$2 h$

Show Answers Only

$B$

Show Worked Solution

$\text{Inverse will not exist if }\ 4h-8g=0$

$\text{i.e. when}\ g=\dfrac{h}{2}$

$\Rightarrow B$

Recursion and Finance, GEN1 2024 VCAA 22-23 MC

Stewart takes out a reducing balance loan of \$240 000, with interest calculated monthly.

Stewart makes regular monthly repayments.

Three lines of the amortisation table are shown below.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Payment} & \textbf {Payment} & \textbf {Interest} &\textbf{Principal reduction} & \textbf{Balance}\\
\rule[-1ex]{0pt}{0pt}\textbf{number} & \textbf{($$$)} & \textbf{($$$)}& \textbf{($$$)}& \textbf{($$$)}\\
\hline
\rule{0pt}{2.5ex} 0 & 0.00 & 0.00 & 0.00 & 240000.00 \\
\hline
\rule{0pt}{2.5ex} 1 & 2741.05 & 960.00 & 1781.05 & 238218.95 \\
\hline
\rule{0pt}{2.5ex} 2 & 2741.05 & & & \\
\hline
\end{array}

Part A

The principal reduction associated with Payment number 2 is closest to

$1773.93
$1781.05
$1788.17
$2741.05

Part B

The number of years that it will take Stewart to repay the loan in full is closest to

Show Answers Only

Part A

$C$

Part B

$A$

Show Worked Solution

Part A

$\text{Interest}$	$=\dfrac{960}{240\,000}\times 238\,218.95$
	$=$952.88$

$\text{Principal Reduction}$	$=2741.05-952.88$
	$=$1788.17$

$\Rightarrow C$

Part B

$\text{Using CAS, number of repayments }= 108\ \text{months} = 9\ \text{years}$

$\Rightarrow A$

Recursion and Finance, GEN1 2024 VCAA 21 MC

Lee took out a loan of $121 000, with interest compounding monthly. He makes monthly repayments of $2228.40 for five years until the loan is repaid in full.

The total interest paid by Lee is closest to

$4434
$5465
$10539
$12 704

Show Answers Only

$D$

Show Worked Solution

$\text{Repayments}$	$ =2228.40\times 12\times 5$
	$=$133\,704$
$\text{Interest}$	$=$133\,704-$121\,000$
	$=$12\,704$

$\Rightarrow D$

Recursion and Finance, GEN1 2024 VCAA 20 MC

Dainika invested $2000 for three years at 4.4% per annum, compounding quarterly.

To earn the same amount of interest in three years in a simple interest account, the annual simple interest rate would need to be closest to

4.60%
4.68%
4.84%
4.98%

Show Answers Only

$B$

Show Worked Solution

$FV$	$=2000\left(1-\dfrac{0.044}{4}\right)^{12}$
	$=$2280.57$
$\text{Interest pa}$	$=\dfrac{280.57}{3}=$93.52$

$\text{SI rate}$	$=\dfrac{93.52}{2000}\times 100\%$
	$=4.676\dots\%$

$\Rightarrow B$

Recursion and Finance, GEN1 2024 VCAA 18 MC

Trevor took out a reducing balance loan of $400 000, with interest calculated weekly. The balance of the loan, in dollars, after $n$ weeks, $T_n$, can be modelled by the recurrence relation

$T_0=400\,000, \quad T_{n+1}=1.00075 T_n-677.55$

Assume that there are exactly 52 weeks in a year.

The interest rate, per annum, for this loan is

0.75%
3.9%
4.5%
7.5%

Show Answers Only

$B$

Show Worked Solution

$\text{Interest rate}\ = 0.00075\times 52 =0.039=3.9\%$

$\Rightarrow B$

Data Analysis, GEN1 2024 VCAA 16 MC

The table below shows the seasonal indices for the monthly takings of a bistro.

The seasonal indices for months 3 and 6 are missing.

\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{Seasonal}& 1.08 & 1.13 & & 0.92 & 0.67 & & 1.09 & 1.35 & 0.82 & 0.88 & 1.01 & 0.98 \\
\textbf{index} \\
\hline
\end{array}

The seasonal index for month 3 is twice the seasonal index for month 6 .

The seasonal index for month 3 is closest to

0.69
1.04
1.38
2.07

Show Answers Only

$C$

Show Worked Solution

$\text{Let}\ x=\text{seasonal index for month 6}$

$\rightarrow 2x=\text{seasonal index for month 3}$

$\dfrac{3x+1.08+1.113+0.92+0.67+1.09+1.35+0.82+0.88+1.01+0.98}{12}=1$

$x=\dfrac{12-9.93}{3}=0.69$

$\therefore\ \text{Seasonal Index month 3}\ =2\times 0.69=1.38$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 13-14 MC

A school runs an orientation program for new staff each January.

The time series plot below shows the number of new staff, $new$, for each year, $year$, from 2011 to 2022 (inclusive).

Part A

The time series is smoothed using seven-median smoothing.

The smoothed value of $new$ for the $year$ 2016 is

Part B

The number of new staff in 2023 is added to the total number of new staff from the previous 12 years. For these 13 years, the mean number of new staff is 11 .

The number of new staff in 2023 is

Show Answers Only

Part A

$A$

Part B

$B$

Show Worked Solution

Part A

$\text{Ordered scores}$

\begin{array} {|c|c|c|c|c|c|}
\hline 2018 & 2014 & 2013 & 2019 & 2015 & 2017 & 2016 \\
\hline 6 & 6 & 7 & 10 & 11 & 12 & 13 \\
\hline
\end{array}

$\text{Smoothed value of }new\ \text{for the }year\ 2016\ \text{is 10.}$

$A$

Part B

$\text{mean}$	$=\dfrac{\Sigma\text{scores}}{\text{Number of scores}}$
$11$	$=\dfrac{x+14+11+7+6+11+13+12+6+10+15+12+10}{13}$
$x+127$	$=143$
$x$	$=16$

$B$

Data Analysis, GEN1 2024 VCAA 9-10 MC

The least squares equation for the relationship between the average number of male athletes per competing nation, $males$, and the number of the Summer Olympic Games, $number$, is

$males =67.5-1.27 \times number$

Part A

The summary statistics for the variables number and males are shown in the table below.

The value of Pearson's correlation coefficient, $r$, rounded to three decimal places, is closest to

$-0.569$
$-0.394$
$0.394$
$0.569$

Part B

At which Summer Olympic Games will the predicted average number of $males$ be closest to 25.6 ?

31st
32nd
33rd
34th

Show Answers Only

Part A

$A$

Part B

$C$

Show Worked Solution

Part A

$b$	$=r\dfrac{s_y}{s_x}$
$-1.27$	$=r\times\dfrac{19}{8.51}$
$\therefore\ r$	$=\dfrac{-1.27\times 8.51}{19}$
	$=-0.5688\dots$

$\Rightarrow A$

Part B

$males$	$=67.5-1.27\times number$
$25.6$	$=67.5-1.27\times number$
$number$	$=\dfrac{67.5-25.6}{1.27}$
	$=32.992\dots$
	$\approx 33$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 8 MC

The scatterplot below displays the average number of female athletes per competing nation, $females$, against the number of the Summer Olympic Games, $number$, from the first Olympic Games, in 1896, to the 29th Olympic Games, held in 2021.

A least squares line has been fitted to the scatterplot.

The equation of the least squares line is closest to

$females =-4.87+1.02 \times number$
$ females =-3.39+0.91 \times number$
$number =-3.39+0.91 \times females$
$number =-0.91+3.39 \times females$

Show Answers Only

$B$

Show Worked Solution

$\text{LSRL is in the form}\ y=a+bx\ \rightarrow \ \text{Eliminate C and D}$

$\text{Using points }(29, 23) (7, 3)$

$\text{Gradient}$	$=\dfrac{23-3}{29-7}$
	$=\dfrac{10}{11}$
	$\approx 0.91$

$\Rightarrow B$

Data Analysis, GEN1 2024 VCAA 7 MC

Fiona plays nine holes of golf each week, and records her score.

Her mean score for all rounds in 2024 is 55.7

In one round, when she recorded a score of 48 , her standardised score was $z=-1.75$

The standard deviation for score in 2024 is

1.1
2.3
4.4
6.95

Show Answers Only

$C$

Show Worked Solution

$z$	$=\dfrac{x-\overline{x}}{s_x}$
$-1.75$	$=\dfrac{48-55.7}{s_x}$
$s_x$	$=\dfrac{48-55.7}{-1.75}$
	$=4.4$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 6 MC

More than 11000 athletes from more than 200 countries competed in the Tokyo Summer Olympic Games.

An analysis of the number of athletes per country produced the following five-number summary.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Minimum} \rule[-1ex]{0pt}{0pt}& \textbf{First quartile } & \textbf{Median } & \textbf{Third quartile} & \textbf{Maximum } \\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& 5 & 11 & 48 & 613 \\
\hline
\end{array}

The smallest number of athletes per country that would display as an outlier on a boxplot of this data is

Show Answers Only

$C$

Show Worked Solution

$IQR=48-5=43$

$\text{Lower limit not applicable as < 0.}$

$\text{Upper boundary}$	$=Q_3+1.5\times IQR$
	$=48+1.5\times 43$
	$=112.5$

$\therefore\ \text{Smallest number of athletes to show as an outlier = 113.}$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 3-4 MC

The histogram below displays the population density, in people per $km ^2$, of the 27 countries in the European Union in 2021. The histogram has a logarithmic (base 10) scale.

Part 1

The median value occurs in a column with a frequency of

Part 2

There is one outlier at the upper end of the histogram. This value could be

330
1330
2030
2730

Show Answers Only

Part 1

$D$

Part 2

$C$

Show Worked Solution

Part 1

$\text{There are 27 scores }\rightarrow\ \text{median is 14th score.}$

$\text{Column frequency }= 9$

$D$

Part 2

$10^{3.2} <\log\text{(Population Density)}<10^{3.4}$

$1585 <\log\text{(Population Density)}<2512$

$\Rightarrow C$

Complex Numbers, EXT2 N2 2024 HSC 9 MC

Consider the solutions of the equation $z^4=-9$.

What is the product of all of the solutions that have a positive principal argument?

$3$
$-3$
$3 i$
$-3 i$

Show Answers Only

$B$

Show Worked Solution

$z^4=-9$

$\text{Convert}\ z^4 \ \text{to Mod/Arg form:}$

$\left|z^4\right|=9, \ \ \arg \left(z^4\right)=\pi \ \text{(\(-9$ is on negative real axis})\)

$\text{By De Moivre:}$

$\abs{z}=\sqrt[4]{9}=\sqrt{3}$

$\arg (z)=\dfrac{\pi}{4}$

$\text{Roots are} \ \ \dfrac{\pi}{2} \ \ \text{rotations of}\ \ z=\sqrt{3} \, \text{cis}\left(\dfrac{\pi}{4}\right)$

$z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{\pi}{4}\right), z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{3 \pi}{4}\right)$

$\sqrt{3}\, \text{cis}\left(\dfrac{\pi}{4}\right) \cdot \sqrt{3} \, \text{cis}\left(\dfrac{3 \pi}{4}\right)=3 \, \text{cis}(\pi)=-3$

$\Rightarrow B$

Complex Numbers, EXT2 N2 2024 HSC 7 MC

It is given that $\abs{z-1+i}=2$.

What is the maximum possible value of $\abs{z}$?

$\sqrt{2}$
$\sqrt{10}$
$2+\sqrt{2}$
$2-\sqrt{2}$

Show Answers Only

$C$

Show Worked Solution

$\abs{z-1+i}=2 \ \Rightarrow \ \text{circle centre} \ \ (1,-i), \ \ \text {radius}=2$

$OA=\sqrt{1^2+1^2}=\sqrt{2}$

$AB=2 \ \ \text{(radius)}$

$\therefore \abs{z}_{\text{max}}=2+\sqrt{2}$

$\Rightarrow C$

Mechanics, EXT2 M1 2024 HSC 6 MC

A light string passes over a smooth pulley. Attached to the ends of the string are masses of 9 kg and 5 kg , as shown.

The acceleration due to gravity is $g$ m s$^{-2}$.

What is the acceleration of the 9 kg mass?

$\dfrac{2}{7}g$
$1 g$
$\dfrac{7}{2}g$
$4 g$

Show Answers Only

$A$

Show Worked Solution

$\text{Net force on 9 kg mass}\ = 9g-T\ \text{(down)}$

$\text{Net force on 5 kg mass}\ = 5g-T\ \text{(up)}$

$\text{Using}\ \ F=ma: $

$9 \times a$	$=9g-T\ …\ (1)$
$5 \times a$	$=-(5g-T)\ …\ (2) $

$\text{Adding (1) + (2):}$

$14a$	$=4g$
$a$	$=\dfrac{2}{7}g$

$\Rightarrow A$

Complex Numbers, EXT2 N2 2024 HSC 11f

Sketch the region defined by $|z|<3$ and $0 \leq \arg (z-i) \leq \dfrac{\pi}{2}$. (3 marks)

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

Show Answers Only

Show Worked Solution

$\text {Region: }\abs{z}<3\ \ \text{and}\ \ 0 \leqslant \arg (z-i) \leqslant \dfrac{\pi}{2}$

Mechanics, EXT2 M1 2024 HSC 15c

A bar magnet is held vertically. An object that is repelled by the magnet is to be dropped from directly above the magnet and will maintain a vertical trajectory. Let $x$ be the distance of the object above the magnet.

The object is subject to acceleration due to gravity, $g$, and an acceleration due to the magnet $\dfrac{27 g}{x^3}$, so that the total acceleration of the object is given by

$a=\dfrac{27 g}{x^3}-g$

The object is released from rest at $x=6$.

Show that $v^2=g\left(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\right)$. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Find where the object next comes to rest, giving your answer correct to 1 decimal place. (2 marks)

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

Show Answers Only

Show Worked Solution

i. $a=\dfrac{27 g}{x^3}-g$

$\dfrac{d}{dx}(\frac{1}{2}v^{2})$	$= \dfrac{27g}{x^3}-g$
$\dfrac{1}{2} v^2$	$=-\dfrac{27 g}{2 x^2}-g x+c$

$\text{When}\ \ x=6, v=0:$

$0$	$=-\dfrac{27g}{2 \times 6^2}-6g+c$
$c$	$=\dfrac{459 g}{72}=\dfrac{51 g}{8}$

	$\dfrac{1}{2} v^2$	$=-\dfrac{27 g}{2 x^2}-g x+\dfrac{51 g}{8}$
	$v^2$	$=-\dfrac{27 g}{x^2}-2 g x+\dfrac{51g }{4}$
		$=g\left(\dfrac{51}{4}-2 x-\dfrac{27}{x^2}\right)$

ii. $\text{Find \(x$ when $v=0$:}\)

$\dfrac{51}{4}-2 x-\dfrac{27}{x^2}$	$=0$
$51 x^2-8 x^3-108$	$=0$
$8 x^3-51 x^2+108$	$=0$

$\text{Given \(x=6$ is a root:}\)

$8 x^3-51 x^2+108=(x-6)\left(8 x^2-3 x-18\right)$

$\text{Other roots:}$

	$x$	$=\dfrac{3 \pm \sqrt{9-4 \cdot 8 \cdot 18}}{2 \times 8}$
		$=\dfrac{3 \pm \sqrt{585}}{16}$
		$=\dfrac{3+3 \sqrt{65}}{16} \quad(x>0)$
		$=1.7 \ \text{units (1 d.p.)}$

$\therefore \ \text{Object next comes to rest at \(x=1.7$ units}\)

Vectors, EXT2 V1 2024 HSC 14e

The diagram shows triangle $O Q A$.

The point $P$ lies on $O A$ so that $O P: O A=3: 5$.

The point $B$ lies on $O Q$ so that $O B: O Q=1: 3$.

The point $R$ is the intersection of $A B$ and $P Q$.

The point $T$ is chosen on $A Q$ so that $O, R$ and $T$ are collinear.

Let $\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}$ and $\overrightarrow{P R}=k \overrightarrow{P Q}$ where $k$ is a real number.

Show that $\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}$. (2 marks)

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

Writing $\overrightarrow{A R}=h \overrightarrow{A B}$, where $h$ is a real number, it can be shown that $\overrightarrow{O R}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b}$. (Do NOT prove this.)

Show that $k=\dfrac{1}{6}$. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find $\overrightarrow{O T}$ in terms of $\underset{\sim}{a}$ and $\underset{\sim}{b}$. (2 marks)

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

Show Answers Only

i. $\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}$

$\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} $

$\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} $

$\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}$

	$\overrightarrow{O R}$	$=\overrightarrow{O P}+k \overrightarrow{P Q}$
		$=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})$
		$=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}$
		$=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}$

ii. $\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}$

$\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}$

$\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}$

$\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}$.

$\text {Equating coefficients:}$

$\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)$

$h=3 k\ \ldots\ (2)$

$\text{Substituting (2) into (1)}$

	$\dfrac{3}{5}(1-k)$	$=1-3 k$
	$3-3 k$	$=5-15 k$
	$k$	$=\dfrac{1}{6}$

iii. $\overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})$

Show Worked Solution

i. $\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}$

$\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} $

$\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} $

$\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}$

	$\overrightarrow{O R}$	$=\overrightarrow{O P}+k \overrightarrow{P Q}$
		$=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})$
		$=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}$
		$=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}$

ii. $\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}$

$\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}$

$\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}$

$\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}$.

$\text {Equating coefficients:}$

$\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)$

$h=3 k\ \ldots\ (2)$

$\text{Substituting (2) into (1)}$

	$\dfrac{3}{5}(1-k)$	$=1-3 k$
	$3-3 k$	$=5-15 k$
	$k$	$=\dfrac{1}{6}$

iii. $\overrightarrow{O T}=\lambda \overrightarrow{O R}$

$\text {Using parts (i) and (ii):}$

$\overrightarrow{O R}=\dfrac{3}{5}\left(1-\dfrac{1}{6}\right)\underset{\sim}{a}+3\left(\dfrac{1}{6}\right) \underset{\sim}{b}=\dfrac{1}{2}\left(\underset{\sim}{a}+\underset{\sim}{b}\right)$

$\overrightarrow{O T}=\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})$

$\text {Find } \lambda:$

	$\overrightarrow{O T}$	$=\overrightarrow{O A}+\mu \overrightarrow{A Q}$
	$\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})$	$=\underset{\sim}{a}+\mu(3 \underset{\sim}{b}-\underset{\sim}{a}) \quad \Big(\text{noting}\ \overrightarrow{A Q}=\overrightarrow{O Q}-\overrightarrow{O A}=3 \underset{\sim}{b}-\underset{\sim}{a}\Big)$
		$=\underset{\sim}{a}(1-\mu)+3 \mu \underset{\sim}{b}$

$\text {Equating coefficients:}$

$\dfrac{\lambda}{2}=3 \mu \ \Rightarrow \ \mu=\dfrac{\lambda}{6}$

	$1-\dfrac{\lambda}{6}$	$=\dfrac{\lambda}{2}$
	$6-\lambda$	$=3 \lambda$
	$\lambda$	$=\dfrac{3}{2}$

$\therefore \overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})$

Proof, EXT2 P1 2024 HSC 14d

The following argument attempts to prove that $0=1$.

We evaluate $\displaystyle\int \frac{1}{x}\, d x$ using the method of integration by parts.

	$\displaystyle \int \frac{1}{x}\,d x$	$=\displaystyle \int \frac{1}{x} \times 1\, d x$
		$=\displaystyle\frac{1}{x} \times x-\int-\frac{1}{x^2} x\, d x$
		$=1+\displaystyle\int \frac{1}{x}\, d x$

We may now subtract $\displaystyle \int \frac{1}{x}\,d x$ from both sides to show that $0=1$.

Explain what is wrong with this argument. (2 marks)

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

Show Answers Only

$\text {Consider the line in proof }$

$\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}$

$\Rightarrow \text{ Each integral will have its own constant }$

$\Rightarrow \text{If constants are \(c_1$ and $c_2, \abs{c_1-c_2}=1$ in this proof.}\)

$\Rightarrow \text{Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}$

Show Worked Solution

$\text {Consider the line in proof }$

$\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}$

$\Rightarrow \text{ Each integral will have its own constant }$

$\Rightarrow \text{If constants are \(c_1$ and $c_2, \abs{c_1-c_2}=1$ in this proof.}\)

$\Rightarrow \text{Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}$

Proof, EXT2 P2 2024 HSC 14b

Use mathematical induction to prove that ${ }^{2 n} C_n<2^{2 n-2}$, for all integers $n \geq 5$. (3 marks)

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

Show Answers Only

$\text{Prove \({ }^{2 n} C_n<2^{2 n-2}$ for $n \geqslant 5$.}\)

$\text {If}\ \ n=5:$

$\text{LHS}={ }^{10} C_5=252$

$\text{RHS}=2^8=256>\text {LHS }$

$\therefore\ \text{True for}\ \ n=5$

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

${ }^{2 k} C_k<2^{2 k-2}\ \ldots\ (1)$

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

$\text{i.e. } {}^{2 k+2}C_{k+1}<2^{2 k}$

	$\text{LHS}$	$=\dfrac{(2 k+2)!}{(k+1)!(k+1)!}$
		$=\dfrac{(2 k)!}{k!k!} \times \dfrac{(2 k+1)(2 k+2)}{(k+1)(k+1)}$
		$={ }^{2 k} C_k \times 2 \times \dfrac{2 k+1}{k+1}$
		$< 2^{k-2} \times 2 \times \dfrac{2 k+1}{k+1}$
		$<2^k \ \ \ \left(\text{since}\ \dfrac{2 k+1}{k+1}=2-\dfrac{1}{k+1}<2\right)$

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

$\therefore \text{ Since true for \(n=5$, by PMI, true for integers $n \geqslant 5$.}\)

Show Worked Solution

$\text{Prove \({ }^{2 n} C_n<2^{2 n-2}$ for $n \geqslant 5$.}\)

$\text {If}\ \ n=5:$

$\text{LHS}={ }^{10} C_5=252$

$\text{RHS}=2^8=256>\text {LHS }$

$\therefore\ \text{True for}\ \ n=5$

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

${ }^{2 k} C_k<2^{2 k-2}\ \ldots\ (1)$

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

$\text{i.e. } {}^{2 k+2}C_{k+1}<2^{2 k}$

	$\text{LHS}$	$=\dfrac{(2 k+2)!}{(k+1)!(k+1)!}$
		$=\dfrac{(2 k)!}{k!k!} \times \dfrac{(2 k+1)(2 k+2)}{(k+1)(k+1)}$
		$={ }^{2 k} C_k \times 2 \times \dfrac{2 k+1}{k+1}$
		$< 2^{k-2} \times 2 \times \dfrac{2 k+1}{k+1}$
		$<2^k \ \ \ \left(\text{since}\ \dfrac{2 k+1}{k+1}=2-\dfrac{1}{k+1}<2\right)$

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

$\therefore \text{ Since true for \(n=5$, by PMI, true for integers $n \geqslant 5$.}\)

Proof, EXT2 P1 2024 HSC 14a

Prove that if $a$ is any odd integer, then $a^2-1$ is divisible by 8. (2 marks)

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

Show Answers Only

$\text {Prove if \(a$ is odd, $a^2-1$ is divisible by 8.}\)

$\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}$

	$a^2-1$	$=(2 n+1)^2-1$
		$=4 n^2+4 n+1-1$
		$=4 n(n+1)$

$\text{If \(n$ is odd ($n=2 k+1, k \in Z)$:}\)

$a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8$

$\text{If \(n$ is even $(n=2 k)$:}\)

$a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8$

$\therefore \text{ If \(a$ is odd, $a^2-1$ is divisible by 8}\)

Show Worked Solution

$\text {Prove if \(a$ is odd, $a^2-1$ is divisible by 8.}\)

$\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}$

	$a^2-1$	$=(2 n+1)^2-1$
		$=4 n^2+4 n+1-1$
		$=4 n(n+1)$

$\text{If \(n$ is odd ($n=2 k+1, k \in Z)$:}\)

$a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8$

$\text{If \(n$ is even $(n=2 k)$:}\)

$a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8$

$\therefore \text{ If \(a$ is odd, $a^2-1$ is divisible by 8}\)

Mechanics, EXT2 M1 2024 HSC 13c

A particle of unit mass moves horizontally in a straight line. It experiences a resistive force proportional to $v^2$, where $v$ m s$^{-1}$ is the speed of the particle, so that the acceleration is given by $-k v^2$.

Initially the particle is at the origin and has a velocity of 40 m s$^{-1}$ to the right. After the particle has moved 15 m to the right, its velocity is 10 m s$^{-1}$ (to the right).

Show that $v=40 e^{-k x}$. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that $k=\dfrac{\ln 4}{15}$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
At what time will the particle's velocity be 30 m s$^{-1}$ to the right? (3 marks)

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

Show Answers Only

Show Worked Solution

i. $\ddot{x}=v \cdot \dfrac{d v}{d x}=-k v^2$

	$\dfrac{d v}{d x}$	$=-k v$
	$\dfrac{d x}{d v}$	$=-\dfrac{1}{k v}$
	$\displaystyle\int \frac{1}{v}\, d v$	$=\displaystyle -\int k\, d x$
	$\ln \abs{v}$	$=-k x+c$
	$v$	$=e^{-k x+c}$
		$=A e^{-k x}\ \ (\text{where } A=e^c)$

$\text {When } x=0, v=40:$

$40=A e^{\circ} \ \Rightarrow \ A=40$

$\therefore V=40 \, e^{-k x}$

ii. $\text {Show}\ \ k=\dfrac{\ln 4}{15}$

$\text {When } x=15, v=10:$

	$10$	$=40 e^{-15 k}$
	$e^{-15 k}$	$=\dfrac{1}{4}$
	$-15 k$	$=\ln \left(\dfrac{1}{4}\right)$
	$15 k$	$=\ln 4$
	$k$	$=\dfrac{\ln 4}{15}$

iii. $\text {Find}\ t\ \text {when}\ \ v=30:$

	$\dfrac{d v}{d t}$	$=-k v^2$
	$\dfrac{d t}{d v}$	$=-\dfrac{1}{k v^2}$
	$t$	$=\displaystyle -\int \dfrac{1}{k v^2}\, d v=\dfrac{1}{k v}+c$

$\text {When}\ \ t=0, v=40:$

$0=\dfrac{1}{40 k}+c \ \Rightarrow \ c=-\dfrac{1}{40 k}$

$\text{Find \(t$ when $v=30$:}\)

	$t$	$=\dfrac{1}{30k}-\dfrac{1}{40k}$
		$=\dfrac{1}{120k}$
		$=\dfrac{15}{120\, \ln 4}$
		$=\dfrac{1}{8\,\ln 4}\ \text{seconds}$

Vectors, EXT2 V1 2024 HSC 13a

The point $A$ has position vector $8 \underset{\sim}{i}-6 \underset{\sim}{j}+5 \underset{\sim}{k}$. The line $\ell$ has vector equation

$x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}=t(\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})$.

The point $B$ lies on $\ell$ and has position vector $p \underset{\sim}{i}+p \underset{\sim}{j}+2 p \underset{\sim}{k}$.

Show that $\abs{A B}^2=6 p^2-24 p+125$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Hence, or otherwise, determine the shortest distance between the point $A$ and the line $\ell$. (2 marks)

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

Show Answers Only

i. $A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)$

$\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)$

	$\abs{AB}^2$	$=(p-8)^2+(p+6)^2+(2 p-5)^2$
		$=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25$
		$=6 p^2-24 p+125$

ii. $\abs{AB}_{\text {min}}=\sqrt{101} \text { units}$

Show Worked Solution

$\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)$

	$\abs{AB}^2$	$=(p-8)^2+(p+6)^2+(2 p-5)^2$
		$=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25$
		$=6 p^2-24 p+125$

ii. $\text{Find shortest distance between \(A$ and $\ell$.}\)

$\Rightarrow \text { Find \(p$ when $\abs{A B}$ is a minimum:}\)

$\text{Minimum occurs when}\ \ p=\dfrac{-b}{2 a}=\dfrac{24}{2 \times 6}=2$

$\therefore \abs{AB}_{\text {min}}=\sqrt{6(2)^2-24(2)+125}=\sqrt{101} \text { units}$

Vectors, EXT2 V1 2024 HSC 12e

The line $\ell$ passes through the points $A(3,5,-4)$ and $B(7,0,2)$.

Find a vector equation of the line $\ell$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Determine, giving reasons, whether the point $C(10,5,-2)$ lies on the line $\ell$. (2 marks)

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

Show Answers Only

Show Worked Solution

i. $A(3,5,-4), \quad B(7,0,2)$

$\overrightarrow{A B}=\left(\begin{array}{l}7 \\ 0 \\ 2\end{array}\right)-\left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)=\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)$

$\therefore \text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )$

ii. $\text{If \(C$ lies on the line}, \ \exists \lambda \in \mathbb{R}\ \ \text{such that:}\)

$\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)$

$\text{Find \(\lambda$ for $x$-component: }\)

$10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}$

$\text{Check \(\lambda=\dfrac{7}{4}$ for $y$-component:}\)

$5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}$

$\therefore C\ \text{does not lie on the line.}$

Complex Numbers, EXT2 N1 2024 HSC 12c

Consider the equation $\abs{z}=z+8+12 i$, where $z=a+b i$ is a complex number and $a, b$ are real numbers.

Explain why $b=-12$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find $z$. (2 marks)

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

Show Answers Only

i. $\abs{z}=z+8+12i$

$z=a+b i\ \ \Rightarrow\ \ \abs{z}=\sqrt{a^2+b^2}$

$\text{Equating moduli:}$

$\sqrt{a^2+b^2}=a+b i+8+12 i=a+8+(b+12) i$

$\text{Since } \sqrt{a^2+b^2} \in \mathbb{R}:$

$b+12=0 \ \Rightarrow \ b=-12$

ii. $z=5-12 i$

Show Worked Solution

i. $\abs{z}=z+8+12i$

$z=a+b i\ \ \Rightarrow\ \ \abs{z}=\sqrt{a^2+b^2}$

$\text{Equating moduli:}$

$\sqrt{a^2+b^2}=a+b i+8+12 i=a+8+(b+12) i$

$\text{Since } \sqrt{a^2+b^2} \in \mathbb{R}:$

$b+12=0 \ \Rightarrow \ b=-12$

ii.	$\abs{a-12 i}$	$=a+8$
	$\sqrt{a^2+144}$	$=a+8$
	$a^2+144$	$=a^2+16 a+64$
	$16a$	$=80$
	$a$	$=5$

$\therefore z=5-12 i$

Calculus, EXT2 C1 2024 HSC 12b

Use partial fractions to find $\displaystyle \int \frac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}\, d x$ (3 marks)

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

Show Answers Only

$I=3 \ln (x-1)+2 \tan ^{-1}(x)+c$

Show Worked Solution

$\displaystyle\int \dfrac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}\, d x$

$\dfrac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}=\dfrac{A}{(x-1)}+\dfrac{B x+C}{x^2+1}$

	$3 x^2+2 x+1$	$=A\left(x^2+1\right)+(x-1)(B x+C)$
		$=A x^2+A+B x^2+C x-B x-C$
		$=(A+B) x^2+(C-B) x+A-C$

$\text {If } x=1:$

$3+2+1=2 A \ \Rightarrow \ A=3$

$A+B=3 \quad \quad \ \Rightarrow \ B=0$

$A-C=1 \quad \quad \ \Rightarrow \ C=2$

	$\therefore I$	$=\displaystyle \int \frac{3}{x-1}+\frac{2}{x^2+1}\, d x$
		$=3 \ln (x-1)+2 \tan ^{-1}(x)+c$

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 $\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$.

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

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

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

Show Answers Only

i. $(-1,0,2) $

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

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

$ \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$

$\therefore\ \text{Vectors are perpendicular.}$

Calculus, EXT2 C1 2024 HSC 11d

Evaluate $\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{\sin \theta+1}\, d \theta$. (3 marks)

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

Show Answers Only

$1$

Show Worked Solution

$t=\tan (\frac{\theta}{2}), \ \sin \theta=\dfrac{2 t}{1+t^2}$

$d t=\dfrac{1}{2} \sec ^2 (\frac{\theta}{2})\, d \theta \ \Rightarrow \ d \theta=\dfrac{2}{1+\tan ^2 (\frac{\theta}{2})}\, d t=\dfrac{2}{1+t^2}\, d t$

$\text {When}\ \ \theta=\dfrac{\pi}{2}\ \ \Rightarrow\ \ \tan (\frac{\theta}{2})=1$

$\text {When}\ \ \theta=0\ \ \Rightarrow\ \ \tan(\frac{\theta}{2})=0$

$\displaystyle{\int}_0^{\frac{\pi}{2}} \dfrac{1}{\sin \theta+1}\, d \theta$	$=\displaystyle{\int}_0^1 \frac{1}{\dfrac{2 t}{1+t^2}+1} \cdot \frac{2}{1+t^2}\, d t$
	$=\displaystyle{\int}_0^1 \dfrac{2}{2 t+1+t^2}\, d t$
	$=\displaystyle{\int}_0^1 \dfrac{2}{(t+1)^2}\, d t$
	$=-\left[\dfrac{2}{t+1}\right]_0^1$
	$=-\left(\dfrac{2}{2}-2\right)$
	$=1$

CHEMISTRY, M2 EQ-Bank 3 MC

The following equation represents the reaction of calcium disilicide $\ce{(CaSi2)}$ with antimony trichloride $\ce{(SbCl3)}$ to produce calcium chloride $\ce{(CaCl2)}$, silicon $\ce{(Si)}$, and antimony $\ce{(Sb)}$:

$\ce{a CaSi2 + b SbCl3 -> c Si + d Sb + e CaCl2}$

What are the stoichiometric values for , , , $d$ , and in the balanced equation?

2, 3, 6, 3, 2
3, 2, 2, 6, 3
3, 3, 6, 2, 3
3, 2, 6, 2, 3

Show Answers Only

$D$

Show Worked Solution

→ Balance the chlorine atoms to begin:

$\ce{a CaSi2 + 2SbCl3 -> c Si + d Sb + 3CaCl2}$

→ As $\ce{Si}$ and $\ce{Sb}$ are by themselves on the right hand side they are be easily balanced last.

→ Next, balance the $\ce{Ca}$ atoms on the left hand side:

$\ce{3CaSi2 + 2SbCl3 -> c Si + d Sb + 3CaCl2}$

→ The values for $c$ and $d$ can be added to balance the remaining atoms.

$\ce{3CaSi2 + 2SbCl3 -> 6 Si + 2 Sb + 3CaCl2}$

$\Rightarrow D$

Calculus, EXT1 C2 2024 HSC 13b

Show that $\cos ^4 x+\sin ^4 x=\dfrac{1+\cos ^2 2 x}{2}$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, evaluate $\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x$. (3 marks)

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

Show Answers Only

i.	$\text{LHS}$	$=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2$
		$=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)$
		$=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)$
		$=\dfrac{1+\cos ^2(2 x)}{2}$

ii. $\dfrac{3 \pi}{16}$

Show Worked Solution

i.	$\text{LHS}$	$=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2$
		$=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)$
		$=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)$
		$=\dfrac{1+\cos ^2(2 x)}{2}$

ii.	$\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x$
		$=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\cos ^2(2 x) d x$
		$=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\dfrac{1}{2}(1+\cos (4 x)) d x$
		$=\dfrac{1}{2}\left[\dfrac{3}{2}x +\dfrac{1}{8} \sin (4 x)\right]_0^{\frac{\pi}{4}}$
		$=\dfrac{1}{2}\left[\dfrac{3}{2} \times \dfrac{\pi}{4}+\dfrac{1}{8} \sin \pi-0\right]$
		$=\dfrac{3 \pi}{16}$

Calculus, EXT1 C3 2024 HSC 13a

In an experiment, the population of insects, $P(t)$, was modelled by the logistic differential equation

$\dfrac{d P}{d t}=P(2000-P)$

where $t$ is the time in days after the beginning of the experiment.

The diagram shows a direction field for this differential equation, with the point $S$ representing the initial population.

Explain why the graph of the solution that passes through the point $S$ cannot also pass through the point $T$. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Clearly sketch the graph of the solution that passes through the point $S$. (1 mark)

--- 5 WORK AREA LINES (style=blank) ---
Find the predicted value of the population, $P(t)$, at which the rate of growth of the population is largest. (2 marks)

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

Show Answers Only

i. $\text{Any solution that passes through}\ S\ \text{will follow the}$

$\text{slope field (not crossing any lines) and approach a horizontal}$

$\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).$

ii.

iii. $P= 1000$

Show Worked Solution

i. $\text{Any solution that passes through}\ S\ \text{will follow the}$

$\text{slope field (not crossing any lines) and approach a horizontal}$

$\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).$.

ii.

iii. $\dfrac{d P}{d t}=P(2000-P)$

$\text {Find \(P$ where $\dfrac{d P}{d t}$ is a maximum.}\)

$\text{Consider the graph}\ \ y=P(2000-P): $

$\Rightarrow \ \text {Graph is a concave down quadratic cutting the axis at}\ \ P=0\ \ \text{and}\ \ P=2000$

$\Rightarrow \ \text{Max value of}\ \ P(2000-P)\ \ \Big(\text{i.e.}\ \dfrac{dP}{dt}\Big)\ \ \text{occurs at}\ \ P=1000\ \text{(axis of symmetry).}$

Proof, EXT1 P1 2024 HSC 12d

Use mathematical induction to prove that $2^{3 n}+13$ is divisible by 7 for all integers $n \geq 1$. (3 marks)

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

Show Answers Only

$\text{Proof (See worked solutions)}$

Show Worked Solution

$\text{Prove}\ \ 2^{3 n}+13\ \ \text{is divisible by 7 for}\ \ n \geq 1.$

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

$2^{3 \times 1}+13=21=3 \times 7$

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

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

$2^{3 k}+13=7P \ \text{(where \(P$ is an integer)}\)

$\Rightarrow 2^{3k}=7 P-13\ \ldots\ (1)$

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

	$2^{3(k+1)}+13$	$=2^{3 k} \times 2^3+13$
		$=8\left(2^{3 k}\right)+13$
		$=8(7P-13)+13\ \ \text{(see (1) above)}$
		$=56 P-8 \times 13+13$
		$=7(8 P-13)$

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

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

Statistics, EXT1 S1 2024 HSC 12c

A charity employs a worker to collect donations. There is a 0.31 chance that when the charity worker talks to someone a donation is made to the charity.

Each day the charity worker must talk to exactly 100 people.

Use a standard normal distribution table to approximate the probability that, on a particular day, at least 35% of the people talked to made a donation. (3 marks)

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

Show Answers Only

$P(\hat{p} \geqslant 0.35) =0.1922$

Show Worked Solution

$E(\hat{p})=p=0.31, n=100$

$\sigma^2=\dfrac{p(1-p)}{n}=\dfrac{0.31 \times 0.69}{100}=0.002139$

$\hat{p} \sim N\left(\mu, \sigma^2\right) \sim N(0.31,0.002139)$

	$P(\hat{p} \geqslant 0.35)$	$=P\left(Z \geqslant \dfrac{x-\mu}{\sigma}\right)$
		$=P\left(Z \geqslant \dfrac{0.35-0.31}{\sqrt{0.002139}}\right)$
		$=P(Z \geqslant 0.87)$
		$=1-P(Z<0.87)$
		$=1-0.8078 \text { (from table)}$
		$=0.1922$

Calculus, EXT1 C3 2024 HSC 12b

The region, $R$, is bounded by the function, $y=x^3$, the $x$-axis and the lines $x=1$ and $x=2$.

What is the volume of the solid of revolution obtained when the region $R$ is rotated about the $x$-axis? (3 marks)

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

Show Answers Only

$V=\dfrac{127 \pi}{7} \ \ \text{u}^3$

Show Worked Solution

	$V$	$=\pi \displaystyle \int_1^2 y^2\, d x$
		$=\pi \displaystyle \int_1^2 x^6\,d x$
		$=\pi\left[\dfrac{x^7}{7}\right]_1^2$
		$=\pi\left(\dfrac{2^7-1}{7}\right)$
		$=\dfrac{127 \pi}{7} \ \ \text{u}^3$

Vectors, EXT1 V1 2024 HSC 12a

The vectors $\displaystyle \binom{a^2}{2}$ and $\displaystyle \binom{a+5}{a-4}$ are perpendicular.

Find the possible values of $a$. (3 marks)

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

Show Answers Only

$x=1,-4 \text { or }-2$

Show Worked Solution

$\text{If vectors are }\perp:$

$\displaystyle\binom{a^2}{2} \cdot\binom{a+5}{a-4}=0$

$a^3+5 a^2+2 a-8=0$

$\text{Test for roots:}$

$1^3+5 \times 1^2+2\times 1-8=0 \, \checkmark$

$(a-1) \text{ is a factor.}$

$\text{By polynomial long division:}$

$(a-1)\left(a^2+6 a+8\right)=0$

$(a-1)(a+4)(a+2)=0$

$\therefore x=1,-4 \text { or }-2$

Calculus, EXT1 C3 2024 HSC 11g

The region, $R$, is bounded by the curves $y=\sin x, y=x$ and the line $x=\dfrac{\pi}{2}$ as shown in the diagram.

Find the area of the region $R$. (3 marks)

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

Show Answers Only

$\dfrac{\pi^2}{8}-1\ \ \text{u}^2$

Show Worked Solution

	$R$	$=\displaystyle{\int}_0^{\frac{\pi}{2}} x-\sin x \, d x$
		$=\left[\dfrac{x^2}{2}+\cos x\right]_0^{\frac{\pi}{2}}$
		$=\left[\left(\dfrac{\pi^2}{8}+\cos \dfrac{\pi}{2}\right)-(0+\cos 0)\right]$
		$=\dfrac{\pi^2}{8}-1\ \ \text{u}^2$

Calculus, EXT1 C1 2024 HSC 11f

The volume of a sphere of radius $r$ cm, is given by $V=\dfrac{4}{3} \pi r^3$, and the volume of the sphere is increasing at a rate of $10 \text{ cm}^3 \text{ s}^{-1}$.

Show that the rate of increase of the radius is given by $\dfrac{d r}{d t}=\dfrac{5}{2 \pi r^2} \text{ cm s}^{-1}$. (2 marks)

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

Show Answers Only

	$\dfrac{d r}{d t}$	$=\dfrac{d V}{d t} \times \dfrac{d r}{d V}$
		$=10 \times \dfrac{1}{4 \pi r^2}$
		$=\dfrac{5}{2 \pi r^2} \text{ cm/s}$

Show Worked Solution

$V=\dfrac{4}{3} \pi r^3 \ \Rightarrow \ \dfrac{d V}{d r}=4 \pi r^2$

$\dfrac{d V}{d t}=10$

	$\dfrac{d r}{d t}$	$=\dfrac{d V}{d t} \times \dfrac{d r}{d V}$
		$=10 \times \dfrac{1}{4 \pi r^2}$
		$=\dfrac{5}{2 \pi r^2} \text{ cm/s}$

Calculus, EXT1 C3 2024 HSC 11d

Solve the differential equation $\dfrac{d y}{d x}=x y$, given $y>0$. Express your answer in the form $y=e^{f(x)}$. (2 marks)

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

Show Answers Only

$y=e^{\frac{x^2}{2}}$

Show Worked Solution

	$\dfrac{d y}{d x}$	$=x y$
	$\displaystyle\int \frac{1}{y}\, d y$	$=\displaystyle\int x\, d x$
	$\ln y$	$=\dfrac{1}{2} x^2+c$
	$y$	$=e^{\frac{x^2}{2}+c}$
		$=e^{\frac{x^2}{2}} \cdot e^c$
		$=A e^{\frac{x^2}{2}} \text{ (where \(A=e^c)$}\)

$\therefore y=e^{\frac{x^2}{2}}\ \ \text{is a solution } (A=1)$

CHEMISTRY, M2 EQ-Bank 2

In an experiment, calcium carbonate $\ce{(CaCO3)}$ is heated strongly to produce calcium oxide $\ce{(CaO)}$ and carbon dioxide according to the reaction below:

$\ce{CaCo3(s) -> CaO(s) + CO2(g)}$

A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.

Using the law of conservation of mass, calculate the mass of carbon dioxide gas produced in this reaction. (2 marks)

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

Explain how the law of conservation of mass applies to this reaction. (2 marks)

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

Show Answers Only

a. $22.0\ \text{g}$

b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

Show Worked Solution

a. According to the law of conservation of mass, the total mass of reactants must equal the total mass of products.

Since mass must be conserved:

$m\ce{(CO2)}$	$=m\ce{(CaCO3)}-m\ce{(CaO)}$
	$= 50.0-28.0$
	$=22.0\ \text{g}$

b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

CHEMISTRY, M2 EQ-Bank 1

Balance the following chemical equations:

$\ce{HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$ (1 mark)

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

$\ce{C2H6(l) + O2(g) -> H2O(l) + CO2(g)}$ (1 mark)

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

$\ce{H3PO4(aq) + CuCO3(aq) -> Cu3(PO4)2(aq) + CO2(g) + H2O(l)}$ (1 mark)

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

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

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

Show Answers Only

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

b. $\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$

c. $\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}$

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

Show Worked Solution

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

b. $\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$

c. $\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}$

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

CHEMISTRY, M2 EQ-Bank 10 MC

When measuring very low concentrations of pollutants in water, which unit is most suitable?

Molarity (mol L$^{-1}$)
Parts per million (ppm)
Percentage by mass
Volume percent (% v/v)

Show Answers Only

$B$

Show Worked Solution

→ Parts per million (ppm) is a unit used to express very low concentrations. One ppm means that there is 1 part of solute per 1,000,000 parts of solution. This unit is ideal for detecting small amounts of pollutants in large volumes of water, where concentrations are generally too low to be expressed effectively in molarity or percentage terms.

→ Molarity is a common unit for concentration but is better suited for more moderate to high concentrations in laboratory and industrial contexts, rather than for trace pollutants in water.

→ Percentage by mass and volume percent (% v/v) are typically used for relatively larger concentrations and are not as effective for expressing concentrations as low as parts per million.

$\Rightarrow B$

CHEMISTRY, M2 EQ-Bank 11

A student is investigating the concentration of copper ions in a water sample collected from a local river. They use an instrument to determine that the sample contains copper ions at a concentration level of 1.75 ppm.

Calculate the mass of copper ions in a 2 L sample of water. (2 marks)

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

Explain why parts per million is a suitable unit for measuring low concentrations of ions in environmental samples like river water. (2 marks)

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

Show Answers Only

a. $3.5\ \text{mg}$

b. Benefits of using ppm as a concentration measurement.

→ Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.

→ This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

Show Worked Solution

a. $1\ \text{ppm} = 1\ \text{mg/L}$

$\therefore \ce{[Cu^{2+}]} = 1.75\ \text{mg/L}$

$m\ce{(Cu^{2+})} = 1.75 \times 2 = 3.5\ \text{mg}$

b. Benefits of using ppm as a concentration measurement.

→ This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

Statistics, EXT1 S1 2024 HSC 8 MC

A local council is proposing to ban dog-walking on the beach. It is known that the proportion of households that have a dog is $\dfrac{7}{12}$.

The local council wishes to poll $n$ households about this proposal.

Let $\hat{p}$ be the random variable representing the proportion of households polled that have a dog.

What is the smallest sample size, $n$, for which the standard deviation of $\hat{p}$ is less than 0.06?

$67$
$68$
$94$
$95$

Show Answers Only

$B$

Show Worked Solution

$E(\hat{p}) = p = \dfrac{7}{12} $

$\sigma^{2} = \dfrac{p(1-p)}{n}$

$\text{Find}\ n\ \text{such that}\ \ \sigma \lt 0.06 :$

$\Bigg( \dfrac{6}{100}\Bigg)^{2} $	$ \gt \dfrac{ \frac{7}{12} \times \frac{5}{12}}{n} $
$n$	$ \gt \dfrac{35}{144} \times \dfrac{100^2}{6^2}$
	$ \gt 67.5$

$\Rightarrow B$

Statistics, EXT1 S1 2024 HSC 7 MC

A driver's knowledge test contains 30 multiple-choice questions, each with 4 options. An applicant must get at least 29 correct to pass.

If an applicant correctly answers the first 25 questions and randomly guesses the last 5 questions, what is the probability that the applicant will pass the test?

$\dfrac{1}{256}$
$\dfrac{15}{1024}$
$\dfrac{1}{64}$
$\dfrac{21}{256}$

Show Answers Only

$C$

Show Worked Solution

$\text{For each of the last 5 questions:}$

$P(C) = \dfrac{1}{4}, \ \ P(\bar{C}) = \dfrac{3}{4}$

$P(\text{at least 4 correct})$	$=\ ^5C_4 \Bigg(\dfrac{1}{4}\Bigg)^{4} \Bigg(\dfrac{3}{4}\Bigg)^{1} + \ ^5C_5 \Bigg(\dfrac{1}{4}\Bigg)^{5}\Bigg(\dfrac{3}{4}\Bigg)^{0}$
	$=5 \times \dfrac{1}{256} \times \dfrac{3}{4}+1\times \dfrac{1}{1024} \times 1$
	$=\dfrac{1}{64}$

$\Rightarrow C$

CHEMISTRY, M2 EQ-Bank 9 MC

Which of the following statements best describes a primary standard solution?

It is a solution that must be prepared from a substance with a low molar mass to increase accuracy.
It is a solution prepared using a substance with a precisely known and stable concentration, suitable for use in standardising other solutions.
It is a solution prepared from any chemical, as long as the concentration is measured with a volumetric flask.
It is a solution that can vary in concentration over time and requires frequent standardisation.

Show Answers Only

$B$

Show Worked Solution

→ Primary standards are used to prepare primary solutions with known and accurate concentrations.

→ A primary standard is a substance that is pure, stable, has a high molar mass (to minimize weighing errors), and is not affected by the atmosphere (such as moisture absorption or $\ce{CO2}$ interaction).

$\Rightarrow B$

f.	\(Wbest\)	\(=0.300 +0.86\times 2.02\)
		\(=2.0372\)

f.	\(Wbest\)	\(=0.300 +0.86\times 2.02\)
		\(=2.0372\)

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

\(\text{Repayments}\)	\( =2228.40\times 12\times 5\)
	\(=$133\,704\)
\(\text{Interest}\)	\(=$133\,704-$121\,000\)
	\(=$12\,704\)

\(FV\)	\(=2000\left(1-\dfrac{0.044}{4}\right)^{12}\)
	\(=$2280.57\)
\(\text{Interest pa}\)	\(=\dfrac{280.57}{3}=$93.52\)

\(\text{SI rate}\)	\(=\dfrac{93.52}{2000}\times 100\%\)
	\(=4.676\dots\%\)

\(\text{Interest}\)	\(=\dfrac{960}{240\,000}\times 238\,218.95\)
	\(=$952.88\)

\(\text{Principal Reduction}\)	\(=2741.05-952.88\)
	\(=$1788.17\)

\(\text{mean}\)	\(=\dfrac{\Sigma\text{scores}}{\text{Number of scores}}\)
\(11\)	\(=\dfrac{x+14+11+7+6+11+13+12+6+10+15+12+10}{13}\)
\(x+127\)	\(=143\)
\(x\)	\(=16\)

\(b\)	\(=r\dfrac{s_y}{s_x}\)
\(-1.27\)	\(=r\times\dfrac{19}{8.51}\)
\(\therefore\ r\)	\(=\dfrac{-1.27\times 8.51}{19}\)
	\(=-0.5688\dots\)