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PHYSICS M7 2022 HSC 26

Light of frequency  `7.5 xx10^(14) \ text{Hz}`  is incident on a calcium metal sheet which has a work function of  `2.9 \ text{eV}`. Photoelectrons are emitted.

The metal is in a uniform electric field of `5.2 \ text{NC}^{-1}`, perpendicular to the surface of the metal, as shown.
 


  

  1. Show that the maximum kinetic energy of an emitted photoelectron is  `3.2 xx10^(-20) \ text{J}`.  (3 marks)
  1. Calculate the maximum distance, `d`, an emitted photoelectron can travel from the surface of the metal.  (3 marks)
Show Answers Only
  1. Proof (See Worked Solution)
  2. `0.038  text{m.}`
Show Worked Solution
a.    `K_(max)` `=hf-Phi`
    `=6.626 xx10^(-34)xx7.5 xx10^(14)-2.9xx 1.602 xx10^(-19)`
    `=3.2 xx10^(-20)  text{J}`

 

b.   Maximum Distance:

→ The electric field will do `3.2 xx10^(-20)\ text{J}` of work to stop the electron.

`W` `=qEd`     
`3.2 xx10^(-20)` `=1.602 xx10^(-19) xx 5.2 xx d_max`  
`:.d_max` `=(3.2 xx10^(-20))/(1.602 xx10^(-19)xx 5.2)`  
  `=0.038  text{m.}`  

♦ Mean mark (b) 51%.

PHYSICS M8 2022 HSC 24

The radioactive decay curve for americium-242 is shown.
 


 

  1. Use the graph to find the half-life of Am-242 and hence show that the decay constant, `\lambda`, is `0.043` `text{h}^(-1).`  (2 marks)
  1. Calculate how long it takes until the mass of Am-242 is 8 micrograms.  (2 marks)
Show Answers Only

a.   `0.043  text{h}^(-1)`

b.  `53.55\ text{h}`

Show Worked Solution

a.   The half life of Am-242 is 16 hours.

`lambda` `=(ln 2)/(t_(1//2))`  
  `=(ln 2)/(16)`  
  `=0.043  text{h}^(-1)`  

 

b.    `N` `=N_(0)e^(-lambda t)`
     `8` `=80e^(-0.043 t)`
  `e^(-0.043 t)` `=8/80`
  `-0.043t` `=ln(1/10)`
     `t` `=((-1)/(0.043))ln ((1)/(10))`
    `=53.55\ text{h}`

PHYSICS M7 2022 HSC 23

Outline a method that could be used to determine a value for the speed of light. In your answer, identify ONE factor that would limit the accuracy of the experimental data.  (4 marks)

Show Answers Only

Foucault’s cogwheel experiment (one possible method):

→ Set up a cogwheel and a mirror 8 km apart.

→ Shine a pulse of light through one of the gaps of the cogwheel.

→ Begin to rotate the cogwheel and record the lowest speed at which the returning light is completely blocked.

→ This is the speed at which, on the light’s journey, the cogwheel has rotated through half the distance between adjacent cogs (or the distance between a cog and the adjacent gap).

→ Using the rotation speed of the wheel and the distance between a cog and a gap, calculate the time taken for light to complete its journey.

→ Calculate the speed of light using  `c=(d)/(t)`, where  `d` = 16 km.

Limitation:

→ The accuracy of this experiment is limited by errors in measurement of the distance between the cogwheel and the mirror, as well as errors measuring the speed of rotation of the cogwheel.
 

Other valid methods:

  • Fizeau’s rotating mirror experiment.
  • Newton’s experiment.
  • Ole Romer’s experiment observing moons of Jupiter.
  • Experiments involving resonant cavities.
Show Worked Solution

Foucault’s cogwheel experiment (one possible method):

→ Set up a cogwheel and a mirror 8 km apart.

→ Shine a pulse of light through one of the gaps of the cogwheel.

→ Begin to rotate the cogwheel and record the lowest speed at which the returning light is completely blocked.

→ This is the speed at which, on the light’s journey, the cogwheel has rotated through half the distance between adjacent cogs (or the distance between a cog and the adjacent gap).

→ Using the rotation speed of the wheel and the distance between a cog and a gap, calculate the time taken for light to complete its journey.

→ Calculate the speed of light using  `c=(d)/(t)`, where  `d` = 16 km.

Limitation:

→ The accuracy of this experiment is limited by errors in measurement of the distance between the cogwheel and the mirror, as well as errors measuring the speed of rotation of the cogwheel.
 

Other valid methods:

  • Fizeau’s rotating mirror experiment.
  • Newton’s experiment.
  • Ole Romer’s experiment observing moons of Jupiter.
  • Experiments involving resonant cavities.

Mean mark 56%.

PHYSICS M6 2022 HSC 22

The diagram shows features of a transformer.
 


 

For TWO features of the transformer, describe how each contributes to the transformer's efficiency.  (4 marks)

Show Answers Only

Continuous Iron Core:

→ The continuous iron core of the transformer increases the flux linkage between the primary and secondary coils. This maximises the proportion of magnetic flux produced by the primary coil that passes through the secondary coil.

Laminations:

→ The laminations in the iron core prevent the production of large eddy currents in the iron core. This reduces energy losses in the form of heat.

Show Worked Solution

Continuous Iron Core:

→ The continuous iron core of the transformer increases the flux linkage between the primary and secondary coils. This maximises the proportion of magnetic flux produced by the primary coil that passes through the secondary coil.

Laminations:

→ The laminations in the iron core prevent the production of large eddy currents in the iron core. This reduces energy losses in the form of heat.

PHYSICS M8 2022 HSC 21

The positions of two stars, `X` and `Y`, are shown in the Hertzsprung-Russell diagram.
 

  1. Compare qualitatively the surface temperature and luminosity of `X` and `Y`.  (2 marks)

  1. Identify the elements undergoing fusion in the core of each star, `X` and `Y`.  (2 marks)
Show Answers Only

a.   Surface Temperature: The surface temperature of `X` is greater than the surface temperature of `Y.`

Luminosity: The luminosity of `Y` is greater than the luminosity of `X.`
 

b.   → `X` is a main sequence star so in its core hydrogen is being fused into helium.

→ `Y` is a red giant so in its core helium is being fused into carbon.

Show Worked Solution

a.   Surface Temperature:

→ The surface temperature of `X` is greater than the surface temperature of `Y.`

Luminosity:

→ The luminosity of `Y` is greater than the luminosity of `X.`
 

b.    → `X` is a main sequence star so in its core hydrogen is being fused into helium.

→ `Y` is a red giant so in its core helium is being fused into carbon.

BIOLOGY, M7 2021 HSC 30

A study compared the incidence of disease and survival of 8134 children who had received the measles vaccine with 8134 children from a neighbouring area who remained unvaccinated against measles. Children in each group were matched for age, sex, size of dwelling, number of siblings and maternal education. The graphs show the number of measles cases among the two groups over three years.


 

The table compares the cause of death and number of deaths of the two groups over the same three years.
 

'A vaccine only protects the community against a specific disease.'

Analyse the data with reference to this statement.   (7 marks)

Show Answers Only

→ These two studies show proof that the vaccine protects against measles.

→ The graphs show the incidence of measles in vaccinated children is extremely low, occasionally zero, compared to unvaccinated children. Since these children are matched for age and other sociocultural/socioeconomic factors, it is highly likely the vaccine is responsible for this difference.

→ The table also supports this conclusion, where 40 children who were unvaccinated died due to measles compared to the 2 who died who were vaccinated.

→ The table also provides evidence that the vaccine may protect children from dying from other diseases, contradicting the statement.

→ The table also uses the same groups as the graph, again, keeping cultural/socioeconomic factors similar. It shows that children vaccinated against measles died at half the rate of unvaccinated children due to diarrhoea and dysentery, and at about a quarter of the rate for oedema.

→ However, the small differences between groups with respect to fever and the small sample size for oedema mean that further research is required.

→ In general, the vaccinated group had less than half the mortality of the unvaccinated group, supporting the conclusion that the vaccine provides a wider protection.

→ However, this statement needs qualification as the data is only with respect to the measles vaccine. 

Show Worked Solution

→ These two studies show proof that the vaccine protects against measles.

→ The graphs show the incidence of measles in vaccinated children is extremely low, occasionally zero, compared to unvaccinated children. Since these children are matched for age and other sociocultural/socioeconomic factors, it is highly likely the vaccine is responsible for this difference.

→ The table also supports this conclusion, where 40 children who were unvaccinated died due to measles compared to the 2 who died who were vaccinated.

→ The table also provides evidence that the vaccine may protect children from dying from other diseases, contradicting the statement.

→ The table also uses the same groups as the graph, again, keeping cultural/socioeconomic factors similar. It shows that children vaccinated against measles died at half the rate of unvaccinated children due to diarrhoea and dysentery, and at about a quarter of the rate for oedema.

→ However, the small differences between groups with respect to fever and the small sample size for oedema mean that further research is required.

→ In general, the vaccinated group had less than half the mortality of the unvaccinated group, supporting the conclusion that the vaccine provides a wider protection.

→ However, this statement needs qualification as the data is only with respect to the measles vaccine. 

BIOLOGY, M8 2021 HSC 29

The koala is a mammal that maintains a stable body temperature of close to 36.6°C.

A study was conducted. Koalas were observed in natural forests in south-eastern Australia. Their posture in the tree and the ambient temperature were recorded. Ambient temperatures were divided into two categories, hot and mild.

The graph shows the posture of koalas observed.
 

Explain the adaptations used by the koalas in this study to maintain a stable body temperature. Make reference to the stimulus provided.   (4 marks)

Show Answers Only

→ Posture is a behavioural adaptation that changes the surface area of skin exposed to the air or tree trunk. Posture is often seen associated with hot or mild air temperatures. 

→ In mild conditions, koalas were often seen curling up, minimising heat loss by reducing surface area exposed to air, a behaviour that was never observed during hot conditions.

→ In hotter conditions, the koalas were seen leaning against, and hugging trees much more often than during milder conditions. Since the trees were shown to have lower temperatures, this exposes their ventral surface to the cool trunk and acts as a heat sink.
 

Other answers could include:

→ Reference to the physiological adaptation of directing blood flow to the stomach region to increase cooling when this are is in contact with the tree.

Show Worked Solution

→ Posture is a behavioural adaptation that changes the surface area of skin exposed to the air or tree trunk. Posture is often seen associated with hot or mild air temperatures. 

→ In mild conditions, koalas were often seen curling up, minimising heat loss by reducing surface area exposed to air, a behaviour that was never observed during hot conditions.

→ In hotter conditions, the koalas were seen leaning against, and hugging trees much more often than during milder conditions. Since the trees were shown to have lower temperatures, this exposes their ventral surface to the cool trunk and acts as a heat sink.
 

Other answers could include:

→ Reference to the physiological adaptation of directing blood flow to the stomach region to increase cooling when this are is in contact with the tree.

BIOLOGY, M5 2021 HSC 28a

Describe the role of mRNA in human cells.   (3 marks)

Show Answers Only

→ mRNA carries a complementary copy of a gene, with uracil replacing thymine as a base, into the cytoplasm.

→ At ribosome sites this provides a template; each codon (three nucleotides) results in the addition of a specific amino acid, which forms a polypeptide chain.

Show Worked Solution

→ mRNA carries a complementary copy of a gene, with uracil replacing thymine as a base, into the cytoplasm.

→ At ribosome sites this provides a template; each codon (three nucleotides) results in the addition of a specific amino acid, which forms a polypeptide chain.


Mean mark 58%.

BIOLOGY, M6 2021 HSC 24b

The diagram shows the early stages of embryonic development from a fertilised egg. The developing ball of cells has split and monozygotic (identical) twins have formed. Mutations can occur at different times during embryonic development, for example Mutation `A` would result in both twins having the mutation in all their cells.
 


 

Explain the effects of Mutation `B` and Mutation `C` on each twin and on any offspring that they may have.   (4 marks)

Show Answers Only

→ Mutation B occurred after twin formation but prior to cell differentiation, affecting just Twin 1’s somatic and germ-line cells.

→ Through affecting germ-line cells, it will then affect the offspring.

→ Mutation C occurred after twin formation and after cell differentiation, only affecting Twin 2’s somatic cells.

→ Because it doesn’t affect germ-line cells, this mutation cannot be passed onto offspring.

Show Worked Solution

→ Mutation B occurred after twin formation but prior to cell differentiation, affecting just Twin 1’s somatic and germ-line cells.

→ Through affecting germ-line cells, it will then affect the offspring.

→ Mutation C occurred after twin formation and after cell differentiation, only affecting Twin 2’s somatic cells.

→ Because it doesn’t affect germ-line cells, this mutation cannot be passed onto offspring.

BIOLOGY, M5 2021 HSC 24a

An incidence of an autosomal dominant trait is shown in the pedigree.
 


 

Is this trait likely to be the result of a somatic or a germ-line mutation? Justify your answer.   (3 marks)

Show Answers Only

→ The mutation must be somatic as the trait is not present in the parents.

→ If the mutation were germ-line, occurring in cells that produce gametes, then the trait would most likely be seen in the parents and offspring as it is dominant.

Show Worked Solution

→ The mutation must be somatic as the trait is not present in the parents.

→ If the mutation were germ-line, occurring in cells that produce gametes, then the trait would most likely be seen in the parents and offspring as it is dominant.

ENGINEERING, CS 2021 HSC 21b

A 100 mm long M16 × 2 bolt is used in the concrete decking of a bridge. The bolt has a thread length of 50 mm.

  1. Draw an orthogonal view, to AS 1100, to show the length of the bolt.
  2. Use the centreline provided and do NOT dimension the drawing. Use a scale of 1:1.   (3 marks)
  1. The yield stress of the bolt is 476 MPa and the diameter is 16 mm.
  2. Calculate the maximum load that the bolt can resist, assuming a Factor of Safety of 2.   (3 marks)
Show Answers Only

i. 

    

ii.   `47.9\ text{kN}`

Show Worked Solution

i. 

    

ii.   Find max load of bolt:

`text{Yield stress} = 476\ text{MPa, Bolt diameter = 16 mm}`

`text{Allowable Stress}= 476 -: 2=238\ text{MPa}`

`text{Area (bolt)}\=pir^2=pi xx 8^2=201.1\ text{mm}^2`

`text{Max Load }` `=\ text{allowable stress × area}`  
  `=238 xx 10^6 xx 201.1 xx 10^(-6)`  
  `=47\ 861.8\ text{N}`  
  `=47.9\ text{kN}`  

ENGINEERING, PPT 2021 HSC 4 MC

Which engineering drawing shows a bracket correctly dimensioned to AS 1100 drawing standards?
 

Show Answers Only

`D`

Show Worked Solution

By elimination:

→ Arrowheads are used for measurements instead of crosses (eliminate `A` and `C`).

→ In `B` the radius symbol points to the arrow marking diameter (eliminate `B`).

`=>D`

PHYSICS M8 2022 HSC 16 MC

The binding energy of helium-4 (He-4) is 28.3 MeV and the binding energy of beryllium-6 (Be-6) is 26.9 MeV.

Which of the following rows in the table is correct?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\textbf{A.} & \\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{He-4 requires more energy to separate}\quad & \text{He-4 is less massive than Be-6} \\
\text{into individual protons and neutrons}\rule[-1ex]{0pt}{0pt}&\text{}\\
\hline
\rule{0pt}{2.5ex}\text{He-4 requires less energy to separate}& \text{He-4 is less massive than Be-6}\\
\text{into individual protons and neutrons}\rule[-1ex]{0pt}{0pt}&\text{}\\
\hline
\rule{0pt}{2.5ex}\text{He-4 requires more energy to separate}& \text{He-4 is more massive than Be-6}\\
\text{into individual protons and neutrons}\rule[-1ex]{0pt}{0pt}&\text{}\\
\hline
\rule{0pt}{2.5ex}\text{He-4 requires less energy to separate}& \text{He-4 is more massive than Be-6} \quad \\
\text{into individual protons and neutrons}\rule[-1ex]{0pt}{0pt}&\text{}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

The binding energy of a nucleus is the energy required to separate it into individual particles.

→ He-4 requires more energy to separate into individual protons and neutrons.

→ He-4 has 4 nucleons while Be-6 has 6 nucleons.

→ He-4 is less massive.

\(\Rightarrow A\)

PHYSICS M6 2022 HSC 15 MC

Two wires separated by a distance, `d`, carry equal electric currents producing a magnetic force between them.
  


  

The separation between the wires is increased to `4 d` and the current in each wire is doubled.

What happens to the magnetic force between the wires, compared to the original force?

  1. It does not change.
  2. It increases by a factor of 4 .
  3. It decreases by a factor of 4 .
  4. It decreases by a factor of 8 .
Show Answers Only

`A`

Show Worked Solution

The force per unit length between the wires:

`(F)/(l)=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`
 

By increasing the distance by a factor of four, and doubling both currents, the new force per unit length is given by:

`(F)/(l)` `=(mu_(0))/(2pi)(2I_(1)2I_(2))/(4r)`  
  `=(mu_(0))/(2pi)(I_(1)I_(2))/(r)`  

 
`=>A`

PHYSICS, M7 2022 HSC 14 MC

Line `X` shows the results of an experiment carried out to investigate the photoelectric effect.
 


 

What change to this experiment would produce the results shown by line `Y` ?

  1. Increasing the frequency of the radiation
  2. Using a metal that has a greater work function
  3. Decreasing the intensity of the incident radiation
  4. Decreasing the maximum energy of photoelectrons
Show Answers Only

`B`

Show Worked Solution

→ The work function of the metals is given by the respective y-intercepts of lines `X` and `Y`. 

→ As line `Y` has a y-intercept with greater magnitude, it can be produced by using a metal with a greater work function.

`=>B`

PHYSICS M5 2022 HSC 13 MC

Two satellites share an orbit around a planet. One satellite has twice the mass of the other.
  


  

Which quantity would be different for the two satellites?

  1. Speed
  2. Momentum
  3. Orbital period
  4. Centripetal acceleration
Show Answers Only

`B`

Show Worked Solution

The orbital velocity, `v=sqrt((GM)/(r))` is independent of the mass of the orbiting body.

→ Both satellites have the same orbital velocity.

→ The momentum,  `p=mv` of the satellite with mass 2`m` is twice that of the satellite with mass `m`.

→ The two satellites have different momentum.

`=>B`

PHYSICS, M5 2022 HSC 11 MC

A projectile is launched vertically upwards. The displacement of the projectile as a function of time is shown.
 


 

Which velocity-time graph corresponds to this motion?
 

 

 

Show Answers Only

`B`

Show Worked Solution

By elimination:

Initially, the projectile is moving upwards (positive velocity)

→ Eliminate`A`

At  `t=2` s, the projectile reaches its maximum height and has a velocity of zero.

→ Eliminate `D`

Projectile then moves downwards (negative velocity)

→ Eliminate `C`

`=>B`

PHYSICS M5 2022 HSC 6 MC

The elliptical orbit of a planet around a star is shown.
 


  

Which type of energy is greater at position `P` than at `Q` ?

  1. Kinetic
  2. Nuclear
  3. Potential
  4. Total
Show Answers Only

`A`

Show Worked Solution

As the planet moves from `Q`  to  `P` is converted to kinetic energy.

→ its kinetic energy must be greater at `P`.

`=>A`

PHYSICS, M5 2019 HSC 30

A ball, initially at rest in position `P`, travels along a frictionless track to point `Q` and then falls to strike the floor below.
 

At the instant the ball leaves the track at `Q` it has a velocity of 1.5 m `text{s}^(-1)` at an angle of 50° to the horizontal.

  1. Calculate the difference in height between `P` and `Q`.   (3 marks)
  1. The ball takes 0.5 s to reach the floor after leaving the track at `Q`.
  2. Calculate the height of `Q` above the floor.   (3 marks)
Show Answers Only

a.   `h=`0.11 m

b.   `h=`1.8 m

Show Worked Solution

a. `Delta U` `=Delta E_(k)`
  `mgDeltah` `=(1)/(2)mv^2-(1)/(2)m u^2`
    `=(1)/(2)mv^2\ \ (u=0)`
  `Delta h` `=(v^2)/(2g)`
    `=(1.5^(2))/(2xx9.8)`
    `=0.1145` m
    `=0.11` m


♦ Mean mark part (a) 41%.

 

b. `u_(y)` `=u sin theta`
    `=1.50  sin 50^(@)`
    `=1.15` m `text{s}^(-1)`
  `s_(y)` `=ut+(1)/(2)at^(2)`
    `=1.15 xx0.5+(1)/(2)(9.80)xx0.5^(2)`
    `=` 1.8 m

 
`:.` Height of Q `=` 1.8 m

PHYSICS, M6 2019 HSC 29

A particle having mass `m` and charge `q` is accelerated from rest through a potential difference `V`. Assume that the only force acting on the particle is due to the electric field associated with this potential difference.

Show that the final velocity of the particle is given by  `v = sqrt((2qV)/m)`.   (3 marks)

Show Answers Only
`W` `=Delta E_(k)`  
  `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)mv^2` `(u=0)`
`qV` `=(1)/(2)mv^2` `(W=qV)`
`v^2` `=(2qV)/m`  
`∴v` `=sqrt((2qV)/(m))`  
Show Worked Solution

The work done on the particle is equal to its change in its kinetic energy:

`W` `=Delta E_(k)`  
  `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)mv^2` `(u=0)`
`qV` `=(1)/(2)mv^2` `(W=qV)`
`v^2` `=(2qV)/m`  
`∴v` `=sqrt((2qV)/(m))`  

PHYSICS, M7 2019 HSC 27

  1. Outline a thought experiment that relates to the prediction of time dilation.   (3 marks
  1. Outline experimental evidence that validated the prediction of time dilation.   (3 marks
Show Answers Only

a.    Consider a train moving at a high speed:

→ There are two observers, one on the train and one stationary outside the train. 

→ A pulse of light starts from one side of the carriage and then reflects off a mirror on the other side of the carriage returning to its source. 

 → The stationary observer, outside the train, will observe the light travel in a triangular path. This is longer than the path observed by the observer on the train.

The speed of light is constant for both observers.

→ the observer outside the train measures a longer, dilated time for the light pulse to travel. This demonstrates time dilation. 

 

b.    Experiment predicting time dilation:

→ Muons are particles in the upper atmosphere produced by cosmic rays that travel at a high speed greater than 0.99c and have a short half-life. 

→ The amount of muons striking the ground in a particular area at the top of a mountain was measured.

→ Using this data the number of muons expected to reach the ground at sea level was predicted (assuming no relativistic effects).

→ The actual number observed at sea level was greater than predicted.

→ This is consistent with an increase in the muons half life due to time dilation.
 

Answers could also reference:

→ The Hafele-Keating atomic clock experiment.

→ Evidence from particle accelerators.

Show Worked Solution

a.    Consider a train moving at a high speed:

→ There are two observers, one on the train and one stationary outside the train. 

→ A pulse of light starts from one side of the carriage and then reflects off a mirror on the other side of the carriage returning to its source. 

 → The stationary observer, outside the train, will observe the light travel in a triangular path. This is longer than the path observed by the observer on the train.

The speed of light is constant for both observers.

→ the observer outside the train measures a longer, dilated time for the light pulse to travel. This demonstrates time dilation. 
 

b.    Experiment predicting time dilation:

→ Muons are particles in the upper atmosphere produced by cosmic rays that travel at a high speed greater than 0.99c and have a short half-life. 

→ The amount of muons striking the ground in a particular area at the top of a mountain was measured.

→ Using this data the number of muons expected to reach the ground at sea level was predicted (assuming no relativistic effects).

→ The actual number observed at sea level was greater than predicted.

→ This is consistent with an increase in the muons half life due to time dilation.
 

Answers could also reference:

→ The Hafele-Keating atomic clock experiment.

→ Evidence from particle accelerators.

PHYSICS, M5 2019 HSC 26

A student carried out an experiment to investigate the relationship between the torque produced by a force and the angle at which the force is applied. A 400 N force was applied to the same position on the handle of a spanner at different angles, as shown.
 

A high-precision device measured the torque applied to the bolt.

The data from the experiment is graphed below.
 

The student concluded that the torque `(tau)` was proportional to the angle `(theta)` and proposed the model

`tau=ktheta`

where  `k` = 1.7 Nm/degree.

  1. Justify the validity of the student's model using information from the graph.   (3 marks)
  1. What happens to the accuracy of this model's predictions as the angle increases beyond 25°? Justify your answer with reference to a different model.  (3 marks)
Show Answers Only

a.   → The graph shows a linear relationship.

→ Constructing a line of best fit on the student’s graph shows the the gradient is approximately`=(34-17)/(20-10) =1.7`.

These findings are consistent with the student’s model of `tau=1.7 theta`

⇒ The model is valid.
 

b.   The accuracy of this model decreases as the angle increases beyond 25°.

Torque produced by a force is more accurately described by `tau=rF sin theta.`

At small angles, `sin theta ~~ theta`, so the model is accurate. However at larger angles, `sin theta lt theta`, so the student’s model will predict values of torque greater than those predicted by `tau=rF sin theta`.

Show Worked Solution

a.   → The graph shows a linear relationship.

→ Constructing a line of best fit on the student’s graph shows the the gradient is approximately`=(34-17)/(20-10) =1.7`.

These findings are consistent with the student’s model of `tau=1.7 theta`

⇒ The model is valid.
 

b.   The accuracy of this model decreases as the angle increases beyond 25°.

Torque produced by a force is more accurately described by `tau=rF sin theta.`

At small angles, `sin theta ~~ theta`, so the model is accurate. However at larger angles, `sin theta lt theta`, so the student’s model will predict values of torque greater

BIOLOGY, M5 2021 HSC 22

In a population of rabbits, black fur colour is dominant over white fur. A black rabbit, whose mother has white fur, mates with a white rabbit.

Predict the phenotypic ratio for the offspring of this cross. Show your working.   (3 marks)

 

Show Answers Only

Black : White = 1 : 1

Show Worked Solution

`text{P: Bb × bb}`

\begin{array} {|c|c|c|}
\hline  & \text{b} & \text{b} \\
\hline \text{B} & \text{Bb} & \text{Bb} \\
\hline \text{b} & \text{bb} & \text{bb} \\
\hline \end{array}

 
Black : White = 1 : 1

BIOLOGY, M7 2021 HSC 21

  1. Label TWO features on the diagram below that would help to classify this pathogen as a bacterium.   (2 marks)
     
     

     
  2. A scientist followed Koch's postulates to confirm that this bacterium was causing diarrhoea in pigs on a local farm. 
  3. Complete the boxes in the flowchart provided to show the steps taken by the scientist.    (2 marks)
     
     

  4.  
  5. Two pig farmers on neighbouring farms noticed that their pigs were suffering from diarrhoea and gradually losing weight. The farmers each adopted a different strategy to deal with this disease, as shown in the table.  

\begin{array} {|c|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Farm} \quad \rule[-1ex]{0pt}{0pt} & \quad\quad\quad \quad \textit{Strategy} & \quad\quad\quad \quad \textit{Result}\\
\hline
\rule{0pt}{2.5ex} 1  & \text{Treatment with antibiotics} & \text{All pigs recovered after two}\\
\rule[-1ex]{0pt}{0pt} & \text{} & \text{weeks}\\
\hline
\rule{0pt}{2.5ex} 2  & \text{Elimination of rats and mice} & \text{Decrease in number of sick}\\
& \text{from pig sheds to improve} & \text{animals over three months}\\
\rule[-1ex]{0pt}{0pt} & \text{hygiene} & \\
\hline
\end{array}

Outline ONE benefit and ONE limitation of the strategies used on each farm.   (3 marks)

Show Answers Only

a.    Include two of the following labels:
 
       

b.   Box 2: Bacteria grown in pure culture and identified.

Box 4: Healthy pig became ill with diarrhoea.
 

c.   Benefits and Limitations of the strategies used on each farm.

→ The use of antibiotics on farm 1 has resulted in a rapid elimination of diarrhoea cases, however may induce antibiotic resistance in the
future, rendering the strategy less effective.

→ The removal of rats and mice from pig sheds to increase hygiene on farm 2 is slow to eliminate diarrhoea cases, however provides reassurance to prevent future outbreaks.
 

Other correct answers:

→ Proper hygiene practices can reduce incidence of other diseases, not just diarrhoea.

Show Worked Solution

a.    Include two of the following labels:
 
       
 

b.   Box 2: Bacteria grown in pure culture and identified.

Box 4: Healthy pig became ill with diarrhoea.
 

c.   Benefits and Limitations of the strategies used on each farm.

The use of antibiotics on farm 1 has resulted in a rapid elimination of diarrhoea cases, however may induce antibiotic resistance in the
future, rendering the strategy less effective.

The removal of rats and mice from pig sheds to increase hygiene on farm 2 is slow to eliminate diarrhoea cases, however provides reassurance to prevent future outbreaks.
 

Other correct answers:

  • Proper hygiene practices can reduce incidence of other diseases, not just diarrhoea.

Statistics, EXT1 S1 2022 HSC 13e

A chocolate factory sells 150-gram chocolate bars. There has been a complaint that the bars actually weigh less than 150 grams, so a team of inspectors was sent to the factory to check. They randomly selected 16 bars, weighed them and noted that 8 bars weighed less than 150 grams.

The factory manager claims 80% of the chocolate bars produced by the factory weigh 150 grams or more.

  1. The inspectors used the normal approximation to the binomial distribution to calculate the probability, `cc "P"`, of having at least 8 bars weighing less than 150 grams in a random sample of 16, assuming the factory manager's claim is correct.
  2. Using the attached probability table, calculate the value of `cc "P"`.   (2 marks)

  3. The factory manager disagrees with the method used by the inspectors as described in part (i).
  4. Explain why the method used by the inspectors might not be valid.   (1 mark)

Show Answers Only
  1. `P=0.0013`
  2. `text{See Worked Solutions}`
Show Worked Solution

i.   `text{Let}\ \ X=\ text{number of chocolate bars < 150 grams}`

`X\ ~\ text{Bin}(16, 0.2)`

`mu=np=16xx0.2=3.2`

`sigma^2=np(1-p)=16xx0.2xx0.8=2.56`

`ztext{-score}\ (X=8)=(8-3.2)/sqrt2.56=3`

 

`cc P(X>=8)` `=P(z>=3.0)`  
  `=1-P(z<=3.0)`  
  `=1-0.9987`  
  `=0.0013`  

 

ii.   `np=3.2,\ \ n(1-p)=16xx0.8=12.8`

`text{S}text{ince}\ \ np<5,\ text{the sample size is too small.}`


♦ Mean mark part (i) 37%.
 
Mean mark part (ii) 54%.

Calculus, EXT1 C3 2022 HSC 13b

A solid of revolution is to be found by rotating the region bounded by the `x`-axis and the curve  `y=(k+1) \sin (k x)`, where  `k>0`, between  `x=0`  and  `x=\frac{\pi}{2 k}`  about the `x`-axis.
 

     

Find the value of `k` for which the volume is `pi^2`.  (3 marks)

Show Answers Only

`k=1`

Show Worked Solution

`y=(k+1) \sin (k x)`

`V` `=pi int_0^((pi)/(2k)) (k+1)^2sin^(2)(kx)\ dx`  
  `=pi(k+1)^2 int_0^((pi)/(2k)) 1/2[1-cos(2kx)]\ dx`  
  `=(pi/2)(k+1)^2[x-(sin(2kx))/(2k)]_0^((pi)/(2k)) `  
  `=(pi/2)(k+1)^2[(pi/(2k)- sin(pi)/(2k))-(0-sin0/(2k))]`  
  `=(pi/2)(k+1)^2(pi/(2k))`  
  `=pi^2/(4k)(k+1)^2`  

 
`text{Given}\ \ V=pi^2:`

`pi^2/(4k)(k+1)^2` `=pi^2`  
`(k+1)^2` `=4k`  
`k^2+2k+1` `=4k`  
`k^2-2k+1` `=0`  
`(k-1)^2` `=0`  

 
`:.k=1`

Vectors, EXT1 V1 2022 HSC 13a

Three different points `A, B` and `C` are chosen on a circle centred at `O`.

Let  `underset~a= vec(OA),underset~b= vec(OB)`  and  `underset~c= vec(OC)`. Let  `underset~h=underset~a + underset~b+ underset~c`  and let `H` be the point such that  `vec(OH)= underset~h`, as shown in the diagram.
 

     
 

Show that  `vec(BH)`  and  `vec(CA)`  are perpendicular.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Prove}\ \ vec(BH) ⊥ vec(CA):`

`vec(BH)*vec(CA)=(-underset~b+underset~h)*(-underset~c+underset~a)`
 

`text{Given}\ \ underset~h=underset~a + underset~b+ underset~c\ \ =>\ \ underset~h-underset~b=underset~a + underset~c`

`vec(BH)*vec(CA)` `=(underset~a+underset~c)*(underset~a-underset~c)`  
  `=|underset~a|^2-|underset~c|^2`  

 
`underset~a and underset~c\ \ text{are radii}\ \ => \ \ |underset~a|=|underset~c|`

 
`:.\ text{S}text{ince}\ \ vec(BH)*vec(CA)=0\ \ =>\ \ vec(BH)⊥vec(CA)`

Proof, EXT1 P1 2022 HSC 12f

Use mathematical induction to prove that  `15 ^(n)+6^(2n+1)`  is divisible by 7 for all integers  `n >= 0`.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Prove}\ \ 15 ^(n)+6^(2n+1)\ \ text{is divisible by 7 for}\ \ n>=0`

`text{If}\ \ n=0:`

`15^0+6^1=7`

`:.\ text{True for}\ \ n=0`
 

`text{Assume true for}\ \ n=k`

`text{i.e.}\ \ 15^(k)+6^(2k+1)=7P\ \ text{(where}\ P\ text{is an integer)}`

`=>6^(2k+1)=7P-15^(k)\ \ …\ (1)`
 

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

`15^(k+1)+6^(2k+3)` `=15*15^(k)+6^2*6^(2k+1)`  
  `=15*15^(k)+36(7P-15^(k))\ \ text{(see (1))}`  
  `=15*15^(k)+36*7P-36*15^(k)`  
  `=36*7P-21*15^(k)`  
  `=7(36P-3*15^(k))\ \ text{(which is divisible by 7)}`  

 
`=>\ text{True for}\ \ n=k+1`

`:.\ text{S}text{ince true for}\ \ n=0, text{by PMI, true for integers}\ \ n>=0`

Calculus, EXT1 C1 2022 HSC 12d

In a room with temperature 12°C, coffee is poured into a cup. The temperature of the coffee when it is poured into the cup is 92°C, and it is far too hot to drink.

The temperature, `T`, in degrees Celsius, of the coffee, `t` minutes after it is made, can be modelled using the differential equation  `(dT)/(dt)=k(T-T_(1))`, where `k` is the constant of proportionality and `T_1` is a constant.

  1. It takes 5 minutes for the coffee to cool to a temperature of 76°C.
  2. Using separation of variables, solve the given differential equation to show that  `T=12+80e^((t)/(5)ln((4)/(5)))`.  (3 marks)

  3. The optimal drinking temperature for a hot beverage is 57°C.
  4. Find the value of `t` when the coffee reaches this temperature, giving your answer to the nearest minute.  (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{13 minutes}`
Show Worked Solution

i.   `T_1=12\ \ text{(Room temperature = 12°C)}`

`(dT)/(dt)` `=k(T-12)`  
`(dt)/(dT)` `=1/(k(T-12))`  
`int dt` `=1/k int(1/(T-12))\ dT`  
`t` `=1/k ln|T-12|+c`  

 
`text{When}\ \ t=0,\ \ T=92:`

`0` `=1/k ln|92-12|+c`  
`c` `=-1/k ln(80)`  

 

`t` `=1/k ln|T-12|-1/k ln(80)`  
`kt` `=ln((|T-12|)/80)`  
`T-12` `=80e^(kt)`  
`T` `=12+80e^(kt)`  

 
`text{When}\ \ t=5,\ \ T=76:`

`76` `=12+80e^(5k)`  
`e^(5k)` `=64/80`  
`5k` `=ln(4/5)`  
`k` `=1/5 ln(4/5)`  

  
`:.T=12+80e^((t)/(5)ln((4)/(5)))\ \ \ text{… as required}`
 

ii.   `text{Find}\ \ t\ \ text{when}\ \ T=57:`

`57` `=12+80e^((t)/(5)ln((4)/(5)))`  
`e^((t)/(5)ln((4)/(5)))` `=45/80`  
`t/5ln(4/5)` `=ln(9/16)`  
`t` `=(5ln(9/16))/ln(4/5)`  
  `=12.89…`  
  `~~13\ text{minutes}`  

Calculus, EXT1 C2 2022 HSC 12c

Find the equation of the tangent to the curve  `y=x  text{arctan}(x)`  at the point with coordinates `(1,(pi)/(4))`. Give your answer in the form  `y=mx+c`  (3 marks)

Show Answers Only

`y=((2+pi)/4)x-1/2`

Show Worked Solution
`y` `=xtan^(-1)(x)`  
`dy/dx` `=x xx 1/(1+x^2)+tan^(-1)(x)`  

 
`text{When}\ \ x=1:`

`dy/dx=1/2+tan^(-1)(1)=1/2+pi/4=(2+pi)/4`
 

`text{Equation of tangent}\ \ m=(2+pi)/4,\ text{through}\ \ (1,(pi)/(4)):`

`y-pi/4` `=(2+pi)/4 (x-1)`  
`y` `=((2+pi)/4)x-(2+pi)/4+pi/4`  
`y` `=((2+pi)/4)x-1/2`  

Trigonometry, EXT1 T3 2022 HSC 11e

Express  `sqrt3sin(x)-3cos(x)`  in the form  `R sin(x+alpha)`.  (3 marks)

Show Answers Only

`2sqrt3 sin(x-pi/3)`

Show Worked Solution

`R sin(x+alpha)=sqrt3sin(x)-3cos(x)`

`R sin(x)cos alpha +Rcos(x)sin alpha=sqrt3sin(x)-3cos(x)` 

`text{Equating coefficients:}`

`R cos alpha=sqrt3, \ \ R sin alpha=-3`

`R^2` `=(sqrt3)^2+(-3)^2=12`  
`:.R` `=sqrt12=2sqrt3`  

 
`text{Given}\ \ sin alpha<0\ \ text{and}\ \ cos alpha>0`

`=> alpha\ \ text{is in the 4th quadrant}`

`sqrt12 cos alpha` `=sqrt3`  
`cos alpha` `=sqrt3/sqrt12=1/2`  
`:.alpha` `=-pi/3\ \ (-pi<=alpha<=pi)`  

 
`:.sqrt3sin(x)-3cos(x)=2sqrt3 sin(x-pi/3)`

Combinatorics, EXT1 A1 2022 HSC 11c

Find the coefficients of `x^(2)` and `x^(3)` in the expansion of `(1-(x)/(2))^(8)`.  (2 marks)

Show Answers Only

`x^2: 7, \ x^3:-7`

Show Worked Solution

`text{General Term:}\ (1-(x)/(2))^(8) `

`T_k` `=((8),(k))(1)^(8-k)(- x/2)^k`  
  `=((8),(k))(- 1/2)^k x^k`  

 
`text{Coefficient of}\ \ x^2 = ((8),(2))(- 1/2)^2=7`

`text{Coefficient of}\ \ x^3 = ((8),(3))(- 1/2)^3=-7`

Functions, EXT1 F1 2022 HSC 5 MC

A curve is defined in parametric form by  `x=2+t`  and  `y=3-2t^(2)`  for `-1 <= t <= 0`.

Which diagram best represents this curve?
 


 

Show Answers Only

`B`

Show Worked Solution

`x=2+t\ \ =>\ \ t=x-2`

`text{Substitute into}\ \ y=3-2t^(2)`

`y=3-2(x-2)^2`

`text{Concave down parabola with maximum at (2, 3)}`

`=>B`

Trigonometry, EXT1 T1 2022 HSC 1 MC

It is given that  `cos((23 pi)/(12))=(sqrt6+sqrt2)/(4)`.

Which of the following is the value of  `cos^(-1)((sqrt6+sqrt2)/(4))`?

  1. `{23pi}/{12}`
  2. `{11pi}/{12}`
  3. `pi/12`
  4. `- {11pi}/{12}`
Show Answers Only

`C`

Show Worked Solution

`(23pi)/12\ \ =>\ \ text{4th quadrant}`

`cos((23pi)/12)=cos(2pi-(23pi)/12)=cos(pi/12)`

`:. cos^(-1)((sqrt6+sqrt2)/(4))=pi/12`

`=>C`


Mean mark 54%.

Financial Maths, 2ADV M1 2022 HSC 32

In a reducing-balance loan, an amount `$P` is borrowed for a period of `n` months at an interest rate of 0.25% per month, compounded monthly. At the end of each month, a repayment of `$M` is made. After the `n`th repayment has been made, the amount owing, `$A_n`, is given by

`A_(n)=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`

(Do NOT prove this.)

  1. Jane borrows $200 000 in a reducing-balance loan as described.

  2. The loan is to be repaid in 180 monthly repayments.

  3. Show that  `M` = 1381.16, when rounded to the nearest cent.  (2 marks)

  4. After 100 repayments of $1381.16 have been made, the interest rate changes to 0.35% per month.

  5. At this stage, the amount owing to the nearest dollar is $100 032. (Do NOT prove this.)

  6. Jane continues to make the same monthly repayments.

  7. For how many more months will Jane need to make full monthly payments of $1381.16?  (3 marks)

  8. The final payment will be less than $1381.16.
  9. How much will Jane need to pay in the final payment in order to pay off the loan?   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `83`
  3. `$931.54`
Show Worked Solution

a.   `text{Show}\ \ M=$1381.16`

`A_(n)` `=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`  
`0` `=200\ 000(1.0025)^180-M underbrace((1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^179))_(text(GP where)\ a = 1,\ r = 1.0025,\ n = 180)`  
`0` `=200\ 000(1.0025)^180-M((1(1.0025^180- 1))/(1.0025-1))`  

`M((1.0025^180-1)/(1.0025-1))=200\ 000(1.0025)^180`

`:.M` `=200\ 000(1.0025)^180 -: (1.0025^180-1 )/(1.0025-1)`   
  `=1381.163…`  
  `=$1381.16\ \ text{… as required}`  

 

b.   `P=$100\ 032,\ \ r=1.0035  and M=$1381.16`

`text{Find}\ \ n\ \ text{when}\ \ A_n=0:`

`A_(n)` `=P(1.0035)^(n)-1381.16(1+(1.0035)^(1)+(1.0035)^(2)+cdots+(1.0035)^(n-1))`  
`0` `=100\ 032(1.0035)^n-1381.16 ((1.0035^n- 1)/(1.0035-1))`  
`0` `=100\ 032(1.0035)^n-1381.16/0.0035 (1.0035^n- 1)`  
  `=100\ 032(1.0035)^n-394\ 617(1.0035)^n- 394\ 617`  
`294\ 585(1.0035)^n` `=394\ 617`  
`1.0035^n` `=(394\ 617)/(294\ 585)`  
`n` `=ln((394\ 617)/(294\ 585))/ln1.0035`  
  `=83.674…`  
  `=83\ text{more months with full payment}`  

 


♦♦ Mean mark part (b) 38%.

c.   `text{Find}\ \ A_83:`

`A_83` `=100\ 032(1.0035)^83-1381.16 ((1.0035^83- 1)/(1.0035-1))`  
  `=928.291…`  

 
`text{Interest will be added for the last month:}`

`:.\ text{Final payment}` `=928.291… xx 1.0035`  
  `=$931.54`  

♦♦♦ Mean mark part (c) 14%.

Calculus, 2ADV C3 2022 HSC 27

Let  `f(x)=xe^(-2x)`.

It is given that  `f^(′)(x)=e^(-2x)-2xe^(-2x)`.

  1. Show that  `f^(″)(x)=4(x-1)e^(-2x)`.  (2 marks)

  2. Find any stationary points of  `f(x)`  and determine their nature.  (2 marks)

  3. Sketch the curve  `y=x e^{-2 x}`, showing any stationary points, points of inflection and intercepts with the axes.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Max S.P. at}\ \ (1/2, 1/(2e))`
  3.  
Show Worked Solution

a.   `f^(′)(x)=e^(-2x)-2xe^(-2x)`

`f^(″)(x)` `=-2e^(-2x)-[2x(-2)e^(-2x)+2e^(-2x)]`  
  `=-2e^(-2x)+4xe^(-2x)-2e^(-2x)`  
  `=4xe^(-2x)-4e^(-2x)`  
  `=4(x-1)e^(-2x)\ \ text{… as required}`  

 

b.   `text{S.P.’s occur when}\ \ f^(′)(x)=0`

`e^(-2x)-2xe^(-2x)` `=0`  
`e^(-2x)(1-2x)` `=0`  

 
`e^(-2x)=0\ \ =>\ \ text{No solution}`

`1-2x=0\ \ =>\ \ x=1/2`

`f^(″)(1/2)` `=4(1/2-1)e^(-2xx1/2)`  
  `=-2e^(-1)<0\ \ =>\ text{MAX}`  

 
`f(1/2)=1/2e^(-1)=1/(2e)`
 

c.   `text{POI occurs when}\ \ f^(″)(x)=0`

`f^(″)(x)=4(x-1)e^(-2x)=0\ \ =>\ \ x=1`

`text{Test for change in concavity:}`

`f^(″)(0)=4(0-1)e^0=-4`

`f^(″)(2)=4(2-1)e^(-4)>0\ \ =>\ text{Concavity changes}`

`:.\ text{POI exists at}\ \ (1,1/e^2)`

`text{As}\ \ x->oo\ \ =>\ \ y->0^+`
 


Mean mark part (c) 54%.

Calculus, 2ADV C4 2022 HSC 29

  1. The diagram shows the graph of  `y=2^{-x}`. Also shown on the diagram are the first 5 of an infinite number of rectangular strips of width 1 unit and height  `y=2^{-x}`  for non-negative integer values of `x`. For example, the second rectangle shown has width 1 and height `(1)/(2)`. 
     


 

  1. The sum of the areas of the rectangles forms a geometric series.
  2. Show that the limiting sum of this series is 2. (1 mark)

  3. Show that `int_(0)^(4)2^(-x)\ dx=(15)/(16 ln 2)`.  (2 marks)

  4. Use parts (a) and (b) to show that  `e^(15) < 2^(32)`.  (2 marks)

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

a.   `text{Consider the rectangle heights:}`

`2^0=1, \ 2^(-1)=1/2, \ 2^(-2)= 1/4, \ 2^(-3)= 1/8, …`

`=>\ text{Rectangle Areas}\ = 1, \ 1/2, \  1/4, \ 1/8, …`

`a=1,\ \ r=1/2`

`S_oo=a/(1-r)=1/(1-1/2)=2\ \ text{… as required}`
 

b.   `text{Show}\ \ int_0^4 2^(-x)\ dx = 15/(16ln2)`

`int_0^4 2^(-x)\ dx ` `=(-1)/ln2[2^(-x)]_0^4`  
  `=(-1)/ln2(1/16-1)`  
  `=1/ln2-1/(16ln2)`  
  `=(16-1)/(16ln2)`  
  `=15/(16ln2)\ \ text{… as required}`  

 


Mean mark (b) 56%.

c.   `text{Show}\ \ e^15<2^32`

`text{Area under curve < Sum of rectangle areas}`

`15/(16ln2)` `<2`  
`15` `<32ln2`  
`15/32` `<ln2`  
`e^(15/32)` `<e^(ln2)`  
`root(32)(e^15)` `<2`  
`e^15` `<2^32\ \ text{… as required}`  

♦♦♦ Mean mark (c) 9%.

Calculus, 2ADV C4 2022 HSC 28

The graph of the circle  `x^2+y^2=2`  is shown.

The interval connecting the origin, `O`, and the point `(1,1)` makes an angle `theta` with the positive `x`-axis.
 

  1. By considering the value of `theta`, find the exact area of the shaded region, as shown on the diagram.  (2 marks)

Part of the hyperbola  `y=(a)/(b-x)-1`  which passes through the points `(0,0)` and `(1,1)` is drawn with the circle  `x^2+y^2=2`  as shown.
 

  1. Show that  `a=b=2`.  (2 marks)

  2. Using parts (a) and (b), find the exact area of the region bounded by the hyperbola, the positive `x`-axis and the circle as shown on the diagram.   (3 marks)

Show Answers Only
  1. `(pi-2)/4\ text{u}^2`
  2. `text{Proof (See Worked Solutions)}`
  3. `(8ln2+pi-6)/4\ text{u}^2`
Show Worked Solution

a.   `tan theta=1\ \ =>\ \ theta = pi/4`

`text{Using Pythagoras,}`

`r=sqrt(1^2+1^2)=sqrt2`

`text{Shaded Area}` `=A_text{sector}-A_Delta`  
  `=(pi/4)/(2pi) xx pi r^2-1/2 xx b xx h`  
  `=1/8xxpixx(sqrt2)^2-1/2xx1xx1`  
  `=pi/4-1/2`  
  `=(pi-2)/4\ text{u}^2`  

 


Mean mark (a) 54%.

b.   `text{Show}\ \ a=b=2`

`y=(a)/(b-x)-1\ \ text{passes through}\ \ (0,0):`

`0` `=a/(b-0)-1`  
`a/b` `=1`  
`a` `=b`  

 
`y=(a)/(b-x)-1\ \ text{passes through}\ \ (1,1):`

`1` `=a/(b-1)-1`  
`a/(b-1)` `=2`  
`a` `=2(b-1)`  
`a` `=2b-2`  
`b` `=2b-2\ \ text{(using}\ a=b)`  
`b` `=2`  

 
`:.a=b=2`
 


♦ Mean mark (b) 47%.
c.    `int_0^1 2/(2-x)-1\ dx` `=int_0^1 -2 xx (-1)/(2-x)-1\ dx`
    `=[-2ln|2-x|-x]_0^1`
    `=[(-2ln1-1)-(-2ln2-0)]`
    `=-1+2ln2`

 

`:.\ text{Total Area}` `=2ln2-1 + (pi-2)/4`  
  `=(8ln2-4+pi-2)/4`  
  `=(8ln2+pi-6)/4\ text{u}^2`  

♦♦ Mean mark (c) 36%.

Calculus, 2ADV C2 2022 HSC 25

Let  `f(x)=sin(2x)`.

Find the value of `x`, for  `0 < x < pi`, for which  `f^(′)(x)=-sqrt3`  AND  `f^(″)(x)=2`.  (3 marks)

Show Answers Only

`(7pi)/12`

Show Worked Solution

`f^(′)(x)=2cos(2x)`

`2cos(2x)` `=-sqrt3`  
`cos(2x)` `=- sqrt3/2`  
`2x` `=pi-pi/6,\ \ pi+pi/6`  
  `=(5pi)/6,\ \ (7pi)/6`  
`x` `=(5pi)/12,\ \ (7pi)/12`  

 
`f^(″)(x)=-4sin(2x)`

`-4sin(2x)` `=2`  
`sin(2x)` `=- 1/2`  
`2x` `=pi+pi/6,\ \ 2pi-pi/6`  
  `=(7pi)/6,\ \ (11pi)/6`  
`x` `=(7pi)/12,\ \ (22pi)/12`  

 
`:.x=(7pi)/12\ \ text{(satisfies both equations)}`


Mean mark 53%.

Trigonometry, 2ADV T3 2022 HSC 23

The depth of water in a bay rises and falls with the tide. On a particular day the depth of the water, `d` metres, can be modelled by the equation

`d=1.3-0.6 cos((4pi)/(25)t)`,

where `t` is the time in hours since low tide.

  1. Find the depth of water at low tide and at high tide.  (2 marks)

  2. What is the time interval, in hours, between two successive low tides?  (1 mark)

  3. For how long between successive low tides will the depth of water be at least 1 metre?  (3 marks)

Show Answers Only
  1. `text{Low: 0.7 m,  High: 1.9 m}`
  2. `25/2\ text{hours}`
  3. `25/3\ text{hours}`
Show Worked Solution

a.   `text{S}text{ince}\ \ –1<=cos((4pi)/(25)t)<=1:`

`text{Low Tide}\ =1.3-0.6(1)=0.7\ text{m}`

`text{High Tide}\ =1.3-0.6(-1)=1.9\ text{m}`
 

b.   `text{Time between two low tides = Period of equation}\ (n)`

`(2pi)/n` `=(4pi)/25`  
`n/(2pi)` `=25/(4pi)`  
`n` `=25/2\ text{hours}`  

 


♦ Mean mark part (b) 47%.

c.   `text{Find}\ \ t\ \ text{when}\ \ d=1:`

`1.3-0.6 cos((4pi)/(25)t)` `=1`  
`-0.6 cos((4pi)/(25)t)` `=-0.3`  
`cos((4pi)/(25)t)` `=1/2`  

 

`(4pi)/(25)t` `=pi/3,\ \ (5pi)/3`  
`t` `=25/12,\ \ 125/12`  

 
`:.\ text{Time between low tides where water depth}\ >= 1\ text{m}`

`=125/12-25/12`

`=100/12`

`=25/3\ text{hours}`


♦ Mean mark part (c) 46%.

Calculus, 2ADV C3 2022 HSC 22

Find the global maximum and minimum values of  `y=x^(3)-6x^(2)+8`, where  `-1 <= x <= 7`.   (4 marks)

Show Answers Only

`text{Global max = 57}`

`text{Global min = – 24}`

Show Worked Solution
`y` `=x^3-6x^2+8`  
`dy/dx` `=3x^2-12x`  
`(d^2y)/(dx^2)` `=6x-12`  

 
`text{SP’s when}\ \ dy/dx=0:`

`3x^2-12x` `=0`  
`3x(x-4)` `=0`  

 
`x=0\ \ text{or}\ \ 4`

`text{When}\ \ x=0,\ \ y=8,\ \ (d^2y)/(dx^2)<0`

`=>\ text{Local Max at}\ \ (0,8)`

`text{When}\ \ x=4,\ \ y=4^3-6(4^2)+8=-24,\ \ (d^2y)/(dx^2)>0`

`=>\ text{Local Min at}\ \ (4,-24)`
 

`text{Check ends of domain:}`

`text{When}\ \ x=-1,\ \ y=-1-6+8=1`

`text{When}\ \ x=7,\ \ y=7^3-6(7^2)+8=57`

`:.\ text{Global max = 57}`

`:.\ text{Global min = – 24}`

Calculus, 2ADV C4 2022 HSC 18

  1. Differentiate  `y=(x^(2)+1)^(4)`.  (2 marks)

  2. Hence, or otherwise, find `int x(x^(2)+1)^(3)dx`.  (1 mark)

Show Answers Only
  1.  `dy/dx=8x(x^2+1)^3`
  2.  `1/8(x^2+1)^4+C`
Show Worked Solution

a.   `y=(x^2+1)^4`

`text{Using chain rule:}`

`dy/dx` `=4 xx 2x(x^2+1)^3`  
  `=8x(x^2+1)^3`  

 

b.    `intx(x^2+1)^3\ dx` `=1/8 int8x(x^2+1)^3\ dx`
    `=1/8(x^2+1)^4+C`

Calculus, 2ADV C4 2022 HSC 16

The parabola `y=x^2` meets the line `y=2 x+3` at the points `(-1,1)` and `(3,9)` as shown in the diagram.
 

Find the area enclosed by the parabola and the line.  (3 marks)

Show Answers Only

`32/3\ text{u}^2`

Show Worked Solution
`A` `=int_(-1)^(3) 2x+3-x^2\ dx`  
  `=[x^2+3x-1/3 x^3]_(-1)^3`  
  `=(9+9-27/3)-(1-3+1/3)`  
  `=9-(-5/3)`  
  `=32/3\ text{u}^2`  

Probability, 2ADV S1 2022 HSC 15

In a bag there are 3 six-sided dice. Two of the dice have faces marked  1, 2, 3, 4, 5, 6. The other is a special die with faces marked  1, 2, 3, 5, 5, 5.

One die is randomly selected and tossed.

  1. What is the probability that the die shows a 5?  (1 mark)

  2. Given that the die shows a 5, what is the probability that it is the special die?  (1 mark)

Show Answers Only
  1. `5/18`
  2. `3/5`
Show Worked Solution
a.    `P(5)` `=1/3 xx 1/6 + 1/3 xx 1/6 + 1/3 xx 1/2`
    `=1/18+1/18+1/6`
    `=5/18`

 

b.   `Ptext{(special die given a 5 is rolled)}`

`=(Ptext{(special die)} ∩  P(5))/(P(5))`

`=(1/3 xx 1/2)/(5/18)`

`=1/6 xx 18/5`

`=3/5`


Mean mark (a) 53%.
 
♦ Mean mark (b) 40%.

Trigonometry, 2ADV T3 2022 HSC 14

The graph of  `y=k sin (ax)`

What are the values of `k` and `a`?  (2 marks)

Show Answers Only

`k=4, \ \ a=1/3`

Show Worked Solution

`text{Amplitude = 4}`

`=> k=4`

`text{Period} = 6pi`

`(2pi)/a` `=6pi`  
 `6pia` `=2pi`  
 `=>a` `=1/3`  

Calculus, 2ADV C4 2022 HSC 13

Use two applications of the trapezoidal rule to find an approximate value of `int_(0)^(2)sqrt(1+x^(2))\ dx`. Give your answer correct to 2 decimal places.  (2 marks)

Show Answers Only

`3.03`

Show Worked Solution
`A` `~~h/2(y_0+2y_1+y_2)`  
  `~~1/2(1+2 xx sqrt2+sqrt5)`  
  `~~3.03\ \ text{(to 2 d.p.)}`  

Mean mark 59%.
COMMENT: A surprisingly low mean mark warrants attention.

Calculus, 2ADV C4 2022 HSC 8 MC

The graph of the even function  `y=f(x)`  is shown.

The area of the shaded region `A` is `1/2` and the area of the shaded region `B` is `3/2`.

What is the value of `int_(-2)^(2)f(x)\ dx`?

  1. `4`
  2. `2`
  3. `–2`
  4. `–4`
Show Answers Only

`C`

Show Worked Solution

`text{Areas under the}\ x text{-axis are negative}`

`int_0 ^2 f(x)\ dx = 1/2-3/2=-1`
 

`text{S}text{ince}\ \ f(x)\ \ text{is even:}`

`int_(-2) ^2 f(x)\ dx = 2int_0 ^2 f(x)\ dx = -2`

`=>C`


Mean mark 52%.

Functions, 2ADV F1 2022 HSC 12

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.    (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}`.   (2 marks)
     

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M &  &  &  \\
\hline \end{array}

 
                   

Show Answers Only

a.    `M=180/T`

 b.    

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}       

 

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`
 


♦ Mean mark (a) 49%.

b.   

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}

Calculus, 2ADV C1 2022 HSC 5 MC

Let  `h(x)=(f(x))/(g(x))`, where

`{:[f(1)=2, qquad f^(')(1)=4],[g(1)=8, qquad g^(')(1)=12]:}`

What is the gradient of the tangent to the graph of  `y=h(x)`  at  `x=1` ?

  1. `-8`
  2. `qquad 8`
  3. `- 1/8`
  4. `qquad 1/8`
Show Answers Only

`D`

Show Worked Solution

`text{Using the quotient rule:}`

`h′(1)` `=(g(1)\ f′(1)-f(1)\ g′(1))/[g(1)]^2`  
  `=(8xx4-2xx12)/(8^2)`  
  `=(32-24)/64`  
  `=1/8`  

 
`=>D`

Trigonometry, 2ADV T1 2022 HSC 3 MC

A tower `B T` has height `h` metres.

From point `A`, the angle of elevation to the top of the tower is `26^@` as shown.
 


 

Which of the following is the correct expression for the length of `A B` ?

  1. `h  tan 26^(@)`
  2. `h  cot 26^(@)`
  3. `h  sin 26^(@)`
  4. `h\ text{cosec}\ 26^(@)`
Show Answers Only

`B`

Show Worked Solution
`tan 26^@` `=h/(AB)`  
`AB` `=h/(tan26^@)`  
  `=h\ cot26^@`  

 
`=>B`

Statistics, 2ADV S3 2022 HSC 26

The life span of batteries from a particular factory is normally distributed with a mean of 840 hours and a standard deviation of 80 hours.

It is known from statistical tables that for this distribution approximately 60% of the batteries have a life span of less than 860 hours.

What is the approximate percentage of batteries with a life span between 820 and 920 hours?  (3 marks)

Show Answers Only

`44text{%}`

Show Worked Solution

`mu=840, \ sigma=80`

`ztext{-score (860)}\ = (x-mu)/sigma=(860-840)/80=0.25` 

`ztext{-score (820)}\ =(820-840)/80=-0.25` 

`ztext{-score (920)}\ =(920-840)/80=1`
 

`text{50% of batteries have a life span below 840 hours (by definintion)}`

`=>\ text{10% lie between 840 and 860 hours}`

`=>\ text{By symmetry, 10% lie between 820 and 840 hours}`

`=> P(-0.25<=z<=0)=10text{%}`
 

`:.\ text{Percentage between 820 and 920}`

`=P(-0.25<=z<=1)`

`=P(-0.25<=z<=0) + P(0<=z<=1)`

`=10+34`

`=44text{%}`

Statistics, 2ADV S2 2022 HSC 24

Jo is researching the relationship between the ages of teenage characters in television series and the ages of actors playing these characters.

After collecting the data, Jo finds that the correlation coefficient is 0.4564.

A scatterplot showing the data is drawn. The line of best fit with equation  `y=-7.51+1.85 x`, is also drawn.
 


 

Describe and interpret the data and other information provided, with reference to the context given.  (4 marks)

Show Answers Only

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`
Show Worked Solution

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`

Financial Maths, 2ADV M1 2022 HSC 21

Eli is choosing between two investment options.

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

  1. What is the value of Eli's investment after 10 years using Option 1 ?  (2 marks)

  2. What is the difference between the future values after 10 years using Option 1 and Option 2?  (2 marks)

Show Answers Only
  1. `$45\ 097.17`
  2. `$40.87`
Show Worked Solution

a.   `text{Monthly r/i}\ = 1.2/12=0.1text{%}\ \ =>\ \ r= 0.001`

`text{Compounding periods}\ (n)=12xx10=120`

`FV` `=PV(1+r)^n`  
  `=40\ 000(1+0.001)^120`  
  `=$45\ 097.17`  

 

b.   `text{Quarterly r/i}\ = 2.4/4=0.6text{%}\ \ =>\ \ r= 0.006`

`text{Compounding periods}\ (N) =4xx10=40`

`text{Annuity factor (from table) = 45.05630}`

`FV` `=1000xx45.05630`  
  `=45\ 056.30`  

 

`text{Difference}` `=45\ 097.17-45\ 056.30`  
  `=$40.87`  

Statistics, 2ADV S2 2022 HSC 11

The table shows the types of customer complaints received by an online business in a month.

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

  2. The data from the table are shown in the following Pareto chart.
     
     
  3. The manager will address 80% of the complaints.
  4. Which types of complaints will the manager address?  (1 mark)

Show Answers Only
  1. `A=160, \ B=96`
  2. `text{Stock shortages and delivery fees.}`
Show Worked Solution

a.   `A=98+62=160`

`text{% Damaged items}\ = 8/200 xx 100 = 4text{%}`

`text{Cumulative % after damaged items = 96%}`

`B = 92+4=96`
 

b.   `text{The right hand side cumulative frequency percentage}`

`text{shows that 80% of all complaints received concern}`

`text{stock shortages and delivery fees.}`

`:.\ text{The manager will address stock shortages and delivery fees.}`

Financial Maths, STD2 F4 2022 HSC 36

Frankie borrows $200 000 from a bank. The loan is to be repaid over 23 years at a rate of 7.2% per annum, compounded monthly. The repayments have been set at $1485 per month.

The interest charged and the balance owing for the first three months of the loan are shown in the spreadsheet below.
 

  1. What are the values of `A` and `B`?  (2 marks)
  2. After 50 months of repaying the loan, Frankie decides to make a lump sum payment of $ 40 000 and to continue making the monthly repayments of $1485. The loan will then be fully repaid after a further 146 monthly repayments.
  3. How much less will Frankie pay overall by making the lump sum payment?  (3 marks)
Show Answers Only
  1. `A=$1198.29,\ \ B=$199\ 139.86`
  2. `$78\ 800`
Show Worked Solution

a.   `text{Monthly interest rate}\ =7.2/12=0.6text{%}`

`A` `=199\ 715 xx 0.6/100`  
  `=$1198.29`  

 

`B` `=P+I-R`  
  `=199\ 428.29 + 1196.57-1485`  
  `=$199\ 139.86`  

 

b.   `text{Total payments if lump sum not paid}`

`= (23xx12) xx 1485`

`=$409\ 860`
 

`text{Total payments if lump sum paid}`

`=40\ 000 + (50 + 146) xx 1485`

`=$331\ 060`
 

`text{Savings by paying the lump sum}`

`=409\ 860-331\ 060`

`=$78\ 800`


♦♦♦ Mean mark (b) 17%.
 

Measurement, STD2 M1 2022 HSC 32

The diagram shows a park consisting of a rectangle and a semicircle. The semicircle has a radius of 100 m. The dimensions of the rectangle are 200 m and 250 m.

A lake occupies a section of the park as shown. The rest of the park is a grassed section. Some measurements from the end of the grassed section to the edge of the lake are also shown.
 

  1. Using two applications of the trapezoidal rule, calculate the approximate area of the grassed section.  (2 marks)

  2. Hence calculate the approximate area of the lake, to the nearest square metre.   (2 marks)

Show Answers Only
  1. `35\ 500\ text{m}^2`
  2. `30\ 208\ text{m}^2`
Show Worked Solution

a.   `h=100\ text{m}`

`A` `~~h/2(x_1+x_2)+h/2(x_2+x_3)`  
  `~~100/2(160+150)+100/2(150+250)`  
  `~~50xx310+50xx400`  
  `~~35\ 500\ text{m}^2`  

 

b.    `text{Total Area}` `=\ text{Area of Rectangle + Area of Semi-Circle}`
    `= (250 xx 200) + 1/2 xx pi xx 100^2`
    `=65\ 707.96\ text{m}^2`

 

`text{Area of Lake}` `=67\ 708-35\ 500`  
  `=30\ 208\ text{m}^2`  

♦ Mean mark (b) 44%.