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PHYSICS, M7 2023 HSC 23b

The James Webb Space Telescope (JWST) is sensitive to wavelengths from  6.0 \(\times\) 10\(^{-7}\) m  to  2.8 \(\times\) 10\(^{-5}\) m.

What is the minimum photon energy that it can detect?  (3 marks)

Show Answers Only

\(E_{\text{min}}= 7.1 \times 10^{-21}\) \( \text{J}\)

Show Worked Solution

→ The minimum photon energy corresponds to the minimum frequency.

→ Minimum frequency occurs at the maximum wavelength as frequency and wavelength are inversely proportional.

\(E_{\text{min}}\) \(=hf\)  
  \(=\dfrac{hc}{ \lambda}\)  
  \(=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.8 \times 10^{-5}} \)  
  \(=7.1 \times 10^{-21}\) \(\text J\)  

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, M5 2023 HSC 14 MC

Planet \(X\) has a mass 4 times that of Earth and a radius 3 times that of Earth. The escape velocity at the surface of Earth is 11.2 km s\(^{-1}\).

What is the escape velocity at the surface of planet \(X\) ?

  1. 8.40 km s\(^{-1}\) 
  2. 9.70 km s\(^{-1}\)
  3. 12.9 km s\(^{-1}\)
  4. 14.9 km s\(^{-1}\)
Show Answers Only

\(C\)

Show Worked Solution

→ The escape velocity of Earth, \( v_\text{esc}=  \sqrt{ \dfrac{2GM}{r}} \)

→ The escape velocity of planet \(X\)

\(v_\text{esc}\) \(=\sqrt{ \dfrac{2G \times 4M}{3r}} \)  
  \(=\dfrac{2}{\sqrt{3}} \times \sqrt{ \dfrac{2GM}{r}} \)  
  \(=\dfrac{2}{\sqrt{3}} \times 11.2 \)  
  \(= 12.9\ \text{km s}^{-1}\)  

 

\(\Rightarrow C\)

PHYSICS, M7 2023 HSC 9 MC

The graph shows the relationship between radiation intensity and wavelength for a black body at 4500 K.
 

Which statement describes the expected difference in the graph for a black body at 4000 K?

  1. Intensity at all wavelengths will be less.
  2. Intensity at all wavelengths will be greater.
  3. The peak intensity will occur at a higher frequency.
  4. The peak intensity will occur at a shorter wavelength.
Show Answers Only

\(A\)

Show Worked Solution

→ The total power output of a black body diminishes with decreasing temperature, resulting in lower intensity across all wavelengths for the 4000 K curve.

\(\Rightarrow A\)

PHYSICS, M5 2023 HSC 8 MC

A ball is launched from a platform at position \(A\) with velocity \(u\). It lands in the position shown.
 


 

The ball could be made to land at position \(B\) by increasing the

  1. velocity \(u\).
  2. launch angle.
  3. mass of the ball.
  4. height of the platform.
Show Answers Only

\(B\)

Show Worked Solution

→ As the launch angle increases, the horizontal velocity of the ball decreases.

→ At a steep launch angle, the ball will travel high into the air but have a short range, thus it could be made to land at position \(B\) as seen in the diagram below.
 

\(\Rightarrow B\)

Mean mark 56%.

PHYSICS, M8 2023 HSC 7 MC

A proton and a neutron travel at the same speed.

Which statement correctly explains the difference between their de Broglie wavelengths?

  1. The proton has a longer wavelength because its mass is greater.
  2. The proton has a shorter wavelength because its mass is smaller.
  3. The neutron has a shorter wavelength because its mass is greater.
  4. The neutron has a longer wavelength because its mass is smaller.
Show Answers Only

\(C\)

Show Worked Solution

→ de Broglie wavelengths are given by the equation  \( \lambda = \dfrac{h}{mv}\)

→ \( \lambda \propto \dfrac{1}{m} \)

∴ Since the mass of a neutron is slightly greater than the mass of a proton, the neutron will have a shorter wavelength.

\(\Rightarrow C\)

PHYSICS, M5 2023 HSC 5 MC

An exoplanet is in an elliptical orbit, moving in the direction shown. The distances between consecutive positions \(P, Q, R\) and \(S\) are equal.
 


 

Between which two points is the exoplanet's travel time the greatest?

  1. \(P\) and \(Q\)
  2. \(Q\) and \(R\)
  3. \(R\) and \(S\)
  4. \(S\) and \(P\)
Show Answers Only

\(D\)

Show Worked Solution

→ As the exoplanet is in an elliptical orbit it will travel the slowest when it is the furthest away from the star.

→ As all distances between the points are equal, the exoplanet will have the greatest travel time between \(S\) and \(P\) where it moves the slowest.

\(\Rightarrow D\)

PHYSICS, M7 2023 HSC 3 MC

A diagram representing a double slit experiment using light is shown.
  

Which of the following best represents the expected pattern on the screen?
 

Show Answers Only

\(C\)

Show Worked Solution

→ Light waves diffract when they pass through a double slit and produce an interference pattern.

→ The central maximum will occur in the centre of the screen where the path difference is 0 and where the path lengths differ by integral wavelengths.

\(\Rightarrow C\)

BIOLOGY, M4 EQ-Bank 24

Explain ONE example of a mutual symbiotic relationship.   (3 marks)

Show Answers Only

→ A mutual symbiotic relationship is one in which both organisms involved benefit.

→ One example would be the relationship between kangaroos and their gut bacteria.

→ Kangaroos cannot individually produce the enzyme cellulase that is able to digest cellulose, however, the bacteria that exist in their gut are able to.

→ This means that the kangaroo can break down the cell wall of plant cells (made of cellulose) and access the nutrients inside, while the bacteria can also obtain a food supply from the kangaroo as well as a safe place to live.

Other answers could include

→ Termites have protozoans that can digest cellulose.

→ Birds that sit atop cows eat ticks that may get stuck on the cow

Show Worked Solution

→ A mutual symbiotic relationship is one in which both organisms involved benefit.

→ One example would be the relationship between kangaroos and their gut bacteria.

→ Kangaroos cannot individually produce the enzyme cellulase that is able to digest cellulose, however, the bacteria that exist in their gut are able to.

→ This means that the kangaroo can break down the cell wall of plant cells (made of cellulose) and access the nutrients inside, while the bacteria can also obtain a food supply from the kangaroo as well as a safe place to live.

Other answers could include

→ Termites have protozoans that can digest cellulose.

→ Birds that sit atop cows eat ticks that may get stuck on the cows.

BIOLOGY, M4 EQ-Bank 7 MC

A tapeworm is an animal that gains it's nutrients while living inside the host, often making the host seriously ill.

What kind of symbiotic relationship is this an example of?

  1. Predation.
  2. Parasitism.
  3. Mutualism.
  4. Commensalism.
Show Answers Only

\(B\)

Show Worked Solution

→ A tapeworm is an example of a parasite, it lives at the expense of a host.

\(\Rightarrow B\)

BIOLOGY, M5 2023 HSC 23

The following graph outlines some hormonal changes during pregnancy.
 

Complete the table for TWO of the hormones graphed.  (4 marks)

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Hormone name} \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \textit{Function in pregnancy} \quad \quad & \quad \textit{Trimester where}\quad \\
\text{} \rule[-1.5ex]{0pt}{0pt} & \text{} & \textit{peak occurs}\\
\hline
\text{} & \text{} & \text{} \\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{} \\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\hline
\text{} & \text{} & \text{} \\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{} \\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\text{} & \text{} & \text{}\\
\hline
\end{array}

Show Answers Only

Answers could include two of the following:

Hormone Name Function in Pregnancy Trimester where Peak Occurs

Progesterone

Initially, progesterone thickens the uterine lining to support the developing foetus. It is also produced in the placenta, helping to prevent lactation and contractions until full development. 

It will slowly increase until a peak in the third trimester when the placenta is fully developed. 

Oestrogen

Oestrogen is needed for stimulation of the development of the placenta, increasing the size of the uterus and helping the development of foetal organs.

Oestrogen will also gradually increase over time and peak in the final trimester.

Human Chorionic Gonadotrophin (hCG)

As soon as the blastocyst is implanted in the body, hCG will keep the corpus luteum active so it can produce oestrogen and progesterone.

hCG will peak in the first trimester.
Show Worked Solution

Answers could include two of the following:

Hormone Name Function in Pregnancy Trimester where Peak Occurs

Progesterone

Initially, progesterone thickens the uterine lining to support the developing foetus. It is also produced in the placenta, helping to prevent lactation and contractions until full development. 

It will slowly increase until a peak in the third trimester when the placenta is fully developed. 

Oestrogen

Oestrogen is needed for stimulation of the development of the placenta, increasing the size of the uterus and helping the development of foetal organs.

Oestrogen will also gradually increase over time and peak in the final trimester.

Human Chorionic Gonadotrophin (hCG)

As soon as the blastocyst is implanted in the body, hCG will keep the corpus luteum active so it can produce oestrogen and progesterone.

hCG will peak in the first trimester.

BIOLOGY, M7 2023 HSC 22a

Describe how phagocytes help protect against pathogens.  (2 marks)

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→ Phagocytes are white blood cells which can identify and engulf foreign material.

→ Upon detection of foreign material, phagocytes can engulf them forming a vesicle which fuses with a lysosome and allows them to be digested.

Show Worked Solution

→ Phagocytes are white blood cells which identify and engulf foreign material.

→ Upon detection of foreign material, phagocytes can engulf them forming a vesicle which fuses with a lysosome and allows them to be digested.

BIOLOGY, M6 2023 HSC 20 MC

The diagram shows the karyotype of a normal female Tasmanian devil cell and the karyotype of a Tasmanian devil facial tumour cell.

M1, M2, M3 and M4 are marker chromosomes. These are chromosomes of unknown origin additional to the normal chromosomes found in the cells of Tasmanian devil facial tumour disease.

What can be deduced from the karyotypes?

  1. The karyotype of the tumour cells shows trisomy.
  2. The karyotype of tumour cells contains multiple chromosomal inversions.
  3. The karyotype of tumour cells contains both chromosomal insertions and deletions.
  4. The karyotype of the tumour cells contains more chromosomes than the karyotype of the normal Tasmanian devil cells.
Show Answers Only

\(C\)

Show Worked Solution

→ The karyotype of the tumour cells contains a deletion of the chromosome set 2 and addition of marker chromosomes.

\(\Rightarrow C\)

BIOLOGY, M6 2023 HSC 19 MC

The following are the five steps in the process of gene cloning.

  1. Selection of organisms containing recombinant DNA sequences
  2. Creation of recombinant DNA joined using DNA ligase
  3. Introduction of recombinant DNA into host organism
  4. Extraction and amplification of DNA to be cloned
  5. Choice of host organism and cloning vector

Which is the correct order for this process?

  1. 5, 2, 1, 3, 4
  2. 5, 4, 2, 3, 1
  3. 3, 2, 4, 5, 1
  4. 4, 5, 1, 2, 3
Show Answers Only

\(B\)

Show Worked Solution

By Elimination:

→ The choice of host organism and cloning vector (5) must come first (Eliminate C and D).

→ The extraction/amplification of the DNA segment of interest must come before the creation of the recombinant DNA (Eliminate A).

\(\Rightarrow B\)

BIOLOGY, M5 2023 HSC 17 MC

In humans, blood groups are produced by combinations of three alleles \(I^A, I^B\) and \(i\).

\begin{array}{|c|l|}
\hline \rule{0pt}{2.5ex}\textit{Blood type} \rule[-1ex]{0pt}{0pt}& \textit{Genotype(s)}\\
\hline \rule{0pt}{2.5ex}\text{A} \rule[-1ex]{0pt}{0pt}& I^A I^A \  \text{or}\  I^A i \\
\hline \rule{0pt}{2.5ex}\text{B} \rule[-1ex]{0pt}{0pt}& I^B I^B \ \text{or}\  I^B i \\
\hline \rule{0pt}{2.5ex}\text{AB} \rule[-1ex]{0pt}{0pt}& I^A I^B \\
\hline \rule{0pt}{2.5ex}\text{O} & i i \\
\hline
\end{array}

A mother has blood type \(O\) and her child has blood type \(A\).

Which of the following includes all possible genotype(s) of the father?

  1. \(I^A I^A\)
  2. \(I^A I^A\)  or  \(I^A i\)
  3. \(I^A I^A\)  or  \(I^A i\)  or  \(I^A I^B\)
  4. \(I^A I^A\)  or  \(I^A i\)  or  \(I^A I^B\)  or  \(i i\)
Show Answers Only

\(C\)

Show Worked Solution

→ To have blood type \(A\), you can have either \(I^A I^A\) or \(I^A i\) genotypes.

→ As the mother has genotype \(i  i\), the child with blood type \(A\) can have a father that has a genotype with at least one \(I^A\) allele.

→ Option C includes every possible genotype with at least one \(I^A\) allele.

\(\Rightarrow C\)

Mean mark 56%.

BIOLOGY, M7 2023 HSC 15 MC

A pharmaceutical was tested for its effectiveness in treating a viral infection. A symptom severity score of zero indicates no symptoms.
 

What conclusion can be drawn from the graph?

  1. The pharmaceutical tested had no antiviral properties.
  2. The pharmaceutical tested did not reduce the severity of symptoms.
  3. The pharmaceutical tested increased symptoms in patients in the study.
  4. Pharmaceuticals do not reduce symptoms in patients suffering from viral infections.
Show Answers Only

\(B\)

Show Worked Solution

By Elimination:

→ Option C is an incorrect statement as symptoms did decrease (Eliminate C).

→ This test cannot be used to make a generalised statement about the ineffectiveness of all pharmaceuticals (Eliminate D).

→ The tested pharmaceutical may have antiviral properties against other viruses (Eliminate A).

→ The similarity between administering the antiviral and not treating someone were very similar, and therefore the drug did not reduce the severity of symptoms.

\(\Rightarrow B\)

♦ Mean mark 49%.

BIOLOGY, M8 2023 HSC 12 MC

The following diagram shows the regulation of blood calcium \(\ce{(Ca^{2+})}\) levels in the body.
 

 

A blood test shows a person has a blood calcium level of 6 mg/100 mL.

What will occur in the body to restore homeostasis?

  1. Calcium will be deposited in the bones.
  2. Calcium will be removed from the bone matrix.
  3. The thyroid gland will release the hormone thyroxine.
  4. The thyroid gland will release the hormone calcitonin.
Show Answers Only

\(B\)

Show Worked Solution

→ 6 mg/100 mL is lower than the homeostasis concentration range.

→ This means that the parathyroid gland will release PTH causing \(\ce{Ca^{2+}}\) ions to be removed from the bone matrix.

\(\Rightarrow B\)

Mean mark 58%.

BIOLOGY, M5 2023 HSC 11 MC

The diagram shows part of a protein molecule.
 

Which row of the table is correct?

\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}X \rule[-1ex]{0pt}{0pt}& \textit{Level of protein structure} \\
\hline \rule{0pt}{2.5ex} \text {Amino Acid} \rule[-1ex]{0pt}{0pt}& \text {Primary} \\
\hline \rule{0pt}{2.5ex} \text {Amino Acid} \rule[-1ex]{0pt}{0pt}& \text {Secondary} \\
\hline \rule{0pt}{2.5ex} \text {Polypeptide} \rule[-1ex]{0pt}{0pt}& \text {Primary} \\
\hline \rule{0pt}{2.5ex} \text {Polypeptide} \rule[-1ex]{0pt}{0pt}& \text {Secondary} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ The diagram shows that \(X\) is an amino acid which is part of the polypeptide chain, and therefore the level of protein structure is primary.

\(\Rightarrow A\)

BIOLOGY, M6 2023 HSC 10 MC

The diagram shows a bacterial cell that will be used to produce a protein.
 

Which component would be inserted into gap \(X\) ?

  1. A cell
  2. A gene
  3. An amino acid
  4. A chromosome
Show Answers Only

\(B\)

Show Worked Solution

→ A gene will be inserted into the bacteria's DNA so that it can produce the associated protein of interest.

\(\Rightarrow B\)

BIOLOGY, M5 2023 HSC 9 MC

Sheep have 54 chromosomes, while goats have 60 chromosomes. The hybrid offspring of a sheep-goat pairing is called a geep.

During fertilisation, an egg from the sheep is fertilised by a sperm from the goat, resulting in a geep.

Which of the following will be correct for the geep?

  1. \(n\) = 29 chromosomes
  2.  \(n\) = 52 chromosomes
  3. \(2n\) = 57 chromosomes
  4. \(2n\) = 114 chromosomes
Show Answers Only

\(C\)

Show Worked Solution

→ \(54 + 60 = 114\)

→ \(114\ ÷\ 2 = 57\)

\(\Rightarrow C\)

BIOLOGY, M7 2023 HSC 8 MC

Neutrophils and T-cells are cells of the human immune system. After an infection, the concentration of these types of cells in infected tissue was plotted as a function of time.
 

Based on the data provided, neutrophils are part of which human immune system?

  1. Acquired
  2. Adaptive
  3. Innate
  4. Primary
Show Answers Only

\(C\)

Show Worked Solution

→ The relatively quick spike in neutrophil concentration when compared to the T-cells (which are part of the specific immune response) would indicate they are part of the innate/non-specific immune response.

\(\Rightarrow C\)

BIOLOGY, M6 2023 HSC 4 MC

The following diagram shows a summary of the process of artificial pollination.
 

The purpose of Process \(B\) is to

  1. produce seeds.
  2. collect the pollen.
  3. fertilise the flower.
  4. prevent self-pollination.
Show Answers Only

\(D\)

Show Worked Solution

→ Process \(B\) shows removal of the anthers of the plant.

→ The anthers contain the pollen and male gametes of the plant, hence by removing them, self-pollination is prevented.

\(\Rightarrow D\)

BIOLOGY, M8 2023 HSC 1 MC

A cochlear implant is a device that is used to assist with hearing loss.

What does the cochlear implant electrode array stimulate?

  1. Hairs
  2. Ossicles
  3. Oval window
  4. Auditory nerve
Show Answers Only

\(D\)

Show Worked Solution

→ The cochlear implant is able to bypass a dysfunctional cochlear and directly stimulate the auditory nerve.

\(\Rightarrow D\)

Vectors, EXT2 V1 2023 HSC 15b

On the triangular pyramid \(A B C D, L\) is the midpoint of \(A B, M\) is the midpoint of \(A C, N\) is the midpoint of \(A D, P\) is the midpoint of \(C D, Q\) is the midpoint of \(B D\) and \(R\) is the midpoint of \(B C\).
 

Let  \(\underset{\sim}{b}=\overrightarrow{A B}, \underset{\sim}{c}=\overrightarrow{A C}\)  and  \(\underset{\sim}{d}=\overrightarrow{A D}\).

  1. Show that \(\overrightarrow{L P}=\dfrac{1}{2}(-\underset{\sim}{b}+\underset{\sim}{c}+\underset{\sim}{d})\).  (1 mark)

  2. It can be shown that

\(\overrightarrow{M Q}=\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})\)  and

\(\overrightarrow{N R}=\dfrac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\).   (Do NOT prove these.) 

  1. Prove that

  \( \Big{|}\overrightarrow{A B}\Big{|}^2+\Big{|}\overrightarrow{A C}\Big{|}^2+\Big{|}\overrightarrow{A D}\Big{|}^2+\Big{|}\overrightarrow{B C}\Big{|}^2+\Big{|}\overrightarrow{B D}\Big{|}^2+\Big{|}\overrightarrow{C D}\Big{|}^2 \)

\(=4\left(\Big{|}\overrightarrow{L P}\Big{|}^2+\Big{|}\overrightarrow{M Q}\Big{|}^2+\Big{|}\overrightarrow{N R}\Big{|}^2\right)\)  (3 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{Proof (See Worked Solutions)}\)
Show Worked Solution
i.     \(\overrightarrow{LP}\) \(=\overrightarrow{LA}+\overrightarrow{AC}+\overrightarrow{CP} \)
    \(=\dfrac{1}{2} \overrightarrow{BA}+\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CD} \)
    \(=-\dfrac{1}{2} \underset{\sim}b +\underset{\sim}c + \dfrac{1}{2}(\underset{\sim}d-\underset{\sim}c) \)
    \(=-\dfrac{1}{2} \underset{\sim}b +\underset{\sim}c + \dfrac{1}{2}\underset{\sim}d-\dfrac{1}{2}\underset{\sim}c\)
    \(=\dfrac{1}{2} (-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) \)

 

ii.   \(\overrightarrow{M Q}=\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})\)

\(\overrightarrow{N R}=\dfrac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\).

\(\text{RHS}\) \(=4\left(\Big{|}\overrightarrow{L P}\Big{|}^2+\Big{|}\overrightarrow{M Q}\Big{|}^2+\Big{|}\overrightarrow{N R}\Big{|}^2\right)\)  
  \(=4\left(\overrightarrow{L P}\cdot \overrightarrow{L P} +\overrightarrow{MQ}\cdot \overrightarrow{MQ} +\overrightarrow{NR}\cdot \overrightarrow{NR}\right)\)  
  \(=4\Bigg{(}\dfrac{1}{4}(-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) \cdot(-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) +  \)  
  \( \dfrac{1}{4}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) \cdot(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) +  \)  
  \( \dfrac{1}{4}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) \cdot(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) \Bigg{)}\)  
  \(=(\underset{\sim}c+(\underset{\sim}d-\underset{\sim}b)) \cdot(\underset{\sim}c+(\underset{\sim}d-\underset{\sim}b))+(\underset{\sim}d+(\underset{\sim}b-\underset{\sim}c))\cdot (\underset{\sim}d+(\underset{\sim}b-\underset{\sim}c)) + \)  
  \( (\underset{\sim}b+(\underset{\sim}c-\underset{\sim}d))\cdot (\underset{\sim}b+(\underset{\sim}c-\underset{\sim}d))  \)  
  \(=|\underset{\sim}c|^2+2\underset{\sim}c(\underset{\sim}d-\underset{\sim}b) + |\underset{\sim}d-\underset{\sim}b|^2 + \)  
  \(|\underset{\sim}d|^2+2\underset{\sim}d(\underset{\sim}b-\underset{\sim}c) + |\underset{\sim}b-\underset{\sim}c|^2 + \)  
  \( |\underset{\sim}b|^2+2\underset{\sim}b(\underset{\sim}c-\underset{\sim}d) + |\underset{\sim}c-\underset{\sim}d|^2 \)  
  \(=|\underset{\sim}b|^2+|\underset{\sim}c|^2+|\underset{\sim}d|^2+|\underset{\sim}b-\underset{\sim}c|^2+|\underset{\sim}d-\underset{\sim}b|^2+|\underset{\sim}c-\underset{\sim}d|^2 + \)  
  \( 2(\underset{\sim}c \cdot\underset{\sim}d-\underset{\sim}c \cdot\underset{\sim}b+\underset{\sim}d \cdot\underset{\sim}b-\underset{\sim}d \cdot\underset{\sim}c+\underset{\sim}b \cdot\underset{\sim}c-\underset{\sim}b \cdot\underset{\sim}d) \)  
  \(=\Big{|}\overrightarrow{A B}\Big{|}^2+\Big{|}\overrightarrow{A C}\Big{|}^2+\Big{|}\overrightarrow{A D}\Big{|}^2+\Big{|}\overrightarrow{B C}\Big{|}^2+\Big{|}\overrightarrow{B D}\Big{|}^2+\Big{|}\overrightarrow{C D}\Big{|}^2 + 0\)  
  \(=\ \text{LHS} \)  

♦♦ Mean mark (ii) 34%.

BIOLOGY, M3 EQ-Bank 23

Explain how the finches on the Galapagos Islands support Darwin's theory of Natural Selection.   (5 marks)

Show Answers Only

→ The diversity between finches that exist on seperate islands in the Galapagos can be explained by Darwin’s theory of evolution and Natural Selection.

→ It is believed that a few million years ago, one species of finch migrated to the rocky Galapagos from Central or South America. This finch species can be considered the common ancestor.

→ As the finch species spread out between the islands, the ecological niches of each island placed different selection pressures that pushed the two finch populations in different directions.

→ Some random variations that provided a breeding or survival advantage became more prevalent after generations because finches with the advantage reproduced more.

→ This process resulted in the development of different finch species between the islands, each with varying beak sizes and shapes which were better suited to the different diets on each island. For example, one species that eat grubs developed longer beaks to poke into holes and extract the grubs.

→ In this way, the varying finch species observed by Darwin provides evidence for his theory of Natural Selection.

Show Worked Solution

→ The diversity between finches that exist on seperate islands in the Galapagos can be explained by Darwin’s theory of evolution and Natural Selection.

→ It is believed that a few million years ago, one species of finch migrated to the rocky Galapagos from Central or South America. This finch species can be considered the common ancestor.

→ As the finch species spread out between the islands, the ecological niches of each island placed different selection pressures that pushed the two finch populations in different directions.

→ Some random variations that provided a breeding or survival advantage became more prevalent after generations because finches with the advantage reproduced more.

→ This process resulted in the development of different finch species between the islands, each with varying beak sizes and shapes which were better suited to the different diets on each island. For example, one species that eat grubs developed longer beaks to poke into holes and extract the grubs.

→ In this way, the varying finch species observed by Darwin provides evidence for his theory of Natural Selection.

BIOLOGY, M3 EQ-Bank 5-6 MC

The following diagram shows features which relate to the goanna.
 

Question 5

Which of the following features is a behavioural adaptation?

  1. Skin colour related to the environment.
  2. The base of the tail becoming thick to store fat.
  3. Long, sharp claws that are used for tearing at prey.
  4. Rapid movements of the gular region to cool down.

 
Question 6

Which of the following features is a physiological adaptation?

  1. Skin colour related to the environment.
  2. The base of the tail becoming thick to store fat.
  3. Long, sharp claws that are used for tearing at prey.
  4. Rapid movements of the gular region to cool down.
Show Answers Only

\(\text{Question 5:}\ D\)

\(\text{Question 6:}\ B\)

Show Worked Solution

\(\text{Question 5}\)

→ A behavioural adaptation is an action that is carried out by the organism. This would be the action of moving the gular region to cool down.

\(\Rightarrow D\)
 

\(\text{Question 5}\)

→ A physiological adaptation is a process which occurs within the organism. This would be the thickening of the tail to store fat.

\(\Rightarrow B\) 

BIOLOGY, M2 EQ-Bank 21

Describe two key differences between colonial and multicellular organisms.   (2 marks)

Show Answers Only

→ Colonial organisms do not exhibit specialisation. Each cell in the colony is the same but they are connected in such a way to enhance processes such as food gathering and production.

→ Multicellular organisms do consist of specialised cells, each of which will carry out one specific function which will benefit the whole organism.

Show Worked Solution

→ Colonial organisms do not exhibit specialisation. Each cell in the colony is the same but they are connected in such a way to enhance processes such as food gathering and production.

→ Multicellular organisms do consist of specialised cells, each of which will carry out one specific function which will benefit the whole organism.

BIOLOGY, M2 EQ-Bank 4 MC

What component of a mature red blood cell relates to its function?

  1. Its biconcave shape allows them to move quicker.
  2. Its biconcave shape allows them to more effectively carry antibodies.
  3. The reduction of the nucleus allows for more effective duplication via mitosis.
  4. The reduction of the nucleus allow them to carry more haemoglobin and hence more \( \ce{O_2}\).
Show Answers Only

\(D\)

Show Worked Solution

By Elimination

→ The biconcave shape allows for greater manoeuvrability, especially through tight areas like capillaries (Eliminate A and B).

→ The reduction of a large nucleus allows more space for haemoglobin molecules (Eliminate C).

\(\Rightarrow D\)

BIOLOGY, M2 EQ-Bank 3 MC

Which of the following does NOT occur in the mouth?

  1. Lubrication of food.
  2. Breakdown of starch.
  3. Beginning of protein digestion.
  4. Breakdown of food into smaller fragments.
Show Answers Only

\(C\)

Show Worked Solution

By Elimination

→ Saliva in the mouth lubricates food to help in pass through the oesophagus easier (Eliminate A).

→ The enzyme amylase, found in saliva, is able to breakdown starch into maltose (Eliminate B).

→ Chewing breaks down food into smaller pieces which both allows for easier movement through the rest of the digestive system and increasing surface area to allow enzymes to work more effectively (Eliminate D).

→ The breakdown of proteins begins in the stomach as it contains the adequate enzymes.

\(\Rightarrow C\)

BIOLOGY, M1 EQ-Bank 2 MC

Which of the following best describes the fluid-mosaic model of the cell membrane?

  1. Phospholipid bilayer with a protein coating.
  2. Phospholipid bilayer with embedded proteins.
  3. Phospholipid monolayer with protein coating.
  4. Phospholipid monolayer with embedded proteins.
Show Answers Only

\(B\) 

Show Worked Solution

By Elimination

→ The model has it’s phospholipids arranged in a bilayer, with the heads facing outwards and the tails facing each other inward (Eliminate C and D).

→ The model’s proteins do not form a coat around the membrane, but are rather embedded within different locations within the bilayer depending on their function (Eliminate A).

\(\Rightarrow B\)

Calculus, EXT2 C1 2023 HSC 15a

  1. Let  \(J_n= {\displaystyle \int_0^{\frac{\pi}{2}}} \sin ^n \theta \ d \theta\)  where \(n \geq 0\) is an integer. 
  2. Show that  \(J_n=\dfrac{n-1}{n} J_{n-2}\)  for all integers \(n \geq 2\).  (3 marks)

  3. Let  \(I_n={\displaystyle \int_0^1 x^n(1-x)^n}\, dx \)  where \( n \)  is a positive integer.
  4. By using the substitution  \(x=\sin ^2 \theta\), or otherwise,
  5. show that  \( I_n=\dfrac{1}{2^{2 n}} {\displaystyle \int_0^{\frac{\pi}{2}}} \sin ^{2 n+1} \theta \ d \theta \).  (4 marks)

  6. Hence, or otherwise, show that  \(I_n=\dfrac{n}{4 n+2} I_{n-1}\), for all integers  \(n \geq 1\).   (2 marks)
Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution
i.    \(J_n\) \(= \displaystyle \int_0^{\frac{\pi}{2}} \sin^{n}\, \theta\, d\theta \)  
    \(= \displaystyle \int_0^{\frac{\pi}{2}} \sin^{n-1}\, \theta \cdot \sin\,\theta\, d\theta \)  

 
\(\text{Integrating by parts:} \)

\(u\) \(=\sin^{n-1}\, \theta \) \(u^{′}=(n-1) \cos\, \theta \cdot \sin^{n-2}\,\theta \)
\(v\) \(=-\cos\,\theta \) \(v^{′}=\sin \, \theta\)
\(J_n\) \(= \big{[} -\cos\,\theta \cdot \sin^{n-1}\,\theta \big{]}_0^{\frac{\pi}{2}} + (n-1) \displaystyle \int_0^{\frac{\pi}{2}} \cos^{2} \theta \cdot \sin^{n-2}\theta\, d\theta\)  
  \(=(n-1) \displaystyle \int_0^{\frac{\pi}{2}} (1-\sin^{2} \theta) \sin^{n-2}\theta\, d\theta\)  
  \(=(n-1) \displaystyle \int_0^{\frac{\pi}{2}} \sin^{n-2}\theta-\sin^{n}\theta\, d\theta\)  
  \(=(n-1)J_{n-2}-(n-1)J_n \)  
\(J_n+(n-1)J_n \) \(=(n-1)J_{n-2} \)  
\(J_n(1+n-1) \) \(=(n-1) J_{n-2} \)  
\(J_n\) \(= \dfrac{n-1}{n} J_{n-2} \)  

 

ii.   \(I_n= \displaystyle \int_0^1 x^{n}(1-x)^n\,dx \)

\(\text{Let}\ \ x=\sin^{2}\,\theta \)

\(\dfrac{dx}{d\theta}=2\sin\,\theta\, \cos\,\theta \ \Rightarrow \ dx=2\sin\,\theta \,\cos\,\theta\,d\theta \)

\(\text{When}\ \ x=1, \ \theta=\dfrac{\pi}{2}; \ \ x=0, \ \theta=0 \)

\(I_n\) \(= \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n}\,\theta(1-\sin^{2}\,\theta)^{n} \cdot 2\sin\,\theta \,\cos\,\theta\,d\theta \)  
  \(= 2 \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}\,\theta \cdot \cos^{2n+1}\,\theta\,d\theta \)  
  \(=\dfrac{2}{2^{2n+1}} \displaystyle \int_0^{\frac{\pi}{2}} (2\sin\,\theta\,\cos\,\theta)^{2n+1} d\theta \)  
  \(=\dfrac{1}{2^{2n}} \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}(2\theta)\, d\theta \)  

 
\(\text{Let}\ \ u=2\theta \)

\(\dfrac{du}{d\theta}=2\ \ \Rightarrow\ \ \dfrac{du}{2}=d\theta\)

\(\text{When}\ \ \theta=\dfrac{\pi}{2}, \ u=\pi; \ \theta=0, \ u=0 \)

\(I_n=\dfrac{1}{2^{2n+1}} \displaystyle \int_0^{n} \sin^{2n+1} u\, du\)

 
\(\text{Since}\ \ \sin^{2n+1} u\ \ \text{is symmetrical about}\ \ u=\dfrac{\pi}{2} \)

\(\Rightarrow \displaystyle \int_0^{\pi} \sin^{2n+1} u\,du=2\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} u\,du \)

\(I_n\) \(= \dfrac{1}{2^{2n+1}} \cdot 2\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} u\,du \)  
  \(= \dfrac{1}{2^{2n}} \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} u\,du \)  

 
\(\text{Which can be expressed as:}\)

\(I_n= \dfrac{1}{2^{2n}}  \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} \theta\, d\theta \)
 

♦♦ Mean mark (ii) 34%.
iii.    \(I_n\) \(=\dfrac{1}{2^{2n}}  \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} \theta\, d\theta \)
    \(=\dfrac{1}{2^{2n}} J_{2n+1} \ \ \ \text{(by definition)}\)
    \(=\dfrac{1}{2^{2n}} \cdot \underbrace{\dfrac{2n}{2n+1} \cdot J_{2n+1}}_{\text{using part i}}  \)
    \(= \dfrac{n}{2(2n+1)} \cdot \dfrac{2^2}{2^{2n}}\cdot J_{2n-1} \)
    \(= \dfrac{n}{4n+2} \cdot \underbrace{\dfrac{1}{2^{2n-2}}\cdot J_{2n-1}}_{=I_{n-1} \ \ \text{(using part ii)}} \)
    \(=\dfrac{n}{4 n+2} I_{n-1}\)
♦♦ Mean mark (iii) 28%.

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

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

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

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

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

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

\(2^k \geq k^2-2 \ \ \ …\ (*) \)
 

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

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

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

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

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

Calculus, EXT2 C1 2023 HSC 13a

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

Show Answers Only

\(I=\sqrt{5-4x-x^2} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)

Show Worked Solution

\(I\) \(={\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(= \dfrac{1}{2} {\displaystyle \int \frac{2-2x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(=\dfrac{1}{2} {\displaystyle \int \frac{-4-2x}{\sqrt{5-4 x-x^2}}\ dx} + \dfrac{1}{2} {\displaystyle \int \frac{6}{\sqrt{5-4 x-x^2}}\ dx}\)  

 
\(\text{Let}\ \ u=5-4x-x^2 \)

\( \dfrac{du}{dx}=-4-2x\ \ \Rightarrow\ \ du=-4-2x\ dx \)

\(I\) \(= \dfrac{1}{2} {\displaystyle \int u^{-\frac{1}{2}}\ du} + {\displaystyle 3 \int \frac{1}{\sqrt{9-(x+2)^2}}\ dx}\)  
  \(= u^{\frac{1}{2}} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  
  \(=  \sqrt{5-4x-x^2}+ 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  

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.    \(P(z)=z^4-3 z^3+c z^2+d z-30\)

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

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

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

 
\(\text{Product of roots:} \)

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

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

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

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

Complex Numbers, EXT2 N2 2023 HSC 12d

Find the cube roots of  \(2-2 i\). Give your answer in exponential form.  (3 marks)

Show Answers Only

\(\sqrt[3]{2-2i}= \sqrt2 e^{- \frac{i\pi}{12}}, \sqrt2 e^{\frac{i7\pi}{12}}, \sqrt2 e^{- \frac{i3\pi}{4}} \)

Show Worked Solution

\(\text{Express}\ \ 2-2i\ \ \text{in exponential form:}\)
 

\(\big{|}2-2i\big{|}=2\sqrt2 \)

\(\arg(2-2i) = – \dfrac{\pi}{4} \)

\(2-2i=2\sqrt2 e^{-\frac{i\pi}{4}} \)
 

\(\text{Find}\ \ z=re^{i\theta},\ \ \text{where}\ \ z^3=r^3e^{i 3\theta}=2\sqrt2 e^{-\frac{i\pi}{4}} \)

\(r^3=2\sqrt2\ \ \Rightarrow\ \ r=\sqrt2 \)

\( i3\theta\) \(= -\dfrac{i\pi}{4} \)  
\(\theta_1\) \(=-\dfrac{\pi}{12} \)  

 
\(\text{Since roots are found symmetrically around Argand diagram:}\)

\(\theta_2=-\dfrac{\pi}{12}+\dfrac{2\pi}{3}=\dfrac{7\pi}{12} \)

\(\theta_3=-\dfrac{\pi}{12}-\dfrac{2\pi}{3}=-\dfrac{3\pi}{4} \)

\(\therefore \sqrt[3]{2-2i}= \sqrt2 e^{- \frac{i\pi}{12}}, \sqrt2 e^{\frac{i7\pi}{12}}, \sqrt2 e^{- \frac{i3\pi}{4}} \)

Mechanics, EXT2 M1 2023 HSC 12c

An object with mass \(m\) kilograms slides down a smooth inclined plane with velocity \( \underset{\sim}{v}(t)\), where \(t\) is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle \(\theta\) with the horizontal, as shown in the diagram. The normal reaction force is \(\underset{\sim}{R}\). The acceleration due to gravity is \(\underset{\sim}{g}\) and has magnitude \(g\). No other forces act on the object.

The vectors \(\underset{\sim}{i}\) and \( \underset{\sim}{j} \) are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram.
 

  1. Show that the resultant force on the object is  \(\underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i}\).  (2 marks)

  2. Given that the object is initially at rest, find its velocity \(\underset{\sim}{v}(t)\) in terms of \(g\), \(\theta, t\) and \(\underset{\sim}{i}\).  (2 marks)

Show Answers Only

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

ii.   \(\underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)

Show Worked Solution

i.       
         

\(\text{Resolving forces in}\ \underset{\sim}{j} \ \text{direction:} \)

\( {\underset{\sim}{F}}_\underset{\sim}{j} = \underset{\sim}{R} + m\underset{\sim}{g}\ \cos \theta = 0\ \ \text{(in equilibrium)} \)

\(\text{Resolving forces in}\ \underset{\sim}{i} \ \text{direction:} \)

\( {\underset{\sim}{F}}_\underset{\sim}{i} = -m\underset{\sim}{g} \ \sin \theta \ \ \ \text{(down slope)} \)

\(\therefore \text{Resultant force:}\ \ \underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i} \)
 

♦ Mean mark (i) 50%.

ii.   \(\text{Using}\ \ \underset{\sim}{F}=m \underset{\sim}{a}: \)

\(m \underset{\sim}{a}\) \(=-mg\ \sin \theta \ \underset{\sim}{i} \)  
\(\underset{\sim}{a}\) \(=-g\ \sin \theta \ \underset{\sim}{i} \)  
\(\underset{\sim}{v}\) \(= \displaystyle \int \underset{\sim}{a}\ dt \)  
  \(=-gt\ \sin \theta +c \)  

 
\(\text{When}\ \ t=0,\ \ \underset{\sim}{v}=0\ \ \Rightarrow \ \ c=0 \)

\(\therefore \underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)

Proof, EXT2 P1 2023 HSC 12b

Prove that for all real numbers \(x\) and \(y\), where \(x^2+y^2 \neq 0\),

\(\dfrac{(x+y)^2}{x^2+y^2} \leq 2 \text {. }\)  (2 marks)

Show Answers Only

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

Show Worked Solution
\((x-y)^2 \) \(\geq 0 \)  
\(x^2-2xy+y^2 \) \(\geq 0\)  
\(x^2+y^2\) \(\geq 2xy \)  
\( \dfrac{2xy}{x^2+y^2}\) \( \leq 1\ \text{… (1)}\)  

 

\(\dfrac{(x+y)^2}{x^2+y^2}\) \(=\dfrac{x^2+2xy+y^2}{x^2+y^2}\)  
  \(=\dfrac{x^2+y^2}{x^2+y^2}+\underbrace{\dfrac{2xy}{x^2+y^2}}_{\text{see (1) above}} \)  
  \(\leq 1+1\)  
  \(\leq 2\)  

Proof, EXT2 P1 2023 HSC 12a

Prove that \(\sqrt{23}\) is irrational.  (3 marks)

Show Answers Only

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

Show Worked Solution

\(\text{Proof by contradiction:} \)

\(\text{Assume}\ \sqrt{23}\ \text{is rational} \)

\( \sqrt{23} = \dfrac{p}{q}\ \ \text{where}\ p, q \in \mathbb{Z}\ \ \text{with no common factor except 1} \)

\(23\) \(= \dfrac{p^2}{q^2} \)  
\(23q^2\) \(=p^2\)  

 
\(\Rightarrow \text{23 is a factor of}\ p^2 \)

\(\Rightarrow \text{23 is a factor of}\ p \)
 

\( \exists k \in \mathbb{Z}\ \ \text{such that}\ \ p=23k \)

\(23q^2\) \(=(23k)^2 \)  
\(q^2\) \(=23k^2 \)  

 
\(\Rightarrow \text{23 is a factor of}\ q^2 \)

\(\Rightarrow \text{23 is a factor of}\ q \)

\(\therefore \text{HCF}\ \geq 23 \)

\(\therefore \text{By contradiction,}\ \sqrt{23}\ \text{is rational} \)

Measurement, STD1 M2 2023 HSC 27

When it is 10 am in Town \(A\), the time in Town \(B\) is 2 pm on the same day.

Joe lives in Town \(A\) and wishes to watch a live soccer game played in Town \(B\). The game commences at 11:30 am, Town \(B\) local time, and lasts for 2 hours.

What time will it be in Town \(A\) when the game finishes?  (2 marks)

Show Answers Only

\(9:30\ \text{am}\)

Show Worked Solution

\(11:30-4+2=9:30\ \text{am}\)

\(\therefore\ \text{The game finishes in Town } A\ \text{at}\ 9:30\ \text{am.}\)

Vectors, EXT2 V1 2023 HSC 11d

The quadrilaterals \(A B C D\) and \(A B E F\) are parallelograms.

By considering \(\overrightarrow{A B}\), show that \(C D F E\) is also a parallelogram.  (2 marks)

Show Answers Only

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

Show Worked Solution

\(\text{Parrallelogram}\ \Rightarrow\ \text{Show opposite sides are equal} \)

\( \Big{|}\overrightarrow{A B} \Big{|} = \Big{|}\overrightarrow{CD} \Big{|} \ \ (ABCD\ \text{is a parallelogram}) \)

\( \Big{|}\overrightarrow{A B} \Big{|} = \Big{|}\overrightarrow{EF} \Big{|} \ \ (ABEF\ \text{is a parallelogram}) \)

\( \Rightarrow \Big{|}\overrightarrow{CD} \Big{|} = \Big{|}\overrightarrow{EF} \Big{|} \)
 

\( \Big{|}\overrightarrow{CE} \Big{|} = \Big{|}\overrightarrow{BE} \Big{|}-\Big{|}\overrightarrow{BC} \Big{|} \)

\(\text{Similarly,} \)

\( \Big{|}\overrightarrow{DF} \Big{|} = \Big{|}\overrightarrow{AF} \Big{|}-\Big{|}\overrightarrow{AD} \Big{|} \)

\( \text{Since}\ \ \Big{|}\overrightarrow{BE} \Big{|} = \Big{|}\overrightarrow{AF} \Big{|}\ \ \text{and}\ \ \Big{|}\overrightarrow{BC} \Big{|} = \Big{|}\overrightarrow{AD} \Big{|}\)

\( (ABCD\ \text{and}\ ABEF\ \text{are parallelograms}) \)

\( \Rightarrow\ \Big{|}\overrightarrow{DF} \Big{|} = \Big{|}\overrightarrow{CE} \Big{|} \)

\(\therefore CDFE\ \text{is a parallelogram} \)

Functions, EXT1 F2 2023 HSC 14b

Consider the hyperbola  \(y=\dfrac{1}{x}\)  and the circle  \((x-c)^2+y^2=c^2\), where \(c\) is a constant.

  1. Show that the \(x\)-coordinates of any points of intersection of the hyperbola and circle are zeros of the polynomial  \(P(x)=x^4-2 c x^3+1\).  (1 mark)

  2. The graphs of  \(y=x^4-2 c x^3+1\)  for  \(c=0.8\)  and  \(c=1\) are shown.
     

  1. By considering the given graphs, or otherwise, find the exact value of  \(c>0\)  such that the hyperbola  \(y=\dfrac{1}{x}\)  and the circle  \((x-c)^2+y^2=c^2\)  intersect at only one point.  (3 marks)

Show Answers Only

  1. \(\text{See Worked Solutions}\)
  2. \(\sqrt[4]{\dfrac{16}{27}}\approx 0.877\)

Show Worked Solution

i.     \(y=\dfrac{1}{x}\ …\ (1) \)

\((x-c)^2+y^2=c^2\ …\ (2) \)

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

\((x-c)^2+\Big{(}\dfrac{1}{x}\Big{)}^2 \) \(=c^2\)  
\(x^2-2cx+c^2+\dfrac{1}{x^2}\) \(=c^2\)  
\(x^4-2cx^3+1\) \(=0\)  
Mean mark (i) 53%.

ii.    \(\text{The two graphs show that for some value of}\ \ 0.8 \leq c \leq 1,\)

\(P(x)\ \text{has a minimum that touches the}\ x\text{-axis once.}\)

\(P(x)\) \(=x^4-2cx^3+1\)  
\(P^{′}(x)\) \(=4x^3-6cx^2\)  

 
\(\text{Find}\ x\ \text{when}\ P^{′}(x)=0: \)

\(4x^3-6cx^2\) \(=0\)  
\(2x^2(2x-3c)\) \(=0\)  
\(x\) \(=\dfrac{3c}{2}\ \ (x \neq 0)\)  

 
\(\text{Find}\ c\ \text{when}\ P(\frac{3c}{2})=0: \)

\(\Big{(} \dfrac{3c}{2} \Big{)}^4-2c\Big{(} \dfrac{3c}{2} \Big{)}^3+1 \) \(=0\)  
\(\dfrac{81c^4}{16}-\dfrac{54c^4}{8}+1\) \(=0\)  
\(\dfrac{(108-81)c^4}{16}\) \(=1\)  
\(\dfrac{27c^4}{16}\) \(=1\)  
\(c^4\) \(=\dfrac{16}{27}\)  
\(c\) \(=\sqrt[4]{\dfrac{16}{27}} \)  
  \(\approx 0.877\)  
Mean mark (ii) 19%.

Calculus, EXT1 C2 2023 HSC 14a

Let  \(f(x)=2 x+\ln x\), for \(x>0\).

  1. Explain why the inverse of \(f(x)\) is a function.  (1 mark)

  2. Let  \(g(x)=f^{-1}(x)\). By considering the value of \(f(1)\), or otherwise, evaluate \(g^{\prime}(2)\).  (2 mark)

Show Answers Only

i.     \(f(x)=2 x+\ln x\)

\(f^{′}(x)=2+\dfrac{1}{x} \)

\(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \)

\(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \)
 

ii.    \(\dfrac{1}{3}\)

Show Worked Solution

i.     \(f(x)=2 x+\ln x\)

\(f^{′}(x)=2+\dfrac{1}{x} \)

\(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \)

\(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \)

Mean mark (i) 54%.

 
ii.    \(g(x)=f^{-1}(x) \)

\(f(g(x))=x\)

\(\text{Differentiate both sides:}\)

\(g^{′}(x)\ f^{′}(g(x))\) \(=1\)  
\(g^{′}(x)\) \(=\dfrac{1}{f^{′}(g(x))}\)  
\(g^{′}(2)\) \(=\dfrac{1}{f^{′}(g(2))}\)  

 
\(f(1)=2 \times 1 + \ln1 = 2 \)

\(\Rightarrow g(2)=1 \ \text{(by inverse definition)}\)

\(\therefore g^{′}(2)\) \(= \dfrac{1}{f^{′}(1)} \)  
  \(=\dfrac{1}{2+\frac{1}{1}}\)  
  \(=\dfrac{1}{3} \)  
♦♦ Mean mark (ii) 32%.

Vectors, EXT1 V1 2023 HSC 13b

Particle \(A\) is projected from the origin with initial speed \(v\) m s\(^{-1}\) at an angle \(\theta\) with the horizontal plane. At the same time, particle \(B\) is projected horizontally with initial speed \(u\) ms\(^{-1}\) from a point that is \(H\) metres above the origin, as shown in the diagram.
  

The position vector of particle \(A, t\) seconds after it is projected, is given by

\[\textbf{r}_A(t)=\left(\begin{array}{c}v t\ \cos \theta \\vt\ \sin\theta-\dfrac{1}{2} g t^2\end{array}\right) \text{. (Do NOT prove this.)}\]

The position vector of particle \(B, t\) seconds after it is projected, is given by

\[\textbf{r}_B(t)=\left(\begin{array}{c}u t \\H-\dfrac{1}{2} g t^2\end{array}\right) \text{. (Do NOT prove this.)}\]

The angle \(\theta\) is chosen so that  \(\tan \theta=2\).

The two particles collide.

  1. By first showing that  \(\cos \theta=\dfrac{1}{\sqrt{5}}\), verify that  \(v=\sqrt{5} u\).   (2 marks)

  2. Show that the particles collide at time  \(T=\dfrac{H}{2 u}\).   (1 mark)

When the particles collide, their velocity vectors are perpendicular.

  1. Show that  \(H=\dfrac{2 u^2}{g}\).   (3 marks)

  2. Prior to the collision, the trajectory of particle \(A\) was a parabola. (Do NOT prove this.)  
  3. Find the height of the vertex of that parabola above the horizontal plane. Give your answer in terms of \(H\).   (2 marks)

Show Answers Only

  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
  4. \(H\)

Show Worked Solution

i.    \(\text{Given}\ \ \tan \theta =2\)

\(\cos \theta = \dfrac{1}{\sqrt 5} \)

\(\text{Since particles collide, for some}\ t: \)

\(vt\ \cos \theta\) \(=ut\)  
\(v \cdot \dfrac{1}{\sqrt 5} \) \(=u\)  
\(v\) \(=\sqrt5 u\)  


ii.    \(\text{Equating y-components of}\ \textbf{r}_A\ \text{and}\ \textbf{r}_B : \)

\(vt\ \sin \theta-\dfrac{1}{2}gt^2\) \(=H-\dfrac{1}{2}gt^2\)  
\(vt\ \sin \theta\) \(=H\)  
\(u \sqrt{5} \times t \times \dfrac{2}{\sqrt5} \)  \(=H\)  
\(t\) \(= \dfrac{H}{2u} \)  

 
iii.  \(\text{Velocity vectors:} \)

\[\textbf{v}_A(t)=\left(\begin{array}{c}v\ \cos \theta \\v\ \sin\theta-gt\end{array}\right)=\left(\begin{array}{c} u \\ 2u-gt \end{array}\right)\]

\[\textbf{v}_B(t)=\left(\begin{array}{c}u \\-gt \end{array}\right)\]

\(\text{Since particles are perpendicular at collision:}\)

\(\textbf{v}_A \cdot \textbf{v}_B=0\)

♦♦ Mean mark (iii) 37%.
\(u^2+(-gt)(2u-gt)\) \(=0\)  
\(u^2-2gtu+g^2t^2\) \(=0\)  
\((u-gt)^2\) \(=0\)  
\(gt\) \(=u\)  
\(t\) \(=\dfrac{u}{g}\)  
\(\dfrac{H}{2u}\) \(=\dfrac{u}{g}\ \ \text{(see part (ii))}\)  
\(\therefore H\) \(=\dfrac{2u^2}{g}\)  

 
iv.   \(\text{Height of vertex}\ \ \Rightarrow \ \text{Find}\ t\ \text{when y-component of}\ \textbf{v}_A=0 \)

\(v\ \sin \theta-gt\) \(=0\)  
\(t\) \(=\dfrac{v\ \sin \theta}{g} \)  

 
\(\text{Height of vertex}\ =\ \text{y-component of}\ \textbf{r}_A\ \text{when}\ \ t= \dfrac{v\ \sin \theta}{g} \)

\(\text{Height}\) \(=vt\ \sin \theta-\dfrac{1}{2}gt^2 \)  
  \(=\dfrac{v^2\sin^2 \theta}{g}-\dfrac{1}{2}g\Big{(} \dfrac{v^2\sin^2 \theta}{g^2}\Big{)} \)  
  \(=\dfrac{v^2 \sin^2 \theta}{2g} \)  
  \(=\dfrac{(2u)^2}{2g} \)  
  \(=\dfrac{2u^2}{g} \)  
  \(=H\)  
♦♦ Mean mark (iv) 34%.

Algebra, STD1 A3 2023 HSC 26

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)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ 400\ \ \  & \ \ 1000\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & & 290 \\
\hline
\end{array}

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 on 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.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ 400\ \ \  & \ \ 1000\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & \textbf{140} & 290 \\
\hline
\end{array}

b.

c.    \(400\text{ kWh}\)

d.    \(A\text{ is cheaper by }$40.\)

Show Worked Solution

a.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ 400\ \ \  & \ \ 1000\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & \textbf{140} & 290 \\
\hline
\end{array}

b.   


♦♦ Mean mark (b) 40%.

c.    \(\text{Same charge when Provider }A = \text{Provider } B \text{ i.e. where the lines intersect}\)

\(=400\text{ kWh (see graph above)}\)

d.    \(\text{When kWh}= 800, \ \ A=$240 \text{ and } B=$280\)

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

Financial Maths, STD1 F2 2023 HSC 25

An artwork is currently valued at $ 15000. It appreciates at a rate of 5.3% per annum.

What will the value of the artwork be in 8 years time?  (2 marks)

Show Answers Only

\($22\ 673.48\text{ (2 d.p.)}\)

Show Worked Solution

\(PV=$15\ 000,\ \ r=\dfrac{5.3}{100},\ \ n=8\)

\(FV\) \(=PV(1+r)^n\)
  \(=15\ 000(1+\dfrac{5.3}{100})^8\)
  \(=$22\ 673.48242\)
  \(=$22\ 673.48\text{ (2 d.p.)}\)

Financial Maths, STD1 F2 2023 HSC 24

Bobby invested $5000.

The table shows the progress of his investment over the first 4 months.

 

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

  2. Bobby could have earned simple interest on the investment at 0.62% per month.
  3. How much interest would Bobby have earned over 4 months by choosing this option?  (2 marks)

Show Answers Only

a.    \(A=$30.54,\ \ B=$5121.08\)

b.    \($124\)

Show Worked Solution

a.    \(A\) \(=5090.54\times\dfrac{0.6}{100}\)
    \(=$30.54\text{ (2 d.p.)}\)
     
  \(B\) \(=5090.54+30.54\)
    \(=$5121.08\)

b.    \(P=$5000,\ \ r=\dfrac{0.62}{100},\ \ n=4\)

\(I\) \(=Prn\)
  \(=5000\times \dfrac{0.62}{100}\times 4\)
  \(=$124\)


♦♦ Mean mark (b) 31%.

Financial Maths, STD1 M4 2023 HSC 23

The table compares the fuel costs of a petrol car with an electric car.

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Petrol car} & \textit{Electric car}\\
\hline
\rule{0pt}{2.5ex}\text{Fuel consumption}\rule[-1ex]{0pt}{0pt} & \text{8.6 L/100 km} & \text{18 kWh/100 km} \\ \hline \rule{0pt}{2.5ex}\text{Cost of fuel}\rule[-1ex]{0pt}{0pt} & \text{\$1.87/L} & \text{\$0.25/kWh} \\ \hline \end{array}

Jun travels on average 35 000 km per year.

How much will he save on fuel costs in a year by using an electric car?  (3 marks)

Show Answers Only

\($4053.70\)

Show Worked Solution

\(\text{Petrol car fuel costs}\)

\(\text{Litres used}=\dfrac{8.6\times 35\ 000}{100}=3010\text{ L}\)

\(\text{Cost of fuel}=3010\times $1.87 = $5628.70\)

  
\(\text{Electric car power costs}\)

\(\text{kWh}=\dfrac{18\times 35\ 000}{100}=6300\text{ kWh}\)

\(\text{Cost of power}=6300\times $0.25 = $1575\)

  
\(\text{Saving}=$5628.70-$1575=$4053.70\)

Financial Maths, STD1 F1 2023 HSC 22

The hours worked last week by Rose are shown. Her normal rate of pay per hour is $24.05.

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex} \text{Monday to Friday (normal rate)} \rule[-1ex]{0pt}{0pt} & \text{8:30 am – 12:30 pm} \\
\hline
\rule{0pt}{2.5ex} \text{Saturday (time-and-a-half)} \rule[-1ex]{0pt}{0pt} & \text{9:00 am – 11:30 am} \\
\hline\rule{0pt}{2.5ex} \text{Sunday (double time)} \rule[-1ex]{0pt}{0pt} & \text{9:00 am – noon} \\
\hline
\end{array}

How much did Rose earn last week?  (4 marks)

Show Answers Only

\($745.55\)

Show Worked Solution
\(\text{Normal Pay}\) \(=5\times 4\times $24.05\) \(=$481.00\)
\(\text{Time-and-a-half}\) \(=2.5\times 1.5\times $24.05\) \(=$90.19\)
\(\text{Double time}\) \(=2\times 3\times $24.05\) \(=$144.55\)
  \(\text{TOTAL}\) \(=$715.49\)

Networks, STD1 N1 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%.

Combinatorics, EXT1 A1 2023 HSC 12d

It is known that  \({ }^n C_r={ }^{n-1} C_{r-1}+{ }^{n-1} C_r\)  for all integers such that  \(1 \leq r \leq n-1\). (Do NOT prove this.)

Find ONE possible set of values for \(p\) and \(q\) such that

\({ }^{2022} C_{80}+{ }^{2022} C_{81}+{ }^{2023} C_{1943}={ }^p C_q\)  (2 marks)

Show Answers Only

\(p=2024, q=81 \)

Show Worked Solution

\({ }^{2022} C_{80}+{ }^{2022} C_{81}+{ }^{2023} C_{1943}={ }^p C_q\)

\(\text{Using the known relationship:} \)

\({ }^{2022} C_{80}+{ }^{2022} C_{81} = { }^{2023} C_{81}\ \ …\ (1)\)

\(\text{Also, since}\ \ { }^n C_r={ }^n C_{n-r} \)

\({ }^{2023} C_{1943} = { }^{2023} C_{2023-1943} = { }^{2023} C_{80}\ \ …\ (2)\) 

\({ }^{2022} C_{80}+{ }^{2022} C_{81}+{ }^{2023} C_{1943}\) \(={ }^{2023} C_{81}+{ }^{2023} C_{1943}\ \ \text{(see (1) above)}\)  
  \(={ }^{2023} C_{81}+{ }^{2023} C_{80}\ \ \text{(see (2) above)} \)  
  \(={ }^{2024} C_{81} \)  

 
\(\therefore p=2024, q=81 \)

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

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

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

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

Statistics, EXT1 S1 2023 HSC 11f

A recent census found that 30% of Australians were born overseas.

A sample of 900 randomly selected Australians was surveyed.

Let \(\hat{p}\) be the sample proportion of surveyed people who were born overseas.

A normal distribution is to be used to approximate \(P(\hat{p} \leq 0.31)\).

  1. Show that the variance of the random variable \(\hat{p}\) is \(\dfrac{7}{30\ 000}\).  (2 marks)

  2. Use the standard normal distribution and the normal distribution table to approximate \(P(\hat{p} \leq 0.31)\), giving your answer correct to two decimal places.  (2 marks)

Show Answers Only

  1. \(\text{See Worked Solutions}\)
  2. \(0.74\)

Show Worked Solution

i.    \(\text{Let}\ X =\ \text{number of people born overseas} \)

\(X\sim \text{Bin}(900,0.3) \)

\(\hat{p} = \dfrac{X}{900}\)

\(\text{Var}(X) = np(1-p)=900 \times 0.3 \times 0.7 \)

\(\text{Var}(\hat{p})\) \(=\dfrac{\text{Var}(X)}{n^2} \)  
  \(=\dfrac{900 \times 0.3 \times 0.7}{900^2}\)  
  \(=\dfrac{7}{30\ 000} \)  

 

ii.    \(\mu = E(\hat{p}) = 0.3,\ \ \sigma(\hat{p}) = \sqrt{\dfrac{7}{30\ 000}} \)

\(z\text{-score (0.31)} \ = \dfrac{0.31-0.3}{\sqrt{\dfrac{7}{30\ 000}}} = 0.6546… \)

\(P(\hat{p} \leq 0.31) \) \(=P(z \leq 0.6546…) \)  
  \(=0.74\ \ \text{(to 2 d.p.)} \)  

Trigonometry, EXT1 T3 2023 HSC 11e

Solve  \(\cos \theta+\sin \theta=1\)  for  \(0 \leq \theta \leq 2 \pi\).  (3 marks)

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\(\theta=0, \dfrac{\pi}{2}, 2\pi \)

Show Worked Solution

\(\text{Let}\ \ t=\tan \dfrac{\theta}{2} \)

\(\dfrac{1-t^2}{1+t^2} + \dfrac{2t}{1+t^2} \) \(=1\ \ \text{(see reference sheet)}\)  
\(1-t^2+2t\) \(=1+t^2\)  
\(2t^2-2t\) \(=0\)  
\(2t(t-1) \) \(=0\)  
\(t\) \(= 0 \ \text{or} \ 1\)  

 
\(\text{When}\ \ \tan\dfrac{\theta}{2} = 0: \)

\(\dfrac{\theta}{2} = 0, \pi\ \ \Rightarrow \ \theta=0\ \ \text{or}\ \ 2\pi \)

\(\text{When}\ \ \tan\dfrac{\theta}{2} = 1: \)

\(\dfrac{\theta}{2} = \dfrac{\pi}{4}\ \Rightarrow \ \theta=\dfrac{\pi}{2} \)
 

\(\text{Test}\ \ \theta=\pi: \)

\(\cos\ \pi+\sin\ \pi = -1+0=-1\ \ \text{(not a solution)} \)

\(\therefore \theta=0, \dfrac{\pi}{2}, 2\pi \)

Probability, STD1 S2 2023 HSC 8 MC

Four cards marked with the numbers 1, 2, 3 and 4 are placed face down on a table.

One card is turned over as shown.

What is the probability that the next card turned over is marked with an odd number?

  1. \(\dfrac{1}{4}\)
  2. \(\dfrac{1}{3}\)
  3. \(\dfrac{2}{4}\)
  4. \(\dfrac{2}{3}\)
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\(D\)

Show Worked Solution

\(\text{Sample space} =1,3,4\)

\(P(\text{odd})=\dfrac{2}{3}\)

  
\(\Rightarrow D\)

Financial Maths, STD1 F1 2023 HSC 6 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
\(\text{Value}\) \(=65\ 000-132\ 600\times\frac{22}{100}\)
  \(=$35\ 828\)

\(\Rightarrow A\)