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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 C3 2022 HSC 20

A scientist is studying the growth of bacteria. The scientist models the number of bacteria, `N`, by the equation

`N(t)=200e^(0.013 t)`,

where `t` is the number of hours after starting the experiment.

  1. What is the initial number of bacteria in the experiment?  (1 mark)

  2. What is the number of bacteria 24 hours after starting the experiment?  (1 mark)

  3. What is the rate of increase in the number of bacteria 24 hours after starting the experiment?  (2 marks)

Show Answers Only
  1. `200`
  2. `273`
  3. `3.55\ text{bacteria per hour}`
Show Worked Solution
a.    `N(0)` `=200e^0`
    `=200\ text{bacteria}`

 

b.   `text{Find}\ N\ text{when}\ \ t=24:`

`N(24)` `=200e^(0.013xx24)`  
  `=273.23…`  
  `=273\ text{bacteria (nearest whole)}`  

 

c.    `N` `=200e^(0.013 t)`
  `(dN)/dt` `=0.013xx200e^(0.013t)`
    `=2.6e^(0.013t)`

 
`text{Find}\ \ (dN)/dt\ \ text{when}\ \ t=24:`

`(dN)/dt` `=2.6e^(0.013xx24)`  
  `=3.550…`  
  `=3.55\ text{bacteria/hr (to 2 d.p.)}`  

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`

Financial Maths, 2ADV M1 2022 HSC 17

Cards are stacked to build a 'house of cards'. A house of cards with 3 rows is shown.

A house of cards requires 3 cards in the top row, 6 cards in the next row, and each successive row has 3 more cards than the previous row.

  1. Show that a house of cards with 12 rows has a total of 234 cards.  (2 marks)

  2. Another house of cards has a total of 828 cards.
  3. How many rows are in this house of cards?  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `23`
Show Worked Solution

a.   `a=3, \ d=3`

`S_n` `=n/2[2a+(n-1)d]`  
`S_12` `=12/2(2xx3 + 11xx3)`  
  `=6(6+33)`  
  `=234\ \ text{… as required}`  

 
b.   `text{Find}\ \ n\ \ text{given}\ \ S_n=828:`

`828` `=n/2[6+(n-1)3]`  
`1656` `=n(3+3n)`  
  `=3n^2+3n`  

 

`3n^2+3n-1656` `=0`  
`n^2+n-552` `=0`  
`(n+24)(n-23)` `=0`  

 
`:. n=23\ text{rows}\ \ (n>0)`

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.}`

Functions, 2ADV F1 2022 HSC 1 MC

Which of the following could be the graph of  `y= -2 x+2`?
 

Show Answers Only

`A`

Show Worked Solution

`text{By elimination:}`

`y text{-intercept = 2  →  Eliminate}\ B and C`

`text{Gradient is negative  → Eliminate}\ D`

`=>A`

Statistics, STD2 S4 2022 HSC 23

A teacher surveyed the students in her Year 8 class to investigate the relationship between the average number of hours of phone use per day and the average number of hours of sleep per day.

The results are shown on the scatterplot below.
 

  1. The data for two new students, Alinta and Birrani, are shown in the table below. Plot their results on the scatterplot.  (2 marks)

                               
  2. By first fitting the line of best fit by eye on the scatterplot, estimate the average number of hours of sleep per day for a student who uses the phone for an average of 2 hours per day.  (2 marks)

Show Answers Only
  2. `text{9 hours (see LOBF in diagram above)}`
Show Worked Solution

a.   

b.   `text{9 hours (see LOBF in diagram above)}`

Financial Maths, STD2 F4 2022 HSC 27

A company purchases a machine for $50 000. The two methods of depreciation being considered are the declining-balance method and the straight-line method.

  1. For the declining-balance method, the salvage value of the machine after `n` years is given by the formula
  2.     `S=V_(0)xx(0.80)^(n),`
  3. where `S` is the salvage value and `V_(0)` is the initial value of the asset.
  4.  i. What is the annual rate of depreciation used in this formula?  (1 mark)
  5. ii. Calculate the salvage value of the machine after 3 years, based on the given formula.  (1 mark)
  6. For the straight-line method, the value of the machine is depreciated at a rate of 12.2% of the purchase price each year.
  7. When will the value of the machine, using this method, be equal to the salvage value found in part (a) (ii)?  (2 marks)
Show Answers Only
  1.  i. `20text{%}`
  2. ii. `$25\ 600`
  3. `text{4 years}`
Show Worked Solution

a.i.  `text{Depreciation rate}\ = 1-0.8=0.2=20text{%}`
 

a.ii.  `text{Find}\ \ S\ \ text{when}\ \ n=3:`

`S` `=V_0 xx (0.80)^n`  
  `=50\ 000 xx (0.80)^3`  
  `=$25\ 600`  

 
b.   `text{Using the SL method}`

`S_n` `= 50\ 000-(0.122 xx 50\ 000)xxn`  
  `=50\ 000-6100n`  

 

`text{Find}\ \ n\ \ text{when}\ \ S_n=$25\ 600`

`25\ 600` `=50\ 000-6100n`  
`6100n` `=24\ 400`  
`n` `=(24\ 400)/6100`  
  `=4\ text{years}`  

♦♦ Mean mark (a.i.) 24%.
COMMENT: A poor State result in part (a.i.) that warrants attention.
 
♦ Mean mark part (b) 38%.

Financial Maths, STD2 F1 2022 HSC 21

A real estate agent's commission for selling houses is 2% for the first $800 000 of the sale price and 1.5% for any amount over $800 000.

Calculate the commission earned in selling a house for $1 500 000.  (2 marks)

Show Answers Only

`$26\ 500`

Show Worked Solution
`text{Commission}` `=800\ 000 xx 2text{%} + (1\ 500\ 000-800\ 000) xx 1.5text{%}`  
  `=800\ 000 xx 0.02 + 700\ 000 xx 0.015`  
  `=16\ 000 + 10\ 500`  
  `=$26\ 500`  

Networks, STD2 N2 2022 HSC 20

The table below shows the distances, in kilometres, between a number of towns.
 

  1. Using the vertices given, draw a weighted network diagram to represent the information shown in the table.  (2 marks)
     

     
  2. A tourist wishes to visit each town.
  3. Draw the minimum spanning tree which will allow for this AND determine its length.  (3 marks)
     

Show Answers Only
  1.  
     
     
  2.   
     
  3. `1015\ text{km}`
Show Worked Solution

a. 

 

b.   `text{Using Prim’s algorithm (starting at}\ Y):`

`text{1st edge:}\ YC`

`text{2nd edge:}\ CB`

`text{3rd edge:}\ SB`

`text{4th edge:}\ YM`

`text{Length of minimum spanning tree}`

`=275 + 150+60+530`

`=1015\ text{km}`

Algebra, STD2 A2 2022 HSC 16

Tom is 25 years old, and likes to keep fit by exercising.

  1. Use this formula to find his maximum heart rate (bpm).    (1 mark)
  2.      Maximum heart rate = 220 – age in years

  3. Tom will get the most benefit from this exercise if his heart rate is between 50% and 85% of his maximum heart rate.
  4. Between what two heart rates should Tom be aiming for to get the most benefit from his exercise?  (2 marks)

Show Answers Only

a.   `text{195 bpm}`

b.   `98-166\ text{bpm}`

Show Worked Solution
a.    `text{Max heart rate}` `=220-25`
    `=195\ text{bpm}`

 

b.   `text{50% max heart rate}\ = 0.5 xx 195 = 97.5\ text{bpm}`

`text{85% max heart rate}\ = 0.85 xx 195 = 165.75\ text{bpm}`

`:.\ text{Tom should aim for between 98 and 166 bpm in exercise.}`

Financial Maths, STD2 F4 2022 HSC 10 MC

Alex purchased 800 shares. The total cost was $2.60 per share. Alex sold the shares one year later for $3.40 each and paid a fee of $24.95 for selling the shares.

What profit did Alex make on these shares?

  1. $590.10
  2. $615.05
  3. $640.00
  4. $664.95
Show Answers Only

`B`

Show Worked Solution

`text{C}text{ost of shares}\ = 800 xx 2.60=$2080`

`text{Sale funds}` `=\ text{sale price}\ -\ text{fee}`  
  `=(800 xx 3.40)-24.95`  
  `=2695.05`  

 

`text{Profit}` `=\ text{Sale funds}\ -\ text{C}text{ost}`  
  `=2695.05-2080`  
  `= $615.05`  

 
`=>B`

Algebra, STD2 A4 2022 HSC 9 MC

An object is projected vertically into the air. Its height, `h` metres, above the ground after `t` seconds is given by  `h=-5 t^2+80 t`.
 

For how long is the object at a height of 300 metres or more above the ground?

  1. 4 seconds
  2. 6 seconds
  3. 8 seconds
  4. 10 seconds
Show Answers Only

`A`

Show Worked Solution

`text{Object reaches 300 m when}\ \ t=6\ text{seconds.}`

`text{Object drops back below 300 m when}\ \ t=10\ text{seconds.}`

`text{Time at 300 m or above}\ = 10-6=4\ text{seconds}`

`=>A`

Financial Maths, STD2 F1 2022 HSC 7 MC

Tian is paid $20.45 per hour, as well as a meal allowance of $16.20 per day.

What are Tian's total earnings if she works 9 hours per day for 5 days?

  1. `$329.85`
  2. `$936.45`
  3. `$1001.25`
  4. `$1649.25`
Show Answers Only

`C`

Show Worked Solution
`text{Earnings (5 days)}` `=5 xx [(9 xx 20.45) + 16.20]`  
  `=5 xx 200.25`  
  `=$1001.25`  

 
`=>C`

PHYSICS, M6 2019 HSC 24

A step-up transformer is constructed using a solid iron core. The coils are made using copper wires of different thicknesses as shown.
 

The table shows electrical data for this transformer.
 

  1. Explain how the operation of this transformer remains consistent with the law of conservation of energy. Include a relevant calculation in your answer.   (3 marks)
  1. Explain how TWO modifications to this transformer would improve its efficiency.   (4 marks)
Show Answers Only

a.   The energy input into this transformer is `text{500 J s}^(-1)`

  • The energy output is given by:
  •    `P=V_(s)I_(s)=50xx 9=450\ \text{J s}^(-1)`
  •  This is consistent with the law of conservation of energy as `text{500 J s}^(-1)` of energy is converted into other energy forms such as heat.

b.    Modification 1:

  • Laminating the iron core prevents large eddy currents from being induced in it.
  • This reduces energy loss in the form of heat, increasing efficiency.

Modification 2:

  • Increasing the thickness of the wire in the primary coil.
  • This reduces its resistance, increasing the transformer’s efficiency.
Show Worked Solution

a.   The energy input into this transformer is `text{500 J s}^(-1)`

  • The energy output is given by:
  •    `P=V_(s)I_(s)=50xx 9=450\ \text{J s}^(-1)`
  •  This is consistent with the law of conservation of energy as `text{500 J s}^(-1)` of energy is converted into other energy forms such as heat.

b.    Modification 1:

  • Laminating the iron core prevents large eddy currents from being induced in it.
  • This reduces energy loss in the form of heat, increasing efficiency.

Modification 2:

  • Increasing the thickness of the wire in the primary coil.
  • This reduces its resistance, increasing the transformer’s efficiency.

♦ Mean mark part 48%.

PHYSICS, M7 2019 HSC 2 MC

Two stars were observed from Earth. Their spectra are shown with the wavelength in nanometres.
 

Using these spectra, what can be concluded about the motion of the stars relative to Earth and their chemical compositions?
 

\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}{1.5ex}\textit{Motion relative to Earth}\rule[-0.5ex]{0pt}{0pt}& \textit{Chemical composition} \\
\hline
\rule{0pt}{2.5ex}\text{The same}\rule[-1ex]{0pt}{0pt}&\text{The same}\\
\hline
\rule{0pt}{2.5ex}\text{Different}\rule[-1ex]{0pt}{0pt}& \text{The same}\\
\hline
\rule{0pt}{2.5ex}\text{The same}\rule[-1ex]{0pt}{0pt}& \text{Different} \\
\hline
\rule{0pt}{2.5ex}\text{Different}\rule[-1ex]{0pt}{0pt}& \text{Different} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • Both stars have the same set of spectral lines → Chemical composition is the same.
  • Spectral lines at different wavelengths → Different doppler shift → Different motion relative to the earth.

\(\Rightarrow B\)

PHYSICS, M7 2020 HSC 30b

  1. Calculate the wavelength of a proton travelling at 0.1\(c\).   (2 marks)
  1. Explain the relativistic effect on the wavelength of a proton travelling at 0.95\(c\).   (2 marks)
Show Answers Only

i.   `lambda=1xx10^(-14)` m

ii.  See Worked Solutions

Show Worked Solution
i.     `lambda` `=(h)/(mv)`
    `=(6.626 xx10^(-34))/(1.673 xx10^(-27)xx0.1 xx3xx10^(8))`
    `=1xx10^(-14)\ text{m}`

ii.    Due to momentum dilation:

  • The momentum of the proton is greater than that predicted by classical mechanics. 
  • Since  `lambda=(h)/(mv)\ \ =>\ \ lambda prop 1/(mv)`
  • The wavelength of the proton is shorter than would be predicted by classical mechanics.

♦ Mean mark (ii) 45%.

PHYSICS, M6 2020 HSC 23

The graph shows data for a motor connected to a 240 V power supply.
 

The equation for the torque, `tau`, produced by the motor is  `tau=frac{V I \eta}{\omega}`

where    `tau\ ` `=` torque (N m)
  `V \` `=`  voltage (V)
  `I\ ` `=` current (A)
  `eta\ ` `=` efficiency `=` 0.3
  `omega\ ` `=` angular velocity (rad `text{s}^(-1)`)

 
A circuit breaker cuts the current to the motor if the current exceeds 5  A.

Determine what will happen when the motor produces a torque of 2.95 N m. Show relevant calculations.   (3 marks)

Show Answers Only

`I` = 4.1 A

Current is below 5 A, so the circuit breaker will not cut the current and the motor will continue to run.

Show Worked Solution

From the graph, when `tau =` 2.95, `omega ~~ 100`:

`tau` `=(VI eta)/(omega)`  
`I` `=tau(omega)/(Veta)=2.95 xx(100)/(240 xx0.3)=4.1\ \text{A}`  

 

  • Current is below 5 A, so the circuit breaker will not cut the current and the motor will continue to run.

PHYSICS, M8 2020 HSC 8 MC

A uranium isotope, \(U\), undergoes four successive decays to produce \(Q\).
 

Which row of the table correctly shows the decay process \(R\) and product \(Q\) ?

\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}\textit{Process R} \rule[-1ex]{0pt}{0pt}& \textit{Product Q} \\
\hline \rule{0pt}{2.5ex} \alpha \rule[-1ex]{0pt}{0pt}& \text {Pa-230 } \\
\hline \rule{0pt}{2.5ex} \beta \rule[-1ex]{0pt}{0pt}& \text {Pa-234 } \\
\hline \rule{0pt}{2.5ex} \alpha \rule[-1ex]{0pt}{0pt}& \text {Th-230 } \\
\hline \rule{0pt}{2.5ex} \beta \rule[-1ex]{0pt}{0pt}& \text {Th-234 } \\
\hline
\end{array}
\end{align*} 

Show Answers Only

\(C\)

Show Worked Solution
  • Decay process \(R\)  reduces the number of both protons and neutrons by two → alpha decay.
  • Thorium has atomic number 90.

\(\Rightarrow C\)

PHYSICS, M8 2020 HSC 4 MC

The graph shows the mass of a radioactive isotope as a function of time.
 

What is the decay constant, in `text{years}^(-1)`, for this isotope?

  1. 0.0030
  2. 0.0069
  3. 2.0
  4. 100
Show Answers Only

`B`

Show Worked Solution
`t_((1)/(2))` `~~100` years  
`lambda` `=(ln 2)/(t_((1)/(2)))=(ln 2)/(100)=0.0069  text{years}^(-1)`  

 
`=>B`

PHYSICS, M7 2020 HSC 3 MC

What was the basis for Maxwell's prediction of the velocity of electromagnetic waves?

  1. Experiments using magnetic fields to accelerate particles
  2. Experiments using light and mirrors to establish the finite speed of light
  3. Equations showing how oscillating electric and magnetic fields propagate
  4. Equations showing how electromagnetic waves are affected by gravitational fields
Show Answers Only

`C`

Show Worked Solution
  • Maxwell used equations to show interactions between electric and magnetic fields.

`=>C`

CHEMISTRY, M7 2021 HSC 4 MC

The structure of ethyl pentanoate is shown.
 

Which pair of chemicals would produce ethyl pentanoate by esterification?

  1. Ethene and pentan-1-ol
  2. Ethane and pentanoic acid
  3. Ethanol and pentanoic acid
  4. Ethanoic acid and pentan-1-ol
Show Answers Only

`C`

Show Worked Solution
  • Esterification involves a reaction between a carboxylic aciand an alcohol.
  • Creating ethyl pentanoate requires ethanol (which is an alcohol) and pentanoic acid (which is a carboxylic acid).
  • Ethanol + pentanoic acid  ↔ ethyl pentanoate + water

`=> C`

PHYSICS, M7 2021 HSC 26

A student performs an experiment to measure Planck's constant, \(h\), using a device that emits specific frequencies of light when specific voltages are applied.

The voltage, \(V\), needed to produce each frequency, \(f\), is given by

\(V=\dfrac{h f}{q_e}\)

where `q_e` is the charge on an electron.

Data from the experiment is shown.
 

\begin{array}{|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Data point}\rule[-1ex]{0pt}{0pt}& \textit{Frequency} & \quad \textit{Voltage} \quad  \\
\text{}\rule[-1ex]{0pt}{0pt}& (\times 10^{14}\,\text{Hz})& \text{(V)} \\
\hline
\hline \rule{0pt}{2.5ex}1 \rule[-1ex]{0pt}{0pt}& 3.5 & 1.3 \\
\hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 4.8 & 1.7 \\
\hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 5.3 & 1.9 \\
\hline \rule{0pt}{2.5ex}4 \rule[-1ex]{0pt}{0pt}& 7.0 & 2.6 \\
\hline
\end{array}

  1. Graph this data on the axes provided. Include a line of best fit.   (3 marks)
     
     

       
  2. The student proposes using data point 1 to calculate a value for Planck's constant.
  3. Justify a better method to calculate Planck's constant from the experimental data provided.   (3 marks)

Show Answers Only

a.   
         
 

b.    Determine gradient using the line of best fit of  \(V=\dfrac{h f}{q_e}\)

  • The gradient is equal to \(\dfrac{V}{f}\), and \(h\) can be calculated as \(h=\) gradient  \(\times q_e\).
  • This method increases the accuracy by taking more data points into account.

Show Worked Solution

a.   
         
 

b.    Determine gradient using the line of best fit of  \(V=\dfrac{h f}{q_e}\)

  • The gradient is equal to \(\dfrac{V}{f}\), and \(h\) can be calculated as \(h=\) gradient  \(\times q_e\).
  • This method increases the accuracy by taking more data points into account.

PHYSICS, M6 2021 HSC 21

A DC motor is constructed from a single loop of wire with dimensions 0.10 m × 0.07 m. The magnetic field strength is 0.40 T and a current of 14 A flows through the loop.
 

  1. Calculate the magnitude of the maximum torque produced by the motor.   (2 marks)
  1. Describe how the magnitude of the torque changes as the loop moves through half a rotation from the position shown.   (2 marks)
Show Answers Only

a.   0.04  Nm

b.    Magnitude of torque changes:

  • The torque is initially at a maximum and decreases to zero after a 90º rotation.
  • The torque then increases to a maximum at 180º from the coil’s original position.
Show Worked Solution
a.    `tau_max` `=nIAB sin theta`
    `=1xx14 xx0.1 xx0.07 xx0.40 xx sin(90^(@))`
    `=0.04`  Nm

 

b.    Magnitude of torque changes:

  • The torque is initially at a maximum and decreases to zero after a 90º rotation.
  • The torque then increases to a maximum at 180º from the coil’s original position.

BIOLOGY, M8 2021 HSC 4 MC

Which is the most effective strategy for treating non-infectious diseases?

  1. Hygiene
  2. Pharmaceuticals
  3. Quarantine
  4. Vaccination
Show Answers Only

`B`

Show Worked Solution
  • Pharmaceuticals is the only option that can be used against non-infectious disease.
  • All other options treat infectious diseases.

`=>B`

BIOLOGY, M6 2021 HSC 3 MC

A scientist transferred male gametes from one plant to another to achieve a desired characteristic in the offspring.

Which genetic technology was the scientist using?

  1. Gene cloning
  2. Artificial pollination
  3. Artificial insemination
  4. Whole organism cloning
Show Answers Only

`B`

Show Worked Solution
  • The statement describes the manual transferral of pollen from the anther of one plant to the stigma of another which is known as artificial pollination.

`=>B`

BIOLOGY, M5 2021 HSC 2 MC

Which of the following photographs shows an example of sexual reproduction?
 

 

Show Answers Only

`D`

Show Worked Solution

By Elimination

  • Budding in fungi, binary fission in bacteria and runners in plants are all examples of asexual reproduction (Eliminate A, B and C).
  • Frog spawning requires a female to lay eggs and a male frog to fertilise them with sperm.

`=>D`

Calculus, MET2 2020 VCAA 4

The graph of the function  `f(x)=2xe^((1-x^(2)))`, where  `0 <= x <= 3`, is shown below.
 

  1. Find the slope of the tangent to `f` at  `x=1`.   (1 mark)

  2. Find the obtuse angle that the tangent to `f` at  `x = 1`  makes with the positive direction of the horizontal axis. Give your answer correct to the nearest degree.   (1 mark)

  3. Find the slope of the tangent to `f` at a point  `x =p`. Give your answer in terms of  `p`.   (1 mark)

  4.  i. Find the value of `p` for which the tangent to `f` at  `x=1` and the tangent to `f` at  `x=p`  are perpendicular to each other. Give your answer correct to three decimal places.   (2 marks)

  5. ii. Hence, find the coordinates of the point where the tangents to the graph of `f` at  `x=1`  and  `x=p`  intersect when they are perpendicular. Give your answer correct to two decimal places.   (3 marks)

Two line segments connect the points `(0, f(0))`  and  `(3, f(3))`  to a single point  `Q(n, f(n))`, where  `1 < n < 3`, as shown in the graph below.
 
         
 

  1.   i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
  2.      Find the equation of this line segment in terms of  `n`.   (1 mark)

  3.  ii. The second line segment connects the point `Q(n, f(n))` and the point  `(3, f(3))`, where  `1 < n < 3`.
  4.      Find the equation of this line segment in terms of `n`.   (1 mark)

  5. iii. Find the value of `n`, where  `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
  6.      Give your answer correct to three decimal places.   (3 marks)

Show Answers Only
  1. `-2`
  2. `117^@`
  3. `2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
  4.  i. `0.655`
  5. ii. `(0.80, 2.39)`
  6.   i. `y_1=2e^((1-n^2))x`
  7.  ii. `y_2=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3) + 6e^(-8)`
  8. iii. `n= 1.088`
Show Worked Solution

a.   `f(x)=2xe^((1-x^(2)))`

`f^{′}(1)=-2`

♦ Mean mark part (b) 37%.

 

b.   `text{Solve:}\ tan theta =-2\ \ text{for}\ \ theta in (pi/2, pi)`

`theta = 117^@`
 

c.   `text{Slope of tangent}\ = f^{′}(p)`

`f^{′}(p)=2(1-2p^(2))e^(1-p^(2))\ \ text{or}\ \ (2e-4p^(2)e)e^(-p^(2))`
 

d.i.   `text{If tangents are perpendicular:}`

`f^{′}(p) xx-2 =-1\ \ =>\ \ f^^{′}(p)=1/2`

`text{Solve}\ \ 2(1-2p^(2))e^(1-p^(2))=1/2\ \ text{for}\ p:`

`p=0.655\ \ text{(to 3 d.p.)}`
 

d.ii.  `text{Equation of tangent at}\ \ x=1: \ y=4-2x`

♦ Mean mark part (d)(ii) 41%.

`text{Equation of tangent at}\ \ x=p: \ y= x/2 + 1.991…`

`text{Solve}\ \ 4-2x = x/2 + 1.991…\ \ text{for}\ x:`

`=> x = 0.8035…`

`=> y=4-2(0.8035…) = 2.392…`

`:.\ text{T}text{angents intersect at (0.80, 2.39)}`

♦ Mean mark part (e)(i) 44%.

 
e.i.  `Q (n, 2n e^(1-n^2))`

`m_(OQ) = (2n e^((1-n^2)) – 0)/(n-0) = 2e^((1-n^2))`

`:.\ text{Equation of segment:}\ \ y_1=2e^((1-n^2))x`

♦♦ Mean mark part (e)(ii) 28%.
 

e.ii.  `P(3, f(3)) = (3, 6e^(-8))`

`m_(PQ) = (2n e^((1-n^2))-6e^(-8))/(n-3)`

`text{Equation of line segment:}`

`y_2-6e^(-8)` `=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3)`  
`y_2` `=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3) + 6e^(-8)`  
♦♦ Mean mark part (e)(iii) 28%.

 

e.iii.  `text{Find}\ n\ text{where shaded areas are equal.}`

`text{Solve}\ int_(0)^(n)(f(x)-y_(1))\ dx=int_(n)^(3)(y_(2)-f(x))\ dx\ \ text{for} n:`

`=> n= 1.088\ \ text{(to 3 d.p.)}`

GRAPHS, FUR2 2020 VCAA 2

Kyla organises fundraiser car shows at the business.

  1. Fifteen stallholders pay a set-up fee of $120 each.
  2. Twenty-six competitors pay $30 each to enter their cars to win prizes.
  3. How much money in total is raised from stallholders and competitors at each car show?   (1 mark)
  4. Admission fees for the show are $5 per adult and $2 per child.
  5. $1644 was raised from the 537 people who attended the most recent car show.
  6. How many children attended this car show?   (1 mark)
  7. The table below displays the speed of the car, `s`, in kilometres per hour, and the fuel consumption, `F`, in kilometres per litre, of one of the cars at two particular speeds.
     
       

  8. The equation `F = k/s`, where `k` is the constant of proportionality, is used to model the relationship between fuel consumption and speed of the car.
  9. Determine the fuel consumption of the car, in kilometres per litre, when it is travelling at 80 km/h.   (1 mark)

Show Answers Only
  1. `$2580`
  2. `347`
  3. `7.5 \ text{km/litre}`
Show Worked Solution
a.    `text{Total raised}` `= 15 xx 120 + 26 xx 30`
    `= $2580`

 

b.   `text{Let} \ \ A = text{number of adults}`

`text{Let} \ \ C = text{number of children}`

`A + C = 537 \ …\ (1)`

`5A + 2C = 1644 \ …\ (2)`
 
`text{Multiply} \ (1) xx 2`

`2A + 2C = 1074 \ …\ (3)`

`”Subtract “(2) – (3)`

`3A = 570\ \ =>\ \ A = 190`

`text{Substitute}\ \ A = 190 \ \ text{into (1)}`

`:. \ C = 537 – 190 = 347`

♦ Mean mark part (c) 48%.

 

c.    `F` `= k/s`
  `10` `= k/60`
  `k` `= 600`

 
`text{Find} \ F \ text{when} \ S = 80:`

`F` `= 600/80`
  `= 7.5 \text{km/litre}`

Calculus, MET2 2020 VCAA 1

Let  `f:R rarr R, \ f(x)=a(x+2)^(2)(x-2)^(2)`, where `a in R`. Part of the graph of `f` is shown below. 
 

  1. Show that  `a = 1/4`.    (1 mark)

  2. Express  `f(x)=(1)/(4)(x+2)^(2)(x-2)^(2)`  in the form  `f(x)=(1)/(4)x^(4)+bx^(2)+c`  where `b` and `c` are integers.   (1 mark)

Part of the graph of the derivative function  `f^{′}` is shown below.
 
     
 

  1.  i. Write the rule for `f^{′}` in terms of `x`.   (1 mark)

  2. ii. Find the minimum value of the graph of `f^{′}` on the interval  `x in (0, 2)`.   (2 marks)  

Let  `h:R rarr R, \ h(x)=-(1)/(4)(x+2)^(2)(x-2)^(2)+2`. Parts of the graph of `f` and `h` are shown below.
 
 
       
 

  1. Write a sequence of two transformations that map the graph of `f` onto the graph of `h`.   (1 mark)

     
       
     

  2.   i. State the values of `x` for which the graphs of `f`and `h` intersect.   (1 mark)

  3.  ii. Write down a definite integral that will give the total area of the shaded regions in the graph above.   (1 mark)

  4. iii. Find the total area of the shaded regions in the graph above. Give your answer correct to two decimal places.   (1 mark)

  5. Let `D` be the vertical distance between the graphs of `f`and`h`.
  6. Find all values of `x` for which `D` is at most 2 units. Give your answers correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `f(x)=(1)/(4)x^(4)-2x^(2)+4`
  3.  i. `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
  4. ii. `-(16sqrt3)/(9)`
  5. `text{Reflect in the x-axis then translate 2 units up, or translate 2 units down then reflect in the x-axis.}`
  6.   i. `x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
  7.  ii. `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or “int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
  8. iii. `text(2.72 u²)`
  9. `-2.61 <= x <= -1.08,1.08 <= x <= 2.61`
Show Worked Solution

a.   `text(S)text(ince)\ f(x)\ text{passes through (0, 4):}`

`4` `=a(2)^2(-2)^2`  
`4` `=16a`  
`:.a` `=1/4\ \ text(… as required)`  

 

b.  `f(x)` `=(1)/(4)(x+2)^(2)(x-2)^(2)`  
  `=(1)/(4)x^4-2x^2+4`  

 
c.i.  `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
 

c.ii.  `text{Solve}\ \ f^{″}(x)=0\ \ \text{for}\ x:`

`3x^2-4=0`

`=> \ x=(2sqrt3)/(3)`

`f^{′}((2sqrt3)/(3))=-(16sqrt3)/(9)`
 

d.  `text{Reflect in the x-axis then translate 2 units up, or}`

`text{translate 2 units down then reflect in the x-axis.}`
 

e.i.  `text{By CAS, solve}\ \ f(x)=h(x)\ \ text{for}\ x:}`

`x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
 

e.ii.  `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or”`

`int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
 

e.iii.  `text(Area = 2.72 u²)`

♦♦ Mean mark part (f) 26%.

 

f.   `text{By CAS, solve}\ -2 <= h(x)-f(x) <= 2\ \ \text{for}\ x:`

`-2.61 <= x <= -1.08, \ 1.08 <= x <= 2.61,” or”`

`-sqrt(4+2sqrt2) <= x <= -sqrt(4-2sqrt2), \ sqrt(4-2sqrt2) <= x <= sqrt(4+2sqrt2)`

GRAPHS, FUR2 2020 VCAA 1

Kyla owns and manages a truck and car care business.

After a major repair on a truck, one of the mechanics took it on a long test drive.

The test drive started at 12 noon.

After four hours, the mechanic stopped to rest for one hour and then returned to the workshop.

The graph below shows the truck's distance from the workshop, `d`, in kilometres, and the number of hours of test driving, `n`, after 12 noon. 
 

  1. At what time of the day did the mechanic arrive back at the workshop?  (1 mark)
  2. Find the average speed, in kilometres per hour, of the truck during the first four hours of the test drive.   (1 mark)
  3. On the return trip, the truck travelled at an average speed of 80 km/h.
  4. The equation of the line segment `VW` that represents this part of the test drive is of the form
  5.     `d = k - 80n`
  6. Find the value of `k`.   (1 mark)

Show Answers Only
  1. `8 \ text{pm}`
  2. `60 \ text{km/h}`
  3. `k = 640`
Show Worked Solution

a.    `n = 0 \ text{at} \ 12 \ text{noon}`

`text{Mechanic arrives back at}\ \ n = 8\ \ text{which is} \ 8 text{pm.}`
 

b.    `text{Average speed}` `= text{distance}/text{time}`
    `= 240/4`
    `= 60\ text{km/h}`

 

c.    `text{Substitute (8, 0) into}\ \ d = k – 80n:`

♦ Mean mark 43%.

`0 = k – 80 xx 8`

`k = 640`

Functions, 2ADV F2 SM-Bank 9 MC

The graph of the function  `f(x)=(3x+2)/(5-x)`, has asymptotes at

  1. `x=-5,y=(3)/(2)`
  2. `x=(2)/(3),y=-3`
  3. `x=5,y=3`
  4. `x=5,y=-3`
Show Answers Only

`D`

Show Worked Solution
`f(x)` `=(3x+2)/(5-x)`  
  `=(-(15-3x)+17)/(5-x)`  
  `=-3+17/(5-x)`  

 
`text(Vertical asymptote:)\ \ x=5`

`text(Horizontal asymptote:)\ \ y=-3`

`=>D`

Graphs, MET2 2020 VCAA 5 MC

The graph of the function  `f:D rarr R,f(x)=(3x+2)/(5-x)`, where `D` is the maximal domain, has asymptotes

  1. `x=-5,y=(3)/(2)`
  2. `x=-3,y=5`
  3. `x=(2)/(3),y=-3`
  4. `x=5,y=3`
  5. `x=5,y=-3`
Show Answers Only

`E`

Show Worked Solution
`f(x)` `=(3x+2)/(5-x)`  
  `=(-(15-3x)+17)/(5-x)`  
  `=-3+17/(5-x)`  

 
`text(Vertical asymptote:)\ \ x=5`

`text(Horizontal asymptote:)\ \ y=-3`

`=>E`

Calculus, MET2 2020 VCAA 3 MC

 Let  `f^(′)(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
  5. `f(x)=sqrt(2x-3)`
Show Answers Only

`=>C`

Show Worked Solution
`f^(′)(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3) – 2`

`=>C`

Statistics, 2ADV S3 SM-Bank 8 MC

The heights of females living in a small country town are normally distributed:

    • 16% of the females are more than 160 cm tall.
    • 2.5% of the females are less than 115 cm tall.

The mean and the standard deviation of this female population, in centimetres, are closest to

  1. mean = 135               standard deviation = 15
  2. mean = 135               standard deviation = 25
  3. mean = 145               standard deviation = 15
  4. mean = 145               standard deviation = 20
Show Answers Only

`C`

Show Worked Solution

`160 -> ztext{-score} = 1`

`115 -> ztext{-score} = -2`

`1` `= {160 – mu}/sigma`
`mu + sigma` `= 160\ …\ (1)`
`-2` `= {115-mu}/sigma`
`mu-2 sigma` `= 115\ …\ (2)`

 
`(1)-(2)`

`3sigma` `=45`  
`sigma` `=15`  

 
`text{Substitute}\ \ sigma = 15\ \ text{into (1)}`

`mu = 145` 

`=> C`

Statistics, 2ADV S2 SM-Bank 5 MC

 The stem plot below shows the height, in centimetres, of 20 players in a junior football team.
 

A player with a height of 179 cm is considered an outlier because 179 cm is greater than

  1. 162 cm
  2. 169 cm
  3. 173 cm
  4. 175.5 cm
Show Answers Only

`D`

Show Worked Solution

`Q_1 = (148 + 148)/2 = 148`

`Q_3 = (158 + 160)/2 = 159`

`IQR = 159 – 148 = 11`

`text{Upper fence}` `= Q_3 + 1.5 xx IQR`
  `= 159 + 1.5 xx 11`
  `= 175.5`

 
`=> D`

Calculus, 2ADV C4 SM-Bank 5

Let  `f^(′)(x) = x^3 + x`.

Find  `f(x)`  given that  `f(1) = 2`.  (2 marks)

Show Answers Only

`f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Show Worked Solution
`f^(′)(x)` `= x^3 + x`  
`f(x)` `=int x^3 + x\ dx`  
  `=1/4 x^4 + 1/2 x^2 +c`  

 
`text(Given)\ f(1) = 2:`

`2` `= 1/4 + 1/2 + c`  
`c` `= 5/4`  

 
`:. f(x) =1/4 x^4 + 1/2 x^2 +5/4`

GRAPHS, FUR2 2021 VCAA 2

z

Christy sells blocks of land at the housing estate.

Her pay is based on the number of blocks that she sells each week.

The maximum number of blocks sold each week is 20.

The graph below shows Christy's weekly pay, in dollars, for the number of blocks sold.

  1. What is the minimum number of blocks of land that Chirsty must sell to receive $6000 in one week?   (1 mark)

John also sells blocks of land. He is paid $1000 for each block that he sells.

  1. Write down all values for the number of blocks of land sold for which Christy and John will receive the same weekly pay.  (1 mark)
  2. John sells no more than three blocks of land for every five blocks of land sold by Christy.  (1 mark)
  3. Let  `x`  be the number of blocks that Christy sells.
  4. Let  `y`  be the number of blocks that John sells.
  5. Write the inequality, written as  `y` in terms of  `x`, that represents this situation.
Show Answers Only
  1. `6 \ text{blocks}`
  2. `text{Blocks sold for equal pay: 3, 6, 11, 15, 20}`
  3. `y <= 3/5 x`
Show Worked Solution

a.    `6 \ text{blocks}`
 

b.    `text{Blocks sold for equal pay: 3, 6, 11, 15, 20}`
 

c.     `y <= 3/5 x`

GRAPHS, FUR2 2021 VCAA 3

Lam is a builder constructing a community centre at the new housing estate.

The cost, `C`, in dollars, for Lam to work onsite for `n` weeks is given by

`C = 10 \ 000 + k xx n` 

  1. The cost of 15 weeks of onsite work is $92 500.
  2. Show that the value of `k` is 5500.   (1 mark)

Lam's revenue for this job is $6500 per week

  1. Determine the number of weeks that Lam needs to work onsite to break even.   (1 mark)
  2. An equation for the profit, `P`, in dollars, made by Lam in terms of the number of weeks worked onsite, `n`, can be determined.
  3. Complete this equation by filling in the boxes below.   (1 mark)
`P =` 
 
  `xx n +` 
 
Show Answers Only
  1. `5500`
  2. `10 \ text{weeks}`
  3. `1000 n – 10 \ 000`
Show Worked Solution
a. `C` `= 10 \ 000 + k n`
     `92\ 500` `= 10 \ 000 + 15 k`
  `15 k` `= 82 \ 500`
  `:. k` `= (82\ 500)/15`
    `= 5500`

 

b.   `text{Breakdown occurs when} \ \ R = C`

`6500n` `= 10 \ 000 + 5500 n`
`1000n` `= 10 \ 000`
`:. n` `= {10\ 000}/{1000}`
  `= 10 \ text{weeks}`

 

c.   `P` `= R – C`
    `= 6500n – (10 \ 000 + 5500n)`
    `= 1000n – 10 \ 000`

 

GRAPHS, FUR2 2021 VCAA 1

The graph below shows the height, in metres, of a drone flying above a new housing estate over a six-minute period of time.
 

  1. For what length of time, in minutes, was the height of the drone a least 50 m?   (1 mark)
  2. What was the average rate of change in the height of the drone, in metres per minute, in the first two minutes?  (1 mark)
Show Answers Only
  1. `4 \ text{minutes}`
  2. `100 \ text{metres per minute}`
Show Worked Solution

a.   `4 \ text{minutes}`
 

b.   `text{At} \ t = 0 , \ text{height} = 0 \ text{m}`

`text{At} \ t = 2 , \ text{height} = 200 \ text{m}`
  
`:. \ text{Average rate of change}`

`= {200 – 0}/{2}`

`= 100 \ text{metres per minute}`

GEOMETRY, FUR2 2021 VCAA 3

Players will travel from around the world to a squash competition in New York City (41° N, 74° W).

The top three female players are from three different cities:

  • Wei-Yi is from Shanghai (31° N, 121° E) in China.
  • Camilla is from Durban (30° S, 31° E) in South Africa.
  • Ozlem is from Istanbul (41° N, 29° E) in Turkey.
  1. Which players are from the Northern Hemisphere?   (1 mark)

The diagram below shows the location of Shanghai, Durban and New York City.


           
 

  1. The city Istanbul (41° N, 29° E) is missing from the diagram.
  2. On the diagram above, mark with an `X` the location of Istanbul.   (1 mark)
  3. The flight from Istanbul (41° N, 29° E) to New York City (41° N, 74° W) travels along a small circle.
  4. Assume that the radius of Earth is 6400 km.
  5.  i. Show that the radius of this small circle, rounded to the nearest kilometres, is 4830.  (1 mark)
  6. ii. Calculate the distance that the plane travels between Istanbul (41° N, 29° E) and New York City (41° N, 74° W).
  7.     Round your answer to the nearest kilometre.   (1 mark)
  8. Wei-Yi and Camilla both arrive in New York City on Monday 11 January at 10.00 pm, local time.
  9. Wei-Yi's flight left Shanghai on Monday 11 January at 8.00 pm, local time.
  10. Camilla's flight left Durban on Monday 11 January at 4.00 am, local time.
  11. Assume that the time difference between Shanghai and New York City is 13 hours, and that the time difference between Durban and New York City is seven hours.
  12. How much longer, in hours, was Camilla's flight compared to Wei-Yi's flight?   (1 mark)
Show Answers Only
  1. `text{Wei-Yi and Ozlem}`
  2.  
     

     
  3.  i.  `text(See Worked Solutions)`
  4. ii. `8683 \ text{km (nearesrt km)}`
  5. `10 \ text{hours}`
Show Worked Solution

a.    `text{Northern Hemisphere} \ to \ text{latitude is °N}`

`text{Wei-Yi and Ozlem}`
 

b.    `text{Istanbul} \ to \ text{same latitude as New York}`

`text{Longitude 29° E is just left of Durban (31° E)}`
 

 

c.i.

`cos 41^@` `= r/6400`
`r` `= 6400 xx cos 41^@`
  `= 4830.14 …`
  `= 4830 \ text{km (nearest km)}`

 

c.ii.

    `text{Arc Length}` `= 103/360 xx 2 xx pi xx 4830`
    `= 8682.83`
    `= 8683 \ text{km (nearest km)}`

 

d.    `text{In New York time}`

`text{Wei-Yi departure = 8:00 pm less 13 hours = 7:00 am Monday (NYT)}`

`text{Wei-Yi’s flight time = 7:00 am to 10:00 pm = 15 hours}`

`text{Camilla’s departure = 4:00 am less 7 hours = 9:00 pm Saturday (NYT)}`

`text{Camilla’s flight time =  9:00 pm (Sunday) to 10:00 pm (Monday) = 25 hours}`

`:. \ text{Extra flight time}` `= 25 – 15`
  `= 10 \ text{hours}`

GEOMETRY, FUR2 2021 VCAA 2

The game of squash is played indoors on a court with a front wall, a back wall and two side walls, as shown in the image below.
 

 
Each side wall has the following dimensions.
 

The shaded region in the diagram above is considered part of the playing area.

  1. Calculate the area, in square metres, of the shaded region in the diagram above.
  2. Round your answer to two decimal places.   (1 mark)
  3. Calculate the perimeter, in metres, of the shaded region in the diagram.
  4. Round your answer to one decimal place.   (1 mark)
  5. Two squash players, Wei-Yi and Bao, are playing a match.
  6. The diagram below shows the lines on the court floor.

 
             

  1. Wei-Yi is serving from point `A` and and Bao is standing at point `B`.
  2. Point `A` is 2.7 m from Point `C`.
  3. Point `B` is 3.1 m from Point `C`.
  4. The angle `ACB` is 119°.
  5. Show that the distance between Wei-Yi and Bao is 5 m, rounded to the nearest metre.   (1 mark)
Show Answers Only
  1. `32.66 \ text{m}^2`
  2. `26.5 \ text{m}`
  3. `5 \ text{m}`
Show Worked Solution
a.   `text{Shaded Area}` `= 1/2 xx 9.75 xx (4.57 + 2.13)`
    `= 32.6625`
    `= 32.66 \ text{m}^2 \ text{(to 2 d.p.)}`

 

b.

`x = 4.57 – 2.13 = 2.44 \ text{m}`
  

`text{Using Pythagoras}`

`y` `= sqrt{2.44^2 + 9.75^2}`
  `= 10.050 …\ text{m}`

 

`:. \ text{Perimeter}` `= 4.57 + 9.75 + 2.13 + 10.05`
  `= 26.5 \ text{m}`

 
c. 

`text{Using cosine rule:}`

`AB^2` `=  AC^2 + CB^2 – 2 xx AC xx CB xx cos 119^@`
  `= 2.7^2 + 3.1^2 – 2 xx 2.7 xx 3.1 xx cos119^@`
  `= 25.0157`
`:.AB` `= 5 \ text{m (nearest metre)}`

GEOMETRY, FUR2 2021 VCAA 1

The game of squash is played with a special ball that has a radius of 2 cm.

  1. Show that the volume of one squash ball, rounded to two decimal places, is 33.51 cm3.  (1 mark)

Squash balls may be sold in cube-shaped boxes.

Each box contains one ball and has a side length of 4.1 cm, as shown in the diagram below.
 

  1. Calculate the empty space, in cubic centimetres, that surrounds the ball in the box.
  2. Round your answer to two decimal places.   (1 mark)
  3. Calculate the total surface area, in square centimetres, of one box.   (1 mark)
  4. Retail shops store the cube-shaped boxes in a space within a display unit.
  5. The space has a length of 17.0 cm and a width of 12.5 cm. Due to the presence of a shelf above, there is a maximum height of 8.5 cm available. This is shown in the diagram below.
     

     

  1. Calculate the maximum number of cube-shaped boxes that can fit into the space within the display unit.   (1 mark)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `35.41 \ text{cm}^3`
  3. `100.86 \ text{cm}^2`
  4. `text(24 boxes)`
Show Worked Solution
a.   `V` `= 4/3 pi r^3`
    `= 4/3 xx pi xx 2^3`
    `= 33.5103 …`
    `= 33.51 \ text{cm}^3 \ text{(to 2 d.p.)}`

 

b.  `text{Volume of cube} = 4.1^3 = 68.921 \ text{cm}^3`

     `:. \ text{Empty space}` `= 68.921 – 33.51`
  `= 35.41 \ text{cm}^3 \ text{(to 2 d.p.)}`

 

c.   `text(S.A.)` `= 6 xx 4.1 xx 4.1`
    `= 100.86 \ text{cm}^2`

 

d.  `text{Length} = 17 div 4.1 = 4.146 … \ \ to 4 \ text{boxes}`

`text{Width} = 12.5 div 4.1 = 3.048 … \ to 3 \ text{boxes}`

`text{Height} = 8.5 div 4.1 = 2.073 … \ \ to 2\ text{boxes}`
 

`:. \ text{Max boxes}` `= 4 xx 3 xx 2`
  `= 24`

NETWORKS, FUR2 2021 VCAA 2

George lives in Town `G` and Maggie lives in Town `M`.

The diagram below shows the network of main roads between Town `G` and Town `M`.

The vertices `G, H, I, J, K, L, M, N` and `O` represent towns.

The edges represent the main roads. The number on the edges indicate the distance, in kilometres, between adjacent towns.
 

  1. What is the shortest distance, in kilometres, between Town `G` and Town `M`?   (1 mark)

  2. George plans to travel to Maggie's house. He will pass through all the towns shown above.
  3. George plans to take the shortest route possible.
  4. Which town will George pass through twice?    (1 mark)

Show Answers Only
  1. `86 \ text{km}`
  2. `text{Passes through town} \ K \ text{twice.}`
Show Worked Solution
a.  `text{Shortest route}` `= G \ O \ N \ M`
    `= 28 + 42 + 16`
    `= 86 \ text{km}`

 
b.  `text{Shortest route through all towns} = G \ H \ I \ K \ L \ K \ J\ O \ N \ M`

`:. \ text{Passes through town} \ K \ text{twice}`

NETWORKS, FUR2 2021 VCAA 1

Maggie's house has five rooms, `A, B, C, D` and `E`, and eight doors.

The floor plan of these rooms and doors is shown below. The outside area, `F`, is shown shaded on the floor plan.
 

The floor plan is represented by the graph below.

On this graph, vertices represent the rooms and the outside area. Edges represent direct access to the rooms through the doors.

One edge is missing from the graph.
 

  1. On the graph above, draw the missing edge.   (1 mark)

  2. What is the degree of vertex `E`?   (1 mark)

  3. Maggie hires a cleaner to clean the house.
  4. It is possible for the cleaner to enter the house from the outside area, `F`, and walk through each room only once, cleaning each room as he goes and finishing in the outside area, `F`.
  5.  i. Complete the following to show one possible route that the cleaner could take.   (1 mark)

        
         
    ii. What is the mathematical term for such a journey?   (1 mark)

Show Answers Only
  1.  
  2. `2`
  3.  i. `FABEDCF\ text(or)\ FCDEBAF`
  4. ii. `text{Hamiltonian cycle}`
Show Worked Solution

a.

b.  `text{Degree} = 2`
 

c.i.  `FABEDCF\ text(or)\ FCDEBAF`

c.ii.  `text{Hamiltonian cycle}`

MATRICES, FUR2 2021 VCAA 2

The main computer system in Elena's office has broken down.

The five staff members, Alex (`A`), Brie (`B`), Chai (`C`), Dex (`D`) and Elena (`E`), are having problems sending information to each other.

Matrix `M` below shows the available communication links between the staff members.

`qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

`qquadqquadqquadqquadqquadqquadqquad \ \ \ A \ \ B \ \ C \ \ D \ \ E`

`M= \ text{sender} \ \ {:(A),(B),(C),(D),(E):} [(0,1,0,0,1),(0,0,1,1,0),(1,0,0,1,0),(0,1,0,0,0),(0,0,0,1,0)] `

In this matrix:

  • the '1' in row `A`, column `B` indicates that Alex can send information to Brie
  • the '0' in row `D`, column `C` indicates that Dex cannot send information to Chai.
  1. Which two staff members can send information directly to each other?   (1 mark)

  2. Elena needs to send documents to Chai.
  3. What is the sequence of communication links that will successfully get the information from Elena to Chai?   (1 mark)

  4. Matrix  `M^2` below is the square of `M` and shows the number of two-step communication links between each pair of staff members.
     

    `qquadqquadqquadqquadqquadqquadqquadqquadqquad text(receiver)`

    `qquadqquadqquadqquadqquadqquadqquad \ \ \ A \ \ B \ \ C \ \ D \ \ E`

    `M= \ text{sender} \ \ {:(A),(B),(C),(D),(E):} [(0,0,1,2,0),(0,1,0,1,0),(0,1,0,0,0),(0,0,1,1,0),(0,1,0,0,0)] `

    Only one pair of individuals has two different two-step communication links.

  5. List each two-step communication link for this pair.   (1 mark)

Show Answers Only
  1. `text{B and D}`
  2. `text{Elena} to text{Dex} to text{Brie} to text{Chai}`
  3. `text{Alex} to text{Brie} to text{Dex}`
    `text{Alex} to text{Elena} to text{Dex}`
Show Worked Solution

a.   `text{B (sender) to D (receiver)} => 1`

`text{D (sender) to B (receiver)} => 1`

`:. \ text{B and D can send information to each other}`
 

b.   `text{Elena} to text{Dex} to text{Brie} to text{Chai}`
 

c.   `text{The two 2-step links are from Alex to Dex.}`

`text{These are:}`

`text{Alex} to text{Brie} to text{Dex}`

`text{Alex} to text{Elena} to text{Dex}`

MATRICES, FUR2 2021 VCAA 1

Elena imports three brands of olive oil: Carmani (`C`) Linelli (`L`) and Ohana (`O`).

The number of 1 litre bottles of these oils sold in January 2021 is shown in matrix `J` below.

`J = {:[(2800),(1700),(2400)]:} {:(C),(L),(O):}`

  1. What is the order of matrix `J`?   (1 mark)

  2. Elena expected that in February 2021 the sales of all three brands of olive oil would increase by 5%.
  3. She multiplied matrix `J` by a scalar value, `k` , to determine the expected volume of sales for February.
  4. What is the value of the scalar `k`.   (1 mark)

Show Answers Only
  1. `3 xx 1`
  2. `1.05`
Show Worked Solution

a.      `3 xx 1`
 

b.     `text{All brands} \ ↑ 5%`

 `:. k = 1.05`

CORE, FUR2 2021 VCAA 7

Sienna owns a coffee shop.

A coffee machine, purchased for $12 000, is depreciated in value using the unit cost method.

The rate of depreciation is $0.05 per cup of coffee made.

The recurrence relation that models the year-to-year value, in dollars, of the coffee machine is

`M_0 = 12 \ 000,`     `M_{n+1} = M_n - 1440`

  1. Calculate the number of cups of coffee that the machine produces per year.   (1 mark)

  2. The recurrence relation above could also represent the value of the coffee machine depreciating at a flat rate.
  3. What annual flat rate percentage of depreciation is represented?   (1 mark)

  4. Complete the rule below that gives the value of the coffee machine, `M_n`, in dollars, after `n` cups have been produced. Write your answers in the boxes provided.   (1 mark)

     
       `M_n =`
     
    		  `+`  
     
    		 `xx n`  

Show Answers Only

  1. `2880`
  2. `12text{% p.a.}`
  3. `12 \ 000 + (-0.05) xx n`

Show Worked Solution

a.   `text{Cups of coffee}` `= 1440/0.05`
    `= 28 \ 800`

 

b.   `text{Flat rate (%)}` `= {1440}/{12 \000}`
    `= 12text{% p.a.}`

   
c.  `M_n = 12 \ 000 + (-0.05) xx n`

CORE, FUR2 2021 VCAA 3

The time series plot below shows the winning time, in seconds, for the women's 100 m freestyle swim plotted against year, for each year that the Olympic Games were held during the period 1956 to 2016.

A least squares line has been fitted to the plot to model the decreasing trend in the winning time over this period.
 

The equation of the least squares line is

winning time = 357.1 – 0.1515 × year

The coefficient of determination is 0.8794

  1. Name the explanatory variable in this time series plot.   (1 mark)

  2. Determine the value of the correlation coefficient (`r`).
  3. Round your answer to three decimal places.   (1 mark)

  4. Write down the average decrease in winning time, in seconds per year, during the period 1956 to 2016.   (1 mark)

  5. The predicted winning time for the women's 100 m freestyle in 2000 was 54.10 seconds.
  6. The actual winning time for the women's 100 m freestyle in 2000 was 53.83 seconds.
  7. Determine the residual value in seconds.   (1 mark)

  8. The following equation can be used to predict the winning time for the women's 100 m freestyle in the future.
  9.      winning time =  357.1 – 0.1515 × year
  10.  i. Show that the predicted winning time for the women's 100 m freestyle in 2032 is 49.252 seconds.   (1 mark)

  11. ii. What assumption is being made when this equation is used to predict the winning time for the women's 100 m freestyle in 2032?   (1 mark)

Show Answers Only
  1. `text{year}`
  2. `- 0.938`
  3. `0.1515 \ text{seconds}`
  4. `-0.27`
  5.  i. `49.252 \ text{seconds}`
  6. ii. `text{The same trend continues when the graph is extended beyond 2016.}`
Show Worked Solution

a.      `text{year}`
 

b. `r^2` `= 0.8794 \ text{(given)}`
  `r` `= ± sqrt{0.8794}`
    `= ± 0.938 \ text{(to 3 d.p.)}`

 
`text{By inspection of graph, correlation is negative}`

`:. \ r = -0.938`
 

c.    `text{Average decrease in winning time = 0.1515 seconds}`

`text{(this is given by the slope of the line.)}`
 

d.    `text{Residual Value}` `= text{actual}-text{predicted}`
    `= 53.83-54.10`
    `= -0.27`

 

e.i.      `text{winning time (2032)}` `= 357.1-0.1515 xx 2032`
      `=49.252 \ text{seconds}`

 

e.ii.  `text{The assumption is that the graph is accurate when it is extended}`

  `text{beyond 2016 (i.e decreasing trend continues).}`

CORE, FUR2 2021 VCAA 1

In the sport of heptathlon, athletes compete in seven events.

These events are the 100 m hurdles, high jump, shot-put, javelin, 200 m run, 800 m run and long jump.

Fifteen female athletes competed to qualify for the heptathlon at the Olympic Games.

Their results for three of the heptathlon events – high jump, shot-put and javelin – are shown in Table 1

  1. Write down the number of numerical variables in Table 1.   (1 mark)

  2. Complete Table 2 below by calculating the mean height jumped for the high jump, in metres, by the 15 athletes. Write your answer in the space provided in the table.   (1 mark)

  3. In shot-put, athletes throw a heavy spherical ball (a shot) as far as they can. Athlete number six, Jamilia, threw the shot 14.50 m.
  4. Calculate Jamilia's standardised score (`z`). Round your answer to one decimal place.   (1 mark)

  5. In the qualifying competition, the heights jumped in the high jump are expected to be approximately normally distributed.
  6. Chara's jump in this competition would give her a standardised score of  `z = –1.0`
  7. Use the 68–95–99.7% rule to calculate the percentage of athletes who would be expected to jump higher than Chara in the qualifying competition.   (1 mark)

  8. The boxplot below was constructed to show the distribution of high jump heights for all 15 athletes in the qualifying competition.

 

  1. Explain why the boxplot has no whisker at its upper end.   (1 mark)

  2. For the javelin qualifying competition (refer to Table 1), another boxplot is used to display the distribution of athlete's results.
  3. An athlete whose result is displayed as an outlier at the upper end of the plot is considered to be a potential medal winner in the event.
  4. What is the minimum distance that an athlete needs to throw the javelin to be considered a potential medal winner?   (2 marks)

Show Answers Only

  1. `3`
  2. `1.81`
  3. `0.5 \ text{(to d.p.)}`
  4. `84text(%)`
  5. `text{See Worked Solutions}`
  6. `46.89 \ text{m}`

Show Worked Solution

a.    `3 \ text{High jump, shot-put and javelin}`

 `text{Athlete number is not a numerical variable}`
  

b.     `text{High jump mean}`

`= (1.76 + 1.79 + 1.83 + 1.82 + 1.87 + 1.73 + 1.68 + 1.82 +`

`1.83 + 1.87 + 1.87 + 1.80 + 1.83 + 1.87 + 1.78) ÷ 15`

`= 1.81`
 

c.   `z text{-score} (14.50)` `= {14.50-13.74}/{1.43}`
    `= 0.531 …`
    `= 0.5 \ text{(to 1 d.p.)}`

 
d.  `P (z text{-score} > -1 ) = 84text(%)`
 

e.  `text{If the} \ Q_3 \ text{value is also the highest value in the data set,}`

`text{there is no whisker at the upper end of a boxplot.}`
 

f.  `text{Javelin (ascending):}`

`38.12, 39.22, 40.62, 40.88, 41.22, 41.32, 42.33, 42.41, `

`42.51, 42.65, 42.75, 42.88, 45.64, 45.68, 46.53`

`Q_1 = 40.88 \ \ , \ Q_3 = 42.88 \ \ , \ \ IQR = 42.88-40.88 = 2`

`text{Upper Fence}` `= Q_3 + 1.5  xx IQR`
  `= 42.88 + 1.5 xx 2`
  `= 45.88`

 
`:. \ text{Minimum distance = 45.89 m  (longer than upper fence value)}`

Complex Numbers, SPEC2 2021 VCAA 2

The polynomial  `p(z) = z^3 + alpha z^2 + beta z + gamma`, where  `z ∈ C`  and  `alpha, beta, gamma ∈ R`, can also be written as  `p(z) = (z-z_1)(z-z_2)(z-z_3)`, where  `z_1 ∈ R`  and  `z_2, z_3 ∈ C`.

  1.  i. State the relationship between `z_2` and `z_3`.  (1 mark)

  2. ii. Determine the values of `alpha, beta` and `gamma`, given that  `p(2) = -13, |z_2 + z_3| = 0`  and  `|z_2-z_3| = 6`.  (3 marks)

Consider the point  `z_4 = sqrt3 + i`.

  1. Sketch the ray given by  `text(Arg)(z-z_4) = (5pi)/6`  on the Argand diagram below.  (2 marks)
     
  2. The ray  `text(Arg)(z-z_4) = (5pi)/6`  intersects the circle  `|z-3i| = 1`, dividing it into a major and a minor segment.
  3.  i. Sketch the circle  `|z-3i| = 1` on the Argand diagram in part b.  (1 mark)

  4. ii. Find the area of the minor segment.  (2 marks)

Show Answers Only
    1. `z_2 = barz_3`
    2. `alpha = -3, beta = 9, gamma = -27`
  2.  
     

     

c.i.   

c.ii.   `(2pi-3sqrt3)/12\ text(u²)`

Show Worked Solution

a.i.   `text(By conjugate root theory)`

`z_2 = barz_3`

 

a.ii.   `text(Let)\ \ z_1 = a + bi, \ z_2 = a-bi`

♦♦ Mean mark part (a.ii.) 35%.

`|z_2 + z_3| = |2a| = 0 \ => \ a = 0`

`|z_2-z_3| = |2b| = 6 \ => \ b = ±3`
 

`text(Using)\ \ p(2) = -13`

`(2-z_1)(2-3i)(2 + 3i)` `= -13`
`(2-z_1)(4 + 9)` `= -13`
`2-z_1` `= -1`
`z_1` `= 3`

 

`p(z)` `= (z-3)(z-3i)(z + 3i)`
  `= (z-3)(z^2 + 9)`
  `= z^3-3z^2 + 9z-27`

 
`:. alpha = –3, \ beta = 9, \ gamma = –27`

♦ Mean mark part (b) 45%.
MARKER’S COMMENT: Point of emanation is not part of required ray (see open circle).

 

b. 

 

c.i.  

 

c.ii.   `text(Arg)(z-z_4) = (5pi)/6\ \ text(cuts)\ \ ytext(-axis at angle)\ pi/3`

♦♦ Mean mark part (c.ii.) 30%.

`=>\ text(angle at centre of segment) = pi/3\ (text(equilateral triangle))`

`text(Area)` `= (pi/3)/(2pi) xx pi xx 1^2-1/2 xx 1 xx 1 xx sin (pi/3)`
  `= pi/6-sqrt3/4`
  `= (2pi-3sqrt3)/12\ text(u²)`

Calculus, MET2 2021 VCAA 2

Four rectangles of equal width are drawn and used to approximate the area under the parabola  `y = x^2`  from  `x = 0`  to  `x = 1`.

The heights of the rectangles are the values of the graph of  `y = x^2` at the right endpoint of each rectangle, as shown in the graph below.
 

  1. State the width of each of the rectangles shown above.  (1 mark)
  2. Find the total area of the four rectangles shown above.  (1 mark)
  3. Find the area between the graph of  `y = x^2`, the `x`-axis and the line  `x=1`.  (2 marks)
  4. The graph of `f` is shown below.
     
         

    Approximate  `int_(-2)^2 f(x)\ dx`  using four rectangles of equal width and the right endpoint of each rectangle.  (1 mark)

Parts of the graphs of  `y = x^2`  and  `y = sqrtx`  are shown below.
 
     

  1. Find the area of the shaded region.  (1 mark)
  2. The graph of  `y=x^2` is transformed to the graph of  `y = ax^2`, where  `a ∈ (0, 2]`.
  3. Find the values of `a` such that the area defined by region(s) bounded by the graphs of  `y = ax^2`  and  `y = sqrtx`  and the lines  `x = 0`  and  `x = a`  is equal to `1/3`. Give your answer correct to two decimal places.  (4 marks)
Show Answers Only
  1. `0.25`
  2. `15/32\ text(u)^2`
  3. `1/3\ text(u)^2`
  4. `-2`
  5. `1/3\ text(u)^2`
  6. `1.13`
Show Worked Solution

a.   `0.25`

 

b.    `text{Area}` `= 1/4 (1/16 + 1/4 + 9/16 + 1)`
    `= 15/32\ text(u)^2`

 

c.    `text{Area}` `= int_0^1 x^2 dx`
    `= 1/3\ text(u)^2`

 

d.

♦♦♦ Mean mark part (d) 16%.

`int_-2^2 f(x)\ dx` `≈ 6 + 2 – 4 – 6`
  `≈ -2`

 

e.   `text{Area}` `= int_0^1 (sqrtx – x^2) dx`
    `= 1/3\ text(u)^2`

 
f.   `text{Case 1:} \ a ≤ 1`

♦♦♦ Mean mark part (f) 18%.

`int_0^a (sqrtx – ax^2) dx = 1/3`

`a =  0.77 \ \ text{or} \ \ a = 1.00`
  

`text{Case 2:} \ a > 1`

`sqrtx = ax^2 \ => \ x = a^{- 3/2}`

`text{Solve for}\ a\ text{(by CAS)}:`

`int_0^{a^{- 3/2}} (sqrtx – ax^2) dx + int_{a^{- 3/2}}^a (sqrtx – ax^2) dx = 1/3`

`:. a = 1.13`

Calculus, MET2 2021 VCAA 1

A rectangular sheet of cardboard has a width of `h` centimetres. Its length is twice its width.

Squares of side length `x` centimetres, where  `x > 0`  are cut from each of the corners, as shown in the diagram below.
 

The sides  of this sheet of cardboard are then folded up to make a rectangular box with an open top, as shown in the diagram below.

Assume that the thickness of the cardboard is negligible and that  `V_text{box} > 0`.
 

A box is to be made from a sheet of cardboard with  `h` = 25 cm.

  1. Show that the volume, `V_text{box} > 0`, in cubic centimetres, is given by  `V_text{box} (x) = 2x (25-2x)(25-x)`.   (1 mark)

  2. State the domain of  `V_text{box}`.   (1 mark)

  3. Find the derivative of  `V_text{box}`  with respect to `x`.   (1 mark)

  4. Calculate the maximum possible volume of the box and for which value of `x` this occurs.   (3 marks)

  5. Waste minimisation is a goal when making cardboard boxes.
  6. Percentage wasted is based on the area of the sheet of cardboard that is cut out before the box is made.
  7. Find the percentage of the sheet of cardboard that is wasted when `x = 5`.   (2 marks)

Now consider a box made from a rectangular sheet of cardboard where  `h>0` and the box's length is still twice its width.

    1. Let  `V_text{box}` be the function that gives the volume of the box.
    2. State the domain `V_text{box}` in terms of `h`.   (1 mark)

    3. Find the maximum volume for any such rectangular box, `V_text{box}`, in terms of  `h`.   (3 marks)

  2. Now consider making a box from a square sheet of cardboard with side lengths of `h` centimetres.
  3. Show that the maximum volume of the box occurs when  `x = h/6`.   (2 marks)

Show Answers Only
  1. `text{Show Worked Solutions}`
  2. `(0, 12.5)`
  3. `text{Show Worked Solutions}`
  4. `{15625 sqrt3}/{9}  \ text{or} \ {15625}/{3 sqrt3}`
  5. `8%`
  6. i.  `text{Domain} \ (V_text{box}) = (0, h/2)`
  7. (sqrt3 h^3)/9`
  8. `text(See Worked Solutions)`
Show Worked Solution
a.   `V` `= x (h-2x)(2h-2x)`
    `= x (25-2x)(50-2x)`
    `= 2x (25-2x)(25-x)`

 

b.   `text{Sketch} \ y = 2x (25-2x)(25-x) \ text{by CAS:} `

♦ Mean mark (b) 42%.

`text{Domain is} \ (0, 12.5)`
 

c.    `(dV)/dx = 12x^2-300x + 1250`
 

d.    `text{Solve} \ V^{′}(x) = 0 \ text{for} \ x :`

`x = {-25 (sqrt3-3)}/6 \ or \ x = {25(sqrt3 + 3)}/6`

`:. \ x =  {-25 (sqrt3-3)}/6 \ , \ \ x ∈ (0, 12.5)`

`:. \ V_text{max} = {15\ 625 sqrt3}/9  \ text{or} \  \ {15\ 625}/{3 sqrt3}`
 

e.   `text{Area cut out} = 4 xx 5^2 = 100\ text(cm)^2`

`text{% Wasted}` `= {100}/{25 xx 50} xx 100`
  `= 8text(%)`

♦ Mean mark (f.i.) 33%.

 
f.i.  `text{Domain} \ (V_text{box}) = (0, h/2), \ \ text{(using part b)}`
 

f.ii.  `V = x (h-2x)(2h-2x)`

♦ Mean mark (f.ii.) 45%.

`text{Solve} \ V^{′}(x) = 0 \ text{for} \ x:`

`x = {-h (sqrt3-3)}/{6} \ \ text{or} \ \ x = {h(sqrt{3} + 3)}/{6}`

`V_text{max} \ text{occurs when} \ \ x = {-h(sqrt3-3)}/{6} \ \ text{(see part a)}`

`V_text{max} = (sqrt3 h^3)/9`

 

g.    `V = x(h-2x)(h-2x)`

♦ Mean mark part (g) 40%.

`text{Solve} \ V^{′}(x) = 0 \ text{for} \ x:`

`x = h/2 \ \ text{or} \ \ x =  h/6`

`V (h/6) = {2h^3}/{27} \ , \ V(h/2) = 0`

`:. V_text{max} \ text{occurs when} \ \ x = {h}/{6}`

Calculus, MET2 2021 VCAA 13 MC

The value of an investment, in dollars, after `n` months can be modelled by the function

`f(n) =  2500 xx (1.004)^n`

where  `n ∈ {0, 1, 2, ...}`.

The average rate of change of the value of the investment over the first 12 months is closest to

  1. $10.00 per month.
  2. $10.20 per month.
  3. $10.50 per month.
  4. $125.00 per month.
  5. $127.00 per month.
Show Answers Only

`B`

Show Worked Solution

`text(By CAS):`

`text{Average ROC}` `={f(12) – f(0)}/{12 – 0}`
  `= 10.2229 …`

`=> B`