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PHYSICS, M5 2023 HSC 23a

The James Webb Space Telescope (JWST) has a mass of 6.1 × 10³ kg and orbits the Sun at a distance of approximately 1.52 × 10\(^{11}\) m.

The Sun has a mass of 1.99 × 10\(^{30}\) kg.

Calculate the magnitude of gravitational force the Sun exerts on the JWST.  (2 marks)

Show Answers Only

\(F= 35\) \( \text{N}\)

Show Worked Solution

\(F\) \(=\dfrac{GMm}{r^2}\)  
  \(=\dfrac{ 6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times 6.1 \times 10^3}{(1.52 \times 10^{11} )^2}\)  
  \(=35\) \(\text{N}\)  

PHYSICS, M7 2023 HSC 22

A spacecraft passes Earth at a speed of 0.9\(c\). The spacecraft emits a light pulse every 3.1 \(\times\) 10\(^{-9}\) s, as measured by the crew on the spacecraft.

What is the time between the pulses, as measured by an observer on Earth?  (3 marks)

Show Answers Only

\(7.1 \times 10^{-9}\ \text{s} \)

Show Worked Solution

\(t\) \(=\dfrac{t_o}{\sqrt{(1- \frac{v^2}{c^2})}} \)  
  \(=\dfrac{3.1 \times 10^{-9}}{\sqrt{(1-\frac{(0.9c)^2}{c^2})}}\)  
  \( =\dfrac{3.1 \times 10{-9}}{\sqrt{(1-{0.9}^2)}} \)  
  \(=7.1 \times 10^{-9}\ \text{s} \)  

PHYSICS, M8 2023 HSC 21

A Hertzsprung–Russell diagram is shown.
 

  1. Identify TWO variables that determine the luminosity of a star  (2 marks)
  1. Describe differences between stars \(A\) and \(B\).  (3 marks)
Show Answers Only

a.    The luminosity of a star can be determined by its size and temperature.

Other variables could include:

→ Mass, colour and the power output of a star.
 

b.   Differences:

→ Star A is a main sequence star and is therefore fusing hydrogen to helium in its core via both the proton-proton chain and CNO cycle whereas Star B is a white dwarf star and therefore has no fusion taking place in its core.

→ Star A has a greater luminosity compared to Star B.

→ Star A is a younger star then Star B which is at the end of its lifecycle. 

Other differences could include:

→ Mass

→ Radius or size

Show Worked Solution

a.    The luminosity of a star can be determined by its size and temperature.

Other variables could include:

→ Mass, colour and the power output of a star.
 

b.   Differences between stars:

→ Star A is a main sequence star and is therefore fusing hydrogen to helium in its core via both the proton-proton chain and CNO cycle whereas Star B is a white dwarf star and therefore has no fusion taking place in its core.

→ Star A has a greater luminosity compared to Star B.

→ Star A is a younger star then Star B which is at the end of its lifecycle.
 

Other differences could include:

→ Mass

→ Radius or size

PHYSICS, M8 2023 HSC 11 MC

The chart shows part of a nuclear decay series beginning with uranium.
 


 

Which option correctly identifies \(X\) and \(Y\) and the process by which each was produced?
 

  \(X\) \(Y\)
A.  

\({ }_{\ \ 90}^{234}\text{Th}\)

alpha decay

\({ }_{\ \ 91}^{234}\text{Pa}\)

beta decay
B.

\({ }_{\ \ 90}^{234}\text{Th}\)

alpha decay

\({ }_{\ \ 91}^{234}\text{Pa}\)

alpha decay
C.

\({ }_{\ \ 91}^{234}\text{Pa}\)

beta decay

\({ }_{\ \ 90}^{234}\text{Th}\)

beta decay
D.

\({ }_{\ \ 91}^{234}\text{Pa}\)

beta decay

\({ }_{\ \ 90}^{234}\text{Th}\)

alpha decay
Show Answers Only

\(A\)

Show Worked Solution

→ During the decay process producing \(X\), the Atomic number decreases by 2 to 90 and the Mass number decreases by 4 to 234

→ Alpha decay producing  \({ }_{\ \ 90}^{234}\text{Th}\)

→ During the decay process producing \(Y\), the Atomic number increases by 1 to 91 and the Mass number does not change. 

→ Beta decay producing \({ }_{\ \ 91}^{234}\text{Pa}\)

\(\Rightarrow A\)

BIOLOGY, M8 2023 HSC 14 MC

Minamata Disease is caused by regular consumption of contaminated fish and shellfish. The symptoms include numbness in the hands and feet, muscle weakness, and damage to vision, hearing and speech.

Pellagra is a disease which causes delusions or mental confusion, diarrhoea, weakness and loss of appetite caused by insufficient levels of iron and niacin.

Wildervanck Syndrome is a condition that affects the bones in the neck, the eyes and the ears, and occurs primarily in females.

Given the information above, which row in the table correctly identifies the classification of these diseases?
 

  Minamata Disease Pellagra Wildervanck Syndrome
A.   Genetic Nutritional Environmental
B.   Genetic Environmental Nutritional
C.   Environmental Genetic Nutritional
D.   Environmental Nutritional Genetic
Show Answers Only

\(D\)

Show Worked Solution

By Elimination:

→ Minamata is caused by consumption of contaminated seafood and is therefore environmental, not genetic (Eliminate A and B).

→ Pellagra is caused by insufficient iron and niacin levels, which are a mineral and a vitamin respectively. It is therefore a nutritional disease and not genetic (Eliminate C).

→ As Wildervanck Syndrome occurs primarily in females, there cannot be 100% certainty it is genetic. However, with all other options eliminated, it can be assumed to be the case here.

\(\Rightarrow D\)

BIOLOGY, M7 2023 HSC 5 MC

An experiment was conducted to investigate the rate of binary fission in E.coli. The results of the experiment are shown.
 

Time (minutes) Number of E.coli
0 20
20 40
40 80
60 160
80 320
100 640

 
Which graph represents the data in the table?
 


  

Show Answers Only

\(B\)

Show Worked Solution

By Elimination:

→ The independent variable (time) must be on the \(x\)-axis (Eliminate C and D).

→ The number of E. Coli starts at 20, not 0 (Eliminate A).

\(\Rightarrow B\)

BIOLOGY, M5 2023 HSC 3 MC

A Punnett square is shown.

  \(\text{B}\) \(\text{b}\)
\(\text{B}\) \(1\) \(2\)
\(\text{B}\) \(3\) \(4\)

 
 Which of the following options represents heterozygous offspring?

  1. 1, 2
  2. 1, 3
  3. 2, 4
  4. 3, 4
Show Answers Only

\(C\)

Show Worked Solution

→ A heterozygous genotype is given by Bb, where-as a homozygous genotype is given by BB or bb.

→ Only options 2 and 4 show this.

\(\Rightarrow C\)

BIOLOGY, M5 2023 HSC 2 MC

Which of the following is an advantage of internal fertilisation?

  1. Decreases the risk of gamete dehydration
  2. Increases the number of gametes released
  3. Increases the number of zygotes at one time
  4. Decreases the care provided to gamete and offspring
Show Answers Only

\(A\)

Show Worked Solution

→ Internal fertilisation occurs in a moist environment, decreasing the risk of gamete dehydration.

\(\Rightarrow A\)

Mechanics, EXT2 M1 2023 HSC 14c

A projectile of mass \(M\) kg is launched vertically upwards from the origin with an initial speed \(v_0\) m s\(^{-1}\). The acceleration due to gravity is \( {g}\) ms\(^{-2}\).

The projectile experiences a resistive force of magnitude \(kMv^2\) newtons, where \(k\) is a positive constant and \(v\) is the speed of the projectile at time \(t\) seconds.

  1. The maximum height reached by the particle is \(H\) metres.
  2. Show that  \(H=\dfrac{1}{2 k} \ln \left(\dfrac{k v_0{ }^2+g}{g}\right)\).  (3 marks)

  3. When the projectile lands on the ground, its speed is \(v_1 \text{m} \ \text{s}^{-1}\), where \(v_1\) is less than the magnitude of the terminal velocity.
  4. Show that  \(g\left(v_0^2-v_1^2\right)=k v_0^2 v_1^2\).  (3 marks)

Show Answers Only
  1. \(\text{See Worked Solution}\)
  2. \(\text{See Worked Solution}\)
Show Worked Solution

i.    \(\text{Taking up as positive:}\)

\(M\ddot x\) \(=-Mg-kMv^2\)  
\(\ddot x\) \(=-g-kv^2\)  
\(v \cdot \dfrac{dv}{dx}\) \(=-(g+kv^2) \)  
\(\dfrac{dv}{dx}\) \(=-\dfrac{g+kv^2}{v} \)  
\(\dfrac{dx}{dv}\) \(=-\dfrac{v}{g+kv^2} \)  
\(x\) \(=-\dfrac{1}{2k} \displaystyle \int \dfrac{2kv}{g+kv^2}\, dv \)  
  \(=-\dfrac{1}{2k} \ln |g+kv^2|+c \)  

 
\(\text{When}\ \ x=o, \ v=v_0: \)

\(c=\dfrac{1}{2k} \ln |g+kv_0^2| \)

\(x\) \(=\dfrac{1}{2k} \ln |g+kv_0^2|-\dfrac{1}{2k} \ln |g+kv^2| \)  
  \(=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g+kv_0^2}{g+kv^2} \Bigg{|} \)  

 
\(\text{When}\ \ v=0, x=H: \)

\(H=\dfrac{1}{2k} \ln \Bigg{(} \dfrac{g+kv_0^2}{g} \Bigg{)},\ \ \ \ (k>0) \)
  

ii.   \(\text{When projectile travels downward:} \)

\(M \ddot x\) \(=Mg-kMv^2\)  
\(\ddot x\) \(=g-kv^2\)  
\(v \cdot \dfrac{dv}{dx}\) \(=g-kv^2\)  
\(\dfrac{dx}{dv}\) \(=\dfrac{v}{g-kv^2}\)  
\(x\) \(=-\dfrac{1}{2k} \displaystyle \int \dfrac{-2kv}{g-kv^2}\,dv \)  
  \(=-\dfrac{1}{2k}  \ln|g-kv^2|+c \)  

 
\(\text{When}\ \ x=0, \ v=0: \)

\(c=\dfrac{1}{2k} \ln g \)

\(x=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g}{g-kv^2} \Bigg{|} \)
 

\(\text{When}\ \ x=H, \ v=v_1: \)

\(\dfrac{1}{2k} \ln \Bigg{(} \dfrac{g+kv_0^2}{g} \Bigg{)}\) \(=\dfrac{1}{2k} \ln \Bigg{|} \dfrac{g}{g-kv_1^2} \Bigg{|} \)  
\(\dfrac{g+kv_0^2}{g} \) \(=\dfrac{g}{g-kv_1^2} \)  
\(g^2\) \(=(g+kv_0^2)(g-kv_1^2) \)  
\(g^2\) \(=g^2-gkv_1^2+gkv_0^2-k^2v_0^2v_1^2 \)  
\(gkv_0^2-gkv_1^2 \) \(=k^2v_0^2v_1^2 \)  
\(gk(v_0^2-v_1^2) \) \(=k^2v_0^2 v_1^2 \)  
\(g(v_0^2-v_1^2) \) \(=kv_0^2v_1^2 \)  
♦♦ Mean mark (ii) 31%.

BIOLOGY, M1 EQ-Bank 1 MC

Which of the following eukaryotic organelles is correctly matched with its function?

  \(\textbf{Organelle}\) \(\textbf{Function}\)
\(\text{A.}\)   Golgi body mRNA production
\(\text{B.}\)  Mitochondria ATP synthesis
\(\text{C.}\)    Ribosome Waste storage
\(\text{D.}\)  Lysosome Movement
Show Answers Only

\(B\) 

Show Worked Solution

By Elimination

→ Golgi bodies are involved in the modification, storage and distribution of the lipids, carbohydrates and proteins made in the E.R. (Eliminate A).

→ Ribosomes carry out protein synthesis (Eliminate C).

→ Lysosomes are involved in solid waste processing (Eliminate D).

\(\Rightarrow B\)

Complex Numbers, EXT2 N2 2023 14a

Let \(z\) be the complex number  \(z=e^{\small{\dfrac{i \pi}{6}}} \)  and \(w\) be the complex number  \(w=e^{\small{\dfrac{3 i \pi}{4}}} \).

  1. By first writing \(z\) and \(w\) in Cartesian form, or otherwise, show that
  2.    \(|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}\).  (3 marks)

  3. The complex numbers \(z, w\) and \(z+w\) are represented in the complex plane by the vectors \(\overrightarrow{O A},\overrightarrow{O B}\) and \(\overrightarrow{O C}\) respectively, where \(O\) is the origin.
  4. Show that  \(\angle A O C=\dfrac{7 \pi}{24}\).  (2 marks)

  5. Deduce that  \(\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}\).  (1 mark)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution

i.    \(z=e^{\small\dfrac{i \pi}{6}} = \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i \)

\(w=e^{\small\dfrac{3i \pi}{4}} = \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} \)

\(|z+w|^2\) \(=\Bigg{|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{|}\)  
  \(=\Bigg{|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{|}\)  
  \(=\Bigg{|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{|}\)  
  \(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)  
  \(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)  
  \(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)  
  \(=\dfrac{4-\sqrt6+\sqrt2}{2} \)  

 
ii.   

\(\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} \)

\( |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} \)

\(\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC \)

\(\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB \)

\(\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} \)
  

iii.   \(\text{In}\ \triangle AOC: \)

\( \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} \)

\(\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. \)
 

\(\text{Using the cos rule in}\ \triangle AOC: \)

\(\cos\,\dfrac{7\pi}{24}\) \(=\dfrac{|z|^2+|z+w|^2-|w|^2}{2|z||z+w|}\)  
  \(=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1  \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} \)  
  \(=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} \)  
  \(=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} \)  
  \(=\dfrac{8-2\sqrt6+2\sqrt2}{4} \)  
♦♦ Mean mark (iii) 26%.

Mechanics, EXT2 M1 2023 HSC 13c

A particle of mass 1 kg is projected from the origin with speed 40 m s\( ^{-1}\) at an angle 30° to the horizontal plane.

  1. Use the information above to show that the initial velocity of the particle is
  2.     \(\mathbf{v}(0)={\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \).   (1 mark)

The forces acting on the particle are gravity and air resistance. The air resistance is proportional to the velocity vector with a constant of proportionality 4 . Let the acceleration due to gravity be 10 m s \( ^{-2}\).

The position vector of the particle, at time \(t\) seconds after the particle is projected, is \(\mathbf{r}(t)\) and the velocity vector is \(\mathbf{v}(t)\).
 

  1. Show that  \(\mathbf{v}(t)={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\)  (3 marks)

  2. Show that  \(\mathbf{r}(t)=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)  (2 marks)

  3. The graphs  \(y=1-e^{-4 x}\)  and  \(y=\dfrac{4 x}{9}\) are given in the diagram below.
     
     
  4. Using the diagram, find the horizontal range of the particle, giving your answer rounded to one decimal place.  (2 marks)

Show Answers Only

  1. \(\text{Proof (See Worked Solutions)}\)
  2. \(\text{Proof (See Worked Solutions)}\)
  3. \(\text{Proof (See Worked Solutions)}\)
  4. \(8.7\ \text{metres}\)

Show Worked Solution

i. 

\(\underset{\sim}{v}(0)={\displaystyle\left(\begin{array}{cc} 40 \cos\ 30° \\ 40 \sin\ 30°\end{array}\right)} = {\displaystyle\left(\begin{array}{cc} 40 \times \frac{\sqrt3}{2} \\ 40 \times \frac{1}{2}\end{array}\right)}  = {\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \)
 

ii.   \(\text{Air resistance:} \)

\(\underset{\sim}{F} = -4\underset{\sim}{v} = {\displaystyle\left(\begin{array}{cc} -4\dot{x} \\ -4\dot{y} \end{array}\right)} \)

\(\text{Horizontally:}\)

\(1 \times \ddot{x} \) \(=-4 \dot{x} \)  
\(\dfrac{d\dot{x}}{dt}\) \(=-4\dot{x}\)  
\(\dfrac{dt}{d\dot{x}}\) \(= -\dfrac{1}{4\dot{x}} \)  
\(t\) \(=-\dfrac{1}{4} \displaystyle \int \dfrac{1}{\dot{x}} \ d\dot{x} \)  
\(-4t\) \(=\ln |\dot{x}|+c \)  

 
\(\text{When}\ \ t=0, \ \dot{x}=20\sqrt3 \ \ \Rightarrow\ \ c=-\ln{20\sqrt3} \)

\(-4t\) \(=\ln|\dot{x}|-\ln 20\sqrt3 \)  
\(-4t\) \(=\ln\Bigg{|}\dfrac{\dot{x}}{20\sqrt{3}} \Bigg{|} \)  
\(\dfrac{\dot{x}}{20\sqrt{3}} \) \(=e^{-4t} \)  
\(\dot{x}\) \(=20\sqrt{3}e^{-4t}\)  

 
\(\text{Vertically:} \)

\(1 \times \ddot{y} \) \(=-1 \times 10-4 \dot{y} \)  
\(\dfrac{d\dot{y}}{dt}\) \(=-(10+4\dot{y})\)  
\(\dfrac{dt}{d\dot{y}}\) \(= -\dfrac{1}{10+4\dot{y}} \)  
\(t\) \(=- \displaystyle \int \dfrac{1}{10+4\dot{y}} \ d\dot{y} \)  
\(-4t\) \(=- \displaystyle \int \dfrac{4}{10+4\dot{y}} \ d\dot{y} \)  
\(-4t\) \(=\ln |10+4\dot{y}|+c \)  

 
\(\text{When}\ \ t=0, \ \dot{y}=20 \ \ \Rightarrow\ \ c=-\ln{90} \)

\(-4t\) \(=\ln|10+4\dot{y}|-\ln 90 \)  
\(-4t\) \(=\ln\Bigg{|}\dfrac{10+4\dot{y}}{\ln{90}} \Bigg{|} \)  
\(\dfrac{10+4\dot{y}}{90} \) \(=e^{-4t} \)  
\(4\dot{y}\) \(=90e^{-4t}-10\)  
\(\dot{y}\) \(=\dfrac{45}{2} e^{-4t}-\dfrac{5}{2} \)  

 
\(\therefore \underset{\sim}v={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\) 

 
iii.   \(\text{Horizontally:}\)

\(x\) \(= \displaystyle \int \dot{x}\ dx\)  
  \(= \displaystyle \int 20\sqrt3 e^{-4t}\ dt \)  
  \(=-5\sqrt3 e^{-4t}+c \)  

 
\(\text{When}\ \ t=0, \ x=0\ \ \Rightarrow\ \ c=5\sqrt3 \)

\(x\) \(=5\sqrt3-5\sqrt3 e^{-4t} \)  
  \(=5\sqrt3(1-e^{-4t}) \)  

 
\(\text{Vertically:}\)

\(y\) \(= \displaystyle \int \dot{y}\ dx\)  
  \(= \displaystyle \int \dfrac{45}{2} e^{-4t}-\dfrac{5}{2}\ dt \)  
  \(=-\dfrac{45}{8}e^{-4t}-\dfrac{5}{2}t+c \)  

 
\(\text{When}\ \ t=0, \ y=0\ \ \Rightarrow\ \ c= \dfrac{45}{8} \)

\(y\) \(=\dfrac{45}{8}-\dfrac{45}{8} e^{-4t}-\dfrac{5}{2}t \)  
  \(=\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2} \)  

 
\(\therefore \underset{\sim}{r}=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)
 

iv.   \(\text{Range}\ \Rightarrow\ \text{Find}\ \ t\ \ \text{when}\ \ y=0: \)

\(\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2}t \) \(=0\)  
\(\dfrac{45}{8}(1-e^{-4t}) \) \(=\dfrac{5}{2}t \)  
\(1-e^{-4t}\) \(=\dfrac{4}{9}t \)  

 
\(\text{Graph shows intersection of these two graphs.}\)

\(\Rightarrow \text{Solution when}\ \ t\approx 2.25\)

\(\therefore\ \text{Range}\) \(=5\sqrt3(1-e^{(-4 \times 2.25)}) \)  
  \(=8.659…\)  
  \(=8.7\ \text{metres (to 1 d.p.)}\)  

♦ Mean mark (iv) 43%.
 

Proof, EXT2 P2 2023 HSC 13b

  1. Show that  \(k^2-2 k-3 \geq 0\)  for  \(k \geq 3\).  (1 mark)

  2. Hence, or otherwise, use mathematical induction to prove that
  3.   \(2^n \geq n^2-2\), for all integers  \(n \geq 3\).  (3 marks)

Show Answers Only

i.    \(\text{Proof (See Worked Solutions)} \)

ii.   \(\text{Proof (See Worked Solutions)} \)

Show Worked Solution

i.     \(k^2-2k-3\) \(=0\)
  \( (k-3)(k+1) \) \(=0\)

\(\text{Vertex (min) at}\ (1,-4) \)

\(\text{Quadratic is monotonically increasing for}\ \ k \geq1 \)

\(\text{At}\ \ k=3, \ k^2-2k-3=0 \)

\( \therefore\ \ k^2-2 k-3 \geq 0\ \ \text{for}\ \ k \geq 3\)
 

ii.    \(\text{Prove}\ \ 2^n \geq n^2-2 \)

\(\text{If}\ \ n=3: \)

\(\text{LHS}\ = 2^3=8 \)

\(\text{RHS}\ =3^3-2=7 \leq \text{LHS} \)

\(\therefore \ \text{True for}\ \ n=3. \)
 

\(\text{Assume true for}\ \ n=k: \)

\(2^k \geq k^2-2 \ \ \ …\ (*) \)
 

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

\(\text{i.e.}\ \ 2^{k+1} \geq (k+1)^2-2 \)

\(\text{LHS}\) \(=2^{k+1} \)  
  \(=2 \cdot 2^{k} \)  
  \( \geq 2(k^2-2) \ \ \ \text{(see (*) above)} \)  
  \( \geq 2k^2-4 \)  
  \( \geq k^2 + \underbrace{k^2-2k-3}_{\geq 0\ \  \text{(see part (i))}} +2k-1 \)  
  \(\geq k^2+2k-1 \)  
  \(\geq k^2+2k+1-2 \)  
  \(\geq (k+1)^2-2 \)  

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

\(\therefore \text{Since true for}\ \ n=3,\ \text{by PMI, true for integers}\ \ n\geq3 \)

Complex Numbers, EXT2 N2 2023 HSC 12e

The complex number  \(2+i\)  is a zero of the polynomial

\(P(z)=z^4-3 z^3+c z^2+d z-30\)

where \(c\) and \(d\) are real numbers.

  1. Explain why  \(2-i\)  is also a zero of the polynomial \(P(z)\).  (1 marks)

  2. Find the remaining zeros of the polynomial \(P(z)\).  (2 marks)

Show Answers Only

i.    \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)

\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)

ii.   \(\text{Remaining zeros:}\ \ -3, 2 \)

Show Worked Solution

i.    \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)

\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)

 

ii.    \(P(z)=z^4-3 z^3+c z^2+d z-30\)

\(\text{Let roots be:}\ \ 2+i, 2-i, \alpha, \beta \)

\( \sum\ \text{roots:}\)

\(2+i+2-i+\alpha + \beta\) \(=-\dfrac{b}{a} \)  
\(4+\alpha+\beta\) \(=3\)  
\(\alpha + \beta\) \(=-1\ \ \ …\ (1) \)  

 
\(\text{Product of roots:} \)

\((2+i)(2-i)\alpha\beta \) \(= \dfrac{e}{a} \)  
\(5\alpha\beta\) \(=-30\)  
\(\alpha \beta \) \(=-6\ \ \ …\ (2) \)  

 
\(\text{Substitute}\ \ \beta=-\alpha-1\ \ \text{into (2):} \)

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

 
\(\therefore\ \text{Remaining zeros:}\ \ -3, 2 \)

PHYSICS, M6 2023 HSC 2 MC

Which diagram best represents the transmission of energy from a power station to people's houses?
 

 
 

Show Answers Only

\(C\)

Show Worked Solution

→ Energy from power stations passes through a step-up transformer to increase the voltage and decrease the current.

→ This energy then passes through a step-down transformer to be used in homes.

 \(\Rightarrow C\)

PHYSICS, M8 2023 HSC 4 MC

Caesium-137 has a half-life of 30 years.

What mass of caesium-137 will remain after 90 years, if the initial mass was 120 g?

  1. 4 g
  2. 15 g
  3. 40 g
  4. 60 g
Show Answers Only

\(B\)

Show Worked Solution

\(t_\frac{1}{2} = \text{30 years,}\ \ \lambda =\dfrac{\ln2}{30}\)

\(N\) \(=N_{0}e^{-\lambda t} \)  
  \(=120e^{-\dfrac{\ln2}{30} \times 90} \)  
  \(=15\ \text{g}\)  

 
\(\Rightarrow B\)

Calculus, EXT2 C1 2023 HSC 11f

Find \({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\).  (4 marks)

Show Answers Only

\(-\ln3 \)

Show Worked Solution

\( \text{Let}\ \ \dfrac{5x-3}{(x+1)(x-3)} = \dfrac{A}{x+1}+\dfrac{B}{x-3}\ \ \ \text{for}\ \ A, B \in \mathbb{R} \)

\(\dfrac{5x-3}{(x+1)(x-3)}\) \(=\dfrac{A}{x+1}+\dfrac{B}{x-3} \)  
  \(=\dfrac{A(x-3)+B(x+1)}{(x+1)(x-3)} \)  

 
\(\text{Equating numerators:}\)

\( A(x-3)+B(x+1)=5x-3 \)

\(\text{When}\ \ x=3, 4B=12\ \Rightarrow \ B=3 \)

\(\text{When}\ \ x=-1,  -4A=-8\ \Rightarrow \ A=2 \)

\({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\) \( =\displaystyle \int_0^2 \dfrac{2}{x+1}+\dfrac{3}{x-3}\ dx \)  
  \(= \Big{[} 2\ln|x+1|+3\ln|x-3| \Big{]}_0^2 \)  
  \(= 2\ln3+3\ln1-2\ln1-3\ln3 \)  
  \(=-\ln3 \)  

Mechanics, EXT2 M1 2023 HSC 11e

A particle moves in simple harmonic motion described by the equation

\( \ddot{x}=-9(x-4) . \)

Find the period and the central point of motion.  (2 marks)

Show Answers Only

\(\text{Period}\ = \dfrac{2 \pi}{3} \)

\(\text{Centre of Motion:}\ x=4 \) 

Show Worked Solution

\( \ddot{x}=-9(x-4) \)

\( \Rightarrow\ n=3,\ \ c=4 \)

\(\text{Period}\ = \dfrac{2 \pi}{3} \)

\(\text{Centre of Motion:}\ x=4 \) 

Vectors, EXT2 V1 2023 HSC 11c

Find a vector equation of the line through the points  \(A(-3,1,5)\)  and  \(B(0,2,3)\).  (2 marks)

Show Answers Only

\[\underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]

Show Worked Solution

\[\overrightarrow{AB}=\left(\begin{array}{c} 0-(-3) \\2-1 \\ 3-5 \end{array}\right) = \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right)\]

\[\therefore \underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]

Complex Numbers, EXT2 N2 2023 HSC 11a

Solve the quadratic equation

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

where \(z\) is a complex number. Give your answers in Cartesian form.  (2 marks)

Show Answers Only

\(\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)

Show Worked Solution

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

\(z\) \(=\dfrac{3 \pm \sqrt{(-3)^2-4 \times 1 \times 4}}{2} \)  
  \(=\dfrac{3 \pm \sqrt{-7}}{2} \)  
  \(=\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)  

Calculus, EXT1 C3 2023 HSC 13a

A hemispherical water tank has radius \(R\) cm. The tank has a hole at the bottom which allows water to drain out.

Initially the tank is empty. Water is poured into the tank at a constant rate of  \(2 k R\) cm³ s\(^{-1}\), where \(k\) is a positive constant.

After \(t\) seconds, the height of the water in the tank is \(h\) cm, as shown in the diagram, and the volume of water in the tank is \(V\) cm³.
  

It is known that  \(V= \pi \Big{(} R h^2-\dfrac{h^3}{3}\Big{)}. \)    (Do NOT prove this.)

While water flows into the tank and also drains out of the bottom, the rate of change of the volume of water in the tank is given by  \(\dfrac{d V}{d t}=k(2 R-h)\).

  1. Show that  \(\dfrac{d h}{d t}=\dfrac{k}{\pi h}\).  (2 marks)

  2. Show that the tank is full of water after  \(T=\dfrac{\pi R^2}{2 k}\) seconds.  (2 marks)

  3. The instant the tank is full, water stops flowing into the tank, but it continues to drain out of the hole at the bottom as before.
  4. Show that the tank takes 3 times as long to empty as it did to fill.  (3 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution

i.    \(V=\pi \Big{(}Rh^2-\dfrac{h^3}{3} \Big{)} \)

\(\dfrac{dV}{dh} = \pi(2Rh-h^2) \)

\(\dfrac{dV}{dt} = k(2R-h)\ \ \ \text{(given)} \)

\(\dfrac{dh}{dt}\) \(= \dfrac{dV}{dt} \cdot \dfrac{dh}{dV} \)  
  \(=k(2R-h) \cdot \dfrac{1}{\pi} \cdot \dfrac{1}{h(2R-h)} \)  
  \(= \dfrac{k}{\pi h} \)  

 
ii.    \(\dfrac{dt}{dh} = \dfrac{\pi h}{k} \)

\(t\) \(= \displaystyle \int \dfrac{dt}{dh}\ dh \)  
  \(= \dfrac{\pi}{k} \displaystyle \int h\ dh \)  
  \(= \dfrac{\pi}{k} \Big{[} \dfrac{h^2}{2} \Big{]} +c \)  

 
\(\text{When}\ \ t=0, h=0 \)

\(\Rightarrow c=0 \)

\( t= \dfrac{\pi h^2}{2k} \)

 
\(\text{Tank is full at time}\ T\ \text{when}\ \ h=R: \)

\( T= \dfrac{\pi R^2}{2k}\ \text{seconds} \)

♦ Mean mark (ii) 41%.

iii.   \(\text{Net water flow}\ = k(2R-h)\ \ \text{(given)} \)

\(\text{Flow in}\ =2kR\ \ \text{(given)} \)

\(\text{Flow out}\ = k(2R-h)-2kR=-kh \)
 

\( \dfrac{dh}{dt}= \dfrac{-kh}{\pi h(2R-h)} = \dfrac{-k}{\pi (2R-h)} \)

♦♦♦ Mean mark (iii) 20%.
 

\(\dfrac{dt}{dh}\) \(=\dfrac{- \pi (2R-h)}{k} \)  
\( \displaystyle \int k\ dt\) \(=- \pi \displaystyle \int (2R-h)\ dh \)  
\(kt\) \(=- \pi \Big{(} 2Rh-\dfrac{h^2}{2} \Big{)}+c \)  

 
\(\text{When}\ \ t=0, \ h=R: \)

\(0\) \(=- \pi \Big{(}2R^2-\dfrac{R^2}{2} \Big{)} + c\)  
\(c\) \(= \pi \Big{(} \dfrac{3R^2}{2} \Big{)} \)  

 
\(\text{Find}\ t\ \text{when}\ h=0: \)

\(kt\) \(=- \pi(0) + \pi \dfrac{3R^2}{2} \)  
\(t\) \(= \dfrac{3 \pi R^2}{2k} \)  
  \(= 3 \times \dfrac{\pi R^2}{2k} \)  

 
\(\therefore\ \text{Tank takes 3 times longer to empty than fill.} \)

Networks, STD1 N1 2023 HSC 18

A network of running tracks connects the points \(A, B, C, D, E, F, G, H\), as shown. The number on each edge represents the time, in minutes, that a typical runner should take to run along each track.
 

 

  1. Which path could a typical runner take to run from point \(A\) to point \(D\) in the shortest time?  (2 marks)

  2. A spanning tree of the network above is shown.
     

  1. Is it a minimum spanning tree? Give a reason for your answer.  (2 marks)

Show Answers Only

a.    \(ABFGD\)

b.    \(\text{See worked solutions}\)

Show Worked Solution

a.    \(\text{Using Djikstra’s Algorithm:}\)
 

\(\text{Shortest route}\) \(=ABFGD\)  
  \(=3+1+5+5\)  
  \(=14\)  

 
b.   \(\text{Total time of given spanning tree}\)

\(=3+11+1+2+4+5+5\)

\(=31\)
 

\(\text{Consider the MST below:}\)
 

\(\text{Total time (MST)}= 3+1+2+4+5+5+9=29\)

\(\therefore \text{ Given tree is NOT a MST.}\)

♦♦ Mean mark (b) 21%.

Networks, STD1 N2 2023 HSC 15

The table shows some of the flight distances (rounded to the nearest 10 km between various Australian cities.

  1. Use the information in the table to complete the network diagram where the edges are labelled with distances.  (2 marks)
     


 

  1. Mahsa wants to travel from Hobart to Darwin. She wants to change planes only once.
  2. Using the network diagram, calculate how many kilometres she will travel by plane.  (1 mark)

Show Answers Only

a.  

b.    \(4190\text{ km}\)

Show Worked Solution

a.

b.    \(\text{Changing planes only once}\Rightarrow H\rightarrow S\rightarrow D\)

\(\text{Kilometres travelled}=1040+3150=4190\text{ km}\)

Measurement, STD1 M4 2023 HSC 14

The distance-time graph shows the first two stages of a car journey from home to a holiday house.
  

  1. At what speed, in kilometres per hour, did the car travel during stage \(A\) of the journey?   (1 mark)

  2. For how long did the car stop during stage \(B\) of the journey?   (1 mark)

  3. After stage \(B\), the car continues to travel towards the holiday house at a constant speed of \(50\ \text{km/h}\) for 2 hours. Graph this part of the journey on the grid above.   (2 marks)

Show Answers Only

a.    \(100\text{ km/h}\)

b.    \(30\text{ minutes}\)

c.   

Show Worked Solution

a.   \(S=\dfrac{D}{T}=\dfrac{150}{1.5}=100\ \text{km/h}\)
 

b.    \(\text{Stage}\textit{ B}=1.5\rightarrow 2\text{ hours}=30\text{ minutes}\)
 

c.    \(\text{Position after 2 hours at 50km/h}=(150+2\times 50 , 2 + 2) = (250 , 4)\)


♦ Mean mark (c) 50%.

Measurement, STD1 M5 2023 HSC 12

A floor plan is shown.
 

  1. What are the dimensions, in metres, of the living room?  (2 marks)

  2. The shaded area of the kitchen floor is to be tiled. Each tile is 40 cm by 40 cm. How many tiles are needed to cover the kitchen floor?  (2 marks)

  3. The tiles are supplied in boxes of 10 . Only full boxes can be purchased. How many boxes need to be purchased to tile the kitchen floor?  (1 mark)

Show Answers Only

a.    \(5.2\times 5.94\)

b.    \(72\text{ tiles}\)

c.    \(8\text{ boxes}\)

Show Worked Solution

a.    \(\text{Dimensions}=5.2\times (2.15+0.16+3.6)=5.2\times5.94\)


♦ Mean mark (a) 51%.

b.    \(\text{Method 1}\)

\(\text{Kitchen Area}\) \(=3.2\times 3.6\)  
  \(=11.52\text{ m}^2\)  
\(\text{Tile area}\) \(=0.4\times 0.4\)  
  \(=0.16\text{ m}^2\)  
\(\text{Tiles needed}\) \(=\dfrac{11.52}{0.16}\)  
  \(=72\text{ tiles}\)  

  
\(\text{Method 2}\)

\(\text{Tiles to fit width}\) \(=\dfrac{3.6}{0.4}\)
  \(=9\)
\(\text{Tiles to fit length}\) \(=\dfrac{3.2}{0.4}\)
  \(=8\)
\(\text{Tiles needed}\) \(=9\times 8\)
  \(=72\text{ tiles}\)

♦ Mean mark (b) 43%.

c.    \(\text{Boxes}=\dfrac{72}{10}=7.2\)

\(\therefore\ \text{Boxes }=8\)

Statistics, STD1 S1 2023 HSC 11

A company employs 50 people.

The annual income of the employees is shown in the grouped frequency distribution table.

\begin{array} {|c|c|c|c|}
\hline
\textit{Annual income} & \textit{Class centre} & \textit{Number of} & fx \\ \text{(\$)} & (x) & \textit{employees}\ (f) &  \\
\hline
\rule{0pt}{2.5ex} \text{40 000 – 49 999} \rule[-1ex]{0pt}{0pt} & 45\ 000 & 12 & 540\ 000 \\
\hline
\rule{0pt}{2.5ex} \text{50 000 – 59 999} \rule[-1ex]{0pt}{0pt} & 55\ 000 & 13 & 715\ 000 \\
\hline\rule{0pt}{2.5ex} \text{60 000 – 69 999} \rule[-1ex]{0pt}{0pt} & 65\ 000 & 15 & A \\
\hline\rule{0pt}{2.5ex} \text{70 000 – 79 999} \rule[-1ex]{0pt}{0pt} & 75\ 000 & 7 & 525\ 000 \\
\hline\rule{0pt}{2.5ex} \text{80 000 – 89 999} \rule[-1ex]{0pt}{0pt} & 85\ 000 & 3 & 255\ 000 \\
\hline
\hline\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} &  & \textit{Total}\ = 50 & \textit{Total = B} \\
\hline
\end{array}  

  1. What are the values of \(A\) and \(B\)?  (2 marks)

  2. Find the mean for this distribution.  (1 mark)

Show Answers Only

a.    \(A=$975\ 000\), \(B=$3\ 010\ 000\)

b.    \($60\ 200\)

Show Worked Solution

a.    \(A=65\ 000\times 15 = $975\ 000\)

\(B=540\ 000+715\ 000+975\ 000+525\ 000+255\ 000=$3\ 010\ 000\)

  

b.    \(\text{Mean}=\dfrac{\text{Total }fx}{\text{Total }f}=\dfrac{3\ 010\ 000}{50}=$60\ 200\)

♦♦♦ Mean mark (b) 12%.

Statistics, EXT1 S1 2023 HSC 12c

A gym has 9 pieces of equipment: 5 treadmills and 4 rowing machines.

On average, each treadmill is used 65% of the time and each rowing machine is used 40% of the time.

  1. Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use.  (2 marks)

  2. Find an expression for the probability that, at a particular time, exactly 3 of the 5 treadmills are in use and no rowing machines are in use.  (1 mark)

Show Answers Only

  1. \(P\text{(3 of 5 treadmills in use)}\ =\ ^5C_3 (0.65)^3(0.35)^2 \)
  2. \(P\text{(3 treadmills and no rowing)}\ = \ ^5C_3 (0.65)^3(0.35)^2 \times (0.6)^4 \)

Show Worked Solution

i.    \(P(T)=0.65, \ \ P(\overline{T})=1-0.65=0.35 \)

\(P\text{(3 of 5 treadmills in use)}\ =\ ^5C_3 (0.65)^3(0.35)^2 \)
 

ii.  \(P(R)=0.4, \ \ P(\overline{R})=1-0.4=0.6 \)

\(P\text{(no rowing machines in use)}\ =\ ^4C_0 (0.6)^4(0.4)^0=(0.6)^4 \)

\(\text{Since 2 events are independent:} \)

\(P\text{(3 treadmills and no rowing)}\ = \ ^5C_3 (0.65)^3(0.35)^2 \times (0.6)^4 \)

Proof, EXT1 P1 2023 HSC 12b

Use mathematical induction to prove that

\((1 \times 2)+\left(2 \times 2^2\right)+\left(3 \times 2^3\right)+\cdots+\left(n \times 2^n\right)=2+(n-1) 2^{n+1}\)

for all integers \(n \geq 1\).   (3 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)} \)

Show Worked Solution

\(\text{Prove true for}\ \ n=1: \)

\(\text{LHS}\ = 1 \times 2=2 \)

\(\text{RHS}\ = 2+(1-1)2^2 = 2 = \text{LHS} \)
 

\(\text{Assume true for}\ \ n=k: \)

\((1 \times 2)+\left(2 \times 2^2\right)+\cdots+\left(k \times 2^k\right)=2+(k-1) 2^{k+1}\)
 

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

\((1 \times 2)+\left(2 \times 2^2\right)+\cdots+\left(k \times 2^k\right) + (k+1)2^{k+1}=2+k \times 2^{k+2}\)

\(\text{LHS}\) \(=2+(k-1)2^{k+1} + (k+1) 2^{k+1} \)  
  \(=2+2^{k+1}(k-1+k+1) \)  
  \(=2+2^{k+1}(2k) \)  
  \(=2+k \times 2^{k+2} \)  
  \(=\ \text{RHS} \)  
 

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

\(\therefore\ \text{Since true for}\ \ n=1, \text{by PMI, true for integers}\ \ n \geq 1. \)

Calculus, EXT1 C2 2023 HSC 12a

Evaluate \(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) using the substitution  \(u=x-3\).   (3 marks) 

Show Answers Only

\(\dfrac{56}{15}\)

Show Worked Solution

\(u=x-3\ \ \Rightarrow \ x=u+3 \)

\(\dfrac{du}{dx}=1\ \ \Rightarrow \ du=dx \)

\(\text{When}\ \ x=4, u=1 \)

\(\text{When}\ \ x=3, u=0 \)

\(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) \(=\displaystyle \int_0^1(u+5) \sqrt{u}\ du\)  
  \(=\displaystyle \int_0^1 u^\frac{3}{2} +5u^\frac{1}{2}\ du\)  
  \(=\Big{[}\dfrac{2}{5} \times u^\frac{5}{2} + \dfrac{2}{3} \times 5u^\frac{3}{2}\Big{]}_0^1 \)  
  \(=\Big{[} \Big{(}\dfrac{2}{5} + \dfrac{10}{3}\Big{)}-0\Big{]}\)  
  \(=\dfrac{56}{15}\)  

Functions, EXT1 F2 2023 HSC 11c

Consider the polynomial

\(P(x)=x^3+a x^2+b x-12\),

where \(a\) and \(b\) are real numbers.

It is given that  \(x+1\)  is a factor of \(P(x)\) and that, when \(P(x)\) is divided by  \(x-2\), the remainder is \(-18\) .

Find \(a\) and \(b\).  (3 marks)

Show Answers Only

\(a=2, \ b=-11\)

Show Worked Solution

\((x+1)\ \ \text{is a factor of}\ P(x)\ \ \Rightarrow\ \ P(-1)=0 \)

\(0\) \(=(-1)^3+a-b-12\)  
\(a-b\) \(=13\ \ …\ (1) \)  

 
\(P(x) ÷ (x-2) \ \text{has a remainder of}\ -18 \ \Rightarrow\ \ P(2)=-18 \)

\(-18\) \(=8+4a+2b-12\)  
\(4a+2b\) \(=-14\)  
\(2a+b\) \(=-7\ \ …\ (2) \)  

 
\(\text{Add}\ \ (1) + (2): \)

\(3a=6\ \ \Rightarrow\ \ a=2\)

\(\text{Substitute}\ \ a=2\ \ \text{into (1):} \)

\(2-b=13\ \ \Rightarrow\ \ b=-11 \)

\(\therefore a=2, \ b=-11\)

Functions, EXT1 F1 2023 HSC 11a

The parametric equations of a line are given below.

\begin{aligned}
& x=1+3 t \\
& y=4 t
\end{aligned}

Find the Cartesian equation of this line in the form  \(y=m x+c\).  (2 marks)

Show Answers Only

\(y=\dfrac{4}{3}x-\dfrac{4}{3} \)

Show Worked Solution

\(x=1+3t\ \ \Rightarrow \ \ t=\dfrac{x-1}{3} \)

\(y\) \(=4t\)  
\(y\) \(=4\bigg{(}\dfrac{x-1}{3}\bigg{)} \)  
\(y\) \(=\dfrac{4}{3}x-\dfrac{4}{3} \)  

Calculus, EXT1 C3 2023 HSC 3 MC

The diagram shows the direction field of a differential equation. A particular solution to the differential equation passes through \((-2,1)\).

Where does the solution that passes through \((-2,1)\) cross the \(y\)-axis?

  1. \(y=1.12\)
  2. \(y=1.34\)
  3. \(y=1.56\)
  4. \(y=1.78\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Following gradients → cross y-axis slightly above 1.5}\)

\(\Rightarrow C\)

Proof, EXT2 P1 2023 HSC 2 MC

Consider the following statement.

'If an animal is a herbivore, then it does not eat meat.'

Which of the following is the converse of this statement?

  1. If an animal is a herbivore, then it eats meat.
  2. If an animal is not a herbivore, then it eats meat.
  3. If an animal eats meat, then it is not a herbivore.
  4. If an animal does not eat meat, then it is a herbivore.
Show Answers Only

\(D\)

Show Worked Solution

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

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

\(\Rightarrow D\)

Functions, 2ADV F2 2023 HSC 19

  1. Sketch the graphs of the functions  `f(x)=x-1`  and  `g(x)=(1-x)(3+x)`  showing the `x`-intercepts.  (2 marks)
        

  1. Hence, or otherwise, solve the inequality  `x-1<(1-x)(3+x)`.  (2 marks)

Show Answers Only

a.   
         

b.    `-4<x<1`

Show Worked Solution

a.   `g(x)\ text{cuts}\ xtext{-axis at 1 and}\ -3.`

`g(x)_max=g(-1)=4`

`text{Find intersection of graphs:}`

`(1-x)(3+x)` `=x-1`  
`3+x-3x-x^2` `x-1`  
`x^2+3x-4` `=0`  
`(x+4)(x-1)` `=0`  

 
`text{Intersections at:}\ (1,0), (-4,-5)`
 

 

b.    `text{From the graph:}`

`x-1<(1-x)(3+x)\ \ text{when}\ \ -4<x<1`

`text{Test}\ \ x=0:`

`0-1<(1-0)(3+0)\ \ =>\ \ -1<3\ \ text{(correct)}`

Statistics, 2ADV S2 2023 HSC 18

A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C).

Using `x` as the average daily outside temperature and `y` as the total daily gas usage, the equation of the least-squares regression line was found.

The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW.

The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°, 9°, 9°, 10°,

The total gas usage for the ten weekdays was 1840 MW.

In any bivariate dataset, the least-squares regression line passes through the point `(bar x,bar y)`, where `bar x` is the sample mean of the `x`-values and `bary` is the sample mean of the `y`-values.

  1. Using the information provided, plot the point `(bar x,bar y)` and the `y`-intercept of the least-squares regression line on the grid.  (3 marks)
     

 

  1. What is the equation of the regression line?  (2 marks)

  2. In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23°C.  (1 mark)

Show Answers Only

a.    
       

b.    `y=-10.4x+236`

c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Show Worked Solution

a.    `barx=(0+0+0+2+5+7+8+9+9+10)/10=5^@text{C}`

`bary=1840/10=184`

`text{Regression line passes through:}\ (0,236) and (5,184)`
 

 

b.    `m=(y_2-y_1)/(x_2-x_1)=(184-236)/(5-0)=-10.4`

`text{Equation of line}\ m=-10.4\ text{passing through}\ (0,236):`

`(y-y_1)` `=m(x-x_1)`  
`y-236` `=-10.4(x-0)`  
`y` `=-10.4x+236`  

 
c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Financial Maths, 2ADV M1 2023 HSC 15

A table of future value interest factors for an annuity of $1 is shown.
 

  1. Micky wants to save $450 000 over the next 10 years.
  2. If the interest rate is 6% per annum compounding annually, how much should Micky contribute each year? Give your answer to the nearest dollar.  (2 marks)

  3. Instead, Micky decides to contribute  $8535 every three months for 10 years to an annuity paying 6% per annum, compounding quarterly.
  4. How much will Micky have at the end of 10 years?  (3 marks)

Show Answers Only

  1. `$34\ 140`
  2. `$463\ 177.38`

Show Worked Solution

a.    `text{Applicable interest rate}\ =6%`

`text{Compounding periods}\ =10xx1=10`

`=>\ text{Factor}\ = 13.181`

`:.\ text{Contribution (annual)}` `=(450\ 000)/13.181`  
  `=$34\ 140`  

 
b.    `text{Applicable interest rate}\ =(6%)/4=1.5%\ text{per quarter}`

`text{Compounding periods}\ =10xx4=40`

`=>\ text{Factor}\ = 54.268`

`text{Total (after 10 years)}` `=8535 xx 54.268`  
  `=$463\ 177.38`  

Calculus, 2ADV C1 2023 HSC 14

Find the equation of the tangent to the curve  `y=(2x+1)^3`  at the point `(0,1)`.  ( 3 marks)

Show Answers Only

`y=6x+1`

Show Worked Solution
`y` `=(2x+1)^3`  
`dy/dx` `=3xx2(2x+1)^2`  
  `=6(2x+1)^2`  

 
`text{At}\ x=0\ \ =>\ \ dy/dx=6xx1^2=6`
 

`text{Find equation of line}\ \ m=6,\ text{through}\ (0,1)`

`y-y_1` `=m(x-x_1)`  
`y-1` `=6(x-0)`  
`y` `=6x+1`  

Probability, 2ADV S1 2023 HSC 12

The table shows the probability distribution of a discrete random variable.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 1 & 2 & 3 & 4 \\
\hline
\rule{0pt}{2.5ex} P(X = x) \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ 0.3\ \  & \ \ 0.5\ \  & \ \ 0.1\ \  & \ \ 0.1\ \  \\
\hline
\end{array}

  1. Show that the expected value `E(X)=2`.  (1 mark)

  2. Calculate the standard deviation, correct to one decimal place.  (2 marks)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `0.9`
Show Worked Solution
a.     `E(X)` `=0+1xx0.3+2xx0.5+3xx0.1+4xx0.1`
    `=0.3+1+0.3+0.4`
    `=2`

 

b.     `text{Var}(X)` `=E(X^2)-[E(X)]^2`
    `=(0+1^2xx0.3+2^2xx0.5+3^2xx0.1+4^2xx0.1)-2^2`
    `=(0.3+2+0.9+1.6)-4`
    `=0.8`

 

`:. sigma` `=sqrt(0.8)`  
  `=0.8944…`  
  `=0.9\ \ text{(to 1 d.p.)}`  

Statistics, 2ADV S2 2023 HSC 1 MC

The number of bees leaving a hive was observed and recorded over 14 days at different times of the day.
 

Which Pearson's correlation coefficient best describes the observations?

  1. – 0.8
  2. – 0.2
  3. 0.2
  4. 0.8
Show Answers Only

`D`

Show Worked Solution

`text{Correlation is positive and strong.}`

`text{Best option:}\ r=0.8`

`=>D`

NOTE: Inputting all data points into a calculator is unnecessary and time consuming here.

Algebra, STD2 A4 2023 HSC 21

Electricity provider \(A\) charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

  1. Complete the table showing Provider \(A\)'s monthly charges for different levels of electricity usage.  (1 mark)
     
\(\textit{Electricity used in a
 month (kWh)}\)		 \(\ \ 0\ \ \) \(400\) \(1000\)
\(\textit{Monthly charge (\$)}\) \(40\)   \(290\)
 
Provider \(B\) charges 35 cents per kWh, with no fixed monthly charge. The graph shows how Provider \(B\)'s charges vary with the amount of electricity used in a month.
 

 
  1. On the grid above, graph Provider \(A\)'s charges from the table in part (a).   (1 mark)
  2. Use the two graphs to determine the number of kilowatt hours per month for which Provider \(A\) and Provider \(B\) charge the same amount.   (1 mark)

  3. A customer uses an average of 800 kWh per month.
  4. Which provider, \(A\) or \(B\), would be the cheaper option and by how much?   (2 marks)

Show Answers Only

a. 

\(\textit{Electricity used in a
 month (kWh)}\)		 \(\ \ 0\ \ \) \(400\) \(1000\)
\(\textit{Monthly charge (\$)}\) \(40\) \(140\) \(290\)

b.    
         

c.    \(\text{400 kWh}\)

d.    \(\text{Provider}\ A\ \text{is cheaper by \$40.}\)

Show Worked Solution

a. 

\(\textit{Electricity used in a
 month (kWh)}\)		 \(\ \ 0\ \ \) \(400\) \(1000\)
\(\textit{Monthly charge (\$)}\) \(40\) \(140\) \(290\)

b. 
          
 

c.    \(A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}\)

\(\therefore\ \text{Same charge at 400 kWh}\)
 

d.    \(\text{Cost at 800 kWh:}\)

\(\text{Provider}\ A: \ 40 + 0.25 \times 800 = $240\)

\(\text{Provider}\ B: \ 0.35 \times 800 = $280\)

\(\therefore \text{Provider}\ A\ \text{is cheaper by \$40.}\)

Networks, STD2 N2 2023 HSC 19

A network of running tracks connects the points `A, B, C, D, E, F, G, H`, as shown. The number on each edge represents the time, in minutes, that a typical runner should take to run along each track.


  1. Which path could a typical runner take to run from point `A` to point `D` in the shortest time?  (2 marks)

  2. A spanning tree of the network above is shown.
  3.  
  5. Is it a minimum spanning tree? Give a reason for your answer.  (2 marks)

Show Answers Only

a.    `ABFGD`

b.    `text{See worked solutions}`

Show Worked Solution

a.    `text{Using Djikstra’s Algorithm:}`
 

`text{Shortest route}` `=ABFGD`  
  `=3+1+5+5`  
  `=14`  

 
b.   `text{Total time of given spanning tree}`

`=3+11+1+2+4+5+5`

`=31`
 

`text{Consider the MST below:}`
 

`text{Total time (MST)}\ = 3+1+2+4+5+5+9=29`

`:.\ text{Given tree is NOT a MST.}`

♦ Mean mark (b) 47%.

Algebra, STD2 A4 2023 HSC 20

On another planet, a ball is launched vertically into the air from the ground.

The height above the ground, `h` metres, can be modelled using the function  `h=-6 t^2+24t`, where `t` is measured in seconds. The graph of the function is shown.

  1. Based on the graph, what is the maximum height reached by the ball?  (1 mark)

  2. Based on the graph, at what TWO times is the ball at `3/4` of its maximum height?  (2 marks)

Show Answers Only

a.    `h_max = 24\ text{metres}`

b.    `t=1 and 3\ text{seconds}`

Show Worked Solution

a.    `h_max = 24\ text{metres}`

b.    `3/4 xx h_max = 3/4xx24=18\ text{metres}`

`text{From graph, ball is at at 18 metres when:}`

`t=1 and 3\ text{seconds}`

Networks, STD2 N2 2023 HSC 17

The table shows some of the flight distances (rounded to the nearest 10 km) between various Australian cities.
 

  1. Use the information in the table to complete the network diagram where the edges are labelled with distances.  (2 marks)
     

  1. Mahsa wants to travel from Hobart to Darwin. She wants to change planes only once.
  2. Using the network diagram, calculate how many kilometres she will travel by plane.  (1 mark)

Show Answers Only

a.

 

b.    `4190\ text{km}`

Show Worked Solution

a. 

       

 

b.    `text{Hobart to Darwin (with 1 stop):}`

`text{Distance}` `=H→S + S→D`  
  `=1040+3150`  
  `=4190\ text{km}`  

Measurement, STD2 M7 2023 HSC 16

The graph shows Peta's heart rate, in beats per minute, during the first 60 minutes of a marathon.
 

  1. What was Peta's heart rate 20 minutes after she started her marathon?   (1 mark)

  2. Peta started the marathon at 10 am. At what time would her heart rate first reach 140 beats/minute?   (1 mark)

Show Answers Only

a.    `120\ text{beats/minute}`

b.    `10:30\ text{am}`

Show Worked Solution

a.    `120\ text{beats/minute}`

b.    `10:30\ text{am}`

Financial Maths, STD2 F1 2023 HSC 4 MC

A delivery truck was valued at $65 000 when new. The value of the truck depreciates at a rate of 22 cents per kilometre travelled.

What is the value of the truck after it has travelled a total distance of 132 600 km?

  1. $35 828
  2. $29 172
  3. $14 872
  4. $14 300
Show Answers Only

`A`

Show Worked Solution
`text{Depreciation}` `=0.22 xx 132\ 600`  
  `=$29\ 172`  

 

`text{Truck value}` `=65\ 000-29\ 172`  
  `=$35\ 828`  

 
`=>A`

BIOLOGY, M3 2013 HSC 16 MC

The theory of evolution has been supported by studying the structures of vertebrate forelimbs from the fossil record.

This type of study is best described as

  1. biogeography.
  2. comparative biochemistry.
  3. comparative embryology.
  4. palaeontology.
Show Answers Only

\(D\)

Show Worked Solution

→ Palaeontology is simply, the study of fossils.

\(\Rightarrow D\)

BIOLOGY, M2 2014 HSC 35a

  1. Name a scientist from the 17th or 18th century who contributed to the development of ideas on the structure and function of plants.   (1 mark)

  2. Outline how the findings from a 17th or 18th century experiment informed scientists about plant structure and/or function.   (3 marks)

Show Answers Only

i.    Scientists could include one of:

→ Van Helmont, Hales, Priestley, Ingen-Housz, Senebier, Saussure
 

ii.   Priestly’s Experiment

→ In 1771, Joseph Priestly ignited a candle in a jar with some mint, in which he ignited the candle and then it went out. After a few trials, the candle wouldn’t ignite as all the oxygen in the air had been used (note that oxygen was not yet discovered). However after 27 days, by igniting the candle with a mirror and sunlight as to not open the jar, it reignited.

→ Similarly, by putting mice into a jar with and without a sprig of mint, the jar with the mint allowed the mouse to survive much longer.

→ These observations lead to questions about plant structure, and how they were somehow able to ‘restore air’ that is used by animals breathing and combustion reactions. 

Show Worked Solution

i.    Scientists could include one of:

→ Van Helmont, Hales, Priestley, Ingen-Housz, Senebier, Saussure
 

ii.   Priestly’s Experiment

→ In 1771, Joseph Priestly ignited a candle in a jar with some mint, in which he ignited the candle and then it went out. After a few trials, the candle wouldn’t ignite as all the oxygen in the air had been used (note that oxygen was not yet discovered). However after 27 days, by igniting the candle with a mirror and sunlight as to not open the jar, it reignited.

→ Similarly, by putting mice into a jar with and without a sprig of mint, the jar with the mint allowed the mouse to survive much longer.

→ These observations lead to questions about plant structure, and how they were somehow able to ‘restore air’ that is used by animals breathing and combustion reactions. 

BIOLOGY, M3 2015 HSC 1 MC

What is the name of the theory which describes evolution as patterns of rapid first appearances or extinctions followed by periods of little or no change?

  1. Gradualism
  2. Convergence
  3. Adaptive radiation
  4. Punctuated equilibrium
Show Answers Only

\(D\)

Show Worked Solution

→ Punctuated equilibrium describes when species adapt until they reach a stable or equilibrium state.

\(\Rightarrow D\)

BIOLOGY, M3 2017 HSC 5 MC

Four vertebrate forelimbs are shown.
 

In which area of study do these forelimbs support the theory of evolution?

  1. Biogeography
  2. Comparative anatomy
  3. Comparative embryology
  4. Palaeontology
Show Answers Only

\(B\)

Show Worked Solution

→ Comparative anatomy is the practise of comparing body structures of different species.

→ In this example we see how across vertebrate species there are similar forelimb structures, suggesting divergence from a common ancestor. 

\(\Rightarrow B\)

BIOLOGY, M3 2018 HSC 5 MC

Darwin and Wallace proposed the theory of evolution by natural selection.

Punctuated equilibrium differs from this theory in that punctuated equilibrium

  1. occurs in relatively short bursts of rapid change.
  2. requires genetic variation to exist in a population.
  3. is driven by selection pressures in the environment.
  4. discounts the idea that living organisms share a common ancestor.
Show Answers Only

\(A\)

Show Worked Solution

→ Darwin and Wallace’s theory explains how species adapt overtime to their environment.

→ Punctuated equilibrium is a seperate theory which explains how species adapt until they reach a stable stage but this can be effected by rapid evolutionary change to their environment.

\(\Rightarrow A\)

BIOLOGY, M1 2017 HSC 10 MC

A scaled outline of a cell is shown.
 

What is the diameter of the cell?

  1. \( 7\mu \text{m}\)
  2. \( 10\mu \text{m}\)
  3. \( 12\mu \text{m}\)
  4. \( 15\mu \text{m}\)
Show Answers Only

\(B\)

Show Worked Solution

→ The diameter of the cell is 2 times the length of the 5μm scale reference.

\(\Rightarrow B\)

CHEMISTRY, M2 2005 HSC 25a

A student collected a 500 mL sample of water from a local creek for analysis. It was filtered and the filtrate evaporated to dryness. The following data were collected.

\begin{array} {|l|r|}
\hline
\rule{0pt}{2.5ex}\text{Mass of filter paper}\rule[-1ex]{0pt}{0pt} & \text{0.16 g}\\
\hline
\rule{0pt}{2.5ex}\text{Mass of filter paper and solid}\rule[-1ex]{0pt}{0pt} & \text{0.19 g}\\
\hline
\rule{0pt}{2.5ex}\text{Mass of evaporating basin}\rule[-1ex]{0pt}{0pt} & \text{45.33 g}\\
\hline
\rule{0pt}{2.5ex}\text{Mass of basin and solid remaining}\rule[-1ex]{0pt}{0pt} & \text{45.59 g}\\
\hline
\end{array}

Calculate the percentage of total dissolved solids in the creek sample.  (2 marks)

Show Answers Only

\(89.66\% \)

Show Worked Solution

→ The dissolved solids in the creek sample are contained in the evaporating basin and the undissolved solids are contained in the filter paper.

→ Mass of undissolved solids in filter paper = \(0.19-0.16 = 0.03\ \text{g}\)

→ Mass of dissolved solids in evaporating basin = \(45.59-45.33 = 0.26\ \text{g}\)

→ Total Mass of solids = \(0.29\ \text{g}\)

→ Percentage of dissolved solids = \(\dfrac{0.26}{0.29} \times 100 = 89.66\% \) of total solids in the creek sample

CHEMISTRY, M2 2005 HSC 3 MC

The heat of combustion of butan-1-ol \((\ce{C4H10O})\) is 2676 kJ mol\(^{-1}\).

What is the value of the heat of combustion in kJ g\(^{-1}\) ?

  1. 30.41
  2. 36.10
  3. 44.60
  4. 47.79
Show Answers Only

\(B\)

Show Worked Solution

→ The molar mass of butan-1-ol is \(74.12\ \text{gmol}^{-1}\).

→ The heat of combustion in kJ g\(^{-1} = \dfrac{2676}{74.12}=36.10\)  kJ g\(^{-1}\).

\(\Rightarrow B\)

CHEMISTRY, M2 2006 HSC 18

A student studying the mass change that occurs during fermentation added glucose, water and yeast to a flask and stoppered the flask with some cotton wool.

The student measured the mass of the flask daily for seven days. The table shows the data collected.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \ \textit{Day}\ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \textit{Mass}\ \text{(g)}\ \ \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 381.05\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 376.96\\
\hline
\rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} & 373.42\\
\hline
\rule{0pt}{2.5ex} 4 \rule[-1ex]{0pt}{0pt} & 370.44\\
\hline
\rule{0pt}{2.5ex} 5 \rule[-1ex]{0pt}{0pt} & 370.42\\
\hline
\rule{0pt}{2.5ex} 6 \rule[-1ex]{0pt}{0pt} & 370.40\\
\hline
\rule{0pt}{2.5ex} 7 \rule[-1ex]{0pt}{0pt} & 370.39\\
\hline
\end{array}

  1. Calculate the moles of \(\ce{CO2}\) released between days 1 and 7.  (1 mark)

  2. Calculate the mass of glucose that underwent fermentation between days 1 and 7. Include a balanced chemical equation in your answer.  (3 marks)

Show Answers Only

a.    59.37 moles

b.    5395.92 g

Show Worked Solution

a.    \(\ce{MM (CO2) = 12.01 + 2 \times 16.0 = 44.01 g mol^{-1}}\)

\(\ce{Mass CO2 released = 2613.08 g}\)

\[\ce{n(CO2 released) = \frac{2613.08}{44.01} = 59.37 moles}\]  

b.  \(\ce{C6H12O6 \rightarrow 2C2H6O + 2CO2}\)

→ The moles of \(\ce{CO2}\) released and the reaction’s molar ratio can be used to calculate the mass of glucose that underwent fermentation.

→ Molar ratio between glucose and carbon dioxide = \(1:2\)

\(\ce{n(C6H12O6) = \dfrac{1}{2} \times 59.37 = 29.685\ \text{mol}} \)

\(\ce{MM(C6H12O6) = 180.56\ \text{g mol}^{-1}} \)

\(\ce{m(C6H12O6) = 29.685 \times 180.56 = 5395.92\ \text{g}} \)