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CHEMISTRY, M4 EQ-Bank 17

The decomposition of a metal carbonate is represented by the following equation:

\(\ce{MCO3(s) → MO(s) + CO2(g)}\)

The following data was recorded:

\(\Delta H = +130 \, \text{kJ/mol},\ \ \Delta S = +160 \, \text{J/mol K}\)

  1. Calculate the Gibbs free energy at 350 K.   (2 marks)
  1. Determine if the reaction is spontaneous at this temperature.   (1 mark)
  1. Discuss how both enthalpy and entropy influence the spontaneity of this reaction and predict the temperature range in which the reaction will be spontaneous.   (4 marks)
Show Answers Only

a.    \(\Delta G = +74 \, \text{kJ/mol}\)

b.    The reaction is non-spontaneous at 350 K.

c.   Enthalpy and entropy influence the spontaneity:

  • To be spontaneous, \(\Delta G\) must be negative, where  \(\Delta G = \Delta H- T\Delta S\).
  • The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.
  • However, the positive entropy means the disorder of the system increases, which favours spontaneity.
  • At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.
  • To find the temperature where the reaction becomes spontaneous, set \(\Delta G = 0\):
  •    \(0=\Delta H-T\Delta S\)
  •    \(T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}\)
  • Thus, the reaction will be spontaneous above 812.5 K.
Show Worked Solution

a.    \(\Delta G = \Delta H- T\Delta S\)

Convert \(\Delta S\) to \(\text{kJ/mol K}\)

\(\Delta S = 0.160 \, \text{kJ/mol K}\)

\(\Delta G = 130-(350 \times 0.160) = 130-56 = +74 \, \text{kJ/mol}\)
 

b.    Since \(\Delta G > 0\), the reaction is non-spontaneous at 350 K.
 

c.   Enthalpy and entropy influence the spontaneity:

  • To be spontaneous, \(\Delta G\) must be negative, where  \(\Delta G = \Delta H- T\Delta S\).
  • The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.
  • However, the positive entropy means the disorder of the system increases, which favours spontaneity.
  • At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.
  • To find the temperature where the reaction becomes spontaneous, set \(\Delta G = 0\):
  •    \(0=\Delta H-T\Delta S\)
  •    \(T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}\)
  • Thus, the reaction will be spontaneous above 812.5 K.

v1 Algebra, STD2 A2 2009 HSC 24d

A factory makes both cloth and leather lounges. In any week

• the total number of cloth lounges and leather lounges that are made is 400
• the maximum number of leather lounges made is 270
• the maximum number of cloth lounges made is 325.

The factory manager has drawn a graph to show the numbers of leather lounges (\(x\)) and cloth lounges (\(y\)) that can be made.
 

 

  1. Find the equation of the line \(AD\).   (1 mark)

  2. Explain why this line is only relevant between \(B\) and \(C\) for this factory.     (1 mark)

  3. The profit per week, \($P\), can be found by using the equation  \(P = 2520x + 1570y\).

     

    Compare the profits at \(B\) and \(C\).     (2 marks)

Show Answers Only
  1. \(x+y=400\)
  2. \(\text{Since the max amount of leather lounges}=270\)

     

    \(\rightarrow\ x\ \text{cannot be}\ >270\)

     

    \(\text{Since the max amount of cloth lounges}=325\)

     

    \(\rightarrow\ y\ \text{cannot be}\ >325\)

     

    \(\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.\)

  3. \(\text{The profits at}\ C\ \text{are }$185\ 250\ \text{more than at}\ B.\)
Show Worked Solution

i.   \(\text{We are told the number of leather lounges}\ (x),\)

\(\text{and cloth lounges}\  (y),\ \text{made in any week} = 400\)

\(\rightarrow\ \text{Equation of}\ AD\ \text{is}\ x+y=400\)


♦♦♦ Mean mark part (i) 14%.
Using \(y=mx+c\) is a less efficient but equally valid method, using  \(m=–1\)  and  \(b=400\) (\(y\)-intercept).

ii.   \(\text{Since the max amount of leather lounges}=270\)

\(\rightarrow\ x\ \text{cannot}\ >270\)

\(\text{Since the max amount of cloth lounges}=325\)

\(\rightarrow\ y\ \text{cannot}\ >325\)

\(\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.\)


♦ Mean mark part (ii) 49%.

iii.  \(\text{At}\ B,\ x=75,\ y=325\)

\(\rightarrow\ $P  (\text{at}\ B)\) \(=2520\times 75+1570\times 325\)
  \(=189\ 000+510\ 250\)
  \(=$699\ 250\)

  
\(\text{At}\ C,\ x=270,\ y=130\)

\(\rightarrow\ $P  (\text{at}\ C)\) \(=2520\times 270+1570\times 130\)
  \(=680\ 400+204\ 100\)
  \(=$884\ 500\)

  
\(\text{Difference in profits}=$884\ 500-$699\ 250=$185\ 250\)

\(\text{The profits at}\ C\ \text{are } $185\ 250\ \text{more than at}\ B.\)


♦ Mean mark (iii)40%.

v1 Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.
  

  1. Use the graph to find the tax payable on a taxable income of \($18\ 000\).  (1 mark)

  2. Use suitable points from the graph to show that the gradient of the section of the graph marked  \(A\)  is  \(\dfrac{7}{15}\).    (1 mark)

  3. How much of each dollar earned between  \($18\ 000\)  and  \($33\ 000\) is payable in tax? Give your answer correct to the nearest whole number.   (1 mark)

  4. Write an equation that could be used to calculate the tax payable, \(T\), in terms of the taxable income, \(I\), for taxable incomes between  \($18\ 000\)  and  \($33\ 000\).   (2 marks)

Show Answers Only
  1. \($3000\ \ \text{(from graph)}\)
  2. \(\text{See worked solution}\)
  3. \(46\frac{2}{3}\approx  47\ \text{cents per dollar earned}\)
  4. \(\text{Tax payable →}\ T=\dfrac{7}{15}I-5400\)
Show Worked Solution
i.   

\(\text{Income on}\ $18\ 000=$3000\ \ \text{(from graph)}\)

  

ii.  \(\text{Using the points}\ (18, 3)\ \text{and}\ (33, 10)\)

\(\text{Gradient at}\ A\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)
  \(=\dfrac{10\ 000-3000}{33\ 000-18\ 000}\)
  \(=\dfrac{7000}{15\ 000}\)
  \(=\dfrac{7}{15}\ \ \ \ \text{… as required}\)

♦♦ Mean mark (ii) 25%.

iii.  \(\text{The gradient represents the tax applicable on each dollar}\)

\(\text{Tax}\) \(=\dfrac{7}{15}\ \text{of each dollar earned}\)
  \(=46\frac{2}{3}\approx 47\ \text{cents per dollar earned (nearest whole number)}\)

♦♦♦ Mean mark (iii) 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

iv.  \(\text{Tax payable up to }$18\ 000 = $3000\)

\(\text{Tax payable on income between }$18\ 000\ \text{and }$33\ 000\)

\(=\dfrac{7}{15}(I-18\ 000)\)

\(\therefore\ \text{Tax payable →}\ \ T\) \(=3000+\dfrac{7}{15}(I-18\ 000)\)
  \(=3000+\dfrac{7}{15} I-8400\)
  \(=\dfrac{7}{15}I-5400\)

♦♦♦ Mean mark (iv) 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

v1 Algebra, STD2 A2 2007 HSC 18 MC

Art started to make this pattern of shapes using matchsticks.
  

 

If the pattern of shapes is continued, which shape would use exactly 416 matchsticks?

  1. Shape 83
  2. Shape 103
  3. Shape 104
  4. Shape 138
Show Answers Only

\(D\)

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape}\ \textit(S) \rule[-1ex]{0pt}{0pt}\ \ &\  \ 1\ \ &\ \ 2\ \ &\ \ 3\ \  \\
\hline
\rule{0pt}{2.5ex} \text{Matches}\ \textit(M) \rule[-1ex]{0pt}{0pt} \ \ & \ \ 5\ \ &\ \ 8\ \ &\ \ 11\ \  \\
\hline
\end{array}

\(\text{Equation rule:}\)

\(M=3S+2\)

\(\text{Find}\ \ S\ \text{when}\ \ M=416:\)

\(416\) \(=3S+2\)
\(3S\) \(=414\)
\(S\) \(=138\)

 
\(\therefore\ \text{The 138th shape uses 416 matchsticks.}\)

\(\Rightarrow D\)

v1 Algebra, STD2 A1 2020 HSC 13 MC

When Stuart stops drinking alcohol at 11:30 pm, he has a blood alcohol content (BAC) of 0.08625.

The number of hours required for a person to reach zero BAC after they stop consuming alcohol is given by the formula:

\(\text{Time}=\dfrac{BAC}{0.015}\).

At what time on the next day should Stuart expect his BAC to be 0.05?

  1.  1:33 am
  2.  1:55 am
  3.  2:15 am
  4.  5:15 am
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Time from  0.08625 → 0}\ BAC\)

\(=\dfrac{0.08625}{0.015}\)

\(=5.75\ \text{hours}\)
 

\(\text{Time from  0.08625 → 0.05}\ BAC\)

\(=\dfrac{(0.08625 – 0.05)}{0.08625}\times 5.75\) 

\(=\dfrac{29}{69}\times 5.75\)

\(=2.41\dot{6}=2\ \text{h}\ 25\ \text{min}\)
 

\(\therefore\ \text{Time}\) \(=11:30\ \text{pm} \ + 2 \ \text{h} \ 25 \ \text{min}\)
  \(=1:55\ \text{am}\)

 
\(\Rightarrow B\)


♦♦♦ Mean mark 17%.
COMMENT: The rates aspect of this question proved extremely challenging.

v1 Algebra, STD2 A1 2007 HSC 28b

This shape is made up of two right-angled triangle and a regular hexagon.
 

The area of a regular hexagon can be estimated using the formula  \(A=2.598S^2\)  where \(S\) is the hexagon's side-length.

Calculate the total area of the shape using this formula.   (3 marks)

Show Answers Only

\(619.6\ \text{cm}^2\)

Show Worked Solution

\(\text{Area}=2.598S^2\)

\(\text{Using Pythagoras}\)

\(S^2= 10^2+10^2=200\)

\(S=\sqrt{200}\)

\(A=2.598\times (\sqrt {200})^2=519.6\ \text{cm}^2\)

\(\text{Area of Δ}\ =\dfrac{1}{2}bh=\dfrac{1}{2}\times 10\times 10=50 \ \text{cm}^2\)

\(\therefore\ \text{Total Area}\ =519.6+50+50=619.6\ \text{cm}^2\)

EXAMCOPY Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2.  i. Find `f^{\prime}(0)`.   (2 marks)

  3. ii. State the maximal domain over which `f` is strictly increasing.   (1 mark)

  4. Show that `f(x)+f(-x)=0`.   (1 mark)

  5. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)

  6. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  7. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  8. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  9. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  1. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)

  2. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark)

Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)` `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
  `= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

CHEMISTRY, M8 2022 VCE 5*

A chemist uses spectroscopy to identify an unknown organic molecule, Molecule \(\text{J}\), that contains chlorine.

The \({}^{13}\text{C NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

The infra-red (IR) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. Name the functional group that produces the peak at 168 ppm in the \({}^{13}\text{C NMR}\) spectrum on the first image, which is consistent with the IR spectrum shown above. Justify your answer with reference to the IR spectrum.   (2 marks)

The mass spectrum of Molecule \(\text{J}\) is shown below
 

  1. The molecular mass of Molecule \(\text{J}\) is 108.5
  2.  Explain the presence of the peak at 110 m/z.  (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. The \({ }^1 \text{H NMR}\) spectrum consists of two singlet peaks.
  2. What information does this give about the molecule?   (2 marks)

  3. Draw a structural formula for Molecule \(\text{J}\) that is consistent with the information provided in parts a–c.   (2 marks)

Show Answers Only

a.   Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

  • Molecule \(\text{J}\) must be an ester. 

b.    The peak at 110 m/z:

  • due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35. 

c.    The two singlet peaks indicate:

  • two different hydrogen environments within the molecule.
  • there are no adjacent hydrogen environments.
  • The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3. 

d.    Either of the two molecules shown below are correct:

Show Worked Solution

a.    Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

  • Molecule \(\text{J}\) must be an ester. 
♦ Mean mark (a) 41%.

b.    The peak at 110 m/z:

  • due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35. 

c.    The two singlet peaks indicate:

  • two different hydrogen environments within the molecule.
  • there are no adjacent hydrogen environments.
  • The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3. 
♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.

d.    Either of the two molecules shown below are correct:
 

♦ Mean mark (d) 40%.

CHEMISTRY, M8 2021 VCE 16 MC

Which one of the following statements about IR spectroscopy is correct?

  1. IR radiation changes the spin state of electrons.
  2. Bond wave number is influenced only by bond strength.
  3. An IR spectrum can be used to determine the purity of a sample.
  4. In an IR spectrum, high transmittance corresponds to high absorption.
Show Answers Only

\(C\)

Show Worked Solution
  • Every pure compound has a different fingerprint region on the Infrared spectrum. 
  • Hence the fingerprint region of the sample can be compared against the fingerprint region of the pure substance to determine the purity of the sample.

\(\Rightarrow C\)

♦♦♦ Mean mark 15%.

CHEMISTRY, M8 2023 VCE 7-2*

The infrared (IR) spectrum of the molecule 3-methyl-2-butanone is shown below.
 

Explain why different frequencies of infrared radiation can be absorbed by the same molecule as shown in the spectrum above.   (3 marks)

Show Answers Only

  • As infrared radiation is passed through the molecules, the different bonds within the molecule vibrate at specific wavelengths leading to the absorption of the infrared radiation.
  • In this way, different frequencies of infrared radiation can be absorbed by a molecule as bonds differ in electronegativity, dipole strengths and in the masses of atoms at the end of bonds.  
  • For example, the double bond between the oxygen atom and carbon atom in the given molecule has a greater dipole than the carbon-hydrogen bonds. This causes a transmittance at 1450 whereas \(\ce{C-H}\) bonds have a transmittance at 3000.
  • An oxygen atom has a higher molecular mass than hydrogen atoms and this also leads to different frequencies of infrared radiation being absorbed in the one molecule.

Other possible explanations:

  • Students could have also discussed the strength of bonds, bond length or molecular vibrations.

Show Worked Solution

  • As infrared radiation is passed through the molecules, the different bonds within the molecule vibrate at specific wavelengths leading to the absorption of the infrared radiation.
  • In this way, different frequencies of infrared radiation can be absorbed by a molecule as bonds differ in electronegativity, dipole strengths and in the masses of atoms at the end of bonds.  
  • For example, the double bond between the oxygen atom and carbon atom in the given molecule has a greater dipole than the carbon-hydrogen bonds. This causes a transmittance at 1450 whereas \(\ce{C-H}\) bonds have a transmittance at 3000.
  • An oxygen atom has a higher molecular mass than hydrogen atoms and this also leads to different frequencies of infrared radiation being absorbed in the one molecule.

Other possible explanations:

  • Students could have also discussed the strength of bonds, bond length or molecular vibrations.
♦♦ Mean mark 30%.
COMMENT: A deep understanding of the principles behind analytical techniques required here.

CHEMISTRY, M8 2013 VCE 2

The strength of the eggshell of birds is determined by the calcium carbonate, \(\ce{CaCO3}\), content of the eggshell.

The percentage of calcium carbonate in the eggshell can be determined by gravimetric analysis.

0.412 g of clean, dry eggshell was completely dissolved in a minimum volume of dilute hydrochloric acid.

\(\ce{CaCO3(s) + 2H+(aq)\rightarrow Ca^2+(aq) + CO2(g) + H2O(l)}\)

An excess of a basic solution of ammonium oxalate, \(\ce{(NH4)2C2O4}\), was then added to form crystals of calcium oxalate monohydrate, \(\ce{CaC2O4.H2O}\).

The suspension was filtered and the crystals were then dried to constant mass.

0.523 g of \(\ce{CaC2O4.H2O}\) was collected.

  1. Write a balanced equation for the formation of the calcium oxalate monohydrate precipitate.   (1 mark)

  2. Determine the percentage, by mass, of calcium carbonate in the eggshell.   (3 marks)

Show Answers Only

a.    \(\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l) \rightarrow CaC2O4.H2O(s) + 2NH4^+(aq)}\)

b.    \(86.9\)%

Show Worked Solution

a.    \(\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l)} \rightarrow \)

\(\ce{CaC2O4.H2O(s) + 2NH4^+(aq)}\)
 

♦♦♦ Mean mark (a) 30%.

b.    \(\ce{M(CaC2O4.H2O)= 40.08 + 2(12.01) + 4(16) + 2(1.008) + 16 = 146.116\ \text{g mol}^{-1}}\)

\(\ce{n(CaC2O4.H2O)=\dfrac{0.523}{146.116}=0.003579\ \text{mol}}\)

\(\ce{n(CaC2O4.H2O) = n(Ca^2+)= n(CaCO3) = 0.003579\ \text{mol}}\)

\(\ce{m(CaCO3)=0.003579 \times (40.08 +12.01 + 3(16))= 0.358\ \text{g}}\)

\(\text{% Mass}\ =\dfrac{0.358}{0.412} \times 100 = 86.9\%\)

CHEMISTRY, M7 2018 VCE 1a

Organic compounds are numerous and diverse due to the nature of the carbon atom. There are international conventions for the naming and representation of organic compounds.

  1. Draw the structural formula of 2-methyl-propan-2-ol.   (1 mark)

  2. Give the molecular formula of but-2-yne.   (1 mark)

  1. Give the IUPAC name of the compound that has the structural formula shown above.   (1 mark)

Show Answers Only

i.    
         

ii.    \(\ce{C4H6}\)

  • Structural or semi-structural formulas are not appropriate.
     

iii. •   Longest carbon chain is 6, hence hexane

  • Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl
  • Compound name: 2,3-dibromo-4-methylhexane

Show Worked Solution

i.    
         

ii.    \(\ce{C4H6}\)

  • Structural or semi-structural formulas are not appropriate.
     

iii.    Longest carbon chain is 6, hence hexane

  • Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl
  • Compound name: 2,3-dibromo-4-methylhexane
♦♦♦ Mean mark (c) 27%.

CHEMISTRY, M8 2016 VCE 6

Brass is an alloy of copper and zinc.

To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper\(\text{(II)}\) ions and measured their absorbance using an atomic absorption spectrometer (AAS).

The calibration curve obtained is shown below.
 

  1. A 0.198 g sample of the brass was dissolved in acid and the solution was made up to 100.00 mL in a volumetric flask. The absorbance of this test solution was found to be 0.13
  2. Calculate the percentage by mass of copper in the brass sample.   (3 marks)

  3. If the analyst had made up the solution of the brass sample to 20.00 mL instead of 100.00 mL, would the result of the analysis have been equally reliable? Why?   (2 marks)

  4. Name another analytical technique that could be used to verify the result from part a.   (1 mark)

Show Answers Only

a.    55.6%

b.    No, this would increase the concentration of the copper solution by a factor of 5.

  • The concentration of the solution and absorbance would be too high and outside the range of the calibration curve (it can’t be assumed that the calibration curve remains linear beyond the range of the known data).

c.    Answers could include:

  • UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.
Show Worked Solution

a.    Absorbance of 0.13 → \(\ce{Cu^2+}\) concentration of 1.1 gL\(^{-1}\)  (see graph)

\(\text{In 100 mL:}\)

\(\ce{m(Cu^2+)}=1.1 \times 0.1=0.11\ \text{g}\)

\(\Rightarrow \ce{\% Cu^2+}=\dfrac{0.11}{0.198} \times 100=55.6\%\)
 

b.    No, this would increase the concentration of the copper solution by a factor of 5.

  • The concentration of the solution and absorbance would be too high and outside the range of the calibration curve.
  • It can’t be assumed that the calibration curve remains linear beyond the range of the known data. 
♦♦♦ Mean mark (b) 25%.

c.    Answers could include:

  • UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.

PHYSICS, M8 2019 VCE 17

Students are comparing the diffraction patterns produced by electrons and X-rays, in which the same spacing of bands is observed in the patterns, as shown schematically in the diagram. Note that both patterns shown are to the same scale.
 

The electron diffraction pattern is produced by 3.0 × 10\(^3\) eV electrons.

  1. Explain why electrons can produce the same spacing of bands in a diffraction pattern as X-rays.   (3 marks)
  1. Calculate the frequency of X-rays that would produce the same spacing of bands in a diffraction pattern as for the electrons. Show your working.   (4 marks)
Show Answers Only

a.    Electron vs X-ray wavelength:

  • Electrons with momentum exhibit wave-like properties.
  • In this way, moving electrons produce a de Broglie wavelength.
  • As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same. 

b.    \(1.34 \times 10^{19}\ \text{Hz}\)

Show Worked Solution

a.    Electron vs X-ray wavelength:

  • Electrons with momentum exhibit wave-like properties.
  • In this way, moving electrons produce a de Broglie wavelength.
  • As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same. 
♦ Mean mark (a) 49%.

b.    Find velocity of the electrons using  \(E=\dfrac{1}{2}mv^2 :\)

\(3.0 \times 10^3 \times 1.602 \times 10^{-19}=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times v^2\)

\(v^2\) \(=\dfrac{4.806 \times 10^{-16}}{4.5545 \times 10^{-31}}\)  
\(v\) \(=\sqrt{1.055 \times 10^{15}}\)  
  \(=3.25 \times 10^7\ \text{ms}^{-1}\)  

 

The de Broglie wavelength of the electron is:

\(\lambda\) \(=\dfrac{h}{mv}\)  
  \(=\dfrac{6.626 \times 10^{-34}}{3.25 \times 10^7 \times 9.109 \times 10^{-31}}\)  
  \(=2.24 \times 10^{-11}\ \text{m}\)  

 
Frequency of the X-ray:

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{2.24 \times 10^{-11}}=1.34 \times 10^{19}\ \text{Hz}\)

♦♦♦ Mean mark (b) 27%.
COMMENT: Multi-step solutions require clear and logical working.

PHYSICS, M5 2019 VCE 8

A 250 g toy car performs a loop in the apparatus shown in the diagram below.
 

The car starts from rest at point \(\text{A}\) and travels along the track without any air resistance or retarding frictional forces. The radius of the car's path in the loop is 0.20 m. When the car reaches point \(\text{B}\) it is travelling at a speed of 3.0 m s\(^{-1}\).

  1. Calculate the value of \(h\). Show your working.   (3 marks)

  2. Calculate the magnitude of the normal reaction force on the car by the track when it is at point \(\text{B}\). Show your working.   (3 marks)

  3. Explain why the car does not fall from the track at point \(\text{B}\), when it is upside down.   (2 marks)

Show Answers Only

a.    \(0.86\ \text{m}\)

b.    \(8.8\ \text{N}\)

c.    Method 1

  • During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.
  • This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).

Method 2

  • The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 
\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}=\sqrt{9.8 \times 0.2}=1.4\ \text{ms}^{-1}\)  

 

  • As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).
  • Therefore the car will not fall from the track at point \(\text{B}\).

Show Worked Solution

a.    Using the law of the conservation of energy:

\(mgh_A\) \(=mgh_B +\dfrac{1}{2}mv^2_b\)  
\(0.25 \times 9.8 \times h_A\) \(=0.25 \times 9.8 \times 0.4 + \dfrac{1}{2} \times 0.25 \times 3^2\)  
\(2.45h_A\) \(=2.105\)  
\(h_A\) \(=\dfrac{2.105}{2.45}=0.86\ \text{m}\)  

 

b.     \(N + mg\) \(=\dfrac{mv^2}{r}\)
  \(N\) \(=\dfrac{mv^2}{r}-mg\)
    \(=\dfrac{0.25 \times 3^2}{0.2}- 0.25 \times 9.8\) 
    \(=8.8\ \text{N}\)

♦ Mean mark (b) 53%.

c.    Method 1

  • During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.
  • This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms\(^{-1}\).

Method 2

  • The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force. 
\(\dfrac{mv^2}{r}\) \(=mg\)  
\(v\) \(=\sqrt{gr}=\sqrt{9.8 \times 0.2}=1.4\ \text{ms}^{-1}\)  

  • As 3 ms\(^{-1}\) is greater than the minimum speed required, a normal force downwards will be present at point \(\text{B}\).
  • Therefore the car will not fall from the track at point \(\text{B}\).
♦♦♦ Mean mark (c) 23%.
COMMENT: Normal force is poorly understood here..

PHYSICS, M6 2019 VCE 6

A home owner on a large property creates a backyard entertainment area. The entertainment area has a low-voltage lighting system. To operate correctly, the lighting system requires a voltage of 12 V. The lighting system has a resistance of 12 \(\Omega\).

  1. Calculate the power drawn by the lighting system.   (1 mark)

To operate the lighting system, the home owner installs an ideal transformer at the house to reduce the voltage from 240 V to 12 V. The home owner then runs a 200 m long heavy-duty outdoor extension lead, which has a total resistance of 3 \( \Omega\), from the transformer to the entertainment area.

  1. The lights are a little dimmer than expected in the entertainment area.
  2. Give one possible reason for this and support your answer with calculations.   (4 marks)

  3. Using the same equipment, what changes could the home owner make to improve the brightness of the lights? Explain your answer.   (2 marks)

Show Answers Only

a.    \(12\ \text{W}\)

b.   Potential cause of dimmer lights: 

  • Power loss in the extension lead.
  • The total resistance (\(R_T\)) in the circuit \(=12 + 3=15\ \Omega\)
  • The new current in the circuit \(=\dfrac{V}{R_T}=\dfrac{12}{15}=0.8\ \text{A}\)
  • The power loss across the extension lead, \(P_{\text{loss}}=I^2R_l=0.8^2 \times 3=1.92\ \text{W}\)
  • Therefore, power supplied to lights \(=12.0-1.92=10.08\ \text{W}\).

c.    Changes using the same equipment:

  • The homeowner needs to move the transformer from inside the home (start of the lead) to the outside entertainment area (end of the lead).
  • This would mean a higher voltage/lower current is running through the lead which would decrease the power loss through the lead and improve the brightness of the lights.

Show Worked Solution

a.    \(I=\dfrac{V}{R}=\dfrac{12}{12}=1\ \text{A}\)

\(P=IV=1 \times 12 = 12\ \text{W}\)
 

b.   Potential cause of dimmer lights: 

  • Power loss in the extension lead.
  • The total resistance (\(R_T\)) in the circuit \(=12 + 3=15\ \Omega\)
  • The new current in the circuit \(=\dfrac{V}{R_T}=\dfrac{12}{15}=0.8\ \text{A}\)
  • The power loss across the extension lead, \(P_{\text{loss}}=I^2R_l=0.8^2 \times 3=1.92\ \text{W}\)
  • Therefore, power supplied to lights \(=12.0-1.92=10.08\ \text{W}\).
♦♦♦ Mean mark (b) 23%.
COMMENT: Many students used 1 Amp in their calculations instead of finding the total current in the system.

c.    Changes using the same equipment:

  • The homeowner needs to move the transformer from inside the home (start of the lead) to the outside entertainment area (end of the lead).
  • This would mean a higher voltage/lower current is running through the lead which would decrease the power loss through the lead and improve the brightness of the lights.
♦♦♦ Mean mark (c) 22%.
COMMENT: Students incorrectly added or changed the equipment used in the question.

PHYSICS, M6 2019 VCE 1

A particle of mass \(m\) and charge \(q\) travelling at velocity \(v\) enters a uniform magnetic field \(\text{B}\), as shown in the diagram.
 

  1. Is the charge \(q\) positive or negative? Give a reason for your answer.   (1 mark)
  1. Explain why the path of the particle is an arc of a circle while the particle is in the magnetic field.   (2 marks)

Show Answers Only

a.    Using the right-hand rule:

  • Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).
  • As the force on the charge is down the page (back of the hand), the charge must be negative.

b.    For circular motion to occur:

  • The force must be at right-angles to the velocity of the particle.
  • The force must be of constant magnitude \((F=qvB)\).
  • In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.
  • The particle will therefore follow the arc of a circle while in the magnetic field.

Show Worked Solution

a.    Using the right-hand rule:

  • Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).
  • As the force on the charge is down the page (back of the hand), the charge must be negative.
♦ Mean mark (a) 44%.

b.    For circular motion to occur:

  • The force must be at right-angles to the velocity of the particle.
  • The force must be of constant magnitude \((F=qvB)\).
  • In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.
  • The particle will therefore follow the arc of a circle while in the magnetic field.
♦♦♦ Mean mark (b) 27%.

PHYSICS, M7 2020 VCE 11

An astronaut has left Earth and is travelling on a spaceship at 0.800\(c\) directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth.

  1. How long will the trip take according to a clock that the astronaut is carrying on his spaceship? Show your working.   (2 marks)
  1. Is the trip time measured by the astronaut in part (a) a proper time? Explain your reasoning.   (2 marks)

Show Answers Only

a.    \(6.46\ \text{years}\)

b.    Proper time measurement

  • The trip time of 6.46 years on the spaceship is a proper time.
  • This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.

Show Worked Solution

a.    From the earth’s perspective:

\(\text{Travel time}\ =\dfrac{8.61}{0.8}=10.76\ \text{years}\).

From the astronaut’s perspective:

  • The Earth’s time is going to be dilated compared to his, so the time the astronauts clock will measure is:

\(t=\dfrac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

\(t_0\) \(=t\sqrt{1-\frac{v^2}{c^2}}\)
  \(=10.76 \times \sqrt{1-\frac{(0.8c)^2}{c^2}}\)
  \(=10.76 \times \sqrt{1-0.8^2}\)
  \(=6.46\ \text{years}\)

♦♦♦ Mean mark (a) 29%.
COMMENT: Light years is a measure of distance, not time!

b.    Proper time measurement

  • The trip time of 6.46 years on the spaceship is a proper time.
  • This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.
♦ Mean mark (b) 44%.

PHYSICS, M6 2020 VCE 6

Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
 

  1. Will the magnetic flux through the coil increase, decrease or stay the same as the students change the shape of the coil?   (1 mark)
  1. Explain, using physics principles, why the ammeter registered a current in the coil and determine the direction of the induced current.   (3 marks)
  1. The students then push each side of the coil together, as shown in Figure 6a, so that the coil returns to its original circular shape, as shown in Figure 6b, and then changes to the shape shown in Figure 6c. 
     

  1. Describe the direction of any induced currents in the coil during these changes. Give your reasoning.   (2 marks)

Show Answers Only

a.    Decrease 

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
Show Worked Solution

a.    Magnetic flux will decrease.

  • The magnetic flux is proportional to how many field lines are passing through a given area.
  • As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.

b.    Ammeter registers a current:

  • By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 
  • This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.
  • Hence by using the right-hand grip rule, the current will run clockwise through the coil.
♦ Mean mark (b) 42%.

c.    There will be an induced current from 6a to 6b.

  • As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.
  • By the right-hand grip rule, the current will flow anti-clockwise around the coil.
  • There will also be an induced current from 6b to 6c.
  • The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.
  • Using the right-hand grip rule, the current through the coil will flow clockwise.
♦♦♦ Mean mark (c) 28%.
COMMENT: Lenz’s law was poorly understood in this question.

PHYSICS, M5 2020 VCE 4*

The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit. 

  1. Calculate the orbital radius of the ICON satellite.   (1 mark)

  2. Calculate the orbital period of the ICON satellite correct to three significant figures. Show your working.   (4 marks)

  3. Explain how the ICON satellite maintains a stable circular orbit without the use of propulsion engines.   (2 marks)

Show Answers Only

a.    \(r_{orbit}= 6.971 \times 10^6\ \text{m}\)

b.    \(5780\ \text{s}\)

c.   Stable circular orbit:

  • The icon satellite is only subject to the gravitational force of the Earth.
  • This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 
Show Worked Solution

a.    Radius of orbit is equal to the altitude plus the radius of the Earth.

\(r_{orbit}=600\ 000\ \text{m} + 6.371 \times 10^6\ \text{m} = 6.971 \times 10^6\ \text{m}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 \times (6.971 \times 10^6)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}\)
    \(=5780\ \text{s}\ \ \text{(3 sig.fig.)}\)

 

c.   Stable circular orbit:

  • The icon satellite is only subject to the gravitational force of the Earth.
  • This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 
♦♦♦ Mean mark (c) 23%.

PHYSICS, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).

Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
 

  1. Show that the velocity of an electron that travels straight through the aperture to point \(\text{Y}\) is given by  \( v_{0} \) = \( \dfrac{E}{B}\).   (1 mark)

  2. Calculate the magnitude of the velocity, \(v_0\), of an electron that travels straight through the aperture to point \(\text{Y}\) if  \(E\) = 500 kV m\(^{-1}\)  and  \(B\) = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – \(\text{X, Y}\), or \(\text{Z}\) – in Figure 2 could electrons travelling faster than \(v_0\) arrive?   (1 mark)

  4. ii. Explain your answer to part c.i.   (2 marks)

Show Answers Only

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  

 
b.    \(v_0=2 \times 10^6\ \text{ms}^{-1}\)

c.i.    Point \(\text{Z.}\)

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
Show Worked Solution

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  
♦ Mean mark (a) 46%.

b.    \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
 

c.i.    Point \(\text{Z.}\)
 

c.ii.  When electrons travel faster than \(v_0\):

  • The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 
  • Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.
  • Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)
♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)

b.    \(2.06 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
 

b.   \(\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}\)

\(\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}\)

\(\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}\)

\(\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}\)

\(\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.

CHEMISTRY, M3 EQ-Bank 11

A student stirs 2.45 g of copper \(\text{(II)}\) nitrate powder into 200.0 mL of 1.25 mol L\(^{-1}\) sodium carbonate solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)

b.    \(1.61 \text{ g}\)

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.
Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)
 

b.   \(\ce{MM(Cu(NO3)2) = 63.55 + 2 \times (14.01 + 16.00 \times 3) = 187.57}\)

\(\ce{n(Cu(NO3)2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.45}{187.57} = 0.01306 \text{ mol}}\)

\(\ce{n(Na2CO3) = c \times V = 1.25 \times 0.200 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCO3) = n(Cu(NO3)2) = 0.01306 \text{ mol}}\)

\(\ce{m(CuCO3) = n \times MM = 0.01306 \times (63.55 + 12.01 + 16.00 \times 3) = 1.61 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

  • Presence of excess solution in the filter paper or incomplete drying.
  • Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

  • Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.
  • Dry precipitate completely in the incubator before weighing.

Calculus, MET2 2023 SM-Bank 1

The function \(g\) is defined as follows.

\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)

  1. Sketch the graph of \(g\) on the axes below. Label the vertical asymptote with its equation, and label any axial intercepts, stationary points and endpoints in coordinate form, correct to three decimal places.   (3 marks)

     

     

  2.  i. Find the equation of the tangent to the graph of \(g\) at the point where \(x=1\).   (1 mark)

  3. ii. Sketch the graph of the tangent to the graph of \(g\) at \(x=1\) on the axes in part a.   (1 mark)

Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).

  1. Find the value of \(x_1\).   (1 mark)

  2. Find the horizontal distance between \(x_3\) and the closest \(x\)-intercept of \(g\), correct to four decimal places.   (1 mark)

  3.  i. Find the value of \(k\), where \(k>1\), such that an initial estimate of  \(x_0=k\)  gives the same value of  \(x_1\)  as found in part \(c\). Give your answer correct to three decimal places.   (2 marks)

  4. ii. Using this value of \(k\), sketch the tangent to the graph of \(g\) at the point where  \(x=k\)  on the axes in part a.   (1 mark)

Show Answers Only
a.    
  
b.i

 \(y=2x-3\)
  
b.ii
  
c. \(\dfrac{3}{2}=1.5\)
  
d. \(0.0036\)
  
e.i \(k=2.397\)
  
e.ii
Show Worked Solution

a.

b.i    \(g(x)\) \(=3\log_{e}x-x\)
  \(g(1)\) \(=3\log_{e}1-1=-1\)
  \(g^{\prime}(x)\) \(=\dfrac{3}{x}-1\)
  \(g^{\prime}(1)\) \(=\dfrac{3}{1}-1=2\)

  
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)

\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)

b.ii

 
c.    \(\text{Newton’s Method}\)

\(x_1\) \(=x_0-\dfrac{g(x)}{g'(x)}\)
  \(=1-\left(\dfrac{-1}{2}\right)\)
  \(=\dfrac{3}{2}=1.5\)

\(\text{Using CAS:}\)
  
 

d.    \(\text{Using CAS}\)

\(x\text{-intercept}:\ x=1.85718\)

\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)

 

e.i.  \(\text{Using CAS}\)

\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) \(=1.5\)
\(k>1\ \therefore\ \ k\) \(=2.397\)

 
e.ii

CHEMISTRY, M2 EQ-Bank 2

During a laboratory experiment, a gas is collected in a sealed syringe. Initially, the gas has a volume of 5.0 litres and a pressure of 1.0 atmosphere. 

  1.  Calculate the new pressure inside the syringe when the volume is decreased to 3.0 litres, assuming no temperature change.   (2 marks)

  2. After reaching the pressure calculated in part a, the volume is further decreased so that the pressure inside the syringe doubles. Calculate the final volume of the gas.   (2 marks)

  3. Discuss two potential experimental errors that could affect the accuracy of the observed results compared to the theoretical predictions of Boyle's Law.   (2 marks)

Show Answers Only

a.    \(1.67\ \text{atm}\)

b.    \(1.5\ \text{L}\)

c.    Potential experimental errors could include:

  • Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.
  • Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.
  • Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.
Show Worked Solution

a.    Using Boyle’s Law  \( P_1V_1 = P_2V_2 \):

\( P_1 = 1.0 \, \text{atm}, \quad V_1 = 5.0 \, \text{L}, \quad V_2 = 3.0 \, \text{L} \)

\(P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.0 \times 5.0}{3.0} = 1.67 \, \text{atm} \)
 

b.    To find the new volume when pressure doubles:

\( P_3 = 2 \times P_2 = 2 \times 1.67 \, \text{atm} = 3.34 \, \text{atm}\)

\( P_1V_1 = P_3V_3 \ \  \Rightarrow \ \  V_3 = \dfrac{P_1 \times V_1}{P_3} = \dfrac{1.0  \times 5.0 }{3.34} \approx 1.5 \, \text{L}\)
 

c.    Potential experimental errors could include:

  • Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.
  • Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.
  • Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.

CHEMISTRY, M1 EQ-Bank 8

Explain why the boiling points of hydrogen halides \(\ce{(HF, HCl, HBr}\), and \(\ce{HI)}\) increase from \(\ce{HF}\) to \(\ce{HI}\).

Include in your answer the types of intermolecular forces involved and how molecular mass affects these boiling points.   (3 marks)

Show Answers Only
  • \(\ce{HF}\) exhibits hydrogen bonding while \(\ce{HCl, HBr}\), and \(\ce{HI}\) experience dipole-dipole interactions and dispersion forces.
  • As molecular mass increases from \(\ce{HCl}\) to \(\ce{HI}\), the number of electrons increases, enhancing the dispersion forces.
  • Despite \(\ce{HF}\) having hydrogen bonds, the significantly greater molecular mass of \(\ce{HI}\) results in stronger dispersion forces that lead to a higher boiling point.
Show Worked Solution
  • \(\ce{HF}\) exhibits hydrogen bonding while \(\ce{HCl, HBr}\), and \(\ce{HI}\) experience dipole-dipole interactions and dispersion forces.
  • As molecular mass increases from \(\ce{HCl}\) to \(\ce{HI}\), the number of electrons increases, enhancing the dispersion forces.
  • Despite \(\ce{HF}\) having hydrogen bonds, the significantly greater molecular mass of \(\ce{HI}\) results in stronger dispersion forces that lead to a higher boiling point.

Calculus, SPEC2 2022 VCAA 3

A particle moves in a straight line so that its distance, \(x\) metres, from a fixed origin \(O\) after time \(t\) seconds is given by the differential equation \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\), where  \(x=0\)  when  \(t=0\).

  1.  i. Express the differential equation in the form \(\displaystyle \int g(x)dx=\int f(t)dt\).   (1 mark)

  2. ii. Hence, show that  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\).   (2 marks)

  3. The graph of  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\) has a horizontal asymptote.
    1. Write down the equation of this asymptote.   (1 mark)

    2. Sketch the graph of  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\) and the horizontal asymptote on the axes below. Using coordinates, plot and label the point where  \(t=10\), giving the value of \(x\) correct to two decimal places.   (2 marks)

  1. Find the speed of the particle when  \(t=3\). Give your answer in metres per second, correct to two decimal places.   (1 mark)

Two seconds after the first particle passed through \(O\), a second particle passes through \(O\).

Its distance \(x\) metres from \(O, t\) seconds after the first particle passed through \(O\), is given by  \(x=\log _e\left(\tan ^{-1}(3 t-6)+1\right).\)

  1. Verify that the particles are the same distance from \(O\) when  \(t=6\).   (1 mark)

  2. Find the ratio of the speed of the first particle to the speed of the second particle when the particles are at the same distance from \(O\). Give your answer as \(\dfrac{a}{b}\) in simplest form, where \(a\) and \(b\) are positive integers.   (2 marks)

Show Answers Only

a.i.  \(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)

a.ii.  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right) \)

b.i.   \(\text{Asymptote: }\  x=\log _e\left(\dfrac{\pi}{2}+1\right)\)

b.ii. 
           

c.    \(0.02\)

d.   \(\text {Particle positions at}\ \ t=6:\)

\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)

\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)

e.    \(\dfrac{v_1}{v_2}=\dfrac{2}{3}\)

Show Worked Solution

a.i.  \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\)

 \(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
  

a.ii. \(e^x=\tan ^{-1}(2 t)+c\)

\(\text {When}\ \ t=0, \ x=0 \ \Rightarrow \ c=1\)

\begin{aligned}
e^x& =\tan ^{-1}(2 t)+1 \\
x&=\log _e\left(\tan ^{-1}(2 t)+1\right)
\end{aligned}

 
b.i. \(\text {As}\ \ t \rightarrow \infty, \ \tan ^{-1}(2 t) \rightarrow \dfrac{\pi}{2}\)

\(x \rightarrow \log _e\left(\dfrac{\pi}{2}+1\right)\)

\(\therefore \text{Asymptote: }\  x=\log _e\left(\dfrac{\pi}{2}+1\right)\)

♦♦♦ Mean mark 23%.

 
b.ii. \(\log _e\left(\dfrac{\pi}{2}+1\right) \approx 0.944\)

\(\text{When}\ \ t=10 :\)

\(x=\log _e\left( \tan ^{-1}(20)+1\right)=0.92\ \text{(2 d.p.)}\)
 

c.    \(\text {At}\ \  t=3, x=\log _e\left(\tan ^{-1}(6)+1\right)\)

\(\text {Substitute } t \text { and } x \text { into } \dfrac{d x}{d t}:\)

\(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}=0.02\, \text{ms} ^{-1}\ \text{(2 d.p.)}\)
 

d.   \(\text {Particle positions at}\ \ t=6:\)

\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)

\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
 

e.    \(e^x=e^{\log _e\left(\tan ^{-1}(12)+1\right)}=\tan ^{-1}(12)+1\)

\(\text {At}\ \ t=6:\)

\(\dfrac{d x_1}{d t}=\dfrac{2}{1+4 t^2} \times \dfrac{1}{\tan ^{-1}(2 t)+1}=\dfrac{2}{145\left(\tan ^{-1}(12)+1\right)}\)

\(\dfrac{d x^2}{d t}=\dfrac{3}{1+(3 t-6)^2} \times \dfrac{1}{\tan ^{-1}(3 t-6)-1}=\dfrac{3}{145\left(\tan ^{-1}(12)+1\right)}\)

\(\therefore \dfrac{v_1}{v_2}=\dfrac{2}{3}\)

Calculus, MET2 2022 VCAA 5

Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.

Use the following table of values for `f` and `f^{\prime}`.

`\quad x \quad` `\quad\quad 1/2\quad\quad` `\quad\quad(sqrt{2})/2\quad\quad` `\quad\quad(sqrt{3})/2\quad\quad`
`f(x)` `-2` `5` `3`
`\quad\quad f^{prime}(x)\quad\quad` `7` `0` `1/9`

 

  1. Find the value of `g\left(\frac{\pi}{6}\right)`.   (1 mark)

The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.

  1. Show that `g^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{9}`.   (1 mark)

  2. Find the equation of the tangent to `g` at `x=\frac{\pi}{6}`.   (2 marks)

  3. Find the average value of the derivative function `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`.   (2 marks)

  4. Find four solutions to the equation `g^{\prime}(x)=0` for the interval `x \in[0, \pi]`.   (3 marks)

Show Answers Only

a.    `3`

b.    `1/9`

c.    `y=1/9x+3-pi/54`

d.    `-48/pi`

e.    ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

Show Worked Solution
 

a.  `g(pi/6)`
 `= f(sin(pi/3))`  
  `= f(sqrt3/2)`  
  `= 3`  

 

b.  `g\ ^{prime}(x)` `= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))`  
`g\ ^{prime}(pi/6)` `= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)`  
  `= 1/9`  

 

c.   `m = 1/9`  and  `g(pi/6) = 3`

`y  –  y_1` `= m(x-x_1)`  
`y  –  3` `= 1/9(x-pi/6)`  
`y` `= 1/9x + 3-pi/54`  

♦♦ Mean mark (c) 45%.
MARKER’S COMMENT: Some students did not produce an equation as required.

  
d.   The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`

Average `= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x`  
  `=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}`  
  `= 24/pi \cdot(f(sqrt3/2)-f(sqrt2/2))`  
  `= 24/pi (3-5) = -48/pi`  

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted `g^{\prime}(x)`, not `g(x)`.

e.   `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`

`:.\   2 \cos (2 x) = 0\ ….(1)`  or  ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`

(1):   ` 2 \cos (2 x)`  `= 0`      `x \in[0, \pi]`
`\cos (2 x)` `= 0`      `2 x \in[0,2 \pi]`
`2x` `= pi/2 , (3pi)/2`  
`x` `= pi/4 , (3pi)/4`  
     
(2): ` f^{\prime}(\sin (2 x)) ` `= sqrt2/2`  
`2x` `= pi/4 , (3pi)/4`  
`x` `= pi/8 , (3pi)/8`  

  
`:. \  x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`


♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find `pi/4, (3pi)/4`. Some solved `2 cos(2x)=0` or `f^{\prime}(sin(2x))=0` but not both.

Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2.  i. Find `f^{\prime}(0)`.   (2 marks)

  3. ii. State the maximal domain over which `f` is strictly increasing.   (1 mark)

  4. Show that `f(x)+f(-x)=0`.   (1 mark)

  5. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)

  6. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  7. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  8. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  9. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  2.  i. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)
  3. ii. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark
Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)`

`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
`= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

    1. Find \(\text{Pr}(X=5)\).   (1 mark)
    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)
    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

    1. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)
    2. Given that  \(\text{Pr}(H \leq 2)=0.35\)  and  \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

    3. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
    4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
    5. Find the values of \(r\) and \(s\).   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate \(p\).
  3. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)
    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)
    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`   

b. ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`

`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

Calculus, SPEC2 2022 VCAA 14 MC

A particle moving in a straight line with constant acceleration has a velocity of 7 ms\(^{-1}\) at point \(A\) and 17 ms\(^{-1}\) at point \(B\).

The velocity of the particle, in metres per second, at the midpoint of \(AB\) is

  1. \(\sqrt{119}\)
  2. \(11\)
  3. \(12\)
  4. \(13\)
  5. \(\sqrt{240}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Using}\ \ v^2=u^2+2as:\)

\(17^2\) \(=7^2+2as\)  
\(as\) \(=120\)  

 
\(\text{Displacement (halfway)}\ = \dfrac{s}{2} \)

\(v^2\) \(=7^2+2a \times \dfrac{s}{2} \)  
  \(=49+120\)  
  \(=169\)  
\(\therefore v\) \(=13\)  

 
\(\Rightarrow D\)

♦♦♦ Mean mark 28%.

PHYSICS, M6 2020 VCE 7 MC

An ideal transformer has an input DC voltage of 240 V, 2000 turns in the primary coil and 80 turns in the secondary coil.

The output voltage is closest to

  1. 0 V
  2. 9.6 V
  3. 6.0 × 10\(^{3}\) V
  4. 3.8 × 10\(^{7}\) V
Show Answers Only

\(A\)

Show Worked Solution
  • Transformers require an AC current to provide the change in flux that generates an EMF in the secondary coil from the primary coil.
  • The question refers to a DC current. There will therefore be no change in flux in the transformer and no output voltage.

\(\Rightarrow A\)

♦♦♦ Mean mark 16%.
COMMENT: Attention! DC current will not work on a transformer.

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.

  1. Show that `a=900` and `b=\frac{\pi}{80}`.   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of `y=\sin(t)` under the transformation `Q`. Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

`s(t)=1700cdote^(-0.003t)cdot sin((pit)/80)+2500,\qquad text(for all)\ t>=0`

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where `t>40`, in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)

Show Answers Only

ai.   `r(0)=2500`

aii.  Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

aiii. `160`  weeks

b.    See worked solution.

c.    `~~ 5339` (nearest whole number)

d.    Weeks between the periods is 160

e.    `~~ 4142` (nearest whole number)

f.    Average rate of change `=-3.6` rabbits/week (1 d.p.)

g.    `t = 156` weeks (nearest whole number)  

h.    ` s → 2500`

Show Worked Solution

ai.  Initial population of rabbits

From graph when `t=0, \ r(0) = 2500`

Using formula when `t=0`

`r(t)` `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500`  
`r(0)` `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500`  rabbits  


aii. From graph,

Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

OR

Using formula

Minimum is when `t = 120`

`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`

Maximum is when `t = 40`

`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`

 
aiii. Number of weeks between maximum populations of rabbits `= 200-40 = 160`  weeks

 
b.  Period of foxes = period of rabbits = 160:

`frac{\2pi}{b} = 160`

`:.\  b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.

Using the point `(100 , 2500)`

Amplitude when `b = frac{\pi}{80}`: 

`f(t)` `=a \ sin (pi/80(t-60))+1600`  
`f(100)` `= 2500`  
`2500` `= a \ sin (pi/80(100-60))+1600`  
`2500` `= a \ sin (pi/2)+1600`  
`a` `= 2500-1600 = 900`  

 
`:.\  f(t)= 900 \ sin (pi/80)(t-60) + 1600`

  

c.    Using CAS find `h(t) = f(t) + r(t)`:

`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`

  
`text{fMax}(h(t),t)|0 <= t <= 160`      `t = 53.7306….`
  

`h(53.7306…)=5339.46`

  
Maximum combined population `~~ 5339` (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

`text{fMax}(h(t),t)|0 <= t <= 320`     `t = 213.7305…`
 
`h(213.7305…)=5339.4568….`
 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

`t^{\prime} = frac{90}{pi}t + 60`   →   `t = frac{pi}{90}(t^{\prime}-60)`

`y^{\prime} = 900y+1600`   →   `y = frac{1}{900}(y^{\prime}-1600)`

`frac{y^{\prime}-1600}{900} = sin(frac{pi(t^{\prime}-60)}{90})`

`:.\  f(t) = 900\ sin\frac{pi}{90}(t-60) + 1600`

  
Average combined population  [Using CAS]   
  
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
  

`= 4142.2646…..  ~~ 4142` (nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted `r(t)`. Others used average rate of change instead of average value.

f.   Using CAS

`s(t):= 1700e^(-0.003t) dot\sin\frac{pit}{80} + 2500`
 

`text{fMax}(s(t),t)|0<=t<=320`        `x = 38.0584….`
 

`s(38.0584….)=4012.1666….`
 

`text{fMax}(s(t),t)|160<=t<=320`     `x = 198.0584….`
 

`s(198.0584….)=3435.7035….`
 

Av rate of change between the points

`(38.058 , 4012.167)`  and  `(198.058 , 3435.704)`

`= frac{4012.1666….-3435.7035….}{38.0584….-198.0584….} =-3.60289….`
 

`:.` Average rate of change `=-3.6` rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
`frac{s(200)-s(40)}{200-40}` was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

`s^(primeprime)(t) = 0` , `t = 80(n-0.049) \ \forall n \in Z`

After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2` 

`t` `= 80(n-0.049)`  
  `= 80(2-0.049)`  
  `= 156.08`  

 
`:. \ t = 156` weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved `frac{ds}{dt}=0`.
A common answer was 41.8, as `s(156.11…)=41.79`.
Another common incorrect answer was 76 weeks.

h.   As `t → ∞`, `e^(-0.003t) → 0`

`:.\ s → 2500`

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
 

  1. State the equation of the axis of symmetry of the graph of `f`.   (1 mark)

  2. State the derivative of `f` with respect to `x`.   (1 mark)

The tangent to `f` at point `M` has gradient `-2` .

  1. Find the equation of the tangent to `f` at point `M`.   (2 marks)

The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
 

 

  1.  i. Find the equation of the line perpendicular to the tangent passing through point `M`.   (1 mark)

  2. ii. The line perpendicular to the tangent at point `M` also cuts `f` at point `N`, as shown in the diagram above.
  3.     Find the area enclosed by this line and the curve `y=f(x)`.   (2 marks)

  4. Another parabola is defined by the rule `g(x)=\frac{x^2}{4 a^2}`, where `a>0`.
  5. A tangent to `g` and the line perpendicular to the tangent at `x=-b`, where `b>0`, are shown below.

  1. Find the value of `b`, in terms of `a`, such that the shaded area is a minimum.   (4 marks)

Show Answers Only

a.    `x=0`

b.    ` f^{\prime}(x)=1/6x`

c.    `x=-12`

di.    `y=1/2x + 18`

dii.   Area`= 375` units²

e.    `b = 2a^2`

Show Worked Solution

a.   Axis of symmetry:  `x=0`
  

b.    `f(x)`  `=\frac{x^2}{12}`  
  ` f^{\prime}(x)` `= 1/6x`  

  
c.   At `M` gradient `= -2`

`1/6x` `= -2`  
`x` `= -12`  

 
When  `x = -12`
 

`f(x) = (-12)^2/12 = 12`
 
Equation of tangent at `(-12 , 12)`:

`y-y_1` `=m(x-x_1)`  
`y-12` `= -2(x + 12)`  
`y` `= -2x-12`  


d.i   Gradient of tangent `= -2`

`:.`  gradient of normal `= 1/2` 

Equation at `M(- 12 , 12)`

`y -y_1` `=m(x-x_1)`  
`y-12` `= 1/2(x + 12)`  
`y` `=1/2x + 18`  


d.ii  Points of intersection of `f(x)` and normal are at  `M` and `N`.

So equate ` y = x^2/12` and  `y = 1/2x + 18` to find `N`

`x^2/12` `=1/2x + 18`  
`x^2-6x-216` `=0`  
`(x + 12)(x-18)` `=0`  

   
`:.\  x = -12` or `x = 18`

Area `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x`  
  `= [x^2/4 + 18x-x^3/36]_(-12)^18`  
  `= [18^2/4 +18^2-18^3/36] – [12^2/4 + 18 xx (-12)-(-12)^3/36]`  
  `= 375` units²  

 

e.   `g(x) = x^2/(4a^2)`   `a > 0`

At `x = -b`   `y = (-b)^2/(4a^2) = b^2/4a^2`

`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`

Gradient of tangent `= (-b)/(2a^2)`

Gradient of normal `= (2a^2)/b`

Equation of normal at `(- b , b^2/(4a^2))`

`y-y_1` `= m(x-x_1)`  
`y-b^2/(4a^2)` `= (2a^2)/b(x-(-b))`  
`y` `= (2a^2x)/b + 2a^2 + b^2/(4a^2)`  
`y` `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)`  

 
Points of intersection of normal and parabola (Using CAS)

solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`

`x =-b`  or  `x = (8a^4+b^2)/b`

 
Calculate area using CAS

`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2}-frac{x^2}{4a^2} \right) dx`

`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
 
Using CAS Solve derivative of `A = 0`  with respect to `b` to find `b`

solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`

 
`b =-2a^2`   and  `b = 2a^2`

Given `b > 0`

`b = 2a^2`


♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found `\int_{-b}^{\frac{8 a^4+b^2}{b}}\left(y_n\right) d x` and failed to subtract `g(x)` or had incorrect terminals.

Probability, MET2 2022 VCAA 20 MC

A soccer player kicks a ball with an angle of elevation of `theta` °, where `theta` is a normally distributed random variable with a mean of 42° and a standard deviation of 8°.

The horizontal distance that the ball travels before landing is given by the function `d=50 \ sin (2\theta)`.

The probability that the ball travels more than 40 m horizontally before landing is closest to

  1. 0.969
  2. 0.937
  3. 0.226
  4. 0.149
  5. 0.027
Show Answers Only

`A`

Show Worked Solution

Using CAS

Solve:  `50\ sin(2\theta)>=40`   for    `0<= theta <= 90`°

`26.565 \ldots \leq \theta \leq 63.434 \ldots`
 

`\theta \~ N\left(42°,8^2°\right)`

`\text{Pr}(26.565 \leq \theta \leq 63.435)~~0.96947`
 
`=>A`


♦♦ Mean mark 30%.
MARKER’S COMMENT: 24% incorrectly chose B and 25% incorrectly chose C.

Calculus, SPEC2 2022 VCAA 10 MC

Consider the curve given by  `5 x^2 y-3 x y+y^2=10`.

The equation of the tangent to this curve at the point `(1, m)`, where `m` is a real constant, will have a negative gradient when

  1. `m \in R \backslash[-1,0]`
  2. `m=-\sqrt{11}-1 \ text {only}`
  3. `m \in R \backslash(-1,0]`
  4. `m=\sqrt{11}-1 \ text[only]`
  5. `m=-\sqrt{11}-1 or m=\sqrt{11}-1`
Show Answers Only

`E`

Show Worked Solution

`5 x^2 y-3 x y+y^2=10`

`text{At}\ (1,m):\ \ 5m-3m+m^2=10`

`text{Solve (by CAS):}\ \ m=-1+-sqrt{11}`

`text{Check the value of the derivative function at both values of}\ m\ (x=1)`

`text{By CAS, both values (gradients) are negative.}`

`=>E`

Calculus, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g^{'}(x)\), can be expressed in the form \(g^{'}(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
  4.  i. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)

    ii. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

  1.  

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
Show Answers Only

a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g^{\prime}(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h^{\prime}(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h^{”}(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h^{”}(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h^{\prime}(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h^{\prime}(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h^{\prime}(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h^{\prime}(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f^{\prime}(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

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

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

PHYSICS, M6 2021 VCE 5

The digram shows shows a stationary electron (e\(^{-}\)) in a uniform magnetic field between two parallel plates. The plates are separated by a distance of 6.0 × 10\(^{-3}\) m, and they are connected to a 200 V power supply and a switch. Initially, the plates are uncharged. Assume that gravitational effects on the electron are negligible.
 

  1. Explain why the magnetic field does not exert a force on the electron. Justify your answer with an appropriate formula.   (2 marks)

The switch is now closed.

  1. Determine the magnitude and the direction of any electric force now acting on the electron. Show your working.   (3 marks)

  1. Ravi and Mia discuss what they think will happen regarding the size and the direction of the magnetic force on the electron after the switch is closed.

    Ravi says that there will be a magnetic force of constant magnitude, but it will be continually changing direction.

    Mia says that there will be a constantly increasing magnetic force, but it will always be acting in the same direction.

    Evaluate these two statements, giving clear reasons for your answer.   (4 marks)

Show Answers Only

a.   Magnetic force does not exert force on electron:

  • Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).
  • Electron is stationary (\(v=0\))
  • The force on the electron = \(0\).

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

  • Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).

c.    Both statements contain a correct assertion and incorrect assertion.

  • Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.
  • Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.
  • This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 
  • Mia was incorrect to say that the direction of the magnetic force on the electron would not change.
  • Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.
  • This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.

Show Worked Solution

a.   Magnetic force does not exert force on electron:

  • Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).
  • Electron is stationary (\(v=0\))
  • The force on the electron = \(0\).
♦♦ Mean mark (a) 35%. 
COMMENT: Many students confused the forces of electric (not applicable here) and magnetic fields.

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

  • Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).

c.    Both statements contain a correct assertion and incorrect assertion.

  • Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.
  • Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.
  • This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 
  • Mia was incorrect to say that the direction of the magnetic force on the electron would not change.
  • Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.
  • This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.
♦♦♦ Mean mark (c) 18%.
COMMENT: Students need well planned responses to properly evaluate the physics principles required for 4 marks.

PHYSICS, M6 2021 VCE 2

A schematic side view of one design of an audio loudspeaker is shown in Diagram 1 below. It uses a current carrying coil that interacts with permanent magnets to create sound by moving a cone in and out.
 

Diagram 2 shows a schematic view of the loudspeaker from the position of the eye shown in Diagram 1. The direction of the current is clockwise, as shown.
 

   

  1. Draw four magnetic field lines on Diagram 2, showing the direction of each field line using an arrow.   (1 mark)

  2. Which one of the following gives the direction of the force acting on the current carrying coil shown in Diagram 2?   (1 mark)
A. left B. right
C. up the page D. down the page
E. into the page F. out of the page
  1. The current carrying coil has a radius of 5.0 cm and 20 turns of wire, and it carries a clockwise current \((I)\) of 2.0 A. Its magnetic field strength \((B)\) is 200 mT.
  2. Calculate the magnitude of the force, \(F\), acting on the current carrying coil. Show your working.   (2 marks)

Show Answers Only

a.   
         

b.    \(E\)

c.    \(2.51\ \text{N}\)

Show Worked Solution

a.    Magnetic fields run from the north pole to the south pole.
 

   

 

b.    Apply the right-hand rule:

  • Thumb to the right and fingers down, the force on the current carrying coil must be into the page.

\(\Rightarrow E\)
 

c.    \(l= 2\pi \times 5 \times 10^{-2} = 0.1\pi\)

\(\therefore F=nlIB=20 \times 0.1\pi \times 2 \times 0.2=2.51\ \text{N}\)

♦♦♦ Mean mark 24%.

Functions, MET2 2023 VCAA 2

The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
 

Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.

  1. Show that \(b=\dfrac{\pi}{15}\) and \(c=75\).   (2 marks)
  2. Find the average height of a pod on the wheel as it travels from point \(A\) to point \(B\).
  3. Give your answer in metres, correct to two decimal places.   (2 marks)
  4. Find the average rate of change, in metres per minute, of the height of a pod on the wheel as it travels from point \(A\) to point \(B\).   (1 mark)

After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.

The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
 

\(w(t) = \begin {cases}
h(t)         &\ \ 0 \leq t < 15 \\
k         &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)

 
where \(k\geq 0, m\geq 0\) and \(n \in R\).

  1.   i.State the values of \(k\) and \(m\).   (1 mark)
       
     ii. Find all possible values of \(n\).   (2 marks)

      

    iii. Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints.   (3 marks)


     
  1.  
Show Answers Only

a.    \(\text{See worked solution}\)

b.    \(\approx 36.80\ \text{m}\)

c.    \(8\)

d.i.  \(k=135, m=2\)

d.ii. \(n=30p+5,\ p\in Z\)

d.iii. 

Show Worked Solution
a.    \(\text{Period:}\) \(\dfrac{2\pi}{b}\) \(=30\)
    \(\therefore\ b\) \(=\dfrac{\pi}{15}\)

  
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)

\(-60\ \cos(0)+c\)  \(=15\)
\(c\) \(=75\)

  \(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)

b.     \(\text{Average height}\) \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\)
    \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\)
    \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\)
    \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\)
    \(=36.802\dots\approx 36.80\ \text{m}\)
 
♦♦ Mean mark (b) 45%.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.

c.   \(\text{Av rate of change of height}\)

  \(=\dfrac{h(7.5)-h(0)}{7.5}\)
  \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\)
  \(=\dfrac{75-15}{7.5}=8\)
 
♦ Mean mark (c) 50%.
MARKER’S COMMENT: Common incorrect answer was – 8.
di.    \(\text{Period is 30 minutes, so after 15 minutes pod}\)
\(\text{is at the top of the wheel.}\)
  \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\)

  
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)

\(\therefore\ \text{Period}\rightarrow\) \(\dfrac{2\pi}{bm}\) \(=15\)
  \(bm\) \(=\dfrac{2\pi}{15}\)
  \(\dfrac{\pi}{15}\cdot m\) \(=\dfrac{2\pi}{15}\)
  \(m\) \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\)
    \(=2\)
 
♦♦ Mean mark (d)(i) 40%.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
dii.   \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\)
  
\(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\)

   
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)

\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)

\(\therefore\ n=30p+5,\ p\in Z\)

 
♦♦♦ Mean mark (d)(ii) 20%.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.

diii.   

 
♦♦♦ Mean mark (d)(iii) 40%.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.

Calculus, MET2 2023 VCAA 1

Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.

  1. State the coordinates of all axial intercepts of \(f\).   (1 mark)
  2. Find the coordinates of the stationary points of \(f\).   (2 marks)
    1. Let \(g:R\rightarrow R, g(x)=x-2\).
    2. Find the values of \(x\) for which \(f(x)=g(x)\).   (1 mark)
    1. Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\).  (2 marks)
    2. Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places.   (1 mark)
  1. Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
  2. Find the possible values of \(a\) and \(b\).   (4 marks)
Show Answers Only

a.    \((-1, 0), (0, 0), (2, 0)\)

b.    \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)

c.i.  \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

 \(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)

c.iii. \(5.95\)

d.   \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

b.    \(\text{Using CAS solve for}\ x:\)

\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)

\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)

\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)

\(\text{The stationary points of }f\ \text{are}:\)

\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
  

ci    \(\text{Given }f(x)=g(x)\)

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

  
\(\therefore\ \text{Using CAS: } \)

\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

cii  \(\text{Area of bounded region:}\)

\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
  

ciii  \(\text{Solve the integral in c.ii above using CAS:}\)
  \(\text{Total area}=5.946045..\approx 5.95\)

  

d.   \(\text{Method 1 – Equating coefficients}\)

\((x-a)(x-b)^2=x(x-2)(x+1)+k\)

\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)

\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)

\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)

\(2ab+b^2=-2\ \ …(2)\)

\(\text{Substitute (1) into (2) and solve for }b.\)

\(2b(1-2b)+b^2\) \(=-2\)
\(3b^2-2b-2\) \(=0\)
\(b\) \(=\dfrac{1\pm \sqrt{7}}{3}\)
\(\text{When }b\) \(=\dfrac{1+\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\)
\(\text{When }b\) \(=\dfrac{1-\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\)

  

\(\text{Method 2 – Using transformations}\)

\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)

\(\text{shows that the turning point is on the }x\ \text{axis}.\)

\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)

\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

 

PHYSICS, M5 2022 VCE 2

There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. The diagram shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator.
 

  1. Explain why geostationary satellites must be vertically above the equator to remain stationary relative to Earth's surface.   (2 marks)
  1.  Using  \(G=6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}, M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}\)  and  \(R_{\text{E}} = 6.37 \times 10^{6}\ \text{m}\), show that the altitude of a geostationary satellite must be equal to \(3.59 \times 10^{7}\ \text{m}\).   (4 marks)
  1. Calculate the speed of an orbiting geostationary satellite.   (3 marks)

Show Answers Only

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

  • So that it is in the same rotational plane/axis of the Earth. 
  • The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.
  • As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.

b.    See worked solutions

c.    3076 ms\(^{-1}\)

Show Worked Solution

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

  • So that it is in the same rotational plane/axis of the Earth. 
  • The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.
  • As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.
♦♦♦ Mean mark (a) 19%.
b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(r\) \(=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}\)
    \(=\sqrt[3]{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (86\ 400)^2}{4\pi^2}}\)
    \(=42.250 \times 10^6\ \text{m}\)

 
\(\therefore \ \text{Altitude}\ =42.250 \times 10^6 -6.37 \times 10^6 =3.59 \times 10^7\ \text{m}\)
 

c.     \(v\) \(=\sqrt{\dfrac{GM}{r}}\)
    \(=\sqrt{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42.25 \times 10^6}}\)
    \(=3073\ \text{ms}^{-1}\)

Complex Numbers, SPEC2 2023 VCAA 2

Let \(w=\text{cis}\left(\dfrac{2 \pi}{7}\right)\).

  1. Verify that \(w\) is a root of  \(z^7-1=0\).   (1 marks)
  1. List the other roots of  \(z^7-1=0\)  in polar form.   (1 mark)
  1. On the Argand diagram below, plot and label the points that represent all the roots of  \(z^7-1=0\).   (2 marks)

  1.  i. On the Argand diagram below, sketch the ray that originates at the real root of  \(z^7-1=0\)  and passes through the point represented by \( \text{cis}\left(\dfrac{2 \pi}{7} \right)\).   (1 mark)

     

  1. ii. Find the equation of this ray in the form \(\text{Arg}\left(z-z_0\right)=\theta\), where \(z_0 \in C\), and \(\theta\) is measured in radians in terms of \(\pi\).   (1 mark)

  2. Verify that the equation  \(z^7-1=0\)  can be expressed in the form 
  3.     \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\).   (1 mark)
  4.   i. Express  \(\text{cis}\left(\dfrac{2 \pi}{7}\right)+\operatorname{cis}\left(\dfrac{12 \pi}{7}\right)\) in the form \(A \cos (B \pi)\), where \(A, B \in R^{+}\).   (1 mark)

  5. ii. Given that  \(w=\operatorname{cis}\left(\dfrac{2 \pi}{7}\right)\)  satisfies  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\),

use De Moivre's theorem to show that

      1. \(\cos \left(\dfrac{2 \pi}{7}\right)+\cos \left(\dfrac{4 \pi}{7}\right)+\cos \left(\dfrac{6 \pi}{7}\right)=-\dfrac{1}{2}\).   (2 marks)

Show Answers Only

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Show Worked Solution

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Probability, MET2 2023 VCAA 4

A manufacturer produces tennis balls.

The diameter of the tennis balls is a normally distributed random variable \(D\), which has a mean of 6.7 cm and a standard deviation of 0.1 cm.

  1. Find  \(\Pr(D>6.8)\), correct to four decimal places.   (1 mark)
  2. Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
  3. Give your answer in centimetres, correct to two decimal places.   (1 mark)

Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.

  1. Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. In a random selection of 4 tennis balls, find the probability that at least 3 balls can fit through the opening at the top of the container.
  4. Give your answer correct to four decimal places.   (2 marks)

A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.

  1. Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
  2. Give your answer correct to four decimal places.   (2 marks)
  3. The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
  4. Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places.   (2 marks)
  5. An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
  6. The confidence interval is (0.7382, 0.9493), correct to four decimal places.
  7. Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer.   (2 marks)

A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, \(V\), with the probability density function
 

\(f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)         &\ \ 30 \leq v \leq 3\pi^2+30 \\
0         &\ \ \text{elsewhere}
\end{cases}\)
 

  1. Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. Find the exact mean serving speed for grade A balls, in metres per second.   (1 mark)

The serving speed of a grade B ball is given by a continuous random variable, \(W\), with the probability density function \(g(w)\).

A transformation maps the graph of \(f\) to the graph of \(g\), where \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).

  1. If the mean serving speed for a grade B ball is \(2\pi^2+8\) metres per second, find the values of \(a\) and \(b\).   (2 marks)
Show Answers Only

a.    \(0.1587\)

b.    \(d\approx6.83\)

c.    \(0.9938\)

d.    \(0.9998\)

e.    \(0,8960\)

f.    \(0.06\)

g.    \(90\%\)

h.    \(0.1345\)

i.    \(3(\pi^2+4)\)

j.    \(a=\dfrac{3}{2}, b=\dfrac{2}{3}\)

Show Worked Solution

a.    \(\text{Normal distribution:}\to \mu=6.7, \sigma=0.1\)

\(D\sim N(6.7, 0.1^2)\)

\(\text{Using CAS: }[\text{normCdf}(6.8, \infty, 6.7, 0.1)]\)

\(\Pr(D>6.8)=0.15865…\approx0.1587\)
 

b.    \(\Pr(D<d)=0.90\)  

\(\Pr\Bigg(Z<\dfrac{d-6.7}{0.1}\Bigg)=0.90\)

\(\text{Using CAS: }[\text{invNorm}(0.9, 6.7, 0.1)]\)

\(\therefore\ d=6.828155…\approx6.83\ \text{cm}\)

 

c.   \(\text{Normal distribution:}\to \mu=6.7, \sigma=0.1\)

\(D\sim N(6.7, 0.1^2)\)

\(\text{Using CAS: }[\text{normCdf}(-\infty, 6.95, 6.7, 0.1)]\)

\(\Pr(D<6.95)=0.99379…\approx0.9938\)
 

d.    \(\text{Binomial:}\to  n=4, p=0.99379…\)

\(X=\text{number of balls}\)

\(X\sim \text{Bi}(4, 0.999379 …)\)

\(\text{Using CAS: }[\text{binomCdf}(4, 0.999379 …, 3, 4)]\)

\(\Pr(X\geq 3)=0.99977 …\approx 0.9998\)

 

e.    \(\Pr(\text{Grade A|Fits})\) \(=\Pr(6.54<D<6.86|D<6.95)\)
    \(=\dfrac{\Pr(6.54<D<6.86)}{\Pr(D<6.95)}\)
  \(\text{Using CAS: }\) \(=\Bigg[\dfrac{\text{normCdf}(6.54, 6.86, 6.7, 0.1)}{\text{normCdf}(-\infty, 6.95, 6.7, 0.1)}\Bigg]\)
    \(=\dfrac{0.89040…}{0.99977…}\)
    \(=0.895965…\approx 0.8960\)

  

f.    \(\text{Normally distributed → symmetrical}\)

\(\text{Pr ball diameter outside the 99% interval}=1-0.99=0.01\)

\(D\sim N(6.7, \mu^2)\)

\(\Pr(6.54<D<6.86)>0.99\)

\(\therefore\ \Pr(D<6.54)<\dfrac{1-0.99}{2}\)

\(\text{Find z score using CAS: }\Bigg[\text{invNorm}\Bigg(\dfrac{1-0.99}{2},0,1\Bigg)\Bigg]=-2.575829…\)

\(\text{Then solve:} \ \dfrac{6.54-6.7}{\sigma}<-2.575829…\)

\(\therefore\ \sigma<0.0621\approx 0.06\)


♦♦ Mean mark (f) 35%.
MARKER’S COMMENT: Incorrect responses included using \(\Pr(D<6.86)=0.99\). Many answers were accepted in the range \(0<\sigma\leq0.06\) as max value of SD was not asked for provided sufficient working was shown.

g.   \(\hat{p}=\dfrac{0.7382+0.9493}{2}=0.84375\)

\(\text{Solve the following simultaneous equations for }z, \hat{p}=0.84375\)

\(0.84375-z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.7382\)      \((1)\)
\(0.84375+z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.9493\) \((2)\)
\(\text{Equation}(2)-(1)\)    
\(2z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.2111\)  
\(z=\dfrac{0.10555}{\sqrt{\dfrac{0.84375(1-0.84375)}{32}}}\) \(=1.64443352\)  

 
\(\text{Alternatively using CAS: Solve the following simultaneous equations for }z, \hat{p}\)

\(\hat{p}-z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\) \(=0.7382\)
\(\text{and}\)  
\(\hat{p}+z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\) \(=0.9493\)

\(\rightarrow \ z=1.64444…\ \text{and}\ \hat{p}=0.84375\)
 

\(\therefore\ \text{Using CAS: normCdf}(-1.64443352, 1.64443352, 0 , 1)\)

\(\text{Level of Confidence} =0.89999133…=90\%\)


♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: \(\hat{p}\) was often calculated incorrectly with \(\hat{p}=0.8904\) frequently seen.

h.   \(\text{Using CAS: Evaluate }\to \displaystyle \int_{50}^{3\pi^2+30} \frac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=0.1345163712\approx 0.1345\)

i.   \(\text{Using CAS: Evaluate }\to \displaystyle \int_{30}^{3\pi^2+30} v.\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=3(\pi^2+4)=3\pi^2+12\)

j.  \(\text{When the function is dilated in both directions, }a\times b=1\)
 

\(\text{Method 1 : Simultaneous equations}\)

\(g(w) = \begin {cases}
\dfrac{b}{6\pi}\sin\Bigg(\sqrt{\dfrac{\dfrac{w}{b}-30}{3}}\Bigg)         &\ \ 30b \leq v \leq b(3\pi^2+30) \\
\\ 0         &\ \ \text{elsewhere}
\end{cases}\)

\(\text{Using CAS: Define }g(w)\ \text{and solve the simultaneous equations below for }a, b.\)
 

\(\displaystyle \int_{30.b}^{b.(3\pi^2+30)} g(w)\,dw=1\)

\(\displaystyle \int_{30.b}^{b.(3\pi^2+30)} w.g(w)\,dw=2\pi^2+8\)

\(\therefore\ b=\dfrac{2}{3}\ \text{and}\ a=\dfrac{3}{2}\)

 
\(\text{Method 2 : Transform the mean}\)

 

\(\text{Area}\) \(=1\)
\(\therefore\ a\) \(=\dfrac{1}{b}\)
\(\to\ b\) \(=\dfrac{E(W)}{E(V)}\)
  \(=\dfrac{2\pi^2+8}{3\pi^2+12}\)
  \(=\dfrac{2(\pi^2+4)}{3(\pi^2+4)}\)
  \(=\dfrac{2}{3}\)
 
\(\therefore\ a\) \(=\dfrac{3}{2}\)

♦♦♦ Mean mark (j) 10%.

Calculus, MET2 2023 VCAA 5

Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).

  1. Complete a possible sequence of transformations to map \(f\) to \(g\).   (2 marks)
    •    Dilation of factor \(\dfrac{1}{2}\) from the \(x\) axis.

Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.

  1. Give the domain and range for the inverse of \(g_1\).   (2 marks)

Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).
 

The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).

    1. Find the coordinates of \(P\) and \(Q\), correct to two decimal places.   (1 mark)
    1. Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
    2. Give your answer correct to two decimal places.   (2 marks)

Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).

  1. The turning point of \(h\) always lies on the graph of the function \(y=2x^n\), where \(n\) is an integer.
  2. Find the value of \(n\).  (1 mark)

Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).

The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)

  1. What is the smallest value of \(k\) such that \(h\) will intersect with the inverse of \(h_1\)?\
  2. Give your answer correct to two decimal places.   (1 mark)

It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).

  1. Find the area of the region bound by the graphs of \(h\) and the inverse of \(h_1\), where \(k=5\).
  2. Give your answer correct to two decimal places.   (2 marks)

Show Answers Only

a.    \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b.    \(\text{See worked solution}\)

c.i.  \(P(1.27, 1.27\), Q(4.09, 4.09)\)

c.ii. \(2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx=5.56\)

d.    \(n=-1\)

e. \(k\approx 1.27\)

f. \(A=43.91\)

Show Worked Solution

a.    \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b.    \(\text{Some options include:}\)

•  \(\text{Domain}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)
\(\text{or}\)
•  \(\text{Domain}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)
 

\(\text{If functions split at turning point } (2, 1)\ \text{then possible options:}\)

• \(\text{Domain}\ g_1\to [2, \infty)\ \ \ \text{Range}\to [1, \infty)\)
   \(\text{and }\)
  \(\text{Domain}\ g_1^{-1}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)

\(\text{or}\)

•  \(\text{Domain}\ g_1\to (2, \infty)\ \ \ \text{Range}\to (1, \infty)\)
   \(\text{and }\)
   \(\text{Domain}\ g_1^{-1}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)


♦ Mean mark 50%.
MARKER’S COMMENT: Note: maximal domain not requested so there were many correct answers. Many students seemed to be confused by notation. Often domain and range reversed.

c.i.  \(\text{By CAS:}\  P(1.27, 1.27), Q(4.09, 4.09)\)

c.ii.   \(\text{Area}\) \(=2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx\)
    \(=2\displaystyle \int_{1.27…}^{4.09…} \Bigg(x-\dfrac{1}{2}\Big(e^{2-x}+e^{x-2}\Big)\Bigg)\,dx\)
    \(\approx 5.56\)

♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students set up the integral but did not provide an answer.
Others did not double the integral. Some incorrectly used \(\displaystyle \int_{1.27…}^{4.09…}(g_1^-1-g(x))\,dx\).
Others had correct answer but incorrect integral.

d.    \((0, 2)\ \text{turning pt of }f(x)\)

\(\text{and }h(x) \text{is the transformation from }f(x)\)

\(\therefore \Bigg(k, \dfrac{2}{k}\Bigg)\ \text{turning tp of }h(x)\)

\(\text{so the coordinates have the relationship}\)

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

\(\therefore\ n=-1\)


♦♦♦ Mean mark (d) 10%.
MARKER’S COMMENT: Question not well done. \(n=1\) a common error.

e.    \(\text{Smallest }k\ \text{is when inverse of the curves }h_1\ \text{and}\ h\)

\(\text{touch with the line }y=x.\)

\(\therefore\ h(x)\) \(=\dfrac{1}{k}\Big(e^{k-x}+e^{x-k}\Big)\)
  \(=x\)

 

\(\text{and}\ h'(x)\) \(=\dfrac{1}{k}\Big(-e^{k-x}+e^{x-k}\Big)\)
  \(=1\)

\(\text{at point of intersection, where }k>0.\)

\(\text{Using CAS graph both functions to solve or solve as simultaneous equations.}\)

\(\rightarrow\ k\approx 1.2687\approx 1.27\)


♦♦♦♦ Mean mark (e) 0%.
MARKER’S COMMENT: Question poorly done with many students not attempting.

f.    \(\text{When}\ k=5\)

\(h_1^{-1}(x)=\log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5\)

\(\text{and }h(x)=\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\)

\(\text{Using CAS: values of }x\ \text{at of intersection of fns are}\)

\(x\approx 1.45091…\ \text{and}\ 8.78157…\)

\(\text{Area}\) \(=\displaystyle \int_{1.45091…}^{8.78157…}h_1^{-1}(x)- h(x)\,dx\)
  \(=\displaystyle \int_{1.45091…}^{8.78157…} \log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5-\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\,dx\)
  \(=43.91\) 

♦♦♦ Mean mark (f) 15%.
MARKER’S COMMENT: Errors were made with terminals. Common error using \(x=2.468\), which was the \(x\) value of the pt of intersection of \(h\) and \(y=x\).

Calculus, MET2 2023 VCAA 11 MC

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-2)=-7,\ g(-2)=8\)  and  \(f^{′}(-2)=3,\ g^{′}(-2)=2\).

The gradient of the graph  \(y=f(x)\times g(x)\)  at the point where  \(x=-2\)  is

  1. \(-10\)
  2. \(-6\)
  3. \(0\)
  4. \(6\)
  5. \(10\)
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Using the Product Rule when}\ \ x=-2:\)

\(\dfrac{d}{dx}(f(x)\times g(x))\) \(=f(x)g^{′}(x)+g(x)f^{′}(x)\)
  \(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\)
  \(=-7\times 2+8\times 3\)
  \(=10\)

 
\(\Rightarrow E\)


♦♦♦ Mean mark 22%.
MARKER’S COMMENT: 51% of students incorrectly chose D (6).

Calculus, MET1 2022 VCAA 8

Part of the graph of `y=f(x)` is shown below. The rule `A(k)=k \ sin(k)` gives the area bounded by the graph of `f`, the horizontal axis and the line `x=k`.
 

  1. State the value of `A\left(\frac{\pi}{3}\right)`.   (1 mark)

  2. Evaluate `f\left(\frac{\pi}{3}\right)`.   (2 marks)

  3. Consider the average value of the function `f` over the interval `x \in[0, k]`, where `k \in[0,2]`.
  4. Find the value of `k` that results in the maximum average value.   (2 marks)

Show Answers Only

a.    ` (sqrt(3)pi)/6`

b.    `(3sqrt(3) +pi)/6`

c.    `k = pi/2`

Show Worked Solution
a.  `A(pi/3)` `= pi/3sin(pi/3)`  
  `= pi/3 xx sqrt(3)/2`  
  `= (sqrt(3)pi)/6`  
b.   `f(k)` `= A^{\prime}(k)`  
`f(k)` `=d/dx(k\ sin\ k)`  
  `= sin\ k + k\ cos\ k`  
`f(pi/3)` `= sin\ pi/3 + pi/3\ cos\ pi/3`  
  `= sqrt(3)/2 + pi/3 xx 1/2`  
  `= sqrt(3)/2 + pi/6`  
  `=(3sqrt(3) +pi)/6`  

♦♦♦ Mean mark (b) 25%.
MARKER’S COMMENT: Common error was giving derivative of `A(k)` as `k\ cos (k)`.
c.   Average value `= (A(k))/k`  
  `= 1/k(int_0^k f(x) d x)`  
  `= 1/k[x\ sin (x)]_0^k`  
  `= 1/k[k\ sin (k)] = sin\ k`  

 
`:.` Average value has a maximum value of 1 when `k = pi/2`


♦♦♦ Mean mark (c) 25%.
MARKER’S COMMENT: Students often chose unnecessarily complicated methods or inconsistently applied nomenclature when attempting this question.

Calculus, MET1 2022 VCAA 7

A tilemaker wants to make square tiles of size 20 cm × 20 cm.

The front surface of the tiles is to be painted with two different colours that meet the following conditions:

  • Condition 1 - Each colour covers half the front surface of a tile.
  • Condition 2 - The tiles can be lined up in a single horizontal row so that the colours form a continuous pattern.

An example is shown below.
 

There are two types of tiles: Type A and Type B.

For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.

The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
 

  1.  i. Find the area of the front surface of each tile.   (1 mark)

    ii. Find the value of `a` so that a Type A tile meets Condition 1.   (1 mark)


Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
 

  1. Show that a Type B tile meets Condition 1.   (3 marks)

  2. Determine the endpoints of `f(x)` and `g(x)` on each tile. Hence, use these values to confirm that Type A and Type B tiles can be placed in any order to produce a continuous pattern in order to meet Condition 2.   (2 marks)

Show Answers Only

a.i.    `400` cm²

a.ii.   `a = 10`

b.     `200` cm²

c.     See worked solution

Show Worked Solution
a.i   Area `= 20 xx 20`  
  `= 400` cm²  

  
a.ii  `a = 10`
 

b.   Area `=\int_0^{20} \frac{-x^3}{100}+\frac{3 x^2}{10}-2 x+10\ d x`  
  `=\left[\frac{-x^4}{400}+\frac{x^3}{10}-x^2+10 x\right]_0^{20}`  
  `=\left[-\frac{20^4}{400}+\frac{20^3}{10}-20^2+10 xx 20\right]-\left[0\right]`  
  `= – 400 +800 -400 +200`  
  `= 200` cm²  

 
`:.` The area of the coloured section of the Type B tile is 200 cm² which is half the 400 cm² area of the tile.
 

c.   
`f(0)` `=4 \sin \left(\frac{\pi(0)}{10}\right)+10 = 10`  
  `f(20)` `=4 \sin (2 \pi)+10 = 10`  
  `g(0)` `=\frac{-0}{100}+\frac{3(0)}{10}-2(0)+10=10`  
  `g(20)` `=-\frac{8000}{100}+\frac{200}{10}-2(20)+10 = 10`  

 
→`\ f(0) = f(20) = g(0) = g(20) =10`
 

`:.`    The endpoints for `f(x)` are `(0,10)` and `(20,10)` and for `g(x)` are also `(0,10)` and `(20,10)`.

So the tiles can be placed in any order to make the continuous pattern.


♦♦ Mean mark (c) 30%.
MARKER’S COMMENT: Students often only found `f(20)` and `g(20)`, however, `f(0)` and `g(0)` also needed to be found to verify the pattern match.

Graphs, MET1 2022 VCAA 6

The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.
 

  1. On the axes above, draw the graph of `y=g(x)`, where `g(x)` is the reflection of `f(x)` in the horizontal axis.   (2 marks)
  2. Find all values of `k` such that `f(k)=0` and `k \in[0,2 \pi]`.   (3 marks)
  3. Let `h: D \rightarrow R, h(x)=2 \sin(2x)-1`, where `h(x)` has the same rule as `f(x)` with a different domain.
  4. The graph of `y=h(x)` is translated `a` units in the positive horizontal direction and `b` units in the positive vertical direction so that it is mapped onto the graph of `y=g(x)`, where `a, b \in(0, \infty)`.
    1. Find the value for `b`.   (1 mark)

    2. Find the smallest positive value for `a`.   (1 mark)

    3. Hence, or otherwise, state the domain, `D`, of `h(x)`.   (1 mark)

Show Answers Only

a.    Graph `y=g(x)`

b.    `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`

c.i.    `b=2`

cii.    `=\frac{\pi}{2}`

ciii.   `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

Show Worked Solution

a.

b.  `2 \sin (2 k)-1` `=0`       `0<=k<=2\pi`  
`sin (2 k)` `=1/2`  
`2k` `=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}`  
`k` `=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`  

 

c.i   ` 2 \sin 2(x-a)-1+b` `=-(2 \sin 2 x-1)`  
`:.\ -1+b` `=1`  
`b` `=2`  

 

c.ii  `2 \sin 2(x-a)` `=-(2 \sin 2 x-1)`  
`\sin (2 x-2 a)` `=-\sin 2 x`  
`\therefore 2 a` `=\pi`  
`a` `=\frac{\pi}{2}`  

♦ Mean mark (c.ii) 50%.
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`

c.iii The domain for `f(x)` is `[0,2pi]`

`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

 
♦♦♦ Mean mark (c.iii) 10%.
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`

Calculus, MET1 2023 VCAA 9

The shapes of two walking tracks are shown below.
 

 

Track 1 is described by the function  \(f(x)=a-x(x-2)^2\).

Track 2 is defined by the function  \(g(x)=12x-bx^2\).

The unit of length is kilometres.

  1. Given that \(f(0)=12\)  and  \(g(1)=9\), verify that \(a=12\)  and  \(b=-3\).   (1 mark)
  2. Verify that \(f(x)\) and \(g(x)\) both have a turning point at \(P\).
  3. Give the co-ordinates of \(P\).  (2 marks)
  4. A theme park is planned whose boundaries will form the triangle \(\Delta OAB\) where \(O\) is the origin, \(A\) is at \((k, 0)\) and \(B\) is at \((k, g(k))\), as shown below, where \(k \in (0, 4)\).
  5. Find the maximum possible area of the theme park, in km².   (3 marks)
     

Show Answers Only

a.    \(\text{See worked solution}\)

b.    \(P(2, 12)\ \text{Also see worked solution}\)

c.    \(\dfrac{128}{9}\)

Show Worked Solution
a.    \(f(0)\) \(=a-0(0-2)^2=12\)
  \(\therefore\ a\) \(=12\)

 

\(g(1)\) \(=12\times 1+b\times 1^2\)
\(12+b\) \(=9\)
\(\therefore b\) \(=-3\)

 

b.    \(f(x)\) \(=12-x(x-2)^2\)
    \(=-x^3+4x^2-4x+12\)
  \(f^{′}(x)\) \(=-(3x^2-8x+4)\)

 

\(\text{For turning point }f^{′}(x)=0:\)

\( -(3x^2-8x+4)\) \(=0\)
\(-(3x-2)(x-2)\) \(=0\)

 
\(\therefore\ x=\dfrac{2}{3}\ \text{or}\ 2\)
 

\(f(2)\) \(=12-2(2-2)^2=12\)
\(f\Big{(}\dfrac{2}{3}\Big{)}\) \(=12-\dfrac{2}{3}\Big(\dfrac{2}{3}-2\Big)^2<12\)

 
\(\therefore\ \text{Given graph of }f(x)\ \text{relevant turning point is }(2, 12).\)
 

\(g(x)=12x-3x^2\ \Rightarrow\ g^{′}(x)=12-6x\)

\(\text{For turning point}\ \ g^{′}(x)=0:\)

\(12-6x=0\ \ \Rightarrow \ x=2\)

\(g(x)\) \(=12x-3x^2\)
\(g(2)\) \(=12\times 2-3\times 2^2=12\)

 
\(\therefore\ g(x)\ \text{also has a turning point at }(2, 12).\)
 


♦ Mean mark (b) 48%.

c.   \(\text{Area of triangle }\Delta OAB\)

\(A(k)\) \(=\dfrac{1}{2}\times k\times g(k)\)
  \(=\dfrac{k}{2}\times (12k-3k^2)\)
  \(=6k^2-\dfrac{3k^3}{2}\)

 
\(\text{Max area when }A^{′}(k)=0:\)

\(A^{′}(k)=12k-\dfrac{9k^2}{2}=\dfrac{1}{2}(24k-9k^2)\)

\(\dfrac{1}{2}k(24-9k)=0\)

\(\therefore\ k=\dfrac{24}{9}=\dfrac{8}{3}\)

 
\(\text{Maximum area occurs when}\ k=\dfrac{8}{3}:\)

\(A(k)\) \(=\dfrac{1}{2}k\times (12k-3k^2)\)
\(A\bigg(\dfrac{8}{3}\bigg)\) \(=\dfrac{1}{2}\times\dfrac{8}{3}\times\bigg (12\bigg(\dfrac{8}{3}\bigg)-3\bigg(\dfrac{8}{3}\bigg)^2\bigg)\)
  \(=\dfrac{4}{3}\bigg(32-\dfrac{64}{3}\bigg)\)
  \(=\dfrac{4}{3}\bigg(\dfrac{96}{3}-\dfrac{64}{3}\bigg)\)
  \(=\dfrac{4}{3}\times\dfrac{32}{3}\)
  \(=\dfrac{128}{9}\)

♦♦♦ Mean mark (c) 27%.
MARKER’S COMMENT: Many students made arithmetic errors substituting the fractional values of \(k\) into \(A(k)\) to find max area.

Statistics, MET1 2023 VCAA 6

Let \(\hat{P}\) be the random variable that represents the sample proportion of households in a given suburb that have solar panels installed.

From a sample of randomly selected households in a given suburb, an approximate 95% confidence interval for the proportion \(p\) of households having solar panels installed was determined to be (0.04, 0.16).

  1. Find the value of \(\hat{p}\) that was used to obtain this approximate 95% confidence interval.   (1 mark)

Use \(z=2\) to approximate the 95% confidence interval.

  1. Find the size of the sample from which this 95% confidence interval was obtained.   (2 marks)
  2. A larger sample of households is selected, with a sample size four times the original sample.
  3. The sample proportion of households having solar panels installed is found to be the same.
  4. By what factor will the increased sample size affect the width of the confidence interval?   (1 mark)
Show Answers Only

a.    \(\hat{p}=0.1\)

b.    \(n=100\)

c.    \(\text{The confidence interval width is halved.}\)

Show Worked Solution

a.    \(\text{The span of a confidence interval is symmetrical about}\ \hat{p}.\)

\(\hat{p}=\dfrac{0.04+0.16}{2}=0.1\)


♦ Mean mark (a) 52%.
b.    \(0.06\) \(=2\times\sqrt{\dfrac{0.1\times0.9}{n}}\)
  \(0.03\) \(=\sqrt{\dfrac{0.1\times0.9}{n}}\)
  \((0.03)^2\) \(=\dfrac{0.1\times0.9}{n}\)
  \(0.0009\) \(=\dfrac{9}{100n}\)
  \(n\) \(=\dfrac{9}{100 \times 0.0009}\)
  \(n\) \(=100\)

♦♦ Mean mark (b) 38%.

c.    \(\text{C.I. width} \propto \dfrac{1}{\sqrt{n}}\)

→ \(\text{If}\ \ n\ \Rightarrow \ 4n,\ \ \text{C.I. width}\ \propto \dfrac{1}{\sqrt{4n}} = \dfrac{1}{2\sqrt{n}}\)

→ \(\text{The confidence interval is halved.}\)


♦♦♦ Mean mark (c) 23%.

Probability, MET1 2023 VCAA 8

Suppose that the queuing time, \(T\) (in minutes), at a customer service desk has a probability density function given by
 

\(f(t) = \begin {cases}
kt(16-t^2)         &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}\)

 
for some  \(K \in R\).

  1. Show that  \(k=\dfrac{1}{64}\).   (1 mark)
  2. Find  \(\text{E}(T)\).   (2 marks)
  3. What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?   (3 marks)
Show Answers Only
a.    \(\displaystyle \int_{0}^{4} (16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

b.    \(E(T)=\dfrac{32}{15}\)

c.    \(\dfrac{16}{25}=0.64\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{4} kt(16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

♦ Mean mark (a) 43%.
b.    \(E(T)\) \(=\dfrac{1}{64}\displaystyle \int_{0}^{4} (16t^2-t^4)\,dt\)
    \(=\dfrac{1}{64}\left[\dfrac{16t^3}{3}-\dfrac{t^5}{5}\right]_0^4\)
    \(=\dfrac{1}{64}\Bigg(\dfrac{1024}{3}-\dfrac{1024}{5}-0\Bigg)\)
    \(=\dfrac{1}{64}\times\dfrac{2048}{15}\)
    \(=\dfrac{32}{15}\)

♦♦ Mean mark (b) 38%.
c.    \(\text{Pr}(2<T<4|T>1)\) \(=\dfrac{\text{Pr}(2<T<4)}{\text{Pr}(T>1)}\)
    \(=\dfrac{\dfrac{1}{64}\displaystyle \int_{2}^{4} (16t-t^3)\,dt}{\dfrac{1}{64}\displaystyle \int_{1}^{4} (16t-t^3)\,dt}\)
    \(=\dfrac{\left[8t^2-\dfrac{t^4}{4}\right]_2^4}{\left[8t^2-\dfrac{t^4}{4}\right]_1^4}\)
    \(=\dfrac{(64-(32-4))}{\bigg(64-\bigg(8-\dfrac{1}{4}\bigg)\bigg)}\)
    \(=\dfrac{36}{\bigg(\dfrac{225}{4}\bigg)}=\dfrac{144}{225}=\dfrac{16}{25}=0.64\)

♦♦♦ Mean mark (c) 24%.
MARKER’S COMMENT: Simplifying fractions caused problems. Cancelling factors will assist with calculations.

Networks, GEN2 2023 VCAA 14

One of the landmarks in state \(A\) requires a renovation project.

This project involves 12 activities, \(A\) to \(L\). The directed network below shows these activities and their completion times, in days.
 

The table below shows the 12 activities that need to be completed for the renovation project.

It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.

The immediate predecessor(s) for activity \(I\) and the EST for activity \(J\) are missing.

\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
&  & & \textbf{predecessor(s)}  \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}

  1. Write down the immediate predecessor(s) for activity \(I\).  (1 mark)

  2. What is the earliest start time, in days, for activity \(J\) ?  (1 mark)

  3. How many activities have a float time of zero?  (1 mark)

The managers of the project are able to reduce the time, in days, of six activities.

These reductions will result in an increase in the cost of completing the activity.

The maximum decrease in time of any activity is two days.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad A \quad & \quad B  \quad&  \quad F \quad  &  \quad H \quad &  \quad I  \quad &  \quad K \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} &  1500 & 2000 & 2500 & 1000 & 1500 & 3000  \\
\hline
\end{array}

  1. If activities \(A\) and \(B\) have their completion time reduced by two days each, the overall completion time of the project will be reduced.
  2. What will be the maximum reduction time, in days?  (1 mark)
  3. The managers of the project have a maximum budget of $15 000 to reduce the time for several activities to produce the maximum reduction in the project's overall completion time.
  4. Complete the table below, showing the reductions in individual activity completion times that would achieve the earliest completion time within the $ 15 000 budget.  (1 mark)

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

Show Answers Only

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

b.    \(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)

c.    \(\text{5 activities have a float time of zero.}\)

d.    \(\text{Maximum reduction time = 2 days}\)

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

\(\text{Dummy activity before activity}\ C\ \text{does not effect this.}\)
 

b.    \(\text{Scan network:}\)

\(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)
 

c.    \(\text{Critical Path:}\ A\ C\ I\ K\ L\)

\(\text{Activities on the critical path have a float time of zero.}\)

\(\Rightarrow \ \text{5 activities have a float time of zero.}\)
 

d.    \(\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}\)

\(\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}\)

\(\text{Maximum reduction time = 2 days}\)

♦♦ Mean mark (c) 35%.
♦♦ Mean mark (d) 33%.

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

♦♦♦ Mean mark (e) 9%.

Networks, GEN2 2023 VCAA 13

The state \(A\) has nine landmarks, \(G, H, I, J, K, L, M, N\) and \(O\).

The edges on the graph represent the roads between the landmarks.

The numbers on each edge represent the length, in kilometres, along each road.
 

 

Three friends, Eden, Reynold and Shyla, meet at landmark \(G\).

  1. Eden would like to visit landmark \(M\).
  2. What is the minimum distance Eden could travel from \(G\) to \(M\) ?  (1 mark)

  3. Reynold would like to visit all the landmarks and return to \(G\).
  4. Write down a route that Reynold could follow to minimise the total distance travelled.  (1 mark)

  5. Shyla would like to travel along all the roads.
  6. To complete this journey in the minimum distance, she will travel along two roads twice.
  7. Shyla will leave from landmark \(G\) but end at a different landmark.
  8. Complete the following by filling in the boxes provided.  
  9. The two roads that will be travelled along twice are the roads between:  (1 mark)

Show Answers Only

a.    \(\text{Shortest path:}\ G\ K\ I\ M\)

\(\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}\)
 

b.   \(\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:\)

\(G\ H\ K\ I\ J\ M\ O\ L\ N\ G\)

\(G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}\)
 

c.   \(\text{vertex}\ L\ \text{and vertex}\ N\)

\(\text{vertex}\ J\ \text{and vertex}\ M\)

Show Worked Solution

a.    \(\text{Shortest path:}\ G\ K\ I\ M\)

\(\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}\)
 

b.   \(\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:\)

\(G\ H\ K\ I\ J\ M\ O\ L\ N\ G\)

\(G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}\)

♦ Mean mark (b) 44%.

c.   \(\text{Using an educated guess and check methodology:}\)

\(\text{vertex}\ L\ \text{and vertex}\ N\)

\(\text{vertex}\ J\ \text{and vertex}\ M\)

♦♦♦ Mean mark (c) 12%.

Financial Maths, GEN2 2023 VCAA 7

Arthur takes out a new loan of $60 000 to pay for an overseas holiday.

Interest on this loan compounds weekly.

The balance of the loan, in dollars, after \(n\) weeks, \(V_n\), can be determined using a recurrence relation of the form

\(V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d\)

  1. Show that the interest rate for this loan is 7.8% per annum.   (1 mark)

  2. Determine the value of \(d\) in the recurrence relation if
    1. Arthur makes interest-only repayments   (1 mark)
    2. Arthur fully repays the loan in five years. Round your answer to the nearest cent.   (1 mark)
  3. Arthur decides that the value of \(d\) will be 300 for the first year of repayments.  
  4. If Arthur fully repays the loan with exactly three more years of repayments, what new value of \(d\) will apply for these three years? Round your answer to the nearest cent.   (1 mark)
  5. For what value of \(d\) does the recurrence relation generate a geometric sequence?   (1 mark)

Show Answers Only

a.    \( 7.8\%\)

b.i.  \(d=90\)

b.ii. \(d= $278.86\)

c.    \( d= $350.01\)

d.    \(d=0\)

Show Worked Solution


a.   \(\text{Weekly interest rate factor}\ = 1.0015-1 = 0.0015 = 0.15\% \)

\(I\%(\text{annual}) = 52 \times 0.15 = 7.8\%\)



♦♦ Mean mark (a) 32%.



 
b.i.  \(d= 0.15\% \times 60\ 000 = \dfrac{0.15}{100} \times 60\ 000 = 90\)
 

b.ii. \(\text{By TVM solver:} \)

\(N\) \(=5 \times 52 = 260\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -60\ 000\)  
\(PMT\) \(= ?\)  
\(FV\) \(=0\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow PMT = d= $278.86\)
 



♦ Mean mark (b)(i) 47%.
♦ Mean mark (b)(ii) 40%.



c.    \(d=300\ \text{for the 1st 52 weeks.}\)

\(\text{Find}\ FV\ \text{after 52 weeks:}\)

\(N\) \(=52\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -60\ 000\)  
\(PMT\) \(= 300\)  
\(FV\) \(=?\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow FV = $48\ 651.67\)

 
\(\text{Find}\ PMT\ \text{given}\ FV=0\ \text{after 156 more weeks:}\)

\(N\) \(=156\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -48\ 651.67\)  
\(PMT\) \(= ?\)  
\(FV\) \(=0\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow PMT = d= $350.01\)
 



♦♦ Mean mark (c) 32%.



d.    \(\text{Geometric sequence when}\ \ d=0\)

\(V_0=60\ 000, V_1=60\ 000(1.0015), V_2=60\ 000(1.0015)^2, …\)



♦♦♦ Mean mark (d) 11%.



Matrices, GEN1 2022 VCAA 8 MC

Two types of computers - laptops `(L)` and desktops `(D)` - can be serviced by Henry `(H)`, Irvine `(I)` or Jean `(J)`.

Matrix `N` shows the time, in minutes, it takes each person to service a laptop and a desktop.

`{:(qquadqquadquad\ LquadqquadD),(N = [(18,8),(10,17),(12,9)]{:(H),(I),(J):}):}`

Matrix `Q` shows the number of laptops and desktops in four different departments: marketing `(M)`, advertising `(A)`, publishing `(P)` and editing `(E)`.

`{:(qquadqquadquad\ LquadqquadD),(Q = [(6,8),(4,7),(5,5),(10,12)]{:(M),(A),(P),(E):}):}`

A calculation that determines the total time that it would take each of Henry, Irvine or Jean, working alone, to service all the laptops and desktops in all four departments is

  1. `[1\ 1\  1\  1]×(Q×N^T)`
  2. `(Q×N^T)×[(1),(1),(1)]`
  3. `(N×Q^T)×Q`
  4. `[(1,0,0),(0,1,0),(0,0,1)]×N×Q^T`
  5. `[1\ 1\ 1\ 1]×Q×N^T×[(1),(1),(1)]`
Show Answers Only

`A`

Show Worked Solution

`text{Required information requires a 3 × 1 or a 1 × 3 matrix.}`

`text{By elimination}`

`text{Option A: [1 × 4] × ([4 × 2][2 × 3]) → [1 × 3] }`

`text{Option B: ([4 × 2][2 × 3]) × [3 × 1] → [4 × 1] (eliminate B)}`

`text{Option C: ([3 × 2][2 × 4]) × [4 × 2] → [3 × 2] (eliminate C)}`

`text{Option D: [3 × 3][3 × 2][2 × 4] → [3 × 4] (eliminate D)}`

`text{Option E: [1 × 4][4 × 2][2 × 3][3 × 1] → [1 × 1] (eliminate E)}`

`=>A`

CHEMISTRY, M7 2020 VCE 16 MC

The following table provides information about three organic compounds, \(\text{X, Y}\) and \(\text{Z}\).

Which one of the following is the best estimate for the boiling point of Compound \(\text{Z}\)?

  1. 31 °C
  2. 101 °C
  3. 114 °C
  4. 156 °C
Show Answers Only

\(A\)

Show Worked Solution
  • Boiling points of molecular compounds are related to the strength of their intermolecular bonding.
  • All given molecules are polar molecules. Each intermolecular bonding will have contributions from both dispersion forces and dipole-dipole bonding.
  • Since the molar mass of each compound is the same, the contribution from dispersion forces will be similar for all.
  • Compounds \(\text{X}\) and \(\text{Y}\) will both have hydrogen bonding, due to the presence of the \(\ce{O-H}\) functional group.
  • Compound \(\text{Z}\) does not have intermolecular hydrogen bonding and therefore would have the lowest boiling point.

\(\Rightarrow A\)

♦♦♦ Mean mark 20%.

Complex Numbers, EXT2 N2 2023 HSC 16c

The complex numbers \(w\) and \(z\) both have modulus 1, and  \(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{z}{w}\Big{)} \lt \pi\), where \(\text{Arg}\) denotes the principal argument.

For real numbers \(x\) and \(y\), consider the complex number \(\dfrac{xz+yw}{z}\).

On an \(xy\)-plane, clearly sketch the region that contains all points \((x,y)\) for which \(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{xz+yw}{z}\Big{)} \lt \pi\).

Show Answers Only

Show Worked Solution

\(\text{One strategy:}\)

\(\text{Let}\ \ u=\dfrac{z}{w},\ \text{where}\ \abs{z} = \abs{w} = 1\ \ \Rightarrow \ \ \abs{u}=1. \)

\(\dfrac{\pi}{2} \lt \text{Arg} (u) \lt \pi\ \text{(given)}\ \ \Rightarrow \text{Re}(u) \lt 0\ \ \text{and}\ \ \text{Im}(u) \gt 0 \)

\(\text{Using}\ \ \dfrac{z}{w} = \dfrac{1}{u} = \dfrac{\bar u}{\abs{u}} = \bar u, \)

\(\text{find all real numbers}\ x\ \text{and}\ y\ \text{such that:} \)

\(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{xz+yw}{z}\Big{)} \lt \pi\)

\(\dfrac{\pi}{2} \lt \text{Arg} (x+ \bar y u) \lt \pi\)

\(\text{i.e.}\ \ \text{Re}(x+y \bar u) = x + y \times \text{Re}(u) \lt 0\ \ \text{and}\ \ \text{Im}(y \bar u) \gt 0 \)

  \(\therefore y \lt 0\ \ \text{and} \)

\(x \lt -y \times \text{Re}(u) = -y \times \cos (\text{Arg}(u)) = -y \times \cos \Bigg{(}\text{Arg}\Big{(} \dfrac{z}{w} \Big{)} \Bigg{)} \)

\(\text{Region is all real}\ (x,y)\ \text{that lie below the}\ x\text{-axis and to the} \)

\(\text{left of the line}\ \ y=-\sec \Bigg{(}\text{Arg}\Big{(} \dfrac{z}{w} \Big{)} \Bigg{)}x \)

♦♦♦ Mean mark 3%.
COMMENT: On the podium for the toughest questions in Ext2 in the last decade.

Vectors, EXT2 V1 2023 HSC 10 MC

Consider any three-dimensional vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\)  and  \(\underset{\sim}{c}=\overrightarrow{O C}\)  that satisfy these three conditions

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=1\)

\(\underset{\sim}{b} \cdot \underset{\sim}{c}=2\)

\(\underset{\sim}{c} \cdot \underset{\sim}{a}=3\).

Which of the following statements about the vectors is true?

  1. Two of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) could be unit vectors.
  2. The points \(A, B\) and \(C\) could lie on a sphere centred at \(O\).
  3. For any three-dimensional vector \(\underset{\sim}{a}\), vectors \(\underset{\sim}{b}\) and \(\underset{\sim}{c}\) can be found so that \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfy these three conditions.
  4. \(\forall \ \underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfying the conditions, \(\exists \ r, s\) and \(t\) such that \(r, s\) and \(t\) are positive real numbers and  \(r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\).
Show Answers Only

\(B\)

Show Worked Solution

\(\text{By elimination}\)

\(\text{Option}\ A:\)

\(\text{If}\ \underset{\sim}{a}\ \text{and}\ \underset{\sim}{b}\ \text{are the two unit vectors,}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b} = \abs{\underset{\sim}{a}} \abs{\underset{\sim}{b}} \cos\,\theta = \cos\,\theta\)

\(-1 \leq \cos\,\theta \leq1\ \ \Rightarrow \ \ -1 \leq \underset{\sim}{a} \cdot \underset{\sim}{b} \leq1 \)

\(\text{Given}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b}=1\ \Rightarrow\ \underset{\sim}{a} = \underset{\sim}{b}\ \Rightarrow\ \underset{\sim}{b} \cdot \underset{\sim}{c} = \underset{\sim}{c} \cdot \underset{\sim}{a}\)

\(\text{Contradicts}\ \ \underset{\sim}{b} \cdot \underset{\sim}{c} = 2\ \ \text{and}\ \ \underset{\sim}{c} \cdot \underset{\sim}{a} = 3\)

\(\text{Similar reasoning rules out any pair satisfying all conditions (eliminate}\ A). \)
 

\(\text{Option}\ C:\ \text{If}\ \ \underset{\sim}{a} = \underset{\sim}{0}, \  \underset{\sim}{a} \cdot \underset{\sim}{b} = 0 \neq 1\ \text{(eliminate}\ C). \)

\(\text{Option}\ D:\)

\(\text{Consider the vectors below that satisfy the conditions,}\)

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right), \ \  \underset{\sim}{c}=\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right) \]

\(\text{However,}\ r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\ \text{requires}\ \ r=s=t=0\ \ \text{which are not}\)

\(\text{positive constants (eliminate}\ D).\)

\(\Rightarrow B\)

♦♦♦ Mean mark 15%.