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PHYSICS, M8 2025 HSC 3 MC

The diagram shows lines in the emission spectrum of hydrogen.
 

The production of this spectrum can be explained by applying the atomic model developed by which scientist?

  1. Balmer
  2. Bohr
  3. Planck
  4. Rutherford
Show Answers Only

\(B\)

Show Worked Solution
  • Bohr’s atomic model correctly explains hydrogen’s emission spectrum by proposing that electrons occupy quantised energy levels and emit photons when transitioning between them.

\(\Rightarrow B\)

PHYSICS, M6 2025 HSC 2 MC

An ideal transformer converts 240 V to 2200 V.

Which row in the table best describes the transformer?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
 \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Type of}& \text{Number of turns in}& \text{Number of turns in} \\
\quad \text{transformer}\quad \rule[-1ex]{0pt}{0pt}& \text{primary coil}& \text{secondary coil} \\
\hline
\rule{0pt}{2.5ex}\text{Step up}\rule[-1ex]{0pt}{0pt}&\text{120}&\text{1100}\\
\hline
\rule{0pt}{2.5ex}\text{Step up}\rule[-1ex]{0pt}{0pt}& \text{1100}&\text{120}\\
\hline
\rule{0pt}{2.5ex}\text{Step down}\rule[-1ex]{0pt}{0pt}& \text{12}&\text{1100} \\
\hline
\rule{0pt}{2.5ex}\text{Step down}\rule[-1ex]{0pt}{0pt}& \text{1100} &\text{120}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • Since the voltage increases from the primary to the secondary coil, it is a step up transformer (eliminate \(C\) and \(D\)).
  • Using  \(\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}\ \ \Rightarrow\ \ \dfrac{240}{2200}=\dfrac{120}{1100}\).

\(\Rightarrow A\)

PHYSICS, M7 2025 HSC 1 MC

Which of the following did Maxwell contribute to the understanding of the nature of light?

  1. Explanation of atomic emission spectra
  2. Prediction of the speed of electromagnetic waves
  3. Experimental support for the particle model of light
  4. Experimental confirmation of light beyond the visible spectrum
Show Answers Only

\(B\)

Show Worked Solution
  • Maxwell’s contribution was the prediction of the speed of electromagnetic waves, showing that light is an electromagnetic wave.

\(\Rightarrow B\)

PHYSICS, M7 2025 HSC 25

A student conducts an experiment to determine the work function of potassium.

The following diagram depicts the experimental setup used, where light of varying frequency is incident on a potassium electrode inside an evacuated tube.
 

For each frequency of light tested, the voltage in the circuit is varied, and the minimum voltage (called the stopping voltage) required to bring the current in the circuit to zero is recorded.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{Frequency of incident light} \ \ & \quad \quad \quad \textit{Stopping voltage} \quad \quad \quad \\
\left(\times 10^{15} \ \text{Hz}\right) \rule[-1ex]{0pt}{0pt}& \text{(V)}\\
\hline
\rule{0pt}{2.5ex}0.9 \rule[-1ex]{0pt}{0pt}& 1.5 \\
\hline
\rule{0pt}{2.5ex}1.1 \rule[-1ex]{0pt}{0pt}& 2.0 \\
\hline
\rule{0pt}{2.5ex}1.2 \rule[-1ex]{0pt}{0pt}& 2.5 \\
\hline
\rule{0pt}{2.5ex}1.3 \rule[-1ex]{0pt}{0pt}& 3.0 \\
\hline
\rule{0pt}{2.5ex}1.4 \rule[-1ex]{0pt}{0pt}& 3.5 \\
\hline
\rule{0pt}{2.5ex}1.5 \rule[-1ex]{0pt}{0pt}& 4.0 \\
\hline
\end{array}

  1. Construct an appropriate graph using the data provided, and from this, determine the threshold frequency of potassium.   (3 marks)

  1. Using the particle model of light, explain the features shown in the experimental results.   (3 marks)

Show Answers Only

a.    
         
          

b.   Features shown in experiment results:

  • Light is made of discrete energy packets called photons.
  • A photon’s energy is directly proportional to its frequency: \(E = hf\).
  • When photons hit a metal surface, they transfer their energy to electrons.
  • If the photon energy is greater than the metal’s work function \((\phi)\), the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation  \(KE = hf-\phi\).
  • Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
  • A larger stopping voltage is therefore needed to halt these more energetic electrons.
Show Worked Solution

a.    
         
       

b.   Features shown in experiment results:

  • Light is made of discrete energy packets called photons.
  • A photon’s energy is directly proportional to its frequency: \(E = hf\).
  • When photons hit a metal surface, they transfer their energy to electrons.
  • If the photon energy is greater than the metal’s work function \((\phi)\), the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation  \(KE = hf-\phi\).
  • Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
  • A larger stopping voltage is therefore needed to halt these more energetic electrons.

BIOLOGY, M7 2025 HSC 21

The diagram shows components of the innate immune system in humans.
   

State the role of TWO components that protect against infection.   (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \textit{How it protects against infection}\quad \quad \rule[-1ex]{0pt}{0pt}\\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\end{array}

Show Answers Only

Any TWO of the following components

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \textit{How it protects against infection}\quad\rule[-1ex]{0pt}{0pt}\\
\hline
\text{Skin} & \text{Acts as a physical barrier}\\
\ & \text{preventing pathogen entry into} \\
\ & \text{tissue}\\
\hline
\text{Stomach} &\text{Destroys ingested pathogens} \\
\text{acid} & \text{through low pH chemical}\\
\ & \text{environment.}\\
\hline
\text{Mucus} &\text{Traps pathogens and prevents} \\
\text{lining} & \text{their entry into underlying}\\
\ & \text{tissues.}\\
\hline
\text{Nasal} &\text{Filters and traps airborne} \\
\text{hair} & \text{pathogens, preventing}\\
\ & \text{respiratory entry.}\\
\hline
\text{Tear glands} &\text{Produce lysozyme enzyme that} \\
\ & \text{destroys bacterial cell walls.}\\
\hline
\text{Urinary} &\text{Flushes pathogens from the} \\
\text{tract} & \text{urethra preventing infection.}\\
\hline
\end{array}

Show Worked Solution

Any TWO of the following components

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \textit{How it protects against infection}\quad\rule[-1ex]{0pt}{0pt}\\
\hline
\text{Skin} & \text{Acts as a physical barrier}\\
\ & \text{preventing pathogen entry into} \\
\ & \text{tissue}\\
\hline
\text{Stomach} &\text{Destroys ingested pathogens} \\
\text{acid} & \text{through low pH chemical}\\
\ & \text{environment.}\\
\hline
\text{Mucus} &\text{Traps pathogens and prevents} \\
\text{lining} & \text{their entry into underlying}\\
\ & \text{tissues.}\\
\hline
\text{Nasal} &\text{Filters and traps airborne} \\
\text{hair} & \text{pathogens, preventing}\\
\ & \text{respiratory entry.}\\
\hline
\text{Tear glands} &\text{Produce lysozyme enzyme that} \\
\ & \text{destroys bacterial cell walls.}\\
\hline
\text{Urinary} &\text{Flushes pathogens from the} \\
\text{tract} & \text{urethra preventing infection.}\\
\hline
\end{array}

Functions, 2ADV EQ-Bank 08

Solve for \(x\), giving your answers in the simplest form  \(a+b\sqrt{c}\)  where \(a, b\) and \(c\) are real:

\(5 x^2-20 x+4=0\)   (2 marks)

Show Answers Only

\(x=2 \pm \dfrac{4}{5} \sqrt{5}\)

Show Worked Solution

\(5 x^2-20 x+4=0\)

\(x\) \(=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
  \(=\dfrac{20 \pm \sqrt{20^2-4 \times 5 \times 4}}{2 \times 5}\)
  \(=\dfrac{20 \pm \sqrt{320}}{10}\)
  \(=2 \pm \dfrac{8 \sqrt{5}}{10}\)
  \(=2 \pm \dfrac{4}{5} \sqrt{5}\)

Functions, 2ADV EQ-Bank 7 MC

What are the solutions to  \(3x^2+2x-4=0\)?

  1. \(x=\dfrac{-1 \pm \sqrt{13}}{3}\)
  2. \(x=\dfrac{1 \pm \sqrt{13}}{3}\)
  3. \(x=\dfrac{-1 \pm \sqrt{5}}{2}\)
  4. \(x=\dfrac{1 \pm \sqrt{5}}{2}\)
Show Answers Only

\(A\)

Show Worked Solution

\(3 x^2+2 x-4=0\)

\(x\) \(=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2a}\)
  \(=\dfrac{-2 \pm \sqrt{2^2-4 \times 3 \times-4}}{2 \times 3}\)
  \(=\dfrac{-2 \pm \sqrt{52}}{6}\)
  \(=\dfrac{-1 \pm \sqrt{13}}{3}\)

 

\(\Rightarrow A\)

PHYSICS, M6 2025 HSC 22

The diagram represents the parts of the AC system used to transfer energy from a power station to people's houses.
 

Describe the energy transformations that take place in the transformers, and in the transmission line.   (4 marks)

Show Answers Only
  • Transformers \(A\) and \(B\): Some electrical energy is always lost, so each transformer outputs less electrical energy than it receives.
  • Core losses in transformers: Part of the input electrical energy is transformed into heat in the iron core due to eddy currents and magnetic effects.
  • Resistive losses in transformers: Some electrical energy becomes heat in the coils because of their resistance. Small amounts may also become sound/vibration energy.
  • Transmission lines: Electrical energy is also transformed into heat in the wires due to their resistance.
Show Worked Solution
  • Transformers \(A\) and \(B\): Some electrical energy is always lost, so each transformer outputs less electrical energy than it receives.
  • Core losses in transformers: Part of the input electrical energy is transformed into heat in the iron core due to eddy currents and magnetic effects.
  • Resistive losses in transformers: Some electrical energy becomes heat in the coils because of their resistance. Small amounts may also become sound/vibration energy.
  • Transmission lines: Electrical energy is also transformed into heat in the wires due to their resistance.

Functions, 2ADV EQ-Bank 10

  1. Express  \(y=x^2-4 x+6\)  in the form  \(y=(x-a)^2+c\).   (1 mark)

  2. Graph the parabola, labelling its vertex and \(y\)-intercept.   (2 marks)

Show Answers Only

a.   \(y=(x-2)^2+2\)

b.   \(\text {Vertex}=(2,2)\)

\(y \text{-intercept}=(0,6)\)
 

Show Worked Solution
a.     \(y\) \(=x^2-4 x+6\)
    \(=x^2-4 x+4+2\)
    \(=(x-2)^2+2\)

 
b.    \(\text {Vertex}=(2,2)\)

\(y \text{-intercept}=(0,6)\)

Functions, 2ADV EQ-Bank 02

  1. Find the equation of the line that passes through \((2,1)\) and \((-3,4)\).   (2 marks)

  2. Determine whether \((7,-2)\) lies on the line.   (1 mark)

Show Answers Only

a.    \(y=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}\)

 

b.    \(\text {Substitute}\ (7,-2) \ \text{into equation:}\)

\(-2\) \(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\) \(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\) \(=-2 \ \text{(correct)}\)

 

\(\therefore (7,-2) \text{ lies on line.}\)

Show Worked Solution

a.    \((2,1),(-3,4)\)

\(\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}\)

\(\text{Find equation with} \ \ m=-\dfrac{3}{5} \ \ \text{through}\ \ (2,1):\)

\(y-1\) \(=-\dfrac{3}{5}(x-2)\)
\(y\) \(=-\dfrac{3}{5} x+\dfrac{11}{5}\)

 

b.    \(\text {Substitute}\ (7,-2)\ \text{into equation:}\)

\(-2\) \(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\) \(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\) \(=-2 \ \text{(correct)}\)

 

\(\therefore (7,-2) \text{ lies on line.}\)

Functions, 2ADV EQ-Bank 06

Simplify  \(\dfrac{x^2}{x^2-2 x-15}-\dfrac{x}{x+3}\).   (2 marks)

Show Answers Only

\(\dfrac{5 x}{(x+3)(x-5)}\)

Show Worked Solution
\(\dfrac{x^2}{x^2-2 x-15}-\dfrac{x}{x+3}\) \(=\dfrac{x^2}{(x+3)(x-5)}-\dfrac{x}{x+3}\)
  \(=\dfrac{x^2-x(x-5)}{(x+3)(x-5)}\)
  \(=\dfrac{x^2-x^2+5 x}{(x+3)(x-5)}\)
  \(=\dfrac{5 x}{(x+3)(x-5)}\)

PHYSICS, M8 2025 HSC 21

A scientist has two unlabelled sources of radiation. Once source emits alpha particles and the other emits beta particles.

Outline TWO methods that could be used to determine which source is the alpha emitter, and which source is the beta emitter.   (3 marks)

Show Answers Only

Method 1:

  • Pass the radiation through a thin material to a detector.
  • Observe any radiation count changes.
  • The beta source count will be reduced by a lesser amount less than the alpha source count.

Method 2:

  • Pass the two radiation types through a magnetic field.
  • Alpha particles (positive charge) will be deflected slightly in one direction.
  • Beta particles (negative charge) will be deflected strongly in the opposite direction.
  • This difference allows you to identify which source is emitting alphas and which is emitting betas.

Answers could also include:

  • Passing radiation through an electric field.
  • Use of a cloud chamber.
Show Worked Solution

Method 1:

  • Pass the radiation through a thin material to a detector.
  • Observe any radiation count changes.
  • The beta source count will be reduced by a lesser amount less than the alpha source count.

Method 2:

  • Pass the two radiation types through a magnetic field.
  • Alpha particles (positive charge) will be deflected slightly in one direction.
  • Beta particles (negative charge) will be deflected strongly in the opposite direction.
  • This difference allows you to identify which source is emitting alphas and which is emitting betas.

Answers could also include:

  • Passing radiation through an electric field.
  • Use of a cloud chamber.

L&E, 2ADV EQ-Bank 5

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

  1. What is the value of `A`?   (1 mark)

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

Show Answers Only

i.    `1.5\ text(kg)`

ii.   `10text(%)`

Show Worked Solution

i.   `text(When)\ x = 0:`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
ii.   `text(Daily growth rate)\ = 0.1 = 10text(%)`

L&E, 2ADV EQ-Bank 4

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (2 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

Show Answers Only

i.    `text(14%)`

ii.   `714`

iii.  `text(Proof)\ \ text{(See Worked Solutions)}`

iv.   `text(8.4 hours)`

Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`
 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n= 100(1.14)^15= 713.793\ …\ = 714\ \ \ text{(nearest whole)}`
 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after ~ 8.4 hours.)`

Functions, 2ADV EQ-Bank 6

  1. Identify where the graph  \(f(x)=\dfrac{\abs{x}}{x}\)  is not continuous.   (1 mark)

  2. Sketch the graph of \(f(x)\).   (2 marks)

Show Answers Only

a.    \(\text {Denominator} \neq 0\)

\(f(x)\ \text{is not continuous when} \ \ x=0\)
 

b.    \(\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1\)

\(\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1\)
 

Show Worked Solution

a.    \(\text {Denominator} \neq 0\)

\(f(x)\ \text{is not continuous when} \ \ x=0\)
 

b.    \(\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1\)

\(\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1\)
 

Functions, 2ADV EQ-Bank 5

  1. Identify where the graph  \(f(x)=\dfrac{x^2-1}{x-1}\)  is not continuous.   (1 mark)

  2. Sketch the graph of \(f(x)\).   (2 marks)

Show Answers Only

a.    \(f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x+1)(x-1)}{(x-1)}=x+1\)

\(\text{Since denominator} \neq 0\)

\(f(x) \ \ \text{is not continuous when} \ \ x=1.\)
 

b.
       

Show Worked Solution

a.    \(f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x+1)(x-1)}{(x-1)}=x+1\)

\(\text{Since denominator} \neq 0\)

\(f(x) \ \ \text{is not continuous when} \ \ x=1.\)
 

b.
       

Functions, 2ADV EQ-Bank 4

Consider the function  \(y=f(x)\)  where

\(f(x)= \begin{cases}x^2+6, & \text { for } x \leqslant 0 \\ 6, & \text { for } 0<x \leqslant 3 \\ 2^x, & \text { for } x>3\end{cases}\)

  1. Sketch  \(y=f(x)\)   (3 marks)

  2. For what value of \(x\) is  \(y=f(x)\)  NOT continuous?   (1 mark)

Show Answers Only

a.
   

b.    \(f(x)\ \ \text {is NOT continuous at}\ \  x=3.\)

Show Worked Solution

a.
   

b.    \(f(x)\ \ \text {is NOT continuous at}\ \  x=3.\)

Statistics, STD2 EQ-Bank 01

A Physics class of  12 students is going on a 4 day excursion by bus.

The students are asked to each pack one bag for the trip. The bags are weighed, and the weights (in kg) are listed in order as follows:

\(8,\ \ 9, \ \ 10,\ \ 10, \ \ 15, \ \  18, \ \  22, \ \ 25, \ \ 29, \ \ 35, \ \ 38, \ \ 41 \)

  1. Use the above data to produce a five number summary for the weights of the bags.   (2 marks)
  3. Using your five number summary from part (a), calculate the interquartile range of the weights.   (2 marks)
Show Answers Only

a.    \(\text{Five number Summary}\)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} &  8\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} &  20\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 32 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} &  41\\
\hline
\end{array}

b.     \(22\)

Show Worked Solution

a.    \(\text{Five number Summary}\)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} &  8\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} &  20\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 32 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} &  41\\
\hline
\end{array}

b.     \(\text{IQR}\) \(=Q_3-Q_1\)
    \(=32-10=22\)

Measurement, STD2 EQ-Bank 02 MC

Arrange the numbers \(5.6\times 10^{-2}\), \(4.8\times 10^{-1}\), \(7.2\times 10^{-2}\) from smallest to largest.

  1. \(5.6\times 10^{-2}\),  \(7.2\times 10^{-2}\),  \(4.8\times 10^{-1}\)
  2. \(4.8\times 10^{-1}\),  \(5.6\times 10^{-2}\),  \(7.2\times 10^{-2}\)
  3. \(7.2\times 10^{-2}\),  \(5.6\times 10^{-2}\),  \(4.8\times 10^{-1}\)
  4. \(4.8\times 10^{-1}\),  \(7.2\times 10^{-2}\),  \(5.6\times 10^{-2}\)
Show Answers Only

\(A\)

Show Worked Solution

\(5.6\times 10^{-2}=0.052\)

\(4.8\times 10^{-1}=0.48\)

\(7.2\times 10^{-2}=0.072\)

\(\therefore\ \text{Correct order is: }\ 5.6\times 10^{-2}\),  \(7.2\times 10^{-2}\),  \(4.8\times 10^{-1}\)

\(\Rightarrow A\)

Algebra, STD2 EQ-Bank 01

Jerico is the manager of a weekend market in which there are 220 stalls for rent. From past experience, Jerico knows that if he charges \(d\) dollars to rent a stall. then the number of stalls, \(s\), that will be rented is given by:

\(s=220-4d\)

  1. How many stalls will be rented if Jerico charges $7.50 per stall .  (1 mark)

  2. Complete the following table for the function  \(s=220-4d\).   (1 mark)

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad d\quad \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 10\quad\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 30\quad & \rule{0pt}{2.5ex} \quad 50\quad \\
\hline
\rule{0pt}{2.5ex} \quad s\quad \rule[-1ex]{0pt}{0pt} & \ & \ & \\
\hline
\end{array}

  1. Using an appropriate vertical scale and labelled axes, graph the function  \(s=220-4d\) on the grid below.  (2 marks)
  1. Does it make sense to use the formula \(s=220-4d\)  to calculate the number of stalls rented if Jerico charges $60 per stall? Explain your answer.   (2 marks)

Show Answers Only
  

a.    \(190\ \text{stalls will be rented}\)

b.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad d\quad \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 10\quad\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 30\quad & \rule{0pt}{2.5ex} \quad 50\quad \\
\hline
\rule{0pt}{2.5ex} \quad s\quad \rule[-1ex]{0pt}{0pt} & 180 \ & 100 \ & 20  \\
\hline
\end{array}

c.

d.    \(\text{When}\ d=60, s=220-4\times 60=-20\)

\(\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}\)

\(\text{as you cannot have a negative number of stalls.}\)

 

Show Worked Solution
a.     \(s\) \(=220-4d\)
    \(=220-4\times 7.50\)
    \(=190\)

  
\(190\ \text{stalls will be rented}\)

b.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad d\quad \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 10\quad\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 30\quad & \rule{0pt}{2.5ex} \quad 50\quad \\
\hline
\rule{0pt}{2.5ex} \quad s\quad \rule[-1ex]{0pt}{0pt} & 180 \ & 100 \ & 20  \\
\hline
\end{array}

c.

d.    \(\text{When}\ d=60, s=220-4\times 60=-20\)

\(\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}\)

\(\text{as you cannot have a negative number of stalls.}\)

Functions, 2ADV EQ-Bank 02

Rationalise the denominator in \(\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\), and express in the simplest form.   (2 marks)

Show Answers Only

\(\dfrac{\sqrt{15}-\sqrt{6}-\sqrt{10}+2}{3} \)

Show Worked Solution

\(\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}\)

\(=\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{5}-\sqrt{2})}{5-2}\)

\(=\dfrac{\sqrt{15}-\sqrt{6}-\sqrt{10}+2}{3}\)

CHEMISTRY, M7 2025 HSC 24

65.0 g of ethyne gas reacts with an excess of gaseous hydrogen chloride to produce chloroethene.

  1. Draw the full structural formula of ethyne and identify the shape of the molecule.   (2 marks)

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \text{Structural formula } \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad\text{ Shape of molecule } \quad \quad\\
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

  1. The molar masses of the compounds in the reaction are provided.

\begin{array}{|l|c|}
\hline \rule{0pt}{2.5ex}\text{Compound} \rule[-1ex]{0pt}{0pt}& \text{ Molar mass } \\
\hline \rule{0pt}{2.5ex}\text{Ethyne} \rule[-1ex]{0pt}{0pt}& 26.04 \\
\hline \rule{0pt}{2.5ex}\text{Hydrogen chloride} \rule[-1ex]{0pt}{0pt}& 36.46 \\
\hline \rule{0pt}{2.5ex}\text{Chloroethene} \rule[-1ex]{0pt}{0pt}& 62.50 \\
\hline
\end{array}

  1. Calculate the mass of chloroethene produced, using the molar masses provided. Include a relevant chemical equation in your answer.   (3 marks)

Show Answers Only

a.    Structural formula:

   

The shape of ethyne is linear.

b.    156 grams

Show Worked Solution

a.    Structural formula:

   

The shape of ethyne is linear.
 

b.    The chemical equation for the reacton occuring is shown below:

\(\ce{C2H2(g) + HCl(g) -> C2H3Cl(g)}\)

  • \(n\ce{(C2H2)} = \dfrac{m}{MM} = \dfrac{65.0}{26.04} = 2.496\ \text{mol}\)
  • As the reactants and products are in a \(1:1:1\) ratio, \(n\ce{(C2H2)}_{\text{reacted}} = n\ce{(C2H3Cl)}_{\text{produced}}\)
  • \(m\ce{(C2H3Cl)} = n \times MM = 2.496 \times 62.50 = 156\ \text{grams (3 sig.fig.)}\)

CHEMISTRY, M8 2025 HSC 22

Outline why quantitative and qualitative analyses are BOTH important in determining water quality.   (2 marks)

Show Answers Only
  • Qualitative analysis identifies which substances or ions are present in the water, such as heavy metals, nitrates, or bacteria.
  • Quantitative analysis determines how much of each substance is present.
  • Both are important because knowing what contaminants are in the water and in what concentrations allows assessment of whether the water meets safety standards and is suitable for consumption or environmental use.
Show Worked Solution
  • Qualitative analysis identifies which substances or ions are present in the water, such as heavy metals, nitrates, or bacteria.
  • Quantitative analysis determines how much of each substance is present.
  • Both are important because knowing what contaminants are in the water and in what concentrations allows assessment of whether the water meets safety standards and is suitable for consumption or environmental use.

CHEMISTRY, M5 2025 HSC 5 MC

\(\ce{PCl3}\) and \(\ce{Cl2}\) were introduced to an empty sealed vessel.

\(\ce{PCl3}\) reacted with \(\ce{Cl2}\) to produce \(\ce{PCl5}\).

\(\ce{PCl3(g) + Cl2(g) \rightleftharpoons PCl5(g)}\)

Which graph best represents the changing concentration of \(\ce{Cl2}\) as the system approached the equilibrium point?
 

 

Show Answers Only

\(C\)

Show Worked Solution
  • The initial concentration of \(\ce{Cl2}\) is postive it is introduced into the system.
  • The concentration of \(\ce{Cl2}\) will decrease with the rate of decrease slowing down over time as the system reaches a dynamic equilibrium.
  • Hence the concentration of \(\ce{Cl2}\) will be non-zero when the system reaches dynamic equilibrium.

\(\Rightarrow C\)

CHEMISTRY, M8 2025 HSC 4 MC

A student is presented with two clear colourless solutions. One contains \(\ce{Pb^2+}\) and the other \(\ce{Na+}\) ions.

Which ion can be added to the solutions to identify the solutions?

  1. \(\ce{I-}\)
  2. \(\ce{NH4+}\)
  3. \(\ce{NO3-}\)
  4. \(\ce{CH3COO-}\)
Show Answers Only

\(A\)

Show Worked Solution
  • When \(\ce{I-}\) ions are added to a solution containing \(\ce{Pb^2+}\) a bright yellow precipiate is formed:
  • \(\ce{Pb^2+(aq) + I-(aq) -> PbI2(s)}\).
  • When \(\ce{I-}\) ions are added to a solution containing \(\ce{Na+}\) ions the solution is clear. Therefore \(\ce{I-}\) can be used to identify the solutions.
  • Nitrate \((\ce{NO3-})\) and ethanoate \((\ce{CH3COO-})\) ions are soluble with both \(\ce{Pb^2+}\) and \(\ce{Na+}\) ions so they could not be used to identify the solutions.
  • \(\ce{NH4+}\) is a postive cation and so will not form a precipate with other postive cations and so cannot be used to distinguish between the two solutions.

\(\Rightarrow A\)

CHEMISTRY, M6 2025 HSC 1 MC

An aqueous solution of an unknown acid \(\ce{(HA)}\) is represented below.
 

Which row of the table best describes this solution?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Strong}\quad \quad \rule[-1ex]{0pt}{0pt}& \ \ \textit{Concentrated} \ \  \\
\hline
\rule{0pt}{2.5ex}\checkmark\rule[-1ex]{0pt}{0pt}&\checkmark\\
\hline
\rule{0pt}{2.5ex}\checkmark\rule[-1ex]{0pt}{0pt}& \large{\times}\\
\hline
\rule{0pt}{2.5ex}\large{\times}\rule[-1ex]{0pt}{0pt}& \checkmark \\
\hline
\rule{0pt}{2.5ex}\large{\times}\rule[-1ex]{0pt}{0pt}& \large{\times} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • A strong acid is one that fully ionises when placed in solution:
  • \(\ce{HA(aq) -> H^+(aq) + A^-(aq)}\)
  • As there are only a small number of molecules in the solution the acid concentration is dilute.

\(\Rightarrow B\)

Proof, EXT2 P1 2025 HSC 5 MC

Consider the statement:

If  \(x^2-2 x \geq 0\), then  \(x \leq 0\).

Which of the following is the contrapositive of the statement?

  1. If  \(x>0\), then  \(x^2-2 x<0\).
  2. If  \(x \leq 0\), then  \(x^2-2 x \geq 0\).
  3. If  \(x^2-2 x<0\), then  \(x<0\).
  4. If  \(x^2-2 x \leq 0\), then  \(x>0\).
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Statement: If}\ \ p\ \Rightarrow \ q\)

\(\text{Contrapositive statement: If}\ \ \neg\ q\ \Rightarrow \ \neg\ p\)

\(\therefore\ \text{If}\ \ x>0\ \ \Rightarrow\ \ x^2-2 x<0\).

\(\Rightarrow A\)

BIOLOGY, M5 2025 HSC 29

The Varroa mite is an external parasite of European honey bees and considered to be the most serious pest of honey bees worldwide. 

  1. Why is the Varroa mite infection considered to be an infectious disease.   (2 mark2)

  2. In June 2022, the Varroa mite was detected for the first time in Australia at the Port of Newcastle. It then spread to surrounding areas.
  3. Explain TWO procedures that could have been employed to prevent the spread of the Varroa mite in honey bees.   (4 marks)

Show Answers Only

a.    Varroa mite infection classification

  • Varroa mite infection is classified as an infectious disease because the mite acts as a pathogen.
  • The mite transmits from one honey bee host to another through direct contact.
  • This enables the infection to spread between individual bees and across different hive populations.

b.    Procedure 1: Quarantine

  • Quarantine works by isolating infected hives at the Port of Newcastle to prevent mite movement.
  • This stops infected bees from contacting healthy colonies in surrounding areas.
  • Movement restrictions on beekeeping equipment reduce the risk of accidentally transporting mites to new locations

Procedure 2: Surveillance and Destruction

  • Regular inspection of hives allows early detection of Varroa mite presence before widespread establishment.
  • Destruction of heavily infested hives eliminates the source of infection and prevents further transmission.
  • This creates a containment zone that limits geographic spread of the parasite.
Show Worked Solution

a.    Varroa mite infection classification

  • Varroa mite infection is classified as an infectious disease because the mite acts as a pathogen.
  • The mite transmits from one honey bee host to another through direct contact.
  • This enables the infection to spread between individual bees and across different hive populations.

b.    Procedure 1: Quarantine

  • Quarantine works by isolating infected hives at the Port of Newcastle to prevent mite movement.
  • This stops infected bees from contacting healthy colonies in surrounding areas.
  • Movement restrictions on beekeeping equipment reduce the risk of accidentally transporting mites to new locations

Procedure 2: Surveillance and Destruction

  • Regular inspection of hives allows early detection of Varroa mite presence before widespread establishment.
  • Destruction of heavily infested hives eliminates the source of infection and prevents further transmission.
  • This creates a containment zone that limits geographic spread of the parasite.

Proof, EXT2 P1 2025 HSC 13c

  1. For positive real numbers \(a\) and \(b\), prove that  \(\dfrac{a+b}{2} \geq \sqrt{a b}\).   (2 marks)

  2. Hence, or otherwise, show that  \(\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)  for any integer \(n \geq 0\).   (2 marks)

Show Answers Only

 

Show Worked Solution

i.    \(\text {Using}\ \ (\sqrt{a}-\sqrt{b})^2 \geqslant 0:\)

\(a-2 \sqrt{a b}+b\) \(\geqslant 0\)
\(a+b\) \(\geqslant 2 \sqrt{a b}\)
\(\dfrac{a+b}{2}\) \(\geqslant \sqrt{a b}\)

 
ii.    \(\text{Show}\ \ \dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)

\(\text{Using part (i):}\)

\(\text{Let} \ \ a=2 n+1, b=2 n+3\)

\(\dfrac{2 n+1+2 n+3}{2}\) \(\geqslant \sqrt{2 n+1} \sqrt{2 n+3}\)
\(2 n+2\) \(\geqslant \sqrt{2 n+1} \sqrt{2 n+3}\)
\(\dfrac{1}{2 n+2}\) \(\leqslant \dfrac{1}{\sqrt{2 n+1} \sqrt{2 n+3}}\)
\(\dfrac{2 n+1}{2 n+2}\) \(\leqslant \dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)

 

\(\text{Consider part (i):}\ (\sqrt{a}-\sqrt{b})^2 \geqslant 0\)

\(\Rightarrow \ \text{Equality only holds when} \ \ a=b\)

\(\text{Since}\ \ a=2 n+1 \neq b=2 n+3\)

\(\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)

BIOLOGY, M5 2025 HSC 22a

Outline a process used by fungi for reproduction.  (2 marks)

Show Answers Only

Spore production (asexual reproduction):

  • Fungi produce spores within specialised structures called sporangia.
  • Spores are released into the environment and dispersed by wind or water.
  • When conditions are favourable, spores germinate to form new fungal hyphae.
  • This produces genetically identical offspring.

Alternative acceptable response – Budding:

  • A small outgrowth (bud) forms on the parent fungal cell.
  • The bud grows and receives a copy of the nucleus.
  • The bud detaches to become an independent organism.
  • This is common in yeasts.

Show Worked Solution

Spore production (asexual reproduction):

  • Fungi produce spores within specialised structures called sporangia.
  • Spores are released into the environment and dispersed by wind or water.
  • When conditions are favourable, spores germinate to form new fungal hyphae.
  • This produces genetically identical offspring.

Alternative acceptable response – Budding:

  • A small outgrowth (bud) forms on the parent fungal cell.
  • The bud grows and receives a copy of the nucleus.
  • The bud detaches to become an independent organism.
  • This is common in yeasts.

BIOLOGY, M7 2025 HSC 22b

Outline an adaptation in a pathogen that facilitates transmission between hosts.   (2 marks)

Show Answers Only

Answers could include ONE of the following:

Influenza virus:

  • Influenza produces viral particles that are expelled in respiratory droplets when infected individuals cough or sneeze.
  • These droplets can be inhaled by nearby individuals, facilitating airborne transmission.

Plasmodium (malaria parasite):

  • Plasmodium has a complex life cycle adapted to mosquito vectors.
  • The parasite develops specific stages (sporozoites) in mosquito salivary glands.
  • This enables injection into new human hosts during blood feeding.

Bacterial spores:

  • Some bacteria form resistant endospores that survive harsh environmental conditions.
  • Spores can persist on surfaces or in soil for extended periods.
  • This increases opportunity for transmission to new hosts.

Show Worked Solution

Answers could include ONE of the following:

Influenza virus

  • Influenza produces viral particles that are expelled in respiratory droplets when infected individuals cough or sneeze.
  • These droplets can be inhaled by nearby individuals, facilitating airborne transmission.

Plasmodium (malaria parasite):

  • Plasmodium has a complex life cycle adapted to mosquito vectors.
  • The parasite develops specific stages (sporozoites) in mosquito salivary glands.
  • This enables injection into new human hosts during blood feeding.

Bacterial spores:

  • Some bacteria form resistant endospores that survive harsh environmental conditions.
  • Spores can persist on surfaces or in soil for extended periods.
  • This increases opportunity for transmission to new hosts.

Vectors, EXT2 V1 2025 HSC 11d

  1. Force \({\underset{\sim}{F}}_1\) has magnitude 12 newtons in the direction of vector  \(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}\).   
  2. Show that  \({\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\).   (1 mark)

  3. Force \({\underset{\sim}{F}}_1\) from part (i) and a second force,  \({\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}\), both act upon a particle.
  4. Show that the resultant force acting on the particle is given by:
  5.      \({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.\)   (1 mark)

  6. Calculate  \({\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}\), where \({\underset{\sim}{F}}_3\) is the resultant force from part (ii) and  \(\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}\).   (1 mark)

Show Answers Only

i.    \(\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}=\sqrt{2^2+(-2)^2+1^2}=3\)

\({\underset{\sim}{F}}_1=\dfrac{12}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
 

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Show Worked Solution

i.    \(\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}=\sqrt{2^2+(-2)^2+1^2}=3\)

\({\underset{\sim}{F}}_1=\dfrac{12}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
 

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Mechanics, EXT2 M1 2025 HSC 4 MC

A particle in simple harmonic motion has speed \(v \ \text{ms}^{-1}\), given by  \(v^2=-x^2+2 x+8\)  where \(x\) is the displacement from the origin in metres.

What is the amplitude of the motion?

  1. 1 m
  2. 3 m
  3. 6 m
  4. 9 m
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Using} \ \ v^2=n^2\left(a^2-(x-c)^2\right):\)

\(v^2\) \(=-x^2+2 x+8\)
  \(=9-\left(x^2-2 x+1\right)\)
  \(=9-(x-1)^2\)

 
\(\therefore a^2 = 9\ \ \Rightarrow\ \ a=3\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N1 2025 HSC 3 MC

What are the square roots of  \(3-4 i\) ?

  1. \(1-2 i\)  and  \(-1+2 i\)
  2. \(1+2 i\)  and  \(-1-2 i\)
  3. \(2-i\)  and  \(-2+i\)
  4. \(-2-i\)  and  \(2+i\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Let} \ \ z=\sqrt{3-4 i}\) :

\(z^2=3-4 i\)

\(z^2=a^2-b^2+2 a b\,i\)

\(\text{Equate real/imaginary parts:}\)

\(a^2-b^2=3\ \ldots\ (1)\)

\(2 a b=-4 \ \ \Rightarrow \ \ a b=-2\ \ldots\ (2)\)

\(\text{By inspection:}\)

\(a=2, b=-1 \ \Rightarrow \ z_1=2-i\)

\(a=-2, b=1 \ \Rightarrow \ z_2=-2+i\)

\(\Rightarrow C\)

Calculus, EXT2 C1 2025 HSC 12a

Using integration by parts, evaluate \(\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx\).   (3 marks)

Show Answers Only

\(\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx = 1\)

Show Worked Solution
\(u\) \(=x\) \(u^{\prime}\) \(=1\)
\(v^{\prime}\) \(=\sin\,x\) \(v\) \(=-\cos\,x\)
\(\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx\) \(=\Big[-x\,\cos\,x\Big]_0^{\small{\dfrac{\pi}{2}}} + \displaystyle \int_0^{\small{\dfrac{\pi}{2}}} \cos x \, dx\)  
  \(=(0-0)+\Big[\sin\,x\Big]_0^{\small{\dfrac{\pi}{2}}} \)  
  \(=\sin\,\dfrac{\pi}{2}-\sin\,0\)  
  \(=1\)  

Calculus, EXT2 C1 2025 HSC 11f

Find \(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\).   (2 marks)

Show Answers Only

\(5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c\)

Show Worked Solution
\(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\) \(=\displaystyle \int \dfrac{5}{\sqrt{16-(x^2+6x+9)}} \, dx\)  
  \(=\displaystyle \int \dfrac{5}{\sqrt{16-(x+3)^2}} \, dx\)  
  \(=5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c\)  

BIOLOGY, M6 2025 HSC 2 MC

What type of mutagen is UV light?

  1. Biochemical
  2. Biological
  3. Chemical
  4. Electromagnetic
Show Answers Only

\(D\)

Show Worked Solution
  • D is correct: UV light is electromagnetic radiation that causes DNA mutations.

Other Options:

  • A is incorrect: Biochemical mutagens are biological molecules like enzymes or toxins.
  • B is incorrect: Biological mutagens include viruses or transposons that alter DNA.
  • C is incorrect: Chemical mutagens are substances like alkylating agents or base analogs.

BIOLOGY, M5 2025 HSC 1 MC

Amoeba reproduce by the process shown.
  

     

Which of the following is a characteristic of the daughter cells?

  1. Genetically different to parent amoeba
  2. Genetically identical to parent amoeba
  3. More organelles due to the cell division
  4. Fewer chromosomes than the parent amoeba
Show Answers Only

\(B\)

Show Worked Solution
  • B is correct: Asexual reproduction produces genetically identical offspring.

Other Options:

  • A is incorrect: Sexual reproduction produces genetic variation, not asexual.
  • C is incorrect: Cell division distributes existing organelles between daughter cells.
  • D is incorrect: Mitosis maintains the same chromosome number in offspring.

Complex Numbers, EXT2 N1 2025 HSC 11c

The complex number \(z\) is given by  \(x+i y\).

Find, in Cartesian form:

  1. \(z^2\)   (1 mark)

  2. \(\dfrac{1}{z}\).   (2 marks)

Show Answers Only

i.    \(z^2=x^2-y^2+2xy\,i\)

ii.  \(\dfrac{1}{z}=\dfrac{x}{x^2-y^2}-\dfrac{y}{x^2-y^2}\,i\)

Show Worked Solution
i.    \(z^{2}\) \(=(x+iy)^2\)
    \(=x^2-y^2+2xy\,i\)

 

ii.    \(\dfrac{1}{z}\) \(=\dfrac{1}{x+iy}\)
    \(=\dfrac{x-iy}{(x+iy)(x-iy)}\)
    \(=\dfrac{x-iy}{x^2+y^2}\)
    \(=\dfrac{x}{x^2+y^2}-\dfrac{y}{x^2+y^2}\,i\)

Complex Numbers, EXT2 N1 2025 HSC 11b

The complex numbers \(w\) and \(z\) are given by  \(w=2 e^{\small{\dfrac{i \pi}{6}}}\)  and  \(z=3 e^{\small{\dfrac{i \pi}{6}}}\).  Find the modulus and argument of \(w z\).   (2 marks)

Show Answers Only

\(\text{Modulus = 6,  Argument}\ = \dfrac{\pi}{3} \)

Show Worked Solution

\(w=2 e^{\small{\dfrac{i \pi}{6}}}, \ \ z=3 e^{\small{\dfrac{i \pi}{6}}}\)

\(wz=2 e^{\small{\dfrac{i \pi}{6}}} \times 3 e^{\small{\dfrac{i \pi}{6}}} = 6 e^{\small{\dfrac{i \pi}{3}}}\)

\(\text{Modulus = 6,  Argument}\ = \dfrac{\pi}{3} \)

Proof, EXT2 P1 2025 HSC 2 MC

Consider the statement:

\(\exists\, x \in Z\), such that \(x^2\) is odd.

Which of the following is the negation of the statement?

  1. \(\forall\, x \in Z , x^2\)  is odd
  2. \(\forall\, x \in Z , x^2\)  is even
  3. \(x^2\) is even \(\Rightarrow x \in Z\)
  4. \(\exists\, x \in Z\),  such that  \(x^2\) is even
Show Answers Only

\(B\)

Show Worked Solution

\(x^2\ \text{is odd is negated by}\ x^2\ \text{is even.}\)

\(\therefore\ \text{Negation of statement:}\ \forall\, x \in Z , x^2\ \text{is even}\)

\(\Rightarrow B\)

Statistics, STD1 S3 2025 HSC 15

A researcher is using the statistical investigation process to investigate a possible relationship between average number of minutes per day a person spends watching television, and the average number of minutes per day the person spends exercising.

  1. State the statistical question being posed.   (1 mark)

Participants were asked to record the number of minutes they spent watching television each day and the number of minutes they spent exercising each day. The averages for each participant were recorded and graphed, and a line of best fit was included.
 

  1. From the graph, identify the dependent variable.   (1 mark)

  2. Describe the bivariate dataset in terms of its form and direction.   (2 marks)

  3. The points \((0, 70)\) and \((60, 10)\) lie on the line of best fit. By first plotting these points on the graph, find the gradient and the \(y\)-intercept of the line of best fit.   (3 marks)
  4. Explain why it is NOT appropriate to extrapolate the line of best fit to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)
Show Answers Only

a.    \(\text{How does the average daily time spent watching television}\)

\(\text{relate to the average daily time spent exercising?}\)
 

b.    \(\text{Dependent variable: Average minutes per day exercising, or }y.\)
 

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d.   \(\text{Gradient}=-1\)
 

e.    \(\text{The extrapolation of the graph past 70 minutes produces}\)

\(\text{negative average minutes per day exercising (impossible).}\)

Show Worked Solution

a.    \(\text{How does the average daily time spent watching television}\)

\(\text{relate to the average daily time spent exercising?}\)
 

b.    \(\text{Dependent variable:}\)

\(\text{Average minutes per day exercising, or }y.\)
 

c.    \(\text{Form:  Linear}\)

\(\text{Direction:  Negative}\)
 

d. 


 

\(y-\text{intercept = 70}\)

\(\text{Gradient}=\dfrac{\text{rise}}{\text{run}}=\dfrac{-60}{60}=-1\)
 

e.    \(\text{The extrapolation of the graph past 70 minutes produces}\)

\(\text{negative average minutes per day exercising (impossible).}\)

Networks, STD1 N1 2025 HSC 14

The time, in minutes, it takes to travel by road between six towns is recorded and shown in the network diagram below.
 

  1. In this network the shortest path corresponds to the minimum travel time.
  2. What is the minimum travel time between towns \(A\) and \(F\), and what is the corresponding path?   (2 marks)

New roads are built to connect a town \(G\) to towns \(A\) and \(D\). The table gives the times it takes to travel by the new roads.

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Town} \rule[-1ex]{0pt}{0pt} & \textit{Time} \text{(minutes)}  \rule[-1ex]{0pt}{0pt} & \textit{Town} \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 8 \rule[-1ex]{0pt}{0pt} & G \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 22 \rule[-1ex]{0pt}{0pt} & D \\
\hline
\end{array}

  1. Add the new roads and times to the network diagram below.   (2 marks)
     
      

     

  2. Explain whether the path in part (a) is still the shortest path from \(A\) to \(F\) after the new roads are added.   (1 mark)

Show Answers Only

a.    \(\text{Minimum travel time}\ = 15+20+10+5+8=58\ \text{minutes}\)

\(\text{Path: }A → B → C → D → E → F\)
 

b.    \(\text{New roads}\ A  → G\ \text{and }G → D \)
 

c.    \(\text{Using the new roads }A → G\ \text{and }G → D:\)

\(\text{Minimum travel time}\ =8+22+5+8=43\ \text{minutes.}\)

\(\text{Therefore the original path is no longer the shortest path.}\)

Show Worked Solution

a.    \(\text{Minimum travel time}\ = 15+20+10+5+8=58\ \text{minutes}\)

\(\text{Path: }A → B → C → D → E → F\)
 

b.    \(\text{New roads}\ A  → G\ \text{and }G → D \)
 

c.    \(\text{Using the new roads }A → G\ \text{and }G → D:\)

\(\text{Minimum travel time}\ =8+22+5+8=43\ \text{minutes.}\)

\(\text{Therefore the original path is no longer the shortest path.}\)

Measurement, STD1 M4 2025 HSC 2 MC

Mario drives from his home to his friend’s house. He watches a movie at his friend’s house and then drives home.

Which distance−time graph best represents Mario’s complete journey?
 

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Mario’s trip starts at home (zero) and ends at home (zero).}\)

\(\Rightarrow B\)

Measurement, STD2 M1 2025 HSC 26

A toy has a curved surface on the top which has been shaded as shown. The toy has a uniform cross-section and a rectangular base.
 

  1. Use two applications of the trapezoidal rule to find an approximate area of the cross-section of the toy.   (2 marks)

  2. The total surface area of the plastic toy is 1300 cm².
  3. What is the approximate area of the curved surface?   (2 marks)

  4. The measurements shown on the diagram are given to the nearest millimetre.
  5. What is the percentage error of the measurement of 10.2 cm? Give your answer correct to 3 significant figures.   (2 marks)

Show Answers Only

a.   \(34.68 \ \text{cm}^2\)

b.   \(582.64 \ \text{cm}^2\)

c.   \(0.490 \%\)

Show Worked Solution

a.    \(\begin{array}{|c|c|c|c|}
\hline\rule{0pt}{2.5ex} \quad x \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 5.1 \quad & \quad 10.2 \quad\\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 6 & 3.8 & 0 \\
\hline
\end{array}\)

\(A\) \(\approx \dfrac{h}{2}\left(y_0+2y_1+y_2\right)\)
  \(\approx \dfrac{5.1}{2}\left(6+2 \times 3.8+0\right)\)
  \(\approx 34.68 \ \text{cm}^2\)

 

b.    \(\text{Toy has 5 sides.}\)

\(\text{Area of base}=10.2 \times 40=408 \ \text{cm}^2\)

\(\text{Area of rectangle}=6.0 \times 40=240 \ \text{cm}^2\)

\(\text{Approximated areas}=2 \times 34.68=69.36 \ \text{cm}^2\)

\(\therefore \ \text{Area of curved surface}\) \(=1300-(408+240+69.36)\)
  \(=582.64 \ \text{cm}^2\)

 

c.   \(\text{Convert cm to mm:}\)

\(10.2 \ \text{cm}\times 10=102 \ \text{mm}\)

\(\text{Absolute error}=\dfrac{1}{2} \times \text{precision}=0.5 \ \text{mm}\)

\(\% \ \text{error}=\dfrac{0.5}{102} \times 100=0.490 \%\)

Combinatorics, EXT1 A1 2025 HSC 13e

  1. The Pascal's triangle relation can be expressed as
    1. \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.\) (Do NOT prove this.)
  2. Show that \(\displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\).   (1 mark)

  3. Hence, or otherwise, prove that
  4. \(\displaystyle\binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\).   (2 marks)
Show Answers Only

i.    \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)\)

\(\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

\(\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}\)

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)
 

ii.    \(\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\)

\(\text{Using part (i)}:\)

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)
Show Worked Solution

i.    \(\displaystyle \binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\ \ldots\ (1)\)

\(\text{Show:}\ \displaystyle \binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)

\(\text{Substitute} \ \ n=m+1 \ \ \text{and} \ \ r=R+1 \text{ into (1):}\)

\(\displaystyle\binom{m+1}{R+1}=\binom{m}{R}+\binom{m}{R+1}\)

\(\displaystyle\binom{m}{R}=\binom{m+1}{R+1}-\binom{m}{R+1}\)
 

ii.    \(\text{Prove} \ \ \displaystyle \binom{2000}{2000}+\binom{2001}{2000}+\binom{2002}{2000}+\cdots+\binom{2050}{2000}=\binom{2051}{2001}\)

\(\text{Using part (i)}:\)

\(\operatorname{LHS}\) \(=1+\displaystyle \left[\binom{2002}{2001}-\binom{2001}{2001}\right]+\left[\binom{2003}{2001}-\binom{2002}{2001}\right]+\ldots\)
  \(\quad \quad \quad +\displaystyle \left[\binom{2050}{2001}-\binom{2049}{2001}\right]+\left[\binom{2051}{2001}-\binom{2050}{2001}\right]\)
  \(=1-1+\displaystyle \binom{2051}{2001}\)
  \(=\displaystyle \binom{2051}{2001}\)

Functions, EXT1 F2 2025 HSC 11f

The roots of  \(2 x^3+6 x^2+x-1=0\)  are  \(\alpha, \beta\)  and  \(\gamma\).

What is the value of  \(\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}\) ?   (2 marks)

Show Answers Only

\(\dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=-6\)

Show Worked Solution

\(2 x^3+6 x^2+x-1=0\)

\(\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{\alpha+\beta+\gamma}{\alpha \beta \gamma}\)

\(\alpha+\beta+\gamma=-\dfrac{b}{a}=-3\)

\(\alpha \beta \gamma=-\dfrac{d}{a}=\dfrac{1}{2}\)

\(\therefore \dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{-3}{\frac{1}{2}}=-6\)

Trigonometry, EXT1 T2 2025 HSC 11b

Solve  \(\sin 2 \theta-\sin \theta=0\)  for  \(0 \leq \theta \leq \pi\).   (3 marks)

Show Answers Only

\(\theta=0, \dfrac{\pi}{3}\ \text{or}\ \pi\)

Show Worked Solution
\(\sin\,2\theta-\sin\,\theta\) \(=0\)  
\(2\,\sin\,\theta\,\cos\,\theta-\sin\,\theta\) \(=0\)  
\(\sin\,\theta(2\,\cos\,\theta-1)\) \(=0\)  

 
\(\sin\,\theta=0\ \ \Rightarrow\ \ \theta=0, \pi\)

\(2\,\cos\,\theta-1=0\ \ \Rightarrow\ \ \cos\,\theta=\dfrac{1}{2}\ \ \Rightarrow\ \ \theta=\dfrac{\pi}{3}\)

\(\therefore \theta=0, \dfrac{\pi}{3}\ \text{or}\ \pi.\)

Functions, EXT1 F1 2025 HSC 11a

Find the inverse function, \(f^{-1}(x)\), of the function  \(f(x)=1-\dfrac{1}{x-2}\).   (2 marks)

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

Show Worked Solution

\(y=1-\dfrac{1}{x-2}\)

\(\text{Inverse: swap}\ x ↔ y\)

\(x\) \(=1-\dfrac{1}{y-2}\)  
\(\dfrac{1}{y-2}\) \(=1-x\)  
\(y-2\) \(=\dfrac{1}{1-x}\)  
\(y\) \(=2+\dfrac{1}{1-x}\)  

Vectors, EXT1 V1 2025 HSC 2 MC

The projection of \(\underset{\sim}{u}\) onto \(\underset{\sim}{v}\) is given by  \(\left(\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{v}|^2}\right) \underset{\sim}{v}\).

What is the projection of  \(\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}\)  onto  \(\underset{\sim}{v}=2 \underset{\sim}{i}-3 \underset{\sim}{j}\) ?

  1. \(-\dfrac{4}{5}(\underset{\sim}{i}+2 \underset{\sim}{j})\)
  2. \(-\dfrac{4}{13}(2 \underset{\sim}{i}-3 \underset{\sim}{j})\)
  3. \(-\dfrac{4}{\sqrt{5}}(\underset{\sim}{i}+2 \underset{\sim}{j})\)
  4. \(-\dfrac{4}{\sqrt{13}}(2 \underset{\sim}{i}-3 \underset{\sim}{j})\)
Show Answers Only

\(B\)

Show Worked Solution

\(\underset{\sim}{u}=\displaystyle\binom{1}{2},|\underset{\sim}{u}|=\sqrt{1^2+2^2}=\sqrt{5}\)

\(\underset{\sim}{v}=\displaystyle \binom{2}{-3},|\underset{\sim}{v}|=\sqrt{2^2+(-3)^2}=\sqrt{13}\)

\(\operatorname{proj}_{\underset{\sim}{v}}{\underset{\sim}{u}}\) \(=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{v}|^2} \times \underset{\sim}{v}\)
  \(=\dfrac{2-6}{13}(\underset{\sim}{2i}-3\underset{\sim}{j})\)
  \(=-\dfrac{4}{13}(\underset{\sim}{2i}-3\underset{\sim}{j})\)

 
\(\Rightarrow B\)