Band 5

Data Analysis, GEN2 2024 VCAA 2

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

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

--- 2 WORK AREA LINES (style=lined) ---
For this boxplot, what is the smallest possible number of $\textit{Wgold}$ heights lower than 1.85 m (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
i. Using the boxplot, show that the lower fence is 1.565 m and the upper fence is 2.325 m. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
ii. Referring to the boxplot, the lower fence and the upper fence, explain why no outliers exist. (1 mark)

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

Show Answers Only

a. $\text{Negatively skewed}$

b. $1$

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

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

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

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

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

Show Worked Solution

a. $\text{Negatively skewed}$

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

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

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

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

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

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

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

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

Networks, GEN1 2024 VCAA 39 MC

Anush, Blake, Carly and Dexter are workers on a construction site. They are each allocated one task. The time, in hours, it takes for each worker to complete each task is shown in the table below.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& \textbf{Task 1} & \textbf{Task 2} & \textbf{Task 3} & \textbf{Task 4} \\
\hline
\rule{0pt}{2.5ex} \textbf{Anush} \rule[-1ex]{0pt}{0pt}& 12 & 8 & 16 & 9 \\
\hline
\rule{0pt}{2.5ex} \textbf{Blake} \rule[-1ex]{0pt}{0pt}& 10 & 7 & 15 & 10 \\
\hline
\rule{0pt}{2.5ex} \textbf{Carly} \rule[-1ex]{0pt}{0pt}& 11 & 10 & 18 & 12 \\
\hline
\rule{0pt}{2.5ex} \textbf{Dexter} \rule[-1ex]{0pt}{0pt}& 10 & 14 & 16 & 11 \\
\hline
\end{array}

The tasks must be completed sequentially and in numerical order: Task 1, Task 2, Task 3 and then Task 4.

Management makes an initial allocation of tasks to minimise the amount of time required, but then decides that it takes the workers too long.

Another worker, Edgar, is brought in to complete one of the tasks.

His completion times, in hours, are listed below.

When a new allocation is made and Edgar takes over one of the tasks, the minimum total completion time compared to the initial allocation will be reduced by

1 hour.
2 hours.
3 hours.
4 hours.

Show Answers Only

$D$

Show Worked Solution

$\text{Using Hungarian Algorithm on CAS:}$

$\text{Original Allocation:}$

$\text{Minimal Sum}\ =9+7+11+16=43$

$\text{Task Allocation: Anush – Task 4, Blake – Task 2, Carly – Task 1, Dexter – Task 3}$

$\text{Allocation with Edgar included:}$

$\text{Minimal Sum}\ =9+15+10+5=39$

$\text{Task Allocation: Anush – Task 4, Blake – Task 3, Carly – Eliminated, Dexter – Task 1, Edgar – Task 2}$

$\therefore\ \text{The minimum total completion time is reduced from 43 to 39 hours ie 4 hours}\ \text{hours.}$

$\Rightarrow D$

Networks, GEN1 2024 VCAA 38 MC

A connected graph has six vertices and six edges.

How many of the following four statements must always be true?

the graph has no vertices of odd degree
the graph contains a Eulerian trail
the graph contains a Hamiltonian path
the sum of the degrees of the vertices is 12

Show Answers Only

$A$

Show Worked Solution

$\text{Considering each option:}$

$\text{1. Both Graph 1 & 2 above have vertices of odd degree }\rightarrow\ \text{Incorrect} $

$\text{2. Graph 2 has 4 vertices of odd degree}\therefore \text{No Eulerian Trail}\rightarrow \text{Incorrect} $

$\text{3. Path cannot be completed in Graph 2 without repeating a vertex}$

$\therefore\ \text{No Hamiltonian Path}\rightarrow \text{Incorrect} $

$\text{4. The sum of the degrees of the vertices is always 12}\ \rightarrow\ \text{Correct}$

$\therefore\ 1\ \text{of the statements is always true}$

$\Rightarrow A$

Matrices, GEN1 2024 VCAA 37 MC

The network below represents paths through a park from the carpark to a lookout.

The vertices represent various attractions, and the numbers on the edges represent the distances between them in metres.

The shortest path from the carpark to the lookout is 34 m . This can be achieved when

$x=8$ and $y=8$
$x=9$ and $y=7$
$x=10$ and $y=6$
$x=11$ and $y=5$

Show Answers Only

$D$

Show Worked Solution

$\text{Consider Option D}\ \rightarrow\ \ x=11, y=5$

$\Rightarrow D$

This question doesn’t explicitly say that the shortest path includes $x$ or $y$ and is actually a critical path question.

Networks, GEN1 2024 VCAA 35 MC

Consider the following graph.

The number of faces is

Show Answers Only

$B$

Show Worked Solution

$\text{Method 1: Make the diagram planar}$

$\text{Method 2: Euler’s Rule}$

$\text{Vertices}\ =7,\ \text{Edges}\ =11$

$v+f$	$=e+2$
$7+f$	$=11+2$
$f$	$=6$

$\Rightarrow B$

Matrices, GEN1 2024 VCAA 31 MC

The matrix below shows the results of a round-robin chess tournament between five players: $H, I, J, K$ and $L$. In each game, there is a winner and a loser.

Two games still need to be played.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad \ \textit{loser}\\
&\quad \quad\quad \quad \quad \ \ \ H \quad \quad \ \ \ I \quad \quad \quad J \quad \quad \ \ \ K \quad \quad \quad L \\
& \textit{winner} \quad \begin{array}{ccccc}
H\\
\\
I\\
\\
J\\
\\
K\\
\\
L
\end{array}
\begin {bmatrix}
0 \quad & \quad 1 \quad & \quad 0 \quad & \quad 1 \quad & \quad 0 \\
\\
0 \quad& 0 & \ldots & 1 & \quad \ldots \\
\\
1 \quad& \ldots & 0 & 1 & \quad 0 \\
\\
0 \quad& 0 & 0 & 0 & \quad 1 \\
\\
1 \quad& \ldots & 1 & 0 & \quad 0
\end{bmatrix}\\
&
\end{aligned}

A '1' in the matrix shows that the player named in that row defeated the player named in that column. For example, the 1 in row 4 shows that player $K$ defeated player $L$.

A '...' in the matrix shows that the player named in that row has not yet competed against the player in that column.

At the end of the tournament, players will be ranked by calculating the sum of their one-step and two-step dominances.

The player with the highest sum will be ranked first. The player with the second-highest sum will be ranked second, and so on.

Which one of the following is not a potential outcome after the final two games have been played?

Player $I$ will be ranked first.
Player $I$ will be ranked fifth.
Player $J$ will be ranked first.
Player $J$ will be ranked fifth.

Show Answers Only

$D$

Show Worked Solution

$\text{There are 4 possible combinations:}$

$\text{Let the results matrix be}\ D$

$\text{Case 1:}\ D_{23}=1,\ D_{25}=1,\ D_{32}=0,\ D_{52}=0$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 2 \\
\ 2 & 0 & 2 & 2 & 2 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 1 & 1 & 2 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=6\\
\ \ I=8\\
\ \ J=5\\
\ \ K=3\\
\ \ L=6
\end{array}$

$\text{Case 2:}\ D_{23}=0,\ D_{25}=1,\ D_{32}=1,\ D_{52}=0$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 2 \\
\ 1 & 0 & 1 & 1 & 2 \\
\ 1 & 2 & 0 & 3 & 2 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 2 & 1 & 2 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=8\\
\ \ K=3\\
\ \ L=7
\end{array}$

$\text{Case 3:}\ D_{23}=1,\ D_{25}=0,\ D_{32}=0,\ D_{52}=1$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 1 \\
\ 1 & 0 & 1 & 2 & 1 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 2 & 2 & 3 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=5\\
\ \ K=4\\
\ \ L=9
\end{array}$

$\text{Case 4:}\ D_{23}=0,\ D_{25}=0,\ D_{32}=1,\ D_{52}=1$

$D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}+
\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 1 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 2 & 0 & 3 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 3 & 1 & 3 & 0
\end{bmatrix}\begin{array}{ccccc}
\ \ H=4\\
\ \ I=2\\
\ \ J=7\\
\ \ K=4\\
\ \ L=9
\end{array}$

$\text{In all cases Player }K\ \text{will be ranked 5th,}$

$\therefore\ \text{Option D incorrect.}$

$\Rightarrow D$

Matrices, GEN1 2024 VCAA 30 MC

Data has been collected on the female population of a species of mammal located on a remote island.

The female population has been divided into three age groups, with the initial population (at the time of data collection), the birth rate, and the survival rate of each age group shown in the table below.

The Leslie matrix $(L)$ that may be used to model this particular population is

$L=\begin{bmatrix}0 & 1.8 & 0 \\ 0.7 & 0 & 1.2 \\ 0 & 0.6 & 0\end{bmatrix}$
$L=\begin{bmatrix}0 & 1.8 & 1.2 \\ 0.7 & 0 & 0 \\ 0 & 0.6 & 0\end{bmatrix}$
$L=\begin{bmatrix}0 & 1.8 & 1.2 \\ 0.7 & 0.6 & 0 \\ 0 & 0 & 0\end{bmatrix}$
$L=\begin{bmatrix}2100 & 6400 & 4260 \\ 0 & 1.8 & 1.2 \\ 0.7 & 0.6 & 0\end{bmatrix}$

Show Answers Only

$B$

Show Worked Solution

$\text{Row 1 represents the birth rates for the time periods: Eliminate A}$

$\text{The sub-diagonal represents survival rates: Eliminate C and D}$

$\Rightarrow B$

Matrices, GEN1 2024 VCAA 29 MC

A tennis team consists of five players: Quinn, Rosie, Siobhan, Trinh and Ursula.

When the team competes, players compete in the order of first, then second, then third, then fourth.

The fifth player has a bye (does not compete).

On week 1 of the competition, the players competed in the following order.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad\textbf{First} \quad\rule[-1ex]{0pt}{0pt}& \quad \textbf{Second} \quad& \quad\textbf{Third} \quad& \quad\textbf{Fourth}\quad & \quad\textbf{Bye} \quad\\
\hline
\rule{0pt}{2.5ex} \text{Quinn} \rule[-1ex]{0pt}{0pt}& \text {Rosie} & \text {Siobhan} & \text { Trinh } & \text {Ursula} \\
\hline
\end{array}

This information can be represented by matrix $G_1$, shown below.

$G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}$

Let $G_n$ be the order of play in week $n$.

The playing order changes each week and can be determined by the rule $G_{n+1}=G_n \times P$

$\text{where}\quad \\P=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \end{bmatrix}$

Which player has a bye in week 4 ?

Quinn
Rosie
Siobhan
Trinh

Show Answers Only

$C$

Show Worked Solution

$G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}$

$G_2=\begin{bmatrix} S & Q & T & U & R \end{bmatrix}$

$G_3=\begin{bmatrix} T & S & U & R & Q \end{bmatrix}$

$G_4=\begin{bmatrix} U & T & R & Q & S \end{bmatrix}$

$\text{5th player has a bye.}$

$\therefore\ \text{Siobhan does not compete in week 4}$

$\Rightarrow C$

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

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

$\Rightarrow A$

Data Analysis, GEN1 2024 VCAA 11-12 MC

The number of breeding pairs of a small parrot species has been declining over recent years.

The table below shows the number of breeding pairs counted, $pairs$, and the year number, $year$, for the last 12 years. A scatterplot of this data is also provided.

The association between $pairs$ and $year$ is non-linear.

Part A

The scatterplot can be linearised using a logarithmic (base 10) transformation applied to the explanatory variable.

The least squares equation calculated from the transformed data is closest to

$\log _{10}(pairs)=2.44-0.0257 \times year$
$\log _{10}(pairs)=151-303 \times year$
$pairs =274-12.3 \times \log _{10}(year)$
$pairs =303-151 \times \log _{10}( year)$

Part B

A reciprocal transformation applied to the variable $pairs$ can also be used to linearise the scatterplot.

When a least squares line is fitted to the plot of $\dfrac{1}{pairs}$ versus $year$, the largest difference between the actual value and the predicted value occurs at $year$

Show Answers Only

Part A

$D$

Part B

$A$

Show Worked Solution

Part A

$\text{Least squares equation}\rightarrow pairs=303-151\times\log_{10}{(year)}$

$\Rightarrow D$

Part B

$\text{Largest difference occurs at year 1}$

$\Rightarrow A$

Vectors, EXT2 V1 2024 HSC 10 MC

Three unit vectors $\underset{\sim}{a}, \underset{\sim}{b}$ and $\underset{\sim}{c}$, in 3 dimensions, are to be chosen so that $\underset{\sim}{a} \perp \underset{\sim}{b}, \ \underset{\sim}{b} \perp \underset{\sim}{c}$ and the angle $\theta$ between $\underset{\sim}{a}$ and $\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}$ is as small as possible.

What is the value of $\cos \theta$ ?

$0$
$\dfrac{1}{\sqrt{3}}$
$\dfrac{1}{\sqrt{2}}$
$\dfrac{2}{\sqrt{5}}$

Show Answers Only

$D$

Show Worked Solution

$\cos \theta=\dfrac{\underset{\sim}{a} \cdot(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}}$

$\theta_{\text{min}} \ \Rightarrow \ \cos\,\theta_{\text{max}}:$

$\underset{\sim}{a} \cdot\left(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\right)=\abs{\underset{\sim}{a}}^2+\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{a} \cdot \underset{\sim}{c}=1+\underset{\sim}{a} \cdot \underset{\sim}{c}$

$\text{Find}\ \ \abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}:$

	$\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}^2$	$=\abs{\underset{\sim}{a}}^2+\abs{\underset{\sim}{b}}^2+\abs{\underset{\sim}{c}}^2+2\left(\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{c}+\underset{\sim}{a} \cdot \underset{\sim}{c}\right)$
		$=3+2 \underset{\sim}{a} \cdot \underset{\sim}{c}$

$\cos \theta=\dfrac{1+a \cdot c}{\sqrt{3+2 a \cdot c}}$

$\underset{\sim}{a} \cdot\underset{\sim}{c}=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{c}} \cos\, \alpha = \cos\,\alpha $

$0 \leqslant \cos\, \alpha \, \leqslant 1$

$\therefore \cos \theta_{\text{max}}=\dfrac{2}{\sqrt{5}}$

$\Rightarrow D$

Complex Numbers, EXT2 N1 2024 HSC 8 MC

Which of the following is equal to $e^{\large{\bar{z}}}$, where $z=x+i y$ with $x$ and $y$ real numbers?

$\overline{e^{\large{z}}}$
$e^{\large{-z}}$
$e^{\large{2 x}} e^{\large{z}}$
$e^{\large{-2 x}} e^{\large{z}}$

Show Answers Only

$A$

Show Worked Solution

	$e^{\large{\bar{z}}}$	$=e^{x-iy}$
		$=e^x \cdot e^{-iy}$
		$=e^x(\cos (-y)+i \sin (-y))$
		$=e^x(\cos (y)-i \sin (y))$

	$\overline{e^{\large{z}}}$	$=\overline{e^x(\cos (y)+i \sin (y)}$
		$=e^x(\cos (y)-i \sin (y))$
		$=e^{\large{\bar{z}}}$

$\Rightarrow A$

Complex Numbers, EXT2 N2 2024 HSC 16b

The number $w=e^{\small{\dfrac{2 \pi i}{3}}}$ is a complex cube root of unity. The number $\gamma$ is a cube root of $w$.

Show that $\gamma+\bar{\gamma}$ is a real root of $z^3-3 z+1=0$. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---
By using part (i) to find the exact value of $\cos \dfrac{2 \pi}{9} \cos \dfrac{4 \pi}{9} \cos \dfrac{8 \pi}{9}$, deduce the value(s) of $\cos \dfrac{2^n \pi}{9} \cos \dfrac{2^{n+1} \pi}{9} \cos \dfrac{2^{n+2} \pi}{9}$ for all integers $n \geq 1$. Justify your answer. (3 marks)

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

Show Answers Only

i. $w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}$

$\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1$

$\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1$
		$=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1$
		$=w+\bar{w}+1$
		$=0 \quad \text{(Sum of complex roots of 1 = 0)}$

$\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}$

$\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})$

$\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \ z^3-3 z+1=0.$

ii. $e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}$

$\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}$

$\text{Roots of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})_1$	$=2 \cos \left(\dfrac{2 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_2$	$=2 \cos \left(\dfrac{8 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_3$	$=2 \cos \left(\dfrac{4\pi}{9}\right)$

$\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:$

	$2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)$	$=-1$
	$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)$	$=-\dfrac{1}{8}$

$\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):$

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

$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}$

$\text {If} \ \ n=2,$

$\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}$

$\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}$

$\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}$

Show Worked Solution

i. $w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}$

$\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1$

$\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1$
		$=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1$
		$=w+\bar{w}+1$
		$=0 \quad \text{(Sum of complex roots of 1 = 0)}$

$\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}$

$\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})$

$\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \ z^3-3 z+1=0.$

ii. $e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}$

$\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}$

$\text{Roots of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})_1$	$=2 \cos \left(\dfrac{2 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_2$	$=2 \cos \left(\dfrac{8 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_3$	$=2 \cos \left(\dfrac{4\pi}{9}\right)$

$\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:$

	$2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)$	$=-1$
	$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)$	$=-\dfrac{1}{8}$

$\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):$

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

$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}$

$\text {If} \ \ n=2,$

$\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}$

$\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}$

$\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}$

Vectors, EXT2 V1 2024 HSC 16a

Consider the function $y=\cos (k x)$, where $k>0$. The value of $k$ has been chosen so that a circle can be drawn, centred at the origin, which has exactly two points of intersection with the graph of the function and so that the circle is never above the graph of the function. The point $P(a, b)$ is the point of intersection in the first quadrant, so $a>0$ and $b>0$, as shown in the diagram.

The vector joining the origin to the point $P(a, b)$ is perpendicular to the tangent to the graph of the function at that point. (Do NOT prove this.)

Show that $k>1$. (4 marks)

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$y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)$

$P(a, b)=P(a, \cos (k a))$

$\text{Let}\ \ \underset{\sim}{p}=\overrightarrow{OP}.$

$\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :$

	$m_{\text{tang}}$	$=-k \, \sin (ka)$
	$m_{\overrightarrow{OP}}$	$=\dfrac{\cos(ka)}{a}$

$-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1$

	$k$	$=\dfrac{a}{\sin(ka) \cos(ka)}$
		$=\dfrac{2a}{\sin(2ka)}$

$\text{For} \ \ \theta>0, \ \sin \theta<\theta$

$\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1$.

Show Worked Solution

$y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)$

$P(a, b)=P(a, \cos (k a))$

$\text{Let}\ \ \underset{\sim}{p}=\overrightarrow{OP}.$

$\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :$

	$m_{\text{tang}}$	$=-k \, \sin (ka)$
	$m_{\overrightarrow{OP}}$	$=\dfrac{\cos(ka)}{a}$

$-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1$

	$k$	$=\dfrac{a}{\sin(ka) \cos(ka)}$
		$=\dfrac{2a}{\sin(2ka)}$

$\text{For} \ \ \theta>0, \ \sin \theta<\theta$

$\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1$.

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)

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Find where the object next comes to rest, giving your answer correct to 1 decimal place. (2 marks)

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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 \(x=1.7$ units}\)

Calculus, EXT2 C1 2024 HSC 15b

Let $I_n=\displaystyle\int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\, d x$, where $n \geq 0$.

Show that $(2 n+4) I_n=a(2 n+1) I_{n-1}$, for $n>0$. (3 marks)

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$I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,d x \ \ \text{where}\ \ n \geqslant 0$

$\text{Show}\ \ (2 n+4) I_n=a(2 n+1) I_{n-1}\ \ \text{for}\ \ n>0$

$I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}$

$\text{Using integration by parts: }$

$\begin{array}{ll}u=x^{n+\frac{1}{2}} & v^{\prime}=(a-x)^{\frac{1}{2}} \\ u^{\prime}=\left(n+\frac{1}{2}\right) x^{n-\frac{1}{2}} & v=-\dfrac{2}{3}(a-x)^{\frac{3}{2}}\end{array}$

	$I_n$	$=\left[u v\right]_0^a-\displaystyle{\int}_0^a u^{\prime} v\, dx$
		$=\underbrace{\left[-\dfrac{2}{3} x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\right]_0^a}_{=0}+\dfrac{2}{3}\left(n+\frac{1}{2}\right) \displaystyle{\int}_0^a x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\, dx$
		$=\dfrac{2 n+1}{3} \displaystyle{\int}_0^a a x^{n-\frac{1}{2}}(a-x)^{\frac{1}{2}}-x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\, dx$
	$3I_n$	$=(2 n+1)\left[a I_{n-1}-I_n\right]$

	$3 I_n+(2 n+1) I_n$	$=a(2 n+1) I_{n-1}$
	$(2 n+4) I_n$	$=a(2 n+1) I_{n-1}$

Show Worked Solution

$I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,d x \ \ \text{where}\ \ n \geqslant 0$

$\text{Show}\ \ (2 n+4) I_n=a(2 n+1) I_{n-1}\ \ \text{for}\ \ n>0$

$I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}$

$\text{Using integration by parts: }$

$\begin{array}{ll}u=x^{n+\frac{1}{2}} & v^{\prime}=(a-x)^{\frac{1}{2}} \\ u^{\prime}=\left(n+\frac{1}{2}\right) x^{n-\frac{1}{2}} & v=-\dfrac{2}{3}(a-x)^{\frac{3}{2}}\end{array}$

	$I_n$	$=\left[u v\right]_0^a-\displaystyle{\int}_0^a u^{\prime} v\, dx$
		$=\underbrace{\left[-\dfrac{2}{3} x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\right]_0^a}_{=0}+\dfrac{2}{3}\left(n+\frac{1}{2}\right) \displaystyle{\int}_0^a x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\, dx$
		$=\dfrac{2 n+1}{3} \displaystyle{\int}_0^a a x^{n-\frac{1}{2}}(a-x)^{\frac{1}{2}}-x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\, dx$
	$3I_n$	$=(2 n+1)\left[a I_{n-1}-I_n\right]$

	$3 I_n+(2 n+1) I_n$	$=a(2 n+1) I_{n-1}$
	$(2 n+4) I_n$	$=a(2 n+1) I_{n-1}$

Complex Numbers, EXT2 N2 2024 HSC 14c

For the complex numbers $z$ and $w$, it is known that $\arg \left(\dfrac{z}{w}\right)=-\dfrac{\pi}{2}$.

Find $\left|\dfrac{z-w}{z+w}\right|$. (2 marks)

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$\abs{\dfrac{z-w}{z+w}}=1$

Show Worked Solution

$\arg \left(\dfrac{z}{w}\right)=\arg \, z-\arg \, w=-\dfrac{\pi}{2}$

$\text{Graphically, \(\arg \, z$ is a $90^{\circ}$ anticlockwise rotation from $\arg \, w$.}\)

$\abs{z-w}=\abs{z+w} \ \text{(diagonals of rectangle)}$

$\therefore \abs{\dfrac{z-w}{z+w}}=1$

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)

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Show that $k=\dfrac{\ln 4}{15}$. (1 mark)

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At what time will the particle's velocity be 30 m s$^{-1}$ to the right? (3 marks)

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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 \(t$ when $v=30$:}\)

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

Mechanics, EXT2 M1 2024 HSC 13b

A particle is moving in simple harmonic motion, described by $\ddot{x}=-4(x+1)$.

When the particle passes through the origin, the speed of the particle is 4 m s$^{-1}$.

What distance does the particle travel during a full period of its motion? (3 marks)

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Show Worked Solution

	$\ddot{x}$	$=-4(x+1)$
	$\dfrac{d}{dx}\left(\dfrac{1}{2} v^2\right)$	$=-4 x-4$
	$\dfrac{1}{2} v^2$	$=\displaystyle \int-4 x-4\, d x$
	$\dfrac{1}{2} v^2$	$=-2 x^2-4 x+c$
	$v^2$	$=-4 x^2-8 x+c$

$\text {When } x=0, \quad v=4\ \ \Rightarrow \ c=16$

$v^2=-4 x^2-8 x+16$

$\text {Find } x \text { when } v=0:$

$-4 x^2-8 x+16$	$=0$
$x^2+2 x-4$	$=0$

$\Rightarrow\ x=\dfrac{-2 \pm \sqrt{2^2+4 \cdot 1 \cdot 4}}{2}=-1 \pm \sqrt{5}$

$\Rightarrow\ \text{Amplitude}=\sqrt{5}$

$\therefore \text {Distance travelled in full period }=4 \sqrt{5} \text{ units }$

Trigonometry, EXT1 T1 2024 HSC 14c

Explain why the equation $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta$, where $-\pi<\theta<\pi$, has exactly one solution. (1 marks)

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Solve $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}$. (2 marks)

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i. $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi$

$\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\Rightarrow \text { Both are monotonically increasing functions}$

$\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)$

$\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}$

ii. $x=\dfrac{1}{2}$

Show Worked Solution

i. $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi$

$\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\Rightarrow \text { Both are monotonically increasing functions}$

$\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)$

$\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}$

ii. $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}$

$\tan \left(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)\right)=\tan \left(\dfrac{3 \pi}{4}\right)$

$\dfrac{\tan \left(\tan ^{-1}(3 x)\right)+\tan \left(\tan ^{-1}(10 x)\right)}{1-\tan \left(\tan ^{-1}(3 x)\right) \cdot \tan \left(\tan ^{-1}(10 x)\right)}=-1$

	$\dfrac{3 x+10 x}{1-30 x^2}$	$=-1$
	$13 x$	$=30 x^2-1$
	$30 x^2-13 x-1$	$=0$
	$(15 x+1)(2 x-1)$	$=0$

$x=\dfrac{1}{2}\ \ \text {or}\ \ -\dfrac{1}{15}$

$\text {Graph is monotonically increasing through } (0,0) \Rightarrow \ \Big(x \neq -\dfrac{1}{15} \Big)$

$\therefore x=\dfrac{1}{2}$

Calculus, EXT1 C2 2024 HSC 14b

For what values of the constant $k$ would the function $f(x)=\dfrac{k x}{1+x^2}+\arctan x$ have an inverse? (3 marks)

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$f(x)\ \text{has an inverse for}\ \ -1 \leqslant k \leqslant 1$

Show Worked Solution

	$f(x)$	$=\dfrac{k x}{1+x^2}+\arctan x$
	$f^{\prime}(x)$	$=\dfrac{k\left(1+x^2\right)-k x(2 x)}{\left(1+x^2\right)^2}+\dfrac{1}{1+x^2}$
		$=\dfrac{k+k x^2-2 k x^2+1+x^2}{\left(1+x^2\right)^2}$
		$=\dfrac{x^2(1-k)+k+1}{\left(1+x^2\right)^2}$

$\text{Inverse function } \Rightarrow f(x) \text { has no SPs}$

$x^2(1-k)+k+1 \neq 0$

$\text {No solution if } \ \Delta<0:$

	$-4(1-k)(k+1)$	$<0$
	$(1-k)(k+1)$	$>0$

$f(x)\ \text{has an inverse for}\ \ -1 \leq k \leq 1$

Calculus, EXT1 C3 2024 HSC 14a

Find the domain and range of the function that is the solution to the differential equation

$\dfrac{d y}{d x}=e^{x+y}$

and whose graph passes through the origin. (4 marks)

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$\text{Domain:}\ \ x<\ln 2$

$\text{Range:}\ \ y>-\ln 2$

Show Worked Solution

$\dfrac{d y}{d x}=e^{x+y}=e^{x}\cdot e^{y}$

	$\displaystyle\int e^{-y}\, d y$	$=\displaystyle \int e^x\, d x$
	$-e^{-y}$	$=e^x+c$

$\text{Passes through }(0,0):$

$-e^0=e^0+c \ \Rightarrow \ c=-2$

	$-e^{-y}$	$=e^x-2$
	$e^{-y}$	$=2-e^x$
	$-y$	$=\ln \left(2-e^x\right)$
	$y$	$=-\ln \left(2-e^x\right)$

$\text{Since}\ \ 2-e^x>0 \ \Rightarrow \ e^x<2$

$\Rightarrow \ \text{Domain:}\ \ x<\ln 2$

$\text{Since}\ \ e^x>0 \ \Rightarrow \ 2-e^x<2$

$\Rightarrow \ \text{Range:}\ \ y>-\ln 2$

Calculus, EXT1 C2 2024 HSC 13d

Using the substitution $u=e^x+2 e^{-x}$, and considering $u^2$, find $\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x$. (3 marks)

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$\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c$

Show Worked Solution

$u=e^x+2 e^{-x} \ \Rightarrow \ u^2=\left(e^x+2 e^{-x}\right)^2=e^{2 x}+4+4 e^{-2 x}$

$\dfrac{du}{dx}=e^x-2 e^{-x} \ \Rightarrow \ du=\left(e^x-2 e^{-x}\right)\, d x$

	$\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x$
		$=\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}} \times \frac{e^{-2 x}}{e^{-2 x}}\, d x$
		$=\displaystyle \int \frac{e^x-2 e^{-x}}{4 e^{-2 x}+8+e^{2 x}}\, d x$
		$=\displaystyle \int \frac{e^x-2 e^{-x}}{4+\left(e^{2 x}+4+4 e^{-2 x}\right)}\, d x$
		$=\displaystyle \int \frac{1}{4+u^2}\, d u$
		$=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{u}{2}\right)+c$
		$=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c$

Vectors, EXT1 V1 2024 HSC 13c

The vector $\underset{\sim}{a}$ is $\displaystyle \binom{1}{3}$ and the vector $\underset{\sim}{b}$ is $\displaystyle\binom{2}{-1}$.

The projection of a vector $\underset{\sim}{x}$ onto the vector $\underset{\sim}{a}$ is $k \underset{\sim}{a}$, where $k$ is a real number.

The projection of the vector $\underset{\sim}{x}$ onto the vector $\underset{\sim}{b}$ is $p \underset{\sim}{b}$, where $p$ is a real number.

Find the vector $\underset{\sim}{x}$ in terms of $k$ and $p$. (4 marks)

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$\underset{\sim}{x}=\dfrac{5}{7} \displaystyle \binom{2 k+3 p}{4 k-p}$

Show Worked Solution

$\underset{\sim}{a}=\displaystyle \binom{1}{3}, \ \abs{\underset{\sim}{a}}=\sqrt{1^2+3^2}=\sqrt{10}$

$\underset{\sim}{b}=\displaystyle \binom{2}{-1}, \ \abs{\underset{\sim}{b}}=\sqrt{2^2+(-1)^2}=\sqrt{5}$

$\text{Let } \underset{\sim}{x}=\displaystyle \binom{x_1}{x_2}$

$\operatorname{proj}_{\underset{\sim}{a}} \underset{\sim}{x}=\dfrac{\underset{\sim}{x} \cdot \underset{\sim}{a}}{|\underset{\sim}{a}|^2} \underset{\sim}{a}=\dfrac{x_1+3 x_2}{10} \cdot \underset{\sim}{a}$

$k=\dfrac{x_1+3 x_2}{10} \ \Rightarrow \ x_1+3 x_2=10 k\ \ldots\ (1)$

$\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{x}=\dfrac{\underset{\sim}{b} \cdot \underset{\sim}{x}}{|\underset{\sim}{b}|^2} b=\dfrac{2 x_1-x_2}{5} \cdot \underset{\sim}{b}$

$p=\dfrac{2 x_1-x_2}{5} \ \Rightarrow \ 2 x_1-x_2=5 p\ \ldots\\ (2)$

$\text {Multiply } (2) \times 3$

$6 x_1-3 x_2=15 p\ \ldots\ (3)$

$(1)+(3)$

$7 x_1$	$=10 k+15 p$
$x_1$	$=\dfrac{1}{7}(10 k+15)$

$\text {Multiply } (1) \times 2$

$2 x_1+6 x_2=20 k\ \ldots\ (4)$

$\text {Subtract} (4)-(2)$

$7x_2$	$=20 k-5 p$
$x_2$	$=\dfrac{1}{7}(20 k-5 p)$

$\therefore \underset{\sim}{x}=\displaystyle \frac{1}{7}\binom{10 k+15 p}{20 k-5 p}=\frac{5}{7}\binom{2 k+3 p}{4 k-p}$

Calculus, EXT1 C2 2024 HSC 13b

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

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Hence, or otherwise, evaluate $\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x$. (3 marks)

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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)

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Clearly sketch the graph of the solution that passes through the point $S$. (1 mark)

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Find the predicted value of the population, $P(t)$, at which the rate of growth of the population is largest. (2 marks)

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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): $

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

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

Functions, EXT1 F1 2024 HSC 12e

The diagram shows the graph of $y=\dfrac{1}{\abs{x-5}}$.

For what values of $x$ is $\dfrac{x}{6} \geq\dfrac{1}{\abs{x-5}}$ ? (3 marks)

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Show Answers Only

$x \in[2,3] \cup[6, \infty)$

Show Worked Solution

$\dfrac{x}{6} \geqslant \dfrac{1}{|x-5|}$

$x|x-5| \geqslant 6$

$\text{Case 1:}$

$x(x-5) \geqslant 6$

$x^2-5 x-6 \geqslant 0$

$(x-6)(x+1) \geqslant 0$

$x \leqslant-1\ \ \text{or}\ \ x \geqslant 6$

$\text {By inspection of graph} \ \Rightarrow \ x \leqslant -1\ \text{is not a solution}$

$\Rightarrow x \geqslant 6$

$\text {Case 2: }$

$-x(x-5) \geqslant 6$

$-x^2+5 x-6 \geqslant 0$

$x^2-5 x+6 \leqslant 0$

$(x-3)(x-2) \leqslant 0$

$\Rightarrow 2 \leqslant x \leqslant 3$

$\therefore x \in[2,3] \cup[6, \infty)$

CHEMISTRY, M2 EQ-Bank 14

The concentration of a solution of hydrochloric acid $\ce{HCl}$ is 1.80% (w/v). What is the molar concentration produced by diluting 20.0 mL of this solution to a total volume of 200.0 mL with deionised water? (3 marks)

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Show Answers Only

$0.049\ \text{mol/L}$

Show Worked Solution

→ $1.8$% (w/v) means that there is $1.8\ \text{g}$ of solute in $100\ \text{mL}$.

→ In $20\ \text{mL}$, there is $1.8\ \text{g} \times 0.2 = 0.36\ \text{g}$ of $\ce{HCl}$.

$n\ce{(HCl)} = \dfrac{0.36}{1.008 + 35.45} = 0.00987\ \text{mol}$

→ The final concentration after the dilution will be:

$c = \dfrac{n}{V} = \dfrac{0.00987}{0.2} = 0.049\ \text{mol/L}$

CHEMISTRY, M2 EQ-Bank 13

If you have a solution with a concentration of 1.2 mol/L and need 0.6 moles of solute, what volume of solution is required? (1 mark)

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Explain why knowing the volume of a solution is important when preparing specific concentrations in laboratory settings. (2 marks)

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a. $0.5\ \text{L}$

b. The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

Show Worked Solution

a. $V=\dfrac{n}{c}=\dfrac{0.6}{1.2}=0.5\ \text{L}$

b. The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

Statistics, EXT1 S1 2024 HSC 9 MC

A bag contains $n$ metal coins, $n \ge 3$, that are made from either silver or bronze.

There are $k$ silver coins in the bag and the rest are bronze.

Two coins are to be drawn at random from the bag, with the first coin drawn not being replaced before the second coin is drawn.

Which of the following expressions will give the probability that the two coins drawn are made of the same metal?

$\dfrac{k(k-1)+(n-k)(n-k-1)}{n(n-1)}$
$\left(\begin{array}{l}n \\ 2\end{array}\right) \left(\begin{array}{l}n \\ k\end{array}\right) \left(\begin{array}{l}1-\dfrac{k}{n} \end{array}\right)^{n-2} $
$\dfrac{\left(\begin{array}{l}k \\ 2\end{array}\right) + \left(\begin{array}{c}n-k \\ 2\end{array}\right)}{n(n-1)} $
$\dfrac{k^{2}+(n-k)^2}{n^{2}}$

Show Answers Only

$A$

Show Worked Solution

$P(B)=P(\text{Bronze}), \ P(S)=P(\text{Silver}) $

$n=\ \text{total coins},\ k=\ \text{bronze coins},\ n-k=\ \text{silver coins}$

$P(BB \cup SS)$	$=\dfrac{k}{n} \times \dfrac{k-1}{n-1} + \dfrac{n-k}{n} \times \dfrac{n-k-1}{n-1}$
	$=\dfrac{k(k-1)+(n-k)(n-k-1)}{n(n-1)}$

$\Rightarrow A$

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)

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

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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.

CHEMISTRY, M2 EQ-Bank 10

A student is asked to prepare 500.0 mL of a 0.150 mol L$^{-1}$ standard solution of oxalic acid $\ce{(C2H2O4.2H2O)}$, and then to perform a dilution to produce 250.0 mL of a 0.0300 mol L$^{-1}$ solution. Outline and explain each step in this process, including the calculations involved and choice of equipment. (5 marks)

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Justify the procedure in part a by explaining two measures taken to ensure the accuracy of the standard solution and diluted solution produced. (2 marks)

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a. Calculate the mass of oxalic acid:

$MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}$

$n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}$

$m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}$

→ 9.455 g of oxalic acid needs to be weighed using an electronic balance.

Prepare the standard solution:

→ Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.

→ Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.

→ Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

$V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}$

→ Use a pipette to transfer 50.0 mL of the 0.150 mol L$^{-1}$ solution into a 250.0 mL volumetric flask.

→ Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L$^{-1}$.

b. Measures taken to ensure an accurate and precise solution:

→ A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.

→ The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimizing any error in the amount of solute added to the solution.

→ The pipette provides precise measurements crucial for accurate dilutions.

→ Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

Show Worked Solution

a. Calculate the mass of oxalic acid:

$MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}$

$n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}$

$m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}$

→ 9.455 g of oxalic acid needs to be weighed using an electronic balance.

Prepare the standard solution:

→ Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.

→ Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

$V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}$

→ Use a pipette to transfer 50.0 mL of the 0.150 mol L$^{-1}$ solution into a 250.0 mL volumetric flask.

→ Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L$^{-1}$.

b. Measures taken to ensure an accurate and precise solution:

→ The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimizing any error in the amount of solute added to the solution.

→ The pipette provides precise measurements crucial for accurate dilutions.

Functions, EXT1 F1 2024 HSC 6 MC

How many real value(s) of $x$ satisfy the equation

$\abs{b} = \abs{b\,\sin(4x)}$,

where $x \in [0, 2\pi]$ and $b$ is not zero?

$1$
$2$
$4$
$8$

Show Answers Only

$D$

Show Worked Solution

$\abs{b} = \abs{b\,\sin(4x)}\ \ \Rightarrow\ \ \abs{1} = \abs{\sin(4x)}\ \ (b \neq 0)$

$\sin(4x) = \pm 1$

$x \in [0, 2\pi]\ \ \Rightarrow\ \ 4x \in [0, 8\pi]$

$\therefore 8\ \text{real solutions}$

$\Rightarrow D$

Statistics, 2ADV S3 2024 HSC 25

A function $f(x)$ is defined as

$f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
1-\dfrac{x}{h}, & \text { for}\ \ 0 \leq x \leq h, \\
0, & \text { for}\ \ x \gt h \end{array}\right.$

where $h$ is a constant.

Find the value of $h$ such that $f(x)$ is a probability density function. (2 marks)

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By first finding a formula for the cumulative distribution function, sketch its graph. (2 marks)

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Find the value of the median of the probability density function $f(x)$ . Give your answer correct to 3 decimal places. (2 marks)

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a. $h=2$

c. $\text{Median}\ =0.586$

Show Worked Solution

a. $f(x)\ \text{is a PDF if:}$

$\displaystyle \int_{0}^{h} 1-\dfrac{x}{h}\,dx$	$=1$
$\Big[x-\dfrac{x^2}{2h} \Big]_0^{h}$	$=1$
$h-\dfrac{h}{2}$	$=1$
$h$	$=2$

b. $ \displaystyle \int 1-\dfrac{x}{2}\,dx = x-\dfrac{x^2}{4} + c$

$\text{At}\ \ x=0, \ \ F(0)=0,\ \ c=0 $

$f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
x-\dfrac{x^2}{4}, & \text { for}\ \ 0 \leq x \leq 2, \\
1, & \text { for}\ \ x \gt 2 \end{array}\right.$

c. $\text{Let}\ \ m=\ \text{median of}\ f(x) $

$\displaystyle \int_0^{m} 1-\dfrac{x}{2}\,dx$	$=0.5$
$\Big[ x-\dfrac{x^2}{4} \Big]_0^m$	$=0.5$
$m-\dfrac{m^2}{4}$	$=0.5$
$4m-m^2$	$=2$
$m^2-4m+2$	$=0$
$(m-2)^2$	$=2$
$m$	$=2 \pm \sqrt{2}$

$\therefore \ \text{Median}$	$=2-\sqrt{2}\ \ (x \in [0,2]) $
	$=0.586\ \text{(3 d.p.)}$

Statistics, STD1 S1 2024 HSC 24

Students in two classes, Class $A$ and Class $B$, recorded the number of text messages they sent in a day. Each class has 18 students.

The results are shown in the dot plots.

Compare the two datasets by examining the skewness, median and spread of the distributions. (3 marks)

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\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{Range} & \quad Q_1\quad & \text{Median}&\quad Q_3\quad & IQR\\
\hline
\rule{0pt}{2.5ex} \text{Class } A\rule[-1ex]{0pt}{0pt} & 5-1=4&3&4&5&5-3=2\\
\hline
\rule{0pt}{2.5ex} \text{Class } B\rule[-1ex]{0pt}{0pt} & 6-2=4 &2&3&4&4-2=2\\
\hline
\end{array}

$\text{Skewness:}$

$\text{Class A is slightly negatively skewed, while Class B is more symmetrical with}$

$\text{a slightly positive skew.}$

$\text{Median:}$

$\text{Class A has a higher median (4) compared to Class B (3).}$

$\text{Spread:}$

$\text{Both classes have the same range (4) and IQR (2), suggesting similar variability}$

$\text{in the data.}$

$\text{In conclusion, while the spread of data is similar for both classes, Class A
tends to}$

$\text{have a higher number of text messages sent (higher median and negative skew)}$

$\text{compared to Class B, which has a more symmetrical, slightly positive distribution.}$

Show Worked Solution

$\text{Skewness:}$

$\text{Class A is slightly negatively skewed, while Class B is more symmetrical with}$

$\text{a slightly positive skew.}$

$\text{Median:}$

$\text{Class A has a higher median (4) compared to Class B (3).}$

$\text{Spread:}$

$\text{Both classes have the same range (4) and IQR (2), suggesting similar variability}$

$\text{in the data.}$

$\text{In conclusion, while the spread of data is similar for both classes, Class A
tends to}$

$\text{have a higher number of text messages sent (higher median and negative skew)}$

$\text{compared to Class B, which has a more symmetrical, slightly positive distribution.}$

Algebra, STD1 A3 2024 HSC 23

Carrie is organising a fundraiser.

The cost of hiring the venue and the band is $2500. The cost of providing meals is $50 per person.

Complete the table of values to show the total cost of the fundraiser. (1 mark)

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\begin{array} {|l|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 25 & 50 & 75 & 100 & 125 & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Cost} \rule[-1ex]{0pt}{0pt} & & 3750 & 5000 & 6250 & 7500 & 8750 & 10\,000 \\
\hline
\end{array}

Carrie decides that tickets should be sold at $70 per person. The graph shows the expected revenue at this ticket price. Using the information in part (a), plot the line that shows the cost of the fundraiser. (2 marks)
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How many tickets need to be sold for the fundraiser to break even? (1 mark)

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Carrie sold 300 tickets. How much profit did the fundraiser make? (3 marks)

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a. $\text{Cost when 0 people }= $2500$

c. $100\ \text{tickets}$

d. $\text{Profit }=$3500$

Show Worked Solution

a. $\text{Cost when 0 people }= $2500$

c. $\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}$

$\therefore\ \text{Break-even when 100 tickets sold.}$

d. $\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}$

$\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n$

$\text{Find profit}\ (P)\ \text{when}\ \ n=300:$

$P$	$=R-C$
	$=70\times 300-(2500+50\times300)$
	$=21\,000-17\,500$
	$=$3500$

Measurement, STD1 M5 2024 HSC 29

A floor plan for a living area is shown. All measurements are in millimetres.

What is the length and width of the cupboard, in metres? (1 mark)

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The floor of the living area is to be tiled. Tiles will NOT be placed under the cupboard.
Each tile is 0.2 m × 0.5 m. The tiles are supplied in boxes of 15 at a cost of $100 per box. Only full boxes can be purchased.
What is the cost of the tiles for the living area? (4 marks)
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a. $4\ \text{m}\times 0.5\ \text{m}$

b. $$1100$

Show Worked Solution

a. $\text{1 metre = 1000 mm}$

$\text{Cupboard}\ = 4\ \text{m}\times 0.5\ \text{m}$

b. $\text{Method 1}$

$\text{Total area to be tiled}\ =6\times 3-4\times 0.5=16\ \text{m}^2$

$\text{Area of each tile}\ =0.2\times0.5=0.1\ \text{m}^2$

$\text{Tiles needed}\ =\dfrac{16}{0.1}=160\ \text{tiles}$

$\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}$

$\therefore\ \text{Total cost}\ =100\times 11=$1100$

$\text{Method 2}$

$\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 $

$\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 $

$\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160$

$\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}$

$\therefore\ \text{Total cost}\ =100\times 11=$1100$

Measurement, STD1 M3 2024 HSC 25

In a national park, a straight path connects a lookout to a car park. The lookout is 35 m higher than the car park. The path is inclined at an angle of elevation of 4°, as shown.

What is the length of the path, correct to the nearest metre? (2 marks)

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$502\ \text{m (nearest metre)}$

Show Worked Solution

$\sin\ 4^\circ$	$=\dfrac{35}{p}$
$p$	$=\dfrac{35}{\sin\ 4^\circ}$
	$=501.74\dots$
	$= 502\ \text{m (nearest metre)}$

Measurement, STD1 M2 2024 HSC 22

The timetable shows an airline schedule in 24-hour time for a flight from Town $A$ to Town $B$.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Flight} \rule[-1ex]{0pt}{0pt} & \textit{Departure} & \textit{Arrival} \\ & \textit{(local time Town A)} & \textit{(local time Town B)} \\
\hline
\rule{0pt}{2.5ex} \text{Town $A$ to Town $B$} \rule[-1ex]{0pt}{0pt} & 1159 & 1336 \\
\hline
\end{array}

When it is 10 am in Town $A$, the time in Town $B$ is 9 am on the same day.

How long does the flight take to travel from Town $A$ to Town $B$? (2 marks)

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$2\ \text{h}\ 37\ \text{minutes}$

Show Worked Solution

$\text{Town B is 1 hour behind town A.}$

$\Rightarrow \ \text{11:59 in Town A is 10:59 in town B.}$

$\therefore\ \text{Length of flight} =13:36-10:59=2\ \text{h}\ 37\ \text{minutes}$

Financial Maths, STD1 F1 2024 HSC 21

Jan borrowed $1500 at 6% per annum.

Calculate the simple interest for the first three months. (2 marks)

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$$22.50$

Show Worked Solution

$\text{Interest}$	$=Prn$
	$=1500\times 0.06\times \dfrac{3}{12}$
	$=$22.50$

Measurement, STD1 M1 2024 HSC 18

A garden is made up of a right-angled triangle and a semicircle as shown.

What is the area of the garden, correct to the nearest square metre? (3 marks)

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$35\ \text{m}^2$

Show Worked Solution

$\text{Base of triangle }=2\times\ \text{radius} = 6\ \text{metres}$

$\text{Total Area }$	$=\text{ Area of triangle + area of semicircle}$
	$=\dfrac{1}{2}\times 6\times 7 + \dfrac{1}{2}\times \pi\times 3^2$
	$=21+14.137\dots$
	$=35.137\dots$
	$=35\ \text{m}^2\ (\text{nearest m}^{2})$

Probability, STD1 S2 2024 HSC 17

A wheel is shown with the numbers 0 to 19 marked.

A game is played where the wheel is spun until it stops.

When the wheel stops, a pointer points to the winning number. Each number is equally likely to win.

List all the even numbers on the wheel that are greater than 7. (1 mark)

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What is the probability that the winning number is NOT an even number greater than 7? (2 marks)

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a. $8\ ,\ 10\ ,\ 12\ ,\ 14\ ,\ 16\ ,\ 18$

b. $0.7$

Show Worked Solution

a. $8\ ,\ 10\ ,\ 12\ ,\ 14\ ,\ 16\ ,\ 18\ \text{(6 numbers)}$

b. $\text{Total numbers = 20}$

$\text{Numbers not even and > 7}\ = 20-6=14\ \text{numbers}$

$P\text{(not even and > 7)}\ =\dfrac{14}{20}=0.7$

Measurement, STD1 M3 2024 HSC 14

A hotel is located 186 m north and 50 m west of a train station.

What is the straight line distance from the hotel to the train station? Round your answer to the nearest metre. (2 marks)

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What is the bearing of the hotel from the train station? Round your answer to the nearest degree. (2 marks)

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a. $193\ \text{m}$

b. $345^\circ\ \text{(nearest degree)}$

Show Worked Solution

a. $\text{By Pythagoras:}$

	$d^2$	$=50^2+186^2$
	$d^2$	$=37\,096$
	$d$	$=\sqrt{37\,096}$
		$=192.603\dots$
		$=193\ \text{m (nearest metre)}$

b.	$\tan\theta$	$=\dfrac{50}{186}$
	$\theta$	$=15.046\dots^\circ$
		$\approx 15^\circ\ \text{(nearest degree)}$

$\text{Bearing}\ H\ \text{from}\ T =360-15=345^\circ$

Statistics, STD1 S1 2024 HSC 13

Consider the following dataset.

$1, \ 1, \ 2, \ 3, \ 5, \ 7, \ 15$

What is the interquartile range? (1 mark)
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By using the outlier formula, determine whether 15 is an outlier. (2 marks)
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a. $\text{IQR = 6}$

b. $15\ \text{is not an outlier as it is not greater than 16.}$

Show Worked Solution

a. $Q_2=3, \ Q_1=1, \ Q_3=7$

$\therefore\ IQR=7-1=6$

b. $\text{Find upper fence:}$

$Q_3+1.5\times IQR=7 + 1.5\times 6=16$

$\therefore\ \text{15 is not an outlier (15 < 16)}$

Measurement, STD1 M5 2024 HSC 11

A scale drawing of a tree is shown. The scale is $1 : 50$.

What is the actual height of the tree, in metres? (2 marks)

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$4\ \text{m}$

Show Worked Solution

$8\times 1\ \text{cm}\,$	$:\,8\times 50\ \text{cm}$
$8\ \text{cm}\,$	$:\,400\ \text{cm}$
$8\ \text{cm}\,$	$:\,4\ \text{m}$

$\therefore\ \text{The tree is 4 metres tall.}$

Measurement, STD1 M1 2024 HSC 9 MC

An equilateral triangle and an isosceles triangle are shown.

The triangles have the same perimeter.

What is the value of $x$?

$8$
$9$
$11$
$12$

Show Answers Only

$D$

Show Worked Solution

$\text{Perimeter triangle 1 }=3\times 11=33\ \text{m}$

$\text{Perimeter triangle 2 }=2\times x+9=(2x+9)\ \text{m}$

$\therefore\ 2x+9$	$=33$
$2x$	$=24$
$x$	$=12$

$\Rightarrow D$

CHEMISTRY, M2 EQ-Bank 6

Iron $\text{(III)}$ hydroxide can be precipitated from the reaction of iron $\text{(III)}$ nitrate solution with sodium hydroxide solution.

$\ce{Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq)}$

Calculate the mass of precipitate formed when 25.0 mL of 0.150 mol L$^{-1}$ iron $\text{(III)}$ nitrate solution is added to 40.0 mL of 0.250 mol L$^{-1}$ sodium hydroxide solution. (3 marks)

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Calculate the concentration of nitrate ions in the final solution. (2 marks)

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a. $0.356\ \text{g}$

b. $0.173\ \text{mol L}^{-1}$

Show Worked Solution

a. $n\ce{(Fe(NO3)3)} = c \times V = 0.150 \times 0.025 = 0.00375\ \text{mol}$

$n\ce{(NaOH)} = 0.25 \times 0.040 = 0.01\ \text{mol}$

→ The molar ratio of $\ce{Fe(NO3)3:NaOH} = 1:3$

$3 \times 0.00375 = 0.01125\ \text{mol}$ of $\ce{NaOH}$ is required to react with 0.00375 mol of $\ce{Fe(NO3)3}$.

→ Due to there only being 0.01 mol of $\ce{NaOH}$ present, $\ce{NaOH}$ will be the limiting reagent.

$n\ce{(Fe(OH)3)}$ formed $=n\ce{(NaOH)} \times \dfrac{1}{3} = 0.01 \times \dfrac{1}{3} = 0.00333\ \text{mol}$

$m\ce{(Fe(OH)3)}= n \times MM = 0.00333 \times (55.85 + 3(1.008) + 3(16.00))= 0.356\ \text{g}$

b. Volume of final solution $=0.025 + 0.040 = 0.065\ \text{L}$

$n\ce{(NO3^-)}= 3 \times n(Fe^{3+}) = 3 \times 0.00375 = 0.01125\ \text{mol}$

$c\ce{(NO3^-)} = \dfrac{n}{V} = \dfrac{0.01125}{0.065} = 0.173\ \text{mol L}^{-1}$

CHEMISTRY, M2 EQ-Bank 4 MC

A 5.0 L sample of water (density = 1.0 kg L$^{-1}$) is contaminated with 15 ppm of mercury ions. What is the mass of mercury in this sample?

75 mg
7.5 mg
15 mg
150 mg

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$A$

Show Worked Solution

→ The mass of the water is 5 kg.

→ 15 ppm = 15 mg/kg.

→ Mass of mercury in the sample $=15 \times 5 = 75\ \text{mg}$.

$\Rightarrow A$

CHEMISTRY, M2 EQ-Bank 5

The compound potassium nitrate has the formula $\ce{KNO3}$

A student makes a solution of this compound by dissolving 40.0 g in 250.0 mL of distilled water. Calculate the concentration of this solution in mol L$^{-1}$. (2 marks)

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The student now requires 500.0 mL of a 5.0% (w/v) solution. What volume of the solution in part (a) is required to make this? (3 marks)

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a. $1.58\ \text{mol L}^{-1}$

b. $0.156\ \text{L}$

Show Worked Solution

a. $n\ce{(KNO3)}= \dfrac{m}{MM} = \dfrac{40.0}{39.10 + 14.01 + 3(16.00)} =0.396\ \text{mol}$

Concentration of solution $=\dfrac{n}{V} = \dfrac{0.396}{0.25} = 1.58\ \text{mol L}^{-1}$

b. A 5% solution requires 5 grams of solute in 100 mL of solution.

Therefore a 500 mL solution requires 25 grams of solute.

$n\ce{(KNO3)} = \dfrac{25}{101.11} = 0.247\ \text{mol}$

$c\ce{(KNO3)} = \dfrac{n}{V} = \dfrac{0.247}{0.5} = 0.494\ \text{mol L}^{-1}$

$c_1V_1$	$=c_2V_2$
$1.58 \times V_1$	$=0.494 \times 0.5$
$V_1$	$=\dfrac{0.247}{1.58}$
$V_1$	$=0.156\ \text{L}$

Financial Maths, STD1 F2 2024 HSC 28

Alex and Jun each invest $1800 for 5 years.

- Alex's investment earns simple interest at a rate of 7.5% per annum.
- Jun's investment earns interest at a rate of 6.0% per annum, compounding quarterly.

By calculating the interest earned over the 5 years, determine who will have the greater amount. (3 marks)

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$\text {Alex’s investment:}$

$\text{Interest}=Prn=1800 \times 0.075 \times 5=\$ 675$

$\text {Jun’s investment:}$

$r=\dfrac{6.0\%}{4}=1.5 \% \text { per quarter}$

$\text {Compounding periods }=5 \times 4=20$

$F V=P V(1+r)^n=1800(1+0.015)^{20}=\$ 2424.34$

$\text{Total interest}=F V-P V=2424.34-1800=\$ 624.34$

$\text {Alex’s interest }>\text { Jun’s interest.}$

$\Rightarrow \text{ Alex will have a greater amount (since original investment the same)}$

Show Worked Solution

$\text {Alex’s investment:}$

$\text{Interest}=Prn=1800 \times 0.075 \times 5=\$ 675$

$\text {Jun’s investment:}$

$r=\dfrac{6.0\%}{4}=1.5 \% \text { per quarter}$

$\text {Compounding periods }=5 \times 4=20$

$F V=P V(1+r)^n=1800(1+0.015)^{20}=\$ 2424.34$

$\text{Total interest}=F V-P V=2424.34-1800=\$ 624.34$

$\text {Alex’s interest }>\text { Jun’s interest.}$

$\Rightarrow \text{ Alex will have a greater amount (since original investment the same)}$

Financial Maths, STD1 F1 2024 HSC 27

Zazu works a 38-hour week and is paid at an hourly rate of $45. Any overtime hours worked are paid at time-and-a-half.

In a particular week, Zazu worked the regular 38 hours and some overtime hours. In that week Zazu earned $2790.

How many hours of overtime did Zazu work in that week? (3 marks)

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$\text{16 hours of overtime}$

Show Worked Solution

$\text {Let X = overtime hours}$

	$\text{Total pay}$	$=(38 \times 45)+\Big(X \times \dfrac{3}{2}\times 45\Big) $
	$2790$	$=1710+\dfrac{135}{2}X$
	$\dfrac{135}{2}X$	$=1080$
	$X$	$=\dfrac{1080 \times 2}{135}$
		$=16 \text{ hours of overtime}$

Networks, STD1 N1 2024 HSC 20

The diagram shows a network with weighted edges.

Draw a minimum spanning tree for this network and determine its weight. (2 marks)

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Is it possible to find another spanning tree with the same weight? Give a reason for your answer. (1 mark)

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b. $\text{Yes.}$

$\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC\ \text{to create a second MST (with equivalent weight = 24)}$

Show Worked Solution

b. $\text{Yes.}$

$\text{→ The edge}\ FC\ \text{on the MST above could be replaced by the edge}\ BC\ \text{to create a second MST (with equivalent weight = 24)}$

Financial Maths, STD1 F2 2024 HSC 8 MC

Three years ago, the price of a uniform was $180.

Due to inflation, the price increased annually by 2.5%.

What is the price of this uniform now?

$180.14
$ 181.35
$ 193.50
$ 193.84

Show Answers Only

$D$

Show Worked Solution

$r=2.5 \%=\dfrac{2.5}{100}=0.025$

	$FV$	$=PV(1+r)^n$
		$=180(1.025)^3$
		$=193.84$

$\Rightarrow D$

Networks, STD1 N1 2024 HSC 6 MC

The map shows regions within a country.

A network diagram is to be drawn to represent this map. Vertices will be used to indicate each region and edges will be used to represent a border shared between two regions.

How many edges will there be in the network diagram?

$8$
$7$
$6$
$5$

Show Answers Only

$B$

Show Worked Solution

$\text{Network diagram:}$

$\text{Network has 7 edges.}$

$\Rightarrow B$

Probability, 2ADV S1 2024 HSC 9 MC

A bag contains 2 red and 3 white marbles. Jovan randomly selects two marbles at the same time from this bag. The probability tree diagram shows the probabilities for each of the outcomes.

Given that one of the marbles that Jovan has selected is red, what is the probability that the other marble that he has selected is also red?

$\dfrac{1}{10}$
$\dfrac{1}{7}$
$\dfrac{1}{4}$
$\dfrac{7}{10}$

Show Answers Only

$B$

Show Worked Solution

$P(R_2\|_{\text{other is red}})$	$=\dfrac{P(R_1 \cap R_2)}{P\text{(at least 1 red)}}$
	$=\dfrac{\frac{2}{5} \times \frac{1}{4}}{1-P(WW)}$
	$=\dfrac{\frac{2}{5} \times \frac{1}{4}}{1-(\frac{3}{5} \times \frac{2}{4})}$
	$=\dfrac{1}{7}$

$\Rightarrow B$

Calculus, 2ADV C3 2024 HSC 31

Two circles have the same centre $O$. The smaller circle has radius 1 cm, while the larger circle has radius $(1 + x)$ cm. The circles enclose a region $QRST$, which is subtended by an angle $\theta$ at $O$, as shaded.

The area of $QRST$ is $A$ cm$^{2}$, where $A$ is a constant and $A \gt 0$.

Let $P$ cm be the perimeter of $QRST$.

By finding expressions for the area and perimeter of $QRST$, show that $P(x)=2x+\dfrac{2A}{x}$. (3 marks)
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Show that if the perimeter, $P(x)$, is minimised, then $\theta$ must be less than 2. (3 marks)
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a. $\text{Area}\ =\ \text{Sector}\ OQR-\text{Sector}\ OTS$

$A$	$=\dfrac{\theta}{2\pi} \times \pi (1+x)^2-\dfrac{\theta}{2\pi} \times \pi \times 1^2$
$A$	$=\dfrac{\theta}{2} (1+x)^2-\dfrac{\theta}{2}$
$A$	$=\dfrac{\theta}{2}(1+2x+x^2-1)$
$A$	$=\dfrac{\theta}{2}x(x+2)$
$\theta$	$=\dfrac{2A}{x(x+2)}$

$\text{Perimeter}\ QRST$	$=\dfrac{\theta}{2\pi} \times 2\pi(x+1)+\dfrac{\theta}{2\pi} \times 2\pi (1)+2x $
	$=\theta(x+1)+\theta+2x$
	$=\theta(x+2)+2x$
	$=\dfrac{2A}{x(x+2)}(x+2)+2x$
	$=2x+\dfrac{2A}{x}$

b. $P(x)=2x+\dfrac{2A}{x}$

$P^{′}(x) = 2-\dfrac{2A}{x^2}$

$P^{″}(x) = \dfrac{4A}{x^3} \gt0\ \ (x,A \gt 0)$

$\Rightarrow\ \text{MIN when}\ \ P^{′}(x)=0:$

$2-\dfrac{2A}{x^2}$	$=0$
$2x^2$	$=2A$
$x^2$	$=A$

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

$\theta$	$=\dfrac{2A}{x(x-2)}$
	$=\dfrac{2x^2}{x(x+2)}$
	$=\dfrac{2x}{x+2}$
	$=2 \times \dfrac{x}{x+2} $
	$ \lt 2\ \ \Big(\text{since}\ \ \dfrac{x}{x+2} \lt 1 \ \ \text{for all}\ x \Big) $

$\therefore\ \text{If}\ P(x)\ \text{is minimised,}\ \theta \lt 2. $

Show Worked Solution

a. $\text{Area}\ =\ \text{Sector}\ OQR-\text{Sector}\ OTS$

$A$	$=\dfrac{\theta}{2\pi} \times \pi (1+x)^2-\dfrac{\theta}{2\pi} \times \pi \times 1^2$
$A$	$=\dfrac{\theta}{2} (1+x)^2-\dfrac{\theta}{2}$
$A$	$=\dfrac{\theta}{2}(1+2x+x^2-1)$
$A$	$=\dfrac{\theta}{2}x(x+2)$
$\theta$	$=\dfrac{2A}{x(x+2)}$

$\text{Perimeter}\ QRST$	$=\dfrac{\theta}{2\pi} \times 2\pi(x+1)+\dfrac{\theta}{2\pi} \times 2\pi (1)+2x $
	$=\theta(x+1)+\theta+2x$
	$=\theta(x+2)+2x$
	$=\dfrac{2A}{x(x+2)}(x+2)+2x$
	$=2x+\dfrac{2A}{x}$

COMMENT: Sector/arc calculations used in solution are for those who don’t want to remember formulas.

b. $P(x)=2x+\dfrac{2A}{x}$

$P^{′}(x) = 2-\dfrac{2A}{x^2}$

$P^{″}(x) = \dfrac{4A}{x^3} \gt0\ \ (x,A \gt 0)$

$\Rightarrow\ \text{MIN when}\ \ P^{′}(x)=0:$

$2-\dfrac{2A}{x^2}$	$=0$
$2x^2$	$=2A$
$x^2$	$=A$

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

$\theta$	$=\dfrac{2A}{x(x-2)}$
	$=\dfrac{2x^2}{x(x+2)}$
	$=\dfrac{2x}{x+2}$
	$=2 \times \dfrac{x}{x+2} $
	$ \lt 2\ \ \Big(\text{since}\ \ \dfrac{x}{x+2} \lt 1 \ \ \text{for all}\ x \Big) $

$\therefore\ \text{If}\ P(x)\ \text{is minimised,}\ \theta \lt 2. $

Calculus, 2ADV C4 2024 HSC 22

The graph of the function $f(x) = \ln(1 + x^{2})$ is shown.

Prove that $f(x)$ is concave up for $-1 < x < 1$. (3 marks)
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A table of function values, correct to 4 decimal places, for some $x$ values is provided.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
\rule{0pt}{2.5ex} \ln(1+x^2) \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 0.0606 & 0.2231 & 0.4463 & 0.6931 \\
\hline
\end{array}

Using the function values provided and the trapezoidal rule, estimate the shaded area in the diagram. (2 marks)
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Is the answer to part (b) an overestimate or underestimate? Give a reason for your answer. (1 mark)
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a. $f(x)= \ln(1+x^2)$

$f^{′}(x)=\dfrac{2x}{1+x^2}$

$f^{″}(x)$	$=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}$
	$=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}$
	$=\dfrac{2(1-x^2)}{(1+x^2)^2}$

$\text{Consider domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 $

$(1+x^2)^2 \gt 0\ \ \text{for all}\ x$

$\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) $

$\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) $

b. $\text{Total shaded area}\ \approx 0.5383\ \text{(4 d.p.)}$

c. $\text{Trapezoidal rule assumes straight lines join points in the table.}$

$\text{Since the graph is concave up in the given domain, the trapezoidal rule will overestimate the area.}$

Show Worked Solution

a. $f(x)= \ln(1+x^2)$

$f^{′}(x)=\dfrac{2x}{1+x^2}$

$f^{″}(x)$	$=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}$
	$=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}$
	$=\dfrac{2(1-x^2)}{(1+x^2)^2}$

$\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 $

$(1+x^2)^2 \gt 0\ \ \text{for all}\ x$

$\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) $

$\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) $

b. $\text{Total shaded area}$

$\approx 2 \times \dfrac{h}{2}[ y_0 + 2(y_1+y_2+y_3) + y_4] $

$\approx 2 \times \dfrac{0.25}{2}[ 0 + 2(0.0606+0.2231+0.4463) + 0.6931] $

$\approx 0.538275 $

$\approx 0.5383\ \text{(4 d.p.)}$

c. $\text{Trapezoidal rule assumes straight lines join points in the table.}$

$\text{Since the graph is concave up in the given domain, the trapezoidal rule will overestimate the area.}$

Financial Maths, STD2 F4 2024 HSC 21

William has a reducing balance loan on which he owes $5590. He makes monthly repayments of $110.

The loan company charges interest at 24% per annum, compounded monthly.

The spreadsheet shows some of the information for the next two months of the loan.

Complete the entries in the spreadsheet to show the balance owing on the loan at the end of two months. (2 marks)

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Explain why the loan will never be repaid if William continues to make repayments of $110 per month. (1 mark)

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b. $\text{The interest charged exceeds the amount that is repaid.}$

$\text{Interest charges will gradually increase with the monthly repayment staying the same \(\Rightarrow$ loan will never be repaid.}\)

Show Worked Solution

$\text{Calculations:}$

$\text {Cell E2 }= 5590+111.80-110=\$ 5591.80$

$\text{Cell B3 }=\text{ Cell E2}$

$\text{Cell C3}=5591.80 \times 0.02 = \$111.84 \ \text{(monthly r/i = 2%)}$

$\text{Cell E3}=5591.80+111.84-110 = \$5593.64$

b. $\text{The interest charged exceeds the amount that is repaid.}$

$\text{Interest charges will gradually increase with the monthly repayment staying the same \(\Rightarrow$ loan will never be repaid.}\)

Financial Maths, STD2 F5 2024 HSC 20

The table shows the future value for an annuity of $1 for varying interest rates and time periods.

Ken invests $200 at the start of each year for eight years, at an interest rate of 5% per annum.
Calculate the future value of Ken's investment. (1 mark)

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Shay is planning to take a holiday in three years. She needs $4500 for this holiday and will make regular six-monthly payments into an account that earns interest at the rate of 4% per annum, compounded 6 monthly.
What is the minimum amount Shay needs to pay into this account every 6 months? Give your answer to the nearest $10. Support your answer with calculations. (2 marks)

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a. $F V=\$ 2005.32$

b. $\$700$

Show Worked Solution

a. $\text {8 annual periods at 5% p.a.} \Rightarrow \text { Factor}=10.0266$

$F V=200 \times 10.0266=\$ 2005.32$

b. $r=\dfrac{4 \%}{2}=2 \%\ \text{per 6 months}$

$\text {Compounding periods}=3 \times 2=6$

$\Rightarrow \text {Factor }=6.4343$

$4500$	$=\ \text{Annuity} \times 6.4343$
$\text{Annuity}$	$=\dfrac{4500}{6.4343}$
	$=699.38$
	$=\$700\ \text{(nearest \$10)}$

Financial Maths, 2ADV M1 2024 HSC 26

Twenty-five years ago, Phoenix deposited a single sum of money into a new bank account, earning 2.4% interest per annum compounding monthly.

Present value interest factors for an annuity of $1 for various interest rates $(r)$ and numbers of periods $(n)$ are given in the table.

Phoenix made the following withdrawals from this account.

$2000 at the end of each month for the first 15 years, starting at the end of the first month.
$1200 at the end of each month for the next 10 years, starting at the end of the 181st month after the account was opened.

Calculate the minimum sum that Phoenix could have deposited in order to make these withdrawals. (4 marks)

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$\text{Minimum deposit}\ = $391\,344.80$

Show Worked Solution

$\text{1st Annuity}$

$\text{Find PVA for \$2000 paid monthly for 1st 15 years:}$

$r= \dfrac{2.4%}{12} = 0.2\% = 0.002$

$\text{Total payments (to Phoenix)}\ = 15 \times 12 = 180$

$\text{PVA factor (from table)}\ = 151.036$

$\text{PVA (1st annuity)}\ = 2000 \times 151.036 = $302\,072$

$\text{2nd Annuity}$

$\text{Find PVA for \$1200 paid monthly from year 16 to 25:}$

$\text{PVA (2nd annuity) = PVA (25 years) }-\text{ PVA (15 years)}$

$r= \dfrac{2.4%}{12} = 0.2\% = 0.002$

$\text{Total payments (25 years)}\ = 25 \times 12 = 300$

$\text{PVA factors (from table): 225.430 (25 years), 151.036 (15 years)}$

$\text{PVA (2nd annuity)}$	$=(1200 \times 225.430)-(1200 \times 151.036)$
	$=$89\,272.80$

$\therefore\ \text{Minimum deposit}\ = 302\,072+89\,272.80 = $391\,344.80$

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

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

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

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

\(v+f\)	\(=e+2\)
\(7+f\)	\(=11+2\)
\(f\)	\(=6\)

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

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

	\(e^{\large{\bar{z}}}\)	\(=e^{x-iy}\)
		\(=e^x \cdot e^{-iy}\)
		\(=e^x(\cos (-y)+i \sin (-y))\)
		\(=e^x(\cos (y)-i \sin (y))\)

	\(\overline{e^{\large{z}}}\)	\(=\overline{e^x(\cos (y)+i \sin (y)}\)
		\(=e^x(\cos (y)-i \sin (y))\)
		\(=e^{\large{\bar{z}}}\)